Download Principles of Heat Transfer Solutions Manual...
Solutions Manual to Accompany
Principles of Heat Transfer
M. Kaviany [emailprotected]
A WileyInterscience Publication
John Wiley & Sons, Inc.
Outline of Solutions Manual Objectives Literature is strewn with the wreckage of men who have minded beyond reason the opinions of others. V. Woolf The structure of problem statements and problem solutions, the major instructional objectives for each chapter, and a typical syllabus for a 15week term, are given here. The syllabus can change to emphasize and deemphasize topics, per instructors discretion.
1
Problem Statement
The format used in the problem statement is as follows.
1.1 Problem Label The problem label follows the format: Chapter Number.Problem Number.Purpose.Software. The purpose is categorized as follows. (i) Familiarity (FAM) introduces the use of available relations. (ii) Fundamental (FUN) gives further insights into the principles and requires combining some concepts and relations. (iii) Design (DES) uses the available relations and searches for an optimum engineering solution. (iv) Solver option (S) indicates if the problem is intended for use with a solver. For example, PROBLEM 3.5.DES.S indicates an end of chapter problem (as compared to EXAMPLE which is a solved example problem). The problem is in Chapter 3; the problem number is 5; it is in the Design category; and it is intended to be solved using a solver.
1.2 Problem Statement The (i) (ii) (iii)
problem statement gives the following. The thermal problem considered and the knowns. The questions and the unknowns. Any hints on needed simpliﬁcations and assumptions for the analysis.
1.3 Sketch The sketch provides the following. (i) The heat transfer media. (ii) The signiﬁcant variables properly labeled.
2
Problem Solution
The format used for the problem solution is as follows.
2.1 ReState Problem Statement (GIVEN) Reword and restate the problem knowns, assumptions, and simpliﬁcations.
2.2 ReDraw the Physical Problem (SKETCH) Draw the sketch of the thermal problem considered. Show the direction of the heat (and when appropriate the ﬂuid) ﬂow. Identify the mechanisms of heat transfer and mechanisms of energy conversion. Make any other needed additions to the sketch. i
2.3 ReState Questions (OBJECTIVE) Write the objectives of the problem and state the questions asked and the unknowns.
2.4 Solve the Problem (SOLUTION) The solution of the problem includes some or all of the following steps. (i) Control Volume and Control Surface: Mark the appropriate bounding surfaces and deﬁne the control volumes and control surfaces. (ii) Thermal Circuit Diagram: Draw the thermal circuit diagram for the problem, when appropriate. (iii) Energy Equation: Write the appropriate form of the energy equation. (iv) Energy Conversion: Write the appropriate relations for each of the energy conversion mechanisms. (v) Heat Transfer Rates and Thermal Resistances: Write the appropriate relations for the heat transfer rates and the thermal resistances. (vi) Numerical Values: Determine the thermophysical and thermochemical properties from the tables, graphs, and relations. When using tables, make the needed, appropriate interpolations. Always check the units of each parameter and variable. (vii) Solver: When needed, solve algebraic or diﬀerental equations using a solver such as SOPHT. (viii) Final Numerical Solutions: Determine the magnitude of the unknowns.
2.5 Make Additional Comments (COMMENT) Examine the numerical values and compare them to what is avaliable or what is initially expected. State what insights have been gained from the exercise.
3
Major Instructional Objectives
3.1 Chapter 1: Introduction and Preliminaries (i) (ii) (iii) (iv)
Mechanisms of Heat Transfer Qualitative Heat Flux Vector Tracking Qualitative Analysis of Energy Conservation Equation Quantitative Analysis of Energy Conservation Equation
3.2 Chapter 2: Energy Equation (i) (ii) (iii) (iv) (v) (vi) (vii)
Finite and DiﬀerentialLength Energy Equation Divergence of Heat Flux Vector Energy Conversion Mechanisms (to and from Thermal Energy) Chemical and Physical Bonds Energy Conversion Electromagnetic Energy Conversion Mechanical Energy Conversion BoundingSurface Thermal Conditions
3.3 Chapter 3: Conduction (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi)
Physics of Speciﬁc Heat Capacity Physics of Thermal Conductivity Thermal Conduction Resistance and Thermal Circuit Analysis Conduction and Energy Conversion Thermoelectric Cooling Multidimensional Conduction Distributed Transient Conduction and Penetration Depth LumpedCapacitance Transient Conduction Multinodal Systems and FiniteSmall Volume Analysis Conduction and SolidLiquid Phase Change Thermal Expansion and Thermal Stress ii
3.4 Chapter 4: Radiation (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
Surface Emission View Factor for Diﬀuse Gray Enclosures Enclosure Radiation for Diﬀuse Gray Surfaces TwoSurface Enclosures ThreeSurface Enclosures with One Surface ReRadiating Enclosures with Large Number of Surfaces Prescribed Irradiation and Nongray Surfaces Inclusion of Substrate
3.5 Chapter 5: Convection: Unbounded Fluid Streams (i) (ii) (iii) (iv) (v)
ConductionConvection Resistance and P´eclet Number Evaporation Cooling of Gaseous Streams Combustion Heating of Gaseous Streams Joule Heating of Gaseous Streams GasStream Radiation Losses
3.6 Chapter 6: Convection: SemiBounded Fluid Streams (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix)
Laminar Parallel Flow and Heat Transfer: Nusselt, P´eclet, Reynolds, and Prandtl Numbers Average SurfaceConvection Resistance Turbulent, Parallel Flow and Heat Transfer Impinging Jets Thermobuoyant Flows LiquidGas Phase Change Nusselt Number and Heat Transfer for Other Geometries Inclusion of Substrate SurfaceConvection Evaporation Cooling
3.7 Chapter 7: Convection: Bounded Fluid Streams (i) (ii) (iii) (iv) (v) (vi)
Average Convection Resistance, N T U , and Eﬀectiveness Nusselt Number and Heat Transfer for Tubes Nusselt Number and Heat Transfer for Other Geometries Inclusion of Bounding Surface Heat Exchanger Analysis Overall Thermal Resistance
3.8 Chapter 8: Heat Transfer and Thermal Systems (i) Combined Mechanisms of Heat Transfer (ii) Various Energy Conversion Mechanisms (iii) Innovation Applications
iii
4
Typical Syllabus
WEEK
SUBJECT
READING
PROBLEMS
1
Introduction: Control Volume and Surface, Heat Flux Vector and Mechanisms of Heat Transfer, Conservation Equations
1.1  1.9
1.1, 1.4, 1.6, 1.15, 1.18
2
Energy Equation for Diﬀerential Volume, Integral Volume, and Combined Integral and DiﬀerentialLength Volume
2.1  2.2
2.1, 2.5, 2.7, 2.9, 2.11
3
Work and Energy Conversion: Mechanisms of Energy Conversion, BoundingSurface Thermal Conditions, Methodology for Heat Transfer Analysis
2.3  2.6
2.14, 2.17, 2.18, 2.32, 2.35
4
Conduction: Speciﬁc Heat and Thermal Conductivity of Matter; SteadyState Conduction: Conduction Thermal Resistance
3.1  3.3
3.1, 3.3, 3.9, 3.12, 3.13
5
SteadyState Conduction: Composites, Thermal Circuit Analysis, Contact Resistance, Energy Conversion, Thermoelectric Cooling
3.3
3.15, 3.26, 3.27, 3.30, 3.32
6
Transient Conduction: Distributed Capacitance, Lumped Capacitance, Discretization of Medium into SmallFinite Volumes
3.4  3.7, 3.10
3.53, 3.55, 3.63, 3.67, 3.70
7
Radiation: Surface Emission, Interaction of Irradiation and Surface, Thermal Radiometry, Review
4.1  4.3
4.1, 4.4, 4.8, 4.9
EXAM I (Covering Energy Equation and Conduction) 8
Radiation: GrayDiﬀuseOpaque Surface Enclosures, ViewFactor and Grayness Radiation Resistances, Thermal Circuit Analysis, Prescribed Irradiation and Nongray Surfaces, Inclusion of Substrate
4.4  4.7
4.10, 4.19, 4.24, 4.43, 4.49
9
Convection (Unbounded Fluid Streams): ConductionConvection Resistance, P´eclet Number, Combustion Heating of Gaseous Streams
5.1  5.2, 5.4, 5.7
5.1, 5.3, 5.5, 5.19, 5.20
10
Surface Convection (SemiBounded Fluid Streams): Flow and Surface Characteristics, Laminar Parallel Flow over SemiInﬁnite Plate, P´eclet, Prandtl, Reynolds, and Nusselt Numbers, SurfaceConvection Resistance
6.1  6.2
6.1, 6.2, 6.3, 6.4
11
Convection (SemiBounded Fluid Streams): Turbulent Parallel Flow, Perpendicular Flow, Thermobuoyant Flows
6.3  6.5
6.7, 6.9, 6.10, 6.14, 6.18
12
Convection (SemiBounded Fluid Streams): LiquidVapor PhaseChange, Nusselt Number Correlations for Other Geometries, Inclusion of Substrate
6.6  6.8, 6.10
6.19, 6.21, 6.25, 6.40, 6.45
13
Convection (Bounded Fluid Streams): Flow and Surface Characteristics, Tube Flow and Heat Transfer, Average Convection Resistance, Review
7.1  7.2
7.1, 7.3, 7.4
EXAM II (Covering Surface Radiation and Convection: SemiBounded Fluid Streams) 14
Convection (Bounded Fluid Streams): Tube and Ducts, High Speciﬁc Surface Areas, Nusselt Number Correlations, Inclusion of Bounding Surface, Heat Exchangers
7.3  7.7
7.7, 7.12, 7.20, 7.25, 7.33
15
Heat Transfer in Thermal Systems: Thermal Functions, Analysis, and Examples
8.1  8.3
8.1
FINAL EXAM (Comprehensive)
iv
Answers to Problems
1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20
(b) (b) (b) (c) (b) (b)
2.1
(a)
2.3
(a)
2.4 2.5 2.6
(b)
2.7 2.8 2.10 2.11
(a) (c) (a) (a) (a) (a) (c)
2.12 2.13
2.14 2.15
2.16 2.17 2.18 2.19 2.20
(a) (b)
(a) (b) (c) (d) (e) (a) (c) (a) (a) (a)
t = 247 s dT /dt = 1.889 × 10−3 ◦C/s Tg (tf ) = 5,200◦C, (c) η = 2.5% Qku + Qr = 12 kW uF = 0.5904 mm/hr T A = 13.55◦C τ = 744 s = 0.207 hr dTr /dt = 8.696◦C/s lim∆V →0 A q · sn dA/∆V ≡ ∇ · q = 8a dq 2 (qku,o Ro + qku,i Ri ) + k,z = 0 dz Ro2 − Ri2 L = 7.362 m ◦ brake pad region: T (t = 4 s) = 93.50◦C; ∞entire rotor: T (t = 4 s) = 66.94 C m ˙ ph = 5.353 × 1035 photon/m2 s, (b) −∞ S˙ e,σ dt = 0.1296 J S˙ e,σ /V = 2.972 × 1012 J/m3 s˙ e,J = 2.576 × 108 W/m3 , (b) ρe (T ) = 1.610 × 10−5 ohmm ∂T = 0.5580◦C/s, (b) ∂T = 0.7259◦C/s ∂t ∂t ∇ · q = 103 W/m3 3 ◦ qk = −[1.883 × 10 ( C/m) × k(W/m◦C)]sx 2(Ly + Lz ) d2 T qku −k 2 =0 Ly Lz dx s˙ m,µ = 1.66 × 1013 W/m3 ρCH4 = 8.427 × 10−5 g/cm3 , ρO2 = 3.362 × 10−4 g/cm3 m ˙ r,CH4 (without Pt) = 1.598 × 10−7 g/cm2 s, m ˙ r,CH4 (with Pt) = 1.754 × 10−3 g/cm2 s Ts = 1,454 K αe (Ta) = 1.035 × 10−3 1/K, αe (W) = 1.349 × 10−3 1/K ρe (Ta) = 1.178 × 10−6 ohmm, ρe (W) = 7.582 × 10−7 ohmm Re (Ta) = 7.500 ohm, Re (W) = 4.827 ohm Je (Ta) = 3.651 A, Je (W) = 4.551 A ∆ϕ(Ta) = 27.38 V, ∆ϕ(W) = 21.97 V Je = 3.520 A, (b) Pe = 0.2478 W T2 = 2,195 K ∆t1 = 0.4577 hr, (b) ∆t2 = 9.928 hr s˙ m,p = −2 × 108 W/m3 , (c) To − Ti = −42.27◦C s˙ m,µ A = 3.660 × 107 W/m3
v
2
2.21
(a)
2.33 2.36
(b) (d) (a) (b) (c) (a) (a) (d) (a) (c) (a) (c) (b) (b)
2.37 2.38 2.39 2.40
(b) (b) (b) (b)
3.1 3.2 3.3
(a) (a) (c)
2.22 2.23
2.24 2.25 2.26 2.27 2.30
3.9 3.12 3.13
3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.23 3.24 3.25 3.26 3.27 3.29 3.30 3.31 3.33 3.34 3.35 3.36 3.40
(a) (a) (c)
(b) (b) (b) (b) (b) (b) (c) (b)
(b) (b) (a) (c) (a) (c) (c) (b) (b) (b)
S˙ m,F peak = 0.65 Mτuo (1 − τt ) (each of the front brakes) 2 S˙ m,F peak = 0.35 Mτuo (1 − τt ) (each of the rear brakes) S˙ m,F peak = 120.3 kW Ts = 40.49◦C n˙ r,CH4 = −0.9004 kg/m3 s, n˙ r,CH4 = −0.1280 kg/m3 s, n˙ r,CH4 = −0.3150 kg/m3 s M˙ lg = 0.8593 g/s, (b) M˙ r,CH4 = 0.2683 g/s, (c) S˙ e,J = 1.811 × 105 W s˙ e,J = 5.62 × 108 W/m3 , (b) ∆ϕ = 2.18 V, (c) Je = 1.8 A s˙ e,J = 2.81 × 108 W/m3 , ∆ϕ = 3.08 V, Je = 1.3 A qc = 49,808 W/m2 , (b) qh = 56,848 W/m2 Tmax = 45.03◦C at t = 594 s wet alumina: s˙ e,m = 3.338 W/m3 , (b) dry alumina: s˙ e,m = 16.69 W/m3 dry sandy soil: s˙ e,m = 1,446 W/m3 qx (x = x2 ) = 1.378 × 106 W/m2 ˙ S˙ e,α /A = 637.0 W/m2 , S˙ e, /A = −157.8 W/m2 , (c) S/A = 479.2 W/m2 2m ˙ lg −2 2 m ˙ lg = 5.074 × 10 kg/m s, (c) D(t) = D(t = 0) − ρl t, (d) t = 36.57 s qk,e = −8 × 104 W/m2 qk,t = −140 W/m2 qk,s = 8.989 × 109 W/m2 for T = 300 K, kpr = 424.7 W/mK, (b) for T = 300 K, ∆k(%) = 6% k = 0.02392 W/mK, (b) ∆k(%) = 18.20 % (ρcp k)1/2 (argon) = 4.210, (air) = 5.620, (helium) = 11.22, (hydrogen) = 15.42 Ws1/2 /m2 K L = 0.6 nm: k = 0.5914 W/mK, L = 6 nm: k = 1.218 W/mK Qk,21 = −100 W, (b) Qk,21 = −83.3 W, (c) Qk,21 = −82.3 W Ak Rk,12 = 2.5 × 10−5◦C/(W/m2 ), (b) Ak Rk,12 = 7.4 × 10−1◦C/(W/m2 ), T1 = 60.02◦C, (d) Rk value (copper) = 1.4 × 10−4◦F/(Btu/hr), Rk value(silica aerogel) = 4.2◦F/(Btu/hr) ∆Q% = 63.3 % parallel: k = 14.4 W/mK, series: k = 0.044 W/mK, random: k = 0.19 W/mK (i) Q12 = 4.703 W, (ii) Q12 = 4.492 W ˙ lg = 3.960 g/s, (d) T2 = −10.45◦C Qk,2−1 = 8.408 × 102 W, (c) M ◦ (i) Ts = 43.35 C, (ii) Ts = 76.92◦C k = 0.373 W/mK, (c) Tg = 1,643 K, (d) Qg,1 /S˙ r,c = 0.854, Qg,2 /S˙ r,c = 0.146 M˙ lg = 0.051 kg/s, (c) ∆t = 0.01 s Qk,12 = −7.423 × 10−2 W, (d) Qk,12 = −7.635 × 10−2 W for p = 105 Pa, ∆Tc = 4.9◦C, for p = 106 Pa, ∆Tc = 2.7◦C for Ak Rk,c = 10−4 K/(W/m2 ), Th = 105◦C, for Ak Rk,c = 4 × 10−2 K/(W/m2 ), Th = 1,469◦C (i) Qk,2−1 (no blanket) = 1.006 × 103 W, (ii) Qk,2−1 (with blanket) = 6.662 × 102 W L = 5.984 mm Ap = 2.59 × 10−5 m2 , (c) Tmax = 41.8◦C, (d) qc = −5,009 W/m2 , qh = 17,069 W/m2 Qc (2.11 A) = −0.079 W, (b) Qc (1.06 A) = −0.047 W, Qc (4.22 A) = 0.048 W Tc = 231 K, (b) Je = 2.245 A, (c) Tc = 250.2 K T (t → ∞) = −18.40◦C T1 = 71.36◦C Rk,12 = 106.1◦C/W, (c) T1 = 60.34◦C ee,o = 160 V/m T1 = ( ThR+ Tc + Re Je2 )/(2/Rk + αS Je ), (c) Tc (Qc = 0) = 205.4 K k
vi
3.41 3.42 3.44 3.45 3.46 3.47 3.48 3.49 3.50 3.51 3.52 3.53 3.54 3.55 3.56 3.57 3.58 3.59 3.60 3.61 3.62 3.63 3.64 3.65 3.66 3.67 3.68 3.70 3.71 3.72 3.73 3.74 3.75 3.76 3.77 3.79 3.80 3.81 3.82 3.83 4.1 4.2
(b) (c) (b) (b) (a) (a) (b) (a) (a)
(b) (a) (b)
(a) (b) (b) (b) (b) (b) (b) (b) (c) (b) (a) (b) (b) (a) (b) (b) (b) (a) (c) (b) (c) (d)
4.3 4.4 4.7
(a) (a) (c) (a)
Je = 1.210 × 10−2 A, (c) Tc,min = 223.8 K, (d) Qc,max = −5.731 × 10−4 W Qk,12 = 21.81 W ˙ ls = 665.7 g/hr T1 = −19.63◦C, Qk,12 = 61.69 W, (c) M L(t = ∆t) = 6.127 µm, Rk,12 = 145.6 kW t = 0.8378 s δα /[2(ατ )1/2 ] = 0.4310 (i) T (x = 0, t = 10−6 s) = 1.594 × 105 ◦C, (ii) T (x = 0, t = 10−6 s) = 1.594 × 104 ◦C (i) δα (t = 10−6 s) = 6.609 µm, (ii) δα (t = 10−4 s) = 66.09 µm t = 3.872 s, (b) t = 31.53 s (i) t = 2,970 s, (ii) t = 1,458 s, (iii) t = 972.1 s t = 5.6 min, (b) t = 35 min t = 7.7 min t = 24 s ﬁrstdegree burn: x = 8.6 ± 0.4 mm; seconddegree burn: x = 5.8 ± 0.4 mm; thirddegree burn: x = 4.75 ± 0.4 mm T (x = 1 mm) = 215.8 K, (b) T (x = 3 mm) = 291.9 K, (c) qρck = 20, 627 W Ts (x = L, t = 1.5 s) = 71◦C T12 T1 (t = 0) t = 7.8 µs T (x = 4 mm, t = 600 s) = 42.15◦C T (x = 4 mm, t = 600 s) = 64.30◦C t = 7.87 min t = 5.1 ms ub = 51.5 cm/min t = 177.8 s, (c) t = 675.7 s, (d) T1 (t → ∞) = 230.2◦C t = 1.402 µs T (x = 0, t = 2 s) = 91.41◦C T1 (t) − T2 = [T1 (t = 0) − T2 ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ) (i) T1 (t) = 2.6◦C, (ii) T1 (t) = 66.9◦C t = 6.236 s, (c) FoR = 150.3, (d) Nk,1 = 1.803 × 10−3 L = 8.427 m Te (t → ∞) = 1,672 K (i) T1 (t = 4 s) = 66.97◦C, (ii) T1 (t = 4 s) = 254.8◦C kyy = 0.413 W/mK, (b) kyy = 0.8375 W/mK T ∗ (x∗ = 0.125, y ∗ = 0.125) = 0.03044 for N = 21 T1 = 4,804◦C, T2 = 409.8◦C, T3 = 142.9◦C, T4 = 121.3◦C, T5 = 114.4◦C Qk,hc = 4.283 W, (c) k = 42.83 W/mK δα = 47 µm t = 74 s, qk dt = 3.341 × 105 J/m2 , (b) t = 21 s, qk dt = 3.341 × 105 J/m2 t1 = 286.6 s, t2 = 40.31 s t = 31.8 s τrr (r = 0) = τθθ (r = 0) = 1.725 × 108 Pa TR = 550.6◦C Eb = 201,584 W/m2 , (b) Qr, = 82,649 W F0.39T −0.77T = 0.13%, F0.77T −25T = 99.55%, F25T −1000T = 0.32% aluminum: (Qr,ρ )1 = 18,437 W, (Qr,α )1 = 1,823 W, nickel: (Qr,ρ )2 = 13,270 W, (Qr,α )2 = 6,990 W, paper: (Qr,ρ )3 = 1,103 W, (Qr,α )3 = 19,247 W aluminum: (Qr, )1 = 98.94 W, (Qr,o )1 = 18,536 W, nickel: (Qr, )2 = 379.3 W, (Qr,o )2 = 13,649 W, paper: (Qr, )3 = 1,044 W, (Qr,o )3 = 2,057 W aluminum: Qr,1 = −1,724 W, nickel: Qr,2 = −6,611 W, paper: Qr,3 = −18,203 W (r )1 = 0.09, (r )2 = 0.29, (r )3 = 0.65 qr, (visible) = 3.350 × 105 W/m2 , (b) qr, (near infrared) = 2.864 × 106 W/m2 , qr, (remaining) = 8,874 W/m2 T = 477.7 K, (b) λmax = 6.067 µm, (c) Tlg = 373.2 K
vii
4.8 4.9
(a)
4.10
(a) (c) (e) (a) (b) (a) (b)
4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.20
(a) (b) (b) (b) (b)
4.21
(b)
4.22 4.23 4.24 4.25 4.26 4.27
(b) (b) (b) (a)
4.28 4.29 4.30 4.31
(b) (d) (a) (b) (b) (c)
4.32 4.33 4.34
(b) (a) (a)
4.35 4.36 4.37 4.38 4.39 4.40 4.41 4.42 4.43
(a) (a) (c) (d) (a) (a) (a) (a) (b) (b)
4.44
(b)
4.45 4.46 4.47
(b) (a)
4.48 4.49 4.50 4.51
(a) (b) (a) (i) (i)
r (SiC) = 0.8301, r (Al) = 0.008324 (i) white potassium zirconium silicate, (ii) blackoxidized copper, (iii) aluminum foil F12 = 1/9 (b) F11 = 0.7067, F23 = 0.12 F12 = 0.08, F21 = 0.32, (d) F13 = 0.085, F21 = 0.415 F12 = 0.003861, F23 = 0.7529 F12 = 0.2, x∗ = w∗ = 1, w∗ = a∗ = 1, 1/R1 = 1/R2∗ = 1.70, l = 0.9591a for the discs, l = a for the plates r,1 = 1/ {[D/(4L + D) × (1 − r,1 )/r,1 ] + 1} (i) oxygen: qr,12 = 1.17 W/m2 , (i) hydrogen: qr,12 = 0.14 W/m2 (ii) oxygen: qr,12 = 0.596 W/m2 , (ii) hydrogen: qr,12 = 0.0730 W/m2 Qr,1 /S˙ lg = 4.7%, (b) ∆Qr,1 = 4.5% Qr,1 = 82 W Rr,Sigma = 2/(Ar r ) − 1/Ar S˙ e,J = 1,041 W T1 = 808.6◦C, (c) T1 = 806.6◦C r 1 qr,1−2 = (Eb,1 − Eb,2 )/(2[ 1 − r + 2 + r (1 − ) ]). (i) qr,21 = −245.7 W/m2 , (ii) qr,21 = −210.6 W/m2 , (iii) qr,21 = −179.8 W/m2 , (iv) qr,21 = −134.0 W/m2 Qr,12 = 1180 W, (c) Qr,12 = 182.8 W Qr,1−2 = 10,765 W, (c) Qr,2 = 19,300 W F12 = 0.125, F13 = 0.875, F23 = 0.8958; (c) T1 = 1,200 K Qr,12 = −54.13 W, (c) Qr,12 = 114.1 W, (d) T3 = 400 K Qr,2 = −201,322 W Qr,1−2 = 3.860 kW, (c) Qr,1−2 = 0.7018 kW, Qr,1−2 = (Ar,1 /2)(Eb,1 − Eb,2 ) for F1−2 → 0 t = 208.3 s, (b) T (x = 0, t) = 4,741◦C M˙ l = 0.5992 g/s ∆t = 15.36 ns α T2 = σ1SB { 2r,2 [(qr,o )a + (qr,o )b ]}1/4 r,2 (qr,i )f = 9.52 × 104 W/m2 , (c) S˙ e,σ = 2.989 × 105 W, (d) ∆t = 71.14 s (qr,i )f = 1.826 × 104 W/m2 , (b) qr,i = 319.6 W/m2 qku,3 = 6,576 W/m2 , (b) Qr,1 (IR + visible) = 2,835 W, Qr,1 (UV) = 148,960 W, (c) T3,max = 529.9 K (i) Qu L0 = 500 W, (ii) Qu L0 = 362 W, (b) (i) η = 31.22%, (ii) η = 22.60% S˙ e,σ /A = 738 W/m2 , (b) dT L /dt = 40 K/day (i) Qr,1,t /A1 = −1,242 W/m2 , (ii) Qr,1,b /A = −209 W/m2 (i) (dV1 /dt)/A1 = −0.478 µm/s, (ii) (dV1 /dt)/A1 = −0.0805 µm/s Qu L0 = 466.4 W, (b) η = 19.42% Qu L0 = 1,376 W, (b) Q1 = −34.15 W Q1 = 827.7 W Qr,12 = −3.008 W T1 = 860.5 K Q21 = 3,235 W, (c) Q21 = 25.65 W (1 − )(l1 + l2 ) (2 − r )(l1 + l2 ) + Rk + Rr,Σ = Ar ks 4Ar r σSB T 3 l2 = 0.8028, kr = 0.0004399 W/mK at T = 1,000 K t(T1 = 600 K) = 162 s l1 2 − r 1 Rr,Σ = + Ar (1 − )2/3 ks 4Ar (1 − )2/3 σSB T 3 r (i) σex = 86.75 1/m, (ii) σex = 8.551 × 105 1/m (i) σex = 191.0 1/m, (ii) σex = 1.586 × 105 1/m up = 0.48 m/s, (b) Nr = 1.2 × 10−6 < 0.1 q1 = 2.305 W/m2 , (ii) q1 = 2.305 W/m2 Q1−2 = 127.8 W, (ii) Q1−2 = 71.93 W
viii
4.52 4.53 4.54 4.57
(b) (a) (b) (b)
t(T1 = 500◦C) = 5 ms (i) Qk,1−2 /L = −25.7 W/m, (ii) Qr,1−2 /L = −1.743 W, (b) R2 = 2.042 × 108 m t = 181.1 s m ˙ ls = 8.293 g/m2 s
5.1 5.2 5.3 5.5 5.7
(b) (a) (b)
Q(x = 0) = 767.3 W Rk,u /RuL = 0.2586, (b) Rk,u /RuL = 6.535 ×10−3 Tf,2 = 367.9◦C, (c) Tf,2 = 2520◦C (Qk,u )1−2 = 140.3 W (i) (Qk,u )12 = −3.214 W, (qk,u )12 = −4.092 × 106 W/m2 (ii) (Qk,u )12 = −7.855 × 10−2 W, (qk,u )12 = −1.000 × 105 W/m2 ˙ l = 1.160 × 10−7 kg/s, m M ˙ l = 0.1477 kg/sm2 ˙ Mlg = 2.220 g/s, Tf,2 = 186.7◦C Tf,2 = 23.18◦C ˙ l = 0.5476 kg/s M k = 0.63 W/mK, (b) kr = 0.19 W/mK, (c) uf,1 = 1.30 m/s T u = 0.2162 Ze = 8.684, (b) uf = 1.037 m/s Tf,2 = 2,944◦C, (b) uf,1 = 3.744 m/s Ts = Tf,2 = 1,476 K, Qr,2−p = 40, 259 W, (c) η = 37.62% Tf,2 = 197.0◦C, (c) Tf,3 = 2,747◦C, (d) Tf,4 = 2,564◦C, (e) ∆Texcess = 177◦C Tf,2 = 2,039 K (for qloss = 105 W/m2 ) Ts = 1,040 K, (c) η = 60.90% ˙ f = 1.800 g/s M Tf,2 = 2,472◦C Tf,2 = 3,134◦C Tf,2 = 203.7◦C
5.8 5.9 5.10 5.11 5.12 5.13 5.15 5.16 5.17 5.18 5.19 5.21 5.22 5.23 5.24 5.25 6.2
6.3 6.7
6.8 6.9 6.10
(b) (b) (b) (b) (a) (a) (a) (b) (b) (b) (b) (b) (b) (b) (a) (b) (c) (a) (a) (b)
(c) (b) (b) (a) (b)
6.11 6.12 6.13 6.14
(c) (b) (b) (a) (b)
6.15 6.18 6.19
(b) (b) (a) (d)
6.20 6.21
(b)
(i) qku,L = −50,703 W/m2 , (ii) qku,L = −2,269 W/m2 , (iii) qku,L = −39.59 W/m2 (i) δα,L = 5.801 mm, (ii) δα,L = 3.680 mm, (iii) δα,L = 22.39 mm (qku )(i),Pr→0 = −46,115 W/m2 , δα,L,Pr→0 = 7.805 mm. (i) Qku L = 1.502 W, (ii) Qku L = 65.23 W, (b) (i) δα = 14.16 mm, (ii) δα = 3.515 mm (i) NuL = 119.8, (ii) NuL = 378.8, (iii) NuL = 2,335 (i) Aku Rku L = 3.326 × 10−1 ◦C/(W/m2 ), (ii) Aku Rku L = 1.052 × 10−1 ◦C/(W/m2 ), (iii) Aku Rku L = 1.711 × 10−2 ◦C/(W/m2 ) (i) Qku L = 150.3 W, (ii) Qku L = 475.4 W, (iii) Qku L = 2,930 W uf,∞ = 3.78 m/s Tf,∞ = 277.20 K, (c) Qku L = −598.4 W, Qr,1 = 99.57 W, ice would melt. single nozzle: NuL = 46.43, Aku Rku L = 8.413 × 10−2 ◦C/(W/m2 ), Qku L = 406.5 W multiple nozzles: NuL = 28.35, Aku Rku L = 4.593 × 10−2 ◦C/(W/m2 ), Qku L = 744.6 W S˙ m,F = 0.01131 W, Ln = 0.5425 cm t = 21.27 s (i) parallel ﬂow: t = 2.465 s (ii) perpendicular ﬂow: t = 1.123 s vertical: NuL = 43.08, Aku Rku L = 2.232 × 10−1 ◦C/(W/m2 ), Qku L = −7.390 W horizontal: NuD = 18.55, Aku Rku D = 2.073 × 10−1 ◦C/(W/m2 ), Qku D = −7.956 W Qku L = 411.9 W, (c) Qr,sw = 691.0 W, (d) η = 5.656% (i) Qku L = 938.5 W, (ii) Qku L = 1,163 W S˙ e,J = 2,045 W, (b) Qku,CHF = 2,160 W, (c) Ts = 108.9◦C Aku Rku D = 8.575 × 10−6 ◦C/(W/m2 ), NuD = 8,587 Ts,1 = 103.5◦C Qku L = −9,058 × 103 W, (c) M˙ lg = 4.013 g/s
ix
6.22 6.23 6.24 6.25 6.26 6.27 6.28 6.29 6.30 6.31 6.32 6.33 6.34 6.35 6.36 6.37 6.38 6.39 6.40 6.41 6.42 6.43 6.44 6.45 6.46 6.48 6.49 6.50 6.51 6.52 6.53 6.54 7.1 7.2 7.4 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 7.16
(b) (b) (a) (b) (a) (b) (b) (b) (b) (b) (a) (a) (d) (b) (a) (a) (b) (c) (b) (b) (b) (b) (b) (b) (b) (b) (d) (b) (b) (d) (b) (b) (b) (d) (a) (b) (e) (b) (a) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (e) (b) (f)
Qku L = 595.2 W, (c) ∆ϕ = 109.1 V, Je = 5.455 A Qku L = 8.718 × 106 W (i) qku L = 1.099 × 106 W/m2 , (ii) qku L = 2 × 106 W/m2 (i) qku L = 1.896 × 104 W/m2 , (ii) qku L = 9.407 × 105 W/m2 Qku D,s = 195.9 W, (b) Qku D,c = 630.5 W, (c) L = 1.08 cm, (d) T2 = 0.97◦C T2 = 817.3 K. Ts = 1,094 K Tf,∞ − Ts,L = 8.71◦C t = 17.77 s < 20 s (i) Qku D = 378.6 W, (ii) Qku L = 91.61 W, (c) δν /D = 0.6438 < 1.0 rtr = 21.70 cm, (c) Qku L /(Ts − Tf,∞ ) = 19.33 W/◦C, (d) Ts = 1,055◦C t = 71.4 min, (b) t = 6.7 min Ts = 1,146 K, (b) Ts = 722.6 K Ts∗ (r = 0, t) = 0.2779 BiD = 4.121 × 10−4 < 0.1, (c) t = 2.602 ms Qku w = 47.62 W, (b) Qku w = 357.1 W T1 (t = 4 s) = 346.2 K, (b) t = 17.0 min, (c) Bil = 3.89 × 10−3 Rk,sl−b = 1.25◦C/W, Rk,sl−s = 6.25◦C/W, Rku D = 2.54◦C/W Qk,slb = 5.76 W, (d) Qk,sl∞ = 4.23 W, (e) S˙sl = 9.99 W NuD = 413.5, Tr,max = 318 K Tp = 511◦C, (c) Tp = 64.5◦C Bil = 0.0658, (c) T1 = 288.15 K, (d) T1 = 273.76 K BiD = 3.042 × 10−3 , (c) t = 6.685 × 10−3 s Ts = 352.9◦C, l/ks < 1.36 × 10−3 ◦C/(W/m2 ) ηf = 0.9426, (c) Ts = 82.57◦C, (d) Γf = 8.287 T1 (t = t0 = 1 hr) = 18.41◦C Rc = 5.2 mm, (c) Rk,12 = 0.3369◦C/W, (Rku )D,2 = 1.319◦C/W, Rc = 5.241 mm T1 = 54.45◦C, (c) T1 = 50.24◦C ˙ O2 = 0.002639 g/s S˙ r,c = 9,082 W, (c) M˙ O2 = 0.5327 g/s, (d) M Ts = 282.9 K, (e) Qku L = −1,369.30 W, Qk,u = −67.39 W t(droplet vanishes) = 382 s, (c) L = 38.2 m M˙ lg = 3.120 × 10−3 g/s, Ts = 285.6 K, (c) ∆t = 578.2 s Qku L = 2.186 W, (c) M˙ lg = 8.808 × 10−7 kg/s, S˙ lg = −2.034 W dTc /dt = −6.029 × 10−3 ◦C/s 1/2 4/5 4/5 NuL = [0.664ReL,t + 0.037(ReL − ReL,t )]Pr1/3 NuD,H = 38.22, (c) N T U = 0.6901, (d) he = 0.4985 Ru L = 0.1080◦C/W, (f) Qku L−0 = 1,019 W, (g) Tf L = 74.84◦C laminar ﬂow: uf = 0.061 m/s, turbulent ﬂow: uf = 1.42 m/s (i) N T U = 115.5, (ii) N T U = 2.957 × 104 Tf L = −43.73◦C, (c) Qu L0 = 224.4 W Qu L0 = 1.276 W, (c) Qu L0 = 1.276 W Ts = 92.74◦C < Tlg , (c) Ts = 399.0◦C > Tlg Qu L0 = 926.5 W, (c) Tf L = 129.5◦C Qku D,h = −61.80 W, (c) M˙ lg = 4.459 × 10−4 kg/s, xL = 0.05410 ˙ lg = 2.298 × 10−4 kg/s, xL = 0.6298 Qku D = 28.93 W, (c) M ◦ Ts = 560.8 C N T U = 21.57, Tf L = 30◦C for N = 400 N T U = 32.36, Tf L = 30◦C for N = 600 Qku L = 3,022 W, Qu L−0 = 3,158 W NuD,p = 53.10, (c) N T U = 2.770, (d) Ts = 20(◦C) + 1.049S˙ e,J Tf L = 20(◦C) + 0.9832S˙ e,J , (f) Tf L = 870.3◦C k = 0.6521 W/mK, (c) N T U = 3.471, (d) BiL = 1.025 × 104 , (e) τs = 1.159 hr, 4τs Qku dt = 2.562 × 109 J 0
x
7.17 7.18 7.19 7.20 7.21 7.22 7.24 7.25
(b) (b) (b) (b) (e) (a)
7.33 7.34
(b) (b) (b) (d) (b) (b) (b) (b) (f) (b) (b) (i) (ii) (b) (a)
8.1 8.2 8.3
(b) (b) (c)
8.4
(b)
7.26 7.27 7.28 7.29 7.30 7.31 7.32
Ts = 484.6◦C, (c) Qr,s∞ = 822.0 W S˙ e,J = 10.80 W Qc /Ac = 310.2 W/m2 NuD = 85.39, (c) Rku D = 2.527 × 10−5 K/W, (d) Rk,sc = 2.036 × 10−3 K/W ˙ sl = 36.24 g/s Qu L0 = −1.209 × 104 W, M (i) NuD = 14,778 and qku D = 9,946 kW/m2 (ii) NuD = 2,246 and qku D = −237.9 kW/m2 (i) Q1−2 = 681.7 W, (ii) Q1−2 = 545.2 W, (iii) Q1−2 = 567.1 W, (c) la = 2 cm Tf,c 0 = 623.6◦C, (c) N = 5.81 6 Tf,h L = 1,767 K, (c) Tf,h 0 = 450.0 K and Tf,c L = 343.6 K Qu L0 = 1,161 W and η = 89.36% Tf,c L = 27.35◦C, Tf,h L = 27.12◦C, (c) Qu L0 = 171.7 W Rku,c = 0.0250 K/W, Rku,h = 0.2250 K/W, (c) Tf,c L = 31.43◦C Qu L0 = 6,623 W, (c) η = 90.73% N T U = 1.106, (c) Cr = 0.07212, (d) he = 0.655, (e) Tf,c L = 34.13◦C, Tf,h L = 47.8◦C, (g) Qu L0 = 908.4 W N T U = 0.07679, (c) eh = 0.07391, (d) Tf,c L = 307.4 K, (e) Qu L0 = 32.09 kW η = 95.0% ˙ lg = 0.01090 kg/s (b) N T U = 0.8501, (c) Qu L0 = 23,970 W, (d) M ˙ lg = 0.01446 kg/s (b) N T U = 1.422, (c) Qu L0 = 31,823 W, (d) M L = 12.45 m Qu,c L0 = 33.01 W, (b)L = 2.410 m, (c) [∆Tf,c ]max = 66.10◦C Je2 Re,o = 22.18 W Tf 0 = 3.559◦C, (c) S˙ e,J = 5.683 W, (d) Ts = 69.75◦C Qu L0 = −6.47 W, Je = 0.8 A, Tc = 290.5 K, Th = 333.3 K, Tf L = 293.0 K, Rku w = 1.515 K/W, Rk,hc = 1.235 K/W, Ru L = 0.4183 K/W, Qk,hc = 36.65 W, S˙ e,P = −46.02 W, (S˙ e,J )c = 5.04 W Ts,1 (t) = Tf,1 (t) = 400◦C for t > 10 s
xi
Chapter 1
Introduction and Preliminaries
PROBLEM 1.1.FAM GIVEN: Introductory materials, deﬁnitions for various quantities, and the concepts related to heat transfer are given in Chapter 1. OBJECTIVE: Deﬁne the following terms (use words, schematics, and mathematical relations as needed). (a) Control Volume V . (b) Control Surface A. (c) Heat Flux Vector q(W/m2 ). (d) Conduction Heat Flux Vector qk (W/m2 ). (e) Convection Heat Flux Vector qu (W/m2 ). (f) SurfaceConvection Heat Flux Vector qku (W/m2 ). (g) Radiation Heat Flux Vector qr (W/m2 ). (h) Net Rate of Surface Heat Transfer QA (W). (i) Conservation of Energy. SOLUTION: Deﬁne the following terms (use words as well as schematics and mathematical relations when needed). (a) Control Volume V : A control volume is a speciﬁed enclosed region of space, selected based on the information sought, on which the heat transfer analysis is applied. See Figure Ex.1.5 for examples of control volumes. (b) Control Surface A: There are two types of control surfaces. One type of control surface is the closed surface which forms the boundary of the control volume, i.e., separates the interior of the control volume from its surroundings. These are used in volumetric energy conservation analysis. The other type of control surface is the surface containing only the bounding surface of a heat transfer medium (or interface between two media). These control surfaces enclose no mass and are used in surface energy conservation analysis. See Figure 1.2 for an example of a control surface. (c) Heat Flux Vector q(W/m2 ): The heat ﬂux vector is a vector whose magnitude gives the heat ﬂow per unit time and per unit area and whose direction indicates the direction of the heat ﬂow at a given point in space x and instant of time t. The mechanisms contributing to the heat ﬂux vector are the conduction heat ﬂux vector qk (W/m2 ), the convection heat ﬂux vector qu (W/m2 ), and the radiation heat ﬂux vector qr (W/m2 ), i.e., q = qk + qu + qr . (d) Conduction Heat Flux Vector qk (W/m2 ): The conduction heat ﬂux vector is a vector whose magnitude gives the heat ﬂow rate per unit area due to the presence of temperature nonuniformity inside the heat transfer medium and molecular conduction (molecular interaction, electron motion, or phonon motion). (e) Convection Heat Flux Vector qu (W/m2 ): The convection heat ﬂux vector is a vector whose magnitude gives the heat ﬂow rate per unit area due to bulk motion of the heat transfer medium. (f) SurfaceConvection Heat Flux vector qku (W/m2 ): This is a special case of conduction heat transfer from the surface of a stationary solid in contact with a moving ﬂuid. The solid and ﬂuid have diﬀerent temperatures (i.e., are in local thermal nonequilibrium). Since the ﬂuid is stationary at the solid surface (i.e., ﬂuid does not slip on the surface), the heat transfer between the solid and the ﬂuid is by convection, but this heat transfer depends on the ﬂuid velocity (and other ﬂuid properties). Therefore the subscripts k and u are used to emphasize ﬂuid conduction and convection respectively. (g) Radiation Heat Flux Vector qr (W/m2 ): The radiation heat ﬂux vector is a vector whose magnitude gives the heat ﬂow rate per unit area in the form of thermal radiation (a part of the electromagnetic radiation spectrum).
2
(h) Net Rate Of Surface Heat Transfer QA (W): The net rate of surface heat transfer is the net heat transfer rate entering or leaving a control volume. The relation to the heat ﬂux vector is QA = (q · sn )dA, A
where sn is the control surface normal vector (pointing outward) and the integration is done over the entire control surface A. See (1.9) for the sign convention for the net rate of surface heat transfer. (i) Conservation Of Energy: The conservation of energy equation is the ﬁrst law of thermodynamics and states that the variation of the total energy of a system (which includes kinetic, potential, and internal energy) is equal to the sum of the net heat ﬂow crossing the boundaries of the system and the net work performed inside the system or at its boundaries. The integralvolume energy equation can be written as ∂E ˙ p A + W ˙ µ A + W ˙ g,e V + S˙ e V . − E˙ u A + W QA = − ∂t V See (1.22) for the description of the various terms. COMMENT: The mechanisms of heat transfer (conduction, convection, and radiation), and the energy equation (including various energy conversion mechanisms) are the central theme of heat transfer analysis. In Chapter 2, a simpliﬁed form of the energy equation will be introduced.
3
PROBLEM 1.2.FUN GIVEN: An automobile radiator is a crossﬂow heat exchanger (which will be discussed in Chapter 7) used to cool the hot water leaving the engine block. In the radiator, the hot water ﬂows through a series of interconnected tubes and loses heat to an air stream ﬂowing over the tubes (i.e., air is in cross ﬂow over the tubes), as shown in Figure Pr.1.2(a). The airside heat transfer is augmented using extended surfaces (i.e., ﬁns) attached to the outside surface of the tubes. Figure Pr.1.2(b) shows a twodimensional close up of the tube wall and the ﬁns. The hot water convects heat qu (W/m2 ) as it ﬂows through the tube. A portion of this heat is transferred to the internal surface of the tube wall by surface convection qku (W/m2 ). This heat ﬂows by conduction qk (W/m2 ) through the tube wall, reaching the external tube surface, and through the ﬁns, reaching the external surface of the ﬁns. At this surface, heat is transferred to the air stream by surface convection qku (W/m2 ) and to the surroundings (which include all the surfaces that surround the external surface) by surface radiation qr (W/m2 ). The heat transferred to the air stream by surface convection is carried away by convection qu (W/m2 ). SKETCH: Figures Pr.1.2(a) to (c) shows an automobile radiator and its various parts. OBJECTIVE: On Figure Pr.1.2(c), track the heat ﬂux vector, identifying various mechanisms, as heat ﬂows from the hot water to the air. Assume that the radiator is operating in steady state. SOLUTION: Figure Pr.1.2(d) presents the heat ﬂux vector tracking for the control volume shown in Figure Pr.1.2(c). The mechanisms of heat transfer are identiﬁed and the thickness of the heat ﬂux vector is proportional to the magnitude of the heat transfer rate. COMMENT: (i) Note that at each interface the heat ﬂux vectors entering and leaving are represented. This facilitates the application of the energy equation for control surfaces and control volumes enclosing interfaces or parts of the system. (ii) The temperature along the ﬁn is not axially uniform; in general, it is also not laterally uniform. The temperature at the base is higher than that at the tip. Thus, the conduction heat transfer rate is larger near the base and decreases toward the tip. As the temperature ﬁeld is twodimensional, the conduction heat ﬂux vector is not normal to the surface. For ﬁns of highly conducting materials or small aspect ratios (small thickness to length ratio), the lateral variation in temperature is generally neglected. Finally, for suﬃciently long ﬁns, the temperature at the tip approaches that of the air ﬂow and under this condition there is no heat transfer through the ﬁn tip. However, for weight and cost reductions, ﬁn lengths are generally chosen shorter than this limit. (iii) The direction of the convection heat ﬂux vector depends on the direction of the complicated ﬂow ﬁeld around the ﬁn and tube. The ﬂow ﬁeld is usually threedimensional and the convection heat ﬂux vector is usually not normal to the surface. (iv) The direction of the radiation heat ﬂux vector depends on the position and temperature of the surfaces surrounding the ﬁn and tube surfaces. These include the other surfaces in the radiator, external surfaces, etc. In this problem such surfaces have not been directly identiﬁed. (v) The conduction heat ﬂux along the tube wall can be neglected for suﬃciently thin tube walls. (vi) The use of ﬁns is justiﬁable when the heat ﬂux through the ﬁns is larger than the heat ﬂux through the bare surface. (vii) Heat transfer with extended surfaces (i.e., ﬁns) is studied in Chapter 6. In that chapter, the surface convection for semibounded ﬂows (i.e., ﬂows over the exterior of solid bodies) is presented. In Chapter 7, the surface convection for bounded ﬂows (i.e., tube ﬂow) is studied. Surface radiation is studied in Chapter 4.
4
Water Inlet
(a) Car Radiator
Air Flow
Air Flow
Water Outlet
(b) Closeup of Tube Wall and Fin
Air Flow
Symmetry Axes
Tube Water Flow
(d) Heat Flux Vector Tracking around a TubeFin Region
qu
Fin
qu qu
Control Volume for the Heat Flux Vector Tracking
(c) Control Volume for the Heat Flux Vector Tracking
qku
qku
qk
qk qk
Symmetry Axes Water Flow
qu
qr
qk qku qu
Water Flow
qk q ku
Tube Wall
qr
qr qk q ku
Air Flow Fin
qr qk q ku
qk qku qu
qu
qr
Figure Pr.1.2(a), (b), and (c) An automobile radiator shown at various length scales. (d) Heat ﬂux vector tracking around a tubeﬁn region.
5
PROBLEM 1.3.FUN GIVEN: A ﬂatplate solar collector [Figure Pr.1.3(a)] is used to convert thermal radiation (solar energy) into sensible heat. It uses the sun as the radiation source and water for storage of energy as sensible heat. Figure Pr.1.3(b) shows a cross section of a ﬂatplate solar collector. The space between the tubes and the glass plate is occupied by air. Underneath the tubes, a thermal insulation layer is placed. Assume that the glass absorbs a fraction of the irradiation and designate this heat absorbed per unit volume as s˙ e,σ (W/m3 ). Although this fraction is small when compared to the fraction transmitted, the glass temperature is raised relative to the temperature of the air outside the solar collector. The remaining irradiation reaches the tube and ﬁn surfaces, raising their temperatures. The temperature of the air inside the solar collector is higher than the glass temperature and lower than the tube and ﬁn surface temperatures. Then the thermobuoyant ﬂow (i.e., movement of the air due to density diﬀerences caused by temperature diﬀerences) causes a heat transfer by surfaceconvection at the glass and tube surfaces. The net heat transfer at the tube surface is then conducted through the tube wall and transferred to the ﬂowing water by surface convection. Finally, the water ﬂow carries this heat away by convection. Assume that the ambient air can ﬂow underneath the solar collector. SKETCH: Figures Pr.1.3(a) to (c) show a ﬂatplate solar collector and its various components.
(a) FlatPlate Solar Collector Solar Irradiation Wind
(b) Section AA Ambient Air
Glass Plate Glass Plate
A
Fin Air Water
.
se,I Water Inlet
Tube
A
Water Outlet
Plastic or Metallic Container
Control Volume for the Thermal Heat Flux Vector Tracking Insulation
(c) Control Surface for the Heat Flux Vector Tracking Glass Plate Air Tube Fin
Water Thermal Insulation Container
Figure Pr.1.3(a), (b), and (c) A ﬂatplate solar collector shown at various length scales.
OBJECTIVE: Track the heat ﬂux vector for this thermal system. Note that the tubes are arranged in a periodic structure and assume a twodimensional heat transfer. Then, it is suﬃcient to track the heat ﬂux vector for a control volume that includes half of a tube and half of a connecting ﬁn, as shown on Figure Pr.1.3(c). SOLUTION: Figure Pr.1.3(d) shows the heat ﬂux vector path for the cross section of the ﬂatplate collector. COMMENT: The radiation absorption in the glass plate depends on the radiation properties of the glass and on the wavelength of the thermal radiation. The tube and ﬁn surfaces also reﬂect part of the incident radiation. The diagram in Figure Pr.1.3(d) represents the net radiation heat transfer between the tube and ﬁn surfaces and the surroundings. Radiation heat transfer will be studied in detail in Chapter 4. 6
qu
qku
qr
qk
se,I qk
qk
Glass Plate
qku qu
qu
Air qr qku Tube
qk qku
qk Water
qr qku
qk
Fin qk
qr
qku
qk
qku
qk Thermal Insulation
qk
qk
qk
qk qr
qku
qku q qr k
qr
qk
Box
qk qr
qku
qku qu
qu
Figure Pr.1.3(d) Heat ﬂux vector tracking around glass, tube, and insulator.
The objective in the ﬂatplate solar collector is to convert all the available thermal irradiation to sensible heat in the water ﬂow. The heat ﬂux vector tracking allows the identiﬁcation of the heat losses, the heat transfer mechanisms associated with the heat losses, and the heat transfer media in which heat loss occurs. Minimizing heat loss is usually done through the suppression or minimization of the dominant undesirable heat transfer mechanisms. This can be achieved by a proper selection of heat transfer media, an active control of the heat ﬂux vectors, or a redesign of the system. Economic factors will ﬁnally dictate the actions to be taken.
7
PROBLEM 1.4.FAM GIVEN: In printedcircuit ﬁeldeﬀect transistors, conversion of electromagnetic energy to thermal energy occurs in the form of Joule heating. The applied electric ﬁeld is time periodic and the heat generated is stored and transferred within the composite layers. This is shown in Figure Pr.1.4(a). The dimensions of the various layers are rather small (measured in submicrons). Therefore, large electrical ﬁelds and the corresponding large heat generation can elevate the local temperature beyond the threshold for damage. SKETCH: Figures Pr.1.4(a) and (b) show the ﬁeldeﬀect transistor.
(b) Dimensions
(a) Physical Description of FieldEffect Transistor ,ϕds , Applied Voltage ,ϕg
Jd Depletion Region
Source
Gate
Active Layer 0.1 µm
Drain
Source 0.2 µm
.
Se,J Gate
Drain
0.1 µm
Silicon Substrate
Joule . Electron Heating Se,J Transfer
SemiInsulating Substrate
Figures Pr.1.4(a) and (b) Fieldeﬀect transistor.
OBJECTIVE: On Figure Pr.1.4(b), track the heat ﬂux vector. Note that the electric ﬁeld is transient. SOLUTION: The electromagnetic energy converted to thermal energy by Joule heating is stored in the device, thus raising its temperature, and is transferred by conduction toward the surface. At the surface the heat is removed by surface convection and radiation. These are shown in Figure Pr.1.4(c). Convection qu
(c) Heat Flow 0.1 µm
Source 0.2 µm
Surface Convection qku Surface Radiation qr
Conduction qk Se,J
Drain
Gate
Storage qk 0.1 µm qk
Symmetry Line
Figure Pr.1.4(c) Heat ﬂux vector tracking in a ﬁeldeﬀect transistor.
8
COMMENT: For a time periodic electric ﬁeld with very high frequency the heat is mostly stored, resulting in large local temperatures. This limits the frequency range for the operation of transistors (because at higher temperature the dopants migrate and the transistor fails). The search for semiconductors that can safely operate at higher temperatures aims at overcoming this limitation. The Joule heating is caused as an electric ﬁeld is applied and the electrons are accelerated and collide with the lattice atoms and other electrons. Since the electrons are at a much higher temperature than the lattice, these collisions result in a loss of momentum and this is the Joule heating.
9
PROBLEM 1.5.FAM GIVEN: The attachment of a microprocessor to a printed circuit board uses many designs. In one, solder balls are used for better heat transfer from the heat generating (Joule heating) microprocessor to the printed circuit board. This is shown in Figure Pr.1.5(a). SKETCH: Figure 1.5(a) shows a solderball attachment of a microprocessor to a printed circuit board.
Physical Model of Microprocessor and Circuit Board with Solder Balls Microprocessor Adhesive
Heat Sink or Coverplate
Thermal Adhesive Solder Balls Printed Circuit Board Se,J
Figure Pr.1.5(a) Solderball connection of microprocessor to the printed circuit board.
OBJECTIVE: Track the heat ﬂux vector from the microprocessor to the heat sink (i.e., bare or ﬁnned surface exposed to moving, cold ﬂuid) and the printed circuit board. SOLUTION: Figure Pr.1.5(b) shows the heat ﬂux vector starting from the microprocessor. Within the solid phase, the heat transfer is by conduction. From the solid surface to the gas (i.e., air), the heat transfer mechanism is surface radiation. If the gas is in motion, heat is also transferred by surface convection.
qku
qu qr qu qr
qu
qku
qu qku
qk
qk
qr qk
qk
qk qu qk
qku
qk
qk qu
Se,J
qr
qku
qk qr
qk
qu
qr qk
qr
qku qu
qu
qk qr
qku qu
qu qr qu
qk
qr
Figure Pr.1.5(b) Heat ﬂux vector tracking in microprocessor and its substrate.
COMMENT: If the heat generation S˙ e,J is large, which is the case for high performance microprocessors, then a heat sink (e.g., a ﬁnned surface) is needed. We will address this in Section 6.8.
10
PROBLEM 1.6.FAM GIVEN: As part of stemcell transplantation (in cancer treatment), the donor stem cells (bone marrow, peripheral blood, and umbilical cord blood stem cells) are stored for later transplants. Cryopreservation is the rapid directional freezing of these cells to temperatures below −130◦C. Cryopreservative agents are added to lower freezing point and enhance dehydration. Cooling rates as high as −dT /dt = 500◦C/s are used (called rapid vitriﬁcation). The cells are frozen and kept in special leakproof vials inside a liquid nitrogen storage system, shown in Figure Pr.1.6(a). At one atmosphere pressure, from Table C.4, Tlg (p = 1 atm) = 77.3 K = −195.9◦C. The storage temperature aﬀects the length of time after which a cell can be removed (thawed and able to establish a cell population). The lower the storage temperature, the longer the viable storage period. In one protocol, the liquid nitrogen level in the storage unit is adjusted such that T = −150◦C just above the stored material. Then there is a temperature stratiﬁcation (i.e., ﬂuid layer formation with heavier ﬂuid at the bottom and lighter ﬂuid on top) with the temperature varying from T = −196◦C at the bottom to T = −150◦C at the top of the unit, as shown in Figure Pr.1.6(a). SKETCH: Figure Pr.1.6(a) shows the storage container and the temperature stratiﬁcation within the container. Vent
Ambient Air Tf, >> Tlg
Insulation
Container Wall
Nitrogen Vapor Tf = 150 C
Vials
Nitrogen Vapor Tf = 178 C
Makeup Liquid Nitrogen g
Liquid Nitrogen Tf = Tlg = 196 C slg < 0
Figure Pr.1.6(a) An insulated container used for storage of cryopreserved stem cells.
OBJECTIVE: Draw the steadystate heat ﬂux vector tracking for the storage container showing how heat transfer by surface convection and then conduction ﬂows through the container wall toward the liquid nitrogen surface. Also show how heat is conducted along the container wall to the liquid nitrogen surface. Note that S˙ lg < 0 since heat is absorbed during evaporation. In order to maintain a constant pressure the vapor is vented and makeup liquid nitrogen is added. SOLUTION: Figure Pr.1.6(b) shows the heat ﬂux vector tracking, starting from the ambient air convection qu , surface radiation qr , and surface convection qku , and then leading to conduction qk through the insulation. Heat is qr qu qk
qku qk
qr
qk qu
slg
qku
Figure Pr.1.6(b) Tracking of the heat ﬂux vector.
11
conducted through the container wall and, due to the higher temperature at the top of the container, heat is also conducted along the wall and toward the liquid nitrogen surface. All the conducted heat is converted into the liquidvapor phase change S˙ lg (which is negative). The upper portion of the container also transfers heat to the liquid surface by surface radiation qr . COMMENT: Note that between the top and bottom portions of the container there is a diﬀerence in temperature ∆T = 46◦C. Since the heavier gas is at the bottom, no thermobuoyant motion will occur. A special insulation is needed to minimize the heat leakage into the container (and thus reduce the needed liquid nitrogen makeup ﬂow rate).
12
PROBLEM 1.7.FAM GIVEN: Inductioncoupling (i.e., electrodeless) Joule heating S˙ e,J of thermal plasmas, which are high temperature (greater than 10,000 K) ionizedgas streams, is used for particle melting and deposition into substrates. Figure Pr.1.7(a) shows a plasma spraycoating system. The powder ﬂow rate strongly inﬂuences particle temperature history Tp (t), i.e., the speed in reaching the melting temperature (note that some evaporation of the particles also occurs). This is called inﬂight plasma heating of particles. To protect the plasma torch wall, a highvelocity sheathgas stream is used, along with liquidcoolant carrying tubes embedded in the wall. These are also shown in Figure Pr.1.7(a). SKETCH: Figure Pr.1.7(a) shows the torch and the (i) plasmagasparticle stream, and (ii) the sheathgas stream. Also shown is (iii) a single particle. Plasma Gas (e.g., Ar, N2, H2, Air) Particles (Powder) and Carrier Gas (e.g., Ar, N2, H2, Air)
Sheath (or Coolant) Gas (e.g., Ar, N2, H2, Air)
(ii) SheathGas Stream
Plasma Torch Wall
Induction Coil Liquid Coolant Coil
(i) PlasmaGasParticles Stream (iii) Particle, Tp(t)
se,J Powder
Cold Gas Stream High Temperature Plasma Envelope
qku + qr Powder Deposit us , Substrate Motion
qk
Substrate
Figure Pr.1.7(a) A plasma spraycoating torch showing various streams, Joule heating, and wall cooling.
OBJECTIVE: (a) Draw the heat ﬂux vector tracking for the (i) plasmagasesparticles, and (ii) sheathgas streams. Allow for conductionconvectionradiation heat transfer between these two streams. Follow the plasma gas stream to the substrate. (b) Draw the heat ﬂux vector tracking for (iii) a single particle, as shown in Figure Pr.1.7(a). Allow for surface convection and radiation and heat storage as −∂E/∂t (this is sensible and phasechange heat storage). SOLUTION: (a) Figure Pr.1.7(b) shows the heat ﬂux vector tracking. We start with the plasma gasparticle stream and, since it is cold at the torch entrance, its convection heat ﬂux qu is shown. Upon Joule heating, its temperature increases and radiation heat transfer also becomes signiﬁcant. As the stream proceeds, it transfers heat (by conduction, convection, and radiation) to the sheathgas stream. The sheathgas stream in turn transfers heat by surface convection to the cold wall. The plasmagasparticle stream reaches the substrate and transfers heat to the substrate by surface convection (this is similar to an impinging jet). (b) Figure Pr.1.7(b) shows the heat ﬂux vector tracking for the particles. Heat is transferred to the particles by surface radiation and surface convection. This heat is stored in the particles as sensible heat (resulting in a rise in its temperature), and as heat of phase change (melting and evaporation). Using (1.22), this is shown as −∂E/∂t.
13
Plasma Torch Wall
qu
qu
Coolant Coil
se,J qr
qk ,qu ,qr qku
qr qr Particle, Tp
qu
(ii)
qu
se,J
(iii) qu
(i)
qu
qr
qu
qr
qku qr  dE dt
Sheath Gas Plasma Gas qu qloss (qk , qu , qr)
qr
qku
qu
qk
Figure Pr.1.7(b) Tracking of the heat ﬂux vector.
COMMENT: In Chapter 5, we will discuss Joule heating of gas streams and plasma generators. In Chapters 4 and 7, we will discuss surfaceradiation and surfaceconvection heating of objects.
14
PROBLEM 1.8.FAM GIVEN: A bounded cold air stream is heated, while ﬂowing in a tube, by electric resistance (i.e., Joule heating). This is shown in Figure Pr.1.8(a). The heater is a solid cylinder (ceramics with the thin, resistive wire encapsulated in it) placed centrally in the tube. The heat transfer from the heater is by surface convection and by surfaceradiation emission (shown as S˙ e, ). This emitted radiation is absorbed on the inside surface of the tube (shown as S˙ e,α ) and then leaves this surface by surface convection. The outside of the tube is ideally insulated. Assume that no heat ﬂows through the tube wall. SKETCH: Figure Pr.1.8(a) shows the tube, the air stream, and the Joule heater. The surface radiation emission and absorption are shown as S˙ e, and S˙ e,α , respectively. Tube Wall Se,= Se,J
Se,
Ideal Insulation (No Heat Flows in Tube Wall)
Bounded Hot Air Stream Out
Bounded Cold Air Stream In
Electric Resistance Heating (Thin Wires Encapsulated in Ceramic Cover)
Figure Pr.1.8(a) A bounded air stream ﬂowing through a tube is heated by a Joule heater placed at the center of the tube.
OBJECTIVE: Draw the steadystate heat ﬂux tracking showing the change in ﬂuid convection heat ﬂux vector qu , as it ﬂows through the tube. SOLUTION: Figure Pr.1.8(b) shows the inlet ﬂuid convection heat ﬂux vector qu entering the tube. The heat transfer by surface convection qku from the heater contributes to this convection heat ﬂux vector. The surface radiation emission from the heater is absorbed by the inner surface of the tube. Since no heat ﬂows in the tube wall, this heat leaves by surface convection and further contributes to the air stream convection heat ﬂux vector. These are shown in Figure Pr.1.8(b).
No Heat Flow S (Surface Radiation Absorption) (Ideal Insulation) e,= q ku
qu qku
qu
qr
qk
Se, (Surface Radiation Emission)
Se,J
Figure Pr.1.8(b) Tracking of heat ﬂux vector.
COMMENT: In practice, the heater may not be at a uniform surface temperature and therefore, heat ﬂows along the heater. The same may be true about the tube wall. Although the outer surface is assumed to be ideally insulated, resulting in no radial heat ﬂow at this surface, heat may still ﬂow (by conduction) along the tube wall.
15
PROBLEM 1.9.FAM GIVEN: Water is bounded over surfaces by raising the substrate surface temperature Ts above the saturation temperature Tlg (p). Consider heat supplied for boiling by electrical resistance heating (called Joule heating) S˙ e,J in the substrate. This is shown in Figure Pr.1.9(a). This heat will result in evaporation in the form of bubble nucleation, growth, and departure. The evaporation site begins as a bubble nucleation site. Then surfaceconvection heat transfer qku is supplied to this growing bubble (i) directly through the vapor (called vapor heating), (ii) through a thin liquid ﬁlm under the bubble (called micro layer evaporation), and (iii) through the rest of the liquid surrounding the vapor. Surfaceconvection heat transfer is also supplied (iv) to the liquid (resulting in slightly superheated liquid) and is moved away by liquid motion induced by bubble motion and by thermobuoyancy. SKETCH: Figure Pr.1.9(a) shows the heat supplied by Joule heating within the substrate and a site for bubble nucleation, growth, and departure. Liquid Surface Vapor Escape g
Pool of Liquid, Tlg (p) Circulating (Cellular) Liquid Motion Induced by Bubbles and by Thermobuoyancy
Departure Nucleation and Growth of Bubble (Vapor)
Liquid Supply
Slg , Evaporation Ts > Tlg (p)
Heat Substrate
Liquid Microlayer Se,J , Joule Heating
Figure Pr.1.9(a) The nucleate pool boiling on a horizontal surface. The Joule heating results in raising the surface temperature above the saturation temperature Tlg , and bubble nucleation, growth, and departure.
OBJECTIVE: Track the heat ﬂux vector starting from the Joule heating site S˙ e,J within the substrate and show the surfaceconvection heat transfer, (i) to qku (iv). Also follow the heatﬂux vector to the liquid surface. Assume a timeaveraged heat transfer in which the bubbles are formed and depart continuously. SOLUTION: In Figure Pr.1.9(b), starting from S˙ e,J , the heat ﬂows by conduction to the substrate surface. There is also conduction away from the nucleate pool boiling surface and this is labeled as the heat loss. Heat is transferred from the solid by surface convection qku to the vapor, to the thin liquid microlayer, to the liquid surrounding the bubble, and to the bulk liquid phase. These are shown in Figure Pr.1.9(b). The heat is in turn removed by the departing bubbles and by liquid convection qu to the surface resulting in vapor escape and further evaporation. COMMENT: The relative magnitudes of qku (i) to qku (iv) are discussed in Section 6.6.1 and Figure 6.17 gives additional descriptions of the nucleate pool boiling.
16
Slg
Vapor Escape qu
qu
Departing Bubble qu
qu
Slg qku (iv) qk
Se,J
qku (i) qku (ii) qku (iii) qk (Heat Loss)
Figure Pr.1.9(b) Tracking of the heat ﬂux vector, starting from the Joule heating location.
17
PROBLEM 1.10.FAM GIVEN: Deep heat mining refers to harvesting of the geothermal energy generated locally by radioactive decay S˙ r,τ and transferred by conduction Qk from the earth mantle [shown in Figure Ex.1.2(a)]. Mining is done by the injection of cold water into fractured rocks (geothermal reservoir) followed by the recovery of this water, after it has been heated (and pressurized) by surfaceconvection qku in the fractures, through the production wells. These are shown in Figure Pr.1.10(a). The heated water passes through a heat exchanger and the heat is used for energy conversion or for process heat transfer. SKETCH: Figure Pr.1.10(a) shows a schematic of deep heat mining including cold water injection into hot, fractured rocks and the recovery of heated (and pressurized) water. The heat generation by local radioactive decay S˙ r,τ and by conduction from the earth mantle are also shown. Makeup Water Reservoir Cooling
Surface Heat Exchanger Central Monitoring
Power Generation
Observation Borehole
Heat Distribution
Pump u
u
u
Sedime
nts
Production Well (Hot Water) Injection Well (Cold Water)
400060
Observation Borehole
Crystall
ine Roc
ks
00 m Stimulated Fracture System
01
50 00
sr,J
0m
5001
000 m
Q = Qk Condution Heat Transfer at Control Surface
Figure Pr.1.10(a) Deep heat mining by injection of cold water into hot rocks and recovery of heated water.
OBJECTIVE: Starting from the energy conversion sources S˙ r,τ and the heat conduction from lower section Q = Qk , draw the steadystate heat ﬂux vector tracking and show the heat transfer to the cold stream by surface convection qku . Note that heat is ﬁrst conducted through the rock before it reaches the water stream. Follow the returning warm water stream to the surface heat exchanger.
18
SOLUTION: Figure Pr.1.10(b) shows the heat ﬂux vector tracking. The heat ﬂux emanating from S˙ r,τ and boundary Qk ﬂows into the rock by conduction qk . This heat is then transferred to the water stream by surface convection qku . The heated stream convects heat qu to the surface heat exchanger. The direction of qu is the same as u (ﬂuid velocity).
qu qu qu qu
qu
qu
qu
qu qu qu
qku
qu
qk sr,J
qku qk
sr,J
Figure Pr.1.10(b) Tracking of the heat ﬂux vector.
COMMENT: The surfaceconvection heat transfer of bounded ﬂuid streams (such as the water stream in fractured rocks) will be discussed in Chapter 7. There we will show that when a large surface area (per unit volume) exists for surfaceconvection heat transfer, the stream reaches the local bounding solid temperature. Here this temperature can be high, which can cause volumetric expansion (and pressurization) of water.
19
PROBLEM 1.11.FAM GIVEN: In a seabed hydrothermal vent system, shown in Figure Pr.1.11(a), cold seawater ﬂows into the seabed through permeable tissues (fractures) and is heated by the body of magma. The motion is caused by a density diﬀerence, which is due to the temperature variations, and is called a thermobuoyant motion (it will be described in Chapter 6). Minerals in the surrounding rock dissolve in the hot water, and the temperaturetolerant bacteria release additional metals and minerals. These chemical reactions are represented by S˙ r,c (which can be both endo and exothermic). Eventually, the superheated water rises through the vent, its plume forming a “black smoker.” As the hot water cools, its metal content precipitate, forming concentrated bodies of ore on the seabed. SKETCH: Figure Pr.1.11(a) shows the temperature at several locations around the vent and thermobuoyant water ﬂow.
Ambient 2C
Oxyanions (HPO42+, HVO42, CrO42, HAsO42), REE Trace Metals
3
He, Mn2+, H4SiO4 , FeOOH, MnO2 ,
Seawater uf, = 0.1 cm/s Water
Black Smoker
,T, CH4 , Fe2+, FexSy , 222Rn, H2, H2S Basalt
Basalt
g Control Volume, V
er at
+

2+
Precipitation Chimney
ter
Mg2+, SO42
Sea w a
Sedimentation
Hot Plume 350 C Hydrothermal
2+ 3
H , Cl , Fe , Mn , He, H4SiO4 , H2S, CH4 , CO2 , H2 , Ca2+, K+, Li+, Cu2+, Zn2+, Pb2+
E vo l v e d S
ea w
Sr,c 400 C
Magma, 1200 C Sr,J
Figure Pr.1.11(a) A hydrothermal vent system showing the temperature at several locations and the thermobuoyant ﬂow.
OBJECTIVE: Draw the heat ﬂux vector tracking for the volume marked as V . Note that water ﬂows in the permeable seabed, and therefore, convection should be included (in addition to conduction). This is called the intramedium convection (as compared to surface convection) and will be discussed in Chapter 5. SOLUTION: Figure Pr.1.11(b) shows the heat ﬂux vector tracking for the hydrothermal vent. This tracking starts at the location with the highest temperature, which is the magma. The liquid ﬂows toward the high temperature location to be heated and then it rises. Therefore, there is convection toward the magma from the periphery and convection away from the magma toward the vent. The conduction heat transfer is from the magma toward the periphery and the vent. Therefore, the conduction heat ﬂow opposes the convection for the heat ﬂow toward the periphery, but assists it toward the vent. These will be discussed in Chapter 5, where we consider the intramedium conductionconvection. COMMENT: The radiation heat transfer is negligible. Although the visible portion of the thermal radiation will penetrate through the water, the infrared portion will be absorbed over a short distance. Therefore, the meanfree path of photon λph is very short in water. 20
g Thermobuoyant Flow Tlow
Permeable Seabed
Convection Along the Tlow Fluid with Velocity u qu Vent u qu
Tlow qu
Sr,c
qk
qk qk
Conduction Towards LowTemperature Locations
Conduction From Magma Sr,J
Figure Pr.1.11(b) Tracking of the heat ﬂux vector.
21
PROBLEM 1.12.FAM GIVEN: Electric currentcarrying wires are electrically insulated using dielectric material. For low temperature, a polymeric solid (such as Teﬂon) is used, and for high temperature application (such as in top range electrical oven), an oxide ceramic is used. Figure Pr.1.12(a) shows such a wire covered by a layer of Teﬂon. The Joule heating S˙ e,J produced in the wire is removed by a cross ﬂow of air, with air farﬁeld temperature Tf,∞ being lower than the wire temperature Tw . SKETCH: Figure Pr.1.12(a) shows the wire and the cross ﬂow of air.
(a) Air Flow Over Cylinder
(b) CrossSectional View
Cross Flow of Air Tf, , uf,
Insulation (Teflon) Tf, uf,
Joule Heating, Se,J Electrical CurrentConducting Wire Electrical Insulation Coating (Teflon)
Se,J Wire Tw > Tf,
Figure Pr.1.12(a) and (b) An electrical currentcarrying wire is covered with a layer of electrical insulation, and Joule heating is removed by surface convection and surfaceradiation heat transfer.
OBJECTIVE: Draw the steadystate heat ﬂux vector tracking, starting from the heating source, for this heat transfer problem. Allow for surface radiation (in addition to surface convection). SOLUTION: Figure Pr.1.12(c) shows the heat ﬂux vector tracking, starting from the heat source S˙ e,J . Heat is conducted through the wire and electrical insulation (since both media attenuate radiation signiﬁcantly and therefore, radiation heat transfer is neglected). This heat is removed by surface convection and surface radiation. The surfaceradiation heat transfer is to the surroundings (air can be treated as not attenuating the radiation). The surface convection (on the surface, which is by ﬂuid convection but is inﬂuenced by ﬂuid motion) leads to convectionconduction adjacent to the surface and then to convection away from the surface. As the hot air ﬂows downstream from the wire, it loses heat by conduction and eventually returns to its upstream temperature.
qr qu qk qu Air Flow Tf, , uf,
qku qk
qu
Wire
qk
qu qu qr Radiation to qu qk Surroundings
Air Flow
Electrical Insulation (Teflon)
qk
qu qk qku qu
. Se,J
qk
qu qu
qk qku qr Surface Convection
Figure Pr.1.12(c) Tracking of heat ﬂux vector.
22
qu
COMMENT: Note that surface convection that occurs on the solid surface is conduction through the contact of the surface with the ﬂuid molecules, which are nonmoving (from a statistical average view point). But this conduction heat transfer rate is inﬂuenced by the ﬂuid motion (near and far from the surface).
23
PROBLEM 1.13.FAM.S GIVEN: Popcorn can be prepared in a microwave oven. The corn kernels are heated to make the popcorn by an energy conversion from oscillating electromagnetic waves (in the microwave frequency range) to thermal energy designated as s˙ e,m (W/m3 ). With justiﬁable assumptions for this problem, (1.23) can be simpliﬁed to Q A = −ρcv V
dT + s˙ e,m V, dt
integralvolume energy equation,
where the corn kernel temperature T is assumed to be uniform, but time dependent. The control volume for a corn kernel and the associated energy equation terms are shown in Figure Pr.1.13(a). The surface heat transfer rate is represented by Q A =
T (t) − T∞ , Rt
where T∞ is the farﬁeld ambient temperature and Rt (K/W) is the constant heat transfer resistance between the surface of the corn kernel and the farﬁeld ambient temperature. ρ = 1,000 kg/m3 , cv = 1,000 J/kgK, V = 1.13×10−7 m3 , s˙ e,m = 4×105 W/m3 , T (t = 0) = 20◦C, T∞ = 20◦C, Rt = 5 × 103 K/W. SKETCH: Figure Pr.1.13(a) shows the corn kernel and the thermal circuit diagram. Oscillating Electric Field Intensity ee Corn Kernel, T(t)
QA T
V
 ρcvV dT dt
Rt (K/W) se,mV
Figure Pr.1.13(a) Thermal circuit model for a corn kernel heated by microwave energy conversion.
OBJECTIVE: (a) For the conditions given below, determine the rise in the temperature of the corn kernel for elapsed time t up to 5 min. Use a software for the time integration. (b) At what elapsed time does the temperature reach 100◦C? SOLUTION: The energy equation QA =
T − T∞ dT + s˙ e,m V = −ρcv V Rt dt
is an ordinary diﬀerential equation with T as the dependent variable and t as the dependent variable. The solution requires the speciﬁcation of the initial condition. This initial condition is T (t = 0). This energy equation has a steadystate solution (i.e., when the temperature no longer changes). The solution for the steady temperature is found by setting dT /dt = 0 in the above energy equation. Then, we have T − T∞ = s˙ e,m V Rt
or
T = T∞ + s˙ e,m V Rt .
Here we are interested in the transient temperature distribution up to t = 5 min. The solver (such as SOPHT) requires speciﬁcation of the initial condition and the constants (i.e., ρ, cv , V , s˙ e,m and Rt ) and a numerical integration of the transient energy equation. (a) The solution for T = T (t), up to t = 1,000 s, is plotted in Figure Pr.1.13(b). Examination shows that initially (T − T∞ )/Rt is small (it is zero at t = 0) and the increase in T is nearly linear. Later the time rate of increase 24
in T begins to decrease. At steadystate (not shown), the time rate of increase is zero. (b) The time at which T = 100◦C is t = 247 s and is marked in Figure Pr.1.13(b). 220
T, oC
180 140 100 60 20 0
200
400
600
800
1,000
t, s Figure Pr.1.13(b) Variation of corn kernel temperature with respect to time.
COMMENT: The pressure rise inside the sealed corn kernel is due to the evaporation of the trapped water. This water absorbs most of the electromagnetic energy. Once a threshold pressure is reached inside the corn kernel, the sealing membrane bursts.
25
PROBLEM 1.14.FAM GIVEN: In severely cold weathers, an automobile engine block is kept warm heated prior to startup, using a block Joule heater at a rate S˙ e,J with the electrical power provided through the household electrical circuit. This is shown in Figure Pr.1.14(a). The heat generated conducts through the block of mass M and then is either stored within the volume V or lost through the surface A. The energy equation (1.22) applies to the control surface A. Consider that there is no heat transfer by convection across the surface A1 , i.e., Qu = 0. The conduction heat transfer rate (through the fasteners and to the chassis) is Qk , the surfaceconvection heat transfer rate (to the ambient air) is Qku , and the surfaceradiation heat transfer rate (to the surrounding surface) is Qr . In addition, there is a prescribed heat transfer rate Q (not related to any heat transfer mechanism) Q = 20 W, Qku = 80 W, Qk = 30 W, Qr = 15 W, S˙ e,J = 400 W, cv = 900 J/kgK, M = 150 kg. SKETCH: Figure Pr.1.14(a) shows the heated engine block with the Joule heater shown separately. Surface Convection to Ambient Air Qku Surface Radiation Surface A to Ambient Air Qr Volume V Engine Block, Initially at T(t = 0)
,j = 115 V
Assume a Uniform Property
Engine Block Heater Se,J Qk Heat Transfer from Fasterners to Chassis
Q Prescribed Heat Transfer Rate
Figure Pr.1.14(a) A block Joule heater inserted in an automobile engine block.
OBJECTIVE: (a) Draw the heat ﬂux vector tracking starting from the Joule heating site. (b) By applying the energy conservation equation to the control volume surface, determine the rate of change of the block temperature dT /dt, for the following condition. Use (1.22) and set all terms on the righthand side except the ﬁrst and last terms equal to zero. Use ∂E/∂t = M cv dT /dt and the conditions given below. The last term is equal to S˙ e,J . SOLUTION: (a) Figure Pr.1.14(b) shows the heat ﬂux vector tracking starting from the Joule heater. The heat is conducted qk through the block and is either started, −∂E/∂t or conducted qk to the surface. Then it is transferred through surfaceconvection qku , surface radiation qr , conduction qk , or through a prescribed (but not explicitly associated with any heat transfer mechanism) rate q to the surroundings. (b) The energy equation (1.22), applied to the control volume shown in Figure Pr.1.14(b) becomes QA
= Q + Qk + Qu + Qku + Qr ∂E + S˙ e,J = − ∂t dT + S˙ e,J . = −M cv dt
Noting that Qu = 0, and solving for dT /dt, we have S˙ e,J − Q − Qk − Qku − Qr dT = . dt M cv 26
Surface Convection qu A Radiation qku qr Conduction V qk Energy Conversion Se,J qk
Energy Storage Mcv dT =  dE dt dt
qk
q Prescribed Heat Transfer Rate
qk Conduction
Figure Pr.1.14(b) Tracking of heat ﬂux vector.
Now using the numerical values, we have dT dt
=
(400 − 20 − 30 − 80 − 15)(W) 150(kg) × 900(J/kgK)
=
1.889 × 10−3 ◦C/s.
COMMENT: At this rate, to increase the engine block temperature by 10◦C, an elapsed time of ∆T ∆t
=
1.889 × 10−3 ◦C/s
∆t
=
∆T (◦C) = 5,294 s = 1.471 hr, 1.889 × 10−3 (◦C/s)
is needed. In general, Q, Qk , Qku and Qr all change with the engine block temperature T . In Chapters 3 to 7 these surface heat transfer rates are related to heat transfer resistances Rt , which in turn depend on the various heat transfer parameters.
27
PROBLEM 1.15.FAM GIVEN: In sparkignition engines, the electrical discharge produced between the spark plug electrodes by the ignition system produces thermal energy at a rate S˙ e,J (W). This is called the Joule heating and will be discussed in Section 2.3. This energy conversion results in a rise in the temperature of the electrodes and the gas surrounding the electrodes. This hightemperature gas volume V , which is called the plasma kernel, is a mixture of air and fuel vapor. This plasma kernel develops into a selfsustaining and propagating ﬂame front. About A Q A dt = −1 mJ is needed to ignite a stagnant, stoichiometric fuelair mixture of a small surface area A and small volume V , at normal engine conditions. The conventional ignition system delivers 40 mJ to the spark. (ρcv V )g = 2 × 10−7 J/◦C, Tg (t = 0) = 200◦C. SKETCH: Figure 1.15(a) shows the spark plug and the small gas volume V being heated.
Igniting Gas Kernel by Spark Plug
Electrical Insulator
Center Electrode
Insulator Nose
.
Ground Electrode
Plasma Kernel, Se,J
Figure Pr.1.15(a) Ignition of a fuelair mixture by a spark plug in a sparkignition engine. The plasma kernel is also shown.
OBJECTIVE: (a) Draw the heat ﬂux vector tracking for the region around the electrodes marked in Figure Pr.1.15(a). Start from the energy conversion source S˙ e,J . (b) Assume a uniform temperature within the gas volume V . Assume that all terms on the righthand side of (1.22) are negligible, except for the ﬁrst term. Represent this term with dTg ∂E . = (ρcv V )g ∂t V dt Then for the conditions given below, determine the ﬁnal gas temperature Tg (tf ), where the initial gas temperature is Tg (t = 0). (c) What is the eﬃciency of this transient heating process? SOLUTION: (a) Figure Pr.1.15(b) shows the heat ﬂux vector tracking for conduction and heat storage in the electrodes. The gas kernel, where the energy conversion occurs, also stores and conducts heat. (b) Using (1.22), with all the righthand side terms set to zero except for energy storage, we have dTg ∂E . = −(ρcv V )g QA = − ∂t V dt Integrating this with respect to time, from t = 0 to a ﬁnal time where t = tf , we have tf QA dt = −(ρcv V )g [Tg (tf ) − Tg (t = 0)]. 0
28
Plasma Kernal Heated by Joule Heating Qk
 (ρcvV )e dTe dt
Qk Qr
Qk
 (ρcvV )e dTe dt Q u
QA
Qku
 (ρcvV )g dTg dt
Plasma Kernal
Qk
Figure Pr.1.15(b) Thermal circuit diagram.
Solving for Tg (tf ), we have
tf
QA dt Tg (tf ) = Tg (t = 0) −
(ρcv V )g
.
Using the numerical values, we have Tg (tf ) = =
−10−3 (J) 2 × 10−7 (J/◦C) 200(◦C) + 5,000(◦C) = 5,200◦C. 200(◦C) −
(c) The eﬃciency is
tf
QA dt η = 0 tf
= S˙ e,J dt
10−3 (J) = 2.5%. 40 × 10−3 (J)
COMMENT: Very high temperatures are reached for the plasma kernel for a short time. The eﬃciency may even be smaller than 2.5%. There are several regimes in the short sparking period (order of milliseconds). These are breakdown, arc, and glowdischarge regimes. The gas heat up occurs during the glowdischarge regime. Most of the energy is dissipated during the ﬁrst two regimes and does not lead to the gas heat up.
29
PROBLEM 1.16.FAM GIVEN: The temperature distributions for the exhaust gas and the exhaust pipe wall of an automotive exhaust system are shown in Figures Pr.1.16(a) and (b). The exhaust gas undergoes a temperature diﬀerence Tf 0 − Tf L over the upperpipe region (between the exhaust manifold and the catalytic converter). It can be shown that when the energy equation (1.23) is written for this upperpipe region, as shown in the ﬁgure, and under steadystate conditions, the righthand side of this equation is zero. Then the energy equation becomes Q A = 0
integralvolume energy equation.
The surface heat ﬂows are convection on the left and right surfaces and surface convection and radiation from the other sides, i.e., Q A = Qu,L − Qu,0 + Qku + Qr = 0. The convection heat ﬂow rates are written (as will be shown in Chapter 2 and Appendix B) as Qu,L − Qu,0 = M˙ f cp,f (Tf L − Tf 0 ), where M˙ f (kg/s) is the gas ﬂow rate and cp,f (J/kgK) is the speciﬁc heat capacity at constant pressure. SKETCH: Figures Pr.1.16(a) and (b) show the exhaust pipe and its upper portion and temperature distribution along the pipe.
(a) Temperature Distribution Throughout an Exhaust Pipe L 900
Upper Pipe
Manifold 800
Lower Pipe
Tf
Tail Pipe
680oC
700
T, oC
Converter
580oC
600
Ts
500 400 300 0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
x, m (b) Heat Transfer in Upper Pipe
Qku + Qr
Qu,0 Mf Tf
Control Surface A Qu,L
Qu,0
Mf Tf x 0
L Qku + Qr
Figure Pr.1.16(a) and (b) Temperature distribution along a exhaust pipe.
30
L
OBJECTIVE: For M˙ f = 0.10 kg/s and cp,f = 1,000 J/kgK, and using the temperatures given in Figure Pr.1.16(a), determine the sum of the surface convection and radiation heat transfer rates. SOLUTION: The energy equation is Qku + Qr = −Qu,L − Qu,0 = (M˙ cp )f (Tf 0 − Tf L ). From Figure Pr.1.16(a), we have Tf 0 = 800◦C Then
and Tf L = 680◦C.
Qku + Qr = 0.10(kg/s) × 1,000(J/kgK) × (800 − 680)(◦C) = 1.2 × 104 W = 12 kW.
This heat ﬂows out of the control volume (positive). COMMENT: The exhaustgas temperature is most severe in the manifold and upperpipe region of the exhaust line. Most catalytic converters require this temperature drop for safe and eﬀective operation.
31
PROBLEM 1.17.FUN GIVEN: On a clear night with a calm wind, the surface of a pond can freeze even when the ambient air temperature is above the water freezing point (Tsl = 0◦C). This occurs due to heat transfer by surface radiation qr (W/m2 ) between the water surface and the deep night sky. These are shown in Figure Pr.1.17(a). In order for freezing to occur and continue, the net heat ﬂow rate from the ice surface must be enough to cool both the liquid water and the ice layer and also allow for the phase change of the water from liquid to solid. Assume that the ambient temperature is T∞ = 3◦C, the temperature of the deep sky is Tsky = 0 K, the earth atmosphere has an average temperature around Tatm = 230 K, and the temperature of the water at the bottom of the pool is Tl = 4◦C. Then for this transient heat transfer problem between the deep sky, the ambient air, and the water pool: SKETCH: Figure Pr.1.17(a) to (c) show a pond and icelayer growth resulting from the heat losses. (a) Ice Layer Forming on Surface of a Pond Radiation Heat Loss
Wind
Ice Layer
Liquid Water
uF
(b) Control Volume for the Heat Flux Vector Tracking
Air
Asg Ice
x uF Freezing Front Speed
Liquid Water
Als = 1 m2
(c) Control Volume and Control Surface used for Conservation of Energy Analysis Control Surface 1
x
Ice
Control Volume
.
uF
Air
Control Surface 2
Sls
Liquid Water
Figure Pr.1.17(a), (b), and (c) Ice formation at the surface of a pond and the control volume and control surfaces selected for heat transfer analysis.
OBJECTIVE: (a) Track the heat ﬂux vector for the section shown on Figure Pr.1.17(b), (b) For the control volume and control surfaces shown in Figures Pr.1.17(b) and (c) apply the energy conservation equation (1.22). Note that the control volume and surfaces are for only the ice layer, i.e., the control surface 1 includes only the interface between the ice and the ambient air, and control surface 2 includes only the interface between the ice and the liquid water. For this problem the kinetic energy ﬂux and all the work terms in (1.22) are negligible. For control surface 2 (water/ice interface), due to its zero mass (the control surface is wrapped around the interface), the sensible energy storage is zero but there is a latent heat generated due to the phase change from liquid to solid. (Later in Chapter 2, the latent heat will be separated from the sensible heat and treated as an energy conversion mechanism.) Therefore, for control surface 2, (1.22) becomes ˙ ls ∆hls . QA = S˙ ls = −Als m 32
To evaluate QA , use (1.8). (c) For control surface 2 (water/ice interface), at some elapsed time, the following data applies. The conduction heat ﬂux in the ice is qk,x = +250 W/m2 , the surface convection heat ﬂux on the water side is qku,x = −200 ˙ ls ∆hls where the heat of solidiﬁcation W/m2 , the heat absorbed by the interface solidifying is S˙ ls /Als = −m ˙ ls (kg/sm2 ) is the rate of solidiﬁcation. For the density of ice use ρs = 913 ∆hls = −3.34 × 105 J/kg and m kg/m3 . Then determine the speed of the ice/water interface movements uF (m/s). Assume that the heat ﬂux is one dimensional and Als = 1 m2 . SOLUTION: (a) The heat ﬂux vector tracking is shown in Figure Pr.1.17(d). qr qu
Air
qku qk
x
qk qk
Ice
dE dt V qk qku
uF
qls
Liquid Water
dE dt V
qu
Figure Pr.1.17(d) Heat ﬂux vector in the ice layer.
qr
sn
Air
qku
qk
sn
Ice
x
Figure Pr.1.17(e) Airice control surface.
(b) Application of the integralvolume energy conservation equation (1.22) to control surface 1, the control surface wrapped around the ice/air interface [Figure Pr.1.17(e)], and application of (1.8) gives QA = q · sn dA = qr,x=0 Asg − qku,x=0 Asg − qk,x=0 Asg . A
Other terms in (1.22) are
dE ˙ p A = W ˙ µ A = W ˙ g,e V = S˙ e V = 0. = −E˙ u A = W − dt V
Note that the energy storage term is zero because the control surface does not have any mass and no phase change occurs. Therefore, (1.22) becomes Asg (qr,x=0 − qku,x=0 − qk,x=0 ) = 0. (c) For the control volume enclosing the ice layer [Figure Pr.1.17(f)], application of (1.8) gives QA = q · sn dA = qk,x=0 Asg − qk,x=δα Als . A
Other terms in (1.22) are ˙ p A = W ˙ µ A = W ˙ g,e V = S˙ e V = 0. −E˙ u A = W 33
sn
Air qk
x Ice
dE dt V qk
uF
Liquid Water
sn
Figure Pr.1.17(f) Control volume for the ice layer.
The energy storage term is not zero because there is some mass inside the control volume (ice) and the temperature within the control volume changes with respect to time as the ice layer thickens. Therefore, (1.22) becomes ∂E . qk,x=0 Asg − qk,x=δα Als = − ∂t V Note that for this control volume, Asg = Als . For the control surface wrapped around the ice/water interface [Figure Pr.1.17(g)], application of (1.8) gives sn qk qku Sls Als
uF
Ice
Liquid
sn
Figure Pr.1.17(g) Ice water control surface.
q · sn dA = qk,x=δα Als − qku,x=δα Als .
QA = A
Other terms in (1.22) are −E˙ u A dE − dt
˙ p A = W ˙ µ A = W ˙ g,e V = S˙ e V = 0 = W = S˙ ls .
V
The energy required for phase change at the interface is given by the variation of the internal energy of the water as it is transformed from liquid to solid. This is shown in Appendix B. Then, (1.22) becomes qk,x=δα − qku,x=δα = S˙ ls /Als . Noting that S˙ ls /Als = −m ˙ ls ∆hls , the solidiﬁcation mass ﬂux m ˙ ls is determined from qk,x=δα − qku,x=δα = −m ˙ ls ∆hls . Solving for m ˙ ls using the values given −m ˙ ls =
−250(W/m2 ) − (−200)(W/m2 ) = 1.497 × 10−4 kg/m2 s. −3.34 × 105 (J/kg)
The solidiﬁcation mass ﬂux is related to the velocity of the solidiﬁcation front through m ˙ ls = ρs uF . 34
Solving for uF and using the numerical values uF =
1.497 × 10−4 (kg/m2 s) = 1.640 × 10−7 m/s = 0.5904 mm/hr. 913(kg/m3 )
COMMENT: (i) Phase change can be viewed as a form of energy conversion associated with the breaking or formation of physical bonds. Solidiﬁcation involves formation of physical bonds and therefore is associated with a generation of thermal energy. Phase change will be explored in Chapters 2, 3, 6, and 7. (ii) This transient problem also illustrates the role of the sensible heat as an energy storage mechanism. The energy equation for the control volume around the ice layer shows that the conduction heat ﬂow vector at the ice surface includes contributions from the conduction heat ﬂux vector at the bottom of the ice layer and from the sensible heat of the ice layer. Since the temperature of the ice layer decreases with time (i.e., cooling occurs), the energy storage term is positive. (iii) Note that we have assumed that the ice layer is opaque to the thermal radiation. In general, this assumption holds for many solids. Further explanations are given in Chapters 2 and 4.
35
PROBLEM 1.18.FAM GIVEN: The temperature of the earth’s surface and its atmosphere are determined by various electromagnetic energy conversions and, to a smaller extent, by the radioactive decay (within the earth) S˙ r,τ . These are shown in Figure Pr.1.18(a) [which is based on the materials presented in Figures Ex.1.2(a) and (b)]. Starting with solar irradiation (qr,i )s , this irradiation is partly absorbed by the atmospheric gases (S˙ e,τ )s , partly reﬂected (qr,i )ρ , and the remainder is absorbed by the earth’s surface (S˙ e,α )s . The earth’s surface also emits radiation S˙ e, and this mostly infrared radiation is partly absorbed (mostly by the greenhouse gases, such as CO2 ) in the atmosphere (S˙ P e,τ )i and this is in turn reemitted (S˙ e,τ )i = (S˙ e, )i . SKETCH: Figure Pr.1.18(a) shows the various energy conversions. Solar Irradiation (Infrared, Visible, and Ultraviolet) (qr,i)s (Se,=)s
Reflected (qr,i)H Irradiation
(Se,J)i + (Se, )i
Sun
Se, , Surface Emission (Infrared)
(Se,J)s Absorption of Various Gases
(Se,=)i , Surface Absorption qk Se,J
Earth Earth's Atmosphere
Surface Temperature, T
A, Control Surface Wrapped Around the Earth A
Figure Pr.1.18(a) Solar irradiation and internal radiation heating of the earth and its surface and infrared, radiation emission (part of this is absorbed and emitted by the earth atmosphere).
OBJECTIVE: (a) Compute the heat ﬂux vector tracking by drawing the radiation qr and conduction qk heat ﬂux vectors arriving and leaving the earth control surface A, also shown in Figure Pr.1.18(a). Assume a steadystate heat transfer. (b) Starting from (1.22) and assuming a steady state with the lefthand side approximated as Qk = −Aqk , A = 4πR2 , qk = 0.078 W/m2 , and the righthand side approximated by (S˙ e,α )s + S˙ e, + (S˙ e,α )i , where (S˙ e,α )s S˙ e, + (S˙ e,α )i
= A × 172.4(W/m2 ) time and space average solar irradiation = −A(1 − αr,i )σSB T 4A ,
σSB = 5.67 × 10−8 W/m2 K4 ,
determine the timespace averaged earth surface temperature T A for αr,i = 0.55. SOLUTION: (a) The heat ﬂux vector tracking is shown in Figure Pr.1.18(b). Starting from the solar irradiation, this is partly absorbed by the earth’s atmosphere, partly reﬂected, and the remainder is absorbed by the earth’s surface. The earth’s surface emits radiation (in the infrared wavelength range, which will be discussed in Chapter 4), 36
which is partly absorbed in the earth’s atmosphere and then reemitted and part of this is absorbed by the earth’s surface. The radioactive decay releases heat and this will make it turn into the earth’s surface by conduction.
Solar Irradiation (Infrared, Visible, and Ultraviolet)
(qr,i)r
Surface Absorption (Se,=)s qr
(Se,J)s Absorption of Various Gases
qr , To Space
qr
(Se,J)i + (Se, )i Absorption and Emission by Greenhouse Gases
qr
se,J Radiation Decay
qk
qr
se, , Surface Emission (Infrared)
qr To Space
(qr,i)s
Sun
(se,=)i , Surface Absorption
Earth Earth's Atmosphere
A, Control Surface Wrapped Around the Earth
Surface Temperature, T
A
Figure Pr.1.18(b) Tracking of heat ﬂux vector.
(b) The energy equation (1.22) becomes QA
= −Aqk = −A × (0.078)(W/m2 ) = (S˙ e,α )s + S˙ e, + (S˙ e,α )i = A × (172.4)(W/m2 ) − A × (1 − αr,i )σSB T 4A .
Solving for T A , we have
T A
1/4 A × (172.4)(W/m2 ) + A × (0.078)(W/m2 ) A(1 − αr,i )σSB 1/4 (172.4 + 0.078)(W/m2 ) = = 286.7 K = 13.55◦C. (1 − 0.55) × 5.67 × 10−8 (W/m2 K4 ) =
COMMENT: The numerical value αr,i = 0.55 is used here as an approximate representation of the greenhouse eﬀect. As αr,i increases, due to an increase in the concentration of CO2 and other greenhouse gases, T A increases, leading to global warming.
37
PROBLEM 1.19.FAM GIVEN: Sodium acetate (trihydrate) is used as a liquidsolid phasechange heater. It has a heat of melting of ∆hsl = 1.86 × 105 J/kg and melts/freezes at Tls = 58◦C [Table C.5(a)]. It can be kept in a sealed container (generally a plastic bag) as liquid in a metastable state down to temperatures as low as −5◦C. Upon ﬂexing a metallic disk within the liquid, nucleation sites are created at the disk surface, crystallization begins, heat is released, and the temperature rises. Consider a bag containing a mass M = 100 g of sodium acetate. Assume that the liquid is initially at T = 58◦C and that during the phase change the transient surface heat transfer rate (i.e., heat loss) is given by QA = Qo (1 − t/τ ), where Qo = 50 W. This is shown in Figure Pr.1.19. SKETCH: Figure Pr.1.19 shows the liquidsolid phasechange hand warmer.
Plastic Bag Containing PhaseChange (LiquidSolid) Material Sodium Acetate (Trihydrate) ∆hsl = 1.86 x 105 J/kg Sls M = 100 g
Metal Disk
Q A= Qo (1− t / τ) Transient Heat Loss
Figure Pr.1.19 Surface heat transfer from a plastic bag containing phasechange material.
OBJECTIVE: Determine τ , the elapsed time during which all the heat released by phase change will be removed by this surface heat transfer.Start from (1.22) and replace the time rate of change of the internal energy with −S˙ ls = −M˙ ls ∆hls = M˙ ls ∆hsl . This represents isothermal phase change. Then in the absence of any other work and energy conversion, this change in internal energy balances with surface heat losses. SOLUTION: The variation of the total internal energy of the bag is due to phase change only. Then, ∂E ≡ −S˙ ls ∂t V from (1.22) and neglecting all other energy and work form, we have QA = S˙ ls . Now, substituting for QA , we have t Qo 1 − = S˙ ls . τ Next, substituting for S˙ ls = −M˙ ls ∆hls = M˙ ls ∆hsl 38
and integrating over the time interval of interest, we have τ τ t M˙ ls ∆hsl dt Qo 1 − dt = τ 0 0 τ τ t2 ˙ = Mls ∆hsl t . Qo t − 2τ 0 0 Noting that M˙ ls τ = Mls , Qo
τ = Mls ∆hsl . 2
Solving for τ , τ=
2Mls ∆hsl . Qo
Using the numerical values given, τ=
2 × 0.1(kg) × 1.86 × 105 (J/kg) = 744 s = 0.207 hr. 50(W)
COMMENT: In general, since the liquid is critically in a subcooled state, part of the heat released will be used to raise the temperature of the solid formed at the freezing temperature.
39
PROBLEM 1.20.FAM GIVEN: Nearly all of the kinetic energy of the automobile is converted into friction heating S˙ m,F during braking. The front wheels absorb the majority of this energy. Figure Pr.1.20(a) shows a disc brake. This energy conversion raises the rotor temperature Tr and then heat ﬂows from the rotor by conduction (to axle and wheel), by surface radiation to the surroundings and by surface convection to the air. The air ﬂows over the rotor in two parts; one is over the inboard and outboard surfaces, and the other is through the vanes (passages). The air ﬂow is due mostly to rotation of the rotor (similar to a turbomachinary ﬂow). Assume that the rotor is at a uniform temperature (this may not be justiﬁable during rapid braking). Mr = 15 kg, and cv = 460 J/kgK. SKETCH: Figure Pr.1.20(a) shows the disc brake and the air streams. An automobile disc brake is heated by friction heating S˙ m,F , and cooled by various heat transfer mechanisms, and is able to store/release heat. r
Vane Air Flow Brake Fluid
Brake Pad Outboard Surface
Axle
Rotor (Disc), at Uniform Temperature Tr
Rotor Angular Velocity, ω
Caliper Air Flow Over Disc
Sm,F Surface Friction Energy Conversion at Brake PadRotor Interface
Rotationuf Induced Air Flow
Vane Air Flow
RotationInduced Air Flow
Opposite Side of Rotor Figure Pr.1.20(a) An automobile disc brake showing the air ﬂow over the disc and through the rotor vanes.
OBJECTIVE: (a) Draw the heat ﬂux vector tracking for the rotor, by allowing for the heat transfer mechanisms mentioned above. (b) Now consider the heat storage/release mechanism represented by −∂E/∂t, in (1.22). During quick brakes, the rate of heat transfer QA,r is much smaller than ∂E/∂t and S˙ m,F . Assume all other terms on the righthand side of (1.22) are negligible. With no heat transfer, determine the rate of rise in the rotor temperature dT /dt, using dTr ∂E = Mr cv , ∂t dt Mr = 15 kg, and cv = 460 J/kgK. SOLUTION: (a) Figure Pr.1.20(b) shows the various heat transfer from the rotor and the tracking of the heat ﬂux vector. 40
The conduction qk is to the lower temperature axle and wheel (there are various materials, areas, and contacts through which the heat ﬂows). The surface radiation heat transfer qr is to various close and distant surfaces. The surface convection qku is to the air ﬂowing over the rotor (semibounded air streams) and to air ﬂowing through the vanes (bounded ﬂuid stream). The heat is added to these convection streams qu by surface convection qku .
qu (Axle Conduction) qk
(Bounded Fluid Stream) qu Assumed Uniform Rotor Temperature, Tr Sm,F qr − ∂Er ∂t qu qku
qu
qu (SemiBounded Fluid Stream) qk (Wheel Conduction)
qu
qu qu
qr (Surface Radiation)
Figure Pr.1.20(b) Tracking of the heat ﬂux vector.
(b) From (1.22), we have QA,r ∂E ∂t dTr dt
∂E + S˙ m,F ∂t dTr = M˙ r cv,r = S˙ m,F dt S˙ m,F = M˙ r cv,r =
0=−
=
6 × 104 (W) = 8.696◦C/s. 15(kg) × 460(J/kgK)
COMMENT: The heat transfer through the vanes is the most eﬀective during the cooling period. Note that when multiple brakes are applied (as in the downhill driving) the temperature of the rotor can become very large (and damaging to the brake pad). In Chapters 3, 4, 6, and 7, we will address conduction, radiation, and semibounded and bounded ﬂuid stream surface convection.
41
Chapter 2
Energy Equation
PROBLEM 2.1.FAM GIVEN: Consider a steadystate, twodimensional heat ﬂux vector ﬁeld given by q = 3x2 sx + 2xy sy . The control volume is centered at x = a and y = b, with sides 2∆x and 2∆y (Figure Pr.2.1). SKETCH: Figure Pr.2.1 shows a control volume centered at x = a and y = b with side widths of 2∆x and 2∆y.
y sn sn 2,y b sn sn a
x
2, x
Figure Pr.2.1 A ﬁnite control volume in a twodimensional heat transfer medium.
The depth (along z direction) is w. OBJECTIVE: (a) Using the above expression for q show that lim
∆V →0
A
q · sn dA = ∇ · q, ∆V
where the divergence of the heat ﬂux vector is to be evaluated at x = a and y = b. Use a length along z of w (this will not appear in the ﬁnal answers). (Hint: Show that you can obtain the same ﬁnal answer starting from both sides.) (b) If the divergence of the heat ﬂux vector is nonzero, what is the physical cause? (c) In the energy equation (2.1), for this net heat ﬂow (described by this heat ﬂux vector ﬁeld), is the sum of the volumetric terms on the right, causing the nonzero divergence of q, a heat source or a heat sink? Also is this a uniform or nonuniform volumetric source or sink? Discuss the behavior of the heat ﬂux ﬁeld for both positive and negative values of x and y. SOLUTION: (a) To prove the validity of q above for the region shown in Figure Pr.2.1, we calculate separately the lefthand side and the righthand side. For the righthand side we have ∂ ∂ + sy ∇ · q = sx · 3x2 sx + 2xy sy . ∂x ∂y Performing the dot product we have ∂ 3x2 ∂ (2xy) ∇·q= + , ∂x ∂y which results in ∇ · q = 6x + 2x = 8x. 44
Applying the coordinates of the center of the control volume, we have ﬁnally ∇ · q(x=a,
y=b)
= 8a.
The lefthand side can be divided into four integrals, one for each of the control surfaces: (i) Control Surface at x = a − ∆x : The heat ﬂux vector across this control surface and the normal vector are q1 = 3(a − ∆x)2 sx + 2(a − ∆x)y sy sn1 = −sx . The dot product between q and sn is q1 · sn1 = 3(a − ∆x)2 (sx · −sx ) + 2(a − ∆x)y(sy · 0) = −3(a − ∆x)2 . The net heat ﬂow over this control surface is QA1 = (q1 · sn1 ) dA =
b+∆y
−3(a − ∆x)2 dyw = −6(a − ∆x)2 ∆yw.
b−∆y
A1
(ii) Control Surface at x = a + ∆x: The heat ﬂux vector across this control surface and the normal vector are q2 = 3(a + ∆x)2 sx + 2(a + ∆x)y sy sn2 = sx . The net heat ﬂow over this control surface is (q2 · sn2 ) dA = QA2 = A2
b+∆y
3(a + ∆x)2 dyw = 6(a + ∆x)2 ∆yw.
b−∆y
(iii) Control Surface at y = b − ∆y: The heat ﬂux vector across this control surface and the normal vector are q3 = 3x2 sx + 2x(b − ∆y) sy sn3 = −sy . The net heat ﬂow over this control surface is (q3 · sn3 ) dA = QA3 =
a+∆x
−2x(b − ∆y)dxw
a−∆x 2
A3
= −2(b − ∆y)
2
(a + ∆x) − (a − ∆x) w = −4a∆x (b − ∆y) w. 2
(iv) Control Surface at y = b + ∆y: The heat ﬂux vector across this control surface and the normal vector are q4 = 3x2 sx + 2x(b + ∆y) sy sn4 = sy . The net heat ﬂow over this control surface is QA4 = (q4 · sn4 ) dA =
a+∆x
2x(b + ∆y)dxw
a−∆x
A4
2
2
(a + ∆x) − (a − ∆x) w = 4a∆x (b + ∆y) w. 2 Adding up the heat ﬂow across all the surfaces, we have =
QA
2(b + ∆y)
= QA1 + QA2 + QA3 + QA4 = q1 · sn1 dA + q2 · sn2 dA + A1
A2
A3
q3 · sn3 dA +
A4
q4 · sn4 dA
= [−6(a − ∆x)2 ∆y + 6(a + ∆x)2 ∆y − 4a∆x (b − ∆y) + 4a∆x (b + ∆y)]w = (24a∆x∆y + 8a∆x∆y)w = 32a∆x∆yw. 45
Now, applying the limit lim
∆V →0
A
q · sn dA 32a∆x∆yw = lim = lim 8a = 8a, ∆x,∆y→0 (2∆x)(2∆y)w ∆x,∆y→0 ∆V
which is identical to the result found before. These are the two methods of determining the divergence of the heat ﬂux vector for a given location in the heat transfer medium. (b) Since this is a steadystate heat ﬂux vector ﬁeld (i.e., q is not a function of time t), the only reason not to have a divergencefree ﬁeld would be the presence of a heat generation or sink. In this case, the diﬀerentialvolume energy equation is
s˙ i . ∇·q= i
The heat generation or sink is caused by the conversion of work or other forms of energy to thermal energy. In the energy equation, these energy conversions are called source terms. The source terms s˙ i could be due to (i) conversion from physical or chemical bond to thermal energy (ii) conversion from electromagnetic to thermal energy (iii) conversion from mechanical to thermal energy (c) The divergence of the heat ﬂux vector q given above is 8x. For x > 0, this is a positive source term indicating a heat generation. For x < 0, the source term becomes negative indicating a heat sink. Also, since the source term is a function of x, it is a nonuniform source term in the x direction and a uniform source term in the y direction. COMMENT: The application of the divergence operator on the heat ﬂux vector (as in the diﬀerentialvolume energy equation) results in an expression valid for any position x and y while the application of the areaintegral (as in the integralvolume energy equation) results on a number which is valid only for that speciﬁc point in space x = a and y = b. The integral form of the energy equation gives an integral or overall energy balance over a speciﬁed closed region within the medium, while the diﬀerential form is pointwise valid, i.e., is satisﬁed for any point within the medium. For the control surfaces parallel to the x axis, the dot product between the heat ﬂux vector and the surface normal was a function of x (variable). That required the integration along x. The integration is simpliﬁed in the case of a constant heat ﬂux vector normal to the control surface, as obtained for the control surfaces parallel to the y axis. Although the ﬁrst case is more general, here we will mainly deal with situations in which the heat ﬂux normal to the control surface is constant along the control surface. This will allow the use of the thermal resistance concept and the construction of thermal resistance network models, as it will be discussed starting in Chapter 3.
46
PROBLEM 2.2.FUN GIVEN: Figure Pr.2.2(a) shows a ﬂame at the mouth of a cylinder containing a liquid fuel. The heat released within the ﬂame (through chemical reaction) is transferred to the liquid surface by conduction and radiation and used to evaporate the fuel (note that a ﬂame also radiates heat). The ﬂame stabilizes in the gas phase at a location determined by the local temperature and the fuel and oxygen concentrations. The remaining heat at the ﬂame is transferred to the surroundings by convection and by radiation and is transferred to the container wall by conduction and radiation. This heat then conducts through the container wall and is transferred to the ambient, by surface convection and radiation, and to the liquid fuel, by surface convection. The container wall and the liquid fuel also lose some heat through the lower surface by conduction. Figure Pr.2.2(b) shows a cross section of the container and the temperature proﬁles within the gas and liquid and within the container wall. In small and mediumscale pool ﬁres, the heat recirculated through the container wall accounts for most of the heating of the liquid pool. Assume that the liquid pool has a makeup fuel line that keeps the fuel level constant and assume that the system has been operating under steady state (long enough time has elapsed). SKETCH: Figures Pr.2.2(b) and (c) show a crosssectional view of the container and the temperature distribution along the gas and liquid and along the container wall.
(a) SmallScale Pool Fire Hot Air
. Sr,c /V
Flame
Fuel Container
(b) Cutout of Container and Temperature Distributions
(c) Regions of Heat Flow
. Sr,c /V Flame
500
Gas
Liquid
1,500
Region 1: Flame Zone
T, oC
Gas and Liquid Fuel Temperature . Slg /Alg
Container Wall Temperature
Region 2: LiquidGas Interface Region 3: Liquid
Container Wall
Region 4: Container
z
Figure Pr.2.2(a), (b), and (c) A smallscale pool ﬁre showing the various regions.
OBJECTIVE: (a) On Figure Pr.2.2(b) track the heat ﬂux vector, identifying the various mechanisms. (b) For the regions shown in Figure Pr.2.2(c), apply the integralvolume energy equation. Note that region 1 encloses the ﬂame and it is assumed that the fuel vapor burns completely. Region 2 surrounds the liquid/gas interface, region 3 encloses the liquid, and region 4 is the container wall. (c)For each of the regions state whether the areaintegral of the heat ﬂux vector QA is equal to zero or not. 47
SOLUTION: (a) The heat ﬂux vector tracking is shown in Figure Pr.2.2(d). qu qr qu
qr qku
qk
Flame
.
qu
Sr,c / V Gas qr
qk
Liquid Gas and Liquid Fuel Temperature
qr
qr
qku
qu
qku
1,500
T ,OC
qu
qk
qr
qk
1,000
500
qk
qu .
qu Slg / Alg
Container Wall Temperature
qku
qr qku qr
qk
qu
Container
qku
qku
qk
qu qk
qk
z
Figure Pr.2.2(d) Heat ﬂux vector tracking around the container wall.
(b) For each of the regions shown on Figure Pr.2.2(c), the integral energy balances are given below. (i) Region 1: Flame region [Figure Pr.2.2(e)] We assume that the heat ﬂux vectors normal to surfaces 1 and 2 are uniform along those surfaces and then using the notations in Figures Pr.2.2(b) and (c), referencing the products of combustion as (p), and noting that q is positive when pointing away from the surface, we have, (qr,Fa + qu,p − qu,a )A1 + (qk,Fc + qr,Fc + qk,Fl + qr,Fl + qk,Ft + qr,Ft − qu,f g )A2 = S˙ r,c .
Region 1: Flame region qu,p qr,Ft A1
Flame
qu,a Sr,c / V
qk,Ft qr,Ft qk,Fc qr,Fc qr,Fl qk,Fl
A2 qu,Fg
Figure Pr.2.2(e) Heat ﬂux vector tracking in the ﬂame region.
(ii) Region 2: Liquid/gas interface [Figure Pr.2.2(f)] Assuming that the heat ﬂux vectors normal to surfaces 1 and 2 are uniform along those surfaces, (−qr,cl − qk,Fl − qr,Fl + qu,f g )A1 + (−qku,li − qu,Fl )A2 = S˙ lg . 48
Region 2: Liquid/Gas Interface Gas
qr,Fl qk,Fl
qu,Fg qr,cl
A1
Interface (i)
A2
qku,li
qu,Fl Slg / V
Liquid
Figure Pr.2.2(f) Heat ﬂux vector tracking at liquidgas interface.
(iii) Region 3: Liquid [Figure Pr.2.2(g)] The heat ﬂux vector normal to surfaces 1 and 2 will be assumed uniform, while for surface 3 it will be assumed
Region 3: Liquid qku,li
qu,fl
A1
qku,lc
Interface (i)
qku,lc qku,lc A3
Liquid
qku,lc qku,lb
A2
Figure Pr.2.2(g) Heat ﬂowing in and out of the liquid.
nonuniform (i.e., distributed along the height), i.e.,
(qku,lc · sn,3 )dA = 0.
(qku,li + qu,Fl )A1 + (qku,lb )A2 + A3
(iv) Region 4: Container wall [Figure Pr.2.2(h)] The heat ﬂux vector leaving surfaces 1 and 2 will be assumed uniform while the heat ﬂux vectors at surfaces 3,4, and 5 will be assumed nonuniform (i.e., distributed along the wall height), i.e., (qk,Fc · sn,3 )dA + (−qk,ct )A1 + (qk,cb )A2 + A3 (qk,lc · sn,4 )dA + (qk,ca · sn,5 )dA = 0. A4
A5
(c) The divergence of the heat ﬂux vector is zero everywhere inside regions 3 and 4 because no heat sources or sinks are present within these regions. It is greater than zero in region 1, due to the volumetric chemical reaction. At the liquidgas interface (region 2) the integral of the surface heat ﬂow is less than zero, due to surface phase change from liquid to gas. COMMENT: (i) In smallscale pool ﬁres, the heat recirculation from the container wall to the liquid pool accounts for most of the heating and evaporation of the liquid. In largescale pool ﬁres (large diameter containers), the eﬀect of this heat recirculation is small. 49
Region 4: Container Wall qk,tc
A1 qk,Fc
qk,ca
A3 A5
A4 qk,lc
qk,ca qk,lc qk,ca
Container
qk,lc qk,ca qk,lc A2 qk,cb
Figure Pr.2.2(h) Heat ﬂowing in and out of the container wall.
(ii) The chemical reaction inside the ﬂame generates heat and it is a positive source term in the energy equation. The liquid to gas phase change at the liquid surface absorbs heat and is a negative source term for the interfacial, integral energy conservation equation. (iii) For no heat loss from the ﬂame to the ambient or to the liquid surface, there would be a balance between the energy entering the ﬂame by convection carried by the air and the vapor fuel (reactants), the energy leaving the ﬂame by convection carried away by the hot combustion gases (products), and the energy generated inside the ﬂame by the exothermic chemical reaction (combustion). This energy balance determines the adiabatic ﬂame temperature (maximum temperature the combustion gases can reach). This will be discussed in Chapter 5.
50
PROBLEM 2.3.FUN GIVEN: The wall of the burning fuel container is made of a metal, its thickness is small compared to its length, and the surfaceconvection heat ﬂuxes at the inner and outer surfaces of the container wall are designated by qku,o and qku,i . Under these conditions, the temperature variation across the wall thickness is negligibly small, when compared to the axial temperature variation. Also, assume that the heat transfer from within the container is axisymmetric (no angular variation of temperature). Figure Pr.2.3(a) shows the diﬀerential control volume (with thickness ∆z), the inner radius Ri , and outer radius Ro of the container. SKETCH: Figures Pr.2.3(a) shows a control volume with a diﬀerential length along the z direction. Symmetry Axis Ro Ri Fluid
Aku,i
Conduction within Container Wall Aku,o
z Surface Convection qku,o
,z qku,i
Figure Pr.2.3(a) A cylindrical container with a control volume having a diﬀerential length along the z direction.
OBJECTIVE: (a) Apply a combined integral and diﬀerentiallength analysis for the container wall (integral along the radius and polar angle and diﬀerential along the z axis) and derive the corresponding combined integral and diﬀerentiallength energy conservation equation. (b) Sketch the anticipated variations of the conduction heat ﬂux qk,z and the wall temperature T , along the container wall (as a function of z). SOLUTION: (a) The integral and diﬀerentiallength analysis starts with the diﬀerential form of the energy conservation equation (2.9) written as
q · sn dA ∂ = − ρcp T + lim A s˙ i . ∆V →0 ∆V ∂t i For the container walls there is no energy conversion and, as this is a steadystate process, the diﬀerentialvolume energy conservation equation becomes q · sn dA = 0. lim A ∆V →0 ∆V Figure Pr.2.3(b) shows a cross section of the control volume shown in Figure Pr.2.3(a) with the heat ﬂux vectors crossing the control surfaces. For the four control surfaces labeled, the area integral above becomes qk,z+∆z ·sn dA q ·s dA q ·s dA q ·s dA q · sn dA A Ao ku n Ai ku n Az k,z n A = + + + z+∆z ∆V ∆V ∆V ∆V ∆V For this inﬁnitesimal control volume, the heat ﬂux vectors normal to the control surfaces are uniform over each control surface and we have 51
qk,z
sn z
Az sn
sn qku,i ∆z
qku,o Ai
Ao Az+∆z sn
qk,z+∆z
Ri Ro
Figure Pr.2.3(b) Cross section of the control volume.
A
q · sn dA ∆V
qku,o Ao qku,i Ai −qk,z Az qk,z+∆z Az+∆z + + + ∆V ∆V ∆V ∆V qku,o (2πRo ∆z) qku,i (2πRi ∆z) + + π(Ro2 − Ri2 )∆z π(Ro2 − Ri2 )∆z −qk,z π(Ro2 − Ri2 ) qk,z+∆z π(Ro2 − Ri2 ) + π(Ro2 − Ri2 )∆z π(Ro2 − Ri2 )∆z qku,o 2Ro qku,i 2Ri qk,z+∆z − qk,z . + 2 + 2 2 2 Ro − Ri Ro − Ri ∆z
= =
=
Taking the limit as ∆V → 0, this becomes q · sn dA lim A = ∆V →0 ∆V
qku,o 2Ro qku,i 2Ri qk,z+∆z − qk,z + 2 + ∆z→0 Ro2 − Ri2 Ro − Ri2 ∆z qku,i 2Ri dqk,z qku,o 2Ro + 2 + = Ro2 − Ri2 Ro − Ri2 dz
lim
Finally, rearranging the righthand side, the combined integral and diﬀerentiallength energy equation becomes 2 dqk,z =− 2 (qku,o Ro + qku,i Ri ). dz Ro − Ri2 (b) The anticipated variation of the axial conduction heat ﬂux vector along the container wall qk,z as a function of z is given in Figure Pr.2.3(c).
qk,z ,W/m2
500
1,000
T, C
Gas
Liquid
Container Wall Temperature
Axial Conduction Heat Flux through the Container Wall Container
z
z
Figure Pr.2.3(c) Distribution of temperature and heat ﬂux along the container wall.
52
COMMENT: The direction of the axial conduction heat ﬂux vectors qk,z and qk,z+∆z are taken along the direction of the z axis. The direction for the heat ﬂux vectors along the integral length qku,o and qku,i are arbitrary and are conventionally taken as pointing outward from the control surface. The axial conduction heat ﬂux vector along the container wall may be obtained from the solution to the above energy equation, once qku,o (z) and qku,i (z) are known. The axial conduction heat ﬂux vector along the container wall is maximum at some point near the interface level. Above the interface, the wall receives heat from the ﬂame. Below the interface, the wall loses heat to the liquid and the maximum heat loss occurs at the interface location. In Figure Pr.2.3(c), a positive value for qk,z indicates that heat is ﬂowing in the direction of the zaxis. This is in accordance with the reference directions assumed in Figure Pr.2.3(b). Note that the conduction heat ﬂux vector is related to temperature variation through qk = −k∇T . From the temperature distribution shown in Figure Pr.2.3(c), ∇T is negative (T decreases as z increases). Therefore, qk,z is positive everywhere for this temperature distribution, as shown in Figure Pr.2.3(c).
53
PROBLEM 2.4.FUN GIVEN: A nitrogen meat freezer uses nitrogen gas from a pressurized liquid nitrogen tank to freeze meat patties as they move carried by a conveyor belt. The nitrogen ﬂows inside a chamber in direct contact with the meat patties, which move in the opposite direction. The heat transfer mechanism between the nitrogen gas and the meat patties is surface convection. Meat patties are to be cooled down from their processing (initial) temperature of Ti = 10◦C to the storage (ﬁnal) temperature of To = −15◦C. Each meat patty has a mass M = 80 g, diameter D = 10 cm, and thickness l = 1 cm. Assume for the meat the thermophysical properties of water, i.e., speciﬁc heat in the solid state cp,s = 1,930 J/kgK , speciﬁc heat in the liquid state cp,l = 4,200 J/kgK, heat of solidiﬁcation ∆hls = −3.34 × 105 J/kg, and freezing temperature Tls = 0◦C. The average surfaceconvection heat transfer between the nitrogen and the meat patties is estimated as qku = 4,000 W/m2 and the conveyor belt moves with a speed of uc = 0.01 m/s. OBJECTIVE: (a) Sketch the temperature variation of a meat patty as it move along the freezing chamber. (b) Neglecting the heat transfer between the conveyor belt and the meat patties, ﬁnd the length of the freezing chamber. Use the simplifying assumption that the temperature is uniform within the meat patties. This allows the use of a zerothorder analysis (lumpedcapacitance analysis). SOLUTION: (a) The temperature variation of the meat patties as they move along the freezing chamber is given in Figure Pr.2.4.
T , oC Ti
Cooling Solid Regime
Cooling Liquid Regime PhaseChange Regime
= 10 oC
Tls = 0 oC
T0 =  15 oC ti = 0
t2
t1
t0
t,s
Figure Pr.2.4 Variation of meat patty temperature with respect to time.
(b) To calculate the necessary length for the freezing chamber, the cooling process is divided into three regimes (shown in Figure Pr.2.4). (i) Regime 1: Cooling of Liquid During this period of time, the meat patties are cooled from their initial temperature down to the solidiﬁcation temperature. Application of the integralvolume energy equation for a control volume enclosing the meat gives
qku · sn dA = Aku
− V
d ρcp T dt
dV.
Assuming that qku is constant and normal to the surface and that the meat temperature and properties are constant throughout the meat patty (lumpedcapacitance analysis), the energy equation becomes
qku Aku = −ρcp V 54
dT . dt
Integrating the equation above from t = ti = 0 to t = t1 , for a constant qku , gives
t1
qku Aku dt = −
ti =0
t1 = From the data given Aku =
t1 =
πD 2 4
Tls
ρcp V dT Ti
ρcp V (Ti − Tls ) M cp (Ti − Tls ) = qku Aku qku Aku
+ πDl = 0.011 m2 and
0.08(kg) 4,200(J/kgK) × [10(◦C) − 0(◦C)] 2
4,000(W/m ) 0.011(m2 )
= 76.36 s = 1.273 min.
(ii) Regime 2: Solidiﬁcation During this regime the meat patties change phase from liquid to solid. Application of the integralvolume energy equation gives qku · sn dA = s˙ ls dV. As
V
Again, assuming that qku is uniform and normal to the surface and that the meat properties are constant throughout the meat patty (lumpedcapacitance analysis), the energy equation becomes
qku Aku = s˙ ls V, where the volumetric heat consumption due to phase change s˙ ls is obtained from Table 2.1, s˙ ls = −n˙ ls ∆hs . The volumetric solidiﬁcation rate n˙ ls (kg/m3 s) is given by n˙ ls =
m . V (t2 − t1 )
Using the relations above, the energy equation becomes qku Aku = −
m∆hls , (t2 − t1 )
t2 − t1 = −
m∆hls . qku Aku
and solving for t2 − t1 ,
From the values given, t2 − t 1 = −
0.08(kg) (−3.34 × 105 )(J/kg) 2
4,000(W/m ) 0.011(m2 )
= 607.3 s = 10.12 min.
(iii) Regime 3: Cooling of Solid During this period of time, the meat patties are cooled from the melting temperature down to the ﬁnal temperature. Application of the lumpedcapacitance analysis for a control volume enclosing the meat results in an equation similar to t1 = ρcp V (Ti − Tls )/qku Aku = M cp (Ti − Tls )/qku Aku , i.e.,
to − t2 =
ρcp V (Tls − To ) mcp (Tls − To ) = . qku Aku qku Aku 55
From the data given, to − t2 =
0.08(kg) × 1,930(J/kgK) × [0(◦C) − (−15)(◦C)] 2
4,000(W/m ) 0.011(m2 )
= 52.64 s = 0.8773 min.
The total time for cooling of the meat patties is therefore to = t1 + (t2 − t1 ) + (to − t2 ) = 1.273 + 10.12 + 0.8773 = 12.27 min. For the velocity of the conveyor belt uc = 0.01 m/s, the total length necessary is L = uc to = 0.01(m/s) × 12.27(min) × 60(s/min) = 7.362 m. COMMENT: Phase change at constant pressure for a pure substance occurs at constant temperature. The temperature evolution for regimes 1 and 3 are linear because qku has been assumed constant (note that all the properties are treated as constants). In practice, the heat loss by surface convection depends on the surface temperature and therefore is not constant with time when this surface temperature is changing. This will be discussed in Chapter 6. The freezing regime accounts for more than 80 s of the total time, while the cooling of solid accounts for only 7 s of the total time. This is a result of the high heat of solidiﬁcation exhibited by water and the relatively smaller speciﬁc heat capacity of ice compared to liquid water. Liquid water has one of the largest speciﬁc heat capacities among the pure substances. The speciﬁc heat capacity of substances will be discussed in Chapter 3.
56
PROBLEM 2.5.FUN GIVEN: While the integralvolume energy equation (2.9) assumes a uniform temperature and is applicable to many heat transfer media in which the assumption of negligible internal resistance to heat ﬂow is reasonably justiﬁable, the diﬀerentialvolume energy equation (2.1) requires no such assumption and justiﬁcation. However, (2.1) is a diﬀerential equation in space and time and requires an analytical solution. The ﬁnitesmall volume energy equation (2.13) allows for a middle ground between these two limits and divides the medium into small volumes within each of which a uniform temperature is assumed. For a single such volume (2.9) is recovered and for a very large number of such volumes the results of (2.1) are recovered. Consider friction heating of a diskbrake rotor, as shown in Figure Pr.2.5. The energy conversion rate is S˙ m,F . The brake friction pad is in contact, while braking, with only a fraction of the rotor surface (marked by R). During quick brakes (i.e., over less than t = 5 s), the heat losses from the rotor can be neglected. Note that S˙ m,F remains constant, while ∆V changes. SKETCH: Figure Pr.2.5 shows the rotor and the area under the pad undergoing friction heating. Pad Contact Area Sm,F
Hcp,V
l
dT dt Ro
QA= 0
R Ri
T(t = 0) Initial Temperature
Figure Pr.2.5 A discbrake rotor heated by friction heating. The region under the brake pad contact is also shown.
OBJECTIVE: Apply (2.13), with (i) the volume marked as the pad contact region, and (ii) the entire volume in Figure Pr.2.5, and determine the temperature T after t = 4 s for cases (i) and (ii) and the conditions given above. Note that the resulting energy equation, which is an ordinary diﬀerential equation, can be readily integrated. SOLUTION: Starting from (2.13), we have QA 0
d (ρcp T )∆V ∆V + S˙ m,F dt dT + S˙ m,F = −ρcp ∆V dt = −
or by separating the variables, we have dT =
S˙ m,F dt. ρcp ∆V
Using T t=0 = T (t = 0) and integrating from 0 to t, we have T (t) − T (t = 0) =
S˙ m,F (t − 0) ρcp ∆V
or T (t) = T (t = 0) +
57
S˙ m,F t. ρcp ∆V
Then using the numerical values we have T (t = 4 s) =
20(◦C) +
3 × 104 (W) × 4(s) 3.5 × 106 (J/m3 K) × ∆V (m3 )
=
20(◦C) +
3.429 × 10−2 ◦ ( C). ∆V
(i) The smaller volume gives ∆V
= π(Ro2 − R2 )l = π(0.182 − 0.152 )(m2 ) × 0.015(m) = 4.665 × 10−4 m3 T (t = 4 s) = 20(◦C) + 73.50(◦C) = 93.50◦C.
(ii) The larger volume gives ∆V
= π(Ro2 − Ri2 )l = π(0.182 − 0.132 )(m2 ) × 0.015(m) = 7.305 × 10−4 m3 T (t = 4 s)
=
20(◦C) + 46.94(◦C) = 66.94◦C.
COMMENT: For more accurate results, the radial length as well as the length along l are divided into smallﬁnite volumes and then heat transfer is allowed between them. This is discussed in Section 3.7.
58
PROBLEM 2.6.FUN GIVEN: In laserinduced spark ignition, laser irradiation qr,i is used to cause ionization of the fueloxidant mixture at the end of the laser pulse. The ionization is caused by multiphoton ionization. In multiphoton ionization, the ionizing gas molecules absorb a large number of photons. Consider a pulsed laser, emitting a nearinfrared radiation, λ = 1.064 µm, with a timedependent, focused irradiation ﬂux given by 2 2 qr,i (t) = (qr,i )o e−t /τ , where −∞ < t < ∞, (qr,i )o is the peak irradiation, and τ (s) is time constant. Assume that this irradiation ﬂux is uniform over the focal surface. Note that ∞ 2 2 et /τ dt = π 1/2 τ. −∞
(qr,i )o = 1017 W/m2 , T = 106 K, a1 = 0.1645 × 10−42 K1/2 m5 , ne = 1026 (1/m3 ), a2 = 1.35 × 104 K, D = 16.92 µm, L = 194 µm, τ = 3.3 ns, ρr = 0. SKETCH: Figure Pr.2.6 shows the laser irradiation focused on the kernel volume V , where it is partly absorbed. Switched Pulse Laser, l = 1.064 mm Lens Hydrocarbon and Oxygen (e.g., Air)
Pulse Laser Irradiation Focused Beam, qr,i (t) rr qr,i (t)
Ar Focal Volume, V x (Ignition Kernel)
L Se,s
Extinction Coefficient sex = sex(T) D
Figure Pr.2.6 Laserinduced spark ignition of a hydrocarbonoxidizer gaseous mixture.
OBJECTIVE: (a) Using the maximum photon energy given by λ h P f = hP , c where hP is the Planck constant, and f is the frequency, λ is the wavelength, and c is the speed of light, determine the photon ﬂux m ˙ ph (photon/m2 s). Use the speed of light in vacuum c = co . (b) Using a temperaturedependent extinction coeﬃcient σex (1/m) =
a1 n2e T 1/2
1 − e−a2 /T ,
where a1 and a2 are constants and T (K) is the kernel temperature, determine the energy absorbed in the focal volume, shown in Figure Pr.2.6, over the time span, −∞ < t < ∞, i.e., ∞ S˙ e,σ dt. −∞
(c) Express the results of (b) per kernel volume V . 59
SOLUTION: (a) The photon energy ﬂux is (qr,i )o . Then (qr,i )o = m ˙ ph hP co /λ or m ˙ ph =
(qr,i )o λ . hP co
Here we use c = co and co and hP are listed in Table C.1.(b), i.e., hP = 6.626 × 10−34 Js co = 3.000 × 108 m/s. Then m ˙ ph =
1017 (W/m2 ) × 1.064 × 10−6 (m) = 5.353 × 1035 photon/m2 s. 6.626 × 10−34 (Js) × 3 × 108 (m/s)
(b) From (2.43), we have
S˙ e,σ = qr,i (1 − ρr )σex e−σex x .
The extinction coeﬃcient is 0.1645 × 10−42 (K1/2 m5 ) × (1026 )2 (1/m3 )2
σex =
(106 )1/2 (K)1/2
4 6 1 − e−1.35 × 10 /10
= 2.206 × 104 1/m. Here ρr = 0, and upon the time and volume integration, we have
∞
−∞
∞
L
qr,i (t)σex e−σex x dxdt ∞ −σex L − 1) qr,i (t)dt = −Ar (e −∞ ∞ 2 2 −2.206×104 ×194×10−6 = −Ar (e − 1) (qr,i )o e−t /τ dt
S˙ e,σ dt = Ar σex
−∞
−∞
= 0.9861Ar (qr,i )o π 1/2 τ,
πD2 . Ar = 4
Using the numerical values, we have Ar = 2.247 × 10−10 m2 , and ∞ S˙ e,σ dt = 0.9861 × 2.247 × 10−10 (m2 ) × 1017 (W/m2 ) × π 1/2 × 3.3 × 10−9 (s) −∞
= 0.1296 J. (c) Using V = 4.360 × 10−14 m3 , we have S˙ e,σ = σex (qr,i )o π 1/2 τ V = 2.972 × 1012 J/m3 . This is a rather large result. COMMENT: Note that, in practice, the focused laser beam will not be uniform and therefore, a radial average should be taken. Also note that the irradiation ﬂux used is for the focused beam. The beam leaving the laser has a much larger diameter, which makes for a smaller irradiation ﬂux.
60
PROBLEM 2.7.FUN GIVEN: In thermoelectric cooling, a pair of p and ntype semiconductors are jointed at a junction. When an electric current, given as current ﬂux (or current density) je (A/m2 ), passes through their junction, heat is absorbed. This current also produces the undesirable (parasitic) Joule heating. This energy conversion (per unit volume) is given by (2.33) as s˙ e,J = ρe (T )je2 , where ρe (ohmm) is the electrical resistivity and varies with temperature ρe = ρe (T ). Figure Pr.2.7 shows a semiconductor slab (p or ntype), which is a part of a pair. The energy equation (2.8) would be simpliﬁed by assuming that heat ﬂows only in the x direction, that the heat transfer is in a steady state, and that the energy conversion term is given above. A small length ∆x is take along the x direction and the conduction heat ﬂux vectors at x and x + δx are given as qk x and qk x+∆x . qk x = −1.030 × 104 sx W/m2 , qk x+∆x = 1.546 × 104 sx W/m2 , je = 4 × 106 A/m2 , ∆x = 0.1 mm. SKETCH: Figure Pr.2.7 shows the semiconductor slab, the current density je , and the conduction heat ﬂux vectors on both sides of a small length ∆x. Semiconductor Current Density, je
qk x+,x
L
He(T)
,x
sn sn
x
qk x
a
a se,J = He je2
je
Figure Pr.2.7 A semiconductor slab with a onedimensional heat conduction and a volumetric energy conversion (Joule heating). The conduction heat ﬂux vectors are prescribed at locations x and x + ∆x.
OBJECTIVE: (a) Using (2.8), and assuming that the results for the given small ∆x are valid for ∆x → 0, determine the magnitude of s˙ e,J . (b) Using the relationship for s˙ e,J given above, and the value for je given below, determine ρe (T ) and from Tables C.9(a) and (b) ﬁnd a material with this electrical resistivity ρe (T ) (ohmm). SOLUTION: (a) From (2.8), we have (q · sn )dA ∆A
∆V → 0
=−
∂ ρcp T + s˙ i ∂t i
For steadystate, conduction heat transfer only, and Joule heating as the only energy conversion, this becomes (qk · sn )dA ∆A = s˙ e,J , ∆V where from Figure Pr.2.7, we have ∆V
= a2 ∆x
∆A = a2 . 61
Using the values for qx given at locations x and x + ∆x, and noting that for the surface at x we have sn = −sx and for that at x = x + ∆x, we have sn = sx , we have (qk x · sn )∆A + (qk x+∆x · sn )∆A ∆V
= =
{−1.030 × 104 (W/m2 ) × [sx · (−sx )] + 1.546 × 104 (W/m2 ) × [sx · (sx )]}a2 a2 ∆x 4 4 2 (1.030 × 10 + 1.546 × 10 )(W/m ) = 2.576 × 108 W/m3 . 10−4 (m)
(b) Noting that s˙ e,J = ρe (T )je2 , and solving for ρe (T ), we have ρe (T ) = =
s˙ e,J je2 2.576 × 108 (W/m3 ) = 1.610 × 10−5 Wm/A2 = 1.610 × 10−5 ohmm, (4 × 106 )2 (A2 /m4 )
where we note that (W=A2 ohm). Note that in Table C.9(b) at T = 700 K, the ntype silicongermanium alloy has this electrical resistivity. COMMENT: Note that heat leaves both surfaces (x and x + ∆x), as expected from (q · sn )dA ∆A = ∇ · q > 0. ∆V → 0 The products of (q · sn ) at x and x + ∆x are both positive here.
62
PROBLEM 2.8.FUN GIVEN: In some transient heat transfer (i.e., temperature and heat ﬂux vector changing with time) applications, that portion of the heat transfer medium experiencing such a transient behavior is only a small portion of the medium. An example is the seasonal changes of the air temperature near the earth’s surface, which only penetrates a very short distance, compared to the earth’s radius. Then the medium may be approximated as having an inﬁnite extent in the direction perpendicular to the surface and is referred to a semiinﬁnite medium. Figure Pr.2.8 shows such a medium for the special case of a sudden change of the surface temperature from the initial (and uniform throughout the semiinﬁnite medium) temperature of T (t = 0) to a temperature Ts . Under these conditions, the solution for the heat ﬂux is given by qk,x (x, t) =
k[Ts − T (t = 0)] 1/2
(παt)
e−x
2
/4αt
,
where α = k/ρcp is called the thermal diﬀusivity. k = 0.25 W/mK (for nylon), α = 1.29 × 10−5 m2 /s (for nylon), Ts = 105◦C, T (t = 0) = 15◦C, xo = 1.5 cm, to = 30 s. This conduction heat ﬂux changes with time and in space. SKETCH: Figure Pr.2.8 shows the semiinﬁnite slab, the conduction heat ﬂux, and the local energy storage/release. Storage rcp ¶T ¶t x
xo
Dx 0
sn SemiInfinite Slab Initially at Uniform Temperature T = T(t = 0) Constant Surface Temperature, Ts
x
qk,x(x,t)
FarField Temperature T(t,x ) = T(t = 0)
Conduction
Figure Pr.2.8 A semiinﬁnite slab with an initial temperature T (t = 0) has its surface temperature suddenly changed to Ts .
OBJECTIVE: (a) Using (2.1), with no energy conversion and conduction as the only heat transfer mechanism, determine the time rate of change of local temperature ∂T /∂t at location xo and elapsed time to . (b) Determine the location of largest time rate of change (rise) in the temperature and evaluate this for the elapsed time to . SOLUTION: (a) Starting from (2.1) and for no energy conversion and a onedimensional (in the x direction) conduction only, we have ∂ ∂T sx · (qk,x sx ) = −ρcp ∇·q = ∂x ∂t ∂T ∂ qk,x = −ρcp ∂x ∂t or −
∂T 1 ∂ = qk,x . ∂t ρcp ∂x 63
Using the given expression for qk,x (x, t), we have
∂T ∂t
= −
k ρcp
Ts − T (t = 0) (παt)1/2
Ts − T (t = 0)
=
2π 1/2 α1/2 t3/2
−2x 4αt
−x2 e 4αt
−x2 xe 4αt .
Evaluating this at xo and to , we have −x2o Ts − T (t = 0) ∂T = xo e 4αto . ∂t 2π 1/2 α1/2 t3/2 o Using the numerical values, we have ∂T ∂t
(1.5 × 10−2 )2 (m)2 −5 2 × 1.510−2 (m)e 4 × 1.29 × 10 (m /s) × 30(s) −
◦
(105 − 15)( C)
=
2π 1/2 (1.29 × 10−5 )1/2 (m2 /s)1/2 (30)3/2 (s)3/2
=
0.6453(◦C/s) × e−0.1453 = 0.5580◦C/s.
(b) We now diﬀerentiate the above expression for ∂T /∂t, with respect to x, at which we ﬁnd the location of the largest ∂T /∂t occurs. Then by diﬀerentiating and using t = t0 , we have ∂ ∂T ∂x ∂t
= =
∂ 2 qk,x ∂x2
x2 x2 Ts − T (t = 0)  4αto 2x2  4αto e = 0. − e 4αto 2π 1/2 α1/2 t3/2 o
Then 1−
2x2 =0 4αto
or x = (2αto )1/2 . Now using the numerical values, we have x = =
[2 × 1.29 × 10−5 (m2 /s) × 30(s)]1/2 0.02782 m = 2.782 cm.
From part (a), we have ∂T = 1.1968(◦C/s) × e−0.5 = 0.7259◦C/s. ∂t COMMENT: Note that as expected, ∂T /∂t = 0 at x = 0 (because Ts is assumed constant). In Section 3.5.1, we will discuss this transient problem and deﬁne the penetration front as the location beyond which the eﬀect of the surface temperature change has not yet penetrated and this distance is given as x ≡ δα = 3.6(αt)1/2 , as compared to x = 1.414(αt)1/2 for the location of maximum ∂T /∂t.
64
PROBLEM 2.9.FUN GIVEN: A device that allows for heat transfer between two ﬂuid streams of diﬀerent temperatures is called a heat exchanger. In most applications, the ﬂuid streams are bounded by ﬂowing through ducts and tubes and are also kept from mixing with each other by using an impermeable solid wall to separate them. This is shown in Figure Pr.2.9. Assume that radiation and conduction are not signiﬁcant in each stream and that there is steadystate heat transfer and no energy conversion. Then there is only bounded ﬂuid stream convection and surface convection at the separating wall, as shown in Figure Pr.2.9. Assume that the wall has a zero thickness. Also assume convection heat ﬂux qu a uniform (an average) across the surface area for convection Au . The surface area for the surface convection over a diﬀerential length ∆x is ∆Aku . SKETCH: Figure Pr.2.9 shows the two streams and the wall separating them.
Au,1 qu,1
Bounded Fluid Stream 1
sn,2
Thin Impermeable Solid Wall (Negligible Thickness) Bounded Fluid Stream 2
Ideally Insulated Surface
qku Aku = Pku,x
sn,1 qu,2
y x
Au,2 Ideally Insulated Surface
,x 0
Figure Pr.2.9 Two streams, one having a temperature higher than the other, exchange heat through a wall separating them.
OBJECTIVE: Starting from (2.8), write the energy equations for the control surfaces ∆A1 and ∆A2 shown. These control surfaces include both the convection and the surface convection areas. Show that the energy equations become −
Pku d qu,1 = 0, qku + Au,1 dx Pku d qu,2 = 0. qku + Au,2 dx
SOLUTION: Starting from (2.8), we note that ∂/∂t = 0, s˙ = 0, and q = qu along the axis and q = qku along the y axis on the wall surface. Then (2.8) becomes, using ∆V1 = Au,1 ∆x and ∆Aku = Pku ∆x, (q · sn )dA ∆A1
∆V1 → 0
=
0+0
(qku · sn )dA +
=
∆Aku
(qu · sn )dA ∆Au,1
∆V1 → 0 −qku ∆Aku + qu,1 (x + ∆x)Au,1 − qu,1 (x)Au,1 = Au,1 ∆x qku Pku ∆x qu,1 (x + ∆x) − qu,1 (x) + = − Au,1 ∆x ∆x d qu = 0 for ∆x → 0. = −Pku qku + dx 65
Note that qku · sn,1 = −qku , because sn is pointing opposite to the assumed direction for qku . Similarly, (qu · sn )dA qku ∆Aku + qu,2 (x + ∆x)Au,2 − qu,2 (x)Au,2 ∆Au,2 = ∆V2 → 0 Au,2 ∆x d qu,2 = 0 for ∆x → 0. = Pku qku + dx COMMENT: Note that the two energy equations mathematically state what is rendered in Figure Pr.2.9, i.e., heat is convected along each stream and is exchanged through the wall (by surface convection). In Section 7.6.1, we will use these energy equations to determine the total heat transfer (exchange) rate Qku .
66
PROBLEM 2.10.FUN GIVEN: A heat transfer medium with a rectangular control volume, shown in Figure Pr.2.10, has the following uniform heat ﬂuxes at its six surfaces: qx x−∆x/2 qy y−∆y/2 qz z−∆z/2
= −4 W/m2 , = 6 W/m2 , = 2 W/m2 ,
qx x+∆x/2 qy y+∆y/2 qz z+∆z/2
= −3 W/m2 , = 8 W/m2 , = 1 W/m2 .
The uniformity of heat ﬂux is justiﬁable due to the small dimensions ∆x = ∆y = ∆z = 2 mm. SKETCH: Figure Pr.2.10 A rectangular control volume with the heat ﬂux vector on its six surfaces. qz
z+D z/2
qy
DAz
y+Dy/2
Dy Dx qx
qx xD x/2
(x, y, z)
Dz
DAy
x+D x/2
DAx
z
qy yDy/2
y
qz zD z/2
x
Figure Pr.2.10 A control volume in a heat transfer medium.
OBJECTIVE: (a) Assume that ∇V → 0 is approximately valid for this small, but ﬁnite volume and determine the divergence of q for the center of this control volume, located at (x, y, z). (b) Is there a sink or a source of heat in this control volume located at (x, y, z)? (c) What could be the mechanisms for this source or sink of heat? SOLUTION: (a) We have assumed that over each of the six surfaces the heat ﬂux is uniform. Then, we can use (2.8) as lim
∆A
∆V →0
(q · sn )dA ∂(ρcp T ) ˙ Si . ≡∇·q=− + ∆V ∂t i
The divergence of q is given above in terms of the surface integral. This surface integral is expanded using the three components of q and sn in the x, y, and z directions. Using that we have
(q · sn )dA
(qx ·sn,x + qy ·sn,y + qz ·sn,z )dA
=
∆A
∆A
=
(qx+∆x/2 − qx−∆x/2 )∆Ax + (qy+∆y/2 − qy−∆y/2 )∆Ay + (qz+∆z/2 − qz−∆z/2 )∆Az . 67
Then, for the divergence of q we use the deﬁnition and approximations to obtain (q · sn )dA (q · sn )dA ∆A ∇·q ≡ lim ∆A ∆V →0 ∆V ∆V (qy+∆y/2 − qy−∆y/2 )∆Ay (qx+∆x/2 − qx−∆x/2 )∆Ax + + = ∆V ∆V (qz+∆z/2 − qz−∆z/2 )∆Az ∆V (qy+∆y/2 − qy−∆y/2 )∆x∆z (qx+∆x/2 − qx−∆x/2 )∆y∆z + + = ∆x∆y∆z ∆x∆y∆z (qz+∆z/2 − qz−∆z/2 )∆x∆y . ∆x∆y∆z Since ∆x = ∆y = ∆z, we have ∇·q
1 (qx+∆x/2 − qx−∆x/2 + qy+∆y/2 − qy−∆y/2 + qz+∆z/2 − qz−∆z/2 ). ∆x
Now, using the numerical values we have ∇·q=
1 2 3 (+4 − 3 − 6 + 8 − 2 + 1)(W/m ) = 103 W/m . 2 × 10−3 (m)
(b) From (2.2), since ∇ · q is positive, there is a source in the control volume, i.e., there is a source at the location (x, y, z). This is because more heat leaves the control surface than enters it. (c) From the energy equation (2.8), the mechanism for this heat source is storage or energy conversion. There are many energy conversion mechanisms (to and from thermal energy), for example, those listed in Table 2.1. When the temperature within the control volume increases, i.e., ∂T /∂t > 0, heat is being stored in the control volume as sensible heat. Another example for a sink of heat is when there is an endothermic chemical reaction. Then heat is absorbed in the control volume to move the reaction forward (from reactants to products). When there is a gas ﬂow and the gas undergoes expansion as it ﬂows through the control volume, heat is absorbed as the gas performs work (this is called expansion cooling). COMMENT: Here we assumed that q is uniform over the surfaces. This is strictly true when ∆A → 0. The assumption of uniform q over a surface can be justiﬁably made when the heat ﬂow is unidirectional. When the heat ﬂow over a surface is zero, the surface is called adiabatic.
68
PROBLEM 2.11.FUN GIVEN: Although the temperature variation within a heat transfer medium is generally three dimensional, in many cases there is a dominant direction in which the most signiﬁcant temperature variation occurs. Then, the use of a onedimensional treatment results in much simpliﬁcation in the analysis. Consider the steadystate surface temperatures given in Figure Pr.2.11(a), for selected locations on a solid, rectangular piece. The heat ﬂows through the solid by conduction and from its surface to the ambient by surface convection. SKETCH: Figure Pr.2.11(a) shows the rectangular slab and the measured temperature at various locations. qu
Solid
qku
29 C
30 C 29 C
54 C
55 C qk
Lz = 6 mm Lx /2 = 15 mm
54 C
84 C
85 C
84 C
84 C Ly = 50 mm
Lx /2 (0, 0, 0)
Air Tf, = 20 C uf, = 1 m/s
z x y
Figure Pr.2.11(a) Temperature at various locations on a rectangular plate.
OBJECTIVE: (a) By examining the gradient of temperature in each direction, determine the dominant conduction heat ﬂow direction. As an approximation, use ∆Tx ∂T ∆Ty ∂T ∆Tz ∂T , , , ∂x ∆x ∂y ∆y ∂z ∆z where ∆x is the length over which the temperature change ∆Tx occurs. (b) Select a control volume that has a diﬀerential length in the direction of dominant conduction heat ﬂow and an integral length over the other two directions. Schematically show this integraldiﬀerential volume. (c) Write an energy equation for this control volume. SOLUTION: (a) The three principal directions, x, y, and z for the rectangular solid piece are shown in Figure Pr.2.11(b). The heat transfer by conduction is given by Fourier’s law (1.11) and using (1.14) we have qk = −k∇T ≡ −k
∂T ∂T ∂T sx + sy + sz . ∂x ∂y ∂z
We now use the approximation for the gradient of temperature and write these as ∆Tx ∆Ty ∆Tz sx + sy + sz . qk = −k Lx Ly Lz Next we substitute for the temperatures and lengths obtaining 30(◦C) − 85(◦C) 84(◦C) − 85(◦C) 85(◦C) − 84(◦C) sx + sy + sz qk = −k 0.030(m) 0.050(m) 0.006(m) = −k[−1.883 × 103 (◦C/m)sx − 2.000 × 101 (◦C/m)sy + 1.667 × 102 (◦C/m)sz ]. 69
qu qku qk
,Aku ,Ak
qk
sn
x+,x
sn
,x
sn x
Lz Ly
z x y
Figure Pr.2.11(b) The heat ﬂow along the x direction, where a diﬀerential length ∆x is chosen.
We note that the term in x is at least one order of magnitude larger than any of the other two terms. Then, from the order of magnitude of the terms, we can approximate the heat ﬂux by qk [1.883 × 103 (◦C/m) × k(W/m◦C)]sx , which represents a onedimensional conduction heat ﬂow. (b) Figure Pr.2.11(b) shows the diﬀerential length taken along the x direction. (c) Since the temperature ﬁeld is steady and there is no energy conversion, (2.7) becomes (q · sn )dA = 0. ∇ · q ≡ lim ∆A ∆V →0 ∆V As shown in Figure Pr.2.11(b), based on the onedimensional intramedium conduction and surface convection, one can write the limit above as (q · sn )dA (qku ·sn )∆Aku [(qk ·sn )Ak ]x + [(qk ·sn )Ak ]x+∆x ∆A lim = lim + . ∆V →0 ∆V →0 ∆V ∆xLy Lz ∆xLy Lz Here we have Ak = Ly Lz and ∆Aku = 2(Ly + Lz )∆x. Now, noting that sn on the x and x + ∆x surfaces point in opposite directions, we have (q · sn )dA 2(Ly + Lz )∆x (−qk x + qk x+∆x )Ly Lz lim ∆A = lim qku + ∆V →0 ∆V →0 ∆V ∆xLy Lz ∆xLy Lz 2(Ly + Lz ) −qk x + qk x+∆x = qku . + lim ∆V →0 Ly Lz ∆x Now, the limit in the last term is the deﬁnition of a derivative and we have (q · sn )dA 2(Ly + Lz ) dqk lim ∆A = qku + ∆V →0 ∆V Ly Lz dx and, using Fourier’s law of conduction, qk = −k
dT . dx
Assuming that k is constant, we have ﬁnally qku
2(Ly + Lz ) d2 T − k 2 = 0. Ly Lz dx
COMMENT: Note that the gradient of temperature along the x direction is not uniform. Over the ﬁrst half of the length along x, the temperature change is larger than that over the second half. This is a consequence of the local surfaceconvection heat ﬂux being proportional to the diﬀerence between the solid temperature and the farﬁeld ﬂuid temperature. We will discuss this in Section 6.8. 70
PROBLEM 2.12.FUN GIVEN: Many computer disks are read by magnetoresistive transducers. The transducer is located on a thin slider that is situated slightly above the disk, as shown in Figure Pr.2.12. The transistor is developed using the principle that its resistance varies with the variation of the surrounding magnetic ﬁeld. Since its resistance is also temperature dependent, any temperature change will result in a noise in the readout. When the slider and disk are at the same temperature, the viscousdissipation heat generation becomes signiﬁcant in creating this undesired increase in the temperature. Assume the ﬂow of air at T = 300 K between the disk and slider is a Newtonian, onedimensional, Couette ﬂow, as shown in Figure Pr.2.12. The distance between the disk and the slider is L = 20 nm, and the relative velocity is ∆ui = 19 m/s. Use Table C.22, and the relation µf = νf /ρf to determine µf for air at T = 300 K. SKETCH: Figure Pr.2.12 shows the disk and slider, the air ﬂow between them, and the viscous dissipation energy conversion.
Transducer (Reader) Slider (Stationary) y Air
x, ux r
,ui
ux
sm,µ
L
ω (Angular Frequency)
0 ux = Dui 1 
y L
Disk
Figure Pr.2.12 A disk read by a magnetoresistive transducer.
OBJECTIVE: Determine the magnitude of the volumetric viscousdissipation heat generation s˙ m,µ . SOLUTION: The properties of air at T = 300 K, (Table C.22), are νf = 15.66 × 10−6 m2 /s, and ρf = 1.177 kg/m3 . For the onedimensional ﬂow in the x direction, we have from (2.52)
∂ux ∂y
2 ∂ux = µf ∂y y = ∆ui 1 − L −∆ui = L
s˙ m,µ
=
15.66 × 10−6 (m2 /s) × 1.177(kg/m3 ) ×
=
1.66 × 1013 W/m3 .
s˙ m,µ ux
192 (m/s)2 (2 × 10−8 )2 (m2 )
COMMENT: It should be noted that the relation ux = ∆ui (1 − y/L) is derived using the noslip boundary condition. This condition only holds true if the gap L is greater than the meanfree path of the gas λm . For air at STP, we have from Table C.7, λ = 10−7 m, which is greater than the gap L = 2 × 10−8 m. Thus, the ﬂuid velocity should be given in terms of the slip coeﬃcient used for λm > L.
71
PROBLEM 2.13.FAM GIVEN: Catalytic combustion (and catalytic chemical reaction in general) is an enhancement in the rate of chemical reaction due to the physicalchemical mediation of a solid surface. For example, in the automobile catalytic converter the rate of reaction of exhaustgas unburned fuel is increased by passing the exhaust gas over the catalytic surface of the converter. The catalytic converter is a solid matrix with a large surface area over which the exhaust gas ﬂows with a relatively small pressure drop. The catalytic eﬀect is produced by a surface impregnated with precious metal particles, such as platinum. For example, consider the following chemical kinetic model for the reaction of methane and oxygen CH4 + 2O2 → CO2 + 2H2 O stoichiometric chemical reaction 1/2 m ˙ r,CH4 = −ar ρCH4 ρO2 e−∆Ea /Rg T chemical, kinetic model, where m ˙ r,CH4 (kg/m2 s) is the reactionrate per unit surface area. The preexponential factor ar (cm5/2 /sg1/2 ) and the activation energy ∆Ea (J/kmole) are determined empirically. The model is accurate for high oxygen concentrations and for high temperatures. The densities (or concentrations) are in g/cm3 . Consider the following catalytic (in the presence of Pt) and noncatalytic (without Pt) chemical kinetic constants: without Pt with Pt
: ar = 1.5 × 1011 cm5/2 /sg1/2 , : ar = 1.5 × 1012 cm5/2 /sg1/2 ,
∆Ea = 1.80 × 108 J/kmole, ∆Ea = 1.35 × 108 J/kmole.
Use the cgs units (cm, g, s). OBJECTIVE: For a mixture of methane and oxygen at a pressure of 1 atm and a temperature of 500◦C: (a) Determine the densities of CH4 and O2 assuming an idealgas behavior, (b) Determine the rate of reaction per unit surface area m ˙ r,CH4 . (c) Comment on the eﬀect of the catalyst. SOLUTION: (a) For an ideal gas mixture we have p Rg T M νCH4 MCH4 + νO2 MO2 , νCH4 + νO2
ρ = ρCH4 + ρO2 =
M
=
where νCH4 and νO2 are the stoichiometric coeﬃcients for CH4 and O2 in the chemical reaction. Also, ρCH4 νCH4 MCH4 νO 2 M O 2 ρO = , 2 = . ρ νCH4 MCH4 + νO2 MO2 ρ νCH4 MCH4 + νO2 MO2 The molecular weights are found from Table C.4 to be MCH4 = 12.011 + 4 × 1.008 = 16.04 kg/kmole MO2 = 2 × 15.999 = 32.00 kg/kmole. Then M=
1 × 16.04 + 2 × 32.00 = 26.68 kg/kmole. 1+2 72
Using the pressure and temperature given ρ =
ρO2
26.68(kg/kmole) 3
(500 + 273.15)(K)
= 0.4205kg/m
3
0.4205(kg/m ) × 1,000(g/kg) × 10−6 (m3 /cm ) = 4.205 × 10−4 g/cm 1 × 16.04(kg/kmole) 3 3 = 4.205 × 10−4 (g/cm ) = 8.427 × 10−5 g/cm 16.04(kg/kmole) + 2 × 32.00(kg/kmole) 2 × 32.00(kg/kmole) 3 3 = 4.205 × 10−4 (g/cm ) = 3.362 × 10−4 g/cm . 16.04(kg/kmole) + 2 × 32.00(kg/kmole) =
ρCH4
1.013 × 105 (Pa) 8.314×103 (J/kmoleK)
3
3
(b) Now, using − ∆Ea 1/2 m ˙ r,CH4 = ρCH4 ρO2 ar e Rg T the reaction rate without catalytic eﬀect is given by m ˙ r,CH4 (without catalyst)
=
8.427 × 10−5 (g/cm ) × [3.362 × 10−4 (g/cm )]1/2 × 1.5 × 1011 (cm5/2 /g 1.80 × 108 (J/kmole) exp − 8.314 × 103 (J/kmoleK) × (500 + 273.15)(K)
=
1.598 × 10−7 g/cm s.
3
3
1/2
s) ×
2
Using a catalyst, we have m ˙ r,CH4 (with catalyst) =
=
8.427 × 10−5 (g/cm ) × [3.362 × 10−4 ( g/cm )]1/2 × 1.5 × 1012 (cm5/2 /g 1.35 × 108 (J/kmole) exp − 8.314 × 103 (J/kmoleK) × (500 + 273.15)(K) 3
3
1/2
s)
1.754 × 10−3 g/cm s. 2
(c) Due to the presence of the catalyst, the surface reaction rate is increased by a factor m ˙ r,CH4 (with catalyst) 1.754 × 10−3 (g/cm s) = 1.098 × 104 . = 2 m ˙ r,CH4 (without catalyst) 1.598 × 10−7 (g/cm s) 2
There is a substantial increase. Reaction rates that are negligibly small when no catalyst is present can be made substantial with the addition of a catalytic coating. COMMENT: Precious metals are deposited on catalyst surfaces as particles. The relevant linear dimensions are on the order of nanometers. The molecules of the gas are adsorbed on the surface, suﬀer a chemical modiﬁcation, are desorbed, and then react in the gas phase. The total reaction rate for a catalytic converter with surface area Aku is given by M˙ r,CH4 = Aku m ˙ r,CH4 . Using structures such as foams, bundle of tubes, and packed beds, a large surface area per unit volume Aku /V is possible and large reaction rates are achieved.
73
PROBLEM 2.14.FUN GIVEN: In order to produce silicon wafers, singlecrystal silicon ingots are formed by the slow solidiﬁcation of molten silicon at the tip of a cylinder cooled from the base. This was shown in Figure 2.3(c) and is also shown in Figure Pr.2.14. The heat released by solid to ﬂuid phase change S˙ ls (W) is removed from the solidliquid interface Als by conduction through the ingot Qk . The energy equation for the solidﬂuid interface Als (nonuniform temperature in the liquid), as given by (2.9), is Qk = S˙ ls , where the conduction heat ﬂow Qk is given by Qk = Ak k
Tsl− Ts , L
where L and Ts are shown in Figure Pr.2.14 and Tls is the melting temperature. The rate of phasechange energy conversion is S˙ ls = −ρl Asl uF ∆hls = −M˙ ls ∆hls = M˙ ls ∆hsl , where ∆hsl > 0. Assume that the liquid and solid have the same density. L = 20 cm, uF = 4 mm/min. SKETCH: Figure Pr.2.14 shows the cooling of molten silicon and the formation of a crystalline silicon. Surface Temperature Ts D Crystalline Silicon
uF L
BoundingSurface Control Surface, Als Makeup Liquid
qk
Liquid Silicon (Melt)
Sls Tls PhaseChange Temperature Qku Makeup Heat
Figure Pr.2.14 Czochralski method for singlecrystal growth of silicon.
OBJECTIVE: Using the thermophysical properties given in Tables C.2 (periodic table for ∆hsl ) and C.14 (at T = 1,400 K for k), in Appendix C, determine the temperature Ts (at the top of the ingot). SOLUTION: Combining the above equations, we have Ak k
Tsl − Ts = M˙ ls ∆hsl . L
Solving for Ts and using the equation for M˙ ls , we have Ts = Tls −
M˙ ls ∆hls L ρl LuF ∆hls L = Tls − . Ak k Ak k
Now, we obtain the properties from Tables C.2 (periodic table) and C.14 as Tsl = 1,687 K, ρl = ρs = 2,330 kg/m3 , ∆hsl = 1.802 × 106 J/kg, k = 24 W/mK. Note that from Table C.14, k is obtained for T = 1,400 K, 74
which is the highest temperature available. Then, using Ak = Asl = πD2 /4, we have Ts
= Tls −
ρl Asl uF ∆hls L Ak k
2,330(kg/m ) × (4 × 10−4 /60)(m/s) × 1.802 × 106 (J/kg) × 0.2(m) 24(W/mK) = (1,687 − 233.3)(K) = 1,454 K 3
=
1,687(K) −
COMMENT: Note that the thermal conductivity of solid silicon signiﬁcantly decreases as the temperature increases. At room temperature, k is about 150 W/mK, while at T = 1,400 K, it is 24 W/mK. Also, the melt is not at a uniform temperature and is superheated away from Asl . Then, additional heat ﬂows into Asl from the liquid.
75
PROBLEM 2.15.FAM GIVEN: The electrical resistivity of metals increases with temperature. In a surface radiation emission source made of Jouleheating wire, the desired temperature is 2,500◦C. The materials of choice are tantalum Ta and tungsten W. The electrical resistivity for some pure metals, up to T = 900 K, is given in Table C.8. Assume a linear dependence of the electrical resistivity on the temperature, i.e., ρe (T ) = ρe,o [1 + αe (T − To )]. OBJECTIVE: (a) From the data in Table C.8, determine αe for both metals. (b) Using the equation above, determine the metals electrical resistivity at T = 2,500◦C. (c) If the wire has a diameter D = 0.1 mm and a length L = 5 cm (coiled), determine the electrical resistance Re for both metals at T = 25◦C and T = 2,500◦C. (d) If a Joule heating rate of 100 W is needed, what current should be applied at T = 2,500◦C? (e) Determine the voltage needed for this power. SOLUTION: (a) From Table C.8, we choose the last set of data for each metal to determine αe . We also use To = 900 K, since we need to extrapolate beyond 900 K. Then we have tantalum:
=
αe
= tungsten:
αe
= =
1 (40.1 − 31.8) × 10−8 (ohmm) 1 ∆ρe = ρe,o ∆T 40.1 × 10−8 (ohmm) 200(K) 1.035 × 10−3 1/K (21.5 − 15.7) × 10−8 (ohmm ) 1 21.5 × 10−8 ( ohmm) 200(K) 1.349 × 10−3 1/K.
(b) Using ρe = ρe,o [1 + αe (T − To )] and choosing To = 900 K, for T = 2,773 K we have tantalum:
ρe
= =
40.1 × 10−8 (ohmm)[1 + 1.035 × 10−3 (1/K) × (2,773 − 900)(K)] 1.178 × 10−6 ohmm
tungsten:
ρe
= =
21.5 × 10−8 (ohmm)[1 + 1.349 × 10−3 (1/K) × (2,773 − 900)(K)] 7.582 × 10−7 ohmm.
(c) From (2.32), the resistance can be calculated from Re =
ρe L . A
For A = πD2 /4, we have tantalum: tungsten:
4 × 1.178 × 10−6 (ohmm) × 0.05(m) = 7.500 ohm π(10−4 )2 (m2 ) 4 × 7.582 × 10−7 (ohmm) × 0.05(m) = 4.827 ohm. Re = π(10−4 )2 (m2 )
Re =
(d) Substituting (2.31) into (2.28), the Joule heating rate can be written as S˙ e,J = Re Je2 . Solving for Je we have
Je =
S˙ e,J Re 76
1/2 .
Then tantalum: tungsten:
Je Je
= =
[100(W)/7.500(ohm)]1/2 = 3.651 A [100(W)/4.828(ohm)]1/2 = 4.551 A.
(e) The voltage is given by (2.32), i.e., ∆ϕ = Re Je . Then tantalum: tungsten:
∆ϕ = 7.500(ohm) × 3.651(A) = 27.38 V ∆ϕ = 4.828(ohm) × 4.551(A) = 21.97 V.
COMMENT: In order to produce the given Joule heating rate with a given voltage, the diameter and length of the wire are selected accordingly. Care must be taken to keep the wire temperature below the melting point of the wire or insulation. The temperature of the wire will depend on the heat transfer rate (heat losses) at the wire surface. The linear extrapolation of resistivity as a function of temperature, so far from the listed values, is not expected to be very accurate.
77
PROBLEM 2.16.FAM GIVEN: Electrical power is produced from a thermoelectric device. The thermoelectric junctions are heated (heat added) by maintaining the hot junction at Th = 400◦C and cooled (heat removed) by maintaining the cold junction at T = 80◦C. This is shown in Figure Pr.2.16. There are 120 pn pairs. The pairs are p and ntype bismuth telluride (Bi2 Te3 ) alloy with Seebeck coeﬃcients αS,p = 2.30 × 10−4 V/K and αS,n = −2.10 × 10−4 V/K. The resistance (for all 120 pairs) to the electrical current Je produced is Re = 0.02 ohm for the thermoelectric path. For an optimum performance, the external resistance Re,o is also equal to 0.02 ohm. SKETCH: Figure Pr.2.16 shows the pn junctions in a thermoelectric power generator module. The electrical circuit is also shown. Ceramic Plate Semiconductor Je Re,o (External Electrical Resistance) −Qh Th Ceramic Plate
.
n
Se,P
p
n
Semiconductor
p
Electrical Conductor Tc Qc
Je Re,o
Figure Pr.2.16 A thermoelectric generator.
OBJECTIVE: (a) Determine the current produced. (b) Determine the power produced. SOLUTION: (a) The electrical circuit diagram for this device is also shown in the text by Figure 2.16(b). The electrical power generated is given by (2.40). The current can also be found from (2.40) and is given by Je =
(αS,p − αS,n )(Th − Tc ) . Re,o + Re
Using the numerical values, Je =
(2.30 + 2.10) × 10−4 (V/K) × (400 − 80)(K) = 3.520 A. 0.02(ohm) + 0.02(ohm)
(b) The electrical power produced Pe (based on external resistance) is Pe = Je2 Re,o = (3.52)2 (A)2 × 0.02(ohm) = 0.2478 W. COMMENT: The BiTe alloy is not a high temperature thermoelectric alloy. For higher temperatures (i.e., direct exposure to combustion gases), SiGe alloys are used. 78
PROBLEM 2.17.FUN GIVEN: A premixed mixture of methane CH4 and air burns in a Bunsentype burner, as shown in Figure Pr.2.17. Assume that the ﬂame can be modeled as a plane ﬂame. The reactants (methane and air) enter the ﬂame zone at a temperature T1 = 289 K. The concentration of methane in the reactant gas mixture is ρF,1 = 0.0621 kgCH4 /m3 and the heat of reaction for the methane/air reaction is ∆hr,CH4 = −5.55 × 107 J/kgCH4 (these will be discussed in Chapter 5). For both reactants and products, assume that the average density is ρ = 1.13 kg/m3 and that the average speciﬁc heat is cp = 1,600 J/kgK (these are temperatureaveraged values between the temperature of the reactants T1 and the temperature of the products T2 ). SKETCH: Figure Pr.2.17 shows the modeled ﬂame. The ﬂame is at the opening of a tube and is shown to be thin. Products, T2 Flame qu,2
x
Au
T2
δF Tube
Flame Sr,c
δF
T
0 qu,1
T1
Reactants, T1
Figure Pr.2.17 A premixed methaneair ﬂame showing the ﬂame and the temperature distribution across δF .
OBJECTIVE: (a) For the control volume enclosing the ﬂame (Figure Pr.2.17) apply the integralvolume energy conservation equation. Neglect the heat loss by radiation and assume a steadystate condition. (b) Obtain an expression for the heat generation inside the ﬂame S˙ r,c (W) as a function of the cold ﬂow speed uf (m/s), the concentration of methane in the reactant gas mixture ρF,1 (kgCH4 /m3 ), the heat of reaction for the methane/air reaction ∆hr,CH4 (J/kgCH4 ), and the area of the control surface Au (m2 ). Assume a complete combustion of the methane. [Hint: Use the conservation of mass of fuel equation (1.26) to obtain an expression for the volumetric reaction rate n˙ r,F (kg/m3 s).] (c) Using the integralvolume energy conservation equation obtained in item (a) and the expression for the heat generation obtained in item (b) calculate the temperature of the reacted gases (i.e., the adiabatic ﬂame temperature, T2 ). SOLUTION: (a) The heat ﬂux vector tracking is shown in Figure Pr.2.17. The following steps complete the solution. The integralvolume energy equation (2.9) is ∂ q · sn dA = s˙ i dV . − ρcp T dV + ∂t A V V i For this steadystate process the storage term is zero. The only energy conversion present is conversion from chemical bond to thermal energy (chemical reaction). The area integral of the normal component of the heat ﬂux vector over the control surfaces enclosing the ﬂame gives q · sn dA = −qu,1 Au + qu,2 Au , A
79
where Au is the surface area of the ﬂame sheet. Thus, the integralvolume energy conservation equation becomes s˙ r,c dV . −qu,1 Au + qu,2 Au = V
The heat ﬂux by convection is given by (2.1) as qu = ρcp uT. Using this, the energy equation becomes −(ρcp uF T )1 Au + (ρcp uF T )2 Au =
s˙ r,c dV . V
Note that the volumetric heat source term depends on the temperature which is not uniform within the ﬂame (i.e., within the control volume). (b) The volumetric heat source due to chemical reaction s˙ r,c (W/m3 s) can be written as s˙ r,c = n˙ r,F ∆hr,F , where n˙ r,F (kg/m3 s) is the volumetric reaction rate which gives the mass of fuel (methane) burned per unit volume and unit time. As ∆hr,F is constant for this constant pressure process, the rate of heat generation S˙ r,c is ˙ Sr,c = s˙ r,F dV = n˙ r,F ∆hr,F dV = ∆hr,F n˙ r,F dV . V
V
V
To ﬁnd the mass consumption rate, we use the conservation of mass equation for methane. The integralvolume species mass equation (1.26) is ∂MF + m ˙ F · sn dA = − n˙ r,F dV . ∂t A V The variation of the species mass with respect to time is zero for the steadystate process. For the control volume shown in Figure Pr.2.17, the net mass ﬂux, in analogy to the net heat ﬂux, is m ˙ F · sn dA = −m ˙ F,1 Au + m ˙ F,2 Au . A
The mass ﬂux of fuel F is m ˙ F = ρF uF . The mass ﬂux of methane leaving the control volume is zero because all the methane is burned inside the ﬂame. Therefore, the above equation becomes n˙ r,F dV . −ρF uF Au = V
The volumetric variation of the methane mass is due to the chemical reaction only. Thus, using the above equation we have S˙ r,c = s˙ r,F dV = −ρF uF Au ∆hr,F . V
(c) Using this, the integralvolume energy equation becomes −(ρcp uT )1 Au + (ρcp uT )2 Au = −ρF uF Au ∆hr,F . From the conservation of mass of mixture equation (1.25) (ρu)1 = (ρu)2 . By deﬁnition, uF = (ug )1 . Then dividing the energy equation by (ρu)1 Au , we have −(cp T )1 + (cp T )2 = − 80
ρF ∆hr,F . ρ
Solving for T2 , assuming that cp is constant, and using the numerical values given, we have ﬁnally T2 = T1 −
0.0621(kgCH4 /m3 ) × [−5.55 × 107 (J/kgCH4 )] ρF ∆hr,F = 2,195 K. = 289(K) − ρcp 1.13(kg/m3 ) × 1,600(J/kgK)
COMMENT: An expression for the volumetric reaction rate n˙ r,F can also be obtained by looking at the units. The volumetric reaction rate is the mass of methane (kgCH4 ) burned per unit volume (m3 ) per unit time (s). The mass of methane burned is equal to the mass of methane available in the reactants, as all the methane is burned in the ﬂame zone. This mass per unit volume is equal to the density of methane in the reactant mixture ρ. The time it takes to completely burn this mass of methane is equal to the time it takes for this mixture to travel through the reaction region. If the thickness of the reaction region is δF , and the velocity of the gas ﬂow is uF , the time is given by δF /uF . Thus, the volumetric reaction rate of methane becomes n˙ r,F = −ρuF /δF . These are all constant parameters. Therefore, integrating over the volume of the ﬂame (control volume) results in n˙ dV = −ρuF V /δF = −ρuF Au and the rate of heat generation becomes S˙ r,c = −∆hr,F ρuF Au . Note that V r,F the negative sign arises because methane is being consumed (as opposed to produced) in the chemical reaction. The average ρ and cp are temperatureaveraged values calculated over the temperature range between T1 and T2 . Note the high temperatures that are achieved in a ﬂame. The assumption of complete combustion is not true for most combustion processes. Lack of complete mixing of fuel and oxidizer, heat losses, and dissociation of products, all contribute to a lower ﬂame temperature. The ﬂame temperature can be reduced by diluting the reactant mixture, i.e., by reducing the amount of methane as compared to the amount of air. Then the combustion will occur in nonstoichiometric conditions.
81
PROBLEM 2.18.FAM GIVEN: A moistpowder tablet (pharmaceutical product) is dried before coating. The tablet has a diameter D = 8 mm, and a thickness l = 3 mm. This is shown in Figure Pr.2.18. The powder is compacted and has a porosity, i.e., void fraction, Vf /V = 0.4. This void space is ﬁlled with liquid water. The tablet is heated in a microwave oven to remove the water content. The rms of the electric ﬁeld intensity (e2 )1/2 = 103 V/m and the frequency f = 1 GHz. SKETCH: Figure Pr.2.18 shows microwave energy conversion in a moistpowder tablet. Powder Particle
Vl Vs
Water No Surface Heat Transfer (Idealized) Q A= 0 MoistPowder Tablet Microwave Energy Conversion Se,m
Vl = 0.4 Vs + Vl Sensible Heat Storage  ρcp V dT dt PhaseChange Energy Conversion Slg
l
Control Volume Control Surface D
Figure Pr.2.18 Microwave heating of a moistpowder tablet.
OBJECTIVE: (a) Determine the time it takes to heat the water content from the initial temperature of T (t = 0) = 18◦C to the ﬁnal temperature of T = 40◦C, assuming no evaporation. (b) Determine the time it takes to evaporate the water content while the tablet is at a constant temperature T = 40◦C. For the eﬀective (including the liquid and powder) volumetric heat capacity ρcp , which includes both water and powder, use 2 × 106 J/m3 K. For the water density and heat of evaporation, use Table C.27. Assume that the dielectric loss factor for the powder is negligible compared to that for the water. SOLUTION: When the moistpowder tablet is internally heated by microwave electromagnetic energy conversion, the temperature of the tablet increases and the moisture evaporates simultaneously. Here we have assumed that ﬁrst a rise in the temperature occurs, without any evaporation, and then evaporation occurs. (a) For the ﬁrst period, we start with the integralvolume energy equation (2.9) which is QA = −ρcp V
dT ˙ + S. dt
For no heat losses, QA = 0. The energy conversion is due to microwave heating only, i.e., S˙ = S˙ e,m . Then we have dT + S˙ e,m = 0. −ρcp V dt For the conversion from microwave to thermal energy, we have, from (2.49), S˙ e,m = 2πf ec o e2e Vl . The volume of water, which is 40% the total volume of the tablet (Vl = 0.4V ), and the dielectric loss factor for water ec,w will be used. Then we have −ρcp V
dT + 2πf ec,w o e2e (0.4V ) = 0. dt 82
Integrating the equation above assuming constant properties gives dT dt
=
2(0.4)πf ec,w o e2e ρcp
T (t) − T (t = 0)
=
2(0.4)πf ec,w o e2e ∆t1 . ρcp
Solving for ∆t1 we have ∆t1 =
∆T ρcp 2(0.4)πf ec,w o e2e
.
From Table C.10, for water at f = 109 Hz, we have ec,w = 1.2. Using the numerical values we have 3
∆t1
= =
(40 − 18)(K) × 2 × 106 (J/m K) 2
π(0.8) × 109 (1/s) × 1.2 × 8.8542 × 10−12 (A2 s2 /Nm ) × (103 )2 (V/m)2 1.648 × 103 s = 0.4577 hr.
(b) For the second period, we start again with the integral energy equation which is QA = −ρcp V
dT ˙ + S. dt
Again, no heat losses are considered, i.e., QA = 0. The temperature remains constant while the moisture evaporates and thus dT /dt = 0. The energy conversion is due to microwave heating S˙ e,m and to phase change only S˙ lg , i.e., S˙ = S˙ e,m + S˙ lg . Thus, the energy equation becomes S˙ e,m + S˙ lg = 0. Using (2.49) and (2.25) we have S˙ e,m = 2πf ec o e2e Vl S˙ lg = −n˙ lg ∆hlg Vl . 3
The evaporation rate has units of (kg/m s). To evaporate all the water we have M˙ l n˙ lg Vl = , ∆t2 ∆t2
or
n˙ lg =
M˙ l ρl ,ρl = . ∆t2 Vl
Then, using the equations above, the energy equation becomes 2πf ec,w o e2e Vl − Solving for ∆t2 we have ∆t2 =
ρl ∆hlg Vl = 0. ∆t2
ρl ∆hlg 2πf ec,w o e2e
.
From Table C.27, at T = 313.2 K, we have ρl = 991.7 kg/m3 and ∆hlg = 2.406 × 106 J/kg. Then, using the numerical values, we have 3
∆t2
= =
991.7(kg/m ) × 2.406 × 106 (J/kg) 2
2π × 109 (1/s) × 1.2 × 8.8542 × 10−12 (A2 s2 /Nm ) × (103 )2 (V/m)2 3.574 × 104 s = 9.928 hr.
COMMENT: Note that the evaporation period is longer than the sensible heating period. In the sensible heating period, no evaporation was included. In reality, the evaporation occurs simultaneously with the heating due to the diﬀerence in partial pressure of the water inside the powder and outside in the ambient. This evaporation is controlled by the rate of vapor ﬂow out of the powder. The surface heat transfer 83
should also be included if more accurate predictions are required. We have used a constant amount of water to calculate the microwave heating during the evaporation period. In reality, the amount of liquid decreases as the vapor is removed. This, along with the surface heat transfer, should be considered for more accurate predictions. Note also that no resistance to the vapor ﬂow out of the moist powder was considered. This may become the limiting transport rate for very ﬁne powders and the increase in the internal pressure could cause the formation of cracks.
84
PROBLEM 2.19.FUN GIVEN: The automobile airbag deploys when the pressure within it is suddenly increased. This pressure increase is a result of the inﬂow of gaseous products of combustion or pressurized air from an inﬂater connected to the bag. The airbag fabric may be permeable, and this results in expansion of the gas as it ﬂows through the fabric, as rendered in Figure Pr.2.19(a). Assume that the pressure gradient is approximated by po − pi ∂p , ∂x D where po and pi are the external and internal pressure and D is the wovenﬁber diameter. Consider the gas ﬂowing with an average gas velocity uA,x = 2 m/s, and pressures of pi = 1.5 × 105 Pa and po = 1.0 × 105 Pa. The fabric diameter is D = 0.5 mm. Use the expression for the volumetric energy conversion s˙ m,P for the onedimensional ﬂow given in Example 2.14. Use ρcp for air at 300 K (Table C.22). SKETCH: Figure 2.19(a) shows an idealization of airbag fabric and permeation of a gas stream through it.
Airbag
Pore
Side View of Filter Permeable Airbag Filter
pi (Internal Pressure) u
D
po (Ambient Pressure)
A,x
Gas (Products of Combustion or Compressed Air)
Woven Fiber
x
Figure Pr.2.19(a) An automobile airbag system.
OBJECTIVE: (a) Determine the volumetric expansion cooling rate s˙ m,p . (b) Write an integralvolume energy equation for the gas ﬂowing through a pore. Allow for expansion cooling and show the control volume, surface convection, and convection heat ﬂows. Designate the ﬂow area for the pore as Ax , and use D × Ax for the control volume. For the bounding surfaces, choose the two adjacent wovenfabric and imaginarypore walls. (c) For the case of no surface convection heat transfer, determine the drop in temperature Ti − To .
85
SOLUTION: (a) The expression for pressure cooling or heating for a onedimensional ﬂow is the particular form of (2.50) given in Example 2.14. For an ideal gas, this expression is ∂p ∂x po − pi . = uA,x D = uA,x
s˙ m,p
Using the numerical values, we have s˙ m,p = 2(m/s) ×
(1.0 − 1.5) × 105 (Pa) 3 = −2 × 108 W/m . 5 × 10−4 (m)
(b) The control volume and control surface for the gas are shown in Figure Pr.2.19(b). The integral energy equation (2.9) is dT ˙ + S. QA = −ρcp V dt Inlet Conditions: Ti , pi
Qku
Pore Space u
A,x
Outlet Conditions: To , po Control Surface A
 Qu x
Qu x+D
Sm,p
Ax x
Figure Pr.2.19(b) Energy equation for a unit cell containing two adjacent woven fabrics.
For steadystate conditions, dT /dt = 0. Neglecting surface radiation, QA = −Qu x + Qu x+D + Qku . The energy conversion is due to expansion cooling only. Then, the integral energy equation becomes −Qu x + Qu x+D + Qku = s˙ m,p V . We can write −Qu x + Qu x+D in terms of Ti and To as −Qu x + Qu x+D = ρcp uA,x (−Ti + To ). Then, we have ρcp uA,x (−Ti + To ) + Qku = s˙ m,p V . The equation above shows that when solidphase combustion is used in the inﬂater to generate the gas, the gas ﬂowing through the fabric cools down due to surface convection heattransfer (Qku > 0) and expansion cooling (s˙ m,p < 0). (c) For Qku = 0 and solving for (To − Ti ), we have To − Ti =
s˙ m,p Ax D pi − po = . ρcp uA,x ρcp
For air at T = 300 K, from Table C.22, we have ρ = 1.177 kg/m3 and cp = 1,005 J/kgK. Then, using these values we have (1.0 − 1.5) × 105 (Pa) To − Ti = = −42.27◦C. 3 1.177(kg/m ) × 1,005(J/kgK) COMMENT: The reduction in temperature of the gas as it ﬂows through the fabric is due to expansion cooling. The gas velocity is not high enough for viscous heating to become important. Also, during an air bag deployment, both the internal pressure and temperature vary. A typical deployment for a passenger air bag is 80 ms. During this time, the internal pressure and temperature vary from a peak pressure of p = 40 kPa and a temperature of T = 500 K to ambient conditions. 86
PROBLEM 2.20.FUN GIVEN: When high viscosity ﬂuids, such as oils, ﬂow very rapidly through a small tube, large strain rates, i.e., du/dr [where r is the radial location shown in Figure Pr.2.20(a)], are encountered. The high strain rate, combined with large ﬂuid viscosity µf , results in noticeable viscous heating. In tube ﬂows, when the Reynolds number ReD = ρf uA D/µf is larger than 2,300, transition from laminar to turbulent ﬂow occurs. In general, high crosssection averaged ﬂuid velocity uA results in a turbulent ﬂow. The ﬂuid velocity for a laminar ﬂow is shown in Figure Pr.2.20(a). For laminar ﬂow, the centerline velocity is twice the average velocity, while for turbulent ﬂow the coeﬃcient is less than two. Assume that the crosssection averaged viscous heating rate, (2.51), is approximated as s˙ m,µ A = µf
a21 u2A , D2
where a1 = 1 is a constant. SKETCH: Figure Pr.2.20(a) shows viscous heating in ﬂuid ﬂow through a small tube. Laminar Velocity Distribution
Tube
D/2 Liquid Flow uA
r r x
u
D
Au D/2
,x
x
u
A
2 u
A
Figure Pr.2.20(a) Viscous heating of ﬂuid ﬂow inside a small tube.
OBJECTIVE: (a) Determine the volumetric heating rate for engine oil at T = 310 K (Table C.23), noting that µf = νf ρf , uA =10 m/s, and D = 1 mm. (b) Apply (2.8) to a diﬀerential length along the tube. Allow only for surface convection, convection along x, and viscous heating, i.e., similar to (2.11), with added energy conversion. SOLUTION: (a) The crosssectional averaged viscous heating rate is s˙ m,µ A = µf
a1 uA D
2 . 3
The dynamic viscosity of engine oil is given in Table C.23 at T = 310 K as µf = ρf νf = 877.8(kg/m ) × 4.17 × 10−4 (m2 /s) = 0.3660 Pas. Then, using the values given, we have s˙ m,µ A = 0.3660(Pas)
1 × 10(m/s) 10−3 (m)
2
The Reynolds number is ReD =
uA D 10(m/s) × 10−3 (m) = 23.98. = νf 4.17 × 10−4 (m2 /s)
Since ReD < 2300, the ﬂow is in the laminar regime.
87
3
= 3.660 × 107 W/m .
(b) The integral energy equation (2.9) for a steadystate condition and heat conversion due to mechanical friction is (q · sn )dA = s˙ m,µ A . lim ∆A ∆V →0 ∆V Figure Pr.2.20(b) shows various terms in the energy equation applied to the control volume shown. Control Volume ,V = FD2,x/4 Liquid Flow uA
qku
sm,µ
A
r qu x
qu
x
x+,x
D
Flow Contol Surface Au = FD2/4
,x
SurfaceConvection Contol Surface Aku = F D ,x
Figure Pr.2.20(b) Energy equation for viscous heating in ﬂuid ﬂow through the inside of a small tube.
As shown in Figure Pr.2.20(b), the heat transfer occurs along the r and x directions. Then the net heat transfer is (q · sn )dA = qku ∆Aku + (qu x+∆x − qu x )Au ∆A
= qku πD∆x + (qu x+∆x − qu x )
πD2 . 4
Then lim
∆V →0
∆A
(q · sn )dA ∆V
=
lim
qku πD∆x + (qu x+∆x − qu x ) πD 4
∆V →0
πD 2 4 ∆x
4qku (qu x+∆x − qu x ) + = lim ∆V →0 D ∆x dqu 4qku + . = D dx
2
Therefore, the combined integraland diﬀerential length energy equation becomes 4qku dqu + = s˙ m,µ A . D dx COMMENT: Note that for very small D and very large ρf or uA , the viscous heating can be signiﬁcant.
88
PROBLEM 2.21.FUN GIVEN: During braking, nearly all of the kinetic energy of the automobile is converted to frictional heating at the brakes. A small fraction is converted in the tires. The braking time, i.e., the elapsed time for a complete stop, is τ . The automobile mass is M , the initial velocity is uo , and the stoppage is at a constant deceleration (du/dt)o . OBJECTIVE: (a) Determine the rate of friction energy conversion for each brake in terms of M, uo , and τ . The front brakes convert 65% of the energy and the rear brakes convert the remaining 35%. (b) Evaluate the peak energy conversion rate for the front brake using M = 1,500 kg (typical for a midsize car), uo = 80 km/hr, and τ = 4 s. SOLUTION: (a) The total instantaneous friction heating rate S˙ m,F is S˙ m,F = F u, where the force F is
F = −M
Now, using a constant deceleration, we have
du . dt
S˙ m,F = −M
du dt
u, o
du 0 − uo uo ∆u = =− . = dt o ∆t τ τ In order to ﬁnd an expression for u, we integrate the equation above obtaining uo u = − t + a1 . τ For u(t = 0) = uo we have t u = uo 1 − . τ Then, using this we have uo t S˙ m,F = M uo 1 − τ τ 2 M uo t = 1− . τ τ
where
Now, for 65% of the power being dissipated in the front breaks we have for each of the front brakes M u2o t ˙ Sm,F = 0.65 1− τ τ and for each of the rear brakes
2
M uo S˙ m,F = 0.35 τ
t 1− . τ
(b) Using the numerical values given, the peak heating rate (i.e., heating rate at t = 0) is S˙ m,F
M u2o τ 1,500(kg) × (22.22)2 (m/s)2 = 1.203 × 105 W = 120.3 kW. = 0.65 × 4(s)
= 0.65
COMMENT: This is a very large heating rate and its removal from the disc by the heat losses would require a large elapsed time. Therefore, if the brake is applied frequently such that this heat is never removed, overheating of the brake pads occurs. 89
PROBLEM 2.22.FUN GIVEN: In therapeutic heating, biological tissues are heated using electromagnetic (i.e., microwave, and in some cases, Joule heating) or mechanical (i.e., ultrasound heating) energy conversion. In the heated tissue, which may be a sore muscle (e.g., an athletic discomfort or injury), some of this heat is removed through the local blood ﬂow and this is called perfusion heating. Under steady state, the local tissue temperature reaches a temperature where the surface heat transfer from the tissue balances with the energy conversion rate. Consider the therapeutic ultrasound heating shown in Figure Pr.2.22(a). Iac = 5 × 104 W/m2 , σac (from Table C.11, for muscle tissue), V (sphere of R = 3 cm), D = 10−3 m, Aku = 0.02 m2 , NuD = 3.66, kf = 0.62 W/mK (same as water), Tf = 37◦C. SKETCH: Figure Pr.2.22(a) shows the ultrasonic therapeutic heating of a vascular tissue. Acoustic Intensity, Iac f = 106 Hz Acoustic Absorption Coefficient, σac
Blood Vessel with Temperature, Tf
Assume qk = 0 (Also qr = 0)
Tissue with Temperature, Ts
Ak Aku
V
Figure Pr.2.22(a) Therapeutic heating of biological tissue.
OBJECTIVE: (a) Using (2.9) write the integralvolume energy equation that applies to this steadystate heat transfer. Assume no conduction and radiation heat transfer and allow for surface convection through the blood vessels distributed through the tissue with a surfaceconvection area Aku . Draw a schematic showing the various terms in the energy equation. (b) In this energy equation, replace the surfaceconvection heat transfer with Aku qku = Aku NuD
kf (Ts − Tf ), D
where NuD is a dimensionless quantity called the dimensionless surfaceconvection conductance (or Nusselt number), kf is the blood thermal conductivity, D is the average blood vessel diameter, Ts is the tissue temperature, and Tf is the blood temperature. (c) Solve the energy equation for Ts . (d) Using the following numerical values, determine Ts . SOLUTION: (a) The various heat transfer mechanisms and the energy conversion by ultrasound heating are shown in Figure Pr.2.22(b). From (2.9), for steadystate conditions, we have QA = s˙ m,ac V , where we have assumed a uniform s˙ m,ac throughout the volume. From (2.54), we have s˙ m,ac = 2σac Iac . 90
The surface heat transfer is limited to surfaceconvection only, i.e., QA = Aku qku . Then the energy equation becomes Aku qku = 2σac Iac V . Figure Pr.2.22(b) shows the various terms in the energy equation applied to the control volume shown. Acoustic Intensity, Iac f = 106 Hz Acoustic Absorption Coefficient, σac
Blood Vessel with Temperature, Tf
Assume qk = 0 (Also qr = 0)
Tissue with Temperature, Ts
Ak D
Aku
qku
sm,ac
V
Figure Pr.2.22(b) Various terms in the energy equation for therapeutic ultrasound heating.
(b) The surfaceconvection heat transfer is given by Aku qku = Aku
NuD kf (Ts − Tf ). D
Then, the energy equation becomes Aku
NuD kf (Ts − Tf ) = 2σac Iac V . D
The group NuD kf /D is called the (dimensional) surfaceconvection conductance or the heat transfer coeﬃcient. (c) Solving the equation above for Ts , we have Ts = Tf +
2σac Iac V D . NuD kf Aku
(d) From Table C.11, σac = 14 m−1 . Using the numerical values, we have Ts
=
37(◦C) +
=
40.49◦C.
2 × 14(m−1 ) × 5 × 104 (W/m ) × (4/3)π(3 × 10−2 )3 (m3 ) × 10−3 (m) 3.66 × 0.62(W/mK) × 0.02(m2 ) 2
COMMENT The conduction heat losses can be signiﬁcant and should be included. During heating, the blood vessels dilate causing D to increase. This results in a decrease in Ts . The Nusselt number, NuD , will be discussed in Chapter 7.
91
PROBLEM 2.23.FAM GIVEN: Among the normal paraﬃns (nparaﬃns) are the hydrocarbon fuels, e.g., methane CH4 , propane C2 H6 , and butane C4 H10 . Table Pr.2.23 gives two sets of constants for the chemical kinetic model given by n˙ r,F = −ar ρaFF ρaOO e−∆Ea /Rg T , for the CH4 oxidation represented by a singlestep, stoichiometric reaction CH4 + 2O2 → CO2 + 2H2 O. These two sets of parameters are found to give a good agreement between predicted and measured ﬂame speeds as a function of methane/oxygen ratio. OBJECTIVE: Determine the reaction rates n˙ r,F at T = 1,000◦C using the above model (with the constants from Table Pr.2.23). (a) Use a reactantrich condition of ρO2 = 0.9307 kg/m3 , ρCH4 = 0.2333 kg/m3 , and (b) a productrich condition of ρO2 = 0.1320 kg/m3 , ρCH4 = 0.0328 kg/m3 , to represent two locations within the ﬂame. These are characteristics of CH4 reaction with oxygen (called oxyfuel reactions as compared to airfuel reactions) at one atm pressure. (c) Compare the results with the prediction of the zerothorder model given in Example 2.6 by (2.21). Table Pr.2.23 Constants in chemical kinetic model for methane oxidation. ar , s−1 ∆Ea , J/kmole aF aO 1.3 × 108 8.3 × 105 SOLUTION: The two chemical kinetic models are n˙ r,CH4 n˙ r,CH4
2.026 × 108 1.256 × 108
= −ar exp
−∆Ea Rg T
–0.3 –0.3
1.3 1.3
zerothorderkinetics −∆Ea aO F = −ar ρaCH ρ exp ﬁrstorderkinetics. 4 O Rg T
(a) Using the ﬁrst set of constants, the ﬁrstorder model and reactantrich conditions, we have n˙ r,CH4
= −1.3 × 108 (1/s)[0.2333(kg/m )]−0.3 [0.9307( kg/m )]1.3 × −2.026 × 108 (J/kmole) exp 8,314(J/kmoleK)(1,000 + 273.15)(K) 3
3
3
= −0.9004 kg/m s. (b) For the productrich conditions, we have n˙ r,CH4
= −1.3 × 108 (1/s)[0.0328(kg/m )]−0.3 [0.1320( kg/m )]1.3 × −2.026 × 108 (J/kmole) 3 exp = −0.1280 kg/m s 8,314(J/kmoleK)(1,000 + 273.15)(K) 3
3
(c) Using the numerical values from Example 2.6 for the zerothorder model, we have −2.10 × 108 (J/kmole) 3 8 n˙ r,CH4 = −1.3 × 10 (kg/m s) exp 8,314(J/kmoleK)(1,000 + 273.15)(K) 3
= −0.3150 kg/m s. 92
COMMENT: The zerothorder chemical kinetic model is a concentrationindependent averaged model. Its predictions are comparable with the ﬁrstorder chemical kinetic model when the predictions for the reactantrich and the productrich regions of the ﬂame are averaged. The advantage of using the zerothorder chemical kinetic model is its relative mathematical simplicity. This will be further explored in Chapter 5. For more accurate predictions, better models are needed. Some of the more complete models account for hundreds of reactions and tenths of species taking part in the reaction.
93
PROBLEM 2.24.FAM GIVEN: In addition to being abundant and readily available, air is a ﬂuid whose temperature can be raised well above and below room temperature, for usage as a hot or cold stream, without undergoing any phase change. In the high temperature limit, the main constituents of air, nitrogen (N2 ) and oxygen (O2 ), dissociate and ionize at temperatures above T = 2,000 K. In the low temperature limit, oxygen condenses at T = 90.0 K, while nitrogen condenses at T = 77.3 K. Consider creating (a) a cold air stream with T2 = 250 K, (b) a hot air stream with T2 = 1,500 K, and (c) a hot air stream with T2 = 15,000 K. The air stream is at atmospheric pressure, has a crosssectional area Au = 0.01 m2 , an inlet temperature T1 = 290 K, and a velocity u1 = 1 m/s, as shown in Figure Pr.2.24(a). SKETCH: Figure Pr.2.24(a) gives a general control volume through which an air stream ﬂows, while undergoing energy conversion. Assume No Heat Loss Qloss = 0 Inlet Condition: L
Air T1 = 290 K u1 = 1 m/s p1 = 1 atm
Outlet Condition: T2 u2 p2
Au = 0.01 m2
Energy Conversion, S V
Figure Pr.2.24(a) Heating or cooling of an air stream using various energy conversion mechanisms.
OBJECTIVE: For each of the cases above, (i) choose an energy conversion mechanism from Table 2.1 that would provide the required energy conversion mechanisms for heating or cooling, (ii) write the integralvolume energy equations (2.9) for a steadystate ﬂow and heat transfer. Give the amount of ﬂuid, electromagnetic energy, etc., that is needed. SOLUTION: (a) From Table 2.1, we choose phasechange (evaporation) cooling. This is shown in Figure Pr.2.24(b). The integralvolume energy equation for a steadystate condition is QA = S˙ lg .
Evaporation Cooling of Air Q=0 Air T1 = 290 K
T2 = 250 K (Cold, Moist Air)
L
 Qu,1
Qu,2
Au
Slg =  Mlg ,hlg
Liquid
Figure Pr.2.24(b) Evaporation cooling of a gas stream.
The net heat transfer at the control surface is QA = Qu,2 − Qu,1 . For the energy conversion due to phase change, we have S˙ lg = −M˙ lg ∆hlg . Then, the energy equation becomes Qu,2 − Qu,1 = −M˙ lg ∆hlg . 94
The convection heat transfer is given by Qu = Au ρcp T . From the conservation of mass equation (1.25) we have (Au ρu)1 = (Au ρu)2 . Then, we can write the energy equation as Au ρ1 cp,1 u1 (T2 − T1 ) = −M˙ lg ∆hlg . Solving for M˙ lg we have
Au ρ1 cp,1 u1 (T2 − T1 ) M˙ lg = − . ∆hlg
We examine Table C.6 and choose carbon dioxide as the ﬂuid to be evaporated. For this ﬂuid, Tlg = 216.6 K and ∆hlg = 573.2 × 103 J/kg. From Table C.22, we have at T = 290 K, ρ1 = 1.224 kg/m3 . We also assume a constant speciﬁc heat of cp = 1,006 J/kgK. Using the values given, we have 3
M˙ lg
= − =
0.01(m2 ) × 1.224(kg/m ) × 1,006(J/kgK) × 1(m/s) × (250 − 290)(K) 573.2 × 103 (J/kg)
8.593 × 10−4 kg/s = 0.8593 g/s.
(b) For T2 = 1,500 K, we choose combustion and the integralvolume energy equation (2.9) becomes Au ρ1 cp,1 u1 (T2 − T1 ) = −M˙ r,CH4 ∆hr,CH4 . This is shown in Figure Pr.2.24(c). Solving for M˙ r,CH4 we have Au ρ1 cp,1 u1 (T2 − T1 ) M˙ r,CH4 = − . ∆hr,CH4 Combustion Heating of Air Q=0 Air T1 = 290 K
T2 = 1500 K (Hot Flue Gas)
 Qu,1
Qu,2
Au
Sr,c =  Mr,CH4 Dhr,CH4
Gaseous Methane CH4
Figure Pr.2.24(c) Combustion heating of a gas stream.
From Table C.21(a), we have ∆hr,CH4 = −5.553 × 107 J/kg. Then, we have 3
M˙ r,CH4
= − =
0.01(m2 ) × 1.224(kg/m ) × 1,006(J/kgK) × 1(m/s) × (1,500 − 290)(K) −5.553 × 107 (J/kg)
2.683 × 10−4 kg/s = 0.2683 g/s.
(c) From Table 2.1, for T2 = 15,000 K, we choose the Joule heating. After an initial formation of dissociatedionized air by a combustion torch, induction coils are used to heat the charged gas (i.e., the plasma) stream. This is shown in Figure Pr.2.24(d). The integralvolume energy equation becomes Au ρ1 cp,1 u1 (T2 − T1 ) = S˙ e,J . Solving for S˙ e,J we have S˙ e,J
3
= 0.01(m2 ) × 1.224(kg/m ) × 1,006(J/kgK) × 1(m/s) × (15,000 − 290)(K) = 1.811 × 105 W. 95
COMMENT: In all of these examples, we have neglected any heat losses and assumed complete evaporation and complete reaction. These idealizations can be removed by their proper inclusion in the energy equation. The use of a variable cp would require a diﬀerent form of the energy equation, as shown in Appendix B. For most heat transfer analysis, a constant, temperatureaveraged cp is an acceptable approximation. Also, note that other mechanisms of heating/cooling could be used, such as thermoelectric and expansion cooling. Joule Heating of Dissociated, Ionized Air Induction Coil
Q=0 T2 = 15,000 K (Thermal Plasma)
Air T1 = 290 K
Electron e
je
 Qu,1 Au
Qu,2
Se,J = ρe je2 V
Figure Pr.2.24(d) A charged, gas stream heated by induction coils.
96
PROBLEM 2.25.FUN GIVEN: A transparent thinfoil heater is used to keep a liquid crystal display (LCD) warm under cold weather conditions. The thin foil is sandwiched between the liquid crystal and the backlight. A very thin copper foil, with cross section w = 0.4 mm and l = 0.0254 mm and a total length L = 70 cm, is used as the heating element. The foil is embedded in a thin polyester membrane with dimensions W = 10 cm and H = 2 cm, which also acts as an electrical insulator [Figure Pr.2.25(a)]. The thin foil heats the liquid crystal by Joule heating. Assume that the amount of heat ﬂowing to the backlight panel is the same as the amount ﬂowing to the liquid crystal and that the system is operating under a steadystate condition. For the electrical resistivity of copper use ρe = 1.725 × 10−8 ohmm. SKETCH: Figure Pr.2.25(a) shows the thin foil heater and its dimensions. q (W/m2)
()
,ϕ (V) (+)
. Se,J l = 0.0254 mm W = 10 cm w = 0.4 mm H = 2 cm q (W/m2)
Figure Pr.2.25(a) A transparent thinfoil heater.
OBJECTIVE: If the thin foil heater is to provide q = 1,000 W/m2 to the liquid crystal, calculate: (a) The volumetric rate of heat generation in the wire s˙ e,J (W/m3 ), (b) The electrical potential ∆ϕ(V) needed, (c) The current ﬂowing in the wire Je (A), and (d) Recalculate items (a) to (c) for twice the length L. SOLUTION: (a) The volumetric rate of heat generation in the wire s˙ e,J (W/m3 ) is obtained from the integralvolume energy equation. The steps for the solution are (i) Draw the heat ﬂux vector. This is shown in Figure Pr.2.25(b).
q(W/m2)
ThinFoil Heater Af = WH
.
Se,J (W/m3) q(W/m2) Figure Pr.2.25(b) Energy equation for the heater.
97
(ii) Apply the conservation of energy equation. The integralvolume energy equation is d q · sn dA = − (ρcp T ) dV + s˙ i dV. dt V A V i For this steadystate problem the storage term is zero. Solving for the area integral of the normal component of the heat ﬂux vector over the control surfaces gives [see Figure Pr.2.25(b)] q · sn dA = q Ak + q Ak = 2qAk , A
where Ak = W H is the surface area of the thinfoil heater. The only energy conversion taking place inside the control volume is the conversion from electromagnetic to thermal energy by Joule heating. Furthermore, this energy conversion is constant everywhere inside the control volume. Thus, the righthand side of the energy equation becomes s˙ i dV = s˙ e,J dV = s˙ e,J Vl , V
V
i
where Vl = wlL is the volume of the copper foil (heating element). Finally, the energy equation becomes 2qAk = s˙ e,J Vl or 2qW H = s˙ e,J wlL. (iii) Solving the energy equation for s˙ e,J , we have s˙ e,J =
2qW H . wlL
From the numerical values given, we have s˙ e,J =
2 × 1,000(W/m2 ) × 0.02(m) × 0.1(m) = 5.62 × 108 W/m3 . 4 × 10−4 (m) × 2.54 × 10−5 (m) × 0.7(m)
(b) The electrical potential ∆ϕ(V) can be determined from the volumetric Joule heating using (2.32), s˙ e,J =
∆ϕ2 . ρe L2
Solving for ∆ϕ(V) and using the data available, we have ∆ϕ = L(s˙ e,J ρe )1/2 = 0.7(m) × [5.62 × 108 (W/m3 ) × 1.725 × 10−8 (ohmm)]1/2 = 2.18 V. (c) The current Je (A) can be calculated from Ohm’s law, ∆ϕ = Re Je , where the electrical resistance Re is given in (2.32) as Re =
ρe L Aw
and Aw = wl is the wire crosssectional area. Solving for Je and using the data available, we have Je =
2.18(V) × 4 × 10−4 (m) × 2.54 × 10−5 (m) ∆ϕwl = = 1.8 A. ρe L 1.725 × 10−8 (ohmm) × 0.7(m) 98
(d) For twice the wire length 2L, the heat generation, voltage, and current are s˙ e,J (2L)
=
∆ϕ(2L)
=
Je (2L)
s˙ e,J (L) 5.62 × 108 (W/m3 ) 2qW H = = = 2.81 × 108 W/m3 wl2L 2 2 1/2 1/2 s˙ e,J (2L) s˙ e,J (L)ρe 1/2 2L = 2L = (2)1/2 L [s˙ e,J (L)ρe ] ρe 2
=
(2)1/2 ∆ϕ(L) = (2)1/2 2.18(V) = 3.08 V
=
(2)1/2 l∆ϕ(L)wl l∆ϕ(L)wδ l∆ϕ(2L)wl 1.8(A) Je (L) = = = = 1.3 A. = 1/2 1/2 ρe 2L ρe 2L (2) ρe L (2) (2)1/2
COMMENT: Notice the high volumetric energy conversion rate which can be achieved by Joule heating. Doubling the length caused a reduction in power, an increase in voltage, and a reduction in current. A reduction in power leads to a smaller temperature in the heating element. The drawback is the need for a larger voltage.
99
PROBLEM 2.26.FUN GIVEN: A singlestage Peltier cooler/heater is made of Peltier cells electrically connected in series. Each cell is made of p and ntype bismuth telluride (Bi2 Te3 ) alloy with Seebeck coeﬃcients αS,p = 230 × 10−6 V/K and αS,n = −210 × 10−6 V/K. The cells are arranged in an array of 8 by 15 (pairs) cells and they are sandwiched between two square ceramic plates with dimensions w = L = 3 cm [see Figure Pr.2.26(a)]. The current ﬂowing through the elements is Je = 3 A. SKETCH: Figure Pr.2.26(a) shows a thermoelectric module and its various components.  Qc Ac
Ceramic Plate
Tc
Se,P
Ceramic Plate ()
Ah
n
p
n
p
Semiconductor Electrical Conductor
(+) Semiconductor
Th
Se,P Qh
Je
Dϕ , Applied Voltage
Figure Pr.2.26(a) A singlestage Peltier cooler/heater.
OBJECTIVE: (a) If the temperature at the cold junction is Tc = 10◦C, calculate the Peltier heat absorbed at the cold junctions qc (W/m2 ) (per unit area of the ceramic plate). (b) If the temperature of the hot junctions reach Th = 50◦C, calculate the Peltier heat released at the hot junctions qh (W/m2 ) (per unit area of the ceramic plate). SOLUTION: (a) To calculate the heat absorbed at the cold junction we again follow the three steps. (i) Draw the heat ﬂux vector. This is shown in Figure Pr.2.26(b). (ii) Apply the conservation of energy equation. The integralsurface energy equation (2.9) is
S˙ i . q · sn dA = A
i
For this steadystate problem, the storage term is zero. The energy conversion term is due to Peltier cooling only. Then the energy equation becomes −qc As = (S˙ e,P )c , where As = wL is the surface area of the ceramic plate and −Qc is the rate of heat absorbed at the Peltier junctions. (iii) Obtain an expression for the heat absorbed due to Peltier cooling. For 8 × 15 Peltier junctions we have −qc As = 8 × 15 × (S˙ e,P )c where (S˙ e,P )c is the heat absorbed at the Peltier cold junction which is given by (2.44) (S˙ e,P )c = −(αS,p − αS,n )Tc Je . Then, the energy equation becomes −qc As = 15 × 8 × [−(αS,p − αS,n )Tc Je ]. 100
 qc(W/m2) Ceramic Plate
Tc
Metallic Contact
.
Se,P n
Th
p
n
Qh
p
Semiconductor n
p
.
Se,P qh(W/m2)
Figure Pr.2.26(b) The cold and hot surfaces of the thermoelectric module.
(iv) Solve for qc . From the numerical values given, we have (S˙ e,P )c (S˙ e,P )c (15 × 8) qc
= −[230 × 10−6 (V/K) + 210 × 10−6 (V/K)] × 283.15(K) × 3(A) = −0.374 W = −0.374(W) × 120 = −44.8 W −44.8(W) Qc 2 = = 49,808 W/m . = wL [−0.03(m) × 0.03(m)]
(b) For the hot junction, a similar approach is used. The heat released is given by (2.41) as (S˙ e,P )c = (αS,p − αS,n )Th Je = [230 × 10−6 (V/K) + 210 × 10−6 (V/K)] × 323.15(K) × 3(A) = 0.426 W. The total heat generated at the hot junction is then Qh = (S˙ e,P )h (15 × 8) = 0.426(W) × 120 = 51.2 W. The heat ﬂux at the ceramic plate is qh =
51.2(W) Qh 2 = = 56,848 W/m . wL 0.03(m) 0.03(m)
COMMENT: The values calculated above are ideal values for the Peltier heater/cooler. It will be seen in Chapter 3 that both the Joule heating and the heat conduction through the semiconductor legs of the Peltier cell, reduce the amount of heat that can be absorbed by a Peltier cooler. The analysis will lead to the deﬁnition of the ﬁgure of merit which express the eﬃciency of the Peltier cooler.
101
PROBLEM 2.27.FAM.S GIVEN: A pocket combustion heater uses heat released (chemicalbond energy conversion) from the reaction of air with a powder. The powder is a mixture of iron, water, cellulose (a carbohydrate), vermiculite (a clay mineral), activated carbon (made capable of absorbing gases), and salt. Air is introduced by breaking the plastic sealant and exposing the permeable membrane containing the powder to ambient air. Since the air has to diﬀuse through the powder, and also since the powder is not mixed, the heat release rate is time dependent, decreasing with time. We express this as S˙ r,c = S˙ r,o exp(−t/τ ), where τ (t) is called the time constant. The pocket heater has a mass of M = 20 g and a heat capacity of cp = 900 J/kgK. During the usage, heat leaves the pocket heater surface. This heat is expressed as a resistivetype heat transfer and is given by Q = (T − T∞ )/Rt , where T∞ is the ambient temperature and Rt (◦C/W) is the surface heat transfer resistance. Initially the heater is at the ambient temperature, i.e., T (t = 0) = T∞ . This is shown in Figure Pr.2.27(a). SKETCH: Figure Pr.2.27(a) shows the heat transfer model of a combustion pocket heater.
Combustion Handwarmer Qt =
M, cp
T  T Rt
Sr,c
T(t = 0) = T
Figure Pr.2.27(a) A pocket combustion heater and its heat transfer model.
OBJECTIVE: (a) Write the energy equation for the pocket heater. (b) Using a software, plot the temperature of the pocket heater T = T (t) versus time, up to t = τ . (c) What is the maximum heater temperature? SOLUTION: (a) Since we use a uniform temperature for the heater, the energy equation is the integralvolume energy equation (2.9), i.e., Q A
dT + S˙ r,c dt dT + S˙ r,c . = −M cp dt = −ρcp V
Here, the energy conversion term is time dependent, i.e., S˙ r,c = S˙ r,o e−t/τ , where S˙ r,o is a constant and τ is called the time constant. The surface heat transfer rate is given by a surface thermal resistance, i.e., Q A = Qt =
T − T∞ , Rt
where Rt is the heat transfer resistance and T∞ is the ambient temperature. Combining the above equations, we have dT T − T∞ + S˙ r,o e−t/τ . = −M cp Rt dt 102
The initial temperature is T (t = 0) = T∞ . (b) The above energy equation can not be readily integrated to give T = T (t). Here, we use software and provide the constants M, cp , S˙ r,o , τ, and T∞ . The results are plotted in Figure Pr.2.27(b). We note that initially T increases with time. Then it reaches a maximum. Finally, it begins to decrease. During the increase, the energy conversion rate is larger than the surface heat loss term Q A = Qt . At the time of maximum temperature, when dT /dt = 0, the energy conversion and surface heat loss exactly balance. Due to the time dependence of S˙ r,c , the temperature begins to decrease after reaching the maximum and, during the decrease, the energy conversion is less than the surface heat loss. These are also shown in Figure Pr.2.27(b).
(b) Evolution of Handwarmer Temperatures 50 Tmax = 45.03 C 40
Sr,c < Q A Sr,c > Q A
T, C
30
20 T(t = 0) = T
10 t = 594 s 0 0
2,400
4,800
7,200
9,600
τ 12,000
t, s Figure Pr.2.27(b) Variation of the temperature of the pocket heater with respect to time.
(c) The maximum temperature is found to be Tmax = 45.03◦C and occurs at t = 594 s. Note that direct contact of the heater with skin will cause damage. COMMENT: The model for heat release rate is an approximation. By proper design of the powder and its packaging, a uniform heat release rate may be achieved.
103
PROBLEM 2.28.FUN GIVEN: In electrical power generation using thermoelectric energy conversion, the electrical power can be optimized with respect to the external electrical resistance. OBJECTIVE: Starting from (2.40), show that the maximum power generation occurs for Re,o = Re , i.e., when the external electrical resistance is equal to the thermoelectric electrical resistance. SOLUTION: The electrical power generation given by (2.40) is maximized with respect to Re,o by taking the derivative of (2.40) and setting the result equal to zero. This gives 2 αS (Th − Tc )2 ∂ ∂ 2 (J Re,o ) = Re,o = 0 ∂Re,o e ∂Re,o (Re,o + Re )2 which results in 1−2
Re,o =0 Re,o + Re
or Re,o = Re . COMMENT: To prove that this is minimum, we take the second derivative of (2.40) and evaluate it for Re,o = Re . d2 αS2 (Th − Tc )2 1 2 (J R ) = −2 − 2 + 6 × < 0, e e,o 2 dRe,o (Re,o + Re )3 2 and therefore, Re,o = Re results in a minimum in Je2 Re,o .
104
PROBLEM 2.29.FUN GIVEN: The volumetric pressurecompressibility heating/cooling energy conversion s˙ p can be represented in an alternative form using cv instead of cp in the energy equation. OBJECTIVE: Starting from (B.44) in Appendix B, show that for an ideal gas, the volumetric pressurecompressibility energy conversion s˙ p becomes cp − 1 cv ρT ∇ · u. s˙ p = −p∇ · u = cv Use the following relation, derived from combining (1.4), (1.5), and (1.6), ∂p ∂v cp ≡ cv + T . ∂T v ∂T p SOLUTION: Starting from (B.44), we deﬁne s˙ p as s˙ p ≡ −T
∂p ∇ · u. ∂T v
From (1.19), for an ideal gas, we have Rg Rg ρT = T M Mv Rg ρ = p ideal gas. = T M
p ∂p T ∂T v
=
Then s˙ p = −p∇ · u. Also, for ideal gas we have
cp
∂p ∂v ≡ cv + T ∂T v ∂T p Rg Rg Rg . = cv + T = cv + Mv Mp M
Then p
Rg ρT = (cp − cv )ρT M cv − 1 cv ρT = cp
=
or
s˙ p = −p∇ · u =
cv − 1 cv ρT ∇ · u. cp
COMMENT: Note that for an incompressible ﬂuid ﬂow, from (B.40) we have ∇·u=0 Also for an incompressible ﬂuid, we have ∂v = 0, ∂T
incompressible ﬂuid ﬂow.
cp = cv
incompressible ﬂuid.
p
In pressurecompressibility cooling/heating, a large cp /cv (can be optimized by mixing species), a large ∇ · u (would require a large pressure gradient), along with a large cv ρT (high pressure and temperature) would be needed. 105
PROBLEM 2.30.FAM GIVEN: A microwave heater is used to dry a batch of wet alumina powder. The microwave source is regulated to operate at f = 109 Hz and to provide an electrical ﬁeld with a rootmeansquare intensity of (e2 )1/2 = 103 V/m. The eﬀective dielectric loss factor of the alumina powder ec depends on the ﬂuid ﬁlling the pores. For a porosity of 0.4, the eﬀective dielectric loss factor of the completely dry alumina powder is ec = 0.0003 and the eﬀective dielectric loss factor of the completely wet alumina powder is ec = 6.0. Note that although both ec and ec are listed, no distinction is made in Table C.10. OBJECTIVE: (a) Determine the microwave heating s˙ e,m (W/m3 ) for these two cases. (b) Discuss the eﬃciency of the use of microwave heating in drying the alumina powder when the moisture content (i.e., amount of water in the pores) is small. (c) From Table C.10, would a sandy soil dry faster or slower than the alumina powder? SOLUTION: The volumetric energy conversion by microwave heating is given by s˙ e,m = 2πf ec o e2 . (a) For the wet alumina powder, we have s˙ e,m = 2π × 109 (1/s) × 6.0 × 8.8542 × 10−12 (A2 s2 /Nm2 ) × 106 (V/m)2 = 3.338 × 105 W/m . 3
For the dry alumina powder we have s˙ e,m = 2π × 109 (1/s) × 0.0003 × 8.8542 × 10−12 (A2 s2 /Nm2 ) × 106 (V/m)2 = 16.69 W/m . 3
(b) For the same amount of available microwave energy, the wet alumina powder is able to convert 333,795 W/m3 of that energy into volumetric heating. The dry alumina powder only converts 16.7 W/m3 of the available energy into thermal energy. Therefore, the wet alumina powder utilizes microwave heating more eﬃciently in the drying of the powder. (c) For dry sandy soil, we have s˙ e,m = 2π × 109 (1/s) × 0.026 × 8.8542 × 10−12 (A2 s2 /Nm2 ) × 106 (V/m)2 = 1,446 W/m . 3
Assuming that the particle size and porosity of the sandy soil is similar to that of the alumina powder, and since dry sandy soil makes more eﬃcient use of microwave heating than dry alumina powder, we can conclude that wet sandy soil would dry faster. COMMENT: Note the ten thousand fold diﬀerence in the magnitude of the eﬀective dielectric loss of the dry and the wet alumina powder. The dielectric loss factor for water at 25◦C is e,c = 1.2 and the dielectric loss for air is e,c = 0. The dielectric loss for most dry ceramics is small. This explains the small volumetric heating rates in ceramics under low intensity microwave ﬁelds.
106
PROBLEM 2.31.FUN GIVEN: The rangetop electrical heater is shown in Figure Pr.2.31(a). It has electrical elements made of a central electrical conductor (electric current carrying) surrounded by an electrical insulator. The electrical insulator should have a large thermal conductivity to carry the heat generated by Joule heating in the electrical conductor to the surface for surface convectionradiation heat transfer. This is shown in Figure Pr.2.31(b). During the startup and turnoﬀ, the transient heat transfer in the heater becomes signiﬁcant. In order to analyze this transient heating, the temperature distribution in the heater is examined. Since the electric conductor also has a high thermal conductivity, it is treated as having a uniform temperature. However, the electrical insulator (generally an oxide ceramic) has a relatively lower thermal conductivity, and this results in a temperature nonuniformity within it. SKETCH: Figures Pr.2.31(a) and (b) show a rangetop electrical heater and the layers within the heating element.
RangeTop Electrical Heater (a) Physical Model
qku qr
Se,J (W/m) L
Electrical Heater with Coiled Length L
(b) CrossSection of Electrical Heater qr Surface Heat Transfer qku
R3 R2 R1
qk
Sensible Heat Storage Electrical Insulator (but Thermal Conductor)
Se,J (W/m) L
Electrical Conductor (also Thermal Conductor)
Figures Pr.2.31(a) A rangetop electrical heater.(b) The various layers within the heating element.
OBJECTIVE: (a) Divide the volume of the electrical insulator into three regions, as shown in Figure Pr.2.31(b). (b) Select a control volume in the region between r = R1 and r = R2 and render the heat transfer through this control volume. (c) Show that the energy equation for this control volume allowing for conduction and sensible heat storage in the electrical insulator is given by ∂T ∂T2 ∂T . 2R1 − k 2R2 = −(ρcp )2 (R22 − R12 ) k ∂x R1 ∂x R2 ∂t SOLUTION: (a) The control volume and control surface for the volume contained in R1 ≤ r ≤ R2 in the electrical insulator is shown in Figure Pr.2.31(c).
107
Energy Equation for R1 < r < R2 Finite, Small Control Volume ,V2 at Uniform Temperature T2(t) The Length Perpendicular to Page is L
Control Surfaces: Ak = 2pR1L
R2 r
 qk
Ak = 2pR2L R2
R1
To
R1
qk
R1
sr
T1 sn,1
R2
T2 T3
(ρcp V )2 sn,2 Sensible Heat Storage (ρcpV )2 dT2 dt
Figures Pr.2.31(c) A ﬁnitesmall control volume in the heater.
(b) The heat transfer is by conduction only. We begin with (2.13) and write QA = −
d [(ρcp T )∆V2 ∆V2 ] dt
and QA = QA1 + QA2 = [(qk ·sn )Ak ]R1 + [(qk ·sn )Ak ]R2 . From (1.11), the conduction heat transfer qk is related to the temperature gradient. Here we have a onedimensional conduction heat ﬂow in the r direction, and qk = −k
∂T sr . ∂x
Also, the geometric parameters are Ak R1 = 2πR1 L ,
Ak R2 = 2πR2 L ,
∆V2 = π(R22 − R12 )L .
(c) Using these and assuming constant ρcp , similar to (2.15), we then have ∂T ∂T dT2 k (2πR L) − k (2πR2 L) = −(ρcp )2 π(R22 − R12 )L 1 ∂x R1 ∂x R2 dt ∂T dT2 ∂T . 2R1 − k 2R2 = −(ρcp )2 (R22 − R12 ) k ∂x ∂x dt R1
R2
COMMENT: To accurately predict the transient temperature distribution in the electrical insulator, its division (i.e., discretization) into more than three regions is required (as many as twenty regions may be used). This will be discussed in Section 3.7.
108
PROBLEM 2.32.FUN GIVEN: Consider air (ﬂuid) ﬂow parallel to a semiinﬁnite plate (solid, 0 ≤ x ≤ ∞), as shown in Figure Pr.2.32. The plate surface is at a uniform temperature Tsf . The ﬂow is along the x axis. The velocity of air uf at the solid surface is zero. Starting from (2.61) show that at a location L along the plate, the surface energy equation becomes ∂Ts ∂Tf ks − kf = 0 on Asf . ∂y y=0− ∂y y=0+ Neglect surface radiation heat transfer, and use us = 0, and uf = 0 on Asf . There is no surface energy conversion. SKETCH: Figure Pr.2.32 shows the parallel air ﬂow over a semiinﬁnite plate.
Parallel Flow of Air: Far Field Conditions Tf, , uf, = uf, sx Control Surface ,Asf , ,x 0
Fluid Flow
qu,f
Tf = Tf (x,y) Tsf Uniform
uf =us = 0 on Surface w
y
qku sn = sy
,x
x, u
sn = sy y = 0
z L us = 0
y = 0+
Solid Ts = Ts(x,y)
x
qk,s
Figure Pr.2.32 A parallel air ﬂow over a semiinﬁnite plate (0 ≤ x ≤ ∞), with a uniform surface temperature Ts .
OBJECTIVE: Use the deﬁnition of surfaceconvection heat ﬂux qku (positive when leaving the solid toward the gas) given as ∂Ts ∂Tf −ks = −kf ≡ qku on Asf . ∂y y=0− ∂y y=0+
SOLUTION: Starting from (2.62), we have [−k(∇T · sn ) + ρcp T (u · sn ) + qr · sn ]f dAsf + ∆Asf →0
∆Asf →0
˙ [−k(∇T · sn ) + ρcp T (u · sn ) + qr · sn ]s dAsf = S.
Here, qr and u and S˙ are all set to zero (no surface radiation heat transfer, no ﬂuid or solid motion on the surface, and no surface energy conversion). Then −kf (∇Tf · sn )dAsf + −ks (∇Ts · sn )dAf s = 0. ∆Asf →0
∆Asg →0
Now using the surface unit normal vectors shown in Figure Pr.2.32, and noting that ∇Tf · sn =
∂Tf , ∂y
∇Ts · sn = − 109
∂Ts , ∂y
and evaluating these derivatives in their perspective surface (noting that the control surface wraps around the surface), i.e., gas at y ≤ 0+ and solid at y ≤ 0− , we have ∂Tf ∂Ts + ks = 0. −kf ∂y y=0+ ∂y y=0− Since the surfaceconvection heat ﬂux is deﬁned as qku we have
+ ∂Tf ≡ −kf , ∂y y=0
∂Ts ∂Tf −ks = −kf ≡ qku . ∂y y=0− ∂y y=0+
COMMENT: Note that, for example for Tsf > Tf,∞ , heat ﬂows, from the surface to the gas stream. Then ∂Ts /∂y and ∂Tf /∂y both will be negative. Also note that the relationship between the two derivative is given by this energy equation, i.e., ∂Tf /∂yy=0+ ks = . ∂Ts ∂yy=0− kf
110
PROBLEM 2.33.FUN GIVEN: The divergence of the heat ﬂux vector ∇ · q is indicative of the presence or lack of local heat sources (energy storage/release or conversion). This is stated by (2.2). Consider a gaseous, onedimensional steadystate ﬂuid ﬂow and heat transfer with a premixed combustion (exothermic chemical reaction) as shown in Figure Pr.2.33(a). For this, (2.2) becomes ∇·q=
d qx = s˙ r,c (x). dx
Here qx = qk,x + qu,x (assuming no radiation) is idealized with a distribution and the source terms. s˙ r,c (x) = −ρF,1 uf,1 ∆hr,F
1 1/2
σ(2π)
e−(x−xo )
2
/2σ 2
,
where ρF,1 is the ﬂuid density far upstream of the reaction (or ﬂame) region, uf,1 is the ﬂuid velocity there, and ∆hr,F is the heat of combustion (per kg of fuel). The exponential expression indicates that the reaction begins to the left of the ﬂame location xo and ends to its right, with the ﬂame thickness given approximately by 6σ. This is the normal distribution function and represents a chemical reaction that initially increases (as temperature increases) and then decays and vanishes (as products are formed and fuel depletes). SKETCH: Figure Pr.2.33(a) shows the variable source term s˙ r,c (x). The ﬂame thickness δ is approximated as 6σ. sr,c
F,1 uf,1 hr,F Gaseous Fuel u and Oxygen f,1
6 =
xo
Upstream
x Downstream
Flame qx(x1 = )
sr,c
qx(x2 = )
Figure Pr.2.33(a) Variable energy conversion (source) term for combustion in a premixed gaseous ﬂow. The source has a normal distribution around a location xo . The ﬂame thickness is approximated as δ = 6σ.
OBJECTIVE: (a) For σ = δ/6 = 0.1 mm, plot qx /(−ρF,1 uf,1 ∆hr,F ) and s˙ r,c /(−ρF,1 uf,1 ∆hr,F ), with respect to x (use xo = 0 and x1 = −δ < x < xL = δ). Assume qx (x = −δ) = 0. (b) Noting that no temperature gradient is expected at x = x1 = −δ and at x = x2 = δ, i.e., qk,x = 0 at x = x1 and x = x2 , determine qu,x at x = x2 , for ρF,1 = 0.06041 kg/m3 , uf,1 = 0.4109 m/s, and ∆hr,F = −5.553 × 107 J/kg. These are for a stoichiometric, atmospheric airmethane laminar ﬂame. SOLUTION: (a) Using an ordinary diﬀerential equation solver such as SOPHT, we integrate 2 − x2 qx 1 d = e 2σ = s˙ r,c (x) dx (−ρF,1 uf,1 ∆r, F) σ(2π)1/2
for σ = 0.1 mm. The result for −δ ≤ x ≤ δ is plotted in Figure Pr.2.33(b). Also plotted is s˙ r,c (x). We note that qx , which begins as qx = 0 at x = −δ, reaches a maximum value of 1 × (−ρF,1 uf,1 ∆hr,F ) at x = δ, while s˙ r,c peaks at x = 0 and its magnitude is approximately 4 × (−ρF,1 uf,1 ∆hr,F ). 111
sr,c
 rF,1 uf,1 Dhr,F
3
2
(1/mm)
qx
 rF,1 uf,1 Dhr,F 1
sr,c
 rF,1 uf,1 Dhr,F
, 1/mm,
qx
 rF,1 uf,1 Dhr,F
4
d 0 0.6 x1
0.4
0.2
x, mm
0.2
0.4
0.6 x2
Figure Pr.2.33(b) Variation of convection heat ﬂux and the energy conversion with respect to axial location.
(b) Using the value of qx at x = x2 = δ, we have qx (x = x2 )
=
1 × (−ρF,1 uf,1 ∆hr,F )
= −0.06041(kg/m3 ) × 0.4109(m/s) × (−5.553 × 107 )(J/kg) = 1.378 × 106 W/m2 . COMMENT: Note that qx (x) = qk,x (x) + qr,x (x) varies over the ﬂame length. In Chapter 5, we will approximate this conductionconvection region and use a more realistic (but still simple) source term representing the chemical reaction. Also note that qu = (ρcp T u)f .
112
PROBLEM 2.34.FUN GIVEN: A pn junction is shown Figure Pr.2.34(a). The junction (interface) is at temperature Tj . The ends of the two materials are at a lower temperature Tc and a higher temperature Th . SKETCH: Figure Pr.2.34(a) shows the conduction across a slab containing a thermoelectric pn junction. pType SemiConductor (Solid)
nType SemiConductor (Solid) Je (Electric Current)
Tc
Th A Tj
Se,P
Figure Pr.2.34(a) Conduction heat transfer in a slab containing a thermoelectric pn junction.
OBJECTIVE: (a) Starting from (2.62), write the surface energy equation for the interface. Make the appropriate assumptions about the mechanisms of heat transfer expected to be signiﬁcant. (b) Express the conduction heat transfer as Qk,n =
Tj − Th , Rk,n
Qk,p =
Tj − Tc . Rk,p
Comment on the signs of Qk,p and Qk,n needed to absorb heat at the junction to produce electrical potentialcurrent. SOLUTION: (a) From (2.60), using n and p to designate the two media, the surface energy equation is A[(qk · sn )n + (qk · sn )p + (qu · sn )n + (qu · sn )p + (qr · sn )n + (qr · sn )p ] = S˙ e,P . Since both media are not moving, qu = 0. Also, due to the large optical thickness, the radiation heat transfer with both media is expected to be negligible. Then, using S˙ e,P , the surface energy equation becomes A[(qk · sn )n + (qk · sn )p ] = S˙ e,P . This can then be written as Qk,n + Qk,p = S˙ e,P and is shown in Figure Pr.2.34(b). Tj
Rk,n
Tc
Rk,p
Th Qk,n
Qk,p (sn)p
(sn)n Qk,n
Qk,p
A qe,P = Qe,P
Figure Pr.2.34(b) Energy equation for the slab containing the junction.
113
(b) Using the equations for the conduction heat transfer, Qk,n
=
Qk,p
=
Tj − Th Rk,n Tj − Tc . Rk,p
Then, using (2.37) for S˙ e,P (for absorption of energy), the energy equation becomes Tj − Th Tj − Tc + = −αS Je Tj . Rk,n Rk,p The minus sign is used for the energy absorption. (c) Since αS > 0, Je > 0, Tj > 0, (Tj − Th ) < 0, Rk,n > 0, (Tj − Tc ) > 0 and Rk,p > 0, then Qk,n < 0 and Qk,p > 0. In order to produce electrical current, we need to have more conduction heat transfer arriving at the junction than leaving the junction. Therefore, we need Qk,n  > Qk,p  . COMMENT: If we assume that Rk,p = Rk,n , then to have energy conversion we need to have (Th − Tj ) > (Tj − Tc ). Figure Pr.2.34(b) shows the thermal circuit diagram for this problem.
114
PROBLEM 2.35.FUN GIVEN: Below are described two cases for which there is heat transfer and possibly energy conversion on a bounding surface between two media [Figure Pr.2.35 (a) and (b)]. (a) A hot solid surface is cooled by surfaceconvection heat transfer to a cold air stream and by surfaceradiation heat transfer to its surrounding. Note that the air velocity at the surface is zero, ug = 0. Also, assume that the radiation is negligible inside the solid (i.e., the solid is opaque). (b) Two solid surfaces are in contact with each other and there is a relative velocity ∆ui between them. For example, one of the surfaces is a brake pad and the other is a brake drum. SKETCH: Figures Pr.2.35(a) and (b) show a gassolid and a solidsolid interface. (a) SurfaceConvection and SurfaceRadiation Cooling of a Hot Surface
(b) Friction Heating between Two Sliding Solid Surfaces
Air
Solid 2
x
x ug
Asg
∆ui
Solid
A12
Solid 1
Figure Pr.2.35(a) and (b) Two example of bounding surface between two media.
OBJECTIVE: For each of the given cases (a) and (b), apply the boundingsurface energy equation (2.62) to the interface separating the two media. Assume that the surfaces are at uniform temperatures. As a consequence, the heat transfer at the interface is onedimensional and perpendicular to the interface. SOLUTION: (a) The solid surface is cooled by surface convection and by surface radiation. For this solidgas interface, the general boundingsurface energy equation (2.65) is Asg [−ks (∇T · sn )s − kg (∇T · sn )g + (ρcp T u · sn )s + (ρcp T u · sn )g + (qr · sn )s
S˙ i , +(qr · sn )g ] = i
where Asg is the solidgas interfacial area. On the lefthand side of the surface energy equation, the ﬁrst two terms are the conduction heat ﬂux vectors in the solid and in the gas phases normal to the surface, the third and fourth terms are the convection heat ﬂux vectors in the solid and gas phases normal to the surface, and the last two terms are the radiation heat ﬂux vectors in the solid and gas phases normal to the surface. The righthand side accounts for surface energy conversion to thermal energy. At the solidgas interface, the convection heat ﬂuxes are zero because the solid is not moving normal to the control surface and the gas phase velocity at the solid surface is zero (the surface is impermeable to the gas molecules). The radiation heat ﬂux in the solid phase is zero because the solid is assumed to be opaque to thermal radiation. At this bounding surface, there is no energy conversion (at low speeds, the energy production due to viscous heating is negligible). Therefore, the boundingsurface energy equation becomes Asg [−ks (∇T · sn )s − kg (∇T · sn )g + (qr · sn )g ] = 0. For a uniform surface temperature, the conduction and the radiation heat ﬂux vectors are normal to the surface (in the direction of the x axis), i.e., qk,x = −k(∇T · sn ) = −k qr,x = qr · sn = qr 115
dT dx
and the boundingsurface energy equation becomes dTs dTg Asg ks − kg + qr,g = 0. dx dx Note that the conduction term in the solid is positive as the surface normal in that phase is in the negative xdirection. The conduction heat ﬂux on the gas side causes the surfaceconvection heat transfer from the solid surface to the gas stream. This surfaceconvection heat transfer is also inﬂuenced by the velocity of the ﬂow. Therefore, the boundingsurface energy equation can be ﬁnally written as ks
dTs + qku,g + qr,g = 0. dx
(b) For the two solid surfaces, the radiation heat ﬂux vectors are zero. The movement of the surfaces creates a convection heat ﬂux vector in the same direction of the velocity vector. Then, the boundingsurface energy equation becomes
S˙ i . A12 [−k1 (∇T · sn )1 − k2 (∇T · sn )2 + (ρcp T u · sn )1 + (ρcp T u · sn )2 ] = i
Due to surface friction, there is energy conversion at the interface between the two solids (conversion from mechanical to thermal energy) and this energy conversion is assumed uniform along the surface. Therefore,
˙ ˙ Si = Sm,F = qm,F dA = qm,F A12 , A12
i
and, from Table 2.1, qm,F = µF pc ∆ui . The velocity vectors for both surfaces are normal to the normal vectors. Thus, the dot product of the velocity vectors and the normal vectors is zero. As the interface has a uniform temperature, the conduction heat ﬂux at the surface is onedimensional and normal to the surface. Therefore, the boundingsurface energy equation becomes dT1 dT2 − k2 A12 k1 = µF pc ∆ui A12 , dx dx or +k1
dT1 dT2 − k2 = µF pc ∆ui . dx dx
COMMENT: The surface convection heat transfer is transferred from the solid to the ﬂuid by ﬂuid conduction. An enhancement in this heat transfer by conduction leads to an enhancement in the surfaceconvection heat transfer. Consequences and means of enhancing the ﬂuid conduction heat ﬂux at the solid surface will be explored in Chapter 6. The existence of uniform temperature at the bounding surface results in conduction heat transfer normal to the surface (there is no parallel component). The convection heat transfer across the interface exists only when there is ﬂow across the interface.
116
PROBLEM 2.36.FAM GIVEN: An opaque (i.e., a medium that does not allow for any transmission of radiation across it) solid surface is called a selective radiation surface when its ability to absorb radiation is diﬀerent than its ability to emit radiation. This is shown in Figure Pr.2.36. A selective absorber has a higher absorptivity αr compared to its emissivity r . SKETCH: Figure Pr.2.36 shows absorption and emission by an opaque surface. OBJECTIVE: Surface Reflection (1  αr) qr,i
Surface Emission qr, = r σSB T 4 ∋
∋
Surface Irradiation qr,i Surface Radiation Properties: αr (Absorptivity) r (Emissivity)
Surface Temperature Ts
∋
Se, /A =  r σSB T 4 Surface Emission ∋
Se,α /A = αr qr,i Surface Absorption
∋
Figure Pr.2.36 A selective thermal radiation absorber.
(a) From Table C.19, choose four surfaces that are selective absorbers and four that are selective emitters. The data in Table C.19 is for absorption of solar irradiation (a high temperature radiation emission). (b) Using blackoxidized copper, determine the surfaceabsorption heat ﬂux for a solar irradiation of 700 W/m2 and surfaceemission heat ﬂux at surface temperature of 90◦C. (c) Determine the diﬀerence between the heat absorbed and heat emitted. SOLUTION: (a) From Table C.19, we have Table Pr.2.36: Selective absorbers and reﬂectors. Selective Absorber (Good Solar Absorber) Selective Emitter (Good Solar Reﬂector) Material r αr Material r αr Chromium Plate Blackoxidized Copper Nickel (Tabor Solar Absorber) Silicon Solar Cell
0.15 0.16 0.11 0.32
0.78 0.91 0.85 0.94
Reﬂective Aluminum Glass White Epoxy Paint Inorganic Spacecraft Coating
0.79 0.83 0.88 0.89
0.23 0.13 0.25 0.13
(b) From (2.47) and (2.48), we have S˙ e,α A ˙ Se, A
= αr qr,i = − r σSB T 4 .
Using the numerical values for blackoxidized copper, we have 2 2 S˙ e,α /A = 0.91 × 700(W/m ) = 637.0 W/m 2 2 S˙ e, /A = −0.16 × 5.67 × 10−8 (W/m K) × (273.15 + 90)4 (K)4 = −157.8 W/m .
117
(c) The net heat generated at the surface is S˙ e,α S˙ e, S˙ 2 = + = 637 − 157.8 = 479.2 W/m . A A A There is a net heat gained by the surface. COMMENT: The surfaces that have selective behavior, i.e., αr = r , are called nongray surfaces. The gray surfaces are those for which αr = r . We will discuss gray and nongray surfaces in Chapter 4.
118
PROBLEM 2.37.FUN GIVEN: A droplet of refrigeration ﬂuid (refrigerant) R134a, which is used in automobile airconditioning systems, is evaporating. The initial droplet diameter is D(t = 0) and the diameter decreases as heat is absorbed on the droplet surface from the gaseous ambient by surface convection and radiation. SKETCH: Figure Pr.2.37 shows the surface heating and evaporation of a droplet. V = 1 pD3 6 A = pD2
Droplet D(t = 0) = 3 mm
qku,g = 10 kW/m2 Gas
D(t)
ql = 0
Liquid Evaporation Slg A =  mlg Dhlg
sn,l
qr,g = 1 kW/m2
sn,g
Figure Pr.2.37 Droplet evaporation by surface convection and radiation.
OBJECTIVE: (a) Starting from (2.62) and by replacing qk,g by qku,g , and noting that the diﬀerence between the convection terms is represented by S˙ lg , write the appropriate surface energy equation. The radiation heat transfer within the droplet can be neglected. Assume a uniform droplet temperature, i.e., assume the liquid conduction can also be neglected. (b) Using the properties listed in Table C.26 (they are for p = 1 atm) and the heat ﬂux rates given in Figure Pr.2.37, determine the evaporation rate per unit area m ˙ lg . (c) Starting with (1.25) and setting the outgoing mass equal to the evaporation rate, derive an expression giving the instantaneous droplet diameter D(t), as a function of the various parameters. (d) Determine the time needed for the droplet diameter to decrease by a factor of 10. SOLUTION: ˙ (a) From (2.62), with (S/A) = −m ˙ lg ∆hlg from Table 2.1, the surface energy equation becomes ˙ lg ∆hlg . A[(qku · sn )g + (qku · sn )l + (qr · sn )g + (qr · sn )l ] = −Am Since (qku )l and (qr )l are assumed zero, we have A[(qku · sn )g + (qr · sn )g ] (qku · sn )g + (qr · sn )g
= −Am ˙ lg ∆hlg = 0 = −m ˙ lg ∆hlg = 0.
(b) From Table C.26, for R134a at p = 1 atm, we have Tlg = 246.99 K, ρl = 1374.3 kg/m3 and ∆hlg = 2.168×105 J/kg. Solving the energy equation for m ˙ lg , we have m ˙ lg = −
qku,g + qr,g . ∆hlg
Using the numerical values, we then have 2
m ˙ lg = −
(−10,000 − 1,000)(W/m ) 2 = 5.074 × 10−2 kg/m s. 2.168 × 105 (J/kg)
(c) From (1.25), we have d M˙ lg = M˙ A = − dt 119
ρl dV . V (t)
For constant ρl we then have d M˙ A = − (ρl dt
d d dV ) = − (ρl V ) = − dt dt V (t)
1 ρl πD3 6
dD π . = − ρl D2 2 dt
The mass ﬂow rate M˙ A is related to the mass ﬂux by ˙ lg = πD2 m ˙ lg . M˙ A = Am Then, from the equations above we obtain dD 2m ˙ lg =− . dt ρl Integrating this equation we have
D(t)
t
dD = − D(t=o)
2m ˙ lg dt. ρl
For constant rate of phase change D(t) = D(t = 0) −
2m ˙ lg t. ρl
(d) The equation above can be recast as D(t) 2m ˙ lg =1− t. D(t = 0) ρl D(t = 0) Solving for t, we have
t= 1−
D(t) ρl D(t = 0) . D(t = 0) 2m ˙ lg
For D(t)/D(t = 0) = 0.1 and using the other values, we have 1374.3(kg/m )3 × 10−3 (m) 3
t = (1 − 0.1)
2
2 × 5.074 × 10−2 (kg/m s)
= 36.57 s.
COMMENT: (i) This rate of heat ﬂow into the droplet is high, but not very high. In order to evaporate the droplet very rapidly, surface heat transfers of the order of 100 kW/m2 are used. Also, the heat ﬂux changes as the diameter decreases because the area decreases. Thus, the rate of evaporation increases as the diameter decreases. (ii) The droplet evaporation model above is called a heat transfer controlled evaporation. The evaporation can also be mass transfer controlled if the rate of mass transfer of the vapor from the droplet surface to the ambient is slower than the rate of heat transfer. (iii) Refrigerant134a operates under large pressures, both in the evaporator and in the condenser. The heat of evaporation decreases as the critical pressure is approached.
120
PROBLEM 2.38.FUN GIVEN: A thermoelectric element (TE) is exposed at its cold junction surface (at temperature Tc ) partly to an electrical connector (e) and partly to the ambient air (a). This is shown in Figure Pr.2.38. Heat is transferred to the surface through the thermoelectric element qk,T E in addition to a prescribed heat ﬂux, qT E that combines some parasitic heating. These are over the surface area Aa + Ae . Heat is also transferred to the surface from the adjacent air and the connector, over their respective areas Aa and Ae . The area in contact with air undergoes heat transfer by surface convection qku,a and surface radiation qr,a . The connector heat transfer is by conduction qk,e . There is a Peltier energy conversion (S˙ e,J )c /Ae at the surface (and since it occurs where the current passes, it occurs over Ae ). The Joule heating (S˙ e,J )c /Ae is also represented as a surface energy conversion (this presentation will be discussed in Section 3.3.6) and is over the entire element area Aa + Ae . Aa = 10−6 m2 , Ae = 10−6 m2 , qk,T E = −4 × 104 W/m2 , (S˙ e,J )c /A = 2 × 104 W/m2 , (S˙ e,P )c /Ae = −2 × 105 W/m2 , qr,a = 0, qku,a = 0, qT E = 0. Assume quantities are uniform over their respective areas. SKETCH: Figure Pr.2.38 shows the control surface A and the various surface heat transfer and energy conversions. Surface Surface Radiation Convection (Se,J)c , Joule Heating A qr,a qku,a Wrapping Control Surface, A A , Electrical Connector e
Uniform Temperature, Tc Ambient Air, Aa
Thermoelectric Element
sn
sn Aa + Ae Thermoelectric qTE Element qk,TE Prescribed Conduction Heat Flux
Electrical Current, Je qk,e , Conduction (Se,P)c , Peltier Cooling Ae
Figure Pr.2.38 The coldjunction surface of a thermoelectric element showing various surface heat transfer and energy conversions.
OBJECTIVE: (a) Starting from (2.60), write the surface energy equation for the cold junction control surface A. (b) Determine qk,e for the given conditions. SOLUTION: (a) From (2.60), we have (q · sn )dA
=
A
S˙ i
i
(q · sn )dA A
(qT E · sn )dA + (qk,T E · sn )dA + Aa +Ae (qr,a · sn )dA + (qku,a · sn )dA Aa A a
˙ e,P )c ˙ e,J )c ( S ( S S˙ i = (Aa + Ae ). = Ae + Ae A i
(qk,e · sn )dA +
=
Aa +Ae
Ae
Here we have used the appropriate areas for each heat transfer rate and each energy conversion mechanism. Now, since the various heat ﬂux vectors given in Figure Pr.2.38 are all given as leaving the contact surface, we have (S˙ e,J )c (S˙ e,P )c (Aa + Ae )qT E + (Aa + Ae )qk,T E − Ae qk,e + Aa qr,a + Aa qku,a = Ae + (Aa + Ae ) . Ae A 121
(b) Using the numerical values, we have (10−6 + 10−6 )(m2 ) × 0 + (10−6 + 10−6 )(m2 ) × (−4 × 104 )(W/m2 )− 10−6 (m2 ) × qk,e + 10−6 (m2 ) × 0 + 10−6 (m2 ) × 0 =
10−6 (m2 ) × (−2 × 105 )(W/m2 ) + (10−6 + 10−6 )(m2 ) × 2 × 104 (W/m2 )
or qk,e
= (2 × 4 × 104 − 2 × 105 + 4 × 104 )(W/m2 ) = −8 × 104 W/m2
Qk,e
= Ae qk,e = 10−6 (m2 ) × [−8 × 104 (W/m2 )] = −8 × 10−2 W.
or
COMMENT: Note that heat ﬂows into the electric connecter because Qk,e < 0. This is the eﬀective cooling heat rate and the object to be cooled is connected to the electric connector (with a thin layer of electrical insulator between them, in case the object is not a dielectric). In practice, many of these junctions are used to produce the desired cooling rate. This is discussed in Section 3.7.
122
PROBLEM 2.39.FUN GIVEN: When the ambient temperature is high or when intensive physical activities results in extra metabolic energy conversion, then the body loses heat by sweating (energy conversion S˙ lg ). Figure Pr.2.39 shows this surface energy exchange, where the heat transfer to the surface from the tissue side is by combined conduction and convection qk,t , qu,t and from the ambient air side is by conduction, convection, and surface radiation qk,a , qu,a , qr,a . The surface evaporation is also shown as S˙ lg /At where At is the evaporation surface area. The tissue conduction qk,t is signiﬁcant for lowering the body temperature or removing extra metabolic heat generation (i.e., when the heat ﬂow is dominantly from the tissue side). Preventing the high ambient temperature from raising the tissue temperature, however, relies only on intercepting the ambient heat transfer on the surface (i.e., when the heat ﬂow is dominated by the ambient air side and the tissue conduction is not signiﬁcant). Assume that quantities are uniform over their respective surfaces. SKETCH: Figure Pr.2.39 shows the various surface heat transfer mechanisms and the surface energy conversion S˙ lg /At . Tissue qk,t Sweat Glands
qu,t
Ambient Air Wrapping Control Surface, A Slg /At
qu,a
qk,a qr,a Evaporation, mlg Water Vapor
Water sn At
Skin Temperature, Ts sn Aa = At
Figure Pr.2.39 The surface heat transfer, and energy conversion by sweat cooling. across human skin.
OBJECTIVE: (a) Starting from (2.60), write the surface energy equation for the skin control surface A (wrapped around the surface with At = Aa ) in Figure Pr.2.39. (b) For the conditions given below, determine qk,t . SOLUTION: (a) From (2.60), we have QA
= At (qk · sn )t + At (qu · sn )t + Aa (qk · sn )a + Aa (qu · sn )a + Aa (qr · sn )a S˙ lg = At . At
Since the heat ﬂux vectors are deﬁned in Figure Pr.2.39 to be pointing outward from their respective surfaces (i.e., along sn ), then we have S˙ lg At (qk,t + qu,t ) + Aa (qk,a + qu,a + qr,a ) = At At and since all the areas are equal, we have qk,t + qu,t + qk,a + qu,a + qr,a =
S˙ lg . At
(b) Now using the numerical values and qk,a = qku,a , we have qk,t + 0 + qku,a + 0 + qr,a = 123
S˙ lg . At
Solving for qk,t , we have qk,t = −300(W/m2 ) + 150(W/m2 ) + 10(W/m2 ) = −140 W/m2 . This corresponds to heat ﬂowing from the tissue to the surface by conduction at this rate. COMMENT: The convection heat ﬂuxes are negligible due to the small velocities. Also note that we used the surface convection qku,a in place of conduction qk,a because as will be shown in Chapter 6, the air velocity at the surface is zero (neglecting the small water vapor velocity leaving the surface due to the evaporation). Then the heat transfer to the air is by conduction, but inﬂuenced by the air motion.
124
PROBLEM 2.40.FUN GIVEN: In laser materials processingmanufacturing, highpower, pulsed laser irradiation ﬂux qr,i is used and most of this power is absorbed by the surface. Figure Pr.2.40 shows the laser irradiation absorbed S˙ e,α /A = αr qr,i (where αr is the surface absorptivity), the surface radiation emission ﬂux (S˙ e, )/A = r σSB Ts4 , the gasside surface convection qku , and the solid (substrate or working piece) conduction qk,s , over a diﬀerential control surface ∆A → 0. Since the irradiation is time dependent (e.g., pulsed), the heat transfer and energy conversions are all time dependent (and nonuniform over the surface). r = 0.8, αr = 0.9, qr,i = 1010 W/m2 , Ts = 2 × 103 K, qku = 107 W/m2 . Note that the entire surface radiation is represented as energy conversions S˙ e,α and S˙ e, . SKETCH: Figure Pr.2.40 shows the laser irradiated surface, the substrate conduction, surface convection, and surface radiation emission represented as an energy conversion. Laser Generator
Solid Substrate
qr,i(t) sn
DAs
qku(x, t) qk,s(x, t) Se,a + Se,a , where Se,a = ar qr,i(t), Se, = A A A A '
sn
Differential Control Surface Wrapped Around Volume
'
Laser Beam DA 0 Surface Temperature, Ts(x, t), Nonuniform sn and Transient
DAf = DAs
sn
r
sSB Ts4(x, t)
Figure Pr.2.40 Laser irradiation of a substrate and a diﬀerential control surface taken in the laser impingement region.
OBJECTIVE: (a) Starting with (2.58), write the surface energy equation for the diﬀerential control surface ∆A. (b) Determine qk,s for the conditions given. SOLUTION: (a) Starting from (2.58), for diﬀerential control surface ∆A, we have (q · sn )dA = S˙ = S˙ e,α + S˙ e, = ∆Af αr qi − ∆Af r σSB Ts4 , ∆A
where we have used the energy conversion terms given in Figure Pr.2.40 for the righthand side. Then noting that qku and qk,s are along their respective surface normal vectors, we have (q · sn )dA = ∆Af qku + ∆As qk,s = ∆As αr qr,i − ∆As r σSB Ts4 . ∆A
(b) Solving the above equation for qk,s , using ∆As = ∆Af , we have qk,s
= αr qr,i − r σSB Ts4 − qku = 0.9 × 1010 (W/m2 ) − 0.8 × 5.67 × 10−8 ( W/m2 K4 ) × (2 × 103 )4 (K4 ) − 107 (W/m2 ) = (9 × 109 − 8.165 × 105 − 107 )(W/m2 ) = 8.989 × 109 W/m2 .
COMMENT: Note that during the irradiation, the surface radiation emission and surface convection are rather small and negligible.
125
Chapter 3
Conduction
PROBLEM 3.1.FUN GIVEN: Equation (3.25) relates the thermal conductivity to the electrical resistivity of pure solid metals. Values for the electrical resistivity as a function of temperature are listed in Table C.8 for diﬀerent pure metals. OBJECTIVE: (a) Using (3.25), calculate the predicted thermal conductivity of copper kpr for T = 200, 300, 500, and 1,000 K. For T = 1,000 K, extrapolate from the values in the table. (b) Compare the results obtained in (a), for kpr , with the values given in Table C.14, kex . Calculate the percentage diﬀerence from using kpr − kex × 100. ∆k(%) = kex (c) Diamond is an electrical nonconductor (σe 0). However, Figure 3.9(a) shows that the thermal conductivity of diamond is greater than the thermal conductivity of copper for T > 40 K. How can this be explained? SOLUTION: (a) Equation (3.25) relates the electronic contribution in the thermal conductivity k e to the electrical conductivity σe (inverse of resistivity ρe ). Assuming that for copper the electronic contribution is the dominant mechanism for the thermal conductivity (k k e ), (3.25) can be written as k e ρe = 2.442 × 10−8 Wohm/K2 . T From the values of ρe given in Table C.8, the thermal conductivity of copper can be calculated at diﬀerent temperatures. Table Pr.3.1 lists the results. Table Pr.3.1 Thermal conductivity of pure copper. T, K ρe , ohmm kpr , W/mK kex , W/mK ∆k (%) 200 300 500 1,000
1.046 1.725 3.090 6.804
× × × ×
10−8 10−8 10−8 10−8
466.9 424.7 395.2 358.9
413 401 386 352
13 6 2 2
(b) Table Pr.3.1 shows the data obtained from Table C.14 (kex ) and the percentage diﬀerence between kpr and kex . (c) The thermal conduction in diamond occurs dominantly by the mechanism of lattice vibration. The transfer of energy due to lattice vibration is represented by a heat carrier called a phonon and the heat conduction is then said to be due to phonon transport. The phonon transport is more eﬀective at higher temperatures, as shown in Figure 3.7(c). At low temperatures, the heat conduction by electron transport is substantial. Therefore, at low temperatures, copper is a better conductor than diamond. The phonon transport mechanism is also present in copper, but has a relatively smaller contribution. COMMENT: The value of ρe at T = 1,000 K is extrapolated from the values listed in Table C.8.
128
PROBLEM 3.2.FAM GIVEN: An airplane ﬂies at an altitude of about 10 km (32,808 ft). Use the relation for the polyatomic idealgas thermal conductivity given in Example 3.2. OBJECTIVE: Using the relation for the polyatomic ideal gas thermal conductivity given in Example 3.2, (a) Determine the air thermal conductivity at this altitude. Use the thermophysical properties given in Table C.7, and assume that cv and cp are constant. (b) Compare the predicted k with the measured value given in Table C.7. (c) Comment on why k does not change substantially with altitude. SOLUTION: (a) From Example 3.2, we have k=
3cv 9 Rg 5π ρ cv + λ. as 32 4M cp
From Table C.7, we have, for r = 10 km, 3
ρ = 0.41351 kg/m M = 28.965 kg/kmole
Table C.7
as = 299.53 m/s
Table C.7
−7
λ = 1.97 × 10
Table C.7
m
Table C.7.
Also from Example 3.2, we have cv
=
719 J/kgK
cp
=
1,006 J/kgK.
Using the numerical values, we have k
=
k
= =
8,315 9 5π 3 × 0.41351(kg/m ) × 719 + × (J/kgK) × 299.53(m/s) 32 4 28.964 1/2 3 × 719 × 1.97 × 10−7 (m) × 1,006 5π × 0.41351 × 1,365 × 299.5 × 1.464 × 1.97 × 10−7 32 0.02392 W/mK.
(b) The measured k from Table C.7 is k = 0.0201 W/mK
Table C.7.
The diﬀerence, in percentage, is ∆k(%) =
0.02392 − 0.0201 × 100% = 18.20%. 0.0201
This is reasonable, considering that we have a mixture of species and the assumptions made in the kinetic theory. (c) The meanfree path of the air increases with the altitude r, as listed in Table C.7. However, the density decreases with r. These two nearly compensate each other (the changes in the speed of sound is not as substantial as that in λ and ρ), thus making the thermal conductivity not substantially change for 0 < r < 50 km. COMMENT: From Table C.7, note that the air molecular weight does not begin to change substantially until an altitude of about 1,000 km is reached. This is when the air composition begins to change to mostly hydrogen and helium. The temperature at an altitude of 10 km is T = 233.25 K = −39.9◦C, and the pressure is p = 0.026499 MPa = 0.2615 atm. 129
PROBLEM 3.3.FUN GIVEN: Due to their molecular properties, the elemental, diatomic gases have diﬀerent thermodynamic properties, e.g., ρ and cp , and transport properties, e.g., k properties. Consider (i) air, (ii) helium, (iii) hydrogen, and (iv) argon gases at T = 300 K and one atmosphere pressure. OBJECTIVE: (a) List them in order of the increasing thermal conductivity. Comment on how a gas gap used for insulation may be charged (i.e., ﬁlled) with diﬀerent gases to allow none or less heat transfer. (b) List them in the order of the increasing thermal diﬀusivity α = k/ρcp . Comment on how the penetration speed uF can be varied by choosing various gases. (c) List them in order of increasing thermal eﬀusivity (ρcp k)1/2 . Comment on how the transient heat ﬂux qρck (t) can be varied by choosing various gases. SOLUTION: The thermal conductivity, density, and speciﬁc heat capacity for each of the four gases are listed in Table C.22 for p = 1 atm. For T = 300 K, we have (i) air: k = 0.0267 W/mK ρ = 1.177 kg/m3 cp = 1,005 J/kgK (ii) helium: k = 0.1490 W/mK ρ = 0.1624 kg/m3 cp (iii) hydrogen: k
= =
5,200 J/kgK 0.1980 W/mK 3
ρ = 0.0812 kg/m cp = 14,780 J/kgK (iv) argon: k = 0.0176 W/mK ρ = 1.622 kg/m3 cp = 621 J/kgK
Table C.22 Table C.22 Table C.22 Table C.22 Table C.22 Table C.22 Table C.22 Table C.22 Table C.22 Table C.22 Table C.22 Table C.22.
(a) Thermal conductivities in order of increasing magnitude are argon: k
= 0.0176 W/mK
air: k helium: k
= 0.0267 W/mK = 0.1490 W/mK
hydrogen: k
= 0.1980 W/mK.
By changing the gas from argon to hydrogen, the conduction heat transfer rate will be increased by a factor of 11.25. (b) Thermal diﬀusivities α = k/ρcp in order of increasing magnitude are argon: α air: α
= =
1.747 × 10−5 m2 /s 2.257 × 10−5 m2 /s
hydrogen: α helium: α
= =
1.650 × 10−4 m2 /s 1.764 × 10−4 m2 /s.
From (3.154), uF is proportional to α1/2 . Helium has an α that is 10.10 times that of argon. Thus, the penetration speed for helium is 3.178 times larger than that for argon.
130
(c) Thermal eﬀusivities in order of increasing magnitude are argon: (ρcp k)1/2
=
4.210 Ws1/2 /m2 K
air: (ρcp k)1/2
=
5.620 Ws1/2 /m2 K
helium: (ρcp k)1/2
=
11.22 Ws1/2 /m2 K
hydrogen: (ρcp k)1/2
=
15.42 Ws1/2 /m2 K.
From (3.144), we note that qρck (t) is proportional to (ρcp k)1/2 . Hydrogen has an eﬀusivity which is 3.622 times that of argon. Thus, the transient heat ﬂow rate to a semiinﬁnite stagnant gas layer suddenly heated on its bounded surface is 3.662 larger for hydrogen, compared to argon. COMMENT: We have assumed that the gas remains stagnant (i.e., no thermobuoyant motion) while it undergoes heat transfer.
131
PROBLEM 3.4.FUN GIVEN: The bulk (or intrinsic) conductivity refers to the medium property not aﬀected by the size of the medium. In gases, this would indicate that the meanfree path of the gas molecules in thermal motion λm is much smaller than the linear dimension of gas volume L. When the linear dimension of the gas volume is nearly the same as or smaller than the meanfree path, then the gas molecules collide with the bounding surface of the gas with a probability comparable to that of the intermolecular collisions. This will occur either at low pressure or for very small L. There are simple, approximation expressions describing this size (or lowdimensionality) eﬀect. These expressions include parameters modeling the gas moleculebounding surface collision and energy exchange. One of these models that is used to predict the size dependence occurring at low gas pressures is kf (p, T ) =
kf (p = 1 atm, T ) , 4a1 (2 − γ) KnL 1+ γ(cp /cv + 1)
where KnL is the Knudsen number deﬁned in (1.20), i.e., KnL =
λm , L
and λm is given by (1.19). Here 0 ≤ γ ≤ 1 is the accommodation factor and a1 is another semiempirical constant. For example, for nitrogen in contact with ceramic surfaces, a1 = 1.944, cp /cv =1.401, and γ = 0.8. Use Table C.22 for kf (p = 1 atm, T = 300 K). OBJECTIVE: For nitrogen gas with L = 10 µm, use T = 300 K, and dm = 3 × 10−10 m and plot kf /kf (λm L) versus the pressure and the Knudsen number. SOLUTION: From (1.19), we have λm
= =
kB T 1 (1.381 × 10−23 )(J/K) × 300(K) = (3 × 10−10 )2 (m2 ) × p(Pa) 21/2 π d2m p 21/2 π −2 1.037 × 10 (Pam) . p(Pa) 1
From Table C.22, for air at T = 300 K, we have kf (p = 1 atm, T = 300 K) = 0.0267 W/mK. Then kf (p, T = 300 K)
=
=
0.0267(W/mK) 4 × 1.944 × (2 − 0.8) λm (m) 1+ 0.8(1.401 + 1) 10−5 (m) 0.0267(W/mK) . 1 + 4.858 × 105 (1/m) × λm (m)
Figures Pr.3.4(a) and (b) show the variations of kf (p) with respect to p and KnL . COMMENT: Note that the relation used here for kf is an approximation. Also note that, as L becomes very large, the asymptotic value kf (p = 1 atm) is recovered. 132
0.030
0.030
0.0267
0.0267 0.024
kf , W/mK
kf , W/mK
0.024 0.018 0.012
Nitrogen L = 105 m T = 300 K
0.006 0 100
1 x 103
1 x 104
Nitrogen L = 105 m T = 300 K
0.018 0.012 0.006
1 x 105
p, Pa
0 0.01
0.1
KnL =
lm
1
10
L
Figure Pr.3.4 Variation of the gas conductivity with respect to (a) pressure and (b) Knudsen number.
133
PROBLEM 3.5.FUN GIVEN: The lattice (phonon) speciﬁc heat capacity is related to the internal energy e, which in turn is given by the energy of an ensemble of harmonic oscillators as e
=
NA
Ei M i
Ei
=
hP fi np,i 2π
np,i =
1 e −1 xi
xi =
hP fi , 2πkB T
where hP is the Planck constant, kB is the Boltzmann constant, NA is the Avogadro number, M is the molecular weight, and Ei is the average energy per vibrational mode i of each oscillator. This represents the solid as a collection of harmonic oscillators, vibrating over a range of frequencies f , with the number of phonons having a frequency fi given by np,i . Note that from (3.4), Rg ≡ kB NA . OBJECTIVE: Starting from (1.6), and using the above, show that the lattice speciﬁc heat capacity is cv =
Rg x2i exi . M i (exi − 1)2
SOLUTION: The energy per unit mass is e = = = =
NA
Ei M i NA h P M
i
2π
fi np,i
1 NA h P fi M i 2π exi − 1 NA h P 1 fi . hP fi M i 2π exp −1 2πkB T
The speciﬁc heat capacity of a solid at constant volume cv is found by diﬀerentiating with respect to temperature T , i.e., 1 ∂e NA h P ∂ hP fi fi cv = = . ∂T v M i 2π ∂T exp 2πk T − 1 B Letting u(xi ) = exi − 1, and letting w(u) = u−1 , we can simplify the diﬀerentiation on the righthand side as 1 1 1 ∂ ∂ ∂w ∂ hP fi . = = = ∂T exp 2πk T − 1 ∂T exi − 1 ∂T u ∂T B
Applying the chain rule, we obtain ∂w ∂T
= = =
∂w ∂u ∂xi ∂u ∂xi ∂T
hP fi (−u−2 )(exi ) − 2πkB T 2 hP fi exi . 2 xi 2πkB T (e − 1)2 134
Substituting back into the speciﬁc heat expression, we have hP fi exi ∂e NA h P cv = f = i ∂T v M i 2π 2πkB T 2 (exi − 1)2 2 hP fi exi NA kB
= M 2πkB T (exi − 1)2 i =
Rg x2i exi , M i (exi − 1)2
xi =
hP fi . 2πkB T
COMMENT: In practice, the summation is diﬃcult to perform and the Debye approximation given by (3.7) is used instead.
135
PROBLEM 3.6.FUN GIVEN: In the Debye approximation model for the lattice (phonon) speciﬁc heat capacity given by (3.7), the number of vibrational modes or density of state (per unit frequency around a frequency f ) is given by the distribution function 3f 2 V , 2π 2 u3p
P (f ) =
3 V = lm = n−1 ,
where V is the volume, lm is the cubic lattice constant, up is the speed of sound (phonon speed), f is the frequency, and n is the number of oscillators (or atoms) per unit volume. The actual lattice may not be cubic and would then be represented by two or more lattice parameters and, if the lattice is tilted, also by a lattice angle. Using this expression, the lattice speciﬁc heat capacity is approximated (as an integral approximation of the numerically exact summation) as cv =
Rg x2i exi Rg fD x2 ex = P (f )df, M i (exi − 1)2 M 0 (ex − 1)2
x=
hP f . 2πkB T
The Debye distribution function (or density of state), when integrated over the frequencies, gives the total number of vibrational modes (three per each oscillator) 3n =
1 V
fD
P (f )df. 0
OBJECTIVE: (a) Show that fD = (6nπ 2 u3p )1/3 . (b) Using this, derive (3.7), i.e., show that cv = 9
Rg M
T TD
3
TD /T
x4 ex dx, (ex − 1)2
TD =
hP fD . 2πkB
SOLUTION: (a) The Debye cutoﬀ frequency is related to the number of oscillators by 3= 0
fD
3f 2 3 df = 2nπ 2 u3p 2nπ 2 u3p
fD
f 2 df.
Evaluating the integral and solving for fD gives 3 =
fD3 3 2 3 3 2nπ up
fD3
=
6nπ 2 u3p
fD
=
(6nπ 2 u3p )1/3 .
(b) Noting that from the deﬁnition of the Debye temperature we can write fD = (2πkB TD )/(hP ), and recalling that x(T, f ) = (hP f )/(2πkB T ), we can write xD = x(T, f = fD ) as xD =
hP 2πkB TD hP fD TD = . = 2πkB T 2πkB T hP T 136
Substituting the expressions for f and P (f ) into the given integral expression for cv gives cv
≈ = = = = =
Rg M Rg M Rg M Rg M Rg M Rg M
fD
x2 ex P (f )df (ex − 1)2
xD
x2 ex 3f 2 2πkB T dx 2 (e − 1) 2nπ 2 u3p hP
x
TD /T
TD /T
TD /T
TD /T
3f 2 kB T x2 ex dx nπu3p hP (ex − 1)2
2 2πkB T hP 3f 2 kB T x2 ex dx × 2πkB T hP nπu3p hP (ex − 1)2 2 3 3 f hP 12πkB T x2 ex × x dx 3 3 2πkB T nup hP (e − 1)2 3 3 12πkB T x4 ex dx. 3 3 x nup hP (e − 1)2
Substituting fD from part (a) into our expression for TD gives TD =
hP fD hP = (6nπ 2 u3p )1/3 , 2πkB 2πkB
which, after some manipulation, gives 3 kB 1 6π 2 = . nu3p h3P TD3 (2π)3
We then have cv
= = = =
TD /T
3 3 12πkB T x4 ex dx nu3p h3P (ex − 1)2 0 TD /T k3 x4 ex Rg 12πT 3 3B 3 dx M nup hP 0 (ex − 1)2 TD /T 1 6π 2 x4 ex Rg 12πT 3 3 dx M TD (2π)3 0 (ex − 1)2 3 TD /T x4 ex Rg T 9 dx. x M TD (e − 1)2 0
Rg M
COMMENT: The Debye approximation gives a reasonable prediction of cv for both metallic and nonmetallic, crystalline solids.
137
PROBLEM 3.7.FUN GIVEN: A simple approximate expression is found for the lattice thermal conductivity by only considering the normal (i.e., momentum conserving) phonon scattering mechanisms. This is done using the expression for cv , given by (3.7) in the ﬁrst part of the expression for k p given by (3.26), i.e., kp =
1 ρcv up λp , 3
and noting that λ p = u p τp .
OBJECTIVE: As is done in the Debye approximation, use
TD /T
cv λp =
cv (x)λp (x)dx,
hP f 2πkB T
x=
and xD =
TD hP fD = , 2πkB T T
and TD =
hP fD hP = (6nπ 2 u3p )1/3 2πkB 2πkB
to derive an expression for k p as a function of lm as k p = (48π 2 )1/3
3 1 kB T3 2 lm h P T D
TD /T
τp 0
x4 ex dx, (e − 1)2 x
where, for a cubic crystal lattice, lm is a lattice constant related to the number of atoms per unit volume by −1 = n1/3 . From (1.19), use ρRg /M = nkB . lm SOLUTION: Substituting for cv λp into (3.26) and then up τp (x) for λp (x), we obtain kp
= = =
1 ρup 3 1 2 ρu 3 p 3ρ
TD /T
cv (x)λp (x)dx 0
Rg M
TD /T
τp (x)cv (x)dx 0
T TD
3
u2p
TD /T
τp 0
x4 ex dx. (ex − 1)2
From the deﬁnition of the Debye temperature, we have u2p TD2
=
2 (2π)2 kB
h2P (6nπ 2 )2/3
.
Upon substitution for (up /TD )2 in k p , we obtain k
p
= = =
2 Rg T 3 up 3ρ M TD TD2
TD /T
x4 ex dx (ex − 1)2 0 TD /T 2 x4 ex Rg T 3 (2π)2 kB τp x dx 3ρ 2 2 2/3 M TD hP (6nπ ) (e − 1)2 0 TD /T 2 x4 ex Rg T 3 kB (48π 2 )1/3 ρ τ dx. p M TD h2P n2/3 0 (ex − 1)2 τp
138
−1 Noting that ρRg /M = nkB , and that lm = n1/3 , this further simpliﬁes to
kp
= = =
TD /T 2 T 3 kB x4 ex τp x dx 2 2/3 T D hP n (e − 1)2 0 k 2 T 3 1/3 TD /T x4 ex (48π 2 )1/3 kB B n τ dx p h2P TD (ex − 1)2 0 3 3 TD /T x4 ex 2 1/3 1 kB T τ dx. (48π ) p lm h2P TD 0 (ex − 1)2
(48π 2 )1/3 nkB
COMMENT: The total phonon time constant is related to the time constants for the normal (momentum conserving, τp,n ) and the resistive (nonmomentum conserving, τp,r ) processes that work to restore the phonon distribution to equilibrium (i.e., limit the conduction heat ﬂux by damping the phonon propagation). The determination of the lattice thermal conductivity is highly dependent on the manner in which τp , and in turn the various τp,n and τp,r , are evaluated and implemented into the calculation. In the approximate form found here, τp can be evaluated as −1 −1 −1 + τp,r , or for this case in which only normal processes are considered, τp−1 = τp,n . τp−1 = τp,n This is a simple form of (3.26). Most of the resistive relaxations neglected above are not very signiﬁcant at high temperatures (including near room temperature) and therefore, the above simple expression can often be used. The time constant for a normal process can be approximated as τp,n = an 2πf T 4 , where an is a material constant.
139
PROBLEM 3.8.FUN.S GIVEN: The crystal size inﬂuences the phonon thermal conductivity due to phonon scattering caused by variation of phonon propagation properties across the crystal surface (similar to light scattering at the interface of two media of diﬀerent light propagation properties). This boundary scattering is one of the resistive scattering mechanisms included in (3.26). Consider aluminum oxide (AL2 O3 , also called alumina) single crystals at T = 300 K. The eﬀect of crystal size L can be described by a simple relation for the boundary scattering relaxation time constant τb as L , up
τb =
where up is the average phonon velocity. Using the material constants for alumina and at T = 300 K, the lattice conductivity given by (3.26) becomes g22 (x, L) h22 (x, L) p 3 k = bk T g1 (x, L) + = bk h1 (x, L) + , g3 (x, L) h3 (x, L) where bk = 2.240 × 105 W/mK4 and the gi ’s and hi ’s represent integrals as deﬁned below. Some numerical solvers (e.g., SOPHT) have limitations to the size of the numbers which they may use. To avoid this limitation, the T 3 may be taken into the integral by deﬁning θi = τi T 3 and then rewriting the integrals in (3.26) as h1 h3
= g1 T 3 = =
g3 = T3
TD /T
θp
0 TD /T
x4 ex dx, (ex − 1)2
TD /T
h 2 = g2 = 0
θp x4 ex dx, θp,n (1 − ex )2
4 x
θp x e dx, θp,n θp,r (1 − ex )2
where 1 1 1 = + θp θp,n θp,r
1
=
θp,n
+
i
1
θp,r,i
= bn x + 2 × bp x4 + bu x2 +
bb L
,
where TD = 596 K, bn = 3.181 × 103 1/K3 s, bp = 3.596 × 101 1/K3 s, bu = 1.079 × 104 1/K3 s, bb = 2.596 × 10−4 m/K3 s. OBJECTIVE: Use a solver to plot kp versus grain size, L, for 10−9 ≤ L ≤ 10−4 m. SOLUTION: Using a solver such as SOPHT, the integrations are performed numerically. SOPHT is a diﬀerential solver, and therefore the integrals must be transformed into their associated diﬀerential forms. For example, the integral h1 = g1 T 3 =
TD /T
θp 0
x4 ex dx (e − 1)2 x
is transformed to the diﬀerential form dh1 x4 ex = θp x . dx (e − 1)2 Since the lower limit of the integral is zero, the solver can then be used to solve for h1 (x) with the ﬁnal desired answer being h1 = h1 (x = TD /T ). The source code using SOPHT is then
140
h1’=dh1dx h2’=dh2dx h3’=dh3dx x=t L=1e2 //This is manually changed bn=3.181e3 bu=1.079e4 bb=2.596e4 bpv=3.596e1 bk=2.240e5 kern=(xˆ4*exp(x))/(exp(x)1)ˆ2 ithetap=1/theta p ithetapn=1/theta pn ithetapr=1/theta pr ithetap=ithetapn+ithetapr ithetapn=bn*x ithetapr=ithetau+ithetab+ithetapv ithetau=bu*xˆ2 ithetab=bb/L ithetapv=2*(bpv*xˆ4) dh1dx=theta p*kern dh2dx=theta p/theta pn*kern dh3dx=theta p/theta pn/theta pr*kern k=bk*(h1+h2ˆ2/h3)
k p, W/mK
Note that SOPHT solves initial condition diﬀerential equations using t as the independent variable. Here t has been equated to our x. Each execution of SOPHT at diﬀerent input values of L must be done for a range of x = 0 up to xD = TD /T . Note that if the initial conditions for t (i.e., x) or h3 are equal to zero, there will be a division by zero in the ﬁrst iterations of the solver execution resulting in an execution error. To avoid this, initial conditions of 1 × 10−10 were used for t (i.e., x) and h3 , initial conditions of zero were used for h1 and h2 , and the iteration was run for 1,000 steps from a start of t = 1 × 10−10 to an end of t = xD = TD /T = 1.987. Figure Pr.3.8 shows the results. Note that for L ≤ 1µm, the eﬀect of the boundary scattering becomes noticeable.
10
1 Alumina (Al2O3) Single Crystal T = 300 K 0.1 10−10
10−8
10−6
10−4
10−2
1
L, m Figure Pr.3.8 Variation of alumina lattice thermal conductivity with respect to crystal dimension.
COMMENT: Note that 1 ˚ A = 10−10 m = 0.1 nm and as the lattice constant lm = 0.3493 nm (Table 3.1) is reached, this continuum treatment of the lattice vibration will no longer be valid and a direct simulation (e.g., molecular dynamic simulation) is needed.
141
PROBLEM 3.9.FUN.S GIVEN: For thin ﬁlm deposited on surfaces, the thermal conductivity of the ﬁlm becomes ﬁlmthickness dependent, if the ﬁlm thickness L is near or smaller than the heatcarrier, meanfree path. Consider a ceramic, amorphous silicon dioxide (SiO2 , also called silica) where the heat carriers are phonons. This ﬁlmthickness dependence of the thermal conductivity may be approximated as k=
k(L λp ) , 4 λp 1+ 3 L
where k(L λp ) is the bulk (or sizeindependent) thermal conductivity, and λp is the phonon meanfree path. The reduction in the thermal conductivity (as λp /L increases) is due to the scattering of the phonons at the boundaries of the thin ﬁlm. OBJECTIVE: Using Tables 3.1 and C.17, plot the variation of k for amorphous silica for 0.6 ≤ L ≤ 6 nm, for T = 293 K. SOLUTION: From Table 3.1, we have SiO2 :
λp = 0.6 nm at T = 293 K,
Table 3.1.
SiO2 :
k(L λp ) = 1.38 W/mK
Table C.17.
From Table C.17, we have
We note that for λp /L = 0.6(nm)/0.6(nm) = 1, we have k(λp /L = 1) =
1.38(W/mK) = 0.5914 W/mK. 4 1+ ×1 3
For λp /L = 0.6(nm)/6(nm) = 0.1, we have k(λp /L = 0.1) =
1.38(W/mK) = 1.218 W/mK. 4 1 + × 0.1 3
The variation of k as a function of λp /L is shown in Figure Pr.3.9.
k (lp / L), W/mK
k (lp / L 0)
1.380 Silica T = 293 K lp = 0.6 nm
1.104 0.828 0.552 0.276 0 0
0.2
0.4
0.6
0.8
1.0
lp / L Figure Pr.3.9 Predicted variation of the thermal conductivity of a thin ﬁlm, amorphous SiO2 layer as a function of λp divided by the ﬁlm thickness.
COMMENT: Note that 0.6 nm = 6 ˚ A and this means that the meanfree path of the phonon for silicon is only a few lattice lengths. Also note that this conductivity is along the ﬁlm thickness. The conductivity along with the ﬁlm is not the same and it is less aﬀected by the ﬁlm thickness. 142
PROBLEM 3.10.FUN GIVEN: The eﬀective thermal conductivity k is used to describe the conductivity of porous solids (a ﬂuidsolid composite). In many applications requiring a large surface area for surface convection Aku , such as in heat storage in solids, packed bed of particles are used. For example, spherical particles are packed randomly or in an ordered arrangement (e.g., simple, bodycentered or facecentered, cubic arrangement). Figure Pr.3.10(a) shows a simple (also called squarearray) cubic arrangement of particles (porosity = 0.476). Due to their weight or by a contact pressure pc , these elastic particles deform and their contact area changes, resulting in a change in the eﬀective thermal conductivity k. For spheres having a uniform radius R, a Young modulus of elasticity Es , a Poisson ratio νP , and a conductivity ks , the eﬀective conductivity for the case negligible ﬂuid conductivity (kf = 0) and subject to contact pressure pc is predicted as 1/3 (1 − νP2 )pc k = 1.36 . ks Es For aluminum, Es = 68 GPa, νP = 0.25, ks = 237 W/mK. For copper (annealed), Es = 110 GPa, νP = 0.343, ks = 385 W/mK. For magnesium (annealed sheet), Es = 44 GPa, νP = 0.35, ks = 156 W/mK. SKETCH: Figure Pr.3.10(a) shows the particle arrangements, the contact pressure, the equivalent circuit, and the eﬀective thermal conductivity k.
(a) Packed Bed of Spheres in Square Array Arrangement pc
(b) Effective Conductivity, k
T1
Contact Pressure
T1
qk qk
kf = 0
Qk,12 Rk = L Ak k
L
qk T2 Ak
T2
Solid Modulus of Elasticity, Es Young's Modulus, νp Conductivity, ks
Figure Pr.3.10(a) Packed bed of spherical particles with a simple cubic arrangement and a contact pressure pc . (b) The eﬀective thermal conductivity k.
OBJECTIVE: Plot k versus pc for 105 ≤ pc ≤ 109 Pa for packed beds of (i) aluminum, (ii) copper, and (iii) magnesium spherical particles. SOLUTION: Figure Pr.3.10(b) shows the results. Copper has the highest eﬀective conductivity k at any given pressure. COMMENT: Note that relatively high pressures are considered here (105 Pa = 14.7 psi = 1 atm). Similar results are obtained by sintering the particles.
143
Simple Cubic Arrangement of Spheres = 0.476
k , W/mK
100
Cu
10
Al
k
=
Mg
pc1/3
1 105
106
107
108
109
pc , Pa Figure Pr.3.10(b) Variation of the eﬀective conductivity with respect to contact pressure for aluminum, copper, and magnesium.
144
PROBLEM 3.11.FUN.S GIVEN: The crystalline lattice thermal conductivity kp is given by (3.26) as k p = (48π 2 )1/3
3 1 kB T3 g22 (f, T, τp , τp,n ) (f, T, τ ) + g , 1 p lm h2P TD g3 (f, T, τp , τp,n , τp,r )
The integrals gn , gr,1 , and gr,2 and the relaxation times τp,n , τp,r , and τp are deﬁned as
TD /T
g1 =
τp 0
x4 ex dx, g2 = (ex − 1)2
1 1 1 = + , τp τp,n τp,r
TD /T
τp x4 ex dx, g3 = τp,n (ex − 1)2
1
= an
τp,n
2πkB 5 T x, hP
1 τp,r
TD /T
=
τp x4 ex dx, τp,n τp,r (ex − 1)2
i
1 , τp,r,i
where x = (hP f )/(2πkB T ), an is a material constant, and the resistive mechanisms in the summation for tau−1 p,r include the threephonon umklapp processes, τp,r,u , boundary scattering, τp,r,b , point defect scattering, τp,r,p , lattice vacancy scattering, τp,r,v , and phononelectron scattering, τp,r,pe , among others [6]. For alumina (Al2 O3 ), the phononelectron scattering is negligible compared to the other resistive mechanisms, and for simplicity is not considered here. The overall resistive time constant, due to these resistive mechanisms, is then given by 1 τp,r
1
= = A
1
+
+
1
+
1
τp,r,p τp,r,v τp,r,u τp,r,b 4 4 2 2πkB 2πkB 2πkB up T 4 x4 + A T 4 x4 + au T 3 e−TD /(αT ) x2 + , hP hP hP L
where A, au , and α are also material constants, up is the mean phonon velocity, and L is a characteristic length scale of the crystal or grain boundaries. Note that vacancies and point defects behave identically as resistance mechanisms. Consider a single alumina crystal with linear dimension L = 4.12 mm, and the empirically determined material constants, an = 2.7 × 10−13 K−4 , A = 4.08 × 10−46 s3 , au = 1.7 × 10−18 K−1 , and α = 2. Also from Table 3.1, we have TD = 596 K, lm = 0.35 nm, and up = 7,009 m/s. Substituting these values into the expression for the total phonon relaxation time constant, we have 1 τp
=
1 τp,n
+(
1 τp,r
)
= bn T 5 x + (2 × bp T 4 x4 + bu T 3 e−298/T x2 + bb ), where these new bi constants combine the above ai constants with the other coeﬃcients and are bn = 3.535 × 10−2 1/K5 s, bp = 1.199 × 10−1 1/K4 s, bu = 2.914 × 104 1/K3 s, and bb = 1.701 × 106 1/s. Then the expression for the lattice thermal conductivity becomes g 2 (x, T ) k p = bk T 3 g1 (x, T ) + 2 , g3 (x, T ) where bk = 2.240 × 105 W/mK4 . OBJECTIVE: (a) The integrals in the expression for the crystalline lattice thermal conductivity must be evaluated for a given temperature. For various temperatures, between T = 1 and 400 K, use a solver and determine k p , and then plot k p versus T . (b) Compare the result with typical k p vs. T curves for crystalline nonmetals, as shown in Figure 3.7(c). Hint: To avoid overﬂow errors that might occur depending on the solver, factor the T 3 into the brackets containing the gi integrals (i.e., into the bi constants) before solving.
145
SOLUTION: (a) Using a solver, such as SOPHT, the integrations are performed numerically. SOPHT is a diﬀerential solver, and therefore the integrals must be transformed into their associated diﬀerential forms. For example, the integral TD /T x4 ex g1 = τp x dx (e − 1)2 0 is transformed to a diﬀerential as dg1 x4 ex = τp x . dx (e − 1)2 Since the lower limit of the integral is zero, the solver can then be used to solve for g1 (x) with the ﬁnal desired answer being g1 = g1 (x = TD /T ). The source code using SOPHT is then g1’=dg1dx g2’=dg2dx g3’=dg3dx x=t Temp=1 //This is manually changed //Factor in Tempˆ3 from expression for k //Note b’s are in 1/tau’s bn=5.535e2/Tempˆ3 bu=2.914e4/Tempˆ3 bb=1.701e6/Tempˆ3 bpv=1.199e1/Tempˆ3 bk=2.240e5 kern=(xˆ4*exp(x))/(exp(x)1)ˆ2 itaup=1/tau p itaupn=1/tau pn itaupr=1/tau pr itaup=itaupn+itaupr itaupn=bn*Tempˆ5*x itaupr=itauu+itaub+itaupv itauu=bu*Tempˆ3*exp(298/Temp)*xˆ2 itaub=bb itaupv=2*(bpv*Tempˆ4*xˆ4) dg1dx=tau p*kern dg2dx=tau p/tau pn*kern dg3dx=tau p/tau pn/tau pr*kern //Tempˆ3 factored in above k=bk*(g1+g2ˆ2/g3) Note that SOPHT solves initial condition diﬀerential equations using t as the independent variable. Here t has been equated to our x. Each execution of SOPHT at diﬀerent input values of T = Temp must be done for a range of x = 0 up to xD = TD /T . Note that if the initial conditions for t (i.e., x) or g3 are equal to zero, there will be a division by zero in the ﬁrst iterations of the solver execution resulting in an execution error. To avoid this, initial conditions of 1 × 10−10 were used for t (i.e., x) and g3 , initial conditions of zero were used for g1 and g2 , and the iteration was run for 1,000 steps from a start of t = 1 × 10−10 to and end of t = xD = TD /T . For each diﬀerent input value of T , the correct end value of the iteration of t = t(T ) must be entered. Note that for small T (i.e., T < 20 K), x becomes large (i.e., x = t > 30) and the numerator and denominator of kern in the SOPHT program both become large and exceed the capability of the software. Plotting the gi ’s and the k p versus x will show that, for small T , the solution for these variables have already converged to near constant values and that the iterations only need be run to an end of x = t = 25 or 30, instead of x = TD /T , to obtain acceptable predictions. 146
k = k p, W/mK
104 Computed
103 Curve Fit
102
10 SiO2 , Single Crystal 1 1
102
10
103
T, K Figure Pr.3.11 Predicted variation of the lattice thermal conductivity with respect to temperature.
The results are plotted in Figure Pr.3.11 for several temperatures. (b) We note that k p peaks around T = 27 K, where k p is about 6,543 W/mK. Comparing to Figure 3.11(c), the results are similar to those for sodiumﬂuoride, another nonmetal. Note the initial, sharp rise in the lowtemperature region were the interphonon scattering is not signiﬁcant (i.e., τp,r,u → 0). The constants used here slightly underpredict kp at T = 300 K, where the measured value given in Table 3.1 is kp = 36 W/mK and the predicted value is kp = 27.6 W/mK. COMMENT: Note that the boundary scattering will only be signiﬁcant when T is small (since all other scattering mechanisms have a strong, slightly nonlinear dependence on T ). This boundary scattering is addressed in Problem 3.8.FUN.
147
PROBLEM 3.12.FUN GIVEN: Similar to Example 3.6, consider the internal surface of the three surfaces to be covered with an insulation layer of thickness l = 5 cm and thermal conductivity k = 0.1 W/mK. The outside surface is at temperature T1 = 90◦C and the temperature at the inside surface of the insulation is T2 = 40◦C. The surfaces have an outside area A1 = 1 m2 and are of the geometries shown in Figure Pr.3.12, i.e., (a) a planar surface with area A1 = Ly Lz , (b) a cylinder with area A1 = 2πR1 Ly and length Ly = 1 m, and (c) a sphere with surface area A1 = 4πR12 . SKETCH: Figure Pr.3.12 shows the three surfaces to be lined (inside) by insulation. (a) Plane Surface
(b) Cylindrical Surface l
l = Lx
T2
(c) Spherical Surface
l
R1 T1
A1
A1 R1
Ly
T1 T2
Ly
T2
A1
T1 Lz
Figure Pr.3.12 (a), (b), and (c) Three geometries to be lined (inside) with insulation.
OBJECTIVE: For each of these geometries, calculate the rate of heat loss through the vessel surface Qk,2−1 (W). Compare your results with the results of Example 3.6 and comment on the diﬀerences among the answers. Neglect the heat transfer through the ends (i.e., assume a onedimensional heat transfer). SOLUTION: The onedimensional conduction heat ﬂow rate is given by Qk,2−1 =
T2 − T1 . Rk,12
(a) For a plane surface with an area A1 = Ly Lz and thickness l, Rk,12
=
Qk,21
=
0.05(m) l = = 0.5◦C/W Ak k 1(m2 ) × 0.1(W/mK) T2 − T1 40(◦C) − 90(◦C) = −100 W. = Rk,12 0.5(◦C/W)
(b) For a cylinder with an external radius R1 = A1 /2πLy , length Ly = 1 m, and an internal radius R2 = R1 − l Rk,12 =
1 1 )(m)/( 2π − 0.05)(m)] ln[( 2π ln(R1 /R2 ) = 0.600◦C/W = 2πkLy 2π × 0.1(W/mK) × 1(m)
Qk,21 =
T2 − T1 40(◦C) − 90(◦C) = −83.3 W. = Rk,12 0.600(◦C/W) 148
(c) For a sphere with an external radius R1 = (A1 /4π)1/2 and an internal radius R2 = R1 − l Rk,12 =
1/[(4π)−1/2 − 0.05] − 1/(4π)−1/2 1/R2 − 1/R1 = = 0.608◦C/W 4πk 4π × 0.1(W/mK)
Qk,21 =
T2 − T1 40(◦C) − 90(◦C) = −82.3 W. = Rk,12 0.608(◦C/W)
(d) Contrary to results obtained from placing insulation on the outside surface, the rate of heat transfer decreases as the geometry changed from the ﬂat plate to the cylinder and the sphere. The increase in curvature for a cylinder or a sphere, as compared to a ﬂat plate, decreases the available area on the inside surface, for heat transfer, as the radius decreases. As the heat transfer rate Qk,21 (W) is proportional to the heat transfer area Ak , a reduction on the available area for heat transfer causes a reduction on the heat transfer rate. COMMENT: For a given l, as the axis of a cylinder or the center of a sphere is approached, the area for heat transfer decreases, resulting in a largeresistance to heat ﬂow. The heat transfer rate Qk,21 is negative because T2 < T1 . The heat transfer rate Qk,12 has the same magnitude but the opposite sign. The negative sign is mostly a matter of convention and the notation Qk,ij indicates the heat transfer rate from temperature Ti to temperature Tj .
149
PROBLEM 3.13.FUN GIVEN: Consider an inﬁnite plane wall (called a slab) with thickness L = 1 cm, as shown in Figure Pr.3.13. The thermophysical properties of copper and silica aerogel are to be evaluated at 25◦C and 1 atm [Table C.14 and Figure 3.13(a)]. SKETCH: Figure Pr.3.13 shows the onedimensional, steadystate conduction across a slab. L
T1 T2
qk,12 Ak
Figure Pr.3.13 Onedimensional conduction across a slab.
OBJECTIVE: (a) Calculate the conduction thermal resistance Ak Rk,1−2 [◦C/(W/m2 )], if the wall is made of copper. (b) Calculate the conduction thermal resistance Ak Rk,1−2 [◦C/(W/m2 )], if the wall is made of silica aerogel. (c) If the heat ﬂux through the wall is qk,21 = 1,000 W/m2 and the internal wall temperature is T1 = 60◦C, calculate the external wall temperature T2 for the two materials above. (d) Express the results for items (a) and (b) in terms of the Rk value. SOLUTION: (a) From Table C.14, the thermal conductivity of pure copper at 300 K is k = 401 W/mK. The conduction thermal resistance is Ak Rk,12 =
0.01(m) L 2 = = 2.5 × 10−5 ◦C/(W/m ). k 401(W/mK)
(b) From Figure 3.13(a), at p = 1 atm, the thermal conductivity of silica aerogel at 300 K is k = 0.0135 W/mK. This value is for conditions close to our speciﬁed conditions of p = 1 atm and T = 298 K, therefore this would be a better value to use than one linearly extrapolated from Table C.15. The conduction thermal resistance is Ak Rk,12 =
0.01(m) L 2 = = 7.4 × 10−1 ◦C/(W/m ). k 0.0135(W/mK)
(c) The rate of heat ﬂow per unit area through the wall qk,12 (W/m2 ) is qk,12 =
Qk,12 T1 − T2 = . Ak Ak Rk,12
For a heat ﬂow per unit area of qk,12 = 1,000 W/m2 , an internal wall temperature of T2 = 60◦C, the external wall temperature T1 for each of the two cases above is (i) copper T1 = T2 + qk,12 (Ak Rk,12 ) = 60(◦C) + 1,000(W/m ) × 2.5 × 10−5 [◦C/(W/m )] = 60.02◦C, 2
150
2
(ii) silica aerogel T1 = T2 + qk,12 (Ak Rk,12 ) = 60(◦C) + 1,000(W/m ) × 7.4 × 10−1 [◦C/(W/m )] = 800◦C. 2
2
(d) The Rk value for each of the situations above is (i) copper Rk value =
0.0328(ft) L = 1.4 × 10−4 ◦F/(Btu/hr), = 2 Ak k 1(ft ) × 231.9(Btu/hrft◦F)
(ii) silica aerogel Rk value =
0.0328(ft) L = = 4.2◦F/(Btu/hr). 2 Ak k 1(ft ) × 0.00781(Btu/hrft◦F)
COMMENT: The conversion factor for thermal conductivity from W/mK to Btu/hrft◦F is obtained from Table C.1(a).
151
PROBLEM 3.14.FAM GIVEN: A furnace wall (slab) is made of asbestos (ρ = 697 kg/m3 ) and has a thickness L = 5 cm [Figure Pr.3.14(i)]. Heat ﬂows through the slab with given inside and outside surface temperatures. In order to reduce the heat transfer (a heat loss), the same thickness of asbestos L is split into two with an air gap of length La = 1 cm placed between them [Figure Pr.3.14(ii)]. Use Tables C.12 and C.17 to evaluate the conductivity at 273 K or 300 K. SKETCH: Figure Pr.3.14 shows the insulating furnace wall with and without an air gap.
(i) Asbestos Wall
(ii) Asbestos Wall with Air Gap
Asbestos T1
Air T2
Asbestos
T1
qk,12
T2
qk,12
L
L/2
La
L/2
Figure Pr.3.14 A furnace wall. (i) Insulated without an air gap. (ii) With an air gap.
OBJECTIVE: Determine how much the heat ﬂow out of the wall would decrease (show this as a percentage of the heat ﬂow without the air gap). SOLUTION: The conduction heat ﬂux through the wall is Ak qk,12 =
T1 − T2 . Ak Rk,12
The conduction resistance for a plane wall is Ak Rk,12 =
L . ks
From Table C.17, for asbestos with ρ = 697 kg/m3 at T = 273 K, ks = 0.23 W/mK, and the conduction resistance becomes 0.05(m) 2 = 0.217 K/(W/m ). Ak Rk,12 = 0.23(W/mK) For the composite wall, the conduction heat ﬂow through the wall is Ak qk,12 = where Ak (Rk,Σ )12 =
T1 − T2 , Ak (Rk,Σ )12
L/2 La L/2 L La + + = + . ks ka ks ks ka
From Table C.12, for air at T = 300 K, ka = 0.0267 W/mK, and the equivalent conduction resistance becomes 2
Ak Rk,Σ,12 = 0.217[K/(W/m )] +
0.01(m) 2 = 0.591 K/(W/m ). 0.0267(W/mK) 152
The percentage of reduction of the heat ﬂow through the wall is (Qk,12 )i − (Qk,12 )ii × 100 (Qk,12 )i
=
=
1 1 − Ak Rk,12 Ak Rk,Σ,12 1 Ak Rk,12 1 1 − 2 2 0.217[K/(W/m )] 0.591[K/(W/m )] × 100% = 63.3%. 1 2
0.217[K/(W/m )] The heat ﬂow rate through the composite wall is 63.3% lower than the heat ﬂow through the solid wall. COMMENT: Here the airgap resistance is in series with the wall resistance. This is the most eﬀective use of the air gap. When the airgap resistance is placed in parallel, it is not as eﬀective.
153
PROBLEM 3.15.FUN GIVEN: A low thermalconductivity composite (solidair) material is to be designed using alumina as the solid and having the voids occupied by air. There are three geometric arrangements considered for the solid and the ﬂuid. These are shown in Figure Pr.3.15. For all three arrangements , the fraction of volume occupied by the ﬂuid (i.e., porosity) is the same. In the parallel arrangement, sheets of solid are separated by ﬂuid gaps and are placed parallel to the heat ﬂow direction. In the series arrangement, they are placed perpendicular to the heat ﬂow. In the random arrangement, a nonlayered arrangement is assumed with both solid and ﬂuid phase continuous and the eﬀective conductivity is given by (3.28). SKETCH: Figure Pr.3.15 shows the three geometries for solidﬂuid arrangements. Parallel Arrangement Fluid
Composite Insulation x
Solid
T2
Series Arrangement
T1
Fluid
Qk,12(W) Solid Random Arrangement Fluid
L
Solid
Figure Pr.3.15 Three solidﬂuid arrangements for obtaining a low thermal conductivity composite.
OBJECTIVE: (a) Show that the eﬀective thermal conductivity for the parallel arrangement is given by k = kf + ks (1 − ) and that, for the series arrangement, the eﬀective thermal conductivity is given by 1 (1 − ) = + . k kf ks The porosity is deﬁned as the volume occupied by the ﬂuid divided by the total volume of the medium, i.e., =
Vf . Vf + Vs
Also note that 1 − = Vs /(Vf + Vs ). (b) Compare the eﬀective conductivity for the three arrangements for = 0.6 using the conductivity of alumina (Table C.14) and air (Table C.22) at T = 300 K. (c) Comment regarding the design of lowconductivity composites.
154
SOLUTION: (a) (i) Series Arrangement: The equivalent thermal resistance for resistances arranged in series is given by Rk,Σ =
n
Rk,i .
i=1
For n layers of ﬂuid and solid placed in series, the summation becomes Rk,Σ = n(Rk,f + Rk,s ). The solid and ﬂuid resistances are Rk,f =
Lf kf Af
, Rk,s =
Ls . ks As
The equivalent resistance can be expressed in terms of the eﬀective thermal conductivity k as L . kA
Rk,Σ =
From the equations above and noting that Af = As = A and L = n(Lf + Ls ), we obtain Lf + Ls Lf Ls = + , k kf ks or rearranging, we have 1 = k
Lf Lf + Ls
1 + kf
Ls Lf + Ls
1 . ks
The porosity is deﬁned as =
Vf . Vf + Vs
The volumes of the ﬂuid and solid phases can be rewritten as Vf = nAf Lf and Vs = nAs Ls , respectively. Then noting that Af = As , the porosity can be rewritten as =
Lf . Lf + Ls
Also, note that Vs Vs + Vf − Vf Vs + Vf Vf = = − =1− Vf + Vs Vf + Vs Vf + Vs Vf + Vs and using the expressions for the volumes Ls = 1 − . Lf + Ls Therefore, substituting we have 1 1− = + . k kf ks (ii) Parallel Arrangement: The equivalent thermal resistance for resistances arranged in parallel is 1 Rk,Σ
=
n
1 . R k,i i=1
For n layers of ﬂuid and solid arranged in parallel, we have 1 1 1 =n + . Rk,Σ Rf Rs 155
As before, the solid and ﬂuid resistances are Rk,f =
Lf kf Af
,
Rk,s =
Ls ks As
and the equivalent resistance can be expressed in terms of the eﬀective thermal conductivity k as L . kA
Rk,Σ =
For the parallel arrangement Lf = Ls = L and A = n(Af + As ). Then we have k(Af + As ) = kf Af + ks As , which can be rearranged as
k =
Af Af + As
kf +
As Af + As
ks .
The volumes of the ﬂuid and solid phases, as before, can be rewritten as Vf = nAf Lf and Vs = nAs Ls . Then noting that Lf = Ls , we can write =
Af Af + As
1− =
,
As . As + Af
Therefore, from the equations above, we have k = kf + (1 − )ks . (b) The thermal conductivities of alumina and air are alumina: air:
T = 300 K, T = 300 K,
ks = 36 W/mK kf = 0.0267 W/mK
Table C.14 Table C.22.
For each of the arrangements, for = 0.6, the eﬀective thermal conductivities k are Series: 1 0.4 1− 0.6 = + ⇒ k = 0.044 W/mK. + = k kf ks 0.0267 36 Parallel: k = kf + (1 − )ks = 0.6 × 0.0267 + 0.4 × 36 = 14.4 W/mK. Random: Using (3.28) k = kf
ks kf
0.280−0.757 log()−0.057 log(ks /kf ) ,
we obtain k = 0.0267
36 0.0267
0.280−0.757 log(0.6)−0.057 log(36/0.0267) = 0.19 W/mK.
(c) The series arrangement leads to the lowest thermal conductivity possible for a medium composed of solid and ﬂuid thermal resistances and for a given porosity. In the design of an insulating material, one should attempt to approach that limit. COMMENT: The series resistance allows for the high resistance to dominate the heat ﬂow path.
156
PROBLEM 3.16.FAM GIVEN: During hibernation of warmblooded animals (homoisotherms), the heart beat and the body temperature are lowered and in some animals the body waste is recycled to reduce energy consumption. Up to 40% of the total weight may be lost during the hibernation period. The nesting chamber of the hibernating animals is at some distance from the ground surface, as shown in Figure Pr.3.16(a)(i). The heat transfer from the body is reduced by the reduction in the body temperature T1 and by the insulating eﬀects of the body fur and the surrounding air (assumed stagnant). A simple thermal model for the steadystate, onedimensional heat transfer is given in Figure Pr.3.16(a)(ii). The thermal resistance of the soil can be determined from Table 3.3(a). An average temperature T2 is used for the ground surrounding the nest. The air gap size Ra − Rf is an average taken around the animal body. R1 = 10 cm, Rf = 11 cm, Ra = 11.5 cm, T1 = 20◦C, T2 = 0◦C. Evaluate air properties at T = 300 K, use soil properties from Table C.15, and for fur use Table C.15 for hair. SKETCH: Figure Pr.3.16(a) shows a simple thermal model with conduction heat transfer through the fur, air, and surrounding ground.
(i) Diagram of Woodchuck Home
(ii) Simple Thermal Model Surface, T2
Side Entrance Nest Chamber
Soil, ks
Main Entrance
L
T1
Ra
Rf
Woodchuck Body Fur, kf
R1 Air, ka Sr,c r
, T = T
2
Figure Pr.3.16(a) Conduction heat transfer from a warmblooded animal during hibernation. (i) Diagram of woodchuck home. (ii) Thermal model.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine Q12 for (i) L = 2.5Ra , and (ii) L = 10Ra . SOLUTION: (a) Figure Pr.3.16(b) shows the thermal circuit diagram, starting from the body node T1 and after encountering the resistances Rk,1f , Rk,f a , Rk,a2 , node T2 , which is the farﬁeld thermal condition, is reached.
Q12 T1 T2 Rk,a2
Rk,fa
Rk,1f
Sr,c
Figure Pr.3.16(b) Thermal circuit diagram.
(b) From Figure Pr.3.16(b), we have Q12 =
T1 − T2 . Rk,1f + Rk,f a + Rk,a2 157
From Table 3.2, we have 1 1 − R1 Rf Rk,1f = 4πkf 1 1 − Rf Ra Rk,f a = . 4πka From Table 3.3(a), we have Ra 2L Rk,a2 = 4πks Ra 1−
for L > 2Ra .
The thermal conductivities are kf = 0.036 W/mK
Table C.15
ka = 0.0267 W/mK ks = 0.52 W/mK
Table C.15.
Table C.22
Using the numerical values, we have
Rk,1f
=
Rk,f a
=
(i) Rk,a2
=
(ii) Rk,a2
=
1 1 − 0.10(m) 0.11(m) = 2.010 K/W 4π × 0.036(W/mK) 1 1 − 0.11(m) 0.115(m) = 1.178 K/W 4π × 0.0267(W/mK) 1 1− 5 = 1.065 K/W 4π × 0.52(W/mK) × 0.115(m) 1 1− 20 = 1.264 K/W. 4π × 0.52(W/mK) × 0.115(m)
Then (i) Q12
=
(ii) Q12
=
(20 − 0)(K) = 4.703 W (2.010 + 1.178 + 1.065)(K/W) (20 − 0)(K) = 4.492 W. (2.010 + 1.178 + 1.264)(K/W)
There is only a slightly larger Q12 for the nest closer to the surface. COMMENT: Note that the conductivity of fur we used is for the direction perpendicular to the ﬁbers and this is lower than what is expected along the ﬁber (because the ﬁbers have a higher conductivity than the air ﬁlling the space between the ﬁbers). Also note that for L Ra , the results of Tables 3.2 and 3.3(a), for Rk , are identical (as expected).
158
PROBLEM 3.17.FAM GIVEN: A spherical aluminum tank, inside radius R1 = 3 m, and wall thickness l1 = 4 mm, contains liquidvapor oxygen at 1 atm pressure (Table C.26 for T1 = Tlg ). The ambient is at a temperature higher than the liquidgas mixture. Under steadystate, at the liquidgas surface, the heat ﬂowing into the tank causes boil oﬀ at a rate M˙ lg = M˙ g . In order to prevent the pressure of the tank from rising, the gas resulting from boil oﬀ is vented through a safety valve. This is shown in Figure Pr.3.17(a). Then, to reduce the amount of boiloﬀ vent M˙ g (kg/s), insulation is added to the tank. First a low pressure (i.e., evacuated) air gap, extending to location r = R2 = 3.1 m, is placed where the combined conductionradiation eﬀect for this gap is represented by a conductivity ka = 0.004 W/mK. Then a layer of lowweight pipe insulation (slag or glass, Table C.15) of thickness l2 = 10 cm is added. The external surface temperature is kept constant at T2 = 10◦C. Evaluate the thermal conductivity of aluminum at T = 200 K. SKETCH: Figure Pr.3.17(a) shows a tank containing cryogenic liquid oxygen and having heat leaking into the tank from its higher ambient temperature.
Vented Gas, Mg
Cryogenic Tank (Liquid O2) LowPressure Air, Combined RadiationConduction Effect is Shown with ka
Liquid Gas Phase Change Heat Leak into Tank
Tank Pressure, pg
Insulation
Gaseous O2
Aluminum
Liquid O2
Qk,21
r
2'
Slg
2
T1 = Tlg (at pg) Outside Surface Temperature, T2 l1
R1
1 1' R2 l2
Figure Pr.3.17(a) A tank containing a cryogenic liquid and having heat leak to it from a higher temperature ambient.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the rate of heat leak Qk,2−1 . (c) Determine the amount of boil oﬀ M˙ g . (d) Determine the temperature at the innersurface (r = R2 ) of the insulation T2 . SOLUTION: (a) The thermal circuit diagram for this heat ﬂow is shown in Figure Pr.3.17(b). The temperature at the inner surface of the insulation layer is labeled as T2 , and the outer surface of the aluminum shell as T1 . (b) From the diagram, we have
Qk,21 =
T2 − T1 . Rk,11 + Rk,1 2 + Rk,2 2 159
A1 = Alg
Aluminum Shell T1' Rk,11'
T1 = Tlg
Air Shell Rk,1'2'
T2'
Insulation Shell Rk,2'2
T2
Q1 = 0 Slg = − Mlg ∆hlg
Qk,21
Figure Pr.3.17(b) Thermal circuit diagram.
From Table 3.2, we have for a spherical shell
Rk,11
=
Rk,1 2
=
Rk,2 2
=
1 1 − R1 R 1 + l1 4πkAl 1 1 − R 1 + l1 R 2 4πka 1 1 − R 2 R 2 + l 2 . 4πki
From Table C.26, we have T1 = Tlg
=
90.18 K
∆hlg
=
2.123 × 105 J/kg
Table C.26.
From Table C.14, we have (at T = 200 K) kAl = 237 W/mK
Table C.14.
From Table C.15, we have (for low weight pipe insulation) ki = 0.033 W/mK
Table C.15.
Using the numerical values, we have
Rk,11
=
Rk,1 2
=
Rk,2 2
=
1 1 − 3(m) (3 + 0.004)(m) = 1.490 × 10−7 K/W 4π × 237(W/mK) 1 1 − (3 + 0.004)(m) 3.1(m) = 2.052 × 10−1 K/W 4π × 0.004(W/mK) 1 1 − 3.1(m) (3.1 + 0.1)(m) = 2.432 × 10−2 K/W. 4π × 0.033(W/mK)
The largest resistance is that of lowpressure air. Then Qk,21
= =
[(10 + 273.15) − 90.18](K) (1.490 × 10 + 2.052 × 10−1 + 2.432 × 10−2 )(K/W) 192.97 (W) = 8.408 × 102 W. 2.295 × 10−1 −7
(c) The boil oﬀ is determined from the energy equation for the surface T1 . With no other surface heat transfer for surface node T1 , we have (3.87) as Q1 = Qk,21 = −S˙ lg
energy equation for node T1 ,
160
where from Table 2.1, S˙ lg = −m ˙ lg Alg ∆hlg
energy conversion by phase change.
There is a minus sign because heat is absorbed during evaporation. Then from the above two equations, we have Qk,21 M˙ lg = . ∆hlg Using the numerical values, we have 8.408 × 102 (W) M˙ lg = = 3.960 × 10−3 kg/s = 3.960 g/s. 2.123 × 105 (J/kg) (d) The temperature T2 , as shown in Figure Pr.3.17(b), is found by the thermal circuit diagram, i.e., Qk,21 = Qk,22 =
T 2 − T 2 Rk,2 2
or T 2
= T2 − Qk,22 Rk,2 2 = 10(◦C) − 8.408 × 102 (W) × 2.432 × 10−2 (◦C/W) =
10(◦C) − 20.45(◦C) = −10.45◦C.
COMMENT: This heat leak rate, Qk,21 = 840.8 W, is considered large. Additional insulation is required to reduce the heat leak rate.
161
PROBLEM 3.18.FAM GIVEN: A teacup is ﬁlled with water having temperature Tw = 90◦C. The cup is made of (i) porcelain (Table C.15), or (ii) stainless steel 316. The cupwall inside diameter is R and its thickness is L. These are shown in Figure Pr.3.18(a). The water is assumed to be well mixed and at a uniform temperature. The ambient air is otherwise quiescent with a farﬁeld temperature of Tf,∞ , and adjacent to the cup the air undergoes a thermobuoyant motion resulting a surfaceconvection resistance Rku . Tf,∞ = 20◦C, L = 3 mm, Aku Rku = 10−3 K/(W/m2 ). Use L R to approximate the wall as a slab and use Aku = Ak . SKETCH: Figure Pr.3.18(a) shows the cup wall, the uniform water temperature Tw , the farﬁeld temperature Tf,∞ , and the surface convection resistance Rku . Thermobuoyant Air Motion Air
Tf, uf,
g L Tw R
AkuRku
Cup (i) Porcelain Wall (ii) Stainless Steel 316 Ts
Figure Pr.3.18(a) A cup ﬁlled with hot water. The cup is made of (i) porcelain, or (ii) stainless steel 316. The ambient air is otherwise quiescent with a thermobuoyant motion adjacent to the cup wall.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the cup outside surface temperature Ts for cases (i) and (ii). SOLUTION: (a) The thermal circuit diagram for the heat ﬂowing through the cup wall is shown in Figure Pr.3.18(b). Tf,
Rku
Ts
Rk,ws
Tw
Qw
Figure Pr.3.18(b) Thermal circuit diagram.
(b) To determine Ts , we note that the heat ﬂows through the cup wall and the adjacent thermobuoyantmotion resistance, as shown Figure Pr.3.18(b). Then Qw∞ =
Tw − Ts Ts − Tf,∞ = , Rk,ws Rku
or solving for Ts , we have Rku (Tw − Ts ) + Rk,ws (Tf,∞ − Ts ) = 0 or Ts =
Rku Tw + Rk,ws Tf,∞ . Rku + Rk,ws 162
Now, from Table 3.2, we have (for L R) Rk,ws =
L . Ak kw
Then using Aku = Ak , we have L Tf,∞ kw . L Aku Rku + kw
Aku Rku Tw + Ts =
Now, from Tables C.15 and C.16, we have (i) porcelain: kw = (ii) stainless steel 316: kw =
1.5 W/mK 13 W/mK
Table C.15 Table C.16.
Determining Ts , we have 3 × 10−3 (m) × (20 + 273.15)(K) 1.5(W/mK) (10−3 + 3 × 10−3 /1.5)[K/(W/m2 )]
10−3 [K/(W/m2 )] × (90 + 273.15)(K) + (i) Ts
=
Ts
=
(ii) Ts
= =
(3.632 × 10−1 + 5.863 × 10−1 ) = 316.5 K = 43.35◦C 3.000 × 10−3 3 × 10−3 (m) × (20 + 273.15)(K) 10−3 [K/(W/m2 )] × (90 + 273.15)(K) + 13(W/mK) (10−3 + 3 × 10−3 /13)[K/(W/m2 )] 3.632 × 10−1 + 6.765 × 10−2 = 350.1 K = 76.92◦C. 10−3 + 2.308 × 10−4
COMMENT: The temperature sensor at the surface of the human ﬁngers would sense Ts = 43.35◦C as warm and tolerable and Ts = 76.92◦C as hot and intolerable.
163
PROBLEM 3.19.FAM GIVEN: Gaseous combustion occurs between two plates, as shown in Figure Pr.3.19(a). The energy converted by combustion S˙ r,c in the gas ﬂows through the upper and lower bounding plates. The upper plate is used for surface radiation heat transfer and is made of solid alumina (Table C.14). The lower plate is porous and is made of silica (Table C.17, and include the eﬀect of porosity). The porosity = 0.3 and the randomly distributed pores are ﬁlled with air (Table C.22, use T = Ts,2 ). Each plate has a length L, a width w, and a thickness l. The outsides of the two plates are at temperatures Ts,1 and Ts,2 . S˙ r,c = 104 W, Ts,1 = 1,050◦C, Ts,2 = 500◦C, L = 0.3 m, w = 0.3 m, l = 0.02 m. SKETCH: Figure Pr.3.19(a) shows combustion occurring between two plates, one plate is a conductor and the other an insulator. w Uniform Gas Temperature Tg
L Plate 1 (Solid Alumina) Radiant Heating Plate Outside Temperature, Ts,1 Sr,c Plate 2 (Porous Silica) Insulating Plate Outside Temperature, Ts,2
l
Figure Pr.3.19(a) Combustion between two plates; one plate is a conductor while the other is an insulator.
OBJECTIVE: (a) Draw the steadystate thermal circuit diagram. (b) Determine the eﬀective conductivity of the lower plate. (c) Determine the uniform gas temperature Tg . (d) Determine the fraction of heat ﬂow through each plate. SOLUTION: The gas temperature is spatially uniform and the inner walls of plates 1 and 2 are at Tg . There is 1D conduction through the plates. (a) The thermal circuit diagram is shown in Figure Pr.3.19(b). Q1 Ts,1 Rk,g1
Qk,g1 Tg
.
Sr,c Rk,g2
Qk,g2 Ts,2 Q2
Figure Pr.3.19(b) Thermal circuit diagram.
(b) The lower plate has a = 0.3 and consists of solid silica with randomly distributed pores ﬁlled with air. Since the combustion is our source of heat, we expect Ts,2 to be the lowest temperature in plate 2. For lack of a more appropriate value, we evaluate all the properties of the plate at this temperature. We also assume that the air 164
occupying the pore space is in thermal equilibrium with the solid and thus evaluate the gas properties at this temperature as well. Therefore, we have at Ts,2 = 500◦C = 773.15 K, (Tables C.17 and C.22) ks
=
1.38 W/mK
kg
=
0.0544 W/mK
(at T=293 K, only data available) by interpolation
Table C.17 Table C.22.
Using Equation 3.28 for the eﬀective conductivity of a random porous medium, we have ks kg k kg
1.38(W/mK) = 25.37 0.0544(W/mK) 0.280−0.757 log−0.057 log(ks /kf ) ks = kg
k k
= =
=
(0.0544W/mK)(25.37)0.280−0.757 log(0.3)−0.057 log(25.37) 0.373 W/mK.
(c) Applying the conservation of energy around the gas node in the ﬁgure for steady state conditions, we have Qk,g1 + Qk,g2 Tg − Ts,1 Tg − Ts,2 + Rk,1 Rk,2
= −(ρc)g V
dTg + S˙ r,c dt
= S˙ r,c ,
where plate 1 is assumed to be at Ts,1 = 1,323 K, at which k1 = kAl2 O3 = 5.931. Then, the thermal conduction resistance is Rk,1
=
Rk,2
=
0.02(m) ◦ = = 2 = 0.0375 C/W k1 A k1 Lw 5.931(W/mK) × (0.3 × 0.3)(m) 0.02(m) ◦ = = 2 = 0.596 C/W. k2 A kLw 0.373(W/mK) × (0.3 × 0.3)(m)
Substituting these values into the conservation of energy equation and solving for Tg gives Tg − 1,323.15(K) Tg − 773.15(K) + 0.037(◦C/W) 0.596(◦C/W) Tg
=
104 W
=
1,643 K.
(d) The conservation of energy equation says that the fraction of energy going through the top plate Qk,g1 plus the fraction of energy going through the bottom plate Qk,g2 is equal to the amount of energy being generated S˙ r,c . Therefore, Qk,g1 S˙ r,c
=
Qk,g2 S˙ r,c
=
Tg − Ts,1 8539.7(W) Rk,1 = 0.854 of the energy generated ﬂows through the top plate = ˙ 10,000(W) Sr,c Tg − Ts,2 Rk,2 1460.13(W) = 0.146 of the energy generated ﬂows through the bottom plate. = ˙ 10,000(W) Sr,c
COMMENT: The ratio of the two heat ﬂow rates is 5.85. This can be further improved by increasing the porosity and thickness of the insulation.
165
PROBLEM 3.20.FAM GIVEN: In IC engines, during injection of liquid fuel into the cylinder, it is possible for the injected fuel droplets to form a thin liquid ﬁlm over the piston. The heat transferred from the gas above the ﬁlm and from the piston beneath the ﬁlm causes surface evaporation. This is shown in Figure Pr.3.20(a). The liquidgas interface is at the boiling temperature, Tlg , corresponding to the vapor pressure. The heat transfer from the piston side is by onedimensional conduction through the piston and then by onedimensional conduction through the thin liquid ﬁlm. The surfaceconvection heat transfer from the gas side to the surface of the thin liquid ﬁlm is prescribed as Qku . Qku = −13,500 W, ∆hlg = 3.027 × 105 J/kg (octane at one atm pressure, Table C.4), kl = 0.083 W/mK (octane at 360 K, Table C.13), Tlg = 398.9 K (octane at 1 atm pressure, Table C.4), ρl = 900 kg/m3 , ks = 236 W/mK (aluminum at 500 K, Table C.14), T1 = 500 K, L = 3 mm, l = 0.05 mm, D = 12 cm. SKETCH: Figure Pr.3.20(a) shows the liquid ﬁlm being heated by surface convection and by substrate conduction.
Qku
Liquid Film
Tlg , kl L
l
T1
ks
Piston Cylinder
D
Figure Pr.3.20(a) An IC engine, showing liquid ﬁlm formation on top of the piston.
OBJECTIVE: (a) Draw the thermal circuit diagram and write the corresponding energy equation for the liquidgas interface. (b) For the conditions given, determine the rate of evaporation of the liquid ﬁlm, M˙ lg (kg/s). (c) Assuming that this evaporation rate remains constant, determine how long it will take for the liquid ﬁlm to totally evaporate. SOLUTION: (a) The thermal circuit is shown in Figure Pr.3.20(b).
Rk,12
. Slg
Rk,23
Q1
Qku T2 = Ts
T1 Qk,12
T3 = Tlg Qk,23
Figure Pr.3.20(b) Thermal circuit diagram.
The energy equation for node T3 is Qku −
Qk,13 T1 − T3 Qku − Rk,Σ 166
= S˙ lg = −M˙ lg ∆hlg .
(b) M˙ lg will increase as the ﬁlm thickness decreases, since Rk,23 decreases. For the conditions given, we assume a quasisteady state. Then we have from Figure Pr.3.20(b) for Rk,Σ Rk,Σ
= Rk,12 + Rk,23 l 0.003(m) 0.00005(m) L + = + = 2 ks A kl A π × 0.12 π × 0.122 (m2 ) 2 236(W/mK) × ) 0.083(W/mK) × (m 4 4 = 0.001124(K/W) + 0.05326(K/W) = 0.0544 K/W.
Therefore, from the energy equation we have −13,500(W) −
(500 − 398.9)(K) 0.0544(K/W)
= −M˙ lg × 3.027 × 105 (J/kg).
Then M˙ lg = 0.051 kg/s. (c) From (b) we noted that M˙ lg will increase as the ﬁlm thickness decreases. If we assume M˙ lg to be constant, we can ﬁnd an upper limit to the amount of time it would take to completely evaporate the liquid ﬁlm. Then M˙ lg
= =
dM = constant dt ρl V ρl Al Mi − Mf Mi − 0 = = = ti − tf ∆t ∆t ∆t 2
∆t
=
(m2 ) × 0.00005(m) 900(kg/m3 ) × π×0.12 ρl Al 4 = 0.01 s. = 0.051(kg/s) M˙ lg
COMMENT: The heat conduction from the piston is only a small fraction of the heat supplied to the liquid ﬁlm.
167
PROBLEM 3.21.FUN GIVEN: A twodimensional, periodic porous structure has the solid distribution shown in Figure Pr.3.21(a). This is also called a regular lattice. The steadystate twodimensional conduction can be shown with a onedimensional, isotropic resistance for the case of kA kB . Use a depth w (length perpendicular to the page). SKETCH: Figure Pr.3.21(a) shows the solid geometry which is a continuous zigzag arm of thickness l in a periodic structure with length L between each arm.
Unit Cell
Vacuum Q k,12 kA = 0
l L kB Solid
Figure Pr.3.21(a) A twodimensional, periodic structure composite with material A having a conductivity much smaller than B.
OBJECTIVE: (a) Draw the thermal circuit model. (b) Show that for kA /kB 1, the eﬀective thermal conductivity k is k = 1 − 1/2 , kB
kA
1, kB
where is the porosity (void fraction) deﬁned by (3.27). SOLUTION: (a) Figure Pr.3.21(b) shows the thermal circuit model for the unit cell.
(i) Actual Thermal Circuit Model
(ii) Equivalent Thermal Circuit Model
T1' Rk,11'
Qk,12 Rk,1'2
T1
T2
Rk,12
Qk,12
T1 Rk,11'
T2 L Ak k
Rk,1'2 T1'
Figure Pr.3.21(b) Thermal circuit diagram and the equivalent circuit.
168
(b) The overall conduction resistance Rk,Σ for this circuit is, based on (3.78) and (3.82), 1 Rk,12
= =
1 1 + Rk,11 + Rk,1 2 Rk,11 + Rk,1 2 2 , Rk,11 + Rk,1 2
where Ak Rk,11 1 Rk,11
= lw = Rk,1 2 = =
L+l lwkB
lwkB . L+l
The porosity (3.27) can be shown to be =
VA 2L2 L2 = . 2 = VA + VB (L + l)2 2(L + l) 21/2
Now using Rk,12 ≡
1 L+l L1 = = , L1 wk wk lwkB
where L is the arm length. Finally, we have k l = 1 − 1/2 . = kB L+l
COMMENT: Note that for = 0, i.e., L = 0, we recover k = kB and for = 1, i.e., l = 0, recover k = 0 (no heat transfer through vacuum). Also note that this eﬀective resistance is a combination of series and parallel resistance.
169
PROBLEM 3.22.FUN GIVEN: The eﬀective thermal conductivity of twodimensional, periodicstructure (i.e., regular lattice) composites can be estimated using onedimensional resistance models. Figure Pr.3.22 shows a simple, twodimensional unit cell with material B being continuous and material A being the inclusion [similar to the threedimensional, periodic structure of Section 3.3.2(C)]. Use only the porosity (void fraction) and the conductivities kA and kB . Use a depth w (length perpendicular to the page). SKETCH: Figure 3.22(a) shows the composite with material B being continuous and material A being the inclusion.
Unit Cell Qk,12
l Material B Material A
kA L kB
Figure Pr.3.22(a) A twodimensional, periodic structure with material B being continuous.
OBJECTIVE: (a) Derive an expression for the eﬀective conductivity k of this composite using a seriesparallel arrangement of resistances. (b) Derive an expression for the eﬀective conductivity k of this composite using a parallelseries arrangement of resistances. (c) Show that for the case of kA /kB 1, the result for the parallelseries arrangement is k = 1 − 1/2 , kB where is the porosity (this result is also obtained in Problem 3.34). SOLUTION: The seriesparallel and parallelseries resistance arrangements of the Figure Pr.3.22(a) structure are shown in Figure Pr.3.22(b). (a) For the circuit shown in Figure Pr.3.22(b)(i), we have 1
Rk,12 = Rk,11 ,B +
1 Rk,1 2 ,B
+
1
+ Rk,2 2,B .
Rk,1 2 ,A
The individual resistances are Rk,11 ,B
= Rk,2 2,B =
Rk,1 2 ,A
=
Rk,1 2 ,B
=
l (L + 2l)wkB
L LwkA L . 2lwkB
The porosity of the structure is deﬁned as =
VA L2 = . VA + VB (L + 2l)2 170
(i) SeriesParallel Arrangement Rk,1'2',A Rk,11',B
T1
1'
Rk,2'2,B
2'
Qk,12
T2
Rk,1'2',B
(ii) ParallelSeries Arrangement Rk,12,B Qk,12 T1
T1'
Rk,11',B
T2'
Rk,1'2',A
T2
Rk,2'2,B
Figure Pr.3.22(b) Thermal circuit diagram for the two resistance arrangements.
Then L + 2l (L + 2l)wk 2l L + . 2lwkB + LwkA (L + 2l)wkB
≡
Rk,12
= From these, we have
kB 1/2 = + 1 − 1/2 . 1/2 k 1 − + 1/2 (kA /kB ) Note that this does not lead to k/kB = 1 − 1/2 for the case of kA kB . (b) For the circuit shown in Figure Pr.3.22(b)(ii), we have Rk,12 =
1 Rk,1
1 ,B
+R
k,1

2 ,A
+R
k,2
2,B
+
1 Rk,12,B
.
The individual resistances are Rk,11 ,B
=
Rk,1 2 ,A
=
Rk,2 2,B
=
Rk,12,B
=
l LwkB L LwkA l LwkB L + 2l . 2lwkB
The porosity is the same as determined above. Then 1 Rk,12
= =
=
L + 2l (L + 2l)wk 2lwkB 1 + 2l L L + 2l + LwkB LwkA kB 1/2 + (1 − )1/2 . 1/2 + (1 − 1/2 )(kA /kB ) 171
From this, we have kB 1/2 + (1 − 1/2 )(kA /kB ) = . k (1 − 1/2 ) 1/2 + 1/2 (kA /kB ) + (1 − 1/2 )2 (kA /kB ) (c) For the case of the parallelseries arrangement, and for kA /kB 1, the above equation becomes kB 1 = k 1 − 1/2
or
k = 1 − 1/2 . kB
COMMENT: The actual, twodimensional heat ﬂow results in an eﬀective resistance that is between these two eﬀective resistances. As kA /kB becomes closer to unity, the diﬀerence between the models decreases and vice versa for kA /kB far from unity. Note that the threedimensional parallelseries solution given by (3.86) gives, for kA kB , k/kB = (1 − 2/3 )/(1 − 2/3 + ). This gives a higher k/kB , compared to the twodimensional result.
172
PROBLEM 3.23.FUN GIVEN: In the onedimensional, steadystate conduction treatment of Section 3.3.1, for planar geometries, we assumed a constant crosssectional area Ak . In some applications, although the conduction is onedimensional and cross section is planar, the crosssectional area is not uniform. Figure Pr.3.23 shows a rubberleg used for the vibration isolation and thermal insulation of a cryogenic liquid container. The rubber stand is in the form of truncated cone [also called a frustum of right cone, a geometry considered in Table C.1(e)]. Note that ∆V = πR2 (x)∆x, as ∆x → 0. T2 = 20◦C, T1 = 0◦C, L2 = 4 cm, L1 = 10 cm, R1 = 1.5 cm, k = 0.15 W/mK. SKETCH: Figure Pr.3.23 shows the rubber leg, its geometry and parameters, and the onedimensional heat conduction. x qk
T1
L1
Ak(x) R1 R R(x) = 1 x L1
Ideally Insulated q=0 L2
R2 T2 > T1
qk
Rubber Leg
Figure Pr.3.23 Onedimensional, steadystate heat conduction in a variable area rubber leg.
OBJECTIVE: (a) Starting from (3.29), with s˙ = 0, use a variable circular conduction area Ak (x) = πR2 (x), while R(x) varies linearly along the x axis, i.e., R(x) =
R1 x, L1
as shown in Figure Pr.3.23. Then derive the expression for the temperature distribution T = T (x). (b) Using this temperature distribution, determine Qk,12 and Rk,12 , by using (3.46) and noting that there are no lateral heat losses. (c) Evaluate Qk,12 , for the conditions given below. (d) Use a constant surface area with R = (R1 + R2 )/2, and the conduction resistance for a slab, and compare Qk,12 with that from part (c). SOLUTION: (a) Starting from (3.29) with s˙ = 0, we use ∆V = πR2 (x)∆x, and the result is (qk · sn )dA (qk Ak )x+∆x − (qk Ak )x ∆A = = 0. ∆V → 0 πR2 (x)∆x Using Ak = πR2 (x) and R(x) = R1 x/L1 , we have π
R1 L1
2
(qk x2 )x+∆x − (qk x2 )x πR2 (x)∆x
or (qk x2 )x+∆x − (qk x2 )x =0 ∆x → 0 173
=0
or d(qk x2 ) = 0. dx Now using (1.11), and noting that T = T (x), we have d dT −k x2 = 0, dx dx or d dx
dT 2 x dx
= 0.
Integrating once gives dT 2 x = a1 . dx Integrating again gives
T
a1 dx + a2 x2 a1 = − + a2 . x =
Now using the thermal conditions T (x = L1 ) T (x = L1 + L2 )
= T1 = T2 ,
we have T1 T2
a1 + a2 L1 a1 = − + a2 . L1 + L2 = −
Solving for a1 and a2 , we have 1 1 − x L1 T = T1 + (T2 − T1 ) . 1 1 − L1 + L2 L1 (b) Now noting that from (3.46) we have dT Qk,x = Ak (x) −k , dx x and diﬀerentiating T = T (x), we obtain Qk,x
πR2 (x)(−k)(T1 − T2 ) = 1 1 − L1 L1 + L2
1 − 2 . x
Evaluating this at x = L2 , we have Qk,x = Qk,12 = πk
1 R12 (T1 − T2 ) 2 1 1 L1 − L1 L1 + L2 174
or 1 1 + L L1 + L2 Rk,12 = 1 . πkR12 /L21 (c) Using the numerical values, we have Qk,12
= π × 0.15(W/mK) ×
(0.015)2 (m2 ) (0.10)2 (m2 )
1 (0 − 20)K 1 1 − 0.1(m) 0.14(m)
= −7.423 × 10−2 W. (d) Using a constant area with R = =
R1 + R 2 R1 = [L1 + (L1 + L2 )] 2 2L1 0.018 m.
Then from Table 3.2 for a slab we have Qk,12 = πR2 k(T1 − T2 )/L2 = −7.635 × 10−2 W. This is very close to the results for the variable area in part (c). COMMENT: Note that here R1 /L1 = 0.15 and as this ratio becomes smaller, the role of the variable area becomes more signiﬁcant.
175
PROBLEM 3.24.FAM GIVEN: A pair of aluminum slabs with a surface roughness of δ 2 1/2 = 0.25 µm are placed in contact (with air as the interstitial ﬂuid). A heat ﬂux of qk = 4 × 104 W/m2 ﬂows across the interface of the two slabs. OBJECTIVE: Determine the temperature drop across the interface for contact pressures of 105 and 106 Pa. SOLUTION: The heat ﬂux ﬂowing through the contact between the two solids is related to the temperature diﬀerence across the contact by (3.95) Qk,c ∆Tc = . qk,c = Ak Ak Rk,c From Figure 3.25, for a pair of soft aluminum surfaces in air, having rootmeansquare roughness δ 2 1/2 = 0.25 µm, for each of the contact pressures, the contact thermal conductance is p = 105 Pa,
1/Ak Rk,c 8.2 × 103 W/m2 ◦C
Figure 3.25
p = 106 Pa,
1/Ak Rk,c 1.5 × 104 W/m2 ◦C
Figure 3.25.
From Equation (3.95), for qk,c = 4 × 104 W/m2 , the temperature jumps across the contact interface for each contact pressure are ∆Tc = 4.9◦C p = 105 Pa, p = 106 Pa,
∆Tc = 2.7◦C.
COMMENT: A high joint pressure is usually necessary to reduce the contact resistance. The use of thermal conductivity pastes and greases also reduces the contact thermal resistance. This occurs because the air present at the contact is replaced by the more conductive paste. These pastes are usually made of a polymer ﬁlled with submicron metal particles.
176
PROBLEM 3.25.FAM GIVEN: Thin, ﬂat foil heaters are formed by etching a thin sheet of an electrical conductor such as copper and then electrically insulating it by coating with a nonconductive material. When the maximum heater temperature is not expected to be high, a polymer is used as coating. When high temperatures are expected, thin sheets of mica are used. Mica is a mineral silicate that can be cleaved into very thin layers. However, a disadvantage of the use of mica is that the mica surface oﬀers a much higher thermalcontact resistance, thus requiring a larger joint pressure pc . Figure Pr.3.25(a) shows a thin circular heater used to deliver heat to a surface (surface 1). The solid between surface 1 and the heater is aluminum (Table C.14) and has a thickness L1 = 5 mm. In order to direct the heat to this surface, the other side of the heater is thermally well insulated by using a very low conductivity ﬁber insulating board (Table C.15) with thickness L2 = 10 mm. The temperature of the aluminum surface is maintained at T1 = 100◦C, while the outer surface of the thermal insulation is at T2 = 30◦C. The heater generates heat by Joule heating at a rate of S˙ e,J /Ak = 4 × 104 W/m2 and is operating under a steadystate condition. Use the thermal conductivities at the temperatures given in the tables or at 300 K. SKETCH: Figure Pr.3.25(a) shows the layered composite with the energy conversion. Surface 1 at T1 = 100 C L1 = 5 mm
Se,J
Ak
Rk,c Uniform Heater Temperature Th (Nk,h < 0.1) L 2 = 10 mm
Aluminum Mica Heater
Fibrous Insulating Board
T2 = 30 C
Figure Pr.3.25(a) A thinfoil heater encased in mica and placed between an aluminum and an insulation layer.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the heater temperature Th for the case of contact resistances of (i) Ak Rk,c = 10−4 [K/(W/m2 )], and (ii) Ak Rk,c = 4 × 10−2 [K/(W/m2 )]. (c) Comment on the answers obtained above if the heater is expected to fail at Tmax = 600◦C. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.25(b). (b) To ﬁnd the heater temperature, the integralvolume energy equation (3.161) is applied to the heater node Th . Under steadystate conditions we have Qh +
Th − Tj j
Rk,hj
= S˙ e,J .
The temperatures T1 and T2 are known. Therefore, the heat transfer through the thermal resistances are written as functions of these temperatures. As Qh = 0 (there is no prescribed surface heat transfer), we have Th − T1 Th − T2 + = S˙ e,J . (Rk,Σ )h1 (Rk,Σ )h2 For the resistances arranged in series, the overall thermal resistances (Rk,Σ )h1 and (Rk,Σ )h2 are (Rk,Σ )h1
= Rk,c + Rk,i1
(Rk,Σ )h2
= Rk,c + Rk,i2 . 177
T1 Aluminum
Rk,i1
Qi1
T1 = 100 OC
Rk,c
Qh,i,1 Se,J
Ak
Ti,1
L1 = 5 mm
Ti,1
Rk,c
Se,J
Th Ti,2
Rk,c
L2 = 10 mm
Qh,i,2
Mica Heater Interface (i)
Ti,2 Qi2
Uniform Temperature, Th
Rk,i2 T2 = 30 OC T2
Fiber Insulating Board
Figure Pr.3.25(b) Thermal circuit diagram.
The conduction resistances Rk,i1 and Rk,i2 for the slabs are Rk,i1
=
Rk,i2
=
L1 k1 Ak L2 , k2 Ak
and the contact resistance Rk,c is given in the problem statement. The thermal conductivities are needed to calculate the thermal resistances. For each of the materials we have aluminum (Table C.14, T = 300 K) k = 237 W/mK and ﬁber insulating board (Table C.15, T = 294 K) k = 0.048 W/mK . Solving for Th , we have T1 T2 + (Rk,Σ )h1 (Rk,Σ )h2 . 1 1 + (Rk,Σ )h1 (Rk,Σ )h2
S˙ e,J + Th =
(i) For the ﬁrst case Ak Rk,c
=
Ak Rk,i1
=
Ak Rk,i2
=
1 × 10−4 ◦C/(W/m2 ) L1 0.005 = 2.110 × 10−5 ◦C/(W/m2 ) = k1 237 L2 0.01 = 2.083 × 10−1 ◦C/(W/m2 ). = k2 0.048
The equivalent resistances are Ak (Rk,Σ )h1
= Ak Rk,c + Ak Rk,i1 = 1 × 10−4 [◦C/(W/m2 )] + 2.110 × 10−5 [◦C/(W/m2 )] = 1.211 × 10−4 ◦C/(W/m2 )
Ak (Rk,Σ )h2
= Ak Rk,c + Ak Rk,i2 = 1 × 10−4 [◦C/(W/m2 )] + 2.083 × 10−1 [◦C/(W/m2 )] = 2.084 × 10−1 ◦C/(W/m2 ). 178
Dividing all the terms by Ak and solving for Th we have
Th
S˙ e,J T1 T2 + + Ak Ak (Rk,Σ )h1 Ak (Rk,Σ )h2 1 1 + Ak (Rk,Σ )h1 Ak (Rk,Σ )h2 100(◦C) 30(◦C) 4 × 104 (W/m2 ) + + −4 ◦ 2 1.211 × 10 [ C/(W/m )] 2.084 × 10−1 [◦C/(W/m2 )] 1 1 + −4 ◦ −1 ◦ 2 1.211 × 10 [ C/(W/m )] 2.084 × 10 [ C/(W/m2 )] ◦ 105 C.
=
= = (ii) For the second case
Ak Rk,c = 4 × 10−2◦C/(W/m2 ). The conduction thermal resistances remain the same. The equivalent resistances now become Ak (Rk,Σ )h1
= Ak Rk,c + Ak Rk,i1 = 4 × 10−2 [◦C/(W/m2 )] + 2.110 × 10−5 [◦C/(W/m2 )] = 4.002 × 10−2 ◦C/(W/m2 )
Ak (Rk,Σ )h2
= Ak Rk,c + Ak Rk,i2 = 4 × 10−2 [◦C/(W/m2 )] + 2.083 × 10−1 [◦C/(W/m2 )] = 2.483 × 10−1 ◦C/(W/m2 ).
Solving for Th we have
Th
=
= =
S˙ e,J T1 T2 + + Ak Ak (Rk,Σ )h1 Ak (Rk,Σ )h2 1 1 + Ak (Rk,Σ )h1 Ak (Rk,Σ )h2 100(◦C) 30(◦C) 4 × 104 (W/m2 ) + + 4.002 × 10−2 [◦C/(W/m2 )] 2.483 × 10−1 [◦C/(W/m2 )] 1 1 + 4.002 × 10−2 [◦C/(W/m2 )] 2.483 × 10−1 [◦C/(W/m2 )] 1,469◦C.
(c) The heater is rated for 600◦C. Therefore, it would operate normally under case (i) (smaller contact thermal resistance), but it would fail under case (ii) (larger contact thermal resistance). COMMENT: Note that a small air gap present in series with other low resistance layers causes a large decrease in the heat ﬂow rate and a large increase in the temperature drop.
179
PROBLEM 3.26.FAM GIVEN: The automobile exhaust catalytic converter (for treatment of gaseous pollutants) is generally a large surface area ceramic or metallic monolith that is placed in a stainless steel housing (also called can). Figure Pr.3.26(a) shows a ceramic (cordierite, a mineral consisting of silicate of aluminum, iron, and magnesium) cylindrical monolith that is placed inside the housing with (i) direct ceramicstainless contact, and (ii) with a blanket of soft ceramic (vermiculite, a micacious mineral, mat) of conductivity kb placed between them. The blanket is placed under pressure and prevents the gas from ﬂowing through the gap. The direct contact results in a contact resistance similar to that of stainless steelstainless steel with δ 2 1/2 = 1.1 to 1.5 µm and pc = 105 Pa. The soft blanket (vermiculite mat) has a thickness l1 = 3 mm and kb = 0.4 W/mK. Use T1 = 500◦C, T2 = 450◦C, and stainless steel AISI 316 for thermal conductivity. SKETCH: Figure Pr.3.26(a) shows a catalytic converter with and without a ceramic blanket. (i) Without Soft Ceramic Blanket Present
(ii) With Soft Ceramic Blanket Present
T2 = 450oC
T2 R2 = 7 cm R1 = 6.7 cm
o
T1 = 500 C l1 (Contact)= 0 l2 (Stainless steel) = 3 mm
r
Automobile Exhaust
l1'(Blanket) = 3 mm
T1
R2 = 7 cm R1 = 6.4 cm
l2 (Stainless steel) = 3 mm L = 25 cm
Contact Resistance, Rk,c StainlessSteel Housing Cordierite Large Surface Area Monolith (Catalytic Converter)
r L = 25 cm
Automobile Exhaust
Added SoftBlanket kb = 0.4 W/mK
SideView of Catalytic Converter and Housing Housing
CO, HC, NOx , Air
CO2 + H2O + N2 Cordierite Monolith
Figure Pr.3.26(a) An automobile catalytic converter (i) without and (i) with a soft ceramic blanket.
OBJECTIVE: (a) Draw the thermal circuit diagrams. (b) Determine the heat ﬂow between surface at temperature T1 and surface at temperature T2 (i) without and (ii) with the soft ceramic blanket, i.e., Qk,1−2 . SOLUTION: (a) The thermal circuit diagrams for cases (i) and (ii) are shown in Figure Pr.3.26(b). (b) The heat ﬂow rate is written from the thermal circuit model of Figure Pr.3.26(b) as (Qk,12 )without blanket = 180
T1 − T2 . Rk,c + Rk,1 2
Qk,12
(i) − Q1 (From Monolith)
T1
T1'
T2 Q2 (To Ambient)
Rk,c Contact Resistance
Rk,1'2 StainlessSteel Resistance
Qk,12
(ii) − Q1
T1
T1'
T2 Q2 (To Ambient)
Rk,11' SoftBlanket Resistance
Rk,1'2 StainlessSteel Resistance
Figure Pr.3.26(b) Thermal circuit diagrams for cases (i) and (ii).
Here R1 = R2 − l2 = 6.7 cm. From Figure 3.25, for stainlesssteel contact with δ 2 1/2 = 1.1 to 1.5µm and pc = 105 Pa, we have 1 2 = 2 × 102 (W/m )/◦C Ak Rk,c
Figure 3.25.
The area is Ak = 2πR1 L. The stainlesssteel shell resistance is found from Table 3.2, using the geometrical designations of Figure Pr. 3.26(a), to be R2 R1 Rk,1 2 = , 2πLks ln
where from Table C.16, we have ks = 13 W/mK.
Table C.16
(c) From Figure Pr.3.26(b), we have
(Qk,12 )with blanket =
T1 − T2 . Rk,11 + Rk,1 2
Here R1 = R2 − l2 − l1 = 6.4 cm, Rk,1 2 remains the same, and R 1 R1 . Rk,11 = 2πLkb ln
Using the numerical results, we have 181
Rk,c
= =
Rk,1 2
Rk,11
1 2 × 102 (W/m2 −◦C) × 2πR1 L 1 2 × 102 (W/m2 −◦C) × 2π × 0.067(m) × 0.25(m)
= 4.750 × 10−2 ◦C/W 0.07 m ln 0.067 m = 2π × 0.25 m × 13(W/mK) = 2.145 × 10−3 ◦C/W 0.07 m ln 0.064 m = 2π × 0.25 m × 0.4(W/mK) = 7.291 × 10−2 ◦C/W.
For the heat ﬂow, we have (Qk,12 )without blanket
=
(Qk,12 )with blanket
=
(500 − 450)(◦C) = 1.006 × 103 W (4.750 × 10−2 + 2.145 × 10−3 )( ◦C/W) (500 − 450)(◦C) = 6.662 × 102 W. (7.291 × 10−2 + 2.145 × 10−3 )( ◦C/W)
COMMENT: The soft ceramic blanket prevents ﬂow leaks at the housing contact and reduces the heat loss to the housing (Rk,11 is larger than Rk,c ).
182
PROBLEM 3.27.FAM GIVEN: A thermoelectric power generator uses the heat released by gaseous combustion to produce electricity. Since the low temperature thermoelectric materials undergo irreversible damage (such as doping migration) at temperatures above a critical temperature Tcr , a relatively low conductivity material (that withstands the high ﬂame temperature; this is referred to as a refractory material) is placed between the ﬂame and the hot junction, as shown in Figure Pr.3.27(a). Additionally, a copper thermal spreader is placed between the refractory material and the hot junction to ensure even distribution of the heat ﬂux into the thermoelectric device. It is desired to generate 20 W of electricity from the thermoelectric module, where this power is 5% of the heat supplied (−Qh ) at the hot junction Th . The refractory material is amorphous silica with conductivity ks . In addition, there is a contact resistance Rk,c between the copper thermal spreader and the hot junction. The surface area of the hot junction is a × a. Tg = 750◦C, Tcr = 250◦C, ks = 1.36 W/mK, a = 6 cm, Ak Rk,c = 10−4 K/(W/m2 ). SKETCH: Figure Pr.3.27(a) shows the electrical power generation unit with the thermoelectric cooler and the combustion ﬂuegas stream. The lowconductivity (refractory) material used to lower Th (to protect the thermoelectric module from high temperatures) is also shown. Refractory Slab Copper Thermal Spreader Combustion Flue Gas
Thermoelectric Module Spring (For Reduction in Thermal Contact Resistances)
Water Cooled a Heat Sink
Tg . Sr,c
Th = Tcr (Hot Junction) Rk,c (Contact Resitance)
Qh L
Figure Pr.3.27(a) A thermoelectric module, used for power generation, receives heat from a combustion ﬂue gas stream. To reduce the temperature of the hot junction, a refractory slab is used.
OBJECTIVE: (a) Draw the thermal circuit diagram for node Th using the combustion ﬂue gas temperature Tg . (b) Determine the thickness of the refractory material L, such that Th = Tcr . SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.27(b). − Qh
Sr,c Tg
Rk,gh
Rk,c
Th = Tcr
Figure Pr.3.27(b) Thermal circuit diagram.
(b) From Figure Pr.3.27(b), the expression for −Qh in terms of Th and Tg is −Qh =
Th − Tg . Rk,gh + Rk,c 183
Then from Table 3.2, for a slab, we have Rk,gh =
L . Ak ks
Using this, we have −Qh =
Th − Tg L Ak Rk,c + Ak ks Ak
or L Ak Rk,c Tg − Th + = Ak ks Ak Qh or L=
Tg − Th Ak ks − Ak Rk,c ks , Qh
Ak = a2 .
Using the numerical values and noting that Qh =
20(W) P = = 400 W, η 0.05
we have L
=
(750 − 250)(K) × (0.06)2 (m2 ) × 1.36(W/mK) − 10−4 [K/(W/m2 )] × 1.36(W/mK) 400(W)
=
6.120 × 10−3 (m) − 1.36 × 10−4 (m) = 5.984 × 10−3 m = 5.984 mm.
COMMENT: Since active cooling of the cold junction is needed to maintain a low Tc , most of the heat arriving at the hot junction is transferred to the cold junction by conduction. This is the reason for the low eﬃciency.
184
PROBLEM 3.28.FAM GIVEN: In order to reduce the contact resistance, the contacting solids are physically bonded by using high temperatures (as in fusion or sintering) or by using material deposition (as in physical vapor deposition or solidiﬁcation of melts). This creates a contact layer which has a contact thickness Lc (that is nearly twice the rms roughness) and a contact conductivity kc . This contact conductivity is intermediate between the conductivity of the joining materials A and B (with kA < kB ), i.e., kA ≤ kc ≤ kB
for kA < kB .
Consider a contact resistance between a bismuth telluride slab (material A) and copper (material B) slab. This pair is used in thermoelectric coolers, where the semiconductor, doped bismuth telluride is the thermoelectric material and copper is the electrical connector. Then use a general relationship kc = kA +a1 (kB −kA ), 0 ≤ a1 ≤ 1, and plot (semilog scales) the temperature drop across the junction ∆Tc , for the following conditions, as a function of a1 . Here a1 depends on the fabrication method used. kA = 1.6 W/mK, kB = 385 W/mK, Lc = 2δ 2 1/2 = 0.5 µm, qk = 105 W/m2 . SKETCH: Figure Pr.3.28(a) shows the contact region with a physical bonding of materials A and B forming an alloy AB.
Contact Layer in Physical Bonding of Contacting Materials and Contact Conductivity, kc Material A, kA Material B, kB qk Lc = 2 δ2
∆Tc qk
1/2
Alloy AB (Physical Bonding)
Contact Layer with Contact Conductivity kc , and Contact Layer Thickness Lc
Figure Pr.3.28(a) A contact layer in a physical bounding of contacting materials; also shown is the contact conductivity kc .
OBJECTIVE: Use a general relationship kc = kA + a1 (kB − kA ), 0 ≤ a1 ≤ 1, and plot (semilog scales) the temperature drop across the junction ∆Tc , for the following conditions, as a function of a1 . Here a1 depends on the fabrication method used. SOLUTION: From (3.96), we have ∆Tc = qk
Lc kc
and using the expression for kc , we have ∆Tc (K) = qk L =
1 kA + a1 (kB − kA )
105 (W/m2 ) × 5 × 10−7 (m) . [1.6 + a1 (385 − 1.6)](W/mK)
Figure Pr.3.28(b) shows the variation of ∆Tc with respect to a1 . These results show that ∆Tc changes over three orders of magnitude, as a1 is changed from 0 to 1. COMMENT: High kc becomes essential in obtaining temperatures close to the cold junction temperature, when surfaces are brought in contact with the cold junction for heat transfer and cooling. 185
DTc , K
10 kC = kA + a1(kB  kA) kA = 1.6 W/mK kB = 385 W/mK
1
0.1
0.01 0
0.2
0.4
0.6
0.8
1.0
a1 Figure Pr.3.28(b) Variation of the temperature jump across the contact region, as a function of junction conductivity parameter a1 .
186
PROBLEM 3.29.FUN GIVEN: In a thermoelectric cell, the Joule heating that results from the passage of the electric current is removed from the hot and cold ends of the conductor. When the electrical resistivity ρe or the current density je are large enough, the maximum temperature can be larger than the temperature at the hotend surface Th . This is shown in Figure 3.27(b). OBJECTIVE: (a) From the temperature distribution given by (3.104), determine the expression for the location of the maximum temperature in the conductor. (b) Both the p and the ntype legs have the same length L = 2 cm. The crosssectional area of the ntype leg is An = 2.8 × 10−5 m2 . Calculate the crosssectional area for the ptype leg Ap if the ﬁgure of merit is to be maximized. Use the electrical and thermal properties of the ptype and ntype bismuth telluride given in Table C.9(b). (c) Determine the magnitude and the location of the maximum temperature for the ptype leg, when Th = 40◦C, Tc = −2◦C, and Je = 6 A. (d) Determine the rate of heat removal from the hot and cold ends per unit crosssectional area (An + Ap ). SOLUTION: (a) The temperature distribution along a conductor with internal Joule heating is given by (3.104) as x2 x ρe je2 L2 x − 2 . T = Tc + (Th − Tc ) + L 2k L L To ﬁnd the point in which the temperature is maximum, we diﬀerentiate T with respect to x and set the result equal to zero, 1 2x dT ρe je2 L2 1 = (Th − Tc ) + − = 0. dx L 2k L L2 Solving for x gives x=
k (Th − Tc ) L + . 2 ρe je2 L
(b) To maximize the ﬁgure of merit Ze (3.120), for a given temperature diﬀerence Th − Tc , the ratio Re,hc /Rk,hc must be minimized. Minimizing this ratio with respect to the geometric parameters of the Peltier cooler, Ln Ap /Lp An , gives (3.121) 1/2 ρe,p kn Ln Ap = . Lp An ρe,n kp The properties for the p and ntype materials are given in Table C.9(a). For p and ntype bismuth telluride alloys, kp = 1.70 W/mK, ρe,p = 1 × 10−5 ohmm, αS,p = 230 × 10−6 V/◦C, kn = 1.45 W/mK, ρe,n = 1 × 10−5 ohmm, αS,n = −210 × 10−6 V/◦C. For Ln = Lp = 0.02 m and An = 2.8 × 10−5 m2 , the area of the ptype leg is 1/2 1/2 1.45 kn −5 2 = 2.8 × 10 (m ) × = 2.59 × 10−5 m2 . Ap = An kp 1.70 (c) The location in the ptype leg in which the temperature is maximum is given above. The current density is je = Then x(Tmax ) = = =
Je 6(A) 2 5 = A/m . 2 = 2.32 × 10 −5 Ap 2.59 × 10 (m ) k (Th − Tc ) L + 2 ρe je2 L 0.02 1.70(W/mK) (40 + 2)(◦C) (m) + −5 × 2 2 0.02(m) 10 (ohmm)(2.32 × 105 )2 (A ) 0.0167 m = 1.67 cm. 187
The maximum temperature is obtained from the temperature distribution (3.104) x2 x ρe je2 L2 x − Tmax = Tc + (Th − Tc ) + L 2k L L2 0.0167(m) = −2(◦C) + × [40(◦C) + 2(◦C)] 0.02(m) 2 0.0167(m) 10−5 (ohmm) × (2.32 × 105 )2 (A2 ) × 0.022 (m2 ) 0.0167(m) + × − 2 × 1.70(W/mK) 0.02(m) 0.02(m) =
41.8◦C.
(d) The cooling power is given by (3.115) Qc = −αS Je Tc +
Th − Tc 1 + Re,hc Je2 , Rk,hc 2
−1 kp Ap kn An + = 236.3◦C/W Lp Ln ρe,p Lp ρe,n Ln + = 0.0149 ohm. Ap An
where Rk,hc
=
Re,hc
=
Then Qc
= −(230 × 10−6 + 210 × 10−6 )(V/◦C) × 6(A) × 271(◦C) + +
42(◦C) 236.3(◦C/W)
1 × 0.0149(ohm) × 62 (A2 ) = −0.27 W 2
or qc =
−0.27 2 = −5,009 W/m . An + Ap
The heat generated at the hot end is given by (3.125), i.e., Qh Qh
= −Qc + Re,hc Je2 + αS Je (Th − Tc ) = 0.27(W) + 0.01497(ohm) × 62 (A2 ) + 440 × 10−6 (V/◦C) × 6(A) × 42(◦C) = 0.92 W
or qh =
0.92 2 = 17,069 W/m . An + Ap
60 50 40
T,K
30 20 10 0 10 20
0.0
0.2
0.4
0.6
x/L
Cold Junction
0.8
1.0 Hot Junction
Figure Pr.3.29 Temperature distribution along the conductor, showing a maximum near the hot junction.
COMMENT: The temperature distribution along the ptype leg of the Peltier cooler is parabolic, as given by (3.104). Figure Pr.3.29 shows the temperature distribution for the data given in item (c). Notice the maximum occurring near the hot end of the junction. 188
PROBLEM 3.30.FUN GIVEN: A thermoelectric cooler has bismuth telluride elements (i.e., p and ntype pairs) that have a circular cross section of diameter d = dn = dp = 3 mm and a length L = Ln = Lp = 2 cm, as shown in Figure Pr.3.30(a). The temperatures of the hot and cold ends are Th = 40◦C and Tc = −2◦C. SKETCH: Figure Pr.3.30(a) shows the heat ﬂowing into the cold junction of thermoelectric cooler unit.
Qc
Cold Surface, Tc Bismuth Telluride Lp = Ln
Hot Surface, Th
p
n
dp
dn
Je Figure Pr.3.30(a) A thermoelectric cooler unit.
OBJECTIVE: Determine the cooling power for each junction, if the current corresponds to (a) the current that maximizes the cooling power Qc , (b) the current that is half of this optimum current, and (c) the current that is twice the optimum current. SOLUTION: (a) The current that maximizes the cooling power is given by (3.117) as Je (Qc,max ) =
αS Tc . Re,hc
The electrical resistance is given by (3.116) and using ρe,p and ρe,n from Table C.9(a) for bismuth telluride, we have Re,hc =
ρe,p Lp ρe,n Ln L 0.02(m) + = (ρe,p + ρe,n ) = × 2 × 10−5 (ohmm) = 0.0566 ohm. Ap An A π × (0.0015)2 (m2 )
For the n and ptype bismuth telluride, from Table C.9(b), αS = αS,p − αS,n = 440 µV/K. The maximum current is Je (Qc,max ) =
440 × 10−6 (V/K) × 271.15(K) = 2.11 A. 0.0566(ohm)
The cooling power is given by (3.115), Qc = −αS Je Tc +
Th − Tc 1 + Re,hc Je2 . Rk,hc 2 189
The thermal resistance is given by (3.116) and using k from Table C.9(a), we have Rk,hc =
kp Ap kn An + Lp Ln
−1 =
1 0.02(m) L 1 = = 898.2◦C/W. × 2 2 A (kp + kn ) π × (0.0015) (m ) (1.70 + 1.45)(W/m◦C)
Finally, the maximum cooling power is Qc = −(440 × 10−6 )(K/V) × 2.11(A) × 271.15(K) +
0.0566(ohm) × (2.11)2 (A2 ) 42(◦C) + = −0.079 W. 898.2(◦C/W) 2
(b) The cooling power for half the current (1.06 A) is Qc = −(440 × 10−6 )(K/V) × 1.06(A) × 271.15(K) +
×0.0566(ohm) × (1.06)2 (A2 ) 42(◦C) + = −0.047 W. 898.2(◦C/W) 2
(c) The cooling power for twice the current (4.22 A) is Qc = −(440 × 10−6 )(K/V) × 4.22(A) × 271.15(K) +
×0.0566(ohm) × (4.22)2 (A2 ) 42(◦C) + = 0.048 W. 898.2(◦C/W) 2
COMMENT: For a given pair and geometry, the cooling power varies with the applied current. Figure Pr.3.30(b) shows the negative of the cooling power as a function of the current for this Peltier cooler. As the current increases, both the Peltier cooling and the Joule heating increase. However, the Joule heating increases with the square of the current and, for large values of current, its contribution overcomes that of the Peltier cooling. Another possible optimization is the geometric optimization. Figure Pr.3.30(c) shows the negative of the cooling power as a function of the ratio of crosssectional area over length Ak /L for the three diﬀerent currents used in the problem. Note that for each current there is a maximum in the cooling power. For small values of Ak /L, the thermal resistance is large, thus reducing the heat conduction. However, the electrical resistance is also large and that increases the Joule heating. For large values of Ak /L the opposite occurs. The optimum point is given by the balance between the thermal and the electrical resistances. 0.10 0.08
 Qc , W
0.06 0.04 0.02 0.00
 0.02
0.0
1.0
2.0 Je , A
3.0
4.0
Figure Pr.3.30(b) Variation of cooling power (−Qc ) with respect to the current.
190
Je = 1.06 A Je = 2.11 A Je = 4.22 A
0.20
 Qc , W
0.15 0.10 0.05 0.00
 0.05  0.10
0.000
0.001
0.002 A/L , m
0.003
0.004
Figure Pr.3.30(c) Variation of cooling power (−Qc ) with respect to the ratio of conductor cross sectional area to length.
191
PROBLEM 3.31.FUN GIVEN: A thermoelectric device is used for cooling a surface to a temperature Tc . For each bismuth telluride thermoelectric, circular cylinder conductor, use dn = dp = 1.5 mm, and Ln = Lp = 4 mm. The hot junction is at Th = 40◦C. SKETCH: Figure Pr.3.31 shows the thermoelectric cooler unit. Qc
Cold Surface, Tc Bismuth Telluride Lp = Ln
Hot Surface, Th
p
n
dp
dn
Je
Figure Pr.3.31 A thermoelectric cooler unit.
OBJECTIVE: For the conditions given below, determine (a) the minimum Tc , (b) the current for this condition, and (c) the minimum Tc for a current Je = 1 A. SOLUTION: (a) The minimum Tc for a given Th is given by (Th − Tc )max =
αS2 Tc2 αS2 Tc2 = −1 . 2Re,hc /Rk,hc 2Re,hc Rk,h c
The electrical and thermal resistances are given by ρe L ρe L Re,hc = + A n A p kA kA −1 Rk,h = + . c L n L p From the data given, An Ln
π(0.0015)2 (m2 ) πd2n = = 1.767 × 10−6 m2 4 4 = Lp = 0.004 m. = Ap =
For the bismuthtelluride elements, from Table C.9(a), ρe,n = ρe,p = 10−5 ohmm, kn = 1.70 W/mK, kp = 1.45 W/mK, and αS = αS,p − αS,n = 230 × 10−6 (V/K)+ 210 × 10−6 (V/K) = 440 × 10−6 V/K. The resistances then become Re,hc = 2 −1 Rk,h c =
10−5 (ohmm) × 0.004(m) = 0.0453 1.767 × 10−6 (m2 )
1.767 × 10−6 (m2 ) [1.70(W/mK) + 1.45(W/mK)] = 0.00139 0.004(m) 192
ohm W/K.
The minimum Tc is then given by 2 440 × 10−6 (V/K) Tc2 313.15(K) − Tc (K) = (2)0.0453(ohm) × 0.00139(W/K) Tc2 + 6.512 × 102 Tc − 2.038 × 105 = 0 or Tc = 231 K. (b) The current for this temperature is given by Je =
αS Tc 440 × 10−6 (V/K) × 231(K) = 2.245 A. = Re,hc 0.0453(ohm)
(c) The minimum temperature for a current of Je = 1 A is found by setting the cooling power to zero. From (3.115), 1 −1 2 Qc = −αS Je Tc + Rk,h c (Th − Tc ) + 2 Re,hc Je = 0. Solving for Tc gives Tc =
−1 1 2 Rk,h c Th + 2 Re,hc Je
αs Je +
−1 Rk,h c
=
0.00139(W/K) × 313.15(K) + 12 × 0.0453(ohm)12 (A2 ) = 250.2 K. 440 × 10−6 (V/K) × 1(A) + 0.00139(W/K)
COMMENT: Note that for a square cross section, A = d2n = 2.25 × 10−6 m2 , we have Re,hc −1 Rk,h c
= =
0.03556 ohm 0.00177 W/K
Je Tc
= =
2.859 A 258.8 K.
193
PROBLEM 3.32.FUN GIVEN: A thinﬁlm thermoelectric cooler is integrated into a device as shown in Figure Pr.3.32(a). In addition to heat conduction through the p and ntype conductors, heat ﬂows by conduction through the substrate. Assume a onedimensional parallel conduction through the p and ntype conductors and the substrate. Model the conduction through the substrate as two conduction paths (one underneath each of the p and ntype legs). Begin with (3.115) and use the optimum current. SKETCH: Figure Pr.3.32(a) shows the thermoelectric cooler unit, the heat source and sink, and the substrate. Electrical Conductor, Tc Se,J Load
Electrical Conductor, Th Electrical Insulator (−)
Substrate k2
e
yp T yp T p
l2
e
L
n
Je
(+) w l1 kn = kp = k1
Figure Pr.3.32(a) A thinﬁlm thermoelectric cooler placed over a substrate.
OBJECTIVE: (a) Draw the thermal circuit diagram for heat ﬂow between the Th and Tc nodes. (b) Show that the maximum temperature diﬀerence is (Th − Tc )max (Qc = 0) =
Ze Tc2 . l2 k2 2 1+ l1 k1
SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.32(b). The conduction heat ﬂow is through two parallel paths. The energy equation for Tc node is given by (3.115), which we rewrite as 1 1 1 Qc = −αS Je Tc + (Th − Tc ) + + Re,hc Je2 . Rk,hc 1 Rk,hc 2 2 L l1 l2 Substrate
Thermoelectric Element
Se,J + Se,P
Se,J + Se,P
Qc
Qh (Rk,hc)1
(Rk,hc)2
Figure Pr.3.32(b) Thermal circuit diagram.
194
For (Th − Tc )max with Qc = 0 and Je corresponding to the optimum performance as given by (3.117), we have αS Tc Je = R e,hc α2 T 2 α T 1 1 S c Qc = 0 = −αS R + Tc + (Th − Tc ) + 21 Re,hc 2S c (Rk,hc )1 (Rk,hc )2 e,hc Re,hc or 0=−
1 1 αS2 Tc2 1 + (Th − Tc ) + . 2 Re,hc (Rk,hc )1 (Rk,hc )2
(b) Solving for (Th − Tc )max , we have (Th − Tc )max (Qc = 0) =
αS2 TC2 . Re,hc (Rk,hc )1 2 1+ (Rk,hc )1 (Rk,hc )2
Using the thermoelectric ﬁgure of merit given by (3.120), we have (Th − Tc )max (Qc = 0) =
Ze Tc . (Rk,hc )1 2 1+ (Rk,hc )2
From (3.116), and for kp = kn = k1 , we have (Rk,hc )−1 1
=
(Rk,hc )−2 2
=
Ak,1 k1 , L Ak,2 k2 , 2 L 2
Ak,1 = l1 w Ak,2 = l2 w.
Using these, we have (Th − Tc )max (Qc = 0) =
Ze T c . l2 k2 2 1+ l1 k1
COMMENT: The heat conduction through the substrate is not one dimensional. Also, parasitic heat is conducted to the cold junction from locations other than the hot junction.
195
PROBLEM 3.33.DES.S GIVEN: A miniature vapor sensor is cooled, for enhanced performance, by thermoelectric coolers. The sensor and its thermoelectric coolers are shown in Figure Pr.3.33(a). There are four bismuthtelluride thermoelectric modules and each module is made of four pn layers (forming four pn junctions) each p and nlayer having a thickness l, length L, and width w. The sensor and its substrate are assumed to have the ρcp of silicon and a cold junction temperature Tc (t). The hot junction temperature Th (t) is expected to be above the farﬁeld solid temperature T∞ . The conduction resistance between the hot junctions and T∞ is approximated using the results of Table 3.3(b), for steadystate resistance between an ambient placed on the bounding surface of a semiinﬁnite slab (Th ) and the rest of the slab (T∞ ) [shown in Table 3.3(b), ﬁrst entry]. This is 2w 4w ln a , = 2a = πkw πkw ln
Rk,h∞
Qk,h∞ =
Th (t) − T∞ . Rk,h∞
Initially there is a uniform sensor temperature, Tc (t = 0) = Th (t = 0) = T∞ . For heat storage of the thermoelectric modules, divide each volume into two with each portion having temperature Tc or Th . l = 3 µm, w = 100 µm, L = 300 µm, a = 24 µm, T∞ = 20◦C, Je = 0.010 A. SKETCH: Figure 3.33(a) shows the thermoelectrically cooled vapor sensor.
Thin Porous Silicon Layer Vapor Diffusion (Absorber, Sensor) L Thin Capacitance Electrode (Al Grid)
w
Sensor Module, T(t)
T (FarField Temperature)
w l a = 8l
Tc(t) Th(t) Thin Capacitance Electrode (Al) nType Thermoelectric (Bismuth Telluride)
Si
Si
Heat Conduction qk
qk
Thermoelectric Module Si T
pType Thermoelectric (Bismuth Telluride) Electrical Insulator SiO2 Layer (Negligible Thickness) Tc
Th Porous Silicon Layer
Je Electrode
Figure Pr.3.33(a) A thermoelectrically cooled vapor sensor showing the four molecules.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine and plot the sensor temperature Tc . (c) Determine the steadystate sensor temperature Tc (t → ∞). SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.33(b). The sensible heat storage/release in the thermoelectric module is modeled by dividing its volume into two with one portion having the assumed uniform 196
Qk,h
Qk,ch Tc
T
Th
Qc = 0 Rk,ch 1 (ρc V) p TE 2
Rk,h dTh + (Se,J)h + (Se,P)h dt
 [ 14 (ρcpV)S + 12 (ρcpV)TE] dTc + (Se,J)c + (Se,P)c dt
1 of One Module 2
1 of Sensor 4
Figure Pr.3.33(b) Thermal circuit diagram.
temperature Tc (t) and the other one Th (t). The heat conduction from the hot junction to the surrounding silicon is given by the conduction resistance Rk,h∞ . (b) The integralvolume energy equation (2.9) for the cold junction (including the heat storage/release for the sensor and half of the thermoelectric modules) is 1 dTc dTc − ρcp V + (S˙ e,J )c + (S˙ e,P )c cold junctions Qc + Qk,ch = −(ρcp V )s dt 2 T E dt 1 dTh ρcp V + (S˙ e,J )h + (S˙ e,P )h hot junctions, Qk,hc + Qk,h∞ = − 2 T E dt where Qc = 0 c − Th , Qk,ch = TR k,ch
−1 Rk,c h =
Vs = aw2 ,
4 4 + (Rk,ch )p (Rk,ch )n
VT E = awL
(S˙ e,J )c = (S˙ e,J )h = 21 Je2 Re , (S˙ e,P )c = −4αS Je Tc ,
Re,hc = 4(Re,hc )p + 4(Re,hc )n αS = αS,p − αS,n
(S˙ e,P )h = 4αS Je Th L , kp lw ρ L (Re,hc )p = e,p , lw ln(2w/a) . Rk,h∞ = πkw
(Rk,hc )p =
L kn lw ρ L = e,n lw
(Rk,hc )n = (Re,hc )n
Note that we have used one thermoelectric module and 1/4 of the sensor volume in the energy equations. From Table C.9(a), for bismuth telluride, we have αS,n = −210 × 10−6 V/◦C −6
αS,p = 230 × 10
−5
ρe,n = 1.00 × 10
−5
ρe,p = 1.00 × 10
Table C.9(a)
◦
V/ C
Table C.9(a)
ohmm
Table C.9(a)
ohmm
Table C.9(a)
kn = 1.45 W/mK kp = 1.70 W/mK
Table C.9(a) Table C.9(a).
From Table C.2, for bismuth at T = 300 K, we have ρe = 9,790 kg/m3 cp,T E = 122 J/kgK
Table C.2 Table C.2.
197
From Table C.2, for silicon at T = 300 K, we have ρs = 2,330 kg/m3 cp,s = 678 J/kgK
Table C.2
ks = 149 W/mK
Table C.2.
Table C.2
The computed (using a solver, such as SOPHT) cold junction temperature Tc (t) is plotted in Figure Pr.3.33(c), as a function of time. The steady state is reached at an elapsed time at nearly t = 0.06 = 60 ms. Also plotted is Th = Th (t) and the results show that Th remains constant and nearly equal to T∞ . This is due to the rather small resistance between these two modes (i.e., Rk,h∞ Rk,hc ). 300
Th(t) , T
288
T, K
276 Tc(t) 264
Tc(t
) = 254.7 K
252
240 0
0.02
0.04
0.06
0.08
0.1
t, s Figure Pr.3.33(c) Variation of the hot and cold junction temperatures, with respect to time.
(c) The steadystate, cold junction temperature is found from Pr.3.33(c), or by solving the steady state energy equation (3.174). The result is Tc (t → ∞)
=
254.7 K = −18.40◦C.
COMMENT: With a smaller volume for the sensor, the response time can be reduced. Although neglected here, the parasitic heat leaks (i.e., Qc < 0) into the sensor prevent achieving low temperatures and also increases the response time. The electric power is 4Je2 Re,hc = 0.016 W = 16 mW and is considered reasonable for microelectronics. Also, since Th T∞ , the closedform solution to (3.172) can also be used. Note that since (S˙ e,P )c depends on Tc , we should use Qk,ch − (S˙ e,P )c
Tc − Th + αS Je Tc Rk,ch 1 1 = Tc ( + αS Je ) − Th ( ) Rk,ch Rk,ch 1 + αS Je ) + αS Je Th = (Tc − Th )( Rk,ch Tc − Th ≡ + Qc . Rk,c h =
Then we use this newly deﬁned Rk,ch and a1 = −Qc in (3.172).
198
PROBLEM 3.34.FUN GIVEN: A highly localized Joule heating applied to myocardium via a transvenous catheter can destroy (ablate) the endocardial tissue region that mediates lifethreatening arrhythmias. Alternating current, with radiofrequency range of wavelength, is used. This is shown in Figure Pr.3.34(a)(i). The current ﬂowing out of the spherical tip of the catheter ﬂows into the surrounding tissue, as shown in Figures Pr.3.34(a)(ii) and (iii). Due to the rapid decay of the current ﬂux, the Joule heating region is conﬁned to a small region Re ≤ r ≤ R1 adjacent to the electrode tip. The total current Je leaving the spherical tip results in a current density je =
Je . 4πr2
The tissue having a resistivity ρe will have a local energy conversion rate s˙ e,J = ρe je2 . T2 = 37◦C, Re = 1.0 mm, R1 = 1.3 mm, ρe = 2.24 ohmm, Je = 0.07476 A. Use Table C.17 for k of muscle. Assume steadystate heat transfer. SKETCH: Figure Pr.3.34(a)(i) The Joule heating of myocardium region by a spherical electrode tip. (ii) The small heated region. (iii) Heat ﬂow out of the heated region by conduction.
(i) RadioFrequency Catheter Ablation of Myocardium Endocardial Tissue Transvenous Catheter (Electrode) Je Ablation Region
Heart
(ii) JouleHeating Region Re R r 1
Tissue
He
se,J (r) Spherical Tip of Electrode
je(r)
(iii) Temperature Decay Region Re
Tissue k
R1
r
R2
, T2
Se,J (r) Electrode Qk,12
Control Surface A at T1
Region of Joule Heating Region of Temperature Decay
Figure Pr.3.34(a) Radiofrequency catheter ablation of myocardium showing the small a heated region and the conduction heat transfer from this region.
OBJECTIVE: Assuming that all the energy conversion occurs in the region Re ≤ r ≤ R1 , then the heat is conducted from this region toward the remaining tissue. The farﬁeld temperature is T2 , i.e., as r → ∞, T → T2 . (a) Derive the expression for the local S˙ e,J (r) and comment on its distribution. 199
(b) Draw the thermal circuit diagram and write the surface energy equation for its surface located at r = R1 . Use the conduction resistance for a spherical shell (Table 3.2). (c) Determine T1 (R1 ) for the following conditions. SOLUTION: (a) The Joule heating per unit volume is given by (2.33), i.e., s˙ e,J = ρe je2 . The current ﬂux je is related to the total current, for the spherical geometry considered, through je =
Je . 4πr2
Then
Je2 . 16π 2 r4 This shows that the local Joule heating rate drops very quickly as r increases. This is the reason for the small Joule heating region. s˙ e,J = ρe
(b) The thermal circuit diagram for the surface node T1 is shown in Figure Pr.3.34(b). Qk,12 T1
T2
A1
Rk,12
Se,J
Figure Pr.3.34(b) Thermal circuit diagram.
The surface energy equation is
T1 − T2 S˙ e,J = Qk,12 = . Rk,12
From Table 3.2 for R2 → ∞, we have Rk,12 =
1 . 4πR1 k
The integrated energy conversion rate S˙ e,J is S˙ e,J
R1
=
ρe Re
= = =
ρe Je2 4π
Je2 4πr2 dr 16π 2 r4 R1
r−2 dr
Re
1 −ρe Je2 1 − 4π R1 Re 2 1 1 ρe Je − . 4π Re R1
(c) Solving the energy equation for T1 , we have T1
= T2 + S˙ e,J Rk,12 1 1 ρe Je2 1 = T2 + − 4π Re R1 4πR1 k 1 ρe Je2 1 = T2 + − . R1 16π 2 R1 k Re 200
Using the numerical values, we have S˙ e,J
=
1 2.24(ohmm) × 0.074762 (m2 ) 1 × − = 0.230 W. 4π 10−3 (m) 1.3 × 10−3 (m)
From Table C.17 (for muscle tissue), k = 0.41 W/mK. Then T1
= =
0.230(W) 4π × 1.3 × 10 (m) × 0.41(W/mK) 37(◦C) + 34.36(◦C) = 71.36◦C.
37(◦C) +
−3
COMMENT: Since in the Joule heating the local s˙ e,J drops so rapidly with the increase in r, in practice a microwave heater, with a helical coil antenna design is used.
201
PROBLEM 3.35.FUN GIVEN: Refractive surgical lasers are used to correct the corneal refractive power of patients who are nearsighted, farsighted, or astigmatic in their vision. These corneal reshaping procedures can be performed via several mechanisms, including ablation of the corneal surface to change the focal length, removal of a section of the cornea causing reformation, and reshaping of the corneal tissue by thermal shrinkage eﬀects. In order to achieve minimal tissue thermal damage due to the laser ablation, we need to investigate the thermal behavior of the corneal tissue to understand the local thermal eﬀects of laser heating, and to predict the potential for an unintentional injury during laser surgery. Heat transfer in corneal tissue is modeled as a sphere of radius R1 = 5 mm, with the energy source positioned at the center of the sphere [Figure Pr.3.35(a)(i)]. A small diameter laser beam with S˙ e,α = Ar αr qr,i = 220 mW is used. Assume a steadystate conduction heat transfer. SKETCH: Figure Pr.3.35(a) shows the eye laser surgery and the heat transfer model.
(i) Laser Eye Surgery
(ii) Thermal Model
Lens Iris Laser Generator qr,i
SteadyState Conduction
Pupil
Se,α
R2
Beam
2R1
Se,=
k T1 T2
Cornea
Figure Pr.3.35(a)(i) Laser eye surgery showing the absorbed irradiation. (ii) The thermal model.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine Rk,12 . (c) Determine T1 of the laser beam at R1 = 1 mm, using the energy equation. (d) Plot the variation of T with respect to r. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.35(b). Qk,12 T1 Se,α
T2 Rk,12
Figure Pr.3.35(b) Thermal circuit diagram.
(b) Using (3.64), we have
Rk,12 =
1 1 − R1 R2 4πk
=
T1 − T2 . Qk,12
Given that R1 = 1 mm, R2 = 5 mm, and k = 0.6 W/mK, we have 1 1 − 0.001(mm) 0.005(mm) Rk,12 = = 106.1◦C/W. 4π × (0.6W/m◦C) 202
(c) From above, we have Qk,12 = with Qk,12 = S˙ e,α = 220 mW. Solving for T1 , we have
T1 − T2 , Rk,12
T1 = 60.34◦C.
(d) We determine T = T (r) using the diﬀerentialvolume energy equation (3.35). For spherical shells with a constant conductivity k, a dominant radial temperature gradient, and no volumetric conversion, we have −
1 d 2 dT kr = s˙ dr r2 dr
or −
1 d 2 dT r = 0. r2 dr dr
Integrating this once, we have dT a1 = 2. dr r Integrating this once, we have T (r) = −a1
1 + a2 . r
The boundary conditions are T1 = −a1
1 + a2 R1
and T2 = −a1
1 + a2 . R2
Solving for a1 and a2 , we have T2 − T1 1 1 a2 = 1 + T1 . − R1 R1 R2
T2 − T1 a1 = , 1 1 − R1 R2 Then
T2 − T1 1 1 1 − T (r) = T1 + 1 . − R1 r R1 R2
Figure Pr.3.35(c) shows the plot of T (r) versus R1 ≤ r ≤ R2 .
T, C
COMMENT: Note that T (r) drops rapidly as r increases. Also note that T (r = R1 ) = 60.84◦C is far above the reversible temperature limit of 42◦C. The region with T > 42◦C is shown in Figure Pr.3.35(c). 65 60 55 50 45 42 C 40 35 30 R1 = 1 2
Irreversible Damage
3
4
R2 = 5 mm
r, mm Figure Pr.3.35(c) Variation of the tissue temperature with respect to the radial position r.
203
PROBLEM 3.36.FUN GIVEN: Cardiac ablation refers to the technique of destroying heart tissue that is responsible for causing alternations of the normal heart rhythm. A welllocalized region of the endocardial tissue is destroyed by microwave heating. The electric ﬁeld is provided through a catheter that is inserted into the heart through a vein. One example of this cardiac ablation and the microwave catheter design is shown in Figure Pr.3.36(a). The region Re < r < Ro is heated with Re = 1 mm and frequency f = 3.00 × 1011 Hz. A long catheter is assumed, so the heat transfer is dominant in the cylindrical crosssectional plane [shown in Figure Pr.3.36(a)]. The dielectric loss factor for heart tissue is ec = 17.6 and its thermal conductivity is k = 0.45 W/mK. SKETCH: Figure Pr.3.36(a) shows the microwave antenna and the heated tissue region.
Cylindrical Cross Section Torso Tissue
Myocardium
Antenna
Ro
Antenna
r Re
dT = 0 dr To
se,m = 2 F f
2
o ecee
Coaxial Cable
Figure Pr.3.36(a) A microwave catheter used for ablation of cardiac tissue.
OBJECTIVE: (a) Assuming a zero temperature gradient r = Re , start with the onedimensional (radial direction) diﬀerential energy equation with s˙ e,m = s˙ e,m (r), then integrate this diﬀerential equation to obtain the radial distribution of the temperature T = T (r). It is known that e2e = e2e,o (Re /r)2 . In addition to thermal conditions dT /dr = 0 at r = Re use T = To at r = Ro . (b) In order to perform a successful ablation, the microwave antenna needs to produce a temperature Te = 353 K (80◦C) at r = Re . Given that temperature decreases to the normal tissue temperature To = 310.65 K (37.5◦C) at r = 1 cm, determine the required electric ﬁeld intensity ee,o to produce this temperature. (c) Plot the variation of T (r) with respect to r for several values of ee,o . SOLUTION: (a) The energy equation for steady state conduction with an energy conservation in cylindrical coordinate is given by (3.33), i.e.,
∇ · qk s˙ e,m (r)
= − =
dT 1 d rk = s˙ e,m (r), r dr dr
2πf o ec e2e ,
e2e = e2e,o
Then
−
1 d dT rk = 2πf o ec e2e,o r dr dr 204
Re r
2 .
Re r
2 .
Using a = 2πf o ec e2e,o Re2 , we have −
dT 1 d rk r dr dr dT − dr To − dT Te
dT
1 r2 a ln r a1 1 = + k r k r Ro Ro a ln r a1 1 dr + dr = k r Re k r Re = a
a(ln r)2 a1 + ln r + a2 . 2k k
=
To solve for the constants a1 and a2 , we use the following bounding surface thermal conditions:
Then 0=
at r
= Re ,
at r
= Ro ,
a1 a ln r + , k r kr
∂T =0 ∂r T = To . a1 = −a ln Re
and Te − To
=
a2
=
a ln Re ln Ro a (ln Ro )2 − + a2 k 2 k a a (ln Ro )2 ln Re ln Ro − + (Te − To ). k k 2
Substituting for a1 and a2 , the radial temperature distribution is a (ln r)2 (ln Ro )2 − ln Re ln r + ln Re ln Ro − T (r) = To − . k 2 2 (b) Using Te = 355 K and To = 310.65 K (at r = Re = 1 mm and Ro = 1 cm, respectively), we have a1 = 0.047 W/mK, a1 /k = 1.04, and the required ee,o is 160 V/m.
T, K
(c) Figure Pr.3.36(b) shows the variation of temperature T (r) with respect to r, for several values of ee,o . 360 355 350 345 340 335 330 325 320 315 310 305
ee,o = 160 V/m 150 100 40 30 20 10
1
2
3
4
5
6
7
8
9
10
11
r, mm Figure Pr.3.36(b) Variation of T (r) with respect to r for several values of ee,o .
COMMENT: Note that the zero derivative of the temperature at r = Re shows no heat loss across this surface.
205
PROBLEM 3.37.FUN GIVEN: The thermoelectric ﬁgure of merit Ze = αS2 /(Re /Rk ) is maximized by minimizing Re /Rk . OBJECTIVE: Begin with the relations for Re and Rk , and the electrical and thermal properties, i.e., ρe,n , ρe,p . Then calculate kn , kp , and the geometrical parameters Ln , Ak,n , Lp , and Ak,p , as given by (3.116). (a) Then assume that Ln = Lp and minimize Re /Rk with respect to Ak,p /Ak,n . Then show that Ln Ak,p Ak,p = = Lp Ak,n Ak,n
ρe,p kn ρe,n kp
1/2 for optimum Ze ,
which is (3.121). (b) Use this in (3.120), and show that Ze =
αS2 [(kρe )1/2 p
+ (kρe )1/2 n ]
,
optimized ﬁgure of merit.
SOLUTION: (a) From (3.116), we have Re Rk
= =
ρe L Ak
+ p
ρe L Ak
n
(ρe k)p + (ρe k)n + ρe,p kn
Ak k L
+ p
Ak k L
n
Ak,n Lp Ak,p Ln + ρe,n kp . Ak,p Ln Ak,n Lp
Using Ln = Lp , we have Re Ak,n Ak,p = (ρe k)p + (ρe k)n + ρe,p kn + ρe,n kp . Rk Ak,p Ak,n Taking the derivative with respect to Ak,p /Ak,n , and setting the resultant to zero, we have 2 Ak,n d(Re /Rk ) = −ρe,p kn + ρe,n kp = 0, d(Ak,p /Ak,n ) Ak,p or, while remembering that Ln = Lp , Ln Ak,p = Lp Ak,n
ρe,p kn ρe,n kp
1/2 .
(b) Making this substitution, we have Re Rk
ρe,n kp ρe,p kn
1/2
=
(ρe k)p + (ρe k)n + ρe,p kn
+ ρe,n kp
=
(ρe k)p + (ρe k)n + 2(ρe,p kn )1/2 (ρe,n kp )1/2
=
2 [(ρe k)1/2 + (ρe k)1/2 p n ] .
Then (3.120) becomes Ze =
αS2 2 [(kρe )1/2 + (kρe )1/2 p n ]
COMMENT: The length requirement Lp = Ln is indeed a practical necessity. 206
.
ρe,p kn ρe,n kp
1/2
PROBLEM 3.38.FUN GIVEN: Begin with the diﬀerentiallength energy equation (3.102), and use the prescribed thermal boundary conditions (3.103) for a ﬁnite length slab of thickness L. OBJECTIVE: (a) Derive the temperature distribution given by (3.104). (b) Show that the location of the maximum temperature is x(Tmax ) =
L k(Th − Tc ) + . 2 ρe je2 L
(c) Comment on this location for the case of (i) je → 0, and (ii) je → ∞. SOLUTION: (a) Starting with (3.102), i.e., −k
d2 T = ρe je2 , dx2
d2 T ρe je2 2 =− k , dx
and performing the integration once, we have dT ρe j 2 = − e x + a1 . dt k Integrating once more, we have T (x) = −
ρe je2 2 x + a1 x + a2 . 2k
Now, the conditions at x = 0 and x = L are given by (3.103), i.e., T (x = 0) = Tc ,
T (x = L) = Th .
Using the ﬁrst of these conditions, we have T (x = 0) = Tc = −
ρe je2 2 (0) + a1 (0) + a2 2k
or a2 = Tc . Using the second of these conditions, we have T (x = L) = Th = −
ρe je2 2 L + a1 L + Tc 2k
or a1 =
ρe je2 L Th − Tc + . L 2k
Then for a1 and a2 , we have T (x) = Tc +
x ρe je2 (Th − Tc ) + x(L − x). L 2k
(b) By diﬀerentiating the temperature distribution with respect to x, and setting the resultant to zero, we have dT (x) dx
= =
Th − Tc ρe je2 ρe je2 x + (L − x) − L 2k 2k Th − Tc ρe je2 L ρe je2 x + − = 0. L 2k k 207
Solving for x, we have x(Tmax ) =
k(Th − Tc ) L + . 2 ρe je2 L
(c) For (i), where je = 0, we have x(Tmax )je =0 , i.e., there is not a location for the maximum temperature within 0 ≤ x ≤ L. This indicates that no maximum will occurs (i.e., linear temperature distribution). For (ii), where je → ∞, we have x(Tmax ) = L/2, i.e., for a ﬁnite diﬀerence between Th and Tc , the location of the maximum temperature is at the center (for a very large volumetric Joule heating rate). COMMENT: As shown in Figure 3.27, as je is increased, x(Tmax ) moves toward the center. Note that there is current for which x(Tmax ) = L, i.e., the hot surface will be the location of Tmax (with no gradient in temperature, and therefore, no conduction at x = L). This current density is found by x(Tmax ) = L =
L k(Th − Tc ) + 2 ρe je2 L
or
2k(Th − Tc ) je = ρe L2
208
1/2 .
PROBLEM 3.39.FUN GIVEN: The optimum coeﬃcient of performance for the thermoelectric cooler ηcop , given by (3.126), is based on a current that optimizes it. This current is also given in (3.126). OBJECTIVE: Derive the expression for the optimum current, i.e., Je [∂ηcop /∂Je = 0] =
αS (Th − Tc ) Re,hc [(1 + Ze To )1/2 − 1]
,
where Ze is given by (3.120) and To =
Th + Tc . 2
SOLUTION: Starting from (3.126), we diﬀerentiate ηcop with respect to Je and we have 1 −1 2 [αS Tc − Re,h c (Th − Tc ) − 2 Re,hc Je ][2Re,hc Je + αS (Th − Tc )] ∂ηcop αS Je Tc − Re,hc Je = =0 − ∂Je Re,hc Je2 + αS Je (Th − Tc ) [Re,hc Je2 + αS Je (Th − Tc )]2 or 1 −1 2 (αS Tc − Re,hc Je )[Re,hc Je2 + αS Je (Th − Tc )] − [αS Je Tc + Rk,h c (Th − Tc ) + 2 Re,hc Je ][2Re,hc Je + αS (Th − Tc )] = 0 or 2 3 2 2 2 αS Tc Re,hc Je2 − Re,h c Je + α Tc Je (Th − Tc ) − Re,hc Je αS (Th − Tc ) − 2αS Je Tc Re,hc − 1 −1 −1 2 2 3 2 αS2 Je Tc (Th − Tc ) + 2Rk,h c (Th − Tc )Je Re,hc + Rk,hc (Th − Tc ) αS + Re,hc Je + 2 Re,hc Je αS (Th − Tc ) = 0. Then 1 −1 −1 2 −αS Je2 Tc Re,hc − Re,hc Je2 αS2 (Th − Tc ) + 2Rk,h c Re,hc Je (Th − Tc ) + Rk,hc (Th − Tc ) αS = 0. 2
By combining the terms containing Je2 , we have Th − Tc −1 −1 + Tc Je2 − 2Rk,h Re,hc αS c Re,hc (Th − Tc )Je − Rk,hc αS (Th − Tc ) = 0. 2 The acceptable solution to this quadratic equation is 1/2 Th − Tc −1 −1 −1 2 2 2 + T R (T − T ) + [2R R (T − T )] + 4R R α (T − T ) 2Rk,h c c c c c e,hc h k,hc e,hc h k,hc e,hc S h 2 Je = . Th − Tc + Tc 2Re,hc αS 2 Then
Je
1/2 Th − Tc + Tc 1 + 1 + Ze 2 −1 = Rk,h c Re,hc (Th − Tc ) Th − Tc + Tc Re,hc αS 2 −1 = Rk,h c Re,hc (Th − Tc )
=
1 + (1 + Ze To )1/2 Re,hc αS To
−1 1/2 Rk,h c (Th − Tc ) 1 + (1 + Ze To )
αS
To 209
,
To =
Th + Tc , 2
Ze =
αS2 . −1 Rk,h c Re,hc
This can be rearranged by multiplying and dividing to achieve Je
= =
−1 1/2 Rk,h (1 + Ze To )1/2 − 1 c (Th − Tc ) Re,hc 1 + (1 + Ze To ) αS Rk,hc To (1 + Ze To )1/2 − 1 αS (Th − Tc ) Re,hc [(1 + Ze To )1/2 − 1]
.
COMMENT: To arrive at the optimum ηcop , we substitute this current in the deﬁnition for ηcop , i.e., (3.126), and expand the expressions and then recombine them.
210
PROBLEM 3.40.FUN GIVEN: In thermoelectric cooling, the lowest temperature for the cold junction Tc corresponds to Qc = 0 and is given by (3.119). Further lowering of Tc is possible by using thermoelectric units in averaged stages. A twostage unit is shown in Figure Pr.3.40(a). Assume that the temperature drops across the electrical conductor and insulator are negligible such that the top junction is at Tc , the intermediate junction is at T1 , and the lower junction is at Th . Also assume no heat loss at the T1 junction, Q1 = 0, Use the bismuth telluride pn pair and assume Qc = 0. a = 1 mm, L = 1.5 mm, Th = 40◦C, Je = 4 A. SKETCH: Figure Pr.3.40(a) shows the twostage thermoelectric cooler. The temperature drops across the electrical insulators and conductors are assumed to be negligible. Tc
T1
p
n
a
p
a L
a Electrical Insulator
n
a L
a Electrical Conductor
 Qc
a
a
a
Th Qh
Figure Pr.3.40(a) A twostage thermoelectric cooler unit.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Show that Tc (Qc = 0) is given by Th 1 Rk + Re Je2 Rk αS Je + 2 2 Tc = . 1 1 αS Je + 1 − Rk αS Je + 2 Rk Rc Je2 +
(c) Start by writing the junction energy equation (3.115) for the Tc and T1 junctions. Determine Tc (Qc = 0) for the above conditions. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.40(b). Heat losses from junctions Tc and T1 are assumed to be zero. (b) The energy equations for Tc and T1 nodes are found using (3.115), i.e., Tc − T1 Rk T1 − Tc T1 − Th Q1 + + Rk Rk Qc +
1 = −αS Je Tc + Re Je2 2 = −αS Je T1 + Re Je2 ,
where we have used the same Rk and Re for both stages. Noting that Qc = Q1 = 0, we need to eliminate T1 between these two equations and then solve for Tc . From the second equation, we have αS Je T1 +
2T1 Th + Tc = + Re Je2 Rk Rk 211
Qc = 0 Tc
(Se,P)c + (Se,J)c
Ac Rk,1c A1
Qk,1c (Se,P)1 + (Se,J)1
Q1 = 0 T1 Rk,h1
Qk,h1
Th Qh
Figure Pr.3.40(b) Thermal circuit diagram.
or Th + Tc + Re Je2 Rk T1 = . 2 + αS Je Rk Now substituting this in the ﬁrst equation, we have T1 Tc 1 − + Re Je2 = Rk Rk 2 Th + Tc R J2 + 1 e e Tc 1 Rk − + Re Je2 = −αS Je Tc + 2 Rk Rk 2 αS Je + Rk Th Re Je2 + 1 Tc 1 Rk + −1 −αS Je Tc + + Re Je2 . Rk αS Je + 2 Rk αS Je + 2 Rk 2 0 = −αS Je Tc +
Solving for Tc we have Th 1 Rk + Re Je2 Rk αS Je + 2 2 Tc = . 1 1 −1 αS Je − Rk αS Je + 2 Rk Re Je2 +
(c) From Example 3.14, for a bismuth telluride pair and for the given geometry, we have αS
=
4.4 × 10−4 V/K
Re Rk
= =
0.030 ohm 4.762 × 102 K/W.
Then (273.15 + 40)(K) 1 476.2(K/W) + × 0.030(ohm) × 42 (A2 ) 476.2(K/W) × 4.4 × 10−4 (V/K) × 4(A) + 2 2 Tc = . 1 1 4.4 × 10−4 (V/K) × 4(A) − − 1 476.2(K/W) 476.2(K/W) × 4.4 × 10−2 (V/K) × 4(A) + 2 0.030(ohm) × 42 (A2 ) +
212
Tc
=
0.48 + 0.6576 + 0.24 ohmA2 0.8381 +2 1 VA/K 1 −1 1.760 × 10−3 − 0.8381 + 2 476.2 2
= =
V/A A 0.4008 + 0.24 1.760 × 10−3 + 1.360 × 10−3 VA/K 205.4 K.
COMMENT: Note that for a singlestage unit, from (3.115) we have
Tc (Qc = 0)
=
=
=
1 Th Re Je2 + 2 Rk 1 αS Je + Rk 1 313.15 × 0.030 × 42 + 2 476.2 1 −4 4.4 × 10 × 4 + 476.2 0.24 + 0.6576 = 232.5 K. 1.760 × 10−3 + 2.10 × 10−3
As expected, this is higher than Tc found in part (c) for the twostage thermoelectric cooler unit.
213
PROBLEM 3.41.DES GIVEN: An inplane thermoelectric device is used to cool a microchip. It has bismuth telluride elements with dimensions given below. A contact resistance Rk,c = lc /Ak kc , given by (3.94), is present between the elements and the connector. These are shown in Figure Pr.3.41(a). The contact conductivities kc is empirically determined for two diﬀerent connector materials and are kc = 10kT E
copper connector
kc = kT E
solder connector,
where kT E is the average of p and ntype materials. L = 150 µm, w = 75 µm, a = 4 µm, Th = 300 K. SKETCH: Figure Pr.3.41(a) shows the miniaturized thermoelectric cooler unit. Contact Resistance, Rk,c Tc Ts,c
Th
Tc
w lc
Thermoelement
Electrical Connector (Copper or Solder)
Microchip (Cooled Area)
L
1 mm
a
Substrate, Assumed an Ideal Insulator (ks = 0)
Figure Pr.3.41(a) A miniaturized thermoelectric device with a thermal contact resistance between the thermoelectric elements and the connectors.
OBJECTIVE: (a) Draw the thermal circuit diagram showing the contact resistance at each end of the elements. (b) Determine the optimum current for cold junction temperature Tc = 275 K. (c) Determine the minimum Tc [i.e., (3.119)] for these conditions. (d) For Tc = 275 K, determine Qc,max from (3.118). (e) Using this Qc,max and (3.96), determine ∆Tc = Ts,c − Tc and plot Ts,c versus lc for 0 < lc < 10 µm, for both the copper and solder connectors. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.41(b). (b) The current for this temperature is given by (3.117), i.e., Je =
αS Tc . Re,hc 214
.
Se,J = 1 Re Je2 2
.
Control Surface Ac
Se,J = 1 Re Je2 2
Control Surface Ah Ts,h
Je
Th
Ts,c
Tc
Qh
 Qc
Rk,c
Rk,c
.
Se,P = αS Je Th
.
T  Tc Qk,hc = h Rk,hc
Se,P = αS Je Tc
Figure Pr.3.41(b) Thermal circuit diagram.
The electrical resistance is given by (3.116) and, using the ρe values from Table C.9(a), we have αS,p αS,n
Re,hc
= 230 × 10−6 V/K = −210 × 10−6 V/K −5
Table C.9(a) Table C.9(a)
ρe kp
= =
10 ohmm 1.70 W/mK
Table C.9(a)
kn
=
1.45 W/mK
Table C.9(a),
Table C.9(a)
=
ρe,p Lp ρe,n Ln L (ρe,p + ρe,n ) + = Ap An wa
=
1.5 × 10−4 (m) × 2 × (10−5 )(ohmm) = 10 ohm. 7.5 × 10−5 (m) × 4 × 10−6 (m)
Using αS = αS,p − αS,n = 440 µV/K, the current is then Je =
(440 × 10−6 )(V/K) × 275(K) = 1.210 × 10−2 A. 10(ohm)
(c) The minimum Tc for a given Th is given by (3.119) as (Th − Tc )max
=
αS2 Tc2 2Re,hc /Rk,hc
−1 Rk,h c
=
Ak,p kp Ak,n kn wa 7.5 × 10−5 (m) × 4 × 10−6 (m) (kp + kn ) = + = (1.70 + 1.45)(W/mK) Lp Ln L 1.5 × 10−4 (m)
=
6.360 × 10−6 W/K.
This minimum Tc is found from 300(K) − Tc (K)
=
[440 × 10−6 (V/K)]2 Tc2 (K2 ) 2 × 10(ohm) × 6.360 × 10−6 (W/K)
=
1.522 × 10−3 Tc2 (K2 )
=
223.8 K.
or Tc,min
(d) For Tc = 275 K, Qc,max can be found by (3.118) as Qc,max
= −
αS2 Tc2 −1 + Rk,h c (Th − Tc ) 2Re,hc
= −
[440 × 10−6 (V/K)]2 × (275)2 (K2 ) + 6.360 × 10−6 (W/K) × (25)(K) = (−7.321 × 10−4 + 1.590 × 10−4 )(W) 2 × 10(ohm)
= −5.731 × 10−4 W. 215
(e) The value of the average thermoelectric conductivity is kT E =
1.70 + 1.45 kp + kn = = 1.575 W/mK. 2 2
The value of the gap conductivity can be found using the given relationships to be kc
=
kc
= kT E = 1.575 W/mK
10kT E = 15.75 W/mK
copper connector solder connector.
By varying values of lc , the thermal contact resistance can then be found using (3.94), Rk,c =
lc lc . = kc Ak kc wa
Finally, Ts,c is found using (3.127) as Qc =
Tc − Ts,c Rk,c
or Ts,c = Tc − Qc Rk,c = 275(K) + 5.731 × 10−4 (W) ×
lc −10
3 × 10
(m2 )kc
.
The plots of Ts,c versus lc for both the copper and solder connectors are shown in Figures Pr.3.41(c) and (d). Note that while the contact resistance of the copper connector is negligible, that of the solder is not.
(c)
Copper Connector, kc = 10 kTE 276.4 276.2
Ts,c , K
276.0 275.8 275.6 275.4 275.2 275.0 0
(d)
2
4
lc , mm
6
8
10
Solder Connector, kc = kTE 289 287
Ts,c , K
285 283 281 279 277 275 0
2
4
lc , mm
6
8
10
Figure Pr.3.41 Variation of the rise in cold junction temperature across the contact resistance, with respect to the contact gap length, for (c) copper and (d) solder connectors.
216
COMMENT: As lc increases, the value of Ts,c increases linearly. For the copper connectors, and for lc = 10 µm, the temperature rise across the contact is 1.213◦C (4.852 percent the temperature diﬀerence of 25◦C). The cold side temperature is still far enough below Th = 300 K to cool the microchip. If the solder connector is used, then the value of Ts,c will rise about 12.13◦C (48.52 percent the temperature diﬀerence). For this reason, copper or other materials with high thermal and electrical conductivity are used for the connections between the thermoelectric pairs. Note that the substrate thermal conductivity is assumed zero. In practice this has a ﬁnite value and reduces the performance.
217
PROBLEM 3.42.FAM GIVEN: To melt the ice forming on a road pavement (or similarly to prevent surface freezing), pipes are buried under the pavement surface, as shown in Figure Pr.3.42(a). The pipe surface is at temperature T1 , while the surface is at temperature T2 . The magnitude of the geometrical parameters for the buried pipes are given below. Use the thermophysical properties of soil (Table C.17). L = 5 cm, D = 1 cm, l = 2 m, w = 20 cm, T1 = 10◦C, T2 = 0◦C. Assume that the conduction resistances given in Table 3.3(a) are applicable. SKETCH: Figure Pr.3.42(a) shows the hotwater carrying buried pipes.
Melting of Surface Ice by Burried Pipes Carrying Hot Water IceCovered Surface T2 l
L
D Soil, DkE
w Tf = T1 Hot Water Pipe
Qk,12
Figure Pr.3.42(a) Hotwater carrying buried pipes used for melting of an ice layer on a pavement surface.
OBJECTIVE: (a) Draw the thermal circuit diagram for each pipe. (b) From Table 3.3(a), determine the conduction resistance for (i) a single pipe (i.e., cylinder) independent of the adjacent pipes, and (ii) a pipe in a row of cylinders with equal depth and an axial centertocenter spacing w. (c) Determine Qk,12 per pipe for both cases (i) and (ii), and then compare. SOLUTION: (a) Figure Pr.3.42(b) shows the thermal circuit diagram for each pipe. T2
Rk,12
Qk,12
T1
Figure Pr.3.42(b) Thermal circuit diagram.
(b) The conduction resistance for each pipe is given by Table 3.3(a), i.e., 4L ln D for D < L < l. Rk,12 = 2πkl Using (3.128), we have Qk,12 =
T1 − T2 . Rk,12 218
From Table C.17, we have k = 0.52 W/mK = 0.52 W/m◦C
soil: Then
Table C.17
4 × 0.05(m) 0.01(m) 2 × π × 0.52(W/m◦C) × 2(m) ln(20) ◦ ( C/W) = 0.4584◦C/W for each pipe. 6.535 ln
Rk,12
= =
(c) For the heat transfer per pipe, we have Qk,12 for each pipe
=
(10 − 0)(◦C) = 21.81 W. 0.4584(◦C/W)
COMMENT: The centertocenter spacing of adjacent pipes w was assumed to be suﬃciently large such that the heat transfer from each pipe could be assumed independent of the adjacent pipes. As w decreases, the eﬀect of the adjacent pipes on the heat transfer of a single pipe must be considered. The correlation for the thermal resistance from a single cylinder to a row of cylinders at equal depth in a semiinﬁnite solid, as a function of centertocenter spacing w, is given in Table 3.3(a),i.e., 2πL w sinh ln πR w for each cylinder in a row of cylinders. Rk,12 = 2πkl The variation of Rk,12 for each pipe of a row of cylinders as a function of the spacing w is given in Figure Pr.3.42(c). Also shown is the Rk,12 for a single cylinder. Note that the thermal resistance for each pipe in the row increases as w decreases. This is because the adjacent pipes raise the temperature of the solid medium, resulting in a decrease in the heat transfer from each pipe.
Rk,12 , C/W (per cylinder)
2.5 Single Cylinder Row of Cylinders 2
L = 5 cm D = 1 cm l=2m
1.5
Single Cylinder Approximation for Given L, D, and l
1
0.5
0 0
0.1
0.2
0.3
0.4
0.5
0.6
w, m
Figure Pr.3.42(c) Variation of conduction resistance with respect to pipe spacing.
219
PROBLEM 3.43.FUN GIVEN: The steadystate conduction in a twodimensional, rectangular medium, as shown in Figure Pr.3.43, is given by the diﬀerential volume energy equation (B.55), i.e., ∇qk = −k
∂2T ∂2T − k = 0. ∂x2 ∂y 2
This is called a homogeneous, linear, partial diﬀerential equation and a general solution that separates the variable is possible, if the boundary (boundingsurface) conditions can also be homogeneous. This would require that the temperatures on all surfaces be prescribed, or the bounding surface energy equation be a linear resistive type. When the four surface temperatures are prescribed, such that three surfaces have a temperature and different from the fourth, the ﬁnal solution would have a simple form. SKETCH: Figure Pr.3.43 shows the geometry and the prescribed temperatures on the four surfaces. T2 , T* = 1 Lx
Lz
Ly
T1 , T* = 0 T1 , T* = 0 y
T1 , T* = 0 x TwoDimensional Temperature Distribution T = T(x,y)
Figure Pr.3.43 A rectangular, twodimensional geometry with prescribed surface temperatures.
OBJECTIVE: (a) Using the dimensionless temperature distribution T − T1 , T2 − T1 show that the energy equation and boundary conditions become T∗ =
∂2T ∗ ∂2T ∗ + = 0, T ∗ = T ∗ (x, y) ∂x2 ∂y 2 T ∗ (x = 0, y) = 0, T ∗ (x = Lx , y) = 0, T ∗ (x, y = 0) = 0, T ∗ (x, y = Ly ) = 1. (b) Use the method of the separation of variables (which is applicable to this homogeneous diﬀerential equation with all but one boundary conditions being also homogeneous), i.e., T ∗ (x, y) = X(x)Y (y) to show that the energy equation becomes 1 d2 X 1 d2 Y . = 2 X dX Y dy (c) Since the lefthand side is only a function of x and the righthand side is a function of y, then both sides should be equal to a constant. This is called the separation constant. Showing this constant as b2 , show also that −
d2 X + b2 X = 0 dx2 d2 Y + b2 Y = 0. dy 2 220
Then show that the solutions for X and Y are
or
X
= a1 cos(bx) + a2 sin(bx)
Y
= a3 e−by + a4 eby
T ∗ = [a1 cos(bx) + a2 sin(bx)](a3 e−by + a4 eby ).
(d) Apply the homogeneous boundary conditions to show that a1 = 0 a3 = a4 a2 a4 sin(bLx )(eby − e−by ) = 0. Note that the last one would require that sin(bLx ) = 0 or bLx = nπ,
n = 0, 1, 2, 3, · · · .
(e) Using these, show that
nπx T (x, y) = a2 a4 sin (enπy/Lx − e−nπy/Lx ) Lx nπx nπy sinh . ≡ an sin Lx Lx ∗
Since n = 1, 2, 3, · · · and diﬀerential equation is linear, show that ∞
nπx nπy T∗ = an sin sinh . Lx Lx n=0 (f) Using the last (i.e., nonhomogeneous) boundary condition and the orthogonality condition of the special function sin(z), it can be shown that an =
2[1 + (−1)n+1 ] , n = 0, 1, 2, 3, · · · . nπ sinh(nπLy /Lx )
Then express the ﬁnal solution in terms of this and verify that all boundary conditions are satisﬁed. SOLUTION: (a) Using the prescribed temperatures T1 and T2 , we have T ∗ (x, y) =
T (x, y) − T1 T2 − T1
and the energy equation and boundary conditions become ∂ 2 T ∗ (x, y) ∂ 2 T ∗ (x, y) + =0 ∂x2 ∂y 2 T ∗ (x = 0, y) = T ∗ (x = Lx , y) = T ∗ (x, y = 0) = 0 T ∗ (x, y = Ly ) = 1. (b) Using
T ∗ (x, y) = X(x)Y (y)
in the above energy equation, we have Y (y)
d2 X(x) d2 Y (y) + X(x) =0 2 dx dy 2 221
or −
1 d2 X(x) 1 d2 Y (y) = . X(x) dx2 Y (y) dy 2
This results in two ordinary diﬀerential equations being equal, while each side can only be a function of one independent variable. This will only be possible if these two equations are equal to a constant. (c) Using intuition for the form of the solution, we set this constant equal to b2 . Then we have d2 X(x) + b2 X(x) = 0 dx2 d2 Y (y) − b2 Y (y) = 0. dy 2 The solutions to these two diﬀerential equations take the form of special sinusoidal and exponential (or hyperbolic) functions, respectively. These are X(x) = a1 cos(bx) + a2 sin(bx) Y (y) = a3 e−by + a4 eby or T ∗ (x, y) = [a1 cos(bx) + a2 sin(bx)](a3 e−by + a4 eby ). (d) Using T ∗ (x = 0, y) = 0, we have 0 = (a1 cos 0 + a2 sin 0)(a3 e−by + a4 eby ) or a1 = 0. Using Tx,y=0 = 0, we have 0 = a2 sin(bx)(a3 + a4 ) or a3 + a4 = 0,
or
a3 = −a4 .
∗
Using T (x = Lx , y) = 0, we have
0 = a2 sin(bLx )(a3 e−by + a4 eby )
or sin(bLx ) = 0. This would require that b=
nπ , n = 0, 1, 2, 3, · · · . Lx
(e) Now combining these, we have
T ∗ (x, y)
nπy −nπy nπx = a2 a4 sin e Lx − e Lx Lx nπx nπy ≡ an sin sinh , Lx Lx
where we have used an
=
sinh(z) ≡
2a2 a4 ez + e−z . 2
We expect an to depend on n and this will be shown below. 222
Since n = 0, 1, 2, 3, · · · , and since the energy equation used is a linear diﬀerential equation, the sum of the solutions corresponding to n = 0, n = 1, n = 2, etc., is also a solution. Then T ∗ (x, y) =
∞
an sin
n=0
nπx Lx
sinh
nπy Lx
.
(f) Using T ∗ (x, y = Ly ) = 1, we have 1=
∞
n=0
an sin
nπx Lx
sin
nπLy Lx
We now multiply both sides by sin(nπx/Lx ) and then integrate the resultants over 0 ≤ x ≤ Lx . Then it can be shown that 2[1 + (−1)n+1 ] . an = nπ sinh(nπLy /Lx ) Now combining these, we have
T ∗ (x, y) =
2 π
∞
1 + (−1)n+1 sin n n=0
nπy sinh nπx L x . nπLy Lx sinh Lx
COMMENT: The series solution would require a large number of terms in order to obtain a smooth temperature distribution. In a later exercise, we will compare this result with that found from a ﬁnitesmall volume numerical solution.
223
PROBLEM 3.44.FAM GIVEN: In order to maintain a permanent frozen state (permafrost) and a ﬁrm ground, heat pipes are used to cool and freeze wet soil in the arctic regions. Figure Pr.3.44(a) shows a heat pipe, which is assumed to have a uniform temperature T1 , placed between the warmer soil temperature T2 , and the colder ambient air temperature Tf,∞ . The heat transfer between the heat pipe surface and the ambient air is by surface convection and this resistance is given by Aku Rku . The heat transfer between the pipe and soil is by conduction. Assume a steadystate heat transfer. D = 1 m, Lku = 5 m, Lk = 2 m, Tf,∞ = −20◦C, T2 = 0◦C, Aku Rku = 10−1 K/(W/m2 ). Use Table 3.3(b) to determine the resistance Rk,12 . SKETCH: Figure Pr.3.44(a) shows the heat pipe, the farﬁeld ambient air temperature Tf,∞ , and the soil temperature T2 . D
Heat Pipe Ambient Air
Lku
Crossing Air Stream g
FarField Fluid Temperature, Tf, Pipe at Uniform Temperature Tf, < T1 < T2
(Aku Rku)1 Wet Soil
Lk
Rk,12
FarField Soil Temperature, T2
Sls
Figure Pr.3.44(a) A rendering of a heat pipe used for the maintenance of a permafrost layer in the arctic regions.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the heat pipe temperature T1 and the amount of heat ﬂow rate Qk,21 . (c) If this heat is used entirely in phase change (solidiﬁcation of liquid water), determine the rate of ice formation around the buried pipe. SOLUTION: (a) The thermal circuit diagram is shown Figure Pr.3.44(b). From node T1 , heat ﬂows by surface convection and by conduction. Tf, Qku,1
Rku,1 T1
Qk,12
Rk,12 T2 Q2
Sls = Mls ,hls
Figure Pr.3.44(b) Thermal circuit diagram.
(b) From Figure Pr.3.44(b), we have Qk,12 + Qku,1∞ = 0 =
T1 − T2 T1 − Tf,∞ + . Rk,12 Rku,1∞ 224
From Table 3.3(b), for an indented object in a semiinﬁnite medium (soil), we have Rk,12 =
ln(4Lk /D) , 2πkLk
where k is the soil conductivity. From Table C.17, we have k = 0.52 W/mK
Table C.17.
Then 4 × 2(m) 1(m) = 0.3182 K/W = 0.3182◦C/W. 2π × 0.52(W/mK) × 2(m) ln
Rk,12
=
For Rku , we have Rku
=
Aku Rku Aku Rku = Aku πDLku + πD2 /4
=
10−1 [K/(W/m2 )] = 6.063 × 10−3 K/W = 6.063◦C/W. π × 1(m) × 5(m) + π × 12 (m2 )/4
Solving the energy equation for T1 , we have T2
T1
Tf,∞ Rk,12 Rku,1∞ 1 1 + Rk,12 Rku,1∞ −20◦C 0◦C + ◦ 0.3182( C/W) 6.063 × 10−3 (◦C/W) = −19.63◦C. (3.143 + 1.649 × 102 )(W/◦C) T2 − T1 [0 − (−19.63)](◦C) = 61.69 W. = Rk,12 0.3182(◦C/W)
=
= Qk,21
=
+
(c) From Figure Pr.3.44(b), we have Q2 + Qk,21 = S˙ ls . From Table 2.1, we have S˙ ls = −M˙ ls ∆hls . From Table C.4, for water, we have ∆hsl = 3.336 × 105 J/kg = −∆hls , then M˙ ls
=
Qk,21 61.69(W) = ∆hsl 3.336 × 105 (J/kg)
=
1.849 × 10−4 kg/s = 0.1849 g/s = 665.7 g/hr.
COMMENT: Note that since Rku,1∞ Rk,12 , the heatpipe temperature is nearly that of the ambient air, i.e., T1 Tf,∞ .
225
PROBLEM 3.45.FAM.S GIVEN: In scribing of disks by pulsed laser irradiation (also called laser zone texturing) a small region, diameter D1 , is melted and upon solidiﬁcation a protuberance (bump) is formed in this location. The surface of the liquid pool formed through heating is not uniform and depends on the laser energy and its duration, which in turn also inﬂuences the depth of the pool L1 (t). These are shown in Figure Pr.3.45(a). Assume that the irradiated region is already at the melting temperature T1 = Tsl and the absorbed irradiation energy (S˙ e,α )1 is used to either melt the substrate S˙ sl , or is lost through conduction Qk,12 to the substrate. This simple, steadystate thermal model is also shown in Figure Pr.3.45(b). The irradiation is for an elapsed time of ∆t. D1 = 10 µm, (S˙ e,α )1 = 48 W, ∆t = 1.3 × 10−7 s, T2 = 50◦C. Use the temperature, density, and heat of melting of nickel in Table C.2 and, the thermal conductivity of nickel at T = 1, 400 K in Table C.14. The energy conversion rate S˙ sl is given in Table 2.1 and note that M˙ sl = Ml /∆t, where Ml = ρVl (t). Use Table 3.3(b) for the conduction resistance and use L(t = ∆t) for the depth. SKETCH: Figure Pr.3.45(a) and (b) show the irradiation melting and the simple, steadystate heat transfer model.
(a) Laser Zone Texturing Lens Focused Laser Beam Inscription Site
qr,i
FarField Temperature, T2
D1 Hard Disk Drive: NiP (12% W,P)
(b) Simple, SteadyState Thermal Model D1 Constant Melt Temperature, T1 = Tsl
L (t)
Ssl,1 + (Se,=)1 Qk,12
Figure Pr.3.45(a) Laser zone texturing of a disk. (b) The associated simple, steadystate heat transfer model..
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the depth of the melt L1 , after an elapsed time ∆t. SOLUTION: (a) Figure Pr.3.45(c) shows the steadystate thermal circuit diagram. Q1 = 0 Ssl,1 + (Se,=)1
Constant Melt Temperature, T1 = Tsl
Qk,12 Rk,12 T2
Figure Pr.3.45(c) Thermal circuit diagram.
226
(b) From Figure Pr.3.45(c), we have the energy equation Qk,12 = (S˙ e,α )1 + S˙ sl,1 . From Table 3.2, we have Qk,12 =
Tsl − T2 Rk,12
From Table 2.1, we have S˙ sl,1
= −M˙ sl ∆hsl = −M˙ l ∆hsl Vl (t) ∆hsl ∆t L1 (t) ∆hsl = −ρπD12 ∆t = −ρ
Then solving for L1 (t), we have Tsl − T2 L(t) ∆hsl = (S˙ e,α )1 − ρπD12 Rk,12 ∆t or
Tsl − T2 ˙ (Se,α )1 − ∆t Rk,12 L(t = ∆t) = . ρπD12 ∆hsl
From Table 3.3(b), for a cylindrical indentation, we have Rk,12 =
ln(4L/D1 ) . 4πkL
Here we assume that Rk,12 =
ln[4L(t = ∆t)/D1 ] . 4πkL(t = ∆t)
From Table C.14, at T = 1,400 K for nickel, we have k = 80 W/mK
Table C.14.
From Table C.2, for nickel, we have ρ
=
8,900 kg/m3
Table C.2
Tsl ∆hsl
= =
1,728 K 2.91 × 105 J/kg
Table C.2 Table C.2.
Then using the numerical values, we have (1,728 − 323.15)(K) 48(W) − × 1.3 × 10−7 (s) Rk,12 (K/W) L(t = ∆t) = 8,900(kg/m3 ) × π × (10−5 )2 (W2 ) × 2.91 × 105 (J/kg) Rk,12
=
ln[4L(t = ∆t)/10−5 (m)] . 4π × 80(W/mK) × L(t = ∆t)
Solving these using a solver (such as SOPHT), we have 1.826 × 10−4 (Ks) Rk,12 (K/W) −1 8.137 × 10 (J/m)
6.240 × 10−6 (J) − L(t = ∆t)
=
Rk,12
=
ln[4 × 105 × L(t = ∆t)] 1.005 × 103 (W/mK) × L(t = ∆t) 227
or L(t = ∆t) = 6.127 × 10−6 m = 6.127 µm Rk,12 = 145.6 K/W.
COMMENT: The steadystate conduction resistance is not expected to be accurate. The transient resistance is expected to be smaller and thus have a more signiﬁcant rule. The initial heating, to the melting temperature Tsl , can be modeled similarly and its inclusion will reduce the ﬁnal pool depth.
228
PROBLEM 3.46.FUN GIVEN: To estimate the elapsed time for the penetration of a change in the surface temperature of the brake rotor, the results of Table 3.4 and Figure 3.33(a) can be used. Consider the brake rotor shown in Figure Pr.3.46. Use carbon steel AISI 1010 for the rotor at T = 20◦C, and 2L = 3 cm. SKETCH: Figure Pr.3.46 shows the friction heating of the rotor.
Rotor
Assumed Symmetric Line Brake Pad Sm,F
L L
Figure Pr.3.46 Surface friction heating and its penetration into the brake rotor, during the braking period.
OBJECTIVE: (a) For the conditions given above, determine the penetration time. (b) If the brake is on for 4 s, is the assumption of a uniform rotor temperature valid during the braking period ? (c) If the surfaceconvection cooling occurs after braking and over a time period of 400 s, is the assumption of a uniform rotor temperature valid during the cooling period? SOLUTION: (a) From Table C.16, for carbon steel AISI 1010, we have ρ = 7,830 kg/m3 cp = 434 J/kgK
Table C.16
k = 64 W/mK k α= = 18.8 × 10−6 m2 /s. ρcp
Table C.16
Table C.16
Table C.16.
Assuming that results of Figure 3.33(a)(ii) for a ﬁnite slab of thickness 2L apply here, we have for (3.150) FoL,o =
tα = 0.07. L2
Then the elapsed time is t = =
L2 α (1.5 × 10−2 )2 (m2 ) 0.07 = 0.8378 s. 1.88 × 10−5 (m2 /s)
0.07
(b) Figure 3.33(a)(ii) shows that for a nearly uniform temperature, a larger Fourier number is needed. As an approximation from Figure 3.33(a)(ii), we choose FoL = 1.0 to indicate nearly uniform temperature. Then the elapsed time is L2 = 11.97 s. t = 1.0 α This is large compared to the braking time of t = 4 s. Therefore, we cannot justiﬁably assume a uniform rotor temperature during the braking period.
229
(c) During the surfaceconvection cooling of t = 400 s, the changes occurring over the rotor surface will have a suﬃcient time to penetrate through the rotor (only t = 11.97 s is needed for a nearly complete penetration). Then we can justify the use of a uniform temperature assumption during the cooling period. COMMENT: Note that we have used the constant surface temperature results of Table 3.4 for the ﬁnite slab, while in practice, the surface temperature continues to rise during the brake period and drop during the cooling period.
230
PROBLEM 3.47.FUN GIVEN: The timeperiodic variation of the surface temperature of a semiinﬁnite slab (such as that shown in Figure Ex.3.17) can be represented by an oscillating variation Ts = T (t = 0) + ∆Tmax cos(ωt), where ω = 2πf is the angular frequency, f (1/s) is the linear frequency, and ∆Tmax is the amplitude of the surface temperature change. The solution to the energy equation (3.134), with the above used for the ﬁrst of the thermal conditions in (3.135), is ω 1/2 ω 1/2 T (x, t) − T (x, t = 0) = exp −x cos ωt + x . ∆Tmax 2α 2α
OBJECTIVE: Show that the penetration depth δα , deﬁned by T (x, t) − T (x, t = 0) = 0.01, ∆Tmax is given by δα (2αt)1/2
= 1.725,
which is similar to the penetration depth given by (3.148). Evaluate the penetration depth after an elapsed time equal to a period, i.e., t = τ = 1/f , where τ (s) is the period of oscillation. SOLUTION: Upon examining the solution ω 1/2 ω 1/2 T (x, t) − T (x, t = 0) = exp −x cos ωt + x , ∆Tmax 2α 2α we note that the exponential term is a spatial attenuation factor, while the cosine term is a combined temporalspatial phase lag function. For the determination of the penetration depth, the spatial factor can be written as 1/2 1/2 ω 1/2 2πf 2π = exp −δα exp −δα = exp −δα 2α 2α 2ατ π 1/2 = exp −δα . ατ Similarly, π 1/2 ω 1/2 cos ωt + x = cos 2π + δα . 2α ατ From the deﬁnition of penetration depth, we have: π 1/2 π 1/2 cos 2π + δα = 0.01. exp −δα ατ ατ Using a solver (such as SOPHT), this gives δα
π 1/2 = 1.528. ατ 231
To compare with (3.148), we rearrange to obtain δα 1/2
2(ατ )
= 1.528
1 2π 1/2
= 0.4310.
This is similar in form to (3.148), except that the constant is 0.4310 (instead of 1.8). COMMENT: This shows that the assumption made in Example 3.17 is valid, regarding similarity of penetration depths for the periodic and for the sudden (pulsed) change in the surface temperature.
232
PROBLEM 3.48.FAM GIVEN: Pulsed lasers provide a large power qr,i for a short time ∆t. In surface treatment of materials (e.g., lasershock hardening), the surface is heated by laser irradiation using very small pulse durations. During this heating, the transient conduction through the irradiated material can be determined as that of a semiinﬁnite solid subject to constant surface heating −qs = qr,i ; this is shown in Figure Pr.3.48, with the material being a metallic alloy (stainless steel AISI 316, Table C.16). The heated semiinﬁnite slab is initially at T (t = 0). In a particular application, two laser powers (assume all the laser irradiation power is absorbed by the surface), with diﬀerent pulse lengths ∆t, are used. These are (i) −qs = 1012 W/m2 , ∆t = 10−6 s, and (ii) −qs = 1010 W/m2 , ∆t = 10−4 s. SKETCH: Figure Pr.3.48 shows the surface irradiated by a laser and the penetration of the heat into the substrate. Prescribed Laser Irradiation qs (W/m2)
T(t = 0) 0 T(x,t) Stainless Steel Workpiece
δα(t) x
Figure Pr.3.48 Pulsed laser irradiation of a stainless steel workpiece and the anticipated transient temperature distribution within the workpiece.
OBJECTIVE: (a) Determine the surface temperature T (x = 0, t = ∆t) after elapsed time t = ∆t, for cases (i) and (ii). (b) As an approximation, use the same expression for penetration depth δα (t) as that for the semiinﬁnite slabs with a prescribed surface temperature, and determine the penetration depth after the elapsed time t = ∆t, for cases (i) and (ii). (c) Comment on these surface temperatures and penetration depths. SOLUTION: (a) From Table 3.4, for the case of a prescribed surface heat ﬂux qs , we have the expression for T (x, t) as
T (x, t) = T (t = 0) −
qs (4αt)1/2 π 1/2 k
2 qs x − x e 4αt + k
1 − erf
x (4αt)1/2
Evaluating this at x = 0 and t = ∆t, we have T (x = 0, t = ∆t) = T (t = 0) −
qs (4α∆t)1/2 π 1/2 k
From Table C.16, we have, for stainless steel 316 α = 3.37 × 10−6 m2 /s k = 13 W/mK
Table C.16 Table C.16.
233
.
.
Then using the numerical values, we have (i) T (x = 0, t = 10−6 s)
(ii) T (x = 0, t = 10−6 s)
(−1012 )(W/m2 ) × [4 × (3.37 × 10−6 )(m2 /s) × 10−6 (s)]1/2
=
20(◦C) −
=
20(◦C) + 1.594 × 105 (◦C) = 1.594 × 105 ◦C.
=
20(◦C) −
=
20( C) + 1.594 × 10 ( C) = 1.594 × 104 ◦C. ◦
π 1/2 × 13(W/mK) (−1010 )(W/m2 ) × [4 × (3.37 × 10−6 )(m2 /s) × 10−4 (s)]1/2 π 1/2 × 13(W/mK) 4 ◦
These are tremendously large surface temperatures sustained for a very short time (thus the name lasershock harding). (b) From (3.148), for a semiinﬁnite slab with a sudden change in the surface temperature, we have the penetration depth given as δδ = 3.6α1/2 t1/2 . Then using the numerical values, we have (i) δα (t = ∆t)
= 3.6 × (3.37 × 10−6 )1/2 (m2 /s)1/2 × (10−6 )1/2 (s)1/2 = 6.609 × 10−6 m = 6.609 µm
(ii) δα (t = ∆t)
= 3.6 × (3.37 × 10−6 )1/2 (m2 /s)1/2 × (10−4 )1/2 (s)1/2 = 6.609 × 10−5 m = 66.09 µm.
(c) The thin region near the surface is shocked by this large temperature change and this allows for the rearrangement of the molecules. Then upon cooling, any crystalline defects (and also any surface impurities) will be removed. COMMENT: During very fast thermal shocks, the lattice nuclei may not be in thermal equilibrium with their electron clouds. The electrons having a much smaller mass heat up much faster than the nuclei. Also, any phase change occurring during the shock treatment will not follow the equilibrium phase diagrams, which are generally obtained through controlled and much slower heating/cooling processes.
234
PROBLEM 3.49.FUN GIVEN: In a solidiﬁcation process, a molten acrylic at temperature T (t = 0) is poured into a cold mold, as shown in Figure 3.49, to form a clear sheet. Assume that the heat of solidiﬁcation can be neglected. L = 2.5 mm, Tls = 90◦C, T (t = 0) = 200◦C, Ts = 40◦C. SKETCH: The planar mold and the acrylic melt are shown in Figure Pr.3.49.
Melt T(t = 0) = 200 C (Acrylic) Lz >> L
Cooled Mold, T = 40 C Plane of Symmetry z
y
x
Ly >> L
2L
Figure Pr.3.49 Solidiﬁcation of an acrylic melt in a mold having a constant temperature Ts .
OBJECTIVE: (a) Determine the elapsed time for the cooling front to reach the central plane of the melt. (b) Determine the elapsed time for the temperature of the central plane of the melt to reach the glass transition temperature Tls . SOLUTION: From Table C.17, we have for acrylic (at T = 293 K) α = 1.130 × 10−7 m2 /s
Table C.17.
(a) From Figure 3.33(a)(ii), or from (3.151), the time (Fourier number) for the penetration to reach the central plane is FoL = 0.07 =
αt L2
or t
=
(2.5 × 10−3 )2 (m2 ) × 0.07 L2 FoL = = 3.872 s. α 1.130 × 10−7 (m2 /s)
(b) From Figure 3.33(a)(ii) for a ﬁnite slab, we have Tls − T (t = 0) (90 − 200)(◦C) = = 0.6875, Ts − T (t = 0) (40 − 200)(◦C) and by interpolating the value of FoL for x/L = 0, we have FoL = 0.57. Then FoL =
αt L2
235
or t
= =
(2.5 × 10−3 )2 (m2 ) × 0.57 L2 FoL = α 1.130 × 10−7 (m2 /s) 31.53 s
COMMENT: Note that for a signiﬁcant change in the centralplane temperature, an elapsed time is needed which is many times that for just penetrating to the location of the central plane. We can approximately account for the heat of solidiﬁcation ∆hls by adjusting the speciﬁc heat capacity.
236
PROBLEM 3.50.FAM GIVEN: During solidiﬁcation, as in casting, the melt may locally drop to temperatures below the solidiﬁcation temperature Tls , before the phase change occurs. Then the melt is in a metastable state (called supercooled liquid) and the nucleation (short of) of the solidiﬁcation resulting in formation of crystals (and their growth) begins after a threshold liquid supercool is reached. Consider solidiﬁcation of liquid paraﬃn (Table C.5) in three diﬀerent molds. These molds are in the form of (i) a ﬁnite slab, (ii) a long cylinder, and (iii) a sphere, and are shown in Figure Pr.3.50. The melt in the molds in initially at its melting temperature T (t = 0) = Tsl . Then at t = 0 the mold surface is lowered and maintained at temperature Ts . Assume that solidiﬁcation will not occur prior to this elapsed time. Ts = 15◦C, L = R = 2 cm, T = (x = 0, t) = T (r = 0, t) = To = 302. Use the properties of polystyrene (Table C.17). SKETCH: Figure Pr.3.50 shows the three geometries of the paraﬃn mold.
(i) FiniteSlab Mold
Parafin (Wax) Melt x
Initially at T(t = 0) = Tsl
L
T(x = 0, t) = To
Ts > Tsl
(ii) LongCylinder Mold r
R=L
(iii) Sphere Mold r R=L
Figure Pr.3.50 Paraﬃn (wax) melt is cooled in a mold (three diﬀerent geometries) and is gradually solidiﬁed.
OBJECTIVE: Determine the elapsed time needed for the temperature at the center of the mold, i.e., T (x = 0, t) = T (r = 0, t) to reach a threshold value To , for molds (i), (ii) and (iii). Assume that solidiﬁcation will not occur prior to this elapsed time. SOLUTION: We will use the graphical results given in Figures 3.33(a) and (b) to determine t. The center temperature is desired, and noting that T (t = 0) = Tsl , To − Tsl T − T (t = 0) = . Ts − T (t = 0) Ts − Tsl From Table C.5, we have Tsl =310.0 K
Table C.5
∆hsl =2.17 × 10
5
J/kg 237
Table C.5.
Then To − Tsl (302 − 310)(K) = 0.6751. = Ts − Tsl (298.15 − 310)(K) Now from Figures 3.33(a) and (b), we have (i) slab: (ii) cylinder:
x = 0, r = 0,
FoL 0.55 Figure 3.33(a)(ii) FoR 0.27 Figure 3.33(a)(i)
(iii) sphere:
r = 0,
FoR 0.18
Figure 3.33(a)(ii),
where FoL =
αt , L2
FoR =
αt , R2
α=
k . ρcp
From Table C.17, we have (assume the same properties as polystyrene) α = 7.407 × 10−8 m2 /s
Table C.17
Then (i) slab: t
=
0.55 × (2 × 10−2 )2 (m2 ) FoL L2 = = 2,970 s α 7.407 × 10−8 (m2 /s)
(ii) cylinder: t
=
0.27 × (2 × 10−2 )2 (m2 ) FoR R2 = = 1,458 s α 7.407 × 10−8 (m2 /s)
(iii) sphere: t
=
0.18 × (2 × 10−2 )2 (m2 ) FoR R2 = = 972.1 s. α 7.407 × 10−8 (m2 /s)
COMMENT: Note that the slab mold requires three times more elapsed time compared to the sphere. This is due to the monotonically decreasing volume of the sphere as the center is approached, thus requiring a smaller heat (or time) to change the local temperature. Here we used a large supercooling. In practice, solidiﬁcation begins at a smaller supercooling, thus requiring less time. Inclusion of the solidliquid phase change is discussed in Section 3.8.
238
PROBLEM 3.51.FAM GIVEN: An apple (modeled as a sphere of radius R = 4 cm), initially at T (t = 0) = 23◦C is placed in a refrigerator at time t = 0, and thereafter, it is assumed that its surface temperature is maintained at Ts = 4◦C. Use the thermophysical properties of water at T = 293 K from Table C.23. Use the graphical results given in Figure 3.33(b). OBJECTIVE: (a) Determine the elapsed time it takes for the thermal penetration depth to reach the center of the apple. (b) Determine the elapsed time for the center temperature to reach T = 10◦C. SOLUTION: (a) For the penetration depth to reach the center of the apple, assuming that the temperature changes by 1 %, we have T∗ =
T − T (t = 0) = 0.01 Ts − T (t = 0)
at
r = 0. R
From Figure 3.33(b), the Fourier number for these conditions is nearly 0.03. From the deﬁnition of the Fourier number (3.131), we have F oR =
αt = 0.03. R2
From Table C.23, for water at T = 293 K, we have α = 143 × 10−9 m2 /s. Solving for t, we have t=
0.03 × [0.04(m)]2 0.03R2 = = 336 s = 5.6 min. α 143 × 10−9 (m2 /s)
(b) For the center temperature condition T = 10◦C, we have T∗ =
10 − 23 T − T (t = 0) = = 0.68 Ts − T (t = 0) 4 − 23
at
r = 0. R
Again, from Figure 3.33(b) the Fourier number for these conditions is approximately 0.19. From the deﬁnition of the Fourier number and solving for t, we have t=
0.19R2 0.19 × [0.04(m)]2 = = 2126 s = 35 min. α 143 × 10−9 (m2 /s)
COMMENT: It is not an easy task to keep the surface temperature constant. Inside a refrigerator, most likely there is a surfaceconvection boundary condition at the apple surface. This will be studied in Chapter 6.
239
PROBLEM 3.52.FAM GIVEN: In a summer day, the solar irradiation on the surface of a parking lot results in an absorbed irradiation ﬂux qs = −500 W/m2 , as shown in Figure Pr.3.52. The parking lot surface is covered with an asphalt coating that has a softening temperature of 55◦C. The initial temperature is T (t = 0) = 20◦C. Assume that all the absorbed heat ﬂows into the very thick asphalt layer Use the properties of asphalt in Table C.17. SKETCH: Figure Pr.3.52 shows a thick asphalt layer, treated as a semiinﬁnite slab, suddenly heated by solar irradiation. The temperature distribution beneath the surface is rendered for several elapsed times. Prescribed Surface Heat Flux qr,i =  qs (W/m2)
T(t = 0) 0 T(x,t) Asphalt t, Elapsed Time
x
Figure Pr.3.52 Temperature variation within a thick asphalt layer, suddenly heated by solar irradiation.
OBJECTIVE: Determine the elapsed time it takes for the surface temperature of the parking lot to rise to the softening temperature. SOLUTION: For a semiinﬁnite slab with a constant heat ﬂux boundary condition the temperature as a function of x and t is given by Table 3.4 as T (x, t) − T (t = 0) = −
2qs k
αt π
1/2 −x2 x qs x √ e 4αt + . 1 − erf k 2 αt
For the surface of the asphalt layer, x = 0. The solution becomes 2qs Ts (t) − T (t = 0) = − k
αt π
1/2 .
For asphalt, from Table C.17, k = 0.06 W/mK and α = 0.03 × 10−6 m2 /s. Solving for t and using the data given, we have t=
2 2 [55(◦C) − 20(◦C)] × 0.06(W/mK) π (Ts − Ti )k π = = 462 s = 7.7 min. α −2qs −2 × (−500)(W/m2 ) 0.03 × 10−6 (m2 /s)
COMMENT: The solution presented in Table 3.4 applies for a constant (with respect to time), prescribed heat ﬂux qs (W/m2 ).
240
PROBLEM 3.53.FAM GIVEN: In a shaping process, a sheet of Teﬂon (Table C.17) of thickness 2L, where L = 0.3 cm, is placed between two constanttemperature ﬂat plates and is heated. The initial temperature of the sheet is T (t = 0) = 20◦C and the plates are at Ts = 180◦C. This is shown in Figure Pr.3.53. SKETCH: Figure Pr.3.53 shows a Teﬂon sheet heated on its two surfaces.
Mold
Teflon
L Symmetry Line Ts
T (t = 0)
Figure Pr.3.53 Thermal forming of a Teﬂon sheet.
OBJECTIVE: Determine the time it takes for the center of the sheet to reach 20◦C below Ts . SOLUTION: Treating the Teﬂon as a distributed system consisting of a slab bounded on two sides with a prescribed surface temperature Ts , for T ∗ , we have T∗ =
(160 − 20)(◦C) T (t) − T (t = 0) = = 0.875. Ts − T (t = 0) (180 − 20)(◦C)
From Figure 3.33(a), for x/L = 0 (centerline), the Fourier number is Fo =
αt ≈ 0.9. L2
The thermal diﬀusivity of Teﬂon, from Table C.17, is α = 0.34 × 10−6 m2 /s, and therefore t=
(0.9)(0.3 × 10−2 )2 (m2 ) = 24 s. 0.34 × 10−6 (m2 /s)
COMMENT: The assumption of constant mold temperature is good for a preheated metal mold. This is due to the high eﬀusivity (ρcp k)1/2 of metals when compared to those of polymers.
241
PROBLEM 3.54.FUN.S GIVEN: When human skin is brought in contact with a hot surface, it burns. The degree of burn is characterized by the temperature of the contact material Ts and the contact time t. A ﬁrstdegree burn displays no blisters and produces reversible damage. A seconddegree burn is moist, red, blistered, and produces partial skin loss. A thirddegree burn is dry, white, leathery, blisterless, and produces whole skin loss. A pure copper pipe with constant temperature Ts = 80◦C is brought in contact with human skin having ρcp = 3.7×106 J/m3 ◦C, k = 0.293 W/m◦C, and T (t = 0) = 37◦C for a total elapsed time of t = 300 s. Use the solution for transient conduction through a semiinﬁnite slab with prescribed surface temperature to answer the following. SKETCH: Figure Pr.3.54(a) shows how the ﬁrst, second and thirddegree burns of human tissues are deﬁned based on categories.
DegreeTime Relation for Various Burns 120 110
T, C
100
3rd Degree Burn
90 80
2nd Degree Burn
70 60 1st Degree Burn
50 40 0.1
1
10
100
t, s Figure Pr.3.54(a) Thermal damage (burn) to a human skin and the regions of various degrees of burn.
OBJECTIVE: (a) Plot the temperature distribution T (x, t)(◦C) as a function of position x(mm) at elapsed times t = 1, 10, 20, 40, 50, 100, 150, 200, 220, 240, 280, and 300 s. (b) Use this plot, along with the plot shown in Figure Pr.3.54(a), to estimate the maximum depths for the ﬁrst, second, and thirddegree burns, after an elapsed time t = 300 s. SOLUTION: (a) The solution for the transient conduction through a semiinﬁnite slab, (3.142) is used, i.e., x T (x, t) = T (t = 0) + [Ts + T (t = 0)] 1 − erf , 2(αt)1/2 where α=
k 0.293(W/m◦C) = = 7.919 × 10−8 m2 /s, ρcp 3.7 × 106 (J/m3 ◦C)
T (t = 0) = 37◦C,
and Ts = 80◦C.
The temperature distribution is shown in Figure Pr.3.54(b) for several elapsed times. (b) We are interested in ﬁnding the maximum depth of the ﬁrst, second, and thirddegree burns. This is most easily done using the following steps. (i)
Pick the lowest possible temperature that causes a ﬁrstdegree burn using Figure Pr.3.54(a). 242
2nd Degree Burn Regime
3rd Degree Burn Regime 81
1st Degree Burn Regime
Ts = 80 C
77 73 69
150 200 220 240
65
T, C
61
58 C for 12 s, Maximum Depth of 3rd Degree Burn Regime 53 C for 50 s, Maximum Depth of 2nd Degree Burn Regime
57
280 300
53 100
49
45 C for 35 s, Maximum Depth of 1st Degree Burn Regime
50 45
40 10
41
20
t=1s
37 33 0
1
2
3
4
5 4.75
6 5.8
7
8
9
10
11
12
13
14
15
8.6
x, mm Figure Pr.3.54(b) Distribution of the tissue temperature for several elapsed times.
(ii) (iii) (iv) (v) (vi) (vii) (viii) (ix)
Using Figure Pr.3.54(a), ﬁnd the time needed at this temperature to produce ﬁrstdegree burns. Using this temperature and time, ﬁnd a depth from the graph in Figure Pr.3.54(b). Add one degree to the temperature found in step (i). Using the same graph, ﬁnd the time needed at this new temperature to produce a ﬁrstdegree burn. Using this temperature and time, ﬁnd a depth from the graph in Figure Pr.3.54(b). If this depth is greater than the one found in step (iii), go to step (iv). If this depth is less than the one found in step (iii), stop. Repeat this process for the second and thirddegree burns.
From Figure Pr.3.54(b), we ﬁnd that the maximum depth for a ﬁrstdegree burn is x = 8.6 ± 0.4 mm (marked in Figure Pr.3.54(b) at T = 45◦C at t = 35 s). For the seconddegree burn, we have x = 5.8 ± 0.4 mm (at T = 53◦C at t = 50 s). For the thirddegree burn, x = 4.75 ± 0.4 mm (at T = 58◦C at t = 12 s). COMMENT: The above results are conservative estimates. We have assumed a constant temperature between the time of thermal front arrival and the burn initiation, while this location experiences an increase in temperature with time. Therefore, the actual extent of the burn regions could be deeper than indicated. The transient conduction solution also assumes constant thermal properties and that heat is not carried away by the blood perfusion. The thermal properties change with depth, as layers of tissues such as muscle and fat are encountered.
243
PROBLEM 3.55.FAM GIVEN: A hole is to be drilled through a rubber bottle stopper. Starting the hole in a soft roomtemperature rubber often results in tears or cracks on the surface around the hole. It has been empirically determined that the rubber material at the surface can be hardened suﬃciently for crack and tearfree drilling by reducing the surface temperature at x = 1 mm beneath the surface to below T = 220 K . This reduction in temperature can be achieved by submerging the rubber surface into a liquid nitrogen bath for a period of time. This is shown in Figure Pr.3.55. Assume that by submerging, the surface temperature drops from the initial uniform temperature T (x, t = 0) = 20◦C to the boiling temperature of the nitrogen Tlg , and then remains constant. Use the saturation temperature of nitrogen Tlg at one atm pressure (Table C.26) and the properties of soft rubber (Table C.17). SKETCH: Figure Pr.3.55 shows the surface of a bottle stopper suddenly placed in contact with liquid nitrogen. Rubber Bottle Stopper
Liquid Nitrogen x
Ts = Tlg Figure Pr.3.55 A rubber bottle stopper is temporally submerged in a liquid nitrogen bath.
OBJECTIVE: Determine after an elapsed time of t = 10 s, (a) the temperature 1 mm from the surface T (x = 1 mm, t = 10 s), (b) the temperature 3 mm from the surface T (x = 3 mm, t = 10 s), and (c) the rate of heat ﬂowing per unit area out of the rubber surface qs (x = 0, t = 10 s)(W/m2 ). (d) Is t = 10 s enough cooling time to enable crack and tearfree drilling of the hole? SOLUTION: Since we are asked to evaluate conditions after a elapsed time t, this is a transient problem. Since we are given no information about the size of the rubber stopper, we assume we can treat this as a semiinﬁnite medium with a constant surface temperature Ts = Tlg . The solution for the temperature distribution in a semiinﬁnite medium subject to a constant imposed surface Ts temperature is given by (3.140), i.e., T = T (t = 0) + [Ts − T (t = 0)][1 − erf(η)], where η(x) =
x . (4αt)1/2
From Table C.26, we have Tlg = 77.35 K for nitrogen. Initially, the stopper is at a uniform temperature T (x, t = 0) = 20◦C = 293.15 K. From Table C.17, we have α = 0.588 × 10−7 m2 /s and k = 0.13 W/mK for the rubber stopper. (a) x = 1 mm and t = 10 s η(x = 0.001 m)
=
erf(η)
=
0.001(m) 0.001(m) = 0.652 = 1.5336 × 10−3 (m) [4 × 0.588 × 10−7 (m2 /s) × 10(s)]1/2 erf(0.652) = 0.642 Table 3.5
T
=
293.15(K) + (77.35 − 293.15)(K) × (1 − 0.642) = 215.8 K. 244
(b) x = 3 mm and t = 10 s η(x = 0.003 m)
=
erf(η)
=
0.001(m) 0.003(m) = 1.956 = 1.5336 × 10−3 (m) [4 × 0.588 × 10−7 (m2 /s) × 10(s)]1/2 erf(1.956) = 0.994 Table 3.5
T
=
293.15(K) + (77.35 − 293.15)(K) × (1 − 0.994) = 291.9 K.
(c) The solution for the heat ﬂowing out of a semiinﬁnite medium subject to a constant temperature boundary condition is given by (3.145) as qs = qρck
= −
k[Ts − T (t = 0)]
(παt)1/2 0.13(W/mK) × [77.35(K) − 293(K)] = − [π × 0.588 × 10−7 (m2 /s) × 10(s)]1/2 = 20,627 W out of the rubber.
(d) After 10 s, the temperature of the rubber within the ﬁrst millimeter of depth will be 215.84 K or lower. This is below the maximum temperature limit for cracktear free drilling, i.e., T0 = 220 K. Therefore, 10 s is enough cooling time to enable cracktear free drilling of the hole. COMMENT: Note that the temperature at a distance of 1 mm from the surface has been lowered by 76.2 K, while the temperature 3 mm from the surface has been lowered by only 1.3 K. Therefore, as long as the rubber stopper has a thickness greater than 3 mm, the thermal penetration has not traveled across the rubber stopper and our assumption of a semiinﬁnite solid is shown to be valid.
245
PROBLEM 3.56.FAM GIVEN: The friction heat generation S˙ m,F (energy conversion) occurring in grinding ﬂows into a workpiece (stainless steel AISI 316 at T = 300 K) and the grinder (use properties of brick in Table C.17 at T = 293 K). This is shown in Figure Pr.3.56(a). For a thick (i.e., assumed semiinﬁnite) grinder and a thick workpiece, the fraction of the heat ﬂowing into the workpiece a1 can be shown to be 1/2
a1 =
(ρcp k)w 1/2
1/2
,
(ρcp k)w + (ρcp k)g
where the properties of the workpiece are designated by w and that of the grinder by g. t = 15 s, L = 1.5 mm, S˙ m,F /A = 105 W/m2 . SKETCH: Figure Pr.3.56(a) shows the grinding wheel, the workpiece, and the friction heating at the cylindrical surface. Grinder (g) Angular Speed
Sm,F /A Workpiece (w)
L
Location for Temperature Monitoring
Figure Pr.3.56(a) Friction heating of a stainless steel workpiece and the division of the generated heat.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Use this relation for a1 and the transient temperature distribution resulting from the sudden heating of a semiinﬁnite slab at a constant heat ﬂux, given in Table 3.4, to determine the temperature of the workpiece at location L from the interface and after an elapsed time of t. SOLUTION: (a) The thermal circuit is shown in Figure Pr.3.56(b). The friction heating rate splits into two parts ﬂowing into the grinder and the workpiece.
. Sm,F /A RHck(t)
RHck(t) Ts(t)
. (1  a1)Sm,F /A
Tw(t) . a1Sm,F /A
Figure Pr.3.56(b) Thermal circuit diagram.
(b) From Table C.16, for stainless steel AISI 316 at T = 300 K, we have ρw = 8,238 kg/m3 , cp,w = 468 J/kgK, kw = 13 W/mK, and αw = 3.37 × 10−6 m2 /s. From Table C.17 for brick at T = 293 K, we have ρg = 1,925 kg/m3 , cp,g = 835 J/kgK, kg = 0.72 W/mK, and αg = 0.45 × 10−6 m2 /s.
246
From these, we have (ρcp k)w (ρcp k)g
= =
50.12 × 106 W2 s/m4 K2 1.157 × 106 W2 s/m4 K2 .
Then the fraction of heat ﬂowing into the workpiece is 1/2
a1
(ρcp k)w
=
1/2
1/2
(ρcp k)w + (ρcp k)g
(50.12 × 106 )1/2
=
(1.157 × 106 )1/2 + (50.12 × 106 )1/2
= 0.8681.
Applying the surface energy equation (2.62) to node Ts , we have S˙ m,F /A = qw + qg = qtot
to node Ts .
Therefore, for the heat ﬂux ﬂowing out of the surface is qw
= a1 qtot = a1 S˙ m,F /A = (0.8681) × (105 )(W/m2 ) = 86,809 W/m2 .
This corresponds to a prescribed heat ﬂux into the workpiece of qs = −qw . From Table 3.4, the transient temperature distribution in the workpiece is Ts (x, t) = Ts (t = 0) −
qs (4αw t)1/2 π 1/2 kw
x2 x qs x 4α t w + e 1 − erf . kw (4αw t)1/2 −
Then at t = 15 s and x = 0.0015 m, we have 4αw t = 4 × 3.376 × 10−6 (m/s ) × 15(s) = 2.022 × 10−4 m2 x 0.0015(m) erf = erf (4αw t)1/2 (2.022 × 10−4 )1/2 (m) = erf(0.1055) = 0.1185 Table 3.5. 2
Solving for the temperature, we have Ts (x, t) Ts (x = L, t = 1.5 s)
x2 x qs x 4α t w + = Ts (t = 0) − e 1 − erf kw π 1/2 kw (4αw t)1/2 2 (0.0015)2 (m2 ) (−86,809)(W/m ) × (2.022 × 10−4 )1/2 (m) exp − = 300(K) − 2.022 × 10−4 (m2 ) π 1/2 (13)(W/mK) qs (4αw t)1/2
−
2
(−86,809)(W/m ) × (0.0015)(m) [1 − (0.1185)] (13)(W/mK) = 300(K) − [−52.98(K)] + [−8.83(K)] = 344.14 K = 71◦C. +
COMMENT: Note that a large fraction of the heat ﬂows into the workpiece because of its larger eﬀusivity. Also note that if a larger elapsed time is allowed, the temperature will be higher and can reach the damage threshold.
247
PROBLEM 3.57.FUN GIVEN: Two semiinﬁnite slabs having properties (ρ, cp , k)1 and (ρ, cp , k)2 and uniform, initial temperatures T1 (t = 0) and T2 (t = 0), are brought in contact at time t = 0. OBJECTIVE: (a) Show that their contact (interfacial) temperature is constant and equal to 1/2
T12 =
1/2
(ρcp k)1 T1 (t = 0) + (ρcp k)2 T2 (t = 0) 1/2
(ρcp k)1
1/2
.
+ (ρcp k)2
(b) Under what conditions does T12 = T2 (t = 0)? Give an example of material pairs that would result in the limit. SOLUTION: (a)Assuming a constant contact surface temperature, we use (3.140) for the temperature distribution and use Ts = T12 , i.e., for the two semi inﬁnite slabs we have η 2 2 T1 (t) − T1 (t = 0) = 1 − erf(η) = 1 − 1/2 e−η dη T12 − T1 (t = 0) π 0 η 2 T2 (t) − T2 (t = 0) 2 = 1 − erf(η) = 1 − 1/2 e−η dη T12 − T2 (t = 0) π 0 x η= . 2(αt)1/2 Since the heat ﬂowing out of one of the slabs ﬂows into the other, we have qs = q12 = q1 (x = 0) = q2 (x = 0), where x = 0 is the location of the interface. From (3.143), we relate q12 and the derivative of the temperatures, i.e., ∂T1 q12 = −k1 ∂x x=0 ∂T2 = − −k2 . ∂x x=0 We note that from the chain rule for diﬀerentiation, we have ∂ ∂x
∂η ∂ ∂x ∂η 1
= =
Then
∂T1 (t) ∂x x=0
= =
∂T2 (t) ∂x x=0
=
1
∂
2(αt)1/2 ∂η
2
−η 2
.
[T12 − T1 (t = 0)] − 1/2 e 2(α1 t)1/2 π η=0 2 1 − [T12 − T1 (t = 0)] 2(α1 t)1/2 π 1/2 1 2 − [T12 − T2 (t = 0)]. 2(α2 t)1/2 π 1/2
Then using the equality of the surface heat ﬂuxes, we have −1 −1 −k1 − T (t = 0)] = k × [T × [T12 − T2 (t = 0)] 12 1 2 (πα1 t)1/2 (πα2 t)1/2 248
or −
k1 1/2 (k/ρcp )1
[T12 − T1 (t = 0)] =
or
k2 1/2
(k/ρcp )2
1/2
T12 =
[T12 − T2 (t = 0)]
1/2
(ρcp k)1 T1 (t = 0) + (ρcp k)2 T2 (t = 0) 1/2
(ρcp k)1
1/2
.
+ (ρcp k)2
(b) From Tables C.16 and C.17, we can choose material pairs for which one material has a much larger (ρcp k)1/2 , which is called the thermal eﬀusivity. For example, metals have high ρ, low cp , and high k. We choose copper from Table C.16 and wood for Table C.17. Then copper (pure):
wood (pine):
ρ = 8,933 kg/m3
Table C.16
cp = 385 J/kgK k = 401 W/mK
Table C.16
3
Table C.16
ρ = 525 kg/m cp = 2,750 J/kgK
Table C.17
k = 0.12 W/mK
Table C.17.
Table C.17
Then 1/2
= 3.714 × 104 Ws1/2 /m2 K
1/2
= 4.162 × 102 Ws1/2 /m2 K.
copper (pure):
(ρcp k)1
wood (pine):
(ρcp k)2
Using these, we have T12
3.714 × 104 × T1 (t = 0) + 4.162 × 102 × T2 (t = 0) 3.714 × 104 + 4.162 × 102 T1 (t = 0). =
This shows that the contact temperature will be the initial copper temperature T1 (t = 0), i.e., the wood will instantly take on the surface temperature of the copper. COMMENT: Note that we initially assumed that the contact temperature T12 is constant and this allowed us to use the transient solutions for the constant surface temperature.
249
PROBLEM 3.58.FAM.S GIVEN: For thermal treatment, the surface of a thinﬁlm coated substrate [initially at T (t = 0)] is heated by a prescribed heat ﬂux qs = −109 W/m2 (this heat ﬂux can be provided for example by irradiation) for a short period. This heating period to (i.e., elapsed time) is chosen such that only the temperature of the titanium alloy (Ti2 Al2 Mn, mass fraction composition) thin ﬁlm is elevated signiﬁcantly (i.e., the penetration distance is only slightly larger than the thinﬁlm thickness). The thin ﬁlm is depicted in Figure Pr.3.58. SKETCH: Figure Pr.3.58 shows a thin ﬁlm over a semiinﬁnite substrate, is heated by irradiation. Prescribed Surface Heat Flux  qs (W/m2)
l TitaniumAlloy Thin Film x T(t = 0)
T(l,t) Titanium Substrate
Figure Pr.3.58(a) A thinﬁlm coated, semiinﬁnite substrate heated by irradiation.
OBJECTIVE: Determine the required elapsed time to for the temperature of the interface between the thin ﬁlm and the substrate, located at distance l = 5 µm from the surface, to raise by ∆T (l, to ) = T (l, to ) − T (t = 0) = 300 K. Note that this would require determination of t from an implicit relation and would require iteration or use of a software. SOLUTION: Table 3.4 gives the temperature distribution for a semiinﬁnite slab with a prescribed heat ﬂux at the surface. For qs constant on the surface, 2qs T − T (t = 0) = − k
αt π
1/2
x2 qs x e 4αt + k −
1 − erf
x
(4αt)1/2
.
This equation requires k and α. From Table C.16, for a Ti2 Al2 Mn alloy at 300 K, k = 8.4 W/mK and α = 4 × 10−6 m2 /s. From the problem statement, T (l, to ) − T (t = 0) = ∆T = 300 K, qs = −109 W/m2 , and x = l = 5 × 10−6 m. Substituting these values into the equation above, we have 300 = 2.687 × 105 × t1/2 e
−1.563 × 10−6 1.250 × 10−3 t − 5.952 × 102 × 1 − erf . t1/2
The equation is an implicit relation for time. The solution can be obtained from an equation solver software or by hand calculation, iteratively. The method of successive substitutions can be used for the iterative solution. For this method, the above equation is written explicitly for one of the occurrences of the variable time. Choosing the time t appearing in the ﬁrst term in the righthand side, the equation is rewritten as 2 1.250 × 10−3 −3 −3 − 2.216 × 10 erf 3.332 × 10 t1/2 . = −6 −1.563 × 10 t e
tnew
The solution requires guessing a value of t and solving for tnew . The process ends when t and tnew are suﬃciently close. When the initial guess is close to the ﬁnal answer, the convergence is stable and requires only a few 250
iterations. A suitable initial guess can be obtained by calculating the time it takes for the surface temperature to be raised by ∆T . For the surface, x = 0, the equation for T becomes T − T (t = 0) =
−2qs k
αto π
1/2 .
Solving for to gives π to = α
∆T k −2qs
2 .
Using the known values, to =
π
300(K) × 8.4(W/mK)
4 × 10−6 (m2 /s)
2 = 1.25 × 10−6 s = 1.25 µs.
2
2 × 109 (W/m )
The temperature at a depth of 5 µm will take longer to be raised by ∆T . The value of to gives a starting point for the iterations. Table Pr.3.58 shows the iterations. The error function is interpolated from Table 3.5. The solution after 5 iterations is t = 7.8 µs. Table Pr.3.58 Results for successive iterations. t, s
η = 1.250 × 10−3 /t1/2
erf(η)
tnew , s
2 × 10−6 4 × 10−6 6 × 10−6 7.6 × 10−6 7.8 × 10−6
0.884 1.625 0.5103 0.4534 0.4476
0.7880 0.6223 0.5289 0.4775 0.4721
1.201 × 10−5 8.334 × 10−6 7.856 × 10−6 7.801 × 10−6 7.800 × 10−6
COMMENT: The solution found using software [Figure Pr.3.58(b)] is to = 7.785 µs. The result obtained above shows that the use of Table 3.5 for the error function is acceptable. Note also that the same solution could be obtained if the time variable inside the exponential function were chosen as tnew .
T  T (t=0) , K
400 (7.8E6 s, 300.51K)
300
200 100
0 0
2E6
4E6
6E6
8E6
1E5
t, s Figure Pr.3.58(b) Variation of temperature, at a location x = 5 µm from the surface, with respect to time.
251
Figure Pr.3.58(c) shows the temperature distributions as a function of the depth x for diﬀerent values of time. Notice that the derivative dT /dx at the surface is constant, because qs is constant at the surface. Also, the penetration depth increases with time. For the elapsed time of 8 µs, the ∆T at the surface is above 600 K. From Figure Pr.3.58(c) we observe that for t = 8 µs there was substantial heat penetration into the substrate. In order to account for the change of properties between the thin ﬁlm and the substrate, other solution techniques need to be used. One example of such a technique is the use of ﬁnite (i.e., small) volumes, presented in Section 3.7. 1000 800
Thin Film
Substrate
∆T , K
600 400
t = 0.1 µs t = 0.5 µs t = 1 µs t = 2 µs t = 5 µs t = 10 µs
200 0
 200
0.0
1.0 x 10 5 x,m
5.0 x 10 6
1.5 x 10 5
2.0 x 10 5
Figure Pr.3.58(c) Distribution of temperature near the surface, at several elapsed times.
252
PROBLEM 3.59.FAM GIVEN: An automobile tire rolling over a paved road is heated by surface friction, as shown in Figure Pr.3.59. The energy conversion rate divided by the tire surface area is S˙ m,F /At , and this is related to the vehicle mass M and speed uo through S˙ m,F MgµF uo = . At At A fraction of this, a1 S˙ m,F /At , is conducted through the tire. The tire has a covertread layer with a layer thickness L, which is assumed to be much smaller than the tire thickness, but is made of the same hard rubber material as the rest of the tire. The deep unperturbed temperature is T (t = 0). The properties for hard rubber are listed in Table C.17. T (t = 0) = 20◦C, g = 9.807 m/s2 , uo = 60 km/hr, L = 4 mm, At = 0.4 m2 , to = 10 min, µF = 0.015, a1 = 0.1. SKETCH: Figure Pr.3.59 shows friction heating and heat transfer into a tire and the location from the surface.
At
g
x
T(t = 0)
Hard Rubber
CoverTread Layer uo
L
T(x,t)
. S qs = a1 m, F At Uniform Surface Heat Flux
S m,F At
Figure Pr.3.59 Surface friction heating of a tire laminate.
OBJECTIVE: Determine the temperature at this location L, after an elapsed time to , using (a) M = 1,500 kg, and (b) M = 3,000 kg. SOLUTION: (a) Applying the energy equation to the contact area, and assuming all the heat ﬂows into the tire, qs
S˙ m,F At MgµF uo = −a1 At = −a1
2
1,500(kg) × 9.807(m/s ) × 0.015× = −0.1
60,000(m/hr) 3,600(s/hr)
0.4(m2 ) 2
= −919.4 W/m . From Table 3.4, the transient temperature distribution (for prescribed qs ) in the workpiece is given as 2 x q x qs (4αt)1/2 − x 4αt + s e T (x, t) = T (t = 0) − . 1 − erf k π 1/2 k (4αt)1/2 253
For hard rubber, using Table C.17, we have, k = 0.15 W/mK, α = 0.6219 x 10−7 m2 /s. Then at t = 10 × 60 s and at x = 0.004 m, we have erf
4αt = 1.493 × 10−4 m2 x 0.004(m) = erf (4αt)1/2 (1.493 × 10−4 )1/2 (m) = erf (0.3274) = 0.3560 Table 3.5.
Solving for the temperature, we have T (x, t)
= T (t = 0) −
qs (4αt)1/2 π 1/2 k
x2 qs x 4αt + e k −
1 − erf
(4αt)1/2
−919.4(W/m ) × (1.493 × 10−4 )1/2 (m) 2
T (x = 4 mm, t = 600 s)
x
exp −
(0.004)2 (m2 ) 1.493 × 10−4 (m2 )
=
293.15(K) −
=
−919.4(W/m ) × 0.004(m) (1 − 0.3560) 0.13(W/mK) 293.15(K) − [−37.94(K)] + [−15.79(K)] = 315.3 K = 42.15◦C.
π 1/2 × 0.15(W/mK)
2
+
(b) Using M = 3,000 kg, qs = −1,838.8 W/m2 , and solving for temperature, we have −1,838.8(W/m ) × (1.493 × 10−4 )1/2 (m) 2
T (x = 4 mm, t = 600 s) =
293.15(K) −
π 1/2 × 0.15(W/mK)
exp −
(0.004)2 (m2 ) 1.493 × 10−4 (m2 )
2
−1,838.8(W/m ) × 0.004(m) (1 − 0.3560) 0.15(W/mK) 293.15(K) − [−75.88(K)] + [−31.58(K)] = 337.5 K = 64.30◦C. +
=
COMMENT: As the tire heats up, the surface convection heat transfer rate increases. Therefore, the amount of heat conducting through the tire surface decreases with an increase in surface temperature.
254
PROBLEM 3.60.FAM GIVEN: In ultrasonic welding (also called ultrasonic joining), two thick slabs of polymeric solids to be joined are placed in an ultrasonic ﬁeld that causes a relative motion at their joining surfaces. This relative motion combined with a joint pressure causes a surface friction heating at a rate of S˙ m,F /A. This heat ﬂows and penetrates equally into these two similar polymeric solids. The two pieces are assumed to be very thick and initially at a uniform temperature T (t = 0). S˙ m,F /A = 104 W/m2 , Tsl = 300◦C, T (t = 0) = 25◦C, and use the properties of Teﬂon (Table C.17). SKETCH: Figure Pr.3.60(a) shows the two solid surfaces in sliding contact and the friction heat ﬂow into each of the pieces.
q x=0 .
q
Sm,F Ak
Control Surface
Figure Pr.3.60(a) The solids in sliding contact and friction heat ﬂow into both of them.
OBJECTIVE: How long would it take for the contacting surfaces of the two polymers in contact to reach their melting temperature Tsl ? SOLUTION: The two pieces are very thick and therefore are assumed to behave as semiinﬁnite slabs. The surface friction heating occurs uniformly over the entire contact surface and the resulting generated heat ﬂows equally into each piece. The two pieces in contact are drawn schematically in Figure Pr.3.60(a). Performing a surface conservation of energy analysis on the contacting surface and noting that the heat ﬂows equally into both the upper and lower pieces gives qupper + qlower
=
q
=
2q = S˙ m,F /A S˙ m,F . 2A
By symmetry, we only need to analyze one of the pieces to determine when the surface temperature reaches the melting temperature. The control volume for the lower piece is rendered in Figure Pr.3.60(b). Control Surface
q =  qs
x=0 .
Sm,F Ak
Figure Pr.3.60(b) Control volume for the solids in sliding contact.
As shown, the heat ﬂux entering across the control surface into the volume is equal to the negative of that leaving one side of the control surface, i.e. qs = −q. The resulting conservation of energy equation is solved, and 255
the solution for T (x, t) is given in Table 3.4 as T (x, t) = T (t = 0) −
qs x 2qs (αt)1/2 − x2 e 4αt + 1/2 k π k
1 − erf
x 2(αt)1/2
.
At location x = 0, this equation becomes T (x = 0, t) − T (t = 0)
= −
2qs (αt)1/2 π 1/2 k
= Tsl − T (t = 0) = −
∆T (0, t)
2(−
S˙ m,F 1/2 2A )(αt) . π 1/2 k
For Teﬂon at T = 293 K, k = 0.26 W/mK and α = 0.34 × 10−6 m2 /s. Solving for t we have t = = =
∆T π 1/2 k (S˙ m,F /A)α1/2
2 =
∆T k S˙ m,F /A
2
π α
(300◦C − 25◦C)2 × (0.26 W/mK)2 × π (104 W/m2 )2 × (0.34 × 10−6 m2 /s) (300 − 25)2 (◦C)2 × (0.26)2 (W/mK)2 × π = 472.37 s = 7.87 min. (104 )2 (W/m2 )2 × (0.34 × 10−6 )(m2 /s)
COMMENT: The process time is inversely proportional to S˙ m,F to the second power. The process time can be decreased by increasing S˙ m,F .
256
PROBLEM 3.61.FAM GIVEN: A thin ﬁlm is heated with irradiation from a laser source as shown in Figure Pr.3.61(a). Assume that all the radiation is absorbed (i.e., αr,1 = 1). The heat losses from the ﬁlm are by substrate conduction only. The ﬁlm can be treated as having a uniform temperature T1 (t) i.e., Nk,1 < 0.1, and the conduction resistance Rk,12 through the substrate can be treated as constant. SKETCH: Figure Pr.3.61(a) shows the thin ﬁlm, over a substrate, heated by laser irradiation. Laser Source qr,i = 106 W/m2 Film, T1(t) (Nk,1 < 0.1) T1(t = 0) = 20oC
(ρcp)1 = 106 J/m3K αr,1 = 1 Ar,1 = Ak L1 = 10 µm
L2 = 5 mm
Substrate Ts,2 = 20oC k = 1.3 W/mK Figure Pr.3.61(a) Laser radiation heating of a thin ﬁlm over a substrate.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the time needed to raise the temperature of the ﬁlm T1 (t) to 500◦C. SOLUTION: (a) Figure Pr.3.61(b) shows the thermal circuit for the problem. Note that the thin ﬁlm is lumped into a single node and the thick ﬁlm is modeled as a conduction resistance constant with time.
Q1= 0
T1
dT
 (ρcpV)1 dt1 + (Se,=)1 Qk,12
Rk,12
T2 Q2
Figure Pr.3.61(b) Thermal circuit diagram.
(b) To determine the time needed to raise the ﬁlm temperature to 500◦C, the energy equation is applied to the thin ﬁlm. The integral form of the energy equation is QA = − (ρcp V )1 257
dT1 + (S˙ e,α )1 . dt
From Figure Pr.3.61(b), we notice that QA has only a conduction component. The energy source is due to radiation adsorption with αr = 1 and r = 0. Therefore, the energy equation becomes T1 − Ts,2 dT1 + αr,1 qr,i Ar,1 . = − (ρcp V )1 Rk,12 dt The conduction resistance is given by Rk,12 =
L2 5 × 10−3 (m) 3.85 × 10−3 [K/(W/m2 )] = = . ks Ak 1.3(W/mK)Ak Ak
The thermal capacitance is 3 2 (ρcp V )1 = 106 J/m K 10 × 10−6 (m) Ak = 10 J/m ◦C Ak . The source term is
2 S˙ 1 = αr,1 qr,i Ar,1 = (1) × 106 W/m Ar,1 .
The solution for T1 is given by (3.172). Solving for t, we have T1 − Ts,2 − a1 τ1 , t = −τ1 ln T1 (t = 0) − Ts,2 − a1 τ1 where
3.85 × 10−3 K/(W/m2 ) 2 τ1 = (ρcp V )1 Rk,12 = 10 J/m K Ak = 3.85 × 10−2 s Ak 2 106 W/m Ar,1 S˙ 1 a1 = = = 105 K/s 2 (ρcp V )1 10 J/m K Ak
and Ar,1 = Ak has been used. Then
500 (◦C) − 20 (◦C) − 105 (1/s) × 3.85 × 10−2 (s) t = −3.85 × 10 (s) ln 20 (◦C) − 20 (◦C) − 105 (1/s) × 3.85 × 10−2 (s) = 0.0051 s = 5.1 ms. −2
COMMENT: Note that the response is rather fast. The radiation reﬂection from the surface and radiation emission from the surface will be addressed in Chapter 4.
258
PROBLEM 3.62.FUN GIVEN: Carbon steel AISI 4130 spheres with radius R1 = 4 mm are to be annealed. The initial temperature is T1 (t = 0) = 25◦C and the annealing temperature is Ta = 950◦C. The heating is done by an acetylene torch. The spheres are placed on a conveyor belt and passed under the ﬂame of the acetylene torch. The surfaceconvection heat ﬂux delivered by the torch (and moving into the spheres) is given as a function of the position within the ﬂame by (see Figure Pr.3.62) x , qs (t) = qku = qo sin π L where L = 2 cm is the lateral length of the ﬂame and qo = −3 × 106 W/m2 is the heat ﬂux at the center of the ﬂame. Assume that the surface of the spheres is uniformly heated and neglect the heat losses. Use the properties at 300 K, as given in Table C.16. SKETCH: Figure Pr.3.62 shows a sphere placed on a conveyer and passed under a torch. The surface heat ﬂux is also given, as a function of location. AcetyleneOxygen Torch Acetylene Flame
T1(t) Steel Sphere
ub
qs(x)
Conveyor Belt
qo Variation of Heat Flux Within Flame
x L Figure Pr.3.62 Heating of carbon steel spheres placed on a conveyor.
OBJECTIVE: Using a lumpedcapacitance analysis (Nku,1 < 0.1) and starting from (3.160), ﬁnd the speed of the conveyor belt ub needed for heating the spheres from T1 (t = 0) to Ta . SOLUTION: Treating each sphere as a lumpedcapacitance system, the integralvolume energy equation (3.161) becomes dT + S˙ i . (q · sn )dA = −(ρcp V )1 QA = dt A1 As the convection heat ﬂux vector points to the surface and there is no energy conversion inside the spheres, we have (q · sn )dA = −qs (t)A A
S˙ i
=
0.
The energy equation becomes (ρcp V )1
dT1 = −qs (t)A1 . dt 259
Using the equation for qs (t), we have (ρcp V )1
πx dT1 = −qo sin A1 . dt L
Note that the distance traveled along the ﬂame x can be related to the elapsed time t by x = ub t. Now, using this relation we have (ρcp V )1
πu dT1 b = qo sin t A1 . dt L
Separating the variables and integrating gives dT1
=
dT1
=
Ta − T1 (t = 0)
=
Ta
T1 (t=0)
πu −qo A1 b t dt sin (ρcp V )1 L t πu −qo A1 b t dt sin (ρcp V )1 0 L πu " −qo A1 L ! b 1 − cos t . (ρcp V )1 πub L
The ﬁnal time t is the time it takes to travel through the ﬂame and is given by t = L/ub . Then Ta − T (t = 0) =
−2qo A1 L . (ρcp V )1 πub
For spheres, we have 4 πR3 V1 R1 . = 3 21 = A1 4πR1 3
Then Ta − T (t = 0) =
L −6qo . (ρcp )1 R1 πub
The properties for carbon steel AISI 4130 at 300 K, from Table C.16, are ρ1 = 7,840 kg/m3 and cp,1 = 460 J/kgK. Solving for ub gives 2
ub
= =
6 × 3 × 106 (W/m ) × 0.02(m) −6qo L = 3 π(ρcp )1 R1 [Ta − T (t = 0)] π × 7,840(kg/m ) × 460(J/kgK) × 0.004(m) × [950(◦C) − 25(◦C)] 0.008 m/s = 51.5 cm/min.
COMMENT: We would need a smaller speed ub if we considered the presence of heat losses. The lumpedcapacitance analysis is valid when Nku,1 < 0.1 and this will be discussed in Chapter 6, in the context of surfaceconvection heat transfer qku .
260
PROBLEM 3.63.FAM.S GIVEN: An electrical resistance regulator is encapsulated in a rectangular casing (and assumed to have a uniform temperature T1 , i.e., Nk,1 < 0.1) and attached to an aluminum slab with thickness L = 3 mm. The slab is in turn cooled by maintaining its opposite surface at the constant ambient temperature T2 = 30◦C. This slab is called a heat sink and is shown in Figure Pr.3.63(a). Most of the time, the regulator provides no resistance to the current ﬂow, and therefore, its temperature is equal to T2 . Intermittently, the regulator control is activated to provide for an ohmic resistance and then energy conversion from electromagnetic to thermal energy occurs. A joint pressure is exerted to reduce the contact thermal resistance at the interface between the regulator and the heat sink. However, as the regulator temperature reaches a threshold value T1,o = 45◦C, the thermal stresses warp the regulator surface and the contact resistance changes from Ak Rk,c = 10−3 K/(W/m2 ) to a larger value of Ak Rk,c = 10−2 K/(W/m2 ). The regulator has (ρcp V )1 /Ak = 1.3 × 105 J/Km2 and the amount of heat generated by Joule heating is S˙ e,J /Ak = 2 × 104 W/m2 . Neglect the heat losses from the regulator to the ambient. Assume that the conduction resistance in the aluminum slab is steady state (i.e., constant resistance) and the energy storage in the slab is also negligible. Use the thermal conductivity of aluminum at T = 300 K. SKETCH: Figure Pr.3.63(a) shows a regulator subject to the Joule heating and attached to a substrate with a thermal contact resistance. Regulator: Uniform Temperature T1(t) (Nk,1 < 0.1)
Se,J
Ak , (ρcpV)1 Rk,c
Aluminum Slab (Heat Sink) L = 3 mm
T2 = 30 C (+) (−)
Figure Pr.3.63(a) An electrical resistance regulator attached to a substrate with a thermal contact resistance.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Use the lumpedcapacitance analysis to determine the required time to reach the threshold temperature Tc = 45◦C. (c) Starting with T1,o as the initial temperature and using the new contact resistance Ak Rk,c , determine the time required to reach T1 = T2 + 2 Tc . (d) Determine the steadystate temperature. (e) Make a qualitative plot of the regulator temperature versus time, showing (i) the transition in the contact resistance, and (ii) the steadystate temperature. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.63(b). (b) To ﬁnd the switch temperature, the integralvolume energy equation (3.161) is applied to the switch node T1 . For transient conditions, we have
T1 − Tj dT1 + S˙ e,J . = −ρcp V Q1 + R dt k,1 j j 261
Uniform Temperature, T1(t)
(ρCpV)1
dT (ρCpV)1 1 + Se,J dt
T1
Q1i
dT1 + Se,J dt
Rk,c
Ak
Interface (i)
Ti Qi2
L = 3 mm
Rk,c
Rk,i2 T2 = 30 OC
T2
Aluminum Slab (Heat Sink)
Figure Pr.3.63(b) Thermal circuit diagram.
The temperature T2 is known. Therefore, the heat ﬂux through the thermal resistances is written as a function of T2 . The energy equation for node 1 is T1 − T2 dT1 + S˙ e,J . = −ρcp V (Rk,Σ )12 dt For the resistances arranged in series, the overall thermal resistance (Rk,Σ )12 is (Rk,Σ )12 = Rk,c + Rk,i2 . The conduction resistance Rk,i2 for the slab is (Table 3.2) Rk,i2 =
L , k2 Ak
and the contact resistance Rk,c is given in the problem statement. The thermal conductivity k2 is needed to calculate the thermal resistance. For aluminum (Table C.14, T = 300 K) we obtain k2 = 237 W/mK. The solution for T1 (t), given by (3.172), is T1 (t) = T2 + [Ti (t = 0) − T2 ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ), where τ1 = (ρcp V )1 (Rk,Σ )12
a1 =
S˙ e,J − Q1 . (ρcp V )1
For the initial heating period, the conditions are: T1 (t = 0) = 30◦C, T1 (t) = Tc = 45◦C, Ak Rk,c = 10−3 K/(W/m2 ). From the data given, Ak Rk,i2 =
L 0.003 2 = 1.266 × 10−5 K/(W/m ) = k2 237
Ak (Rk,Σ )12 = Ak Rk,c + Ak Rk,i2 = 10−3 + 1.266 × 10−5 = 1.013 × 10−3 K/(W/m ) 2
τ1 = (ρcp V )1 (Rk,Σ )12 =
ρcp V Ak (Rk,Σ )12 = (1.3 × 105 ) × (1.013 × 10−3 ) = 1.316 × 102 s Ak 262
S˙ e,J − Q1 a1 = = (ρcp V )1
S˙ e,J Q1 − Ak Ak
2 × 104 − 0 Ak = = 1.538 × 10−1 K/s. (ρcp V )1 1.3 × 105
Solving for time gives t
T1 (t) − T2 − a1 τ1 T1 (t = 0) − T2 − a1 τ1 45(◦C) − 30(◦C) − (1.538 × 10−1 )(◦C/s) × (1.316 × 102 )(s) = 177.8 s 3.0 min = −1.316 × 102 (s) ln 0 − (1.538 × 10−1 )(◦C/s) × (1.316 × 102 )(s) = −τ1 ln
(c) At T1 = 45◦C, the contact resistance changes to Ak Rk,c = 10−2 K/(W/m2 ). The overall resistance and the time constant τ1 become (note that a1 remains the same) Ak (Rk,Σ )12 = Ak Rk,c + Ak Rk,i2 = 10−2 + 1.266 × 10−5 = 1.001 × 10−2 K/(W/m ) ρcp V τ1 = (ρcp V )1 (Rk,Σ )12 = Ak (Rk,Σ )12 = 1.3 × 105 × 1.001 × 10−2 = 1.302 × 103 s. Ak 2
For this second heating period, T1 (t = 0) = 45◦C and T1 (t) = T2 + 2Tc = 120◦C. Then T1 (t) − T2 − a1 τ1 T1 (t = 0) − T2 − a1 τ1 120(◦C) − 30(◦C) − (1.538 × 10−1 )(◦C/s) × (1.302 × 103 )(s) = 675.7 s 11.3 min. = −1.302 × 103 (s) ln 45(◦C) − 30(◦C) − (1.538 × 10−1 )(◦C/s) × (1.302 × 103 )(s)
t = −τ1 ln
(d) The steadystate temperature is the condition for t → ∞. Setting t → ∞ in (3.172), we obtain T1 (t → ∞) = T2 + a1 τ1 = 30 + (1.538 × 10−1 ) × (1.302 × 103 ) = 230.2◦C. (e) Figure Pr.3.63(c) shows T1 as a function of time. At T1 = 45◦C, the change in the contact resistance causes a change in heat loss, from a smaller heat loss to a larger heat loss. This appears in the graph as an abrupt change in slope, from a smaller to a steeper slope. 300 250
T1 = 230 oC
o
T1 , C
200 150 T1 = 120 oC
100 50 0
T1 = 45 oC
1000
2000 t,s
3000
4000
Figure Pr.3.63(c) Time variation of the regulator temperature for the periods with diﬀerent constant resistance.
COMMENT: Note that the increase in contact resistance results in a much larger regulator temperature. Applying joint pressure or thermal grease will assist in reducing the contact resistance. 263
PROBLEM 3.64.FAM GIVEN: In laser backscribing, a substrate is heated and melted by radiation absorption. Upon solidiﬁcation, a volume change marks the region and this is used for recording. An example is given in Figure Pr.3.64(a), where irradiation is provided through a thick glass layer and arrives from the backside to a thin layer of alumina. The alumina layer absorbs the radiation with an extinction coeﬃcient σex,1 that is much larger than that of glass. Assume that the alumina layer is at a uniform, but timevarying temperature (because of the high thermal conductivity of alumina compared to the glass, i.e., Nk < 0.1). Also assume that the conduction resistance in the glass is constant. a = 100 µm, l1 = 0.6 µm, l2 = 3 µm, qr,i = 3 × 109 W/m2 , σex,1 = 107 1/m, ρr = 0.1, T1 (t = 0) = 20◦C, T2 = 20◦C, Tsl,1 = 1,900◦C. SKETCH: Figure Pr.3.64(a) shows laser backscattering by volumetric absorption of irradiation.
Laser qr,i a
a
ρr qr,i
T2 Glass Alumina (Conductive Oxide) T1(t = 0) Uniform Temperature T1
(1  ρr) qr,i
k2 ,
σex,2
Se,r Scribing Lay
) (ρcp 1
er
Other Layer
=0
, σex,1
l2 l1
s Q1 = 0
Figure Pr.3.64(a) Laser backscribing on a compact disk storage device.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the time it takes to reach the melting temperature of the alumina Tsl,1 . SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.64(b). As stated in the problem, a constant conduction resistance is assumed for the glass layer. The temperature of the scribing layer is assumed uniform and time dependent. The absorbed irradiation is shown with (S˙ e,r )1 . This is the absorption integrated over the layer.
T2 Rk,12
Qk,12
T1 dT
 (ρcpV)1 dt1 + (Se,r)1
Figure Pr.3.64(b) Thermal circuit diagram.
264
(b) The energy absorbed in layer 1 is found from integrating (2.43), i.e., l1 (S˙ e,r /V )dV1 = a2 qr,i (1 − ρr )σex,1 exp(−σex,1 x)dx (S˙ e,r )1 = V1
= a2 qr,i (1 − ρr )[1 − exp(−σex,1 l1 )] = (10−4 )2 (m2 ) × (3 × 109 )(W/m2 ) × (1 − 0.1) × {1 − exp[−107 (1/m) × 6 × 10−7 (m)]} = 26.933 W. The transient temperature of layer 1 is given by (3.172), i.e., T1 (t)
a1
= T2 + [T1 (t = 0) − T2 ] exp(−t/τ1 ) + a1 τ1 [1 − exp(−t/τ1 )] = T2 + a1 τ1 (1 − exp(−t/τ1 ) (S˙ e,r )1 = , τ1 = (ρcp V )1 Rk,12 . (ρcp V )1
From Table C.17, we have for alumina (at T=293 K) ρ1
=
3,975 kg/m3
cp,1
=
765 J/kgK
Table C.17 Table C.17.
From Table C.17, we have for a glass plate (at T = 293 K) k2 = 0.76 W/mK
Table C.17.
The volume is V1 = a2 l1
=
(10−4 )2 (m2 ) × (6 × 10−7 )(m)
=
6 × 10−15 m3 .
The time constant is τ1
=
Rk,12
= = =
(ρcp V )1 Rk,12 l2 l2 = 2 Ak k2 a k2 3 × 10−6 (m) −4 2 (10 ) (m2 ) × 0.76(W/mK) 394.74 K/W.
Note that this large resistance is due to the small conduction area Ak . Then τ1
= =
3,975(kg/m3 ) × 765(J/kgK) × (6 × 10−15 )(m3 ) × 394.74(K/W) 7.202 × 10−6 s
=
7.202 µs.
Also =
a1
26.933(W) = 1.4762 × 109◦C/s. 3,975(kg/m3 ) × 765(J/kgK) × (6 × 10−15 )(m3 )
Solving for t, we have T1 (t) − T2 = 1 − exp(−t/τ1 ) or a1 τ1 or t
T1 (t) − T2 = −τ1 ln 1 − a1 τ1 −6 = −7.202 × 10 (s)ln 1 − =
exp(−t/τ1 ) = 1 −
T1 (t) − T2 a1 τ1
(1,900 − 20)(◦C) 1.4762 × 109 (◦C/s) × 7.202 × 10−6 (s)
1.402 × 10−6 s = 1.402 µs. 265
COMMENT: The assumption of constant conduction resistance in the glass layer can be relaxed by dividing the glass layer into many smaller layers (Section 3.7). In practice, the layer temperature is pulsed and its time variation should by taken into account. This results in a nonuniform distribution of the absorbed energy, once the reﬂections at the various internal boundaries are included.
266
PROBLEM 3.65.FUN GIVEN: In applications such as surface friction heat generation during automobile braking, the energy conversion rate S˙ m,F decreases with time. For the automobile brake, this is modeled as t ˙ ˙ Sm,F (t) = (Sm,F )o 1 − t ≤ to . to For a semiinﬁnite solid initially at T (t = 0), when its surface at x = 0 experiences such timedependent surface energy conversion, the surface temperature is given by the solution to (3.134). The solution, for these initial and boundingsurface conditions, is 1/2 ˙ 5 (Sm,F )o 2t 1/2 (αt) 1− , T (x = 0, t) = T (t = 0) + 4 Ak k 3to where S˙ m,F /Ak is the peak surface heat ﬂux qs . (S˙ m,f )o /Ak = 105 W/m2 , to = 4 s, T (t = 0) = 20◦C. OBJECTIVE: Consider a discbrake rotor made of carbon steel AISI 1010. (a) Plot the surface temperature T (x = 0, t) for the conditions given below and 0 ≤ t ≤ to . (b) By diﬀerentiating the above expression for T (x = 0, t) with respect to t, determine the time at which T (x = 0, t) is a maximum. SOLUTION: (a) From Table C.16 for carbon steel AISI 1010, we have k = 64 W/mK −5
α = 1.88 × 10
Table C.16 2
m /s
Table C.16.
Figure Pr.3.65 shows the variation of surface temperature with respect to time. T (x = 0, t = 2 s) = 91.41oC
T (x = 0,t), C
100 80
(Sm,F)o = 10 5 W/m2 Ak
60 40
t = 2 s, for T (x = 0)max
20 0
0.8
1.6
2.4
3.2
4.0
t, s Figure Pr.3.65 Variation of surface temperature with respect to time.
(b) The diﬀerentiation of T (x = 0, t) with respect to t gives 1/2 ˙ 1/2 1/2 5 1 (Sm,F )o t dT (x = 0, t) 2t 2α = α1/2 t−1/2 1− − dt 4 Ak k 2 2to 3to 1/2 ˙ 5 (Sm,F )o 1/2 1 −1/2 t1/2 2t1/2 α t = − − 4 Ak k 2 3to 3to 1/2 ˙ 5 (Sm,F )o 1/2 1 −1/2 t1/2 α t − = = 0. 4 Ak k 2 to 267
This gives 1 t − = 0 or 2 to
t 1 = . to 2
COMMENT: The maximum surface temperature, occurring at t = to /2, is T (x = 0, t = to /2)
=
1/2 ˙ 5 (Sm,F )o (αt)1/2 = 91.41◦C. 9 Ak k
268
PROBLEM 3.66.FUN GIVEN: The energy equation (3.171) for a lumped capacitance (integralvolume) system with a resistivetype surface heat transfer. The initial condition is T1 = T1 (t = 0) at t = 0. OBJECTIVE: Derive the solution (3.172) for T = T1 (t) starting from the energy equation (3.171), which applies to a lumpedcapacitance system with a resistivetype surface heat transfer. SOLUTION: We start from the energy equation (3.171), i.e., QA,1 = Q1 +
dT1 (t) T1 (t) − T2 + S˙ 1 = −(ρcp V )1 Rt,12 dt
integralvolume(lumped capacitance) energy equation with a resistivetype surface heat transfer,
with T2 being constant and the initial condition of T1 = T1 (t = 0). We rewrite this ordinary diﬀerential equation (initialvalue problem) as S˙ 1 − Q1 1 dT1 + (T1 − T2 ) = dt (ρcp V )1 Rt,12 (ρcp V )1 ˙ S1 − Q1 T1 − T2 dT1 + = a1 , τ1 = (ρcp V )1 Rt,12 , a1 = . dt τ1 (ρcp V )1 Using substitution, we have dθ θ + = a1 , θ = T1 − T2 . dt τ1 First we ﬁnd the homogeneous solution by setting a1 = 0, i.e., dθ θ + = 0, dt τ1
θh = Ae−t/τ1 .
Then we ﬁnd the particular solution by setting dθ/dt = 0, i.e., θ = a1 , τ1 Now combining the solutions, we have
θ = a1 τ1 .
θ = Ae−t/τ1 + a1 τ1 .
Applying the initial condition, we have T1 (t = 0) − T2 = Ae−0/τ1 + a1 τ1 or A = T1 (t = 0) − T2 − a1 τ1 . Then
T1 (t) − T2 = [T1 (t = 0) − T2 ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ),
which is (3.172). COMMENT: Note that for the steadystate solution, i.e., t → ∞, we have T1 (t → ∞)
= T2 + a1 τ1 = T2 + (S˙ 1 − Q1 )Rk,12 .
The steadystate solution is reached when t ≤ 4τ1 .
269
PROBLEM 3.67.FAM GIVEN: Water is heated (and assumed to have a uniform temperature, due to thermobuoyant motion mixing) from T1 (t = 0) to T1 (t = tf ) = Tf by Joule heating in a cylindrical, portable water heater with inside radius R1 and height l, as shown in Figure Pr.3.67(a). The ambient air temperature T2 is rather low. Here we assume that the outside surface temperature (located at outer radius R2 ) is the same as the ambient temperature (i.e., we neglect the resistance to heat transfer between the outside surface and the ambient). Two diﬀerent heater wall designs, with diﬀerent R2 and ks , are considered. R1 = 7 cm, l = 15 cm, T1 (t = 0) = T2 = 2◦C, tf = 2,700 s, S˙ e,J = 600 W. Evaluate the water properties at T = 310 K (Table C.23). Neglect the heat transfer through the top and bottom surfaces of the water heater, and treat the wall resistance as constant. SKETCH: Figure Pr.3.67(a) shows the portable water heater and its Joule heater and side walls.
Water T1(t) T2
R1 R2
l
ks
.
(Se,J)1
Figure Pr.3.67(a) A portable water heater.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the water temperature Tf after an elapsed time tf using, (i) a thin AISI 302 stainlesssteel wall (Table C.16) with outer wall radius R2 = 7.1 cm, and (ii) a thicker nylon wall (Table C.17) with R2 = 7.2 cm. (c) Compare the results of the two designs. SOLUTION: (a) The water has lumped thermal capacitance (uniform temperature) and is losing heat by conduction through the cylindrical side walls of the container. The thermal circuit is shown in Figure Pr.3.67(b).
Qk,12 T1(t) . (HcV)1 dT1 + Se,J dt
T2 Rk,12
Figure Pr.3.67(b) Thermal circuit diagram.
270
(b) The water is a lumped system with a single resistive conduction heat transfer. Applying conservation of energy to node T1 , we have Q A
=
Qk,12
=
T1 − T2 Rk,12
dT1 + S˙ 1 dt dT1 + (S˙ e,J )1 . = −(ρcV )1 dt = −(ρcV )1
˙ is given as The solution to this diﬀerential equation for T1 (t) (with constant S) T1 (t) = T2 + [T1 (t = 0) − T2 ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ), where τ1 = (ρcV )1 Rk,12
and
a1 =
(S˙ e,J )1 . (ρcV )1
Since T1 (t = 0) = T2 , this reduces to T1 (t) = T2 + a1 τ1 (1 − e−t/τ1 ). The volume of the container is V1
= πR12 l = π × 0.072 (m2 ) × 0.15(m) = 2.309 × 10−3 m3 .
From Table C.23 (T1 = 310 K), ρ1 = 995.3 kg/m3 and c = 4,178 J/kgK. The thermal capacitance of the water is then (ρcV )1
= =
995.3(kg/m ) × 4,178(J/kgK) × 2.309 × 10−3 (m3 ) 9,602 J/K. 3
For a cylindrical shell, the conduction resistance is Rk,12 =
ln(R2 /R1 ) . 2πlks
(i) AISI 302 stainless steel, R1 = 0.071 m. From Table C.16, ks = 15 W/mK. Then we have Rk,12
=
a1
=
τ1
=
ln(R2 /R1 ) ln(0.071/0.07) = 1.003 × 10−3 ◦C/W = 2πlks 2 × π × 0.15(m) × 15(W/mK) (S˙ e,J )1 600(W) = 0.0625 ◦C/s, = (ρcV )1 9,601.95(J/K) (ρcV )1 Rk,12 = 9,601.95(J/K) × 1.003 × 10−3 (◦C/W) = 9.631 s.
Upon substitution at t = 2,700 s, we have T1 (t)
= T2 + a1 τ1 (1 − e−t/τ ) = 2◦C + 0.0625(◦C/s) × 9.631(s) × (1 − e−2,700(s)/9.631(s) ) = 2◦C + 0.602(◦C) × (1 − e−280.3 ) = 2.6◦C.
(ii) Nylon, R2 = 0.072 m. From Table C.17, ks = 0.25 W/mK. Then we have Rk,12
=
a1
=
τ1
=
ln(R2 /R1 ) ln(0.072/0.07) = 0.1196◦C/W, = 2πlks 2 × π × 0.15(m) × 0.25(W/m − K) (S˙ e,J )1 600(W) = 0.0625◦C/s, = (ρcV )1 9,601.95(J/K) (ρcV )1 Rk,12 = 9,601.95(J/K) × 0.1196(◦C/W) = 1,148 s. 271
Upon substitution at t = 2,700 s, we have T1 (t)
= T2 + a1 τ1 (1 − e−t/τ ) = 2◦C + 0.0625(◦C/s) × 1,148(s) × (1 − e−2,700(s)/1,148(s) ) = 2◦C + 71.75(◦C) × (1 − e−2.35 ) = 66.9◦C.
(c) For the given external surface temperature, the high thermal conductivity and the wall thickness of the stainless steel produces a wall thermal resistance for design (i) that is too low to allow for the water to be heated higher than 2.6◦C. The low thermal conductivity and larger wall thickness of the Nylon produces a wall thermal resistance nearly 1000× higher than that of design(i), thus allowing the water to be heated to and above the desired temperature. Of the two designs, design (ii) is superior. Increasing the wall thickness and/or decreasing the thermal conductivity of the wall (by selecting an alternate material) would further improve the design. COMMENT: Note that we have neglected heat losses from the bottom and top surface and these can be signiﬁcant.
272
PROBLEM 3.68.FAM GIVEN: A hot lead sphere is cooled by rolling over a cold surface, with a contact resistance (Rk,c )12 , which is approximated by that between a pair of soft aluminum surfaces with δ 2 = 0.25 µm. The surfaceconvection and radiation heat transfer are represented by a constant heat transfer rate Q1 . D1 = 2 mm, T1 (t = 0) = 300◦C, T2 = 30◦C, up = 0.5 m/s, Ak,c = 0.1 mm2 , Q1 = 0.1 W. SKETCH: Figure Pr.3.68(a) shows the rolling sphere and the contact resistance.
Lead Sphere, T1(t)
Q1
−(ρckV)1
T1(t = 0) x
dT1 dt
up D1
Contact Resistance (Rk,c)12 Contact Pressure, pc , Due to Weight, Mg, and Applied Force, F T2 > T1(t = 0)
Figure Pr.3.68(a) A hot lead sphere is cooled by rolling over a cold surface.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Assume a uniform sphere temperature and determine the elapsed time required for the sphere temperature to reach 50◦C above T2 . (c) Use this elapsed time and evaluate the Fourier number FoR for the sphere. From this magnitude and by using Figure 3.33(b)(ii) for estimation, is the assumption of uniform temperature valid? (d) By approximating the internal, steadystate resistance as Rk,1 = (D1 /2)/πD12 k1 = 1/(2πD1 k1 ), evaluate Nk,1 and comment on the validation of uniform sphere temperature assumption from the relative temperature variations inside and outside the object. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.68(b).
(HcpV)1
dT1 dt
Q1 Qk,12
Control Surface, A1 T1(t) T1(t = 0) (Rk,c)12 T2
Figure Pr.3.68(b) Thermal circuit diagram.
(b) From (3.172), we have T1 (t) = T2 + [T1 (t = 0) − T2 ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ), S˙ 1 − Q1 Q1 τ1 = (ρcp V )1 (Rk,c )12 , a1 = =− , (ρcp V )1 (ρcp V )1 since S˙ 1 = 0. 273
From Table C.16, we have for lead ρ1 = 11,340 kg/m3 cp,1 = 129 J/kgK
Table C.16 Table C.16
k1 = 35.3 W/mK −5
α1 = 2.41 × 10
Table C.16 2
m /s
Table C.16.
(3.1) Then using V1 = πD13 /6, we have (ρcp V )1 = 11,340(kg/m3 ) × 129(J/kgK) × π × (2 × 10−3 )3 (m3 )/6 = 6.128 J/K. From Figure 3.25, we have for pc = 105 Pa and soft AlAl surfaces with δ 2 1/2 = 0.25 µm 3 2 (Ak,c Rk,c )−1 12 = 8 × 10 (W/m )/K.
Then (Rk,c )12
τ1 a1
=
(Ak,c Rk,c )12 1 = 3 2 Ak,c 8 × 10 [(W/m )/K] × 10−7 (m2 )
=
1.25 × 103 K/W
6.128(J/K) × 1.25 × 103 (K/W) = 7,659.5 s 0.1(W) = −0.0163 K/s = − 6.128(J/K) =
! " T1 (t) = (30 + 50)(◦C) = 30(◦C) + (300 − 30)(◦C)e−t/7,659.5(s) (−0.0163)(K/s) × 7,659.5(s) × 1 − e−t/7,659.5(s) . Solving for t, we have t L
= 6,239 s = up t = 0.5(m/s) × 6.236(s) = 3,120 m.
(c) From Figure 3.33(b)(ii), the Fourier number is deﬁned as FoR
=
α1 t 4α1 t 4 × 2.41 × 10−5 (m2 /s) × 6.236(s) = = = 150.3. R12 D12 (2 × 10−3 )2 (m2 )
From Figure 3.33(b)(ii), if the surface temperature was suddenly changed, then this elapsed time t, or its dimensionless value FoR , would be suﬃcient to establish a uniform temperature. Here we do not have a constant surface temperature, but we can state that there is a suﬃcient elapsed time to allow for a nearly complete penetration of surface temperature changes, i.e., it is safe to assume a uniform temperature. (d) From (3.161) we have Nk,1
= = =
Rk,1 Rk,1 = Rk,12 (Rk,c )12 1/(2πD1 k1 ) = 1.25 × 103 (K/W) 1 = 1.803 × 10−3 0.1. 2π × 2 × 10−3 (m) × 35.3(W/mK) × 1.25 × 103 (K/W)
This shows that the internal resistance is negligible and a lumped capacitance is also valid from the insideoutside temperature variation point of view. COMMENT: We have veriﬁed that, here, there is a negligible penetration temperature nonuniformity (i.e., large FoR ) and a negligible internal temperature variation compared to the external temperature variation (i.e., small Nk,1 ). As the volume becomes smaller, this assumption becomes more readily satisﬁed. These are the assumptions used in the ﬁnitesmall volume treatment of the heat transfer media (by division of the medium into small, but yet ﬁnite volumes). Also, note that we have not determined pc and that a large pc would require an applied force (other than the particle weight). 274
PROBLEM 3.69.FAM.S GIVEN: A thin, ﬂexible thermofoil (etched foil) heater with a mica (a cleavable mineral) casing is used to heat a copper block. This is shown in Figure Pr.3.69(a). At the surface between the heater and the copper block, there is a contact resistance (Rk,c )12 . Assume that the heater and the copper block both have small internal thermal resistances (Nk,1 < 0.1), so they can be treated as having uniform temperatures T1 (t) and T2 (t), with the initial thermal equilibrium conditions T1 (t = 0) = T2 (t = 0). The other heat transfer rates, from the heater and the copper block are prescribed (and constant) and are given by Q1 and Q2 . If the heater temperature T1 (t) exceeds a threshold value of Tc = 600◦C, the heater is permanently damaged. To avoid this, the thermal contact resistance is decreased by the application of an external pressure (i.e., large contact pressure pc ). For mica, use the density and speciﬁc heat capacity for glass plate in Table C.17. R = 3 cm, L1 = 0.1 cm, L2 = 3 cm, S˙ e,J = 300 W, Q1 = 5 W, Q2 = 50 W, T1 (t = 0) = T2 (t = 0) = 20◦C. SKETCH: Figure Pr.3.69(a) A thermofoil heater is used to heat a copper block. There is a contact resistance between the two. Q2 Copper Block T2(t) R
Contact Resistance (Rk,c)12
L2 pc
Thermofoil Heater with Mica casing T1(t) Joule Heating Se,J
L1 +Q1
Figure Pr.3.69(a) A copper block is heated with a thin heater through a contact resistance.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Plot T1 (t) and T2 (t) with respect to time, for 0 ≤ t ≤ 100 s, for (i) Ak (Rk,c )12 = 10−3 K/(W/m2 ), and (ii) Ak (Rk,c )12 = 10−2 K/(W/m2 ). SOLUTION: (a) The thermal circuit diagram is given in Figure Pr.3.69(b). Q2 (Prescribed and Constant)
(rcpV)2
Copper Block, T2
dT2 dt
(Rk,c)12 Heater, T1
Qk,12
dT1 dt Q1 (Prescribed and Constant) Se,J  (rcpV)1
Figure Pr.3.69(b) Thermal circuit diagram.
(b) This is a twonode thermal system and for each node we use (3.161), i.e., dT1 + S˙ e,J , dt dT2 , Q2 + Qk,21 = −(ρcp V )2 dt
Q1 + Qk,12 = −(ρcp V )1
275
Nk,1 < 0.1
(3.2)
Nk,2 < 0.1,
(3.3)
where = πR2 L1 , V2 = πR2 L2 , T1 (t) − T2 (t) Qk,12 = (Rk,12 ) T1 (t = 0) = T2 (t = 0) = T (t = 0). V1
From Tables C.16 and C.17, we have
ρ1 = 2,710 kg/m3 cp,1 = 837 J/kgK copper: ρ2 = 8,933 kg/m3 mica:
Table C.17 Table C.17 Table C.16
cp,2 = 385 J/kgK
Table C.16.
Then
(ρcp V )1 (ρcv V )2
= =
2,710(kg/m3 ) × 837(J/kgK) × π × (0.03)2 (m2 ) × 10−3 (m) = 6.413 J/K 8,933(kg/m3 ) × 385(J/kgK) × π × (0.03)2 (m2 ) × 3 × 10−2 (m) = 2.918 × 102 J/K.
(Rk,c )12
=
Ak (Rk,c )12 10−2 [K/(W/m2 )] 10−2 [K/(W/m2 )] = = 2 Ak πR π × (0.03)2 (m2 )
= 3.536 K/W, = 0.3536 K/W,
Ak (Rk,c )12 = 10−2 K/(W/m2 )
for for
Ak (Rk,c )12 = 10−3 K/(W/m2 ).
The two energy equations become, for Ak (Rk,c )12 = 10−2 K/(W/m2 ), dT1 (t) T1 (t) − T2 (t) = −6.413 × + 300 3.536 dt T2 (t) − T1 (t) dT2 (t) 50 + = −2.918 × 102 × , 3.536 dt ◦ T1 (t = 0) = T2 (t = 0) = 20 C. 5+
The variations of T1 (t) and T2 (t) with respect to time are plotted in Figure Pr.3.69(c). Due to its smaller mass, the heater heats up quickly and when (Rk,c )12 is large, this results in temperatures in excess of the damaging threshold temperature Tc = 600◦C. COMMENT: Note that the assumption of a uniform temperature may be valid for the copper block, but not for the heater [especially for the smaller (Rk,c )12 ]. In this case, the heater should be divided into segments (along its thickness) and a separate energy equation should be written for each segment (i.e., ﬁnitesmall volume energy equations).
276
1,200 Heater is Permanently Damaged
T1(t), T2(t), oC
960
T1(t), (Ak Rk,c)12 = 102 K/(Wm2)
720 Tc
480
T1(t), (Ak Rk,c)12 = 103 K/(Wm2) T2(t), (Ak Rk,c)12 = 102 K/(Wm2) T2(t), (Ak Rk,c)12 = 103 K/(Wm2)
240
0 0
20
40
60
80
100
t, s Figure Pr.3.69(c) Variation of the two node temperatures with respect to time.
277
PROBLEM 3.70.FAM GIVEN: An initially cold T1 (t = 0), pure aluminum spherical particle is rolling over a hot surface of temperature Ts at a constant speed uo and is heated through a contact conduction resistance Rk,c . This is shown in Figure Pr.3.70(a). T1 (t = 0) = 20◦C, Ts = 300◦C, Tf = 200◦C, D1 = 4 mm, Rk,c = 1,000 K/W, uo = 0.1 m/s. Determine the pure aluminum properties at T = 300 K. SKETCH: Figure Pr.3.70(a) shows the falling spherical particle and the heat transfer by contact conduction.
g
Pure Aluminum Ball, T1(t) Surface Temperature Ts (Uniform)
D1
uo Contact Resistance Rk,c
L
Figure Pr.3.70(a) A pure aluminum ball rolls over a hot surface and is heated by contact conduction.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Is the assumption of a uniform temperature valid? Use Rk,1 = 1/(4πD1 k1 ) for the internal conduction resistance. (c) Determine the length L the ball has to travel before its temperature reaches Tf . Assume that the contact conduction is the only surface heat transfer (surfaceconvection and radiation heat transfer are assumed negligible). SOLUTION: (a) Figure Pr.3.70(b) shows the thermal circuit diagram. Heat transfer to the ball is by contact conduction only.
Qk,1s T1(t) −(ρcpV)1
dT1 dt
Ts Rk,c
Figure Pr.3.70(b) Thermal circuit diagram.
(b) Lumped capacitance treatment is justiﬁed when Rk,i /Rk,ij ≡ Nk,i < 0.1, i.e., (3.161). Using Rk,1 = 1/(4πD1 k1 ), and k1 = 237 W/mK (Table C.16), we have 1 1 4 × π × 0.004(m) × 237(W/mK) 4πD1 k1 = Rk,c 1,000(K/W)
Nk,1
=
Nk,1
= 8.385 × 10−5 0.1, then the uniform temperature assumption is valid. 278
(c) The transient temperature of the ball is given by (3.172), i.e., T1 (t) = Ts + [T1 (t = 0) − Ts ] exp(−t/τ1 ) + a1 τ1 [1 − exp(−t/τ1 )]. Since there is no heat loss and heat generation, i.e., S˙ 1 − Q1 = a1 = 0, this reduces to T1 (t)
= Ts + [T1 (t = 0) − Ts ] exp(−t/τ1 ),
where τ1 = (ρcp V )1 Rk,c . The volume is 4π V1 = 3
D 2
3 =
3 4π 0.004 3 2
= 3.355 × 10−8 m3 . From Table C.16, for pure aluminum, we have ρ1
=
2,702 kg/m3
cp,1 τ1
= =
903 J/kgK 2,702(kg/m3 ) × 903(J/kgK) × 3.355 × 10−8 (m3 ) × 1,000(W/K)
τ1
=
81.85 s.
Then, using the values given, solving for time, −t 473.15(K) = 573.15(K) + [(293.15 − 573.15)(K)] e 81.85 t = 84.27 s. Using L = t uo , with uo = 0.1 m/s, we have L = 8.427 m. COMMENT: The heat transfer to the ambient by surface convection is neglected but becomes increasingly more important as T1 increases. Surface radiation also becomes important. These would increase the required length (time).
279
PROBLEM 3.71.FUN GIVEN: In printedcircuit ﬁeldeﬀect transistors, shown in Figure Pr.3.71(a), the electrons are periodically accelerated in the active layer and these electrons are scattered by collision with the lattice molecules (which is represented as collision with the lattice phonons), as well as collision with the other electrons and with impurities. These collisions result in the loss of the kinetic energy (momentum) of the electrons (represented by the Joule heating) and this energy is transferred to the lattice molecules due to local thermal nonequilibrium between the electrons having temperature Te and the lattice having temperature Tl . An estimate of the electron temperature Te can be made using the concept of relaxation time. As discussed in the footnote on page 288, the energy equation for the electron in a lattice unit cell can be written as Te (t) − Tl dTe (t) − τe dt ue
me,o u2e = ae = 0.1 3kB = µo e,
2 1 − τm τe
where the term on the left is the heat transfer from electron to the lattice, the ﬁrst term on the right is storage, and the second term is energy conversion. Here the coeﬃcient 0.1 in a1 represents that 0.9 of the heat generated is conducted to surroundings. The electron drift velocity is related to the electric ﬁeld and the electron mobility and the mass used is the eﬀective electron mass 0.066me. The two relaxation times are the electron momentum relaxation time τm , and the electronlattice relation time τe . Assume that the lattice temperature is constant (due to the much larger volume of the lattice molecules, compared to the electrons). me,o = 0.066me , me = 9.109 × 10−3 kg, kB = 1.3807 × 10−23 J/K, µo = 0.85 m2 /Vs, e = 5 × 105 V/m, τm = 0.3 ps, τe = 8 ps, Tl = 300 K, Te (t = 0) = Tl . SKETCH: Figure Pr.3.71(a) shows the transient heat generation (Joule heating) in the electron transport layer.
,ϕds , Applied Voltage ,ϕg Loss of Kinetic Energy Due to Collisions
Source
Jd
Qe (A large . Depletion fraction of Se,J) Gate Region e
Active Layer Drain
Electron Te Heat Transfer to Lattice Te  Tl
Drift Velocity ue
Lattice Molecule Tl
Je
Silicon Substrate
Joule . Electron Heating Se,J Transfer
SemiInsulating Substrate
Figure Pr.3.71(a) The printedcircuit ﬁeldeﬀect transistor and heat generation by electron kinetic energy loss due to collisions. The electron heat transfer as a unit cell is also shown.
OBJECTIVE: For the conditions given below, plot the electron temperature, using the solution (3.172), with respect to time, up to an elapsed time of t = 100 ps.
280
SOLUTION: We begin with the solution (3.172) for the electron temperature, i.e., Te (t)
= Tl + [Te (t = 0) − Tl ]e−t/τl + a1 τl (1 − e−t/τl ) = Tl + a1 τl (1 − e−t/τl ),
since Te (t = 0) = Tl . From the problem statement, we have for ae ae ue
me,o u2e = 0.1 3kB = µo e.
2 1 − τm τe
,
and using the numerical values, we have Te (t) = 300(K) + 1.715 × 1014 (K/s) × 8 × 10−12 (s) × [1 − e−t/8×10
−12
(s )
].
Figure Pr.3.71(b) shows the variation of the electron temperature with respect to time. As expected, within 4 time constants 4τe , the electron reaches its steadystate temperature. 2,000 Te (t
) = 1,672 K
Te , K
1,600 1,200 800 400 0
Tl = 300 K = Te (t = 0) 0
20
40
60
80
100
t, ps Figure Pr.3.71(b) Variation of the electron temperature with respect to time.
COMMENT: Note that the steadystate temperature is also found by setting the time derivation equal to zero, i.e., Te (t → ∞) − Tl = ae τe or Te (t → ∞) = Te + a1 τe = 1,672K Depending on the switching time, the electric ﬁeld (applied voltage) is turned oﬀ and therefore, the Joule heating is on for a period. This period is generally longer than τe . Also note that we have allowed for 0.9a1 to leave as Qc (heat loss by conduction to surroundings). By including the conduction through the active and semiinsulating layers, this heat loss can be determined as part of the solution. Since from (3.173), we have ae =
S˙ e − Qe , (ρcp V )e
here, S˙ e /V is rather large.
281
PROBLEM 3.72.FAM GIVEN: Due to defects in the brake pad or the rotor geometry, the friction heat generation S˙ m,F may not have a uniform distribution over the brake padrotor contact surface. This results in a hot spot at the locations of high contact, and due to the thermal expansion, these hot spots continue to have further increase in contact pressure. Eventually very high temperatures and a failure occurs. Consider the friction energy conversion occurring over a rotor surface. The rotor is idealized as a ring of inner radius Ri , outer radius Ro , and thickness l, as shown in Figure Pr.3.72(a). Under normal contact, the energy conversion will be equally distributed over the entire contact surface and a uniform temperature T1 (t) can be assumed. Under hotspot contact, assume that the energy is dissipated over a ring with the inner and outer radii Ri,1 and Ro,1 (with the same thickness l), resulting in a uniform temperature T1 (t) (i.e., Nk,1 < 0.1) and that the rest of the rotor is at a constant temperature T2 with heat ﬂowing by conduction from T1 (t) to T2 with a constant resistance Rk,12 . This is only a very rough approximation. S˙ m,F = 30 kW, Ro = 18 cm, Ri = 13 cm, Ro,1 = 16 cm, Ri,1 = 15 cm, (ρcp )1 = 3.5 × 106 J/m3 K, Rk,12 = 1◦C/W, T1 (t = 0) = 20◦C, T2 = 20◦C. SKETCH: Figure Pr.3.72(a) shows the areas and for the normal and the hotspot braking.
An Idealized DiscBrake Rotor Friction Energy Conversion Sm,F
l
Ro
For HotSpotting, the Rest of Rotor at Constant Temperature, T2
Ri,1 Ri
HotSpot Contact Area, T1(t)
Ro,1
Figure Pr.3.72(a) A discbrake rotor with (i) normal padrotor contact, and (ii) with hotspot partial surface contact.
OBJECTIVE: (a) Draw the thermal circuit diagram for (i) normal contact with no heat transfer, and (ii) hotspot contact with Qk,12 as the heat transfer. (b) Determine the temperature T1 (t) for cases (i) and (ii) after an elapsed time of t = 4 s. (c) Comment on the diﬀerence in T1 (t = 4 s) for cases (i) and (ii). SOLUTION: (a) The thermal circuit diagrams are shown in Figure Pr.3.72(b). The prescribed heat transfer rate Q1 = 0. For case (ii), a resistancetype heat transfer Qk,12 exists and T2 is prescribed and constant.
(i) Normal Contact
(ii) HotSpot Contact T1(t)
T1(t)
Qk,12 T2
Q1 = 0
Q1 = 0 S1 = Sm,F
Rk,12
−(ρcpV)1 dT dt
S1 = Sm,F
−(ρcpV)1 dT dt
Figure Pr.3.72(b) Thermal circuit diagram.
(b) The timedependent, uniform temperature T1 (t) is given by (3.169) for the case of no resistivetype heat transfer and (3.172) for the case with a resistancetype heat transfer. 282
(i) From (3.169), we have T1 (t) = T1 (t = 0) +
S˙ m,F t (ρcp V )1
= π(Ro2 − Ri2 )l.
V1 Using the numerical values, we have
= π(Ro2 − Ri2 )l = π[0.182 (m2 ) − 0.132 (m2 )] × 0.015(m) = 7.300 × 10−4 m3 3 × 104 (W) × 4(s) = 20(◦C) + 6 3.5 × 10 (J/m3 K) × 7.300 × 10−4 (m3 ) = 20(◦C) + 46.97(◦C)
V1
T1 (t = 4 s)
=
66.97◦C.
(ii) From (3.172), we have T1 (t) τ1
= T2 + [T1 (t = 0) − T2 ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ) S˙ m,F = (ρcp V )1 Rk,12 , a1 = . (ρcp V )1
Since T1 (t = 0) = T2 , we have T1 (t = 0)
= T2 + a1 τ1 (1 − e−t/τ1 ) = T2 + S˙ m,F Rk,12 (1 − e−t/τ1 ).
Now noting the smaller volume, and using the numerical values, we have V1
2 2 = π(Ro,1 − Ri,1 )l
= π[0.162 (m2 ) − 0.152 (m2 )] × 0.015(m) = 1.460 × 10−4 m3 τ1
= =
T1 (t = 4s)
= =
3.5 × 106 (J/m3 K) × 1.460 × 10−4 (m3 ) × 1(K/W) 5.110 × 102 s 20(◦C) + 3 × 104 (W) × 1(K/W) × [1 − e−4(s)/511.0(s) ] 20(◦C) + 233.9(◦C) = 253.9◦C.
(c) By comparing the results of (i) and (ii) in (b), we note that for (ii) there is an additional 180◦C rise in the temperature. When multiple braking is made (as in stopandgo or downhill breaking), this increase is compounded each time the brake is applied. Then temperatures above the damage threshold of the brake pad material are reached. COMMENT: Note that since τ1 = 511.0 s is much larger than the elapsed time of t = 4 s, we could have neglected the heat transfer during the brake period. Then we could have used (3.169) also for case (ii). This would give, for case (ii),
=
3 × 104 (W) × 4(s) 3.5 × 106 (J/m3 K) × 1.460 × 10−4 (m3 ) ◦ 20( C) × 234.8(◦C)
=
254.8◦C.
T1 (t) =
20(◦C) +
For longer elapsed times, (3.172) should be used. Note that the uniform rotor temperature assumption may not be valid for such a short elapsed time. Then a distributed, penetration treatment may be made. 283
PROBLEM 3.73.FUN.S GIVEN: A thermal barrier coating in the form of spray deposited, zirconia particles, is used as a thin layer to protect a substrate. Figures Pr.3.73(i) and (ii) show a typical barrier coating and a representative twodimensional conduction network model. The heat conduction through the gas ﬁlling the voids is neglected. The thermal conductivity for the zirconia particles is ks = 1.675 W/mK, and the porosity of the coating is approximately = 0.25. The geometrical properties of the representative network model are given in Table Pr.3.73. SKETCH: Figure Pr.3.73(i) shows the micrograph of the thermal barrier coating, and (ii) shows the twodimensional, representative conduction network model. (i)
Thin Thermal Protection Film
(ii) Pore
21
T2
22
20
18 15 16
Adiabatic
l
L
8
7
T1
10
9
y
6
ks
kf = 0 x
13
Node
Arm
1
14
12
11
y
19
17
4
2
Adiabatic 5
3
x
Substrate
Figure Pr.3.73(i) Micrograph of a thermal barrier coating, and (ii) a twodimensional, representative conduction network model.
OBJECTIVE: (a) Determine the estimated eﬀective thermal conductivity kyy (along the y axis) for the ﬁlm layer by using the twodimensional thermal circuit diagram given in Figure Pr.3.73(ii).
Table Pr.3.73 The geometrical properties of the representative network model.
Arm
l, µm
L, µm
Arm
l, µm
L, µm
17 24 35 45 46 59 68 69 78 711 910 1013 1112
1.5 3 5 3 0.8 5 5 1 2.5 5 3.5 2 2.25
12 7 9 8 10 13 10 12 16 19 11 13 20
1116 1213 1217 1314 1315 1419 1518 1520 1621 1718 1721 1822 1920
3.2 8 4 2 2 3 3.25 1.2 7.5 3 2.5 3 1
10 11 15 6 9 12 13 19 10 6 18 4 6
284
(The network model also represents the thermal circuit diagram for the layer.) Neglect the thermal conductivity of the gas ﬁling the pores (kf = 0). Take the length along the z axis (perpendicular to page) w = 1 m. Use as the temperature at the lower boundary Lz , T1 = 225◦C, and the temperature at the upper boundary, T2 = 400◦C (for eﬀective conductivity is independent of these values). Write onedimensional, steadystate conduction heat ﬂow for each arm and an energy equation for each node. Solve the set of linear algebraic equations for the temperature of each node. Calculate the total heat ﬂux, i.e., Qk , leaving the upper surface and determine the eﬀective thermal conductivity from the expression Qk = kyy Lx Lz
(T2 − T1 ) . Ly
Take Lx = Ly = 100 µm. Note also that left and right boundaries of the network model are adiabatic, i.e., no heat ﬂows across these boundaries. (b) Compare the result of (a) with the analytical result for an isotropic, periodic unitcell model given by k = 1 − 1/2 , kxx = kyy . k SOLUTION: (a) For each arm, a onedimensional steadystate conduction heat ﬂow rate (a total of 26 relations), and for each node an energy equation (a total of 16 equations) are written. For example, for arm 1314, we have T14 − T13 , Ak,1314 = l1314 Lz L1314 Q1314 + Q1315 − Q1213 − Q1013 = 0.
Q1314 = Ak,1314 ks
Since the left and right boundaries of the network model are adiabatic, then for example for node 10, the energy equation becomes Q910 = Q1013 . The set of linear algebraic equations is solved by a solver such as SOPHT. We have ks = 1.675 W/mK, and the result is kyy = 0.413 W/mK. (b) Using the analytical result for an isotropic, periodic unitcell model, we have kyy = ks (1 − 1/2 ) = 1.675(W/mK) × (1 − 0.251/2 ) = 0.8375 W/mK. COMMENT: The analytical result gives us a higher value, since it assumes a geometry for the thermal barrier coating as composed of an isotropic, periodic unit cell. As is evident from the twodimensional network model, the geometry cannot be assumed as isotropic, periodic unit cells. The predictions can be improved by using a larger number of nodes and by including the third dimension.
285
PROBLEM 3.74.FUN GIVEN: Steadystate conduction in a rectangular, twodimensional medium with prescribed temperatures on the bounding surfaces. Lx = Ly , ∆x = ∆y = Lx /N.
SKETCH: Figure Pr.3.74(a) shows the twodimensional medium divided into ﬁnitesmall volumes. T2 , T* = 1 Finite Small Volume
Lx
Lz
,x
T1 , T* = 0 T1 , T* = 0
y
Ly
,y
T(0,1) T1 , T* = 0 x T(1,1) TwoDimensional T(1,0) Temperature Distribution T = T(x,y)
Figure Pr.3.74(a) Twodimensional, steadystate conduction in a rectangular medium. The discretized ﬁnitesmall volumes are also shown.
OBJECTIVE: (a) Determine the temperature distribution for the twodimensional, steadystate conduction in the rectangular geometry shown in Figure Pr.3.74. Use the dimensionless temperature and lengths T ∗ (x, y) =
T (x, y) − T1 ∗ x y , x = , y∗ = . T2 − T1 Lx Ly
(b) Plot the results for N = 3, 15, and 21. (c) Compare the results with the exact series solution ∞ nπx sinh(nπy/Lx ) 2 1 + (−1)n+1 sin T (x, y) = π n=1 n Lx sinh(nπLy /Lx ) ∗
by showing the results on the same plot. Lx = Ly = 20 cm, ∆x = ∆y = Lx /N . SOLUTION: (a) The ﬁnitevolume energy equation for twodimensional heat transfer with the Cartesian coordinates, is given by (3.184), for the interior nodes. This written for ∆x = ∆y, is ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Ti,j Ti,j Ti,j Ti,j − Ti,j − Ti+1,j − Ti,j−1 − Ti,j+1 + + + =0 ∆x ∆x ∆y ∆y k∆yLz k∆yLz k∆xLz k∆xLz
Since we have chosen ∆x = ∆y, we have ∗ ∗ ∗ ∗ ∗ − Ti−1,j − Ti+1,j − Ti,j−1 − Ti,j+1 = 0. 4Ti,j
Here i = 3, · · · N − 1, and j = 2, 3, · · · N − 1, with i designating the xdirection index and j designating the ydirection index. 286
For nodes i = 2 and N and j = 2 and N , we have a diﬀerent energy equation. For example, for i = 2, and 2 < j < N , we have ∗ ∗ ∗ ∗ ∗ − 0.5T1,j − T3,j − T2,j−1 − T2,j+1 = 0. 3.5T2,j These are N 2 interior nodes and the above energy equations are written for each node. ∗ ∗ ∗ ∗ , TN∗ +1,j , Ti,0 and Ti,N The surface nodes T0,j +1 are all prescribed. All are set to zero except Ti,N +1 = 1. A solver, such as SOPHT, is used to determine Ti,j for the interior nodes. (b) Figure Pr.3.74(b) shows the plot of numerical results obtained using N = 3, 15, and 21. Since the plotter uses a curve ﬁt to the discrete data, all results appear the same, although for N = 3, only 3 interior nodes are used. The results for the series solution are also shown and are nearly identical to those obtained numerically (they cannot be distinguished on the ﬁgure). At least 50 terms are needed in the series solution for a converged solution. 1.0
1.0
1.0 0.9 0.8 0.7 0.6 0.5
0.75
0.4
y*
0.3 0.50 0.2 T * = 0.1 0.25
0 0
0.25
0.50
0.75
1.0
x* ∗
∗
∗
Figure Pr.3.74(b) Distribution of T (x , y ) obtained by ﬁnitesmall volume method using N = 3, 15, and 21, and also by the series solution.
The temperature found at (x∗ = 0.125, y ∗ = 0.125) for N = 3 is T ∗ = 0.03571. For N = 15 at the same location, we ﬁnd T ∗ = 0.03051, and for N = 21, T ∗ = 0.03044. The exact solution for this location is T ∗ = 0.03036. This shows that for N > 15 a relatively accurate result is obtained. COMMENT: In practice, the number of increments N is increased until the results no longer change within a small, acceptable criterion.
287
PROBLEM 3.75.FUN GIVEN: A porcelain workpiece (in form of a circular disk) is ablated by laser irradiation. The piece is held inside a cooling ring, as shown in Figure Pr.3.75(a), to maintain its outer surface at a temperature Ts . For the ablation, the temperature of the ceramic much reach a threshold temperature Tsg (i.e., a sublimation temperature), over the area of interest. The radiation is absorbed only over the surface Ar,α , and there is surface convection over the rest of the area Aku . Assume a steadystate heat transfer and a uniform temperature along the z and φ axes [i.e., T = T (r) only]. Assume all irradiation is absorbed on the surface. For the central node, node 1, use Rr /2 as the inner surface location for the determination of the conduction resistance. R = 3 cm, Rr = R/5, l = 3 mm, qr,i = 106 W/m2 , Ts = 90◦C, Aku Rku = 10−3 K/(W/m2 ), Tf,∞ = 120◦C. SKETCH: Figure Pr.3.75(a) shows the workpiece and the cooling ring around it. FarField Air Temperature Tf,
Laser Generator Laser Beam Laser Irradiation, qr,i
Workpiece Aku Rku 2Rr
l
Aku = A  Ar,a
Ar,= R
Ts
Cooling Ring z
r
R
f
2Rr
Node 1
Ts (Maintained Temperature)
Coolant Flow
Workpiece
Figure Pr.3.75(a) A porcelain workpiece is irradiated for ablation. The workpiece is cooled at its periphery by a coolant carrying ring.
OBJECTIVE: Divide the porcelain piece into N segments, i.e., ∆r = R/N , and apply the ﬁnitesmall volume energy equation (3.176) to each segment. (a) Draw the thermal circuit diagram for the entire disk. (b) Determine the segment temperature Ti for N = 5. SOLUTION: (a) Figure Pr.3.75(b) shows the thermal circuit diagram for N = 5 or ∆r = R/5. Note that Q1 = 0, because the only heat transfer from A1 is by conduction Qk,12 . (b) The energy equation (3.176) for V1 , under steadystate conditions, is QA,1 = Qk,12 = (S˙ r,α )1 = Ar,α qr,i . For Qk,12 , from Table 3.2, we have T1 − T2 Qk,12 = , Rk,12
ln Rk,12 =
3∆r/2 Rr /2 . 2πkl
Note that the nodes are located at the center of each segment, except for node 1, where to avoid singularity, we have used Rr /2. The conductivity of porcelain is given in Table C.15, i.e., k = 1.5 W/mK
Table C.15.
288
r R 5 Q1 = 0
2R 5 Qk,12
Qk,23 T2
T1
R
4R 5
3R 5
Cooled Edge
,R
Qk,34 T3
Qk,45 T4
Qk,5s T5 Ts , Prescribed
Se,=
V1
FiniteSmall Volume, V2
Qku,2 Tf,
V3
Qku,3 Tf,
Qku,4 Tf,
V4
Qku,5 Tf,
V5
Figure Pr.3.75(b) Thermal circuit diagram.
Then 3 × 3 × 10−2 (m)/5 × 2 3 × 10−2 (m)/5 × 2 2π × 1.5(W/K) × 3 × 10−3 (m) 38.85 K/W ln
=
Rk,12
= (S˙ r,α )1
(3 × 10−2 )2 (m2 ) × 106 (W/m2 ) 25 1.131 × 102 W.
= πRr2 qr,i = π =
The energy equation (3.176), for V2 , is QA,2 = Qk,21 + Qk,23 + Qku,2∞ = 0, where Qk,21
= −Qk,12 ,
Qk,23
=
Qku,2∞
=
T2 − T3 , Rk,23 T2 − Tf,∞ , Rku,2∞
Rk,21 = Rk,12 5∆r/2 ln 3∆r/2 Rk,23 = 2πkl
Rku,2∞ = π
Aku Rku 2 2 . 2R R − 5 5
Then
Rk,23
=
5 3 = 18.06 K/W 2π × 1.5(W/mK) × 3 × 10−3 (m)
Rku,2∞
=
10−3 [K/(W/m2 )] = 2.947 K/W. 3 × (3 × 10−2 )2 (m2 ) π× 25
ln
289
Similarly, 7 5 = 11.90 K/W 2π × 1.5 × 3 × 10−3 9 ln 7 = 8.887 K/W 2π × 1.5 × 3 × 10−3 1 ln 9/10 = 3.726 K/W 2π × 1.5 × 3 × 10−3 10−3 = 1.768 K/W 5 × (3 × 10−3 )2 π× 25 10−3 = 1.263 K/W 7 × (3 × 10−3 )2 π× 25 10−3 = 0.9822 K/W. 9 × (3 × 10−3 )2 π× 25 ln
Rk,34
=
Rk,45
=
Rk,5∞
=
Rku,3∞
=
Rku,4∞
=
Rku,5∞
=
Using a solver (such as SOPHT), we solve for T1 to T5 , from the ﬁve energy equations. The results are T1
=
4,804◦C
T2 T3
= =
409.8◦C 142.9◦C
T4 T5
= =
121.3◦C 114.4◦C.
COMMENT: By increasing N , a more accurate prediction of T1 is obtained. However, the lack of surface convection and the localized irradiation does result in the desired high temperature T1 (for ablation).
290
PROBLEM 3.76.FUN.S GIVEN: The eﬀective thermal conductivity of some porous media can be determined using the random network model. In one of these models, a regular lattice is used, but the locations of the nodes, within the regular lattice, are generated randomly and then connected, forming a network. The thicknesses (for a two or threedimensional geometry) of these connectors (i.e., arms) are then assigned based on the porosity and any other available information. The network can represent the solid or the ﬂuid part of the medium. A 3 × 3 square unitcell, twodimensional random network model is shown in Figure Pr.3.76(a). This is determined by randomly selecting the location of each node within its unit cell space. The coordinates of each node, and the length and the thickness for each arm are given in Table Pr.3.76. Assume that the heat transfer between adjacent nodes is one dimensional and steady. For the arms, use the thermal conductivity of aluminum (Table C.16). Assume that the thermal conductivity of the ﬂuid is much smaller than that of the solid. The left and right boundaries of the medium are ideally insulated. The temperature for the lower boundary of the medium, [(x, y) = (0.0)], is maintained at Tc = 100◦C, (i.e., Tc = T1 = T6 = T11 = 100◦C), while the temperature for the upper boundary, [(x, y) = (3, 0)], is maintained at Th = 200◦C, (i.e., Th = T5 = T10 = T15 = 200◦C). Table Pr.3.76 Coordinates of each node, and the length and the thickness of each arm, for a twodimensional, random network model. Node
(x, y)
1 2
(0.76,0.00) (0.13,0.34)
3
(0.92,1.71)
4
(0.47,2.14)
5 6 7
(0.83,3.00) (1.73,0.00) (1.17,0.54)
8
(1.93,1.26)
9
(1.44,2.15)
10 11 12 13 14 15
(1.83,3.00) (2.38,0.00) (2.75,0.77) (2.18,1.45) (2.84,2.91) (2.27,3.00)
Li,j (mm)
li,j (mm)
L1,2 = 0.715 L2,3 = 1.581 L2,7 = 1.059 L3,4 = 0.622 L3,8 = 1.105 L4,5 = 0.932 L4,9 = 0.970
l1,2 = 0.414 l2,3 = 0.213 l2,7 = 0.055 l3,4 = 0.152 l3,8 = 0.197 l4,5 = 0.344 l4,9 = 0.376
L6,7 = 0.788 L7,8 = 1.047 L7,12 = 1.596 L8,9 = 1.059 L8,13 = 0.314 L9,10 = 0.935 L9,14 = 1.592
l6,7 = 0.151 l7,8 = 0.232 l7,12 = 0.215 l8,9 = 0.469 l8,13 = 0.125 l9,10 = 0.071 l9,14 = 0.316
L11,12 = 0.854 L12,13 = 0.887 L13,14 = 1.602 L14,15 = 0.577
l11,12 = 0.417 l12,13 = 0.222 l13,14 = 0.118 l14,15 = 0.322
SKETCH: Figure Pr.3.76(a) shows a twodimensional regular (periodic) lattice with random location of nodes within the lattice, making for a random network model. OBJECTIVE: (a) Draw the thermal circuit diagram using the geometrical data. Write the energy equation for each node, along with the conduction heat transfer relation for each arm. (b) Determine the total conduction heat transfer rate, Qk,hc . (c) Using Qk,hc ≡
Ak (Th − Tc )k , L 291
Qk,hc Th = 200 C Node 5
y x
Node 10
Node 15
Rk,54
14
Qk,54
Arm
9
4
1 mm
3
1 mm
13 l23
L23
8
7
12
2 w Tc = 100 C
Node 1
Node 6
Node 11
A 3 x 3 Square UnitCell, TwoDimensional Random Network Model
Figure Pr.3.76(a) The random network model in an ordered lattice.
for L = 10−3 m and Ak = 3 × 10−6 m2 , determine the eﬀective thermal conductivity k. SOLUTION: (a) Figure Pr.3.76(a) shows the 3 × 3 square unit cell and the random network. The porosity is determined by ﬁrst determining the solid fraction by adding the area of all solid arms. Then this area is divided by the total area and then subtracted from unity. (b) For each arm, a, onedimensional steadystate conduction relation (total 18 relations), and for each node an energy equation (total 9 equations) are written. For example, for arm 1  2, we have T2 − T1 L1,2 Q12 − Q23 − Q27 = 0. Q12 = Ak,12 ks
The set of linear algebraic equations is solved by a solver such as SOPHT. From Table C.16, we have ks = 237 W/mK, at T = 300 K. The result is Qk,hc = 4.283 W. (c) Using the above deﬁnition, we have k = 42.83 W/mK.
COMMENT: The eﬀective thermal conductivity for the series arrangement is given by 1 (1 − ) = + . k kf ks The eﬀective conductivity for the parallel arrangement is given by k = kf + ks (1 − ). For the isotropic, unitcell geometry, we have k = ks (1 − 1/2 ). 292
Using these relations, the variation of the dimensionless eﬀective conductivity with respect to porosity is plotted in Figure Pr.3.76(b). Figure Pr.3.76(b) shows that the dimensionless eﬀective conductivity for the random network is lower than that for the unitcell model, for a given porosity.Also note that there are overlaps at the interceptions of the arms. This should be considered in calculating the total solid area. Neglecting this, we use Figure Pr.3.76(b). The curve ﬁtting of Figure 3.76(b) is used to calculate the actual porosity. In Figure 3.76(c), the results for the actual porosity is also shown. 1
0.8 Parallel Arrangement
DkE
/ ks
0.6
Unit Cell Model
0.4
Random Networks (With Overlaps) 0.2 Random Networks (Without Overlaps) Series Arrangement 0 0
0.2
0.4
0.6
0.8
1
∋ Figure Pr.3.76(b) Variation of the dimensionless eﬀective conductivity with respect to porosity.
1
0.8
actual
0.6
∋ 0.4
0.2
0 1
0.5
0.5
1
calculated
∋
Figure Pr.3.76(c) Actual porosity versus the calculated porosity.
293
PROBLEM 3.77.FAM GIVEN: The friction heating during skating over ice layers causes melting, and the thickness of this melt at the end of the blade δα may be estimated when the blade surface temperature Ts = Tl,o is known. Figure Pr.3.77 shows the blade length L in contact with the ice, with the skating speed designated as us . Tl,o = 10◦C, Tsl = 0◦C, L = 0.20 m, up = 2 m/s. Use properties of water given in Tables C.4 and C.27 (at T = 275 K). SKETCH: Figure Pr.3.77 shows the length L for the contact and the blade edge temperature Ts .
g
us Ice δα x
Ts = Tl,o L Length of Blade in Contact with Ice
Melt Formation Under Skate Blade
Figure Pr.3.77 Melt formation during ice skating.
OBJECTIVE: In order to estimate the elapsed time used in determining δα , we can use t = L/us . For the conditions given, determine the liquid ﬁlm thickness δα assuming that the onedimensional melting analysis of Section 3.8 is applicable. SOLUTION: From (3.198), we have δα (t) = 2ηo (αl t)1/2 , where ηo is found from Table 3.7 and depends on the Stefan number given by (3.197), i.e., Stel =
cp,l (Tl,0 − Tsl ) . ∆hsl
From Tables C.4 and C.27 (T = 275K), we have for water ∆hsl = 3.336 × 105 J/kg ρl = 1,000 kg/m3
Table C.4 Table C.27
cp,l = 4,211 J/kg Table C.27 kl = 0.547 W/mK Table C.27 k (0.547)(W/mK) αl = = = 1.299 × 10−7 m2 /s ρcp l 1,000(kg/m3 ) × 4,211(J/kg) 4,211(J/kg) × (10 − 0)(K) Stel = = 0.1262 3.336 × 105 (J/kg) Stel = 0.07121. π 1/2 Then from Table 3.7, we have ηo = 0.2062. 294
The elapsed time is t δα (t = 0.10 s)
=
0.20(m) L = 0.10 s = us 2(m/s)
= 2 × 0.2062 × [1.299 × 10−7 (m2 /s) × 0.10(s)]1/2 = 4.700 × 10−5 m = 47 µm.
COMMENT: The onedimensional analysis overestimates the heat ﬂow rate and the liquid thickness. Also, due to the skater weight, the liquid will be forced out. This is called closecontact melting and this tends to increase the heat transfer rate.
295
PROBLEM 3.78.FUN GIVEN: For the conductionmelting of a semiinﬁnite solid initially at the melting temperature Tsl and suddenly exposed to Tl,o > Ts at its surface (x = 0), the temperature distribution in the melt is given by (3.194). OBJECTIVE: (a) Derive this temperature distribution using the energy equation (3.189) and the thermal conditions at x = 0 and x = δα (t), i.e., as given by (3.190) to (3.191), i.e., Tl (x, t) = a1 + a2 erf(η) Tl (x, t) − Tl,0 erf(η) . = Tsl − Tl,o erf(ηo ) Use the similarity variable (3.195) and an error function solution, i.e., similar to (3.140), with erf(η) deﬁned by (3.141) (b) Using (3.192), show that ηo is determined from (3.196). SOLUTION: (a) The diﬀerential energy equation for the semiinﬁnite medium is given by (3.189), i.e., ∂ 2 Tl 1 ∂Tl 2 − α ∂t = 0. ∂x l Based on the similarity solution for transient conduction in a semiinﬁnite slab, given in Section 3.5.1, we choose the similarity variable (3.136), i.e., η=
x 2(αl t)1/2
,
and a solution of the type Tl (x, t) = a1 + a2 erf(η). Now using (3.190), we have Tl (x = 0, t) = Tl,0 = a1 + a2 erf(0). Using Table 3.5, we note that erf(0) = 0, then a1 = Tl,0 . Next we use (3.191), i.e., Tl [x = δα (t)] = Tsl = Tl,0 + a2 erf(ηo ), where ηo =
δα (t) 2(αl t)1/2
and ηo is a constant. Solving for a2 , we have the melt temperature distribution Tl (x, t) − Tl,0 Tsl − Tl,0 = erf(ηo ) erf(η) erf(η) . = erf(ηo )
a2 = Tl (x, t) − Tl,0 Tsl − Tl,0
296
(b) Now using (3.192) or (3.200), we have the condition for the determination of ηo , i.e., 1/2 αl ∂Tl kl = −ρl ∆hsl uF = −ρl ∆hsl 1/2 ηo . ∂x x=δα (t) t From the deﬁnition of erf(η) given by (3.141), we have ∂ ∂ 2 erf(η) = ∂x ∂x π 1/2
η
e−z dz. 2
We need to use the chain rule, i.e., ∂ ∂η ∂ 1 ∂ = = ∂x ∂x ∂η 2(αl t)1/2 ∂η or 2 2 ∂ 1 erf(η) = 1/2 eη . 1/2 ∂x π 2(αl t)
Then from the interface energy equation, we have e−ηo
2
−kl (Tsl − Tl,o )
π
1/2
1/2
1/2
(αl t)
erf(ηo )
= ρl ∆hsl
αl t
1/2
ηo ,
αl =
kl (ρcp )l
or cp,l
(Tl,0 − Tsl ) π
1/2
∆hsl
2
= ηo eηo erf(ηo ).
This is (3.196). COMMENT: Note that δα (t) = 2ηo (αl t)1/2 , where ηo is the root to (3.196). This relation is in a form similar to (3.148), if we use ηo = 1.8 or Stel /π 1/2 = 45.46. In arriving at (3.148), we assume that T ∗ = 0.01. Then the case of Stel /π 1/2 = 45.46 corresponds to the small conduction heat transfer rate through the surface x = δα (t) which is a result of a very small temperature gradient.
297
PROBLEM 3.79.FAM GIVEN: To remove an ice layer from an inclined automobile windshield, shown in Figure Pr.3.79, heat is supplied by a thinﬁlm Joule heater. The heater maintains the surface temperature of the window at Tl,0 . It is determined empirically that when the melt thickness reaches δα = 1 mm, the ice sheet begins to fall from the window. Assume that the heat transfer through the liquid water is one dimensional and occurs by conduction only, that the ice is at the melting temperature Tls , and neglect the sensible heat of the liquid water. Use the properties of water at T = 273 K from Table C.23 and Table C.6. SKETCH: Figure Pr.3.79 shows an ice layer melting over a glass sheet with the heat provided by a thinﬁlm Joule heater. x
Liquid Water (Initially Ice) Ice Se,J Tls Tl,o
Windshield δα
Figure Pr.3.79 Melting of ice on an automobile windshield.
OBJECTIVE: how long it will take for the ice to be removed t and estimate the amount of thermal energy t Determine 2 qdt(J/m ) required when (a) Tl,0 = 4◦C, and (b) Tl,0 = 15◦C. 0 SOLUTION: The position of the melting front is given by (3.198) δα (t) = 2ηo (αl t)1/2 . The constant ηo is the solution to (3.194), i.e., 2
ao eηo erf(ηo ) =
Stel , π 1/2
where the liquid Stefan number is deﬁned by (3.197) Stel =
cp,l (Tl,0 − Tsl ) . ∆hsl
The total energy per unit area spent to melt the 1 mm ice layer is given by integrating (3.200), i.e., t t ρl ∆hsl ao αl 1/2 −qk dt = dt = 2ρl ∆hsl ηo αl t1/2 = ρl ∆hsl δα (t). t1/2 o o The properties for water are found from Table C.23, T = 273 K, ρl = 1,002 kg/m3 , cp,l = 4,217 J/kgK, αl = 131 × 10−9 m2 /s, from Table C.6, ∆hsl = 333.60 × 103 J/kg, Tsl = 273.2 K. For each of the surface temperatures we have (a) Tl,0 = 4◦C= 277.15 K The Stefan number is Stel =
cp,l (Tl,0 − Tsl ) 4,217(J/kgK) × (277.15 − 273.2)(K) = 0.0499. = ∆hsl 333.60 × 103 (J/kg) 298
For Stel /π 1/2 = 0.0282, from interpolation in Table 3.5 we obtain ηo = 0.16063. Solving for time gives t=
1 αl
δα 2ηo
2 =
1 131 × 10−9
0.001 2 × 0.16063
2 = 74 s 1.2 min.
The total energy per unit area is then t 2 −qk dt = ρl ∆hsl δα (t) = 1,002(kg/m3 ) × 333.60 × 103 (J/kg) × 0.001(m) = 3.341 × 105 J/m . 0
(b) Tl,0 = 15◦C: The Stefan number is Stel =
cp,l (Tl,0 − Tsl ) 4,217(J/kgK) × (288.15 − 273.2)(K) = 0.1890. = ∆hsl 333.60 × 103 (J/kg)
For Stel /π 1/2 = 0.1066, from interpolation in Table 3.5 we obtain ηo = 0.30022. Solving for time gives 1 t= αl
δα 2ηo
2
1 = 131 × 10−9
0.001 2 × 0.30022
2 = 21 s.
As the ice thickness is still the same, the total energy per unit area required remains the same, 2 3.341 × 105 J/m .
t 0
qk dt =
COMMENT: When the liquid water begins to ﬂow due to the action of gravity, the assumption of pure conduction heat transfer is no longer valid. However, the water layer thickness is small enough that the treatment made here will suﬃce.
299
PROBLEM 3.80.DES GIVEN: In a thermostat used to control the passage of coolant through secondary piping leading to the heater core of an automobile, solidliquid phase change is used for displacement of a piston, which in turn opens the passage of the coolant. The thermostat is shown in Figure Pr.3.80(i). The phasechange material is a wax that undergoes approximately 15% volume change upon solidiﬁcation/melting. The response time of the thermostat is mostly determined by the time required for complete melting of the wax. This in turn is determined by the speed of penetration of the melting front into the wax and the time for its complete penetration. Two diﬀerent designs are considered, and are shown in Figures Pr.3.80(ii) and (iii). In the design shown in Figure Pr.3.80(iii), the wax reservoirs have a smaller diameter D2 compared to that of the ﬁrst design, shown in Figure Pr.3.80(ii). Therefore, it is expected that in the threereservoir design the wax will melt faster. To solve the melting problem using the analysis of Section 3.8, we need to assume that the front is planar. Although this will result in an overestimation of the time required for melting, it will suﬃce for comparison of the two designs. Due to unavailability of complete properties for the wax, use the phasechange properties of Table C.5 for paraﬃn and the properties of engine oil in Table C.23 for the wax liquid phase. The melting temperature is a function of pressure and is represented by the ClausiusClapeyron relation (A.14). Here, neglect the pressure variation and use a pressure of one atm and a constant melting temperature. Use the properties of engine oil at T = 310 K. Tl,0 = 80◦C, D1 = 8 mm, D2 = 3 mm. SKETCH: Figure Pr.3.80 shows the coolant thermostat and the (i) single and (ii) three reservoir designs.
(i) Automotive Coolant Thermostat Closed
Hot Water Tf, > Tsl ( pl)
Wax Melting Front δ(t) Q A(t)
Open
Volume Increased Due to Melting Wax Bellow
Piston
(ii) Single Reservoir
(iii) Three Reservoirs Liquid
Wax
Liquid
Solid Solid D1
Tsl δ(t) Tl,0
Ts
High Conductivity Solid
δ(t) D2
Figure Pr.3.80 (i) An automotive coolant thermostat. The solidliquid phase change actuates the piston. (ii) The singlereservoir design. (iii) The threereservoir designs.
OBJECTIVE: Determine the time it takes for the complete melting of the wax in the one and threereservoir designs.
300
SOLUTION: The penetration distance is given by (3.198), i.e., δ(t) = 2ηo (αt)1/2 . For complete penetration, we have δ = D/2, or D = 2ηo (αl t)1/2 2
or
t=
D2 , 16ηo2 αl
where t is the elapsed time needed. Then t1 =
D12 16ηo2 αl
, t2 =
D22 . 16ηo2 αl
From these, the smaller D results in a smaller elapsed time t. The thermophysical properties from Tables C.5. and C.23, are paraﬃn (at 1 atm pressure): Tsl = 310.0 K
Table C.5
∆hsl = 2.17 × 10 J/kg
Table C.5
αl = 8.70 × 10−8 m2
Table C.23
5
engine oil (at T = 310 K):
cp,l = 1,950 J/kgK
Table C.23.
Here ao is found from Table 3.7 and depends on the Stefan number, which is given by (3.197), i.e., Stel
= =
cp,l (Tl,0 − Tsl ) ∆hsl 1,950(J/kgK) × (273.15 + 80 − 310)(K) = 0.3878. 2.17 × 105 (J/kg)
Then Stel π 1/2
= 0.2188.
From Table 3.7, we have ηo = 0.4005
Table 3.7.
Now, for the elapsed times, we have, t1
=
(0.008)2 (m2 ) = 286.6 s 16 × (0.4005)2 × 8.70 × 10−8 (m2 /s)
t2
=
(0.003)2 (m2 ) = 40.31 s. 16 × (0.4005)2 × 8.70 × 10−8 (m2 /s)
COMMENT: Note the much lower response time for the smaller diameter wax reservoir. The planar approximation of the front results in an underestimation of the response time. This is because, in the radial system, as the center is reached, a smaller conduction area becomes available and this increases the penetration speed.
301
PROBLEM 3.81.FAM GIVEN: In a grinding operation, material is removed from the top surface of a small piece of pure copper with dimensions shown in Figure Pr.3.81(a). The grinding wheel is pressed against the copper workpiece with a force Fc = 50 N and there is an interfacial velocity ∆ui = 20 m/s. The coeﬃcient of friction between the two surfaces is µF = 0.4. The copper is initially at temperature T1 (t = 0) = 20◦C. SKETCH: Figure Pr.3.81(a) shows copper being ground and expanding due to friction heating. Workpiece (Sm,F)1 (Due to Grinding) Uniform Ac q1 z Temperature T1(t) (Nt,1 < 0.1) x y l = 0.5 cm L = 5 cm
w = 2.5 cm
Figure Pr.3.81(a) Friction heating of a copper workpiece by grinding resulting in thermal expansion.
OBJECTIVE: (a) Assuming a uniform copper temperature T1 (t), i.e., Nt,1 < 0.1, and a constant surface heat loss rate per unit area q1 = 675 W/m2 , draw the thermal circuit diagram. (b) Assuming an unconstrained expansion, determine the elapsed time needed to cause the copper length L to thermally expand by ∆L = 0.5 mm. Neglect all nonthermally induced stresses and strains. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.3.81(b). Uniform Temperature T1(t)
Workpiece Q1 (constant)
dT −(ρcpV)1 1 dt (constant)
(Sm,F)1 (constant)
Figure Pr.3.81(b) Thermal circuit diagram.
(b) Before using the lumped capacitance, i.e., the singlenode energy equation with a constant surface heat transfer rate (3.169), we determine the strain in the xdirection and the temperature needed to induce the desired strain. From (3.204) we have ∆L∗
= βs (T1 − To )
where βs ∆L∗
= =
1.7 × 10−5 (1/K) and To = 20◦C ∆L/L = 5 × 10−4 (m)/0.05(m) = 0.01.
Then 0.01
=
1.7 × 10−5 × (T1 − 20)(◦C)
T1
=
608.2◦C. 302
From (3.169), we have T1 = T1 (t = 0) +
−q1 A1 + S1 t, (ρcp V )1
where A1
V1
= =
2(wL) + 2(lL) + 2(wl) 2(0.025 × 0.05)(m2 ) + 2(0.005 × 0.05)(m2 ) + 2(0.005 × 0.025)(m2 )
= 0.00325 m2 = wlL 0.025(m) × 0.005(m) × 0.05(m) 6.25 × 10−6 m3 .
= =
The energy conversion by mechanical friction is given in Table C.1(d), i.e., (S˙ m,F )1 /A = µF × pc × ∆ui pc = Fc /Ac = 50(N)/0.00125(m2 ) = 40,000 N/m2 Ac
= Lw = 0.05 × 0.025(m2 ) =
0.00125 m2 .
Then (S˙ m,F )1 /A = 0.4 × 40,000(N/m2 ) × 20(m/s) = 320,000 W/m2 (S˙ m,F )1 = 320,000(W/m2 ) × 0.00125(m2 ) = 400 W. From (3.169), using the properties for pure copper given in Table C.16, ρ = 8933 kg/m3 , cp = 385 J/kgK, k = 401 W/mK, we have [−675(W/m2 ) × (0.00325)(m2 ) + 400(W)] ×t 8,933(kg/m3 ) × 385(J/kgK) × 6.25 × 10−6 (m2 )
608.2(◦C) =
20(◦C) +
588.2(◦C) =
397.8 ◦ ( C/s)t 21.5 31.8 s.
t
=
COMMENT: In order to verify the validity of a uniform temperature (in the presence of energy conversion) within the workpiece, we determine the internal conduction resistance, from Table 3.2, as Rk,1
= L/Ak k = l/Ac k = 0.005(m)/[0.00125(m) × 401(W/mK)] = 0.01 K/W.
For the temperature diﬀerence between the top and bottom surface of the workpiece designated by ∆T , and allowing the entire energy conversion S˙ m,F = 400 W to ﬂow through the workpiece, we have (S˙ m,F )1
=
∆T Rk
or ∆T
= S˙ m,F Rk = 400(W) × 0.01(K/W) = 4 K.
Since we are dealing with a relatively high temporal temperature rise T1 − T1 (t = 0) = 588.2 K, this temperature variation across the workpiece is insigniﬁcant and the two surfaces can be assumed the same temperature. 303
PROBLEM 3.82.FUN GIVEN: The thermal stress in an idealized discbrake rotor can be determined using some simplifying assumptions. The castiron rotor is shown in Figure Pr.3.82(a), along with a prescribed temperature distribution T = T (r). The temperature distribution is steady and one dimensional and the stress tensor would have planar, axisymmetric stresses given by principal components τrr (r), τθθ (r). It can be shown that these stresses are expressed as Es βs a2 r(τ − τo )dr + a1 − 2 r2 r Es βs a2 r(T − To )dr − Es βs (T − To ) + a1 − 2 2 r r 0.
τrr (r)
= −
τθθ (r)
=
τzz (r)
=
For cast iron, Es = 2 × 1011 Pa, and βs is listed in Table C.16. Also R = 17 cm, To = 100◦C, TR = 400◦C. SKETCH: Figure Pr.3.82(a) shows the rotor and its temperature distribution.
DiscBrake Rotor
Jzz = 0 (Planar Stress)
Sm,F r
G
R
Jrr(r) JGG(r) TR 0
r
T0 R
Figure Pr.3.82(a) The temperature distribution within a discbrake rotor and the induced, planar axisymmetric thermal stresses.
OBJECTIVE: (a) Determine the integration constants a1 and a2 using the mechanical conditions. Note that at r = R, there is no radial stress (for surface), and that at r = 0 the stresses should have a ﬁnite magnitude. (b) Plot the distribution of the radial and tangential rotor thermal stresses with respect to the radial location. SOLUTION: (a) Using the temperature distribution r2 T = TR + (To − TR ) 1 − 2 , R We have T − To = (TR − To ) Since at r = 0, the stresses have to be ﬁnite, we have a2 = 0. 304
r2 . R2
Performing the integrations, we have τrr (r)
Es βs r2 − 2 r(TR − To ) 2 dr + a1 r R Es βs (TR − To )r2 − + a1 4R2 Es βs (TR − To )r2 Es βs (TR − To )r2 − + a1 2 4R R2 3Es βs (TR − To )r2 − + a1 4R2
= =
τθθ (r)
= =
Using τrr = 0 at r = R, we have 0=−
Es βs (TR − To )R2 + a1 4R2
or a1 =
Es βs (TR − To ) . 4
Then τrr (r)
=
τθθ (r)
=
r2 1− 2 R Es βs (TR − To ) r2 1−3 2 . 4 R Es βs (TR − To ) 4
(b) From Table C.16, we have carbon steel:
βs = 1.15 × 10−5 1/K.
Using the numerical results, we have Es βs (TR − To ) 4
= =
2 × 1011 (Pa) × 1.15 × 10−5 (1/K) (400 − 100)(K) 4 1.725 × 108 Pa.
Figure 3.82(b) shows the variation of τrr (r) and τθθ (r) with respect to r. Note that at r = 0, the two stresses are equal and at r = R, we have τrr = 0, as expected.
Jrr , JGG , Pa
2.4 x 108 1.2 x 108
Jrr
0 JGG
1.2 x 108 2.4 x 108 3.6 x 108
3.4
6.8
10.2
r, cm
13.6
17 r=R
Figure Pr.3.82(b) Variation of τrr (r) and τθθ (r) with respect to r.
COMMENT: Note that the largest stress is for τθθ and occurs at r = R.
305
PROBLEM 3.83.FUN GIVEN: Consider the thermal stress due to a nonuniform temperature T = T (r) in an aluminum rod encapsulated in a glass shell. This is shown in Figure Pr.3.83. Since the thermal expansion coeﬃcient βs is much smaller for the fused silica glass (Figure 3.43), we assume that the periphery of the aluminum rod is ideally constrained. For the axisymmetric geometry and temperaturestress conditions we have here the stress and strain distributions as given by [5]. Es βs a2 τrr (r) = − 2 r(T − To )dr + a1 + 2 r r Es βs a2 τθθ (r) = − 2 r(T − To )dr − Es βs (T − To ) + a1 − 2 r r βs (1 + νP ) a1 (1 − νP )r (1 + νP )a2 . ∆R(r) = − r(T − To )dr + r Es Es r The temperature distribution within the aluminum rod is estimated as
r2 T (r) = TR + (To − TR ) 1 − 2 R
.
The two constants of integration, a1 and a2 , are determined using the mechanical conditions of a ﬁnite stress at r = 0 and an ideal constraint at r = R. τmax,g = −300 MPa, Es = 68 GPa, νP = 0.25, To = 80◦C, R = 4 cm. SKETCH: Figure Pr.3.83 shows the aluminum rod encapsulated in a low thermal expansion coeﬃcient glass. Fused Silica Glass Shell Aluminum Rod Assume Ideally Constrained Surface
Jrr r
JGG G
R To
T(r)
TR
Figure Pr.3.83 Thermal stress induced in an aluminum rod encapsulated in a fusedsilica glass shell.
OBJECTIVE: (a) Determine the integration constants a1 and a2 and write the expression for τrr (r = R). (b) Using the conditions given, determine the temperature TR at which the ultimate compression stress of the glass τmax,g is reached, i.e., τrr (r = R) = τmax,g . SOLUTION: (a) We rewrite the temperature distribution as
r2 T (r) = TR + (To − TR ) 1 − 2 R
.
Since at r = 0, the stresses have to be ﬁnite, a2 = 0. Next we use the second mechanical condition, i.e., ∆Rr (r = R) = 0, i.e., βs (1 + νP ) 0= R
R
r(TR − To ) 0
306
r2 a1 (1 − νP )R dr + Es R2
or 0=
βs (1 + νP ) 1 a1 (1 − νP )R (TR − To ) R4 + 3 4 Es R
or a1 = −
Es βs (1 + νP )(TR − To ) . 4(1 − νP )
Using this in τrr , we have Es βs R r2 Es βs (1 + νP )(TR − To ) r(TR − To ) 2 dr − 2 4(1 − νP ) R R 0 Es βs (TR − To ) 1 + νP = − . 1+ 4 1 − νP
τrr (r = R) = −
(b) From Table C.16, we have βs = 2.25 × 10−5 1/K
Table C.16.
Using the numerical values, we have −3 × 108 (Pa) = −
TR
(1 + 0.25) 6.8 × 1010 (Pa) × 2.25 × 10−5 (1/K) × (TR − 80)(K) × 1+ 4 (1 − 0.25)
= −6.375 × 105 (TR − 80) = 550.6◦C.
COMMENT: We did not determine τθθ (r), but it can simply be done since a1 and a2 are known. Note that TR is independent of R, but depends on the temperature distribution.
307
Chapter 4
Radiation
PROBLEM 4.1.FUN GIVEN: A piece of polished iron, with a surface area Ar = 1 m2 , is heated to a temperature Ts = 1,100◦C. OBJECTIVE: (a) Determine the maximum amount of thermal radiation this surface can emit. (b) Determine the actual amount of thermal radiation this surface emits. Interpolate the total emissivity from the values listed in Table C.18. (c) What fraction of the radiation energy emitted is in the visible (λ between 0.39 and 0.77 µm), near infrared (λ between 0.77 and 25 µm), and far infrared (λ between 25 and 1,000 µm) ranges of the electromagnetic spectrum? SOLUTION: (a) The maximum amount of thermal radiation that a surface can emit is the blackbody emissive power Eb (Ts ), which is given by (4.6), i.e., Eb (Ts ) = σSB Ts4 . For Ts = (1,100 + 273.15)(K) = 1,373.15 K, we have Eb = 5.67 × 10−8 (W/m2 K4 ) × (1,373.15)4 (K4 ) = 201,584 W/m . 2
(b) For the polished iron, from Table C.18, we have r 0.41. Then the radiation emitted by the surface is given by (4.13), i.e., Qr, = r Ar Eb = 0.41 × 1(m2 ) × 210,584(W/m2 ) = 82,649 W. (c) The fraction of radiation energy emitted in the visible range of the electromagnetic spectrum is given by (4.8), i.e., Fλ1 T λ2 T = F0λ2 T − F0λ1 T . For Ts = 1,373.15 K, with interpolation from Table 4.1, we have λ1 T λ2 T
= =
0.39(µm) × 1,373.15(K) = 535.53 µmK, then F0λ1 T = 0 0.77(µm) × 1,373.15(K) = 1,057.33 µmK, then F0λ2 T = 0.0013.
Then, F0.39T 0.77T = 0.0013 − 0 = 0.0013 = 0.13%. For the fraction of radiation energy emitted in the near infrared range of the electromagnetic spectrum, we have λ1 T λ2 T
= =
0.77(µm) × 1,373.15(K) = 1,057.33 µmK, then F0λ1 T = 0.0013 25(µm) × 1,373.15(K) = 34,329 µmK, then F0λ2 T = 0.9968.
Then, F0.77T 25T = 0.9968 − 0.0013 = 0.9955 = 99.55%. For the fraction of radiation emitted in the far infrared range of the spectrum we have λ1 T
=
25(µm) × 1,373.15(K) = 34,329 µmK,
λ2 T
=
1,000(µm) × 1,373.15(K) = 1,373,000 µmK,
then
F0λ1 T = 0.9968 then
F0λ2 T = 1.
Then, F25T 1000T = 1 − 0.9968 = 0.0032 = 0.32%. COMMENT: The results show that practically 100% of the thermal radiation emitted by a surface at Ts = 1,110◦C is emitted in the near infrared range of the spectrum (0.77 µm ≤ λ ≤ 25 µm). 310
PROBLEM 4.2.FUN GIVEN: A blackbody radiation source at Ti = 500◦C is used to irradiate three diﬀerent surfaces, namely, (i) aluminum (commercial sheet), (ii) nickel oxide, and (iii) paper. The irradiating surfaces have an area Ar = 1 m2 and are assumed gray, diﬀuse, and opaque. Use the emissivities in Table C.18 [for (ii) extrapolate; for others use the available data]. OBJECTIVE: (a) Sketch the radiation heat transfer arriving and leaving the surface, showing the blackbody emitter and the irradiated surface [see Figure 4.9(b)]. Show heat transfer as irradiation Qr,i , absorption Qr,α , reﬂection Qr,ρ , emission Qr, , and radiosity Qr,o . (b) If the three surfaces are kept at Ts = 100◦C, determine the amounts of reﬂected and absorbed energies. (c) Determine the rates of energy emitted and the radiosity for each surface. (d) Determine the net radiation heat transfer rate for each surface. Which surface experiences the highest amount of radiation heating? SOLUTION: (a) Figure Pr.4.2 shows the radiation energy arriving and leaving the surface. From (4.14), the reﬂected radiation is given by Qr,ρ = Ar ρr qr,i , where qr,i is the irradiation ﬂux impinging on the surface. All this irradiation arrives from emission by a blackbody surface at Ti = 773 K. Then using (4.13), we have Qr,ρ = Ar ρr σSB Ti4 . The absorbed energy is given by (4.14), i.e., Qr,α = Ar αr σSB Ti4 . Since the surfaces are opaque (τr = 0), we have from (4.19) αr + ρr = 1. Assuming that the surfaces are gray, we have from (4.20) αr = r . Then the reﬂected and absorbed energy can be rewritten as Qr,ρ = Ar (1 − r )σSB Ti4 Qr,α = Ar r σSB Ti4 . (b) Surface 1: Aluminum, Commercial Sheet From Table C.18, for T = 373 K (it is the only data available), r = 0.09. Thus Qr,ρ
= 1(m2 ) × (1 − 0.09) × 5.67 × 10−8 (W/m2 K4 ) × (773.15)4 (K4 ) = 18,437 W
Qr,α
= 1(m2 ) × 0.09 × 5.67 × 10−8 (W/m2 K4 ) × (773.15)4 (K4 ) = 1,823 W.
Surface 2: Nickel Oxide From Table C.18, for T = 373.15 K (an extrapolation of the data available is possible), r = 0.345. Then Qr,ρ
= 1 × (1 − 0.345) × 5.67 × 10−8 × (773.15)4 = 13,220 W
Qr,α
= 1 × 0.345 × 5.67 × 10−8 × (773.15)4 = 6,990 W. 311
Radiosity Qr,o = Ar qr,o Ti Irradiation Qr,i = Ar qr,i
Reflection Qr,ρ = Ar ρr qr,i Emission Qr, = Ar r Eb(Ts) ∋
Absorption Qr,α = Ar αr qr,i
∋
Ts
Figure Pr.4.2 An opaque, diﬀuse surface at temperature Ts , irradiated by a blackbody surface at temperature Ti .
Surface 3: Paper From Table C.18, for T = 308 K (the data available is limited), r = 0.95. Then Qr,ρ = 1 × (1 − 0.95) × 5.67 × 10−8 × (773.15)4 = 1,012 W Qr,α = 1 × 0.95 × 5.67 × 10−8 × (773.15)4 = 19,247 W. (c) Surface 1: From (4.13), we have the surface emission as Qr,
= Ar r σSB Ts4 = 1(m2 ) × 0.09 × 5.67 × 10−8 (W/m2 K4 ) × (373.15)4 (K4 ) = 98.94 W.
From (4.22), we have the radiosity as Qr,o
= Qr,ρ + Qe, =
18,437(W) + 98.94(W) = 18,536 W.
Surface 2:
Qr, Qr,o
= =
1 × 0.345 × 5.67 × 10−8 × (373.15)4 = 379.3 W 13,270(W) + 379(W) = 13,649 W.
Surface 3:
Qr,
=
1 × 0.95 × 5.67 × 10−8 × (373.15)4 = 1,044 W
Qr,o
=
1,013(W) + 1,044(W) = 2,057 W.
(d) The net radiation heat transfer is given by (4.24), i.e., Qr Qr,i Surface 1:
Qr
=
= Qr,o − Qr,i = Ar qr,i = Ar σSB Ti4
18,536(W) − 1(m2 ) × 5.67 × 10−8 (W/m2 K4 ) × (773.15)4 (K4 ) = −1,724 W.
Surface 2:
Qr
=
13,649(W) − 20,260(W) = −6,611 W.
Surface 3:
Qr
=
2,057(W) − 20,260(W) = −18,203 W.
COMMENT: Examining the net radiation heat transfer to the three surfaces, we note that surface 3 (paper) is the most heated by radiation (net radiation heat transfer into this surface, which is negative, as the largest magnitude). The emissivity data, as a function of temperature is not available for most materials. Usually, it becomes necessary to estimate a value from limited data. This increases the uncertainty in the analysis of radiation heat transfer. For applications that need more accuracy, the measurement of the radiation properties, for the surfaces at the temperatures of interest, may be required. 312
PROBLEM 4.3.FUN GIVEN: Three diﬀerent surfaces are heated to a temperature Ts = 800◦C. The total radiation heat ﬂux leaving these surfaces (i.e., the radiosity) is measured with a calorimeter and the values (qr,o )1 = 6,760 W/m2 , (qr,o )2 = 21,800 W/m2 , and (qr,o )3 = 48,850 W/m2 are recorded. Assume that the reﬂected radiation is negligible compared to surface emission and that the surfaces are opaque, diﬀuse, and gray. OBJECTIVE: (a) Determine the total emissivity for each of these surfaces. (b) Comment on the importance of considering the surface reﬂection in the measurement of surface emissivity r . SOLUTION: The radiant heat ﬂux leaving a surface is the sum of the emitted and the reﬂected radiation, i.e., Ar qr,o = Ar qr, + Ar ρr qr,i . (a) Assuming that the reﬂection part is much smaller than the emitted part, and using the StefanBoltzmann law (4.6), we have qr,o = qr, = r σSB Ts4
for qr, qr,ρ = ρr qr,i .
This equation is then used to ﬁnd r . For each of the surfaces we have Surface 1: (qr,o )1 = 6,760 W/m2
r,1 =
(qr,o )1 6,760 = = 0.09, 2 −8 σSB Ts4 5.67 × 10 (W/m K4 ) × (1,073.15)4 (K4 )
Surface 2: (qr,o )2 = 21,800 W/m2
r,2 =
(qr,o )2 21,800 = = 0.29, 4 −8 σSB Ts 5.67 × 10 (W/m2 K4 ) × (1,073.15)4 (K4 )
Surface 3: (qr,o )2 = 48,850 W/m2
r,3 =
(qr,o )3 48,850 = = 0.65. −8 σSB Ts4 5.67 × 10 (W/m2 K4 ) × (1,073.15)4 (K4 )
(b) When qr,ρ cannot be neglected, then the irradiation ﬂux qr,i as well as the reﬂectivity (ρr = 1 − r , for opaque, gray surfaces) must be included. If the irradiation heat ﬂux qr,i or ρr are large, the above procedure for the determination of r does not lead to accurate results. COMMENT: Emissivities of the order of 0.1 are characteristic of commercial aluminum. The other two emissivities are found, for example, for refractory brick and opaque quartz, both of which are ceramics.
313
PROBLEM 4.4.FAM GIVEN: In an incandescent lamp, the electrical energy is converted to the Joule heating in the thinwire ﬁlament and this is in turn converted to thermal radiation emission. The ﬁlament is at T = 2,900 K, and behaves as an opaque, diﬀuse, and gray surface with a total emissivity r = 0.8. OBJECTIVE: Determine the fractions of the total radiant energy and the amount of emitted energy in the (a) visible, (b) near infrared, and (c) the remaining ranges of the electromagnetic spectrum. SOLUTION: The radiation energy emitted by the ﬁlament is given by (4.13), i.e., qr, = r σSB T 4 . The fraction of radiant energy emitted over a wavelength range between λ1 and λ2 , for a gray surface, is given by (4.8), i.e., (qr, )λ1 λ2 = Fλ1 T λ2 T r σSB T 4 . (a) For the visible range of the spectrum, λ1 = 0.39 µm and λ2 = 0.77 µm, then, Fλ1 T λ2 T = F0λ2 T − F0λ1 T . From Table 4.1, F0λ2 T
= F02233 = 0.10738
F0λ1 T
= F01131 = 0.002766.
Then (qr, )visible
=
(0.10738 − 0.002766) × 0.8 × 5.67 × 10−8 (W/m2 K4 ) × 2,9004 (K4 )
=
3.356 × 105 W/m2 .
(b) For the near infrared range of the spectrum, λ1 = 0.77 µm, and λ2 = 25 µm. Then from Table 4.1 F0λ2 T = F072,500 1.0. Therefore, (qr, )near infrared
=
(1 − 0.10738) × 0.8 × 5.67 × 10−8 (W/m2 K4 ) × 2,9004 (K4 )
=
2.864 × 106 W/m2 .
(c) For the remaining range of the spectrum, we have (qr, )remaining
=
0.002766 × 0.8 × 5.67 × 10−8 (W/m2 K4 ) × 2,9004 (K4 )
=
8,874 W/m2 .
COMMENT: Note that most of the radiant energy is emitted in the near infrared range of the spectrum. A relatively small fraction (10.46 percent) is in the visible portion of the spectrum, and nearly none in the ultraviolet and far infrared ranges.
314
PROBLEM 4.5.FUN GIVEN: The equation of radiative transfer describes the change in the radiation intensity Ir as it experiences local scattering, absorption, and emission by molecules or larger particles. In the absence of a signiﬁcant local emission, and by combining the eﬀects of scattering and absorption into a single volumetric radiation property, i.e., the extinction coeﬃcient σex (1/m), we can describe the ability of the medium to attenuate radiation transport across it. Under this condition, we can write the equation of radiative transport for a onedimensional volumetric radiation heat transfer as dIλ = −σex Iλ dx
radiative transport for nonemitting media
or by integrating this over all the wavelength and solid angles (as indicated in the second footnote of Section 4.1.2), we have arrive at the radiation heat ﬂux qr dqr = −σex qr . dx SKETCH: Figure Pr.4.5 shows the attenuation in a medium with a portion of the incoming radiation qr,i reﬂected on the surface (x = 0) and the remaining entering the medium. Prescribed Irradiation qr,i Reflected Radiation ρr qr,i Absorbed Radiation Se,σ = se,σ V
x qr(x) Local Radiation Intensity
Absorbing and Scattering Medium, σex (1/m)
Figure Pr.4.5 Radiation attenuating (absorbing and scattering) medium with a surface (x = 0) reﬂection.
OBJECTIVE: (a) Integrate this equation of radiative transfer, using qr (x = 0) = qr,i (1 − ρr ), where ρr is the surface reﬂectivity, as shown in Figure Pr.4.5, and show that qr (x = 0) = qr,i (1 − ρr )e−σex x . (b) Starting from (2.1), and assuming onedimensional, volumetric radiation heat transfer only, show that qr,i (1 − ρr )σex e−σex x = −s˙ e,σ = −
S˙ e,σ , V
which is also given by (2.43), when we note that the attenuation of radiation is represented by s˙ e,σ as a source of energy. SOLUTION: (a) We begin with the equation of radiative transfer dqr = −σex qr dx and integrate this once to obtain qr = a1 e−σex x . Then we use the condition at the surface, x = 0, and we have qr (x = 0) = qr,i (1 − ρr ). 315
or qr (x) = qr,i (1 − ρr )e−σex x . (b) Starting from (2.1), for volumetric radiation only and under steadystate heat transfer, we have ∇ · qr =
d qr (x) = s˙ e,σ . dx
using the solution for qr (x) we have d qr,i (1 − ρr )e−σex x = s˙ e,σ dx or qr,i (1 − ρi )
d −σex x e = s˙ e,σ dx
or −qr,i (1 − ρr )σex e−σex x = s˙ e,σ . We note that the attenuation of radiation, which shows a net heat transfer into the diﬀerential volume, is represented on the righthand side of the energy equation as positive s˙ e,σ (i.e., as a source). Therefore, we have used a negative sign in the deﬁnition for s˙ e,σ given in Table 2.1, and s˙ e,σ is positive. COMMENT: Attenuation of radiation heat ﬂux could have been included in the divergence of qr if we had given a more general description of qr . This requires the introduction of a general equation of radiative transfer which is left to the advance studies. Also note that σex = 1/λph , where λph is the phonon meanfree path. The large magnitude for σex (large attenuation and small λph ) gives rise to a signiﬁcant absorption of irradiation. Figure 2.13 gives examples of σex (1/m) for various absorbingscattering media.
316
PROBLEM 4.6.FUN GIVEN: ∗ = σex L, for a heat transfer medium of thickness L and When the optical thickness deﬁned by (2.44), σex extinction coeﬃcient σex is larger than 10, the emission and transfer of radiation can be given by the radiation heat ﬂux as (for a onedimensional heat ﬂow) qr,x = −
16 σSB T 3 dT 3 σex dx
∗ diﬀusion approximation for optically thick (σex > 10) heat transfer media.
This is called the diﬀusion approximation. The equation of radiation transfer, for an emitting medium with a strong absorption, becomes dIr,b = −σex Ir + σex Ir,b , d(x/ cos θ)
πIr,b = Er,b = σSB T 4 ,
where x/ cos θ is the photon path as it travels between surfaces located at x and x + ∆x, as shown in Figure Pr.4.6. The radiation heat ﬂux is found by the integration of Ir over a unit sphere, i.e.,
2π
π
qr =
sIr cos θ sin θdθdφ. 0
Note that this integral is over a complete sphere. Also note that Ir,b is independent of θ and φ. For the x 2π π direction, using Figure Pr.4.6, we have qr,x = 0 0 Ir cos θ sin θdθdφ. SKETCH: Figure Pr.4.6 shows the geometry considered and the angles used. f 2p, 0 q p
0 x + Dx 2
f
x
Ir, s q Local Dx 0 x/cosq Temperature s T(x) ex x  Dx 2 Absorbing/Emitting Medium with Large Optical Thickness * = s L >10 sex ex x
Figure Pr.4.6 Radiation intensity Ir traveling in an optically thick, emitting medium.
OBJECTIVE: Using the equation of radiative transfer and the deﬁnition of qr , both in the given, derive the given expression for qr,x for the diﬀusion approximation. SOLUTION: The equation of radiative transfer can be rearranged as Ir =
dIr,b cos θ + Ir,b . σex dx
Upon integration, we have for qr,x
2π
π
−
qr,x = 0
dIr cos θ + Ir,b cos θd sin θdθdφ. σex dx 317
Now, noting that Ir,b is independent of θ and φ (as discussed in Section 4.1.2) we have qr,x
2π π dIr,b cos2 θ sin θdθdφ σex dx 0 0 2π π cos θ sin θdθdφ + Ir,b
= −
4 dIr,b × π + Ir,b × 0 = − σex dx 3 4π dIr,b = − . 3σex dx Now, using πIr,b = Eb = σSB T14 , we have qr,x
= −
4σSB dT 4 16 σSB T 3 dT 4 dEb =− =− . 3σex dx 3σex dx 3 σex dx
COMMENT: In replacing dIr /dx with dIr,b /dx in the equation of radiative transfer, we are eliminating all the distant radiation heat transfer eﬀects by approximating the local intensity as Ir = Ir,b −
cos θ dIr,b . σex dx
Note that Er is the integral of Ir over a unit hemisphere as discussed in Section 4.1.2.
318
PROBLEM 4.7.FAM GIVEN: The human eye is sensitive to the visible range of the photon wavelength and has a threshold for detection of about Eb,λ = 0.0936 W/mµm at wavelength of λ = 0.77 µm (largest wavelength in the red band, Figure 4.1). This corresponds to the Draper point in Figure 4.2(a). OBJECTIVE: (a) The turtle eye is sensitive to the infrared range and if the threshold for detection is the same, but at λ = 1.5 µm, determine the corresponding temperature at which the turtle can detect blackbody emission. (b) Using this temperature, at what wavelength would Eb,λ peak? (c) Would this turtle eye be able to detect radiation emission from a tank containing liquid water at one atm pressure (Table C.3)? SOLUTION: (a) From (4.2), we have Eb,λ (T, λ)
= =
a1 = 3.742 × 108 Wµm4 /m2 ,
a1 a2 /λT
λ (e − 1) 0.0936 W/m2 µm, 5
a2 = 1.439 × 104 µmK,
ea2 /λT
=
a2 λT
=
T
=
λ = 1.5 µm.
a1 1+ Eb,λ λ5 a1 ln 1 + Eb,λ λ5 a2 . a1 λ ln 1 + Eb,λ λ5
Using the numerical values, we have T
=
= =
1.439 × 104 (µmK) 3.742 × 108 (Wµm4 /m2 ) 1.5(µm) ln 1 + 0.0936(W/m2 µm) × (1.5)5 (µm5 )
9.593 × 103 (K) 9.593 × 103 (K) = ln(1 + 5.265 × 108 ) 2.008 × 101 477.7 K.
(b) From Figure 4.2(a), we have Eb,λ having its maximum value at λmax T = 2,898 µmK. Then λmax
= =
2,898(µmK) 477.7(K) 6.067 µm.
(c) The boiling point is the highest temperature that the liquid will have. From Table C.3, we have for water at one atm pressure, Tlg = 373.2 K < 477.7 K Therefore, the emission from this tank is not detectable by the turtle eye. COMMENT: From Figure 4.2(a), note that for a given Eb,λ and T , there are two wavelength ranges, one is the short and one is the long wavelength range. Here we have selected the short wavelength. Also, it should be kept in mind that λ is multivalued when solving for λ as the unknown. 319
PROBLEM 4.8.FUN GIVEN: Dielectrics, e.g., ceramics such as SiC, have very small extinction index κ and also small refraction index n (optical properties). On the other hand metals have large κ and n. Figure Pr.4.8 shows the interface between two media, 1 and 2, which have diﬀerent optical properties. The normal (i.e., θ = 0) emissivity, for the dielectrics and metals, is predicted using these optical properties and a simple relation given by n2 n1 , r = 2 n2 + 1 + κ2 n1 4
where, as shown in Figure 4.6, 1 and 2 (or i and j) refer to the two media and here we use air as media 1 (with n1 = 1). The measured values of n and κ at λ = 5 µm are given as SiC :
n2 = 2.4,
Al : air :
n2 = 9, n1 = 1.
κ = 0.07, κ = 65,
λ = 5 µm
λ = 5 µm
SKETCH: Figure Pr.4.8 shows the interface between two media with diﬀerent optical properties. Air (Optical Properties: n1 = 1) ∋
Surface Emissivity,
Medium 1
r
Interface Medium 2
SiC or Al (Optical Properties: n2 , k)
Figure Pr.4.8 Interface between two media with diﬀerent optical properties.
OBJECTIVE: (a) Determine the total hemispherical emissivity r for SiC and Al in contact with air. (b) Compare these values with the measured values of r given in Table C.18. SOLUTION: (a) Using the above relation, we have
SiC: r
Al:
r
=
=
2.4 4× 1 = 0.8301 2 2.4 2 + 1 + (0.07) 1 9 4× 1 = 0.008324. 2 9 + 1 + (65)2 1
(b) From Table C.18, we have SiC: Al:
r r
= =
0.83 to 0.96 0.008324 to 0.18
(ﬁne to rough polish).
While the results of SiC are close to what is given in Table C.18, the predicted value for Al is for an ideally polished surface. 320
COMMENT: Although the general trend of high r for dielectrics and smaller r for the metals are predicted, for more accuracy the wavelength dependence of n and κ should be included and then integrated as indicated by (4.11). Also note that for n2 = 1 and κ2 = 0, we obtain r = αr = 1, from the relation given above. This indicates that the photons would continue to travel through the interface when this surface does not mark any change in the optical properties. As stated in Figure 4.6, n and κ are related to the three fundamental electromagnetic properties of matter.
321
PROBLEM 4.9.FAM GIVEN: Some surface materials and coatings are selected for their radiation properties, i.e., their emissivity r and absorptivity αr . Consider the following selections based on surface radiation properties. All surfaces are at Ts = 300 K. (i) Space suit (αr for low heat absorption). (ii) Solar collector surface (αr for high heat absorption and r for low heat emission). (iii) Surface of thermos (αr for low heat absorption). OBJECTIVE: (a) Choose the materials for the applications (i) to (iii) from Table C.19. (b) Determine the emissive power for the selected surfaces (i) to (iii). (c) Determine the surface reﬂectivity for the selected surfaces (i) to (iii), assuming no transmission (opaque surface). SOLUTION: (a) In Table C.19 we search for the closest match. (i) For low solar absorption, from Table C.19, we choose coatings such as white potassium zirconium silicate, αr = 0.13, r = 0.89. (ii) For the solar collector surface coating, from Table C.19, we choose black oxidized copper, r = 0.16, αr = 0.91, a highly nongray (selective) surface. (iii) For the surface of the thermos, we can choose from Table C.19, aluminum foil, r = 0.025, αr = 0.10. (b) The emissive power is given by (4.13) as Er,s = r,s σSB Ts4 . Then (i) white potassium zirconium silicate: Er,s
= 0.89 × 5.67 × 10−8 (W/m2 K4 ) × (300)4 (K4 ) = 408.8 W/m2
Er,s
= 0.16 × 5.67 × 10−8 (W/m2 K4 ) × (300)4 (K4 ) = 73.48 W/m2
Er,s
=
0.025 × 5.67 × 10−8 (W/m2 K4 ) × (300)4 (K4 )
=
11.48 W/m2 .
(ii) blackoxidized copper:
(iii) aluminum foil:
(c) From (4.15), we have for opaque surfaces (written for the total quantities) αr + ρr + τr αr + ρr
= 1 = 1,
for τr = 0
ρr = 1 − αr . Then (i) Write potassium zirconium silicate: ρr = 1 − 0.13 = 0.87 (ii) blackoxidized copper: ρr = 1 − 0.91 = 0.09 322
(opaque surface).
(iii) aluminum foil: ρr = 1 − 0.1 = 0.90. COMMENT: Note that some of these materials are highly nongray (i.e., αr and r are vastly diﬀerent).
323
PROBLEM 4.10.FAM GIVEN: The view factors between two surfaces making up part of an enclosure are given for some geometries in Table 4.2 and Figures 4.11(a) to (e). SKETCH: Figures Pr.4.10(a) to (e) show ﬁve surface pairs for which the view factors are sought.
(a) Inside Surface of Cylinder (Excluding Top Surface) to Top Surface: F12 R
(b) Inside, Side Surfaces of Rectangle to Itself: F11 a
2a
Surface 2
Surface 2 2a Surface 1 (4 sides)
4R Surface 1
Surface 3
(c) Vertical Side of Right Angle Wedge to Its Horizontal Side: F21
(d) Surface of Inner Cylinder to Top Opening of Annulus: F13 2R1
a
R1
Surface 3 (Top Opening)
Surface 2
a/2
Surface 1
4R1
Surface 1
Surface 2 2a Surface 4
(e) Surface of a Sphere Near a Coaxial Disk to Rest of Its Surroudings: F23 Surface 1 (Sphere)
Surrounding Surface 3 R1
2R1 R2 = R1/4 Surface 2 (Disk)
Figure Pr.4.10(a) to (e) View factors between surface pairs for ﬁve diﬀerent surface pairs in diﬀerent geometries.
OBJECTIVE: Determine the view factors (Fij , with i and j speciﬁed for each case on top of the ﬁgures) for the ﬁve surface pairs shown in Figures Pr.4.10(a) to (e). Note that, for the geometry shown in Figure Pr.4.10(a), the view factor can be found using only the summation and the reciprocity rules (4.33) and (4.34), and by using simple inspection (i.e., no tables or ﬁgures are needed) of the limiting view factor (i.e., surfaces that are completely enclosed by another surface). SOLUTION: (a) To determine F11 , we begin by noting that F21 (from the disk to the rest of the cylinder) is equal to unity i.e., F21 = 1. Now we return to surface 1 and use the summation rule (4.33), i.e., F 1 1 + F 1 2 = 1
summation rule
or F 1 1 = 1 − F 1 2 . 324
To determine F12 , we use the reciprocity rule (4.34), i.e., Ar,1 F12 = Ar,2 F21
reciprocity rule
or F 1 2
Ar,2 F 2 1 Ar,1
=
πR2 F 2 1 2πR(4R) + πR2 1 F 2 1 8+1 1 1 ×1= . 9 9
= = = Note that
8 1 = . 9 9
F 1 1 = 1 − Note that the lower disk area is part of surface 1.
(b) To determine F11 , we begin by writing the summation rule for surface 1, i.e., F 1 1 + F 1 2 + F 1 3 = 1
summation rule
or F11 = 1 − 2F12 , where we have used the symmetry to write F12 = F13 . To determine F12 , we use the reciprocity rule F 1 2 =
Ar,2 F 2 1 , Ar,1
reciprocity rule.
To determine F21 , we use the summation rule for surface 2, i.e., F21 + F22 + F23 = 1 summation rule or F 2 1 = 1 − F 2 3 , where F22 = 0, because it is planar. We determine F23 from Figure 4.11(b) for w l F2 3
2a a a 1 = 1, = = 2a l 2a 2 0.12 Figure 4.11(b). =
Then F1 1
=
1 − 2F12 = 1 − 2
=
1−2
Ar,2 Ar,2 F 2 1 = 1 − 2 (1 − F23 ) Ar,1 Ar,1
2a2 1 (1 − 0.12) = 1 − (1 − 0.12) = 0.7067. 3 2(4a ) + 2(2a2 ) 2
(c) To determine F12 , we use Figure 4.11(c) and the reciprocity rule, i.e., Ar,1 F 2 1 = F 1 2 Ar,2 2a × a F12 = 4F12 = 1 a×a 2 1 a 2a l 1 w = = 2, = 2 = a a a a 2 F12 0.08 Figure 4.11(c). 325
Then F21 = 4 × 0.08 = 0.32. (d) To determine F21 , we begin with the summation rule for surface 1, i.e., F 1 1 + F 1 2 + F 1 3 + F 1 4
=
1
1 − F 1 2 1 (1 − F12 ), F 1 3 = 2 where we have used the symmetry condition for surfaces 1 and 3, and observed that F11 = 0 (because it is a convex surface) to determine F12 . We now use the reciprocity rule and Figure 4.11(e) i.e., 2F13
F 1 2 R1 R2 F 2 1
=
Ar,2 F 2 1 Ar,1 l 1 4R1 , = = =2 2 R2 2R1 0.415 Figure 4.11(e). =
Then
1 1 Ar,2 (1 − F12 ) = F 2 1 1− 2 2 Ar,1 1 2π × 2R1 × 4R1 = × 0.415 1− 2 2π × R1 × 4R1 1 = (1 − 0.830) = 0.085. 2 (e) To determine F13 , we begin with the summation rule for surface 2, i.e., F 1 3
=
F 2 1 + F 2 2 + F 2 3 = 1 or F 2 3 = 1 − F 2 1 , where F22 = 0 because it is planar. To determine F21 , we use the reciprocity rule and the results of Table 4.2, i.e., F 2 1
=
R2 l
=
F 1 2
=
=
Ar,1 F 1 2 Ar,2 1 R1 1 4 = 2R1 8 1 1 1− 2 1/2 2 R2 1 + l 1 1 1 = (1 − 0.9923) = 0.003861. 1− 2 1/2 2 2 [1 + 0.125 ]
Then F 2 3
=
1 − F 2 1 = 1 −
=
1−
Ar,1 F 1 2 Ar,2
4πR12 2 F12 = 1 − 64F12 = 0.7529. 1 R1 π 4 326
COMMENT: In general, to determine a view factor, ﬁrst inspect and ﬁnd an available view factor and then work toward the unknown using the summation and reciprocity rules.
327
PROBLEM 4.11.FUN GIVEN: Two planar surfaces having the same area A = A1 = A2 are to have three diﬀerent geometries/arrangements, while having nearly the same view factor F12 . These are coaxial circular disks, coaxial square plates, and perpendicular square plates, and are shown in Figure Pr.4.11. SKETCH: Figure Pr.4.11 shows the three geometries/arrangements.
Configurations of Surfaces A1 and A2 (where A1 = A2) Parallel (Coaxial Circular Disks) A1
Parallel (Coaxial Square Plates) A1 A2
R1 l
Perpendicular (Square Plates) A1
w=a A2
l R2
a
w=a
A2
w=a
a
Figure Pr.4.11 Two planar surfaces having the same area and three diﬀerent geometries/arrangements.
OBJECTIVE: (a) Determine this nearly equal view factor F12 (shared among the three geometries). (b) Under this requirement, are the disks or the plates placed closer together? SOLUTION: (a) The view factors for parallel disks, parallel plates, and perpendicular plates, are given in Figures 4.11(a), (b), and (c). We note that for the perpendicular arrangement, F12 is determined once the plate geometry is known. So, we begin in Figure 4.11(c) and note that for l = a = w, we have l∗ = 1,
w∗ = 1.
Then from Figure 4.11(c), we have F12 0.2
Figure 4.11(c).
Now, this view factor is found in Figure 4.11(b), i.e., for w∗ =
F12 = 0.2,
w a = 1, a∗ = = 1 Figure 4.11(b). l l
Nearly the same view factor is found in Figure 4.11(a), i.e., F12 0.2
for
1 l l = = 1.70 = , R1∗ R1 R2
or
R2∗ =
R2 = 0.588 l
(b) Since the areas are all the same, we have A
= πR2 disk = a2 square plate
or R=
a π
1/2
.
Then we have l
=
1.70R = 1.70
l
= a plates.
a π
1/2
328
= 0.9591a
disks
Figure 4.11(a).
Then the disks are placed slightly closer together, but under the approximations made reading from the graphs, this is negligible. COMMENT: Note that while the parallel arrangements allow for achieving higher F12 by reducing l, the perpendicular arrangement results in a constant F12 , once the plate geometries are ﬁxed.
329
PROBLEM 4.12 FUN GIVEN: The blackbody surface can be simulated using a large cavity (i.e., an enclosure with a small opening). The internal surfaces of the cavity have a total emissivity r,1 which is smaller than unity; however, due to the large cavity surface area, compared to its opening (i.e., mouth), the opening appears as a blackbody surface. To show this, consider the cylindrical enclosure shown in Figure Pr.4.12(a). The surrounding is assumed to be a blackbody at T∞ . SKETCH: Figure Pr.4.12(a) shows the use of apparent emissivity for construction of a blackbody emitter using a deep graybody cavity.
(i) Radiation Exchange with Cavity Surface
(ii) Apparent Radiation Exchange with Cavity Mouth
= 1
Surrounding ∋
Surrounding ∋
r,
T
Qr,1 Cavity
T
Ar, >> Ar,1' F1' = 1
Ar,1' = πD2/4
Cavity Mouth
∋
∋
r,1
T1' = T1
D
L
T1
Ar, >> Ar,1' F1' = 1
Qr,1' = Qr,1
Cavity Mouth Ar,1'
D
= 1
r,
Ar,1 = πDL + πD2/4
r,1
Figure Pr.4.12(a)(i) Surface radiation from a cavity. (ii) The concept of apparent emissivity for the cavity opening.
OBJECTIVE: (a) Equate the net radiation heat transfer Qr,1∞ from the cavity surface in Figure Pr.4.12(a)(i) to that in Figure Pr.4.12(a)(ii). In Figure Pr.4.12(a)(ii), we use the cavity opening area and an apparent emissivity r,1 . Then derive an expression for this apparent emissivity. (b) Show that this apparent emissivity tends to unity for L D. SOLUTION: (a) The thermal circuit diagrams for both cases are shown in Figure Pr.4.12(b). The cavity surface and the surrounding of the opening are treated as blackbody surfaces. The surface radiation heat ﬂow for the twosurface enclosure of Figure Pr.4.12(a)(i) is given by (4.48), i.e., Qr,1∞ = Qr,11
=
Eb,1 (T1 ) − Eb,∞ (T∞ ) . 1 − r,1 1 + +0 Ar r,1 1 Ar,1 F11
Using the reciprocity rule, (4.34), Qr,1∞
=
Eb,1 − Eb,∞ (T∞ ) . 1 − r,1 1 + Ar r,1 1 Ar,1 F1 1 330
Thermal Circuit Model (i)
(ii)
(qr,o) = Eb, (Rr,F)1 =
Qr,1
(Rr,F)1' =
1 Ar,1 F1 Qr,1' = Qr,1
(qr,o)1
(qr,o) = Eb, 1 Ar,1' F1'
(qr,o)1'
(Rr, )1
(Rr, )1'
Eb,1
Eb,1' = Eb,1
T1
T1' = T1
Figure Pr.4.12(b) Thermal circuit diagrams for the twocases.
For surface 1 , from (4.49) and Figure 4.12(b)(ii), we also have Qr,1 ∞
Eb,1 (T1 ) − Eb,∞ (T∞ ) Eb,1 (T1 ) − Eb,∞ (T∞ ) = , 1 1 − r 1 + Ar,1 r,1 Ar r 1 Ar,1 F1 ,∞ ≡ Qr,1∞ . =
for F1 ∞ = 1
with Ar,∞ Ar,1
Solving for r,1 , from these two equations, we have 1 Ar,1 r,1
=
r,1
=
=
1 1 − r,1 1 + Ar,1 r,1 Ar,1 F1 1 1 Ar,1 1 − r,1 Ar,1 1 + Ar,1 r,1 Ar,1 F1 1 1 since F1 1 = 1. Ar,1 1 − r,1 +1 Ar,1 r,1
Now, we use the diameter and length of the cylindrical cavity to have Ar,1 Ar,1
= πD2 /4 = πDL + πD2 /4.
Then, the apparent emissivity of the cavity is r,1 =
1 . D 1 − r,1 +1 4L + D r,1
(b) For large L/D, i.e., a deep cavity, we have, lim
L/D→∞
r,1 =
lim
L/D→∞
1 1 = 1, = 1 − r,1 1 0+1 +1 4L/D + 1 r,1
i.e., to the surrounding, the cavity appears as a blackbody surface. COMMENT: For a given r,1 , the ratio L/D needed to make r,1 near unity, is determined from the above equation. The smaller r,1 , the larger L/D needs to be for creating an apparent blackbody surface. 331
PROBLEM 4.13.FAM GIVEN: Liquid oxygen and hydrogen are used as fuel in space travel. The liquid is stored in a cryogenic tank, which behaves thermally like a thermos. Radiation shields (highly reﬂecting aluminum or gold foils) are placed over the tank to reduce irradiation to the tank surface [Figure Pr.4.13(a)]. The surface of the tank has a total emissivity of r,1 = 0.7 and the shields have an emissivity of r,s = 0.05. Consider placing one [Figure Pr.4.13(a)(i)] and two [Figure Pr.4.13(a)(ii)] radiation shields on the tank. Assume that the surface of the tank is T1 = 80◦C above the saturation temperature of the liquid at one atm pressure, the tank is facing away from the sun, and the deep sky temperature is T2 = 3 K. SKETCH: Figure Pr.4.13(a) shows the idealized tank surface and the radiation shields. (i) One Shield
(ii) Two Shields
Radiation Shield r,s = 0.05
Radiation Shield r,s = 0.05
∋
∋
Liquid H2 or O2
Liquid H2 or O2 Deep Sky T2 = 3 K
Deep Sky T2 = 3 K
Tank Surface r,1 = 0.7 T1 = Tlg + 80 C
Tank Surface r,1 = 0.7 T1 = Tlg + 80 C
∋
∋
Figure Pr.4.13(a) Surfaceradiation heat transfer from a cryogenic liquid tank. (i) With one shield. (ii) With two shields.
OBJECTIVE: (a) Draw the thermal circuit diagrams. (b) Determine the rate of heat ﬂowing out of the tank per unit area for liquid oxygen and liquid hydrogen. SOLUTION: (a) The thermal circuit for one radiation shield is shown in Figure Pr.4.13(b) and for two radiation shields, in Figure Pr.4.13(c). Qr,1s (qr,0)1 (Rr,F)1s
Qr,s2
Eb,s Ts Eb,s
(qr,0)s (Rr, )s
Surface 1
Qr,2
(qr,0)s (Rr, )s
(Rr, )1
Q1
Qr,s
Eb,2 T2
(qr,0)2 (Rr,F)s2
T1 Eb,1
Qr,s
Shield
Q2
(Rr, )2
Qr,1
Surface 2
Figure Pr.4.13(b) Thermal circuit diagram for one shield.
(Rr, )s
(Rr, )s
(Rr,F)ss
Qr,s2
Qr,s Eb,s Ts Eb,s
(qr,0)s
(qr,0)s
Surface 1
(Rr,F)1s
Qr,s
Qr,ss
Qr,s Eb,s Ts Eb,s
(Rr, )s
Shield
Shield
Figure Pr.4.13(c) Thermal circuit diagram for two shields.
332
Qr,2 Eb,2 T2
(qr,0)2
(qr,0)s (Rr, )s
(Rr, )1
Qr,s (qr,0)s
Q1
Qr,1s (qr,0)1
(Rr,F)s2
(Rr, )2
Qr,1 T1 Eb,1
Surface 2
Q2
(b) (i) For one radiation shield, the temperatures T1 and T2 are given (i.e., Eb,1 and Eb,2 are known). From (4.50), the heat transfer rate between surfaces 1 and 2 can then be expressed as a function of the potential diﬀerence Eb,1 − Eb,2 and the overall resistance (Rr,Σ )12 , i.e., Qr,12 =
Eb,1 − Eb,2 . (Rr,Σ )12
The overall resistance, for the radiation resistances arranged in series, is (Rr,Σ )12
=
j
=
1 − r,1 1 1 − r,s 1 − r,s 1 1 − r,2 + + + + + Ar,1 r,1 Ar,1 F1s Ar,s r,s Ar,s r,s Ar,s Fs2 Ar,2 r,2 1 1 − r,s 1 1 − r,2 + + 2 + . + Ar,1 F1s Ar,s r,s Ar,s Fs2 Ar,2 r,2
Rr,j =
1 − r,1 Ar,1 r,1
The view factors for inﬁnite, parallel plates are unity, F1s = Fs2 = 1, and all the surface areas are equal, Ar,1 = Ar,s = Ar . The total emissivities for the surfaces are r,1 = 0.7, r,s = 0.05, and r,2 = 1 (note that the deep sky behaves as a black body). Thus, the equivalent resistance becomes (Rr,Σ )12 =
1 − 0.7 1 1 − 0.05 1 1.43 + 39 40.43 + +2 + +0= = . 0.7Ar Ar 0.05Ar Ar Ar Ar
Oxygen: From Table C.4, we have Tlg = 90 K. Then, the tank surface temperature becomes T1 = 90(K)+80(K) = 170 K. The deep sky temperature is T2 = 3 K. Therefore, heat transfer rate is, Qr,12
= = =
σSB (T14 − T24 ) (Rr,Σ )12
2 5.67 × 10−8 (W/m K4 ) × 1704 (K4 ) − 34 (K4 ) 40.43 Ar 1.17(W/m2 ) × Ar (m2 )
or qr,12 =
Qr,12 2 = 1.17 W/m . Ar
Hydrogen: From Table C.4, we have Tlg = 20.4 K. Then, the tank surface temperature becomes T1 = 20.4(K) + 80(K) = 100.4 K. Therefore, the heat transfer rate is Qr,12
= = =
σSB (T14 − T24 ) (Rr,Σ )12
2 5.67 × 10−8 (W/m K4 ) × 100.44 (K4 ) − 34 (K4 ) 40.43 Ar 0.14(W/m2 ) × Ar (m2 )
or qr,12 =
Qr,12 2 = 0.14 W/m . Ar
(ii) For two radiation shields, the thermal circuit is shown in Figure Pr.4.13(c). For the overall thermal resistance, an equation similar to the one above is obtained. For two radiation shields, with the same surface radiation properties (same r,s ), the term within brackets is multiplied by two. Therefore, we have
1 − r,1 1 1 − r,s 1 1 − r,2 (Rr,Σ )12 = Rr,j = + +2 2 + . + A A F A A F A r,1 r,1 r,1 1 s r,s r,s r,s s 2 r,2 r,2 j 333
From the data available (Rr,Σ )12 =
1 − 0.7 1 1 − 0.05 1 79.43 1.43 + 78 + +2 2 + = . +0= 0.7Ar Ar 0.05Ar Ar Ar Ar
Oxygen: Qr,12
= = =
σSB (T14 − T24 ) (Rr,Σ )12
2 5.67 × 10−8 (W/m K4 ) × 1704 (K4 ) − 34 (K4 ) 79.43 Ar 2 0.5960(W/m ) × Ar (m2 )
or qr,12 =
Qr,12 2 = 0.5960 W/m . Ar
Hydrogen: Qr,12
= = =
σSB (T14 − T24 ) (Rr,Σ )12
2 5.67 × 10−8 (W/m K4 ) × 100.44 (K4 ) − 34 (K4 ) 79.43 Ar 2 2 0.073(W/m ) × Ar (m )
or qr,12 =
Qr,12 2 = 0.073 W/m . Ar
COMMENT: Note that the thermal resistance due to the radiation shield is 39/1.43 = 27 times larger than the resistance due to the surface grayness of the tank alone. If the tank faces the sun, there would be absorption of solar irradiation at the tank surface and this heat would ﬂow into the tank.
334
PROBLEM 4.14.DES GIVEN: A singlejunction thermocouple psychrometer is used to measure the relative humidity in air streams ﬂowing through ducts, as shown in Figure Pr.4.14. In the simplest design, the thermocouple psychrometer consists of a thermocouple bead, which is exposed to the humid air stream, connected simultaneously to a DC power source and a voltmeter. Initially, a voltage is applied to the thermocouple, causing a decrease in the bead temperature, due to the energy conversion from electromagnetic to thermal energy by the Peltier eﬀect. This cooling causes the condensation of the water vapor and the formation of a liquid droplet on the thermocouple bead. When the temperature drops to temperature Tt,o , the power source is turned oﬀ and the voltage generated by the thermocouple is recorded. Since the droplet temperature is lower than the ambient temperature, the droplet receives heat from the ambient by surface convection and surface radiation. This causes the evaporation of the droplet. The voltage measured between the thermocouple leads is related to the temperature of the thermocouple bead/water droplet. An equilibrium condition is reached when the net heat ﬂow at the droplet surface balances with the energy conversion due to phase change. This equilibrium temperature is called the wetbulb temperature for the air stream Twb . Figure Pr.4.14 shows the thermocouple placed in the air stream. The duct diameter is much larger than the thermocouple bead. The water droplet has a diameter Dd = 0.5 mm and its surface is assumed to be a blackbody. The tube surface is opaque, diﬀuse, and gray and has a surface emissivity r,2 = 0.5 and a temperature T2 = 300 K. The evaporation rate of the water is estimated as m ˙ lg = 0.00017 kg/m2 s. SKETCH: Figure Pr.4.14 shows the thermocouple psychrometer with a screen radiation shield. ∋
T2 = 300 K r,2 = 0.5
Thermocouple Psychrometer T1 = 297 K, r,1 = 1
∋
Humid Air Flow
Ds Water Droplet Dd Thermocouple Wires Voltmeter ∆ϕ (V)
Wire Cage (Screen) r,3 = 0.1 ∋
DC Power Source
Figure Pr.4.14 Thermocouple psychrometer with a screen radiation shield.
OBJECTIVE: (a) If the droplet temperature is T1 = 290 K, determine the net heat transfer by surface radiation between the bead and the tube surface and express it as a percentage of the energy conversion due to liquidvapor phase change. (b) If the bead is protected by a porous spherical wire cage with diameter Ds = 3 mm and the ratio between the open area and total area a1 = Avoid /Atotal = 0.7, calculate the reduction in the net heat transfer by surface radiation ∆Qr,1 . The surface of the wires is opaque, diﬀuse, and gray, and has an emissivity r,3 = 0.1. Using the available results for radiation between two surfaces separated by a screen, and for Ar,2 Ar,3 > Ar,1 , the overall radiation resistance is given by (Rr,Σ )12 =
1 − r,1 + r,1 Ar,1
1 a1 Ar,1 +
1 1 +2 Ar,1 (1 − a1 )
335
1 − r Ar r
.
3
(c) In order to reduce the amount of heat transfer between the droplet and the tube surface by surface radiation, should we increase or decrease a1 = Avoid /Atotal and r,3 ? SOLUTION: (a) If the presence of the thermocouple wire is neglected, the droplet and the tube form a twosurface enclosure. The net radiation heat transfer is then given by (4.47) as Qr,1 =
1 − r,1 r,1 Ar,1
σSB (T24 − T14 ) . 1 1 − r,2 + + Ar,1 F12 Ar,2 r,2
The view factor between the droplet and the tube, for a very long tube, is F12 = 1. Also, assuming that Ar,2 Ar,1 , the net radiation heat transfer becomes Qr,1 =
σSB (T24 − T14 ) . 1 r,1 Ar,1
Using the numerical values Qr,1
=
=
5.67 × 10−8 (W/m2 K4 ) × (3004 − 2974 )(K)4 1 1 × π × (0.5 × 10−3 )2 (m)2 1.42 × 10−5 W.
The energy conversion due to phase change is given by S˙ lg = −M˙ lg ∆hlg = −m ˙ lg ∆hlg A1 = −m ˙ lg ∆hlg πDd2 . From Table C.4, ∆hlg = 2,256 kJ/kg, and then, S˙ lg
= −0.00017(kg/m2 s) × 2,256 × 103 (J/kg) × π(0.5 × 10−3 )2 (m) = −3.01 × 10−4 W.
2
The ratio of the radiation heat transfer and the energy conversion due to liquidgas phase change is −Qr,1 −1.42 × 10−5 (W) = = 0.047. −3.01 × 10−4 (W) S˙ lg Therefore, for these conditions, the radiation heat transfer is equal to 4.7% of the heat used for evaporation of the liquid. (b) Using the available result for screens, the net radiation heat transfer between two surfaces separated by a screen with void fraction a1 = Avoid /(Avoid + Asolid ) is, Qr,1 =
σSB (T24 − T14 ) , (Rr,Σ )12
where (Rr,Σ )12 =
1 − r,1 + r,1 Ar,1
1 1 − r,2 + . 1 1 A r,2 r,2 + 1 1 − r,3 1 1 +2 + Ar,1 F12 Ar,1 F13 Ar,3 r,3 Ar,2 F23
The view factors are F12 = a1 and F23 = F13 = (1 − a1 ), and assuming that Ar,2 Ar,3 > Ar1 , we have (Rr,Σ )12 =
1 − r,1 + r,1 Ar,1
1 a1 Ar,1 +
1 2(1 − r,3 ) 1 + Ar,1 (1 − a1 ) Ar,3 r,3
336
.
Using the numerical values, we have (Rr,Σ )12
=
1−1 π × (0.5 × 10−3 )2 (m) × 1 2
1
+
−3 2
0.7 × π × (0.5 × 10
=
2
) (m) +
1 1 2(1 − 0.1) + −3 2 π × (0.5 × 10 ) × 0.3 π × (3 × 10−3 )(m)2 × 0.1
1.325 × 106 1/m2 .
Then, the net heat transfer by surface radiation becomes Qr,1
=
5.67 × 10−8 (W/m2 K4 ) × (3004 − 2974 )(K)4 = 1.37 × 10−5 W. 1.325 × 106 (1/m2 )
The ratio to the energy conversion is −1.37 × 10−5 (W) −Qr,1 = = 0.045. −3.01 × 10−4 (W) S˙ lg The reduction of thermal radiation due to shielding by the screen is only ∆Qr,1 = 4.5%. (c) Increasing the radiation thermal resistance (Rr,Σ )12 causes a decrease in the net heat transfer by radiation. From the expression for (Rr,Σ )12 , to decrease the thermal radiation we should decrease a1 and decrease r,3 (i.e., we should use a polished, metal shield). COMMENT: To allow for the surfaceconvection evaporation of the droplet, it is necessary to use a screen with a large a1 . The reduction of the eﬀect of the surface radiation is desirable.
337
PROBLEM 4.15.DES GIVEN: The polymer coating of an electrical wire is cured using infrared irradiation. The wire is drawn through a circular ceramic oven as shown in Figure Pr.4.15(a). The polymer coating is thin and the drawing speed uw is suﬃciently fast. Under these conditions, the wire remains at a constant and uniform temperature of T1 = 400 K, while moving through the oven. The diameter of the wire is d = 5 mm and its surface is assumed opaque, diﬀuse, and gray with an emissivity r,1 = 0.9. The oven wall is made of aluminum oxide (Table C.18), has a diameter D = 20 cm, and length L = 1 m, and its surface temperature is T2 = 600 K. One of the ends of the furnace is closed by a ceramic plate with a surface temperature T3 = 600 K and a surface emissivity r,3 = 0.5. The other end is open to the ambient, which behaves as a blackbody surface with T4 = 300 K. Ignore the heat transfer by surface convection. SKETCH: Figure Pr.4.15(a) shows the wire and its surface radiation surroundings.
Wire r,1 =
0.9
d
uw (Speed of Wire)
Furnace Wall (Alumina) r,2 , T2 = 600 K ∋
End Plate T3 = 600 K r,1 = 0.9
∋
T1 = 400 K,
D
Ambient: T4 = 300 K r,4 = 1 ∋
L
∋
Figure Pr.4.15(a) A wire drawn through an oven.
OBJECTIVE: (a) Draw the thermal circuit diagram for the foursurface radiation enclosure and write all of the relations for determination of the net heat transfer by radiation to the wire surface Qr,1 . (b) Assuming that the wire exchanges heat by radiation with the tube furnace surface only (i.e., a twosurface enclosure), calculate the net heat transfer by surface radiation to the wire surface Qr,1 . (c) Explain under what conditions the assumption made on item (b) can be used. Does the net heat transfer by surface radiation at the wire surface increase or decrease with an increase in the furnace diameter D (all the other conditions remaining the same)? Explain your answer. SOLUTION: (a) The thermal circuit diagram for the foursurface enclosure is shown in ﬁgure Pr.4.15(b). The energy equations are given below. Surface 1: Eb,1 − (qr,o )1 . −Q1 = Qr,1 = Qr,12 + Qr,13 + Qr,14 , Qr,1 = 1− r,1
r,1 Ar1 Surface 2: −Q2
= Qr,2 = Qr,21 + Qr,23 + Qr,24 ,
Qr,2 =
Eb,2 − (qr,o )2 . 1 − r,2 r,2 Ar2
−Q3
= Qr,3 = Qr,31 + Qr,32 + Qr,34 ,
Qr,3 =
Eb,3 − (qr,o )3 . 1 − r,3 r,3 Ar3
−Q4
= Qr,4 = Qr,41 + Qr,42 + Qr,43 ,
Qr,4 =
Eb,4 − (qr,0 )4 . 1 − r,4 r,4 Ar4
Surface 3:
Surface 4:
338
Q3
T3
T4
Eb,3
Eb,4
(Rr, )3
Qr,34
(qr,o)3
(qr,o)4
Qr,23
Qr,14
(Rr,F)13
(Rr, )1
Qr,24
(Rr,F)24
(Rr,F)21
(Rr, )2
Qr,13
Qr,4
(Rr, )4
Qr,3
Q4
Q1
Q2 T1 Eb,1
(qr,o)1
Eb,2 T2
(qr,o)2
Qr,1
Qr,21
Qr,2
Figure Pr.4.15(b) Foursurface thermal circuit diagram.
The surfaceradiation heat transfer rates are Qr,12
=
Qr,13
=
Qr,14
=
Qr,23
=
Qr,24
=
Qr,34
=
(qr,0 )1 − (qr,o )2 , 1 F12 Ar,1 (qr,0 )1 − (qr,o )3 , 1 F13 Ar,1 (qr,0 )1 − (qr,o )4 , 1 F14 Ar,1 (qr,0 )2 − (qr,o )3 , 1 F23 Ar,2 (qr,0 )2 − (qr,o )4 , 1 F24 Ar,2 (qr,0 )3 − (qr,o )4 , 1 F34 Ar,3
Qr,12 = −Qr,21
Qr,13 = −Qr,31
Qr,14 = −Qr,41
Qr,23 = −Qr,32
Qr,24 = −Qr,42
Qr,34 = −Qr,43
The view factors are F 1 1 + F 1 2 + F 1 3 + F 1 4
= 1
F 2 1 + F 2 2 + F 2 3 + F 2 4 F 3 1 + F 3 2 + F 3 3 + F 3 4
= 1 = 1
F 4 1 + F 4 2 + F 4 3 + F 4 4
= 1
Ar,1 F12 = Ar,2 F21 , Ar,1 F14 = Ar,4 F41 ,
Ar,1 F13 = Ar,3 F31 Ar,2 F23 = Ar,3 F32
Ar,2 F24 = Ar,4 F42 ,
Ar,3 F34 = Ar,4 F43 .
There are 16 view factors and 10 view factor equations. Therefore, 6 view factors need to be determined independently. Considering that F11 = F33 = F44 =0, we need to determine 3 view factors from graphs or equations. The emissive powers are Eb,1 = σSB T14 ,
Eb,2 = σSB T24 ,
Eb,3 = σSB T34 , 339
Eb,4 = σSB T44 .
The variables are T1 , T2 , T3 , T4 , Q1 , Q2 , Q3 , Q4 , Qr,1 , Qr,2 , Qr,3 , Qr,4 , Qr,12 , Qr,13 , Qr,14 , Qr,21 , Qr,23 , Qr,24 , Qr,31 , Qr,32 , Qr,33 , Qr,41 , Qr,42 , Qr,43 , Eb,1 , Eb,2 , Eb,3 , Eb,4 , (qr,o )1 , (qr,o )2 , (qr,o )3 , (qr,o )4 . Therefore, there are 32 unknowns and 28 equations. Four unknowns must then be speciﬁed. For this problem, the 4 temperatures are known. (b) The simpliﬁed formulation assumes a twosurface enclosure formed by the oven and the wire. For this situation, the net heat transfer to the wire surface is −Qr,1 = Qr,21 =
1 − r,1 Ar,1 r,1
σSB (T24 − T14 ) . 1 1 − r,2 + + Ar,1 F1,2 Ar,2 r,2
For this two surface enclosure, F12 = 1. From Table C.18 for alumina at T2 = 600 K, we have r,2 = 0.58. Then, using the values given, −Qr,1
=
=
5.67 × 10−8 (W/m2 K4 ) × (6004 − 4004 )(K4 ) 1 − 0.9 1 1 − 0.58 + + −3 −3 π × (0.2)(m) × 1(m) × 0.58 π × (5 × 10 )(m) × 1(m) × 0.9 π × (5 × 10 )(m) × 1(m) 5.67 × 10−8 (W/m2 K4 )(6004 − 4004 )(K4 ) = 82 W. 7.07 + 63.7 + 1.15
(c) The resistance (Rr, )2 is already small, so a further increase in D2 will increase Qr,1 only by a small amount. COMMENT: The assumption made in item (b) is acceptable when the view factor from the wire to the furnace, F12 , is approximately one. In this case, there is a negligible radiation heat transfer between the wire surface and surfaces 3 and 4. The view factor F12 can be obtained from the relation in Figure 4.11(e) (F12 = F21 Ar,2 /Ar,1 ). Figure Pr.4.15(c) shows the variation of the view factor F12 , with respect to the ratio of radius of the wire R1 to the oven radius R2 , keeping the other dimensions constant. We observe that the view factor is always larger than 0.9 and approaches 1 as R1 approaches R2 . Therefore, with an increase in the furnace diameter, the heat transfer to the wire decreases. 1.0
L=1m
0.8
F12 F13
F12 , F13
0.6
0.4
0.2
0.0 0.0
0.2
0.4
0.6
0.8
1.0
R1 /R2
Figure Pr.4.15(c) Variation of view factor F12 with respect to R1 /R2 .
340
PROBLEM 4.16.FUN GIVEN: As in the application of radiation shields discussed in Section 4.4.5, there are applications where the surface radiation through multiple (thin, opaque solid) layers (or solid slabs) is of interest. This is rendered in Figure Pr.4.16. Since for large N (number of layers) the local radiation heat transfer becomes independent of the presence of the far away layers, we can then use the local (or diﬀusion approximation) approximation of radiation heat transfer and use the temperature diﬀerence (or local temperature gradient) between adjacent layers and write qr,x ≡ −kr
dT T2 − T1 Tc − Th = −kr = −kr , dx l L
where l is the spacing between adjacent layers. This radiant conductivity is
(T12 + T22 )(T1 + T) T = 4
4 e σSB T 3 l kr = , 2 − r
1/3 .
SKETCH: Figure Pr.4.16 shows the multilayer system, where all surfaces have the same emissivity and surface area. The radiant conductivity kr and the radiantconductivity based resistance Rkr are also shown.
(i) SurfaceRadiation in Multiple Parallel Layers (Zero Thickness) and Its Representation by Radiant Conductivity kr r
T2 , Tc ,
r
High Conductivity, Very Thin, Diffuse, Gray, Opaque Solid Slabs
∋
T1 , 1 2
∋
A Medium made of N Parallel Surfaces l= L N +1
(ii) Thermal Circuit Model Using Radiant Conductivity kr
Qr,x
N T1
T2
a r
Rkr
Ar
l , A = a2 r Ar kr
a ∋
Th ,
r
x
l
L
dT qr,x = − kr dx
Figure Pr.4.16(i) Surface radiation in a multilayer (each layer opaque) system. (ii) Its thermal circuit representation by radiant conductivity.
OBJECTIVE: (a) Start from (4.47) for radiation between surfaces 1 and 2 and assume that l a, such that F12 = 1. Then show that qr,12 =
r σSB (T14 − T24 ) . 2 − r
(b) Use that linearization of (4.72) to show that qr,12 =
4 r σSB T 3 (T1 − T2 ) 2 − r
forT1 → T2 ,
i.e., for small diminishing diﬀerence between T1 and T2 . (c) Then using the deﬁnition of qr,x given above, derive the given expression for radiant conductivity kr . SOLUTION: (a) Starting from (4.47), we have for surface radiation between surfaces 1 and 2 Qr,12 =
Eb,1 − Eb,2 , 1 − r 1 1 − r + + Ar r Ar F12 Ar r 341
where we have the same area Ar and emissivity r for both surfaces. Since l a, then from Figure 4.11(b), F12 = 1 and we have qr,12 =
Qr,12 σSB (T14 − T24 ) r σSB (T14 − T24 ) = . = 2 Ar 2 − r −1 r
(b) Using T1 /T2 1, we use the results of (4.72), i.e., (T14 − T24 )
= =
(T12 + T22 )(T12 − T22 ) (T12 + T22 )(T1 + T2 )(T1 − T2 )
≡ 4T 3 (T1 − T2 ) where we have deﬁned (T22 + T12 )(T1 + T2 ) = 4T 3 . Note that for T1 → T2 , i.e., a diminishing temperature diﬀerences between two adjacent layers, we will have T = T1 = T2 . Using the results of (a), we then have qr,12 =
4 r σSB T 3 (T1 − T2 ) . 2 − r
(c) Next, we use the deﬁnition and the results of (b), i.e., qr,x
T2 − T1 l 3 4 r σSB T (T1 − T2 ) T2 − T1 . = −kr 2 − r l
= qr,12 ≡ −kr =
Then solving for kr , we have kr =
4 r σSB T 3 l . 2 − r
COMMENT: Note that we did not allow for any conduction resistance through each layer. This can be signiﬁcant for high emissivity, but low conductivity solids (e.g., polymeric materials such as paper and fabrics).
342
PROBLEM 4.17.FAM GIVEN: A short, oneside closed cylindrical tube is used as a surface radiation source, as shown in Figure Pr.4.17(a). The surface (including the cylindrical tube and the circular closed end) is ideally insulated on its outside surface and is uniformly heated by Joule energy conversion, resulting in a uniform inner surface temperature T1 = 800◦C. The heat transfer from the internal surface to the surroundings is by surface radiation only. T∞ = 100◦C, r,1 = 0.9, D = 15 cm, L = 15 cm. SKETCH: Figure Pr.4.17(a) shows the oneside closed cavity with its wall heated by Joule energy conversion exchanging radiation with surroundings.
r,1
T ,
Qr,1
D
(+)
L
r,
Open End of Cavity Surface 2, T2 = T,
∋
Tc , Closed End
∋
(−)
∋
Se,J
r,2 =
1
Figure Pr.4.17(a) Surface radiation from a oneend closed cavity, to its surroundings.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the required Joule heating rate S˙ e,J . SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.4.17(b).
.
Qr,1
Se,J
Qr,12 (qr,o)1
(qr,o)2
Q1 (Rr, )1
Eb,1
(Rr,F)12
(Rr, )2
T1
Eb,2 T2 = T
Figure Pr.4.17(b) Thermal circuit diagram.
(b) The radiation source here is the inner surface of a oneside closed cylindrical tube. It is at a uniform temperature T1 . Therefore, the inner surface of the tube and the closed end can be modeled as a single node at T1 . This surface exchanges heat by radiation with the surrounding through the open end (surface 2). Surface 2 is a blackbody surface ( r,2 = 1) and we have a twosurface enclosure. The corresponding thermal circuit diagram is shown in Figure Pr.4.17(b). The conservation of energy equation applied to node T1 (for steadystate condition) gives Qr,1∞ + Q1 = Qr,12 + Q1 = S˙ e,J . Since the outer surface is insulated, Q1 → 0, and then Eb,1 − Eb,∞ = S˙ e,J , (Rr, )1 + (Rr,F )12 + (Rr, )2 343
where (Rr, )1
=
(Rr,F )12
=
(Rr, )2
=
1 − r,1 Ar,1 r,1 1 Ar,1 F12 1 − r,2 . Ar,2 r,2
Also Ar,1 =
π × (0.15)2 (m)2 πD2 + πDL = + π × (0.15)(0.15)(m)2 = 0.0884 m2 . 4 4
As noted, all net radiation exchange between surface 1 and the surrounding must pass through the remaining open end. For simplicity, this end can be thought of as an imaginary surface 2 of area Ar,2 = πD2 /4 = 0.01767 m2 and at T2 = T∞ = 373.15 K, that would provide the same eﬀect as the surroundings for radiation heat exchange. This is drawn schematically in the lower part of Figure Pr.4.17(b), where Qr,12 is equal to Qr,1∞ . Note that F21 = 1 by inspection, since the imaginary end surface is a ﬂat end of the tube. Then (Rr, )1
=
(Rr,F )12
=
(Rr, )2
=
1 − r,1 1 − 0.9 = 1.258 m−2 = Ar,1 r,1 0.0884(m2 ) × 0.9 1 1 1 = 56.588 m−2 = = Ar,1 F12 Ar,2 F21 0.01767(m2 ) × 1 1−1 = 0. Ar,∞ × 1
Then the energy equation becomes Qr,1∞ = Qr,12 = S˙ e,J
=
S˙ e,J
=
S˙ e,J
=
S˙ e,J
=
Eb,1 − Eb,2 (Rr, )1 + (Rr,F )12 + (Rr, )2 σSB (T14 − T24 ) (Rr, )1 + (Rr,F )12 + (Rr, )2 5.67 × 10−8 (W/m2 K4 )[(1,073.15 K)4 − (373.15 K)4 ] 2
2
1.258(1/m ) + 56.588(1/m ) + 0 1,041 W.
COMMENT: For cavities, the opening can be treated as a blackbody surface having the temperature of the surrounding. This allows for radiation leaving the cavity to the surrounding with no reﬂection from the opening and also allows for the surrounding to emit into the cavity.
344
PROBLEM 4.18.FAM GIVEN: A cylindrical piece of wood (length L and diameter D) is burning in an oven as shown in Figure Pr.4.18(a). The wood can be assumed to be in the central region of the cube furnace (with each oven side having length a). The internal oven surface temperature is T2 . The burning rate is M˙ r,c , and the heat of combustion is ∆hr,c . Assume that the only surface heat transfer from the wood is by steadystate radiation. T2 = 80◦C, M˙ r,c = 2.9 × 10−4 kg/s, ∆hr,c = −1.4 × 107 J/kg, r,1 = 0.9, r,2 = 0.8, D = 5 cm, L = 35 cm, a = 1 m. Use geometrical relations (not the tables) to determine the view factors. SKETCH: Figure Pr.4.18(a) shows the cylindrical piece of wood.
a a Piece of Wood
L
Sr,c = Mr,c ,hr,c
D
a
r,1 ,
T1
r,2 ,
T2
Figure Pr.4.18(a) A cylindrical piece of wood burning in an oven.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the wood surface temperature T1 . (c) What would T1 be if T2 were lowered by 80◦C? SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.4.18(b). Qr ,12 T1
Eb,1
(qr ,o)1
(qr ,o)2
Eb,2 T2 Q2
(Rr , )1
(Rr ,F)12
(Rr , )2
. Sr,c
Figure Pr.4.18(b) Thermal circuit diagram.
(b) Applying the conservation of energy equation to node T1 , and noting steadystate, we have QA = Qr,12 Eb,1 − Eb,2 Rr,Σ σSB (T14 − T24 ) Rr,Σ
= S˙ r,c = −M˙ r,c ∆hr,c = −M˙ r,c ∆hr,c . 345
The total radiation thermal resistance is then found with
πD2 4 = πD(L + D/2) = π × 0.05(m) × [0.35(m) + 0.05(m)/2]
A1
= Ashaft + Aends = πDL + 2 ×
A2
= 0.0589 m2 = Abox = 6(a × a) = 6 × 1(m) × 1(m)
F 1 2
= =
6 m2 1 by inspection.
Then (Rr, )1
=
(Rr,F )12
=
(Rr, )2
=
1 − r,1 1 − 0.9 = = 1.886 1/m2 A1 r,1 0.0589(m2 ) × 0.9 1 1 = = 16.977 1/m2 A1 F12 0.0589(m2 ) × 1 1 − r,2 1 − 0.8 = = 0.04167 1/m2 A2 r,2 6(m2 ) × 0.8)
or Rr,Σ
= =
(Rr, )1 + (Rr,F )12 + (Rr, )2 1.886(1/m2 ) + 16.977(1/m2 ) + 0.04167(1/m2 )
=
18.90 1/m2 .
Then solving for T1 , we have T1
=
T24
=
Rr,Σ ˙ Mr,c ∆hr,c − σSB 4
1/4
[80 + 273.15]4 (K) −
= 1,081.7 K = 808.6◦C.
1/4 18.90(1/m2 ) −4 7 (kg/s)][−1.4 × 10 (J/kg)] × [2.9 × 10 5.67 × 10−8 (W/m2 K4 )
(c) If T2 is lowered by 80◦C to 0◦C, we have T1
1/4 Rr,Σ ˙ 4 Mr,c ∆hr,c = T2 − σSB 1/4 18.904(1/m2 ) 4 −4 7 = [0 + 273.15] (K) − × [2.9 × 10 (kg/s)][−1.4 × 10 (J/kg)] 5.67 × 10−8 (W/m2 K4 ) = 1,079.8 K = 806.6◦C.
COMMENT: The surfaceconvection heat transfer (due to the thermobuoyant ﬂuid motion) can be signiﬁcant and would tend to reduce the surface temperature.
346
PROBLEM 4.19.FUN GIVEN: Consider two square (each length a) parallel plates at temperatures T1 and T2 and having an equal emissivity r . Assume that the distance between them l is much smaller than a (l a). OBJECTIVE: (a) Show that radiative heat ﬂux between surface 1 and 2 is qr,12 =
r σSB (T14 − T24 ) . 2 − r
(b) Show that if a radiation shield having the same size and emissivity is placed between them, then qr,12 =
r σSB (T14 − T24 ) . 2(2 − r )
SOLUTION: (a) For two, parallel plates placed very close to each other (l a), we have from Figure 4.11(b), for the view factor F12 = 1. Then (4.48) for a twosurface enclosure becomes Qr,12
=
=
=
=
Eb,1 − Eb,2 1 − r 1 1 − r + + Ar r Ar Ar r Eb,1 − Eb,2 2 1 − Ar r Ar Ar (Eb,1 − Eb,2 ) 2 −1 r Ar r σSB (T14 − T24 ) 2 − r
or qr,12 =
Qr,12 r σSB (T14 − T24 ) = . Ar 2 − r
(b) With one shield added, starting from (4.50), we have Qr,12
= =
Eb,1 − Eb,2 2[(Rr, )1 + (Rr,F )12 + (Rr, )2 ] Eb,1 − Eb,2 , 1 − r 1 1 − r 2 + + Ar r Ar Ar r
where again we have used F1s = Fs2 = 1. Then following the steps in part (a), we have qr,12 =
r σSB (T14 − T24 ) . 2(2 − r )
COMMENT: Note that as highly reﬂective ( r → 0) surfaces are used, qr,12 = r σSB (T14 − T24 )/4, which shows a direct proportionality between qr,12 and r . 347
PROBLEM 4.20.FUN GIVEN: Two very large, parallel plates at maintained temperatures T1 and T2 are exchanging surface radiation heat. A third large and thin plate is placed in between and parallel to the other plates [Figure Pr.4.20(a)]. This plate has periodic voids (e.g., as in a screen) and the fraction of void area to total surface area is = Avoids /Atotal . The screen is suﬃciently thin such that its temperature T3 is uniform across the thickness. All plates have opaque, diﬀuse, and gray surfaces with the same total emissivity r . SKETCH: Figure Pr.4.20(a) shows the surfaceradiation heat transfer between two plates separated by a screen.
(i) Physical Model Surface 3 (Screen) at T3 ( r)3 = r
Surface 2 at T2 ( r)2 = r
Surface 1 at T1 ( r)1 = r
(ii) Radiation Heat Flow Paths Solid
Qr,3 Qr,13 Qr,1
Void =
Void Qr,3 Qr,23
Qr,2
Q2
Q1
Avoid Asolid + Avoid
Qr,12 = Qr,21
Surface 1
Surface 3
Surface 2
Figure Pr.4.20(a)(i) and (ii) Surface radiation heat transfer between two plates separated by a screen.
OBJECTIVE: (a) Draw the thermal circuit. (b) Derive the expression for the net heat transfer rate by surface radiation between surfaces 1 and 2, i.e., Qr,12 , given by Qr,12 = Ar
Eb,1 − Eb,2 . 1 − r 1 2 + r 2 + r (1 − )
(Suggestion: Use a threesurface enclosure and allow for heat transfer between surfaces 1 and 2 directly through the screen voids and indirectly through the solid portion of the screen. The screen has radiation exchange on both of its sides, with a zero net heat transfer). (c) Comment on the limits as → 0 and → 1. (d) Would Qr,12 increase or decrease with an increase in the emissivity of the screen? (Suggestion: Analyze Qr,12 in the limits for r → 0 and r → 1.) SOLUTION: (a) The thermal circuit diagram for the problem is shown in Figure Pr.4.20(b). (b) The temperatures T1 and T2 are known. Therefore, the radiation heat transfer rate from surface 1 to surface 2, Qr,12 , is found from Figure Pr.4.20(b) as Qr,12 =
Eb,1 − Eb,2 , (Rr,Σ )12 348
Qr,13
(Rr, )1
Eb,3 T3 Eb,3
(qr,0)3 (Rr,F)13
(Rr, )3
(qr,0)3
Qr,2 Eb,2 T2
(qr,0)2 (Rr,F)32
(Rr, )3
Q1
(qr,0)1
Qr,32
Qr,3
T1 Eb,1
Qr,3
(Rr, )2
Qr,1
Q2
(Rr,F)12 Figure Pr.4.20(b) Thermal circuit diagram.
where the overall resistance for the thermal circuit is 1 − r,1 1 1 − r,2 + + . (Rr,Σ )12 = 1 1 Ar,1 r,1 Ar,2 r,2 + 1 1 − r,3 1 − r,3 1 1 + + + Ar,1 F13 Ar,3 r,3 Ar,3 r,3 Ar,3 F32 Ar,1 F12 Here, surface 3 refers to the solid part of the screen only (Ar,3 = Asolid ). Three view factors are needed, F12 , F13 , and F32 . The view factor from surface 3 to surface 2 is unity, because surface 2 is inﬁnite and parallel to surface 3 (F32 = 1). By symmetry, the view factor from surface 3 to surface 1 is also unity. Applying the reciprocity rule (4.34) to F31 , we obtain Ar,3 F31 = Ar,1 F13 . Solving for F13 gives F 1 3 =
F31 Ar,3 Ar,3 = . Ar,1 Ar,1
Using the relation 1 − = Asolid /(Asolid + Avoid ), we have F13 = 1 − . Applying the summation rule (4.33) to surface 1 gives F11 + F12 + F13 = 1. For surface 1, F11 = 0. Solving for F12 gives F12 = 1 − F13 = 1 . All the surfaces have the same total emissivity r . Then we have (Rr,Σ )12 =
1 − r 1 1 − r + + . 1 1 Ar,1 r Ar,2 r + 1 − r 1 1 1 + + Ar,1 (1 − ) Ar,3 r Ar,3 Ar,1
The surface areas are Ar,1 = Ar,2 = Ar and Ar,3 = (1 − 1 )Ar . Then 1 − r 1 . (Rr,Σ )12 = 2 + 1 Ar r Ar + 2 2 1 − r + Ar (1 − ) 1 − Ar r Finally, Qr,12
=
Eb,1 − Eb,2 (Rr,Σ )12
= 2
1 − r Ar r
Eb,1 − Eb,2
.
1
+ Ar +
349
1 2 2 + Ar (1 − ) 1 −
1 − r Ar r
After dividing by Ar , we have qr,12
=
Qr,12 Ar
=
2
=
1 − r r
Eb,1 − Eb,2
1
+ +
1 2 2 1 − r + 1− 1− r
Eb,1 − Eb,2 . 1 − r 1 2 + r 2 + r (1 − )
(c) It is always a good idea to check the limits of your solution to see whether they agree with your physical understanding of the problem. In the limit when → 1, we have lim qr,12 =
1 →1
Eb,1 − Eb,2 , 2 −1 r
which is the surface radiation heat ﬂux between two inﬁnite, parallel, ﬂat plates. In the limit when → 0, we have lim qr,12 =
1 →0
Eb,1 − Eb,2 , 2 2 −1 r
which is the surface radiation heat ﬂux between two inﬁnite, parallel, ﬂat plates when one radiation shield is placed between them. (d) From the expression for qr,12 , we have for the case of r = 0 qr,12 =
Eb,1 − Eb,2 = 0, ∞
i.e, no surfaceradiation heat transfer occurs. For the case of r = 1, we have qr,12 =
Eb,1 − Eb,2 = (Eb,1 − Eb,2 ). 1
This shows that the radiation heat transfer decreases by a factor of when the surfaces (including the screen) are blackbodies. COMMENT: Note that even a screen with a large 1 can reduce the heat transfer rate between the surfaces.
350
PROBLEM 4.21.FUN GIVEN: In surfaceradiation heat transfer between surfaces 1 and 2, the enclosure geometry dependence of the radiation heat ﬂux qr,21 is examined using four diﬀerent geometries. These are shown in Figures Pr.4.21(a)(i) through (iv) and are: parallel plates, coaxial cylinders, coaxial spheres, and a disk facing an enclosing hemisphere. The plates are assumed to be placed suﬃciently close to each other and the cylinders are assumed to be suﬃciently long, such that for all the four enclosure geometries the radiation is only between surfaces 1 and 2 (i.e., twosurface enclosures). T1 = 120◦C, T2 = 90◦C, r,1 = r,2 = 0.8. SKETCH: Figure Pr.4.21(a) shows the four geometries.
l
qr,21
Ar,1 , T1 ,
r,1
Ar,2 , T2 ,
qr,21
w = 2R1
r,2
a = 2R1
(iii) Coaxial Spheres
Ar,2 , T2 , r,2 Ar,1 , T1 ,
R1
r,2
R2 = 1.2R1
(iv) Circular Disk Surrounded by a Hemisphere
r,1
R2 = R1
Ar,2 , T2 ,
qr,21
qr,21
r,2
Ar,1 , T1 ,
Ar,2 , T2 ,
r,1
Ar,1 , T1 ,
(ii) Coaxial Cylinders l R2
(i) Parallel Plates w ,a l l
r,1
R1 R1
R2 = 1.2R1
Figure Pr.4.21(a)(i) through (iv) Four enclosure geometries used in determining the dependence of qr,21 on the enclosure geometry.
OBJECTIVE: (a) Draw the thermal circuit diagram (one for all geometries). (b) Determine qr,21 = Qr,21 /Ar,2 for the geometries of Figures Pr.4.21(a)(i)(iv), for the given conditions. SOLUTION: (a) Figure Pr.4.21(b) shows the thermal circuit diagram for all the geometries. Qr,1 = Qr,12 = Qr,21 = Qr,2 Q1
Q2 (Rr, )1 (qr,o)1 (Rr,F)12 (qr,o)2
(Rr, )1
T1 Eb,1
Eb,2 T2
Figure Pr.4.21(b) Thermal circuit diagram.
(b) The radiation heat transfer Qr,21 is given by (4.47), i.e., Qr21 =
1 − r,1 Ar,1 r,1
σSB (T24 − T14 ) . 1 1 − r,2 + + Ar,1 F12 Ar,2 r,2 351
(i) For the parallel plate, we have Ar,1 F 1 2
= Ar,2 = wa = 4R12 = 1 (for w∗ → ∞, a∗ → ∞) Figure 4.11(b)
Then Qr,21 = qr,21 Ar,2
=
σSB (T24 − T14 ) 1 − r,1 1 1 − r,2 + + r,1 1 r,2
5.67 × 10−8 (W/m2 K4 )[(363.15)4 − (393.15)4 ](K4 ) 2 × (1 − 0.8) +1 0.8 3.685 × 102 (W/m2 ) = − = −245.7 W/m2 . 0.5 + 1
=
(ii) For long, coaxial cylinders, we have Ar,1 F 1 2
= =
2πR1 l, Ar,2 = 2πR2 l 1 since all radiation leaving surface 1 is assumed to arrive at surface 2.
Then Qr,21 Ar,2
= qr,21 =
R2 R1
σSB (T24 − T14 ) 1 − r,1 R2 1 − r,2 + + r,1 R1 r,2
3.685 × 102 (W/m2 ) 1 − 0.8 1 − 0.8 + 1.2 + (1.2) 0.8 0.8 2 2 −3.685 × 10 (W/m ) = −210.6 W/m2 . 0.3 + 1.2 + 0.25
= −
= (iii) For coaxial spheres,we have Ar,1
=
4πR12 , Ar,2 = 4πR22
F 1 2
=
1
since all radiation leaving surface 1 arrives at surface 2.
Then Qr,21 Ar,2
= qr,21 =
R2 R1
2
σSB (T24 − T14 ) 2 R2 1 − r,1 1 − r,2 + + r,1 R1 r,2
−3.685 × 102 (W/m2 ) 1 − 0.8 1 − 0.8 + (1.2)2 + (1.2)2 0.8 0.8 3.685 × 102 (W/m2 ) = − = −179.8 W/m2 . 0.36 + 1.44 + 0.25
=
(iv) For a disk surrounded by a hemisphere, we have Ar,1
= πR12 , Ar,2 = 2πR22
F 1 2
=
1
since all radiation leaving surface 1 arrives at surface 2. 352
Then Qr,21 Ar,2
= qr,21 = 2
R2 R1
2
σSB (T24 − T14 ) 2 R2 1 − r,1 1 − r,2 +2 + r,1 R1 r,2
−3.685 × 102 (W/m2 ) 1 − 0.8 1 − 0.8 +2+ 2× 0.8 0.8 3.685 × 102 (W/m2 ) = − = −134.0 W/m2 . 0.5 + 2 + 0.25
=
COMMENT: Note that, due to the change in the surface areas, the magnitude of the surface radiation heat ﬂux decreases as we move from the planar surface to the curved surfaces (resulting in an increase in surface area Ar,2 ).
353
PROBLEM 4.22.FAM GIVEN: A hemispherical Joule heater (surface 1) is used for surfaceradiation heating of a circular disk (surface 2). This is shown in Figure Pr.4.22(a). In order to make an eﬃcient use of the Joule heating, a hemispherical cap (surface 3) is placed around the heater surface and is ideally insulated. R1 = 5 cm, R2 = 5R1 , T1 = 1,100 K, T2 = 500 K r,1 = r,2 = 1. Assume that F12 corresponds to that from a sphere to a disk (i.e., assume that the upper hemisphere does not see the disk). SKETCH: Figure Pr.4.22(a) shows the heater, the disk, and the reradiating surface.
r,1 =
1
Reradiating Surface, Similar to Reflector (Surface 3)
l = R2 R1
Surface Radiation Heated Disk (Surface 2)
Qr,12
Ideally Insulated Q3 = 0
R2 T2 ,
T1 ,
Electrical Heater (Surface 1)
Se,J
r,2 =
1
Figure Pr.4.22(a) A Joule heater is used for surfaceradiation heating of a disk. A reradiating hemisphere is used to improve the heating rate.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine Qr,12 for the given conditions. (c) Determine Qr,12 without the reradiator and compare the results with the results in (b). SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.4.22(b).
(qr,o)1 = Eb,1 Qr,12
(qr,o)2 = Eb,2 (Rr,F)12 Qr,12
T1 Se,J
(Rr,A)2 = 0 Eb,2
Eb,1 (Rr,A)1 = 0 (Rr,F)13
T2
(Rr,F)23 (qr,o)3 = Eb,3 T3 Q3 = 0
Figure Pr.4.22(b) Thermal circuit diagram.
(b) Noting that (Rr, )1 = (Rr, )2 = 0, because r,1 = r,2 = 1, (4.60) applies to this threesurface enclosure with one surface reradiating, i.e., Qr,12 =
σSB (T14 − T24 ) . 1 1 Ar,1 F12 + 1 1 + Ar,1 F13 Ar,2 F23 354
Here we obtain F12 from Table 4.2, noting that R2∗ = R2 /l = 1, i.e., 1 1 1 1 = F 1 2 = 1− 1 − 1/2 2 2 (1 + R2 /l)1/2 2 = 0.1464. Then using the summation rule (4.33), we have F11 + F12 + F13 = 1,
F11 = 0 (planar surface)
or F13 = 1 − F12 = 1 − 0.1464 = 0.8536. To ﬁnd F23 , we use the summation rule again, i.e., F21 + F22 + F23 = 1,
F22 = 0 (planar surface)
or F 2 3
=
1 − F 2 1 = 1 −
=
1−
Ar,1 F 1 2 Ar,2
2πR12 2 2 F12 = 1 − 25 × 0.1464 = 0.9883, πR2
where we have used the reciprocity rule (4.34). Now we use these numerical values to evaluate Qr,12 , i.e., Qr,12
σSB (T14 − T24 ) 1
= 2π(0.05)2 (m2 ) × 0.1464 +
=
=
1 1 1 + 2π(0.05)2 (m2 ) × 0.8536 π(0.25)2 (m2 ) × 0.9883
5.67 × 10−8 (W/m2 K4 )[(1,100)4 − (500)4 ](K4 ) 1 1 0.0023(m2 ) + 74.58(1/m2 ) + 5.153(1/m2 ) 7.946 × 104 (W/m2 ) = 1,180 W. 67.379(1/m2 )
(c) When the reradiating surface is not present, then the heat transfer between surface 1 and 2 is found from Figure Pr.4.22(b), where the only resistance between the two surfaces is (Rr,F )12 , i.e., Qk,12
=
=
Eb,1 − Eb,2 1 Ar,1 F12 7.947 × 104 (W/m2 ) = 182.78 W. 1 (1/m2 ) 0.0023
COMMENT: Note that the radiation heat transfer rate has increased by 6.6 folds, when the reradiating surface is used.
355
PROBLEM 4.23.FAM GIVEN: A ﬂat radiation heater is placed along a vertical wall to heat the passing pedestrians who may stop temporarily and face the heater. The heater is shown in Figure Pr.4.23(i) and is geometrically similar to a fullsize mirror. The heater surface is at T1 = 600◦C and the pedestrians have a surface temperature of T2 = 5◦C. Assume that the surfaces are opaque, diﬀuse, and blackbody surfaces (total emissivities are equal to one). Also assume that both the heater and the pedestrian have a rectangular cross section with dimensions a = 50 cm and w = 170 cm and that the distance between them is l = 40 cm, as shown in Figure Pr.4.6(ii). SKETCH: Figure Pr.4.23(a) shows the heater and the pedestrian and the idealized surface radiation geometry. (i) Physical Model
(ii) An Approximation ∋
T1,
r,1 =
1 Reradiating or Open Surfaces
∋
r,2 =
1
Pedestrian r,2 = 1 T2
2 Qr,1
Heater r,1 = 1 T1 ∋
T2,
∋
w
Qr,1
Radiating Surface (Heater)
2
l a
Pedestrian
Figure Pr.4.23(a)(i) Physical model of a radiant wall, pedestrian heater. (ii) Idealized model.
OBJECTIVE: (a) Draw the thermal circuit diagram for a threesurface enclosure (including the surroundings as a blackbody surface). (b) Determine the net radiation heat transfer from the heater to the pedestrian Qr,12 . (c) Determine the net radiation heat transfer to the pedestrian, when a reradiating (i.e., ideally insulated) surface is placed around the heater and pedestrian to increase the radiant heat ﬂow Qr,12 . SOLUTION: (a) A ﬁctitious surface can be wrapped around the open air space between the pedestrian and the heater. This ﬁctitious surface is treated as an additional radiation surface (surface 3) and the problem becomes a threesurface enclosure with diﬀuse, gray surfaces. For these blackbody three surfaces, the thermal circuit is presented in Figure Pr.4.23(b). Q3
T3
(R
) 23
Q1
Qr,23
F
(R r,
Qr,13
r,F ) 13
Eb,3 = (qr,o)3
(Rr,F)12
T1 Eb,1
= (qr,o)1
Qr,12
T2
Q2
Eb,2
= (qr,o)2
Figure Pr.4.23(b) A threesurface blackbody enclosure. Radiation heat transfer ﬂow to surface 2 from surface 1 and when surface 3 is reradiating (Q3 = 0), heat ﬂows indirectly from surface 1 to 2.
356
(b) The surfaceradiation heat transfer rate from surface 1 to surface 2 is found from Figure Pr.4.23(b) as Qr,12 =
Eb,1 − Eb,2 , (Rr,Σ )12
where, for the unity emissivities, we have (Rr,Σ )12 =
1 . Ar,1 F12
The view factor F12 is obtained from Figure 4.11(b). For surfaces 1 and 2, w∗ = w/l = 1.7(m)/0.4(m) = 4.25, a∗ = a/l = 0.5(m)/0.4(m) = 1.25. From Figure 4.11(b) we obtain F12 = 0.39. The area for surface 1 is A1 = aw = 0.5(m) × 1.7(m) = 0.85 m2 . Therefore (Rr,Σ )12 =
1 = 3.03 1/m2 0.85(m2 ) × 0.39
The heat transfer rate is =
Qr,12
=
σSB (T14 − T24 ) (Rr,Σ )12
2 5.67 × 10−8 (W/m K4 ) × 873.154 (K4 ) − 278.154 (K4 ) = 10,765 W. 3.03(1/m2 )
(c) If surface 3 is perfectly insulated, Q3 = 0 (surface 3 is called a reradiating surface). In this case, the overall thermal resistance is found from Figure Pr.4.23(b) as 1 . 1 1 + 1 1 1 + Ar,1 F12 Ar,1 F13 Ar,2 F23
(Rr,Σ )12 =
The viewfactor F12 was obtained above. The view factors F13 and F23 need to be determined. From the summation rule (4.33), we have F11 + F12 + F13 = 1. Since F11 = 0, we have F13 = 1 − F12 = 1 − 0.39 = 0.61. From the reciprocity rule and noting that Ar,2 = Ar,1 , F23 = F13 = 0.61. Then, Ar,2 = Ar,1 = 0.85 m2 and we have (Rr,Σ )12 =
1 = 1.69 1/m2 . 1 1 + 1 2 0.85(m2 ) × 0.61 0.85(m2 ) × 0.39
Finally, the heat transfer rate is Qr,12
= =
σSB (T14 − T24 ) (Rr,Σ )12
2 5.67 × 10−8 (W/m K4 ) 873.154 (K4 ) − 278.154 (K4 ) = 19,300 W. 1.69(1/m2 )
COMMENT: Note that by placing the reradiating surface, the heat transfer from the heater to the pedestrian has nearly doubled. 357
PROBLEM 4.24.FAM GIVEN: A source for thermal irradiation is found by a Joule heater placed inside a solid cylinder of radius R1 and length l. Then a hollow cylinder of radius R2 and length l is placed coaxially around it with this outer cylinder and the top part of the opening ideally insulated. This is shown in Figure Pr.4.24(a) with the radiation leaving through the opening at the bottom spacing between the cylinders (surface 2). This results in surface 1 being the high temperature surface with direct and reradiation exchange with surface 2. T2 = 400 K, r,1 = 0.8, S˙ e,J = 1,000 W, R1 = 1 cm, R2 = 5 cm, l = 10 cm. SKETCH: Figure Pr.4.24(a) shows the heated inner cylinder and the reradiating and the opening surfaces. Solid Cylinder with Embedded Q3 = 0 (Reradiating) Joule Heater Surface 3'' Surface 3 Surface 3' R2 Se,J T1 ,
Radiation Enclosure (Volume between Cylinders and Top and Bottom Surfaces) ∋
R1
l
r,1
∋
T2 , r,2 = 1 Q2 (Radiation Heat Transfer for Process Heating)
Figure Pr.4.24(a) Surface 1 is heated by Joule heating and through direct and reradiation allows for radiation to leave for surface 2.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the view factors F1 2 , F13 , and F23 , using Figures 4.11(d) and (e), and the designations of Figure Pr.4.24(a). (c) Determine the heater surface temperature T1 . SOLUTION: (a) Figure Pr.4.24(b) shows the thermal circuit diagram. Both direct and reradiating radiation are shown.
(Rr,F)13
Qr,1
ReRadiating Surface
T1 , Eb,1
(qr,o)3 = Eb,3
(Rr, )1
Se,J
T3
(Rr,F)32
(Rr,F)13
(qr,o)2 (Rr, )2
 Qr,2
T2 , Eb,2 Q2
Figure Pr.4.24(b) Thermal circuit diagram.
358
(b) Using Figure Pr.4.24(b), and (4.60), we have Qr,1
= S˙ e,J =
1 − r Ar r
σSB (T14 − T24 ) . 1 − r 1 + + 1 Ar r 2 1 Ar,1 F12 + 1 1 + Ar,1 F13 Ar,2 F23
Here we have r,2 = 1, and also Ar,2 = π(R22 − R12 ).
Ar,1 = 2πR1 l,
For the view factors, we begin by using Figure 4.11(e). We deﬁne surface 3 [see Figure Pr.4.24(a)] and the inner surface of the outer cylinder and obtain F13 by using the reciprocity rule (4.34), i.e., F 1  3 =
Ar,3 F 3  1 . Ar,1
To obtain F3 1 , we use Figure 4.11(e) with 1 R∗ l R2
= =
R1 0.01(m) = 0.2 = R2 0.05(m) 0.10(m) =2 0.05(m)
and then we obtain F3 1 0.15. To obtain F13 we use the summation rule (4.33) for surface 1, i.e., F11 + F13 + 2F13 = 1,
where we have noted that the top (3 ) and bottom (2) surfaces between the two cylinders are identical. Here F11 = 0, and we obtain
F 1 2
1 − F 1= = F13 = 2 R2 F 3  1 1− R1 = 2 0.05(m) 1− × 0.15 0.01(m) = 2 = 0.125. 3
1−
Ar,3 F 3  1 Ar,1 2
Then F13 = F13 + F13 = 0.75 + 0.125 = 0.875
view factor between surface 1 and the reradiating surfaces 3.
To determine, F23 , we use the summation rule for surface 2, i.e., F 2 2 + F 2 1 + F 2 3 = 1
359
or F 2 3
= = = = = =
1 − F 2 2 − F 2 1 1 − F 2 1 Ar,1 1− F 1 2 Ar,2 2πR1 l 1− 0.125 π(R22 − R12 ) 2π × 0.01 × 0.1 × 0.125 1− π(0.052 − 0.012 ) 0.8958.
(c) We now solve the energy equation for T1 , i.e., 1 − r + Ar r 1 Ar,1 F12 +
T14
S˙ e,J = T24 + σSB
T14
= (400)4 (K4 ) +
1
1 1 1 + Ar,1 F13 Ar,2 F23
1 − 0.8 1,000(W) + × −8 2 4 5.67 × 10 (W/m K ) 2π × 0.01(m) × 0.1(m) × 0.8
1
2π × 0.01 × 0.1(m2 ) × 0.125 +
1 1 + 1 1 2 2 2 2 2π × 0.01 × 0.1(m ) × 0.875 π(0.05 − 0.01 )(m ) × 0.895
= 2.560 × 1010 (K4 ) + 1.76 × 1010 (K4 m2 ) ×
39.79(1/m2 ) +
T14
1 1 1 −4 2 7.854 × 10 (m ) + + 2 2 2 2 1.819 × 10 (m ) 1.482 × 10 (m ) 1 10 10 4 4 = 2.56 × 10 (K ) + 1.76 × 10 (K ) × 39.79 + 7.854 × 10−4 + 1.225 × 10−2 10 10 = [2.560 × 10 + 1.76 × 10 × (39.79 + 76.71)](K4 )
or T1 = 1,200 K. COMMENT: Note that this is a rather large temperature for the inner cylinder. This is below the melting temperature of oxide ceramics. Note that reradiation by surfaces 3 and 3 reduces T , signiﬁcantly, by reducing the viewfactor resistance from 1/Ar,1 F12 to what was used above.
360
PROBLEM 4.25.FUN.S GIVEN: Consider three opaque, diﬀuse, and gray surfaces with temperatures T1 = 400 K, T2 = 400 K, and T3 = 300 K, with surface emissivities r,1 = 0.2 and r,2 = r,3 = 0.5, and areas Ar,1 = Ar,2 = Ar,3 = 1 m2 . OBJECTIVE: (a) For (i) surfaces 1 and 2 forming a twosurface enclosure (i.e., F12 = 1), and (ii) surfaces 1, 2, and 3 forming a threesurface enclosure (assume a twodimensional equilateral triangular enclosure), is there a net radiation heat transfer rate Qr,12 between surfaces 1 and 2? (b) If there is a nonzero net heat transfer rate, what is the direction of this heat transfer? (c) Would this heat transfer rate change if T3 = 500 K? (d) What is the temperature T3 for which Qr,12 = 0? SOLUTION: (a)(i) Figure Pr.4.25(a) shows the thermal circuit diagram for surfaces 1 and 2 forming a twosurface enclosure. In this case, the net radiation heat transfer rate is
Qr,12 =
σSB (T14 − T24 ) . (Rr,Σ )12
Since T1 = T2 , there is a zero net heat transfer between surfaces 1 and 2, regardless of the value of the surface emissivities. Qr,1
Qr,2
Qr,12 (qr,o)1
(qr,o)2
Q1
Q2 (Rr, )1 ∋
Eb,1
(Rr,F)12
(Rr, )2 ∋
T1
Eb,2 T2
Figure Pr.4.25(a) Thermal circuit diagram for the twosurface enclosure.
(ii) Figure Pr.4.25(b) shows the thermal circuit diagram for surfaces 1, 2 and 3 forming a threesurface enclosure. Then, the net radiation heat transfer rates on surfaces 1, 2 and 3 are Q3
T3 Eb,3 (Rr, )3
Qr,3
(R
) 23
(Rr,F)12
(Rr, )2
(Rr, )1
Qr,23
F
(R r,
Qr,13
r,F ) 13
(qr,o)3
Q1
Q2 T1 Eb,1
(qr,o)1
Qr,1
Eb,2 T2
(qr,o)2
Qr,21
Qr,2
Figure Pr.4.25(b) Thermal circuit diagram for the threesurface enclosure.
361
Qr,1
=
Qr,2
=
Qr,3
=
Eb,1 − (qr,o )1 (qr,o )1 − (qr,o )2 (qr,o )1 − (qr,o )3 = + 1 − r,1 1 1 r,1 Ar,1 F12 Ar,1 F13 Ar,1 Eb,2 − (qr,o )2 (qr,o )2 − (qr,o )1 (qr,o )2 − (qr,o )3 = + 1 − r,2 1 1 r,2 Ar,2 F21 Ar,2 F23 Ar,2 Eb,3 − (qr,o )3 (qr,o )3 − (qr,o )2 (qr,o )3 − (qr,o )2 = + . 1 − r,3 1 1 r,3 Ar,3 F31 Ar,3 F32 Ar,3
Note that, even with Eb,1 = Eb,2 , since r,1 = r,2 , there may be a nonzero net heat transfer rate Qr,12 between surfaces 1 and 2. The view factors and areas are all the same, i.e., F12 = F13 = F23 = 0.5 and Ar,1 = Ar,2 = Ar,3 = 1 m2 . The surface resistances then become (Rr, )1
=
(Rr, )2
=
(Rr, )3
=
1 − r,1 =4 r,1 Ar,1 1 − r,2 =1 r,2 Ar,2 1 − r,3 = 1. r,2 Ar,3
Therefore, the equations for the net heat transfer rates become Eb,1 − (qr,o )1 4 Eb,2 − (qr,o )2 1 Eb,3 − (qr,o )3 1
= = =
(qr,o )1 − (qr,o )2 (qr,o )1 − (qr,o )3 + 2 2 (qr,o )2 − (qr,o )1 (qr,o )2 − (qr,o )3 + 2 2 (qr,o )3 − (qr,o )1 (qr,o )3 − (qr,o )2 + . 2 2
Upon rearranging, we have 1.25(qr,o )1 − 0.5(qr,o )2 − 0.5(qr,o )3 −0.5(qr,o )2 + 2(qr,o )2 − 0.5(qr,o )3 −0.5(qr,o )1 − 0.5(qr,o )2 + 2(qr,o )3
= 0.25Eb,1 = Eb,2 = Eb,3 ,
where Eb,1 = Eb,2 = 1,451.52 W/m2 and Eb,3 = 459.27 W/m2 . Solving the linear system of equations above (e.g., using SOPHT) we obtain, (qr,o )1 = 1,090.7 W/m2 , (qr,o )2 = 1,198.95 W/m2 , and (qr,o )3 = 802.047 W/m2 . Therefore, the net heat transfer rate between surfaces 1 and 2 is Qr,12
(qr,o )1 − (qr,o )2 (1,090.7 − 1,198.95)(W/m2 ) = 1 2(1/m2 ) F12 Ar,1 = −54.13 W.
=
(b) The negative sign indicates that heat is transferred from surface 2 to surface 1, i.e., from the larger to the smaller one. (c) For T3 = 500 K, we have Eb,3 = 3,543.75 W/m2 , and solving the new system of linear equations, we obtain (qr,o )1 = 2,212.33 W/m2 , (qr,o )2 = 1,984.09 W/m2 and (qr,o )3 = 2,820.98 W/m2 , and the net heat transfer rate between surfaces 1 and 2 becomes Qr,12 =
(2,212 − 1,984)(W/m2 ) = 114.1 W. 2(1/m2 )
Note that the heat transfer now occurs from surface 1 to surface 2, i.e., from the smaller to the larger emissivity.
362
(d) For Qr,12 = 0, we need T3 = 400 K. In this case (qr,o )1 = (qr,o )2 = (qr,o )3 = Eb,1 = Eb,2 = Eb,3 = 1,452 W/m2 . COMMENT: When the temperatures are equal, but emissivities or areas are not, the presence of a third surface results in a net radiation heat transfer between these surfaces.
363
PROBLEM 4.26.FAM GIVEN: Surfaceradiation absorption is used to melt solid silicon oxide powders used for glass making. The heat is provided by combustion occurring over an impermeable surface 1 with dimensions a = w = 1 m, as shown in Figure Pr.4.26. The desired surface temperature T1 is 1,600 K. The silicon oxide powders may be treated as a surface 2, with the same area as the radiant heater, at a distance l = 0.25 m away from the heater, and at a temperature T2 = 873 K. The surroundings are at T3 = 293 K. Assume that all surfaces are ideal blackbody surfaces. SKETCH: Figure Pr.4.26(a) shows the radiating surface 1 heating surface 2 in a threesurface enclosure.
Flame l
Radiant Heater T1 = 1,600 K
Ambient T3 = 293 K
a w
Silicon Oxide T2 = 873 K
Figure Pr.4.26(a) Surface radiant heater heated by a ﬂame over it and forming a threesurface radiation enclosure.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the net radiation heat transfer to the silicon oxide surface. SOLUTION: (a) The thermal circuit for a threesurface enclosure is shown in Figure Pr.4.26(b). Q3
T3 Eb,3 (Rr, )3
Qr,3
(R r
) 13
(Rr,F)21
(Rr, )1
(Rr, )2
Qr,13
F
(R r,
Qr,23
,F ) 23
(qr,o)3
Q2
Q1 T2 Eb,2
(qr,o)2
Qr,2
Eb,1 T1
(qr,o)1
Qr,21
Qr,1
Figure Pr.4.26(b) Graybody thermal circuit diagram.
(b) Since all the surfaces are blackbodies ( r,i = 1) the surfacegrayness resistances are zero and the radiosities become equal to the blackbody emissive powers. The thermal circuit then reduces to the one shown in Figure Pr.4.26(c). The net radiation heat transfer rate leaving surface 2 is Qr,2
= Qr,21 + Qr,23 Eb,2 − Eb,1 Eb,2 − Eb,3 = + . (Rr,F )21 (Rr,F )23 364
Q3
T3
Qr,3
(R r , ) 13
Q2
Qr,13
F
(R
Qr,23
r,F ) 23
Eb,3
T2 Eb,2
(Rr,F)21
Eb,1 T1
Qr,2
Qr,21
Qr,1
Q1
Figure Pr.4.26(c) Blackbody thermal circuit diagram.
The viewfactor resistances are (Rr,F )21 =
1 , Ar,2 F21
(Rr,F )23 =
1 . Ar,2 F23
The view factor between surfaces 2 and 1 can be evaluated from Figure 4.11(b). From the dimensions of the two plates and the distance separating them, w∗ = w/l = 1(m)/0.25(m) = 4, a∗ = a/l = 1(m)/0.25(m) = 4. The view factor is then approximately F12 = 0.63. Using the reciprocity rule (4.34), F21 Ar,2 = F12 Ar,1 and as Ar,1 = Ar,2 , F21 = F12 = 0.63. Using the summation rule for surface 2 we have F21 + F22 + F23 = 1. Since F22 = 0 (because surface 2 is ﬂat), we have F23 = 1 − F21 = 0.37. Then, the viewfactor resistances become (Rr,F )21
=
(Rr,F )23
=
1 1 = 1.59 = Ar,2 F21 1 (m) × 1 (m) × 0.63 1 1 = 2.70 = Ar,2 F23 1 (m) × 1 (m) × 0.37
1/m
2
2
1/m .
Finally, the net heat transfer rate leaving surface 2 is σSB T24 − T14 σSB T24 − T34 Qr,2 = + (Rr,F )21 (Rr,F )23 5.67 × 10−8 W/m2 K4 8734 K4 − 1,6004 K4 = + 2 1.59 1/m 5.67 × 10−8 W/m2 K4 8734 K4 − 2934 K4 2 2.70 1/m = −213,353 (W) + 12,031 (W) = −201,322 W. COMMENT: One of the diﬃculties in operating an oven like this is stabilizing the ﬂame over of the ceramic plate. One alternative is to use a porous radiant burner. However, an impermeable ceramic plate prevents the combustion products from contaminating the glass.
365
PROBLEM 4.27.FUN GIVEN: Surfaceradiation emission can be redirected to a receiving surface using reradiating surfaces. Figure Pr.4.27(a) renders such a redirection design using a reradiating surface 3. Surface 3 is ideally insulated and is treated as a single surface having a uniform temperature T3 . Surface 1 has a temperature T1 higher than that of surface 2, T2 . R1 = 25 cm, R2 = 25 cm, F12 = 0.1, r,1 = 1.0, r,2 = 1.0, T1 = 900 K, T2 = 400 K. Note that since surfaces 1 and 2 are blackbody surfaces, (qr,o )1 = Eb,1 and (qr,o )2 = Eb,2 . SKETCH: Figure Pr.4.27(a) shows the two blackbody, surfaceradiation heat transfer surfaces, and the reradiating third surface.
Redirection of Radiation Using a Reradiating Surface ∋
T1 ,
r,1
Qr,12 Direct
R1 Surface 3 Q3 = 0
Qr,12 R2 ∋
T2 ,
Reradiation
r,2
Figure Pr.4.27(a) Two blackbody surfaces that are exchanging surfaceradiation heat and are completely enclosed by a reradiation surface.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine Qr,12 for the given conditions. (c) Compare this with Qr,12 without reradiation. (d) Show the expression for Qr,12 for the case of F12 = 0 and comment on this expression. SOLUTION: (a) The thermal circuit diagram for surface radiation from surface 1 to surface 2, with the presence of the reradiating surface 3, is shown in Figure Pr.4.27(b). (b) The net surface radiation between surface 1 and 2 is determined from Figure 4.27(b) or from (4.60). Then for the case of r,1 = r,2 = 1, we have Qr,12 = Qr,1 =
Eb,1 − Eb,2 , 1 1 Ar,1 F12 + 1 1 + Ar,1 F13 Ar,2 F23
where Ar,1 = Ar,2
= πR2 = π(0.25)2 (m2 ) =
0.1963 m2 .
We use the summation rule (4.33) to ﬁnd F 1 3
=
F 2 3
=
1 − F11 − F12 = 1 − 0 − 0.1 = 0.9 Ar,1 F12 1 − F 2 2 − F 2 1 = 1 − F 2 2 − = 0.9. Ar,2 366
Q1 T1 (qr,o)1 = Eb,1
(Rr,F)13 Q3 = 0
T3
(Rr,F)12
(qr,o)3 = Eb,3
(qr,o)2 = Eb,2
(Rr,F)31 T2
Q2
Figure Pr.4.27(b) Thermal circuit diagram.
Qr,12 with reradiation
=
5.67 × 10−8 (W/m2 K4 ) × (9004 − 4004 )(K4 )
0.1963(m3 ) × 0.1 +
= =
−1
1 1 1 + 3 3 0.1963(m ) × 0.9 0.1963(m ) × 0.9
3.575 × 104 (W/m2 ) [0.01963(m3 ) + 0.08834(m2 )]−1 3.860 kW.
(c) For Qr,12 no reradiation , we have from Figure Pr.4.27(b) Qr,12 no reradiation
=
= =
Eb,1 − Eb,2 1 Ar,1 F12 3.575 × 104 (W/m2 ) [0.01963(m2 )]−1 0.7018 kW.
This is substantially less than the one with reradiation, i.e., thus it is only 18.18% of (b). (d) For the case of F12 → 0, we have F13 → 1, F23 → 1. Note that we still assume that we have a threesurface enclosure. Then (4.60) becomes
Qr,12
= = =
1 1 + Ar,1 F13 Ar,1 F23 Ar,1 F13 (Eb,1 − Eb,2 ) 2 Ar,1 (Eb,1 − Eb,2 ) for F12 → 0. 2
−1
(Eb,1 − Eb,2 )
Examining Figure Pr.4.27(b) shows that for (Rr,F )12 → ∞, the radiation is completely transferred by reradiation, subject to two viewfactor resistances. COMMENT: The reradiating surface 3 does facilitate surfaceradiation heat transfer between surfaces 1 and 2, but there is a ﬁnite geometrical resistance associated with this participation. Note that we have assumed a threesurface enclosure as we allowed F12 → 0, which also allowed for surface radiation heat transfer. In practice, as F12 → 0, these will only by a twosurface enclosure.
367
PROBLEM 4.28.FAM GIVEN: Fire barriers are used to temporarily protect spaces adjacent to ﬁres. Figure Pr.4.28 shows a suspended ﬁre barrier of thickness L and eﬀective conductivity k (and ρ and cp ) subjected to a ﬂame irradiation (qr,i )f . The barrier is a ﬂexible, wirereinforced mat made of a ceramic (high melting temperature, such as ZrO2 ) ﬁbers. The barrier can withstand the high temperatures resulting from the ﬂow of (qr,i )f into the mat until, due to thermal degradation of the ﬁbers and wires, it fails. In some cases the barrier is actively water sprayed to delay this degradation. The transient conduction through the mat, subject to a constant (qr,i )f , can be treated analytically up to the time that thermal penetration distance δα reaches the back of the mat x = L. This is done by using the solution given in Table 3.4 for a semiinﬁnite slab, and by neglecting any surface radiation emission and any surface convection. Assume that these simpliﬁcations are justiﬁable and the transient temperature T (x, t) can then be obtained, subject to a constant surface ﬂux qs = (qr,i )f and a uniform initial temperature T (t = 0). L = 3 cm, k = 0.2 W/mK, ρ = 600 kg/m3 , cp = 1,000 J/kgK, (qr,i )f = −105 W/m2 , T (t = 0) = 40◦C. SKETCH: Figure Pr.4.28 shows the suspended ﬁrebarrier mat. x
Suspended Fire Barrier
(qr,i)f Protected Space Fire T(t = 0)
k , H , cp L
Figure Pr.4.28 A ﬁre barrier is used to protect a space adjacent to a ﬁre.
OBJECTIVE: (a) Determine the elapsed time t for the thermal penetration using (3.148). (b) Determine the surface temperature T (x = 0, t) at this elapsed time. (c) Using the melting temperature of ZrO2 in Figure 3.8, would the mat disintegrate at this surface? SOLUTION: (a) Using the thermal penetration depth given by (3.148), we have δα α
= L = 3.6(αt)1/2 ,
α=
k ρcp
0.2(W/mK) = 3.333 × 10−7 m2 /s. 600(kg/m3 ) × 1,000(J/kgK)
=
Solving for t, we have t
= =
L2 (0.03)2 (m2 ) = 2 2 (3.6) α (3.6) × 3.333 × 10−7 (m2 /s) 208.3 s. 368
(b) Using Table 3.4, for qs = (qr,i )f , and for x = 0 we have the transient surface temperature given by T (x = 0, t)
= T (t = 0) −
(qr,i )f (4αt)1/2 π 1/2 k
(−105 )(W/m2 ) × [4 × 3.333 × 10−7 (m2 /s) × 208.3(s)]1/2
=
40(◦C) −
=
40(◦C) + 4,701(◦C)
=
4,741◦C.
π 1/2 × 0.2(W/mK)
(c) From Table 3.9, for ZrO2 , we have a melting temperature of Tlg = 2,715◦C. We expect the mat to melt (and sublimated) at this surface. COMMENT: The ﬂame irradiation is readily estimated using (4.62), with the known relevant thermalchemicalphysical properties of the ﬂame. Here we did not include the heat of melting (and sublimation) of the mat materials. Inclusion of these reduces the surface temperature. Also by soaking the mat with water, the temperature is further reduced (due to evaporation energy conversion).
369
PROBLEM 4.29.FAM GIVEN: Using reﬂectors (mirrors) to concentrate solar irradiation allows for obtaining very large (concentrated) irradiation ﬂux. Figure Pr.4.29 shows a parabolic concentrator that results in concentration irradiation ﬂux (qr,i )c , which is related to the geometric parameters through the energy equation applied to solar energy, i.e., (qr,i )s wL = (qr,i )c DL, where D is the diameter, DL is the projected crosssectional area of the receiving tube, and wL is the projected concentrator crosssectional area receiving solar irradiation. The concentrated irradiation is used to produce steam from saturated (at T = Tlg ) water, where the water mass ﬂow rate is M˙ l . The absorptivity of the collector is αr,c and its emissivity r,c is lower (nongray surface). In addition to surface emission, the collector loses heat to the ambient through surface convection and is given as a prescribed Qku . Assume that collector surface temperature is Tc = Tlg . (qr,i )s = 200 W/m2 , Qku = 400 W, Tc = Tlg = 127◦C, αr,c = 0.95, r,c = 0.4, D = 5 cm, w = 3 m, L = 5 m. Use Table C.27 for properties of saturated water. SKETCH: Figure Pr.4.29(a) shows the concentration and the steam producing collector. Solar Irradiation (qr,i)s Water, Ml
Qku L
Parabolic Concentrator (Reflector)
(qr,i)s Coaxial Tube, Tc
w
Control Volume Around Collector
(qr,i)c Water Turning to Steam
Steam, Ml
Se,= + Se, + Slg ∋
Figure Pr.4.29(a) A concentratorsolar collector system used for steam production.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the stream production rate M˙ l . SOLUTION: (a) Figure Pr.4.29(b) shows the thermal circuit diagram. The only surface heat transfer is the surface convection and there are three energy conversion mechanisms (because the surface is a nongray surface, radiation absorption and emission are treated as energy conversions).
Tc
Se,a = DL ar,c (qr,i)c Se, =  p DL r,c sSB Tc4 Slg =  Ml Dhlg
'
Ac
'
Qku Figure Pr.4.29(b) Thermal circuit diagram.
(b) The energy equation from Figure Pr.4.29(b), and similarly from (4.60), becomes QA,c Qku
= S˙ r,α + S˙ r, + S˙ lg = DLαr,c (qr,i )c − πDL r,c σSB Tc4 − M˙ l ∆hlg . 370
Note that for the irradiation we have used the projected area DL and for the emission we have used the surface area πDL. Solving for M˙ l , and using DL(qr,i )c = wL(qr,i )s , we have αr,c wL(qr,i )s − πDL r,c σSB Tc4 − Qku M˙ l = . ∆hlg From Table C.27, at T = (273.15 + 127)(K) = 400.15 K, we have ∆hlg = 2.183 × 106 J/kg. Using the numerical values, we have M˙ l
= =
0.95 × 3(m) × 5(m) × 200(W/m2 ) − π × 0.05(m) × 0.4 × 5(m) × 5.67 × 10−8 (W/m2 K4 ) × (400.15)4 (K)4 − 400(W) 2.183 × 106 (J/kg) (2,850 − 456.8 − 400)(W) = 9.131 × 10−4 kg/s = 0.9131 g/s. 2.183 × 106 (J/kg)
COMMENT: Note that from Figure 4.18, the value we used for (qr,i )s is close to the annual average over the earth surface, i.e., (qr,i )s A = 172.4 W/m2 . The seasonal and daily peaks in (qr,i )s lead to much larger instantaneous stream production rates.
371
PROBLEM 4.30.FAM GIVEN: Pulsed lasers may be used for the ablation of livingcell membrane in order to introduce competent genes, in gene therapy. This is rendered in Figure Pr.4.30(a). The ablation (or scissors) laser beam is focused on the cell membrane using a neodymium yttrium aluminum garnet (Nd:YAG) laser with λ = 532 nm = 0.532 µm, and a focus spot with diameter D = 500 µm. There is a Gaussian distribution of the irradiation across D, but here we assume a uniform distribution. Assume a steadystate heat transfer. Although the intent is to sublimate S˙ sg the targeted membrane region for a controlled depth (to limit material removal to the thin, cell membrane), the irradiation energy is also used in some exothermic chemical reaction S˙ r,c and in some heat losses presented by Q (this includes surface emission). D = 500 nm, L = 10 nm, ρ = 2 × 103 kg/m3 , (qr,i )l = 1010 W/m2 , ∆hsg = 3 × 106 J/kg, S˙ r,c = −7 × 10−4 W, Q = 3 × 10−4 W, αr = 0.9. SKETCH: Figure Pr.4.30(a) shows the ablating membrane.
Laser Ablation of Living Stem Cells Deleting Laser Beam Ablation (Scissor Beam) (qr,i)l
Laser Tweezer (Holder Beam)
Competent Gene
Faulty Gene
Laser Tweezer Organelles Cut Volume: D
Membrane
L 10 mm
Figure Pr.4.30(a) A living cell is ablated at a region on its membrane, for introduction of competent genes.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the required duration of the laser pulse ∆t, for the given conditions. SOLUTION: (a) The thermal circuit diagram is given in Figure Pr.4.30(b). Under steady state, the irradiation energy is used for sublimation, endothermic chemical reaction, and heat losses.
T Se,= + Ssg + Sr,c Q
Figure Pr. 4.30(b) Thermal circuit diagram.
(b) From Figure Pr.4.30(b), and similarly from (4.60), the energy equation for the targeted volume is QA = Q = S˙ e,α + S˙ sg + S˙ r,c 372
or πD2 (qr,i )l − M˙ ∆hsg + S˙ r,c 4 πD2 πD2 L (qr,i )l − ρ ∆hsg + S˙ r,c = αr 4 4∆t
Q = αr
Solving for ∆t, we have 1 ∆t
=
=
= ∆t
πD2 (qr,i )l + S˙ r,c − Q 4 πD2 L ∆hsg ρ 4 π(5 × 10−7 )2 (m2 ) × 1010 (W/m2 ) − 7 × 10−4 (W) − 3 × 10−4 (W) 0.9 × 4 2 × 103 (kg/m3 ) × π × (5 × 10−7 )2 (m2 ) × 10−8 (m) × 3 × 106 (J/kg) 4 1.767 × 10−3 (W) − 7 × 10−4 (W) − 3 × 10−4 (W) 1.178 × 10−11 −8 1.536 × 10 s = 15.36 ns. αr
=
COMMENT: Due to lack of speciﬁc data, we have estimated the heat of sublimation based on a physical bond similar to water.
373
PROBLEM 4.31.FUN GIVEN: When thermal radiation penetrates a semitransparent medium, e.g., a glass plate, reﬂection, absorption, and transmission occur at the interface between two adjacent media along the radiation path, e.g., each of the glass/air interfaces for a glass plate surrounded by air. These multiple absorptions and reﬂections result in an overall attenuation, absorption, and reﬂection of the radiation incident in the glass plate. These eﬀects are modeled as overall transmittance, absorptance, and emittance. The glass plate is then assumed diﬀuse and gray. Consider a glass plate bounded by its two inﬁnite, parallel surfaces, as shown in Figure Pr.4.31. The glass plate has a transmittance τr,2 = 0.1 and an absorptance αr,2 = 0.7. Its temperature T2 is assumed uniform across the thickness. The surfaces are opaque, diﬀuse, and gray with total emissivity r,1 = r,3 = 0.5. SKETCH: The semitransparent layer is shown in Figure Pr.4.31, along with the various radiation ﬂux terms. αr,2(qr,i)2 L
Q3
r,3Eb,3
∋
w
(qr,i)3
ρr,3(qr,i)3
r,3
Opaque Surface 3 , T3
∋
(qr,o)a (qr,o)3 ∋
ρr,2(qr,i)a
(qr,i)a
τr,2(qr,i)b
r,2
, τr,2 , αr,2 Surface a
r,2Eb,2
∋ αr,2(qr,i)a
Semitransparent Medium, T2
Q2 αr,2(qr,i)b
αr,1(qr,i)1
(qr,i)b
r,2
, τr,2 , αr,2
(qr,o)1 ρr,1(qr,i)1
Q1
∋
(qr,i)1
ρr,2(qr,i)b
r,1Eb,1
∋
∋
(qr,o)b
Surface b ∋
τr,2(qr,i)a
r,2Eb,2
r,1
, T1
Opaque Surface 1
Figure Pr.4.31 Radiative heat transfer across a glass plate. The glass is placed between two parallel solid surfaces. OBJECTIVE: (a) Show that the net heat transfer by surface radiation from both sides of the glass plate is given by * + Qr,a 1 [ρr,2 (1 − ζ 2 ) − 1](qr,o )a + ζ(qr,o )b + (1 − ζ) r,2 = 2 Ar,2 ρr,2 (1 − ζ ) + * Qr,b 1 [ρr,2 (1 − ζ 2 ) − 1](qr,o )b + ζ(qr,o )a + (1 − ζ) r,2 , = Ar,2 ρr,2 (1 − ζ 2 ) where ζ = ρr,2 /τr,2 . (b) Write a system of equations that would allow for the solution of the problem and identify the set of variables that are being solved for and the number of variables that need to be known. 374
(c) If there is no other heat transfer from the glass plate, write the energy equation to be solved the glass plate temperature T2 and write the expression for T2 . (d) Compare the results of this problem with the results of Problem Pr.4.31. Is there an analogy between the transmittance of the glass plate and the porosity of the screen? SOLUTION: (a) A radiation balance at any surface gives Qr = qr,o − qr,i . A For surface b of the glass plate, the radiosity (qr,o )b is given by (qr,o )b = τr,2 (qr,i )a + ρr,2 (qr,i )b + r,2 Eb,2 . Analogously, the radiosity for surface a is given by (qr,o )a = τr,2 (qr,i )b + ρr,2 (qr,i )a + r,2 Eb,2 . Solving for the irradiation (qr,i )a , we have (qr,o )a r,2 Eb,2 τr,2 − − (qr,o )b . ρr,2 ρr,2 ρr,2
(qr,i )a =
Substituting this into the equation for (qr,o )b and rearranging, we have τr,2 (qr,o )a + (qr,o )b = ρr,2
2 τr,2 ρr,2 − ρr,2
τr,2 (qr,0 )b + 1 − r,2 Eb,2 . ρr,2
Solving for (qr,o )b , we obtain
(qr,o )b = ρr,2
1
1−
τr,2 τr,2 (qr,o )b − (qr,o )a − 1 − r,2 Eb,2 . 2 ρr,2 ρr,2 τr,2 ρ2r,2
Then, substituting this expression into the expression for Qr,b /A, we ﬁnally obtain 1 Qr,b = {[ρr,2 (1 − ζ 2 ) − 1](qr,o )b + ζ(qr,o )a + (1 − ζ) r,2 Eb,2 } A ρr,2 (1 − ζ 2 ) where, ζ = τr,2 / ρr,2 . By symmetry, for surface a we obtain Qr,a 1 = {[ρr,2 (1 − ζ 2 ) − 1](qr,o )a + ζ(qr,o )b + (1 − ζ) r,2 Eb,2 }. A ρr,2 (1 − ζ 2 ) (b) The solution would require energy equations for all surfaces and relations for radiation exchange between the surfaces. Surface 1: Qr,b Ar,1 Qr,1b Ar,1 −Q1 Eb,1
= =
1 ρr,2
[ r,1 Eb,1 − (1 − ρr,1 )(qr,o )1 ]
(qr,o )1 − (qr,o )b
= Qr,1 = Qr,1b = σSB T14 . 375
Surface 3: Qr,3 Ar,3 Qr,3a Ar,3 −Q3 Eb,3
1
=
ρr,3
[ r,3 Eb,3 − (1 − ρr,3 )(qr,o )3 ]
(qr,o )3 − (qr,o )a
=
= Qr,3 = Qr,3a = σSB T34 .
Surface 2: Qr,b Ar,2 Qr,a Ar,2
= =
1 {[ρr,2 (1 − ζ 2 ) − 1](qr,o )b + ζ(qr,o )a + (1 − ζ) r,2 Eb,2 } ρr,2 (1 − ζ 2 ) 1 {[ρr,2 (1 − ζ 2 ) − 1](qr,o )a + ζ(qr,o )b + (1 − ζ) r,2 Eb,2 } ρr,2 (1 − ζ 2 )
−Q2 Qr,b
= −Qr,3a = −Qr,1b = Qr,b + Qr,a = −Qr,b1
Qr,a Eb,2
= −Qr,a3 = σSB T24 .
Qr,a3 Qr,b1
The unknowns are, T1 , T2 , T3 , Q1 , Q2 , Q3 , Qr,1 , Qr,2 , Qr,3 , Qr,1b , Qr,b1 , Qr,3a , Qr,a3 , Eb,1 , Eb,2 , Eb,3 , (qr,o )1 , (qr,o )a , (qr,o )b , (qr,o )3 There are 20 unknowns and 17 equations. Therefore, 3 unknowns need to be speciﬁed. These could be, for example, the temperatures T1 and T3 and the heat transfer rate Q2 could be set to zero. (c) Using the expressions for (qr,o )a and (qr,o )b in the energy equation for the glass plate, we obtain −
1 Q2 = {2(1 − ζ) r,2 Eb,2 − [1 − ζ − ρr,2 (1 − ζ 2 )][(qr,o )a + (qr,o )b ] Ar,2 ρr,2 (1 − ζ 2 )
For the case when Q2 = 0, after solving the equation above for Eb,2 , we have Eb,2
= =
1 − ρr,2 (1 + ζ) [(qr,o )a + (qr,o )b ] 2 r,2 αr,2 [(qr,o )a + (qr,o )b ], 2 r,2
or T2 =
1 σSB
1/4 αr,2 [(qr,o )a + (qr,o )b ] . 2 r,2
(d) The porosity of the screen a1 in Problem 4.5 behaves like the transmittance of the glass plate τr,2 . Note that, in contrast to Problem 4.5, this problem has been solved assuming that the transmitted fraction of the radiosity of surface 1 becomes part of the radiosity of surface a (the same is used for 3 and b). COMMENT Semitransparent layers are treated as shown in Figure Pr.4.31, by allowing the transmitted radiation to interact with the surroundings. For semitransparent thin ﬁlms, similar relations are derived.
376
PROBLEM 4.32.FUN GIVEN: Semitransparent, ﬁreﬁghting foams (closed cell) have a very low eﬀective conductivity and also absorb radiation. The absorbed heat results in the evaporation of water, which is the main component (97% by weight) of the foam. As long as the foam is present, the temperature of the foam is nearly that of the saturation temperature of water at the gas pressure. A foam covering (i.e., protecting) a substrate while being exposed to a ﬂame of temperature Tf is shown in Figure Pr.4.32. The foam density ρ and its thickness L, both decreases as a results of irradiation and evaporation. However, for the sake of simplicity here we assume constant ρ and L. The absorbed irradiation, characterized by the ﬂame irradiation ﬂux (qr,i )f and by the foam extinction coeﬃcient σex , results in the evaporation of foam. The ﬂame is a propaneair ﬂame with a composition given below. ρ = 30 kg/m3 , σex = a1 ρ, a1 = 3 m2 /kg, L = 10 cm, R = 1 m, Tf = 1,800 K, pCO2 = 0.10 atm, pH2 O = 0.13 atm, s = 10−7 , ρr = 0. Assume no heat losses. SKETCH: Figure Pr.4.32 shows the foam layer, the ﬂame, and the protected substrate. Radius of Flame Region R Tf (qr,i)f
Protective ClosedCell Foam
Hr qr,i
x
Flame Irradiation
Iex Volumetric se,I Radiation Absorption
L
Foam
H
Substrate
sij Volumetric Phase Change
Protected Substrate
Figure Pr.4.32 A ﬁreﬁghting foam layer protecting a substrate from ﬂame irradiation. A closeup of the closedcell foam is also shown.
OBJECTIVE: (a) Write the energy equation for the constantvolume foam layer. (b) Determine the ﬂame irradiation ﬂux (qr,i )f impinging on the foam. (c) Determine the rate of irradiation absorbed into the foam layer S˙ e,σ . Use (2.43) and integrate it over the foam thickness L. (d) Assuming that irradiation heat absorbed results in the foam evaporation, determine the elapsed time for the complete evaporation of the foam, Use (2.25) with ∆hlg being that of water at T = 100◦C. SOLUTION: (a) The energy equation for the constant foam volume is the integralvolume energy equation (2.9). The result for a steadystate conduction, and with no heat loss, is QA = 0 = S˙ e,σ + S˙ lg . From (2.43) and (2.25), we have S˙ e,σ
s˙ e,σ dV = πR2
= V
= πR2
L
s˙ e,σ dx o
L
(qr,i )f (1 − ρr )σex e−σex x dx
o
S˙ lg
M ρV ∆hlg , = −N˙ lg ∆hlg = − ∆hlg = − ∆t ∆t 377
where we have assumed a constant evaporation rate M/∆t, and ∆t is the elapsed time for complete evaporation. (b) The ﬂame irradiation ﬂux is determined from (4.62), i.e., (qr,i )f = r,f σSB Tf4 . The propaneair ﬂame considered has the same condition as those of Example 4.11. From Example 4.11, we have (qr,i ) = 9.52 × 104 W/m2 . (c) The integral over L gives S˙ e,σ
L
= πR2 (qr,i )f (1 − ρr )
σex e−σex dx
o
= −πR2 (qr,i )f (1 − ρr )e−σex L 0 = −πR2 (qr,i )f (1 − ρr )(e−σex L − 1) = πR2 (qr,i )f (1 − ρr )(1 − e−σex L ). Note that when σex L → ∞, all the radiation is absorbed in the foam layer. Using the numerical results, we have " ! 2 3 S˙ e,σ = π × 12 (m2 ) × 9.52 × 104 (W/m2 ) × (1 − 0) × 1 − e−3(m /kg)×30(kg/m )×0.1(m) =
2.989 × 105 (W)(1 − 0.0001234)
=
2.989 × 105 W
∗ This shows that all the irradiation has been absorbed by the foam layer, because σex = σex L = 9.
(d) From the energy equation
ρV ∆hlg = S˙ e,σ ∆t
or ∆t =
ρV ∆hlg ρπR2 L∆hlg = . S˙ e,σ S˙ e,σ
The heat of evaporation is obtained from Table C.4., i.e., water:
∆hlg = 2.256 × 106 J/kg
Table C.4.
Using the numerical values, we have ∆t
=
30(kg/m3 ) × π × 12 (m2 ) × 0.1(m) × 2.256 × 106 (J/kg) = 71.14 s, 2.989 × 105 (W)
COMMENT: The decrease in the foam thickness L and density ρ will inﬂuence the absorption of irradiation. However, as long as σex L > 4, nearly all the radiation is absorbed within the foam layer.
378
PROBLEM 4.33.FAM GIVEN: Flame radiation from a candle can be sensed by the temperature sensors existing under the thin, skin layer of the human hands. The closer the sensor (or say hand) is, the higher irradiation ﬂux qr,i it senses. This is because the irradiation leaving the approximate ﬂame surface (qr,i )f Ar,f , Ar,f = 2πR1 L, is conserved. This is rendered in Figure Pr.4.33. Then, assuming a spherical radiation envelope R2 , we have 2πR1 L(qr,i )f = 4πR22 qr,i
R22 R12 .
for
L = 3.5 cm, R1 = 1 cm, R2 = 10 cm, Tf = 1,100 K, pCO2 = 0.15 atm, pH2 O = 0.18 atm, s = 2 × 10−7 . SKETCH: Figure Pr.4.33 shows the candle ﬂame envelope and a hand sensing it a distance R2 from the center. Ar = 4FR22 qr,i
R1
(qr,i)f
R2 Distance to Observer
Approximate Flame Envelope L Wick
Palm of Hand
Melt Candle (Wax)
Figure Pr.4.33 A candle ﬂame and the sensing of its irradiation at a distance R2 from the ﬂame center line.
OBJECTIVE: (a) Determine the ﬂame irradiation ﬂux (qr,i )f at the ﬂame envelope for the given heavysoot condition. (b) Determine qr,i at r = R2 , using the given relation. SOLUTION: (a) The ﬂame irradiation ﬂux is found from (4.62), i.e., (qr,i )f = r,f σSB Tf4 , where for (4.63), we have for ∆ r = 0, r,f = r,CO2 + r,H2 O + r,soot . The emissivity for the CO2 and H2 O band emissions are determined using the partial pressures and the mean beam length λph . The mean beam length is found from Table 4.4, where for a cylindrical ﬂame we have λph = 1.9R1
Table 4.4.
Then for Tf = 1,100 K, we have pCO2 λph
=
r,CO2 pH2 O λph
= 0.002850 atmm 0.028 Figure 4.20(a) = 0.18 × 1.9 × 0.01 = 0.003420 atmm
r,H2 O r,soot r,f (qr,i )f
0.15(atm) × 1.9 × 0.01(m)
0.014 0.220
Figure 4.20(b) Figure 4.20(c)
= =
0.028 + 0.014 + 0.220 = 0.262 0.220 × 5.67 × 10−8 (W/m2 K4 ) × (1,100)4 (K4 )
=
1.826 × 104 W/m2 . 379
(b) Using the radiation heat ﬂow conservation equation given above, and solving for qr,i (r = R2 ), we have qr,i
= = =
2πR1 L (qr,i )f 4πR22 0.01(m) × 0.035(m) × 1.826 × 104 (W/m2 ) 2 × (0.10)2 (m2 ) 319.6 W/m2 .
COMMENT: The irradiation heat ﬂux qr,i drops as the distance to the hand R2 increases. Therefore, to sense the heat, the hand needs to be brought close to the ﬂame envelope.
380
PROBLEM 4.34.DES GIVEN: New coating technologies employ ultraviolet curable coatings and ultraviolet radiation ovens. The coatings contain monomers and oligomers that cross link to form a solid, cured ﬁlm upon exposure to the ultraviolet radiation. The radiation is produced by a mercury vapor or a gallium UV (ultraviolet) lamp. The intensity of the radiation is selected to suit the type of coating applied, its pigmentation, and its thickness. One advantage of the UVcurable coatings is that a smaller amount of solvent is used and discharged to the atmosphere during curing. In a wood coatingﬁnishing process, the infrared fraction of the emitted radiation is undesirable. The infrared radiation can heat the wood panels to a threshold temperature where the resins leach into the coating before it cures, thus producing an inferior ﬁnish. To prevent this, inclined selective surfaces, which reﬂect the ultraviolet fraction of the radiation, are used. Figure Pr.4.34(i) shows a UV oven. A wood panel, with length L1 = 80 cm, and width w = 1 m, occupies the central part of the oven and a bank of UV lamps, with length L2 = 50 cm and width w = 1 m, are placed on both sides of the workpiece. The top surfaces act as selective reﬂecting surfaces, i.e., absorb the infrared radiation and reﬂect the ultraviolet radiation. They are cooled in their back by a low temperature air ﬂow in order to minimize emission of infrared radiation. The UV lamps emit (S˙ e, /Ar )2 = (qr,o )2 = 7×105 W/m2 which is 95% in the ultraviolet range of the spectrum and 5% in the visible and infrared range of the spectrum. The wood boards have a curing temperature T1 = 400 K and behave as a blackbody surface. The selective surfaces have a temperature T3 = 500 K and the emissivity and reﬂectivity shown in Figure Pr.4.34(ii). SKETCH: Figure Pr.4.34(i) and (ii) show the oven and the selective reﬂector.
(i) Oven, lamps, workpiece, and reflectors
(ii) Selective Reflections
Selective Reflecting Surfaces T3 = 500 K
ρr,3
r
ρr ,
DQkuE3
∋
∋
Air Flow
1.0 r,3
0.5
L3 0 103
(Qr,o)2/2
0.4
103
λ, µm
Bank of UV Lamps L2 L1 Wood Slab T1 = 400 K r,1 = 1
L2
w
∋
Figure Pr.4.34 Ultraviolet irradiation. (i) UV oven. (ii) Selective reﬂections.
OBJECTIVE: (a) Determine the amount of heat transfer by surface convection Qku (W) needed to keep the selective surfaces at T3 = 500 K. (b) Determine the radiation heat transfer in the ultraviolet range Qr,1 (UV) and infrared and visible range Qr,1 (IR + V) reaching the workpiece (surface 1). (c) Determine the maximum allowed temperature for the selective surfaces T3,max , such that the amount of infrared and visible radiation reaching the workpiece is less than 3% of the ultraviolet radiation [i.e., Qr,1 (IR + V)/Qr,1 (UV) < 0.03]. SOLUTION: (a) The integralvolume energy equation (4.66), applied to the reﬂection surface at T3 , gives QA,3 = S˙ 3 . 381
Assuming that the only heat transfer from the surface is by surface convection, QA,3 = Qku,3 = qku,3 Aku,3 . For this node, the energy conversions occur by absorption and emission of thermal radiation, i.e., S˙ 3 = (S˙ e,α )3 + (S˙ e, )3 . The energy conversion by radiation absorption is given by (4.64), i.e., ∞ (αr,λ )3 (qr,λ,i )3 dλ, (S˙ e,α )3 = Ar,3 0
where (qr,λ,i )3 is the irradiation on surface 3 and the integration is done over all the wavelengths λ. The irradiation on surface 3 is given by the spectral form of (4.35), i.e., (qr,i,λ )3 Ar,3 = F13 Ar,1 (qr,0,λ )1 + F23 Ar,2 (qr,0,λ )2 . Dividing both sides by Ar,3 and using the reciprocity rule (4.34), we obtain, (qr,i,λ )3 = F31 (qr,0,λ )1 + F32 (qr,0,λ )2 . The radiosity from surface 1 is (qr,0,λ )1
=
(ρr,λ )1 (qr,i,λ )1 + ( r,λ )1 (Eb,λ,1 )1 .
Substituting into the equation for (qr,i,λ )3 we obtain, (qr,i,λ )3
= F31 [(ρr,λ )1 qr,i,λ + ( r,λ )1 Eb,1,λ ] + F32 (qr,o,λ )2 .
From Figure Pr.4.34(ii), the selective surface has a constant reﬂectivity and emissivity for the wavelength intervals 0 to 0.4 µm and 0.4 µm to very large wavelengths. Then the surface absorption is written as ∞ 0.4 ˙ αr,3 (qr,i,λ )3 dλ + (qr,i,λ )3 dλ. (Se,α )3 = Ar,3,λ 0
0.4
From Figure Pr.4.34(ii), αr,λ,3 = 0 for 0 λ < 0.4 µm and αr,λ,3 = r,λ,3 = 1 for 0.4 µm < λ < ∞. Then, using the equation for (qr,i,λ )3 and (ρr,λ )1 = 1 − ( r,λ )1 = 0, we have ∞ (S˙ e,α )3 = Ar,3 (qr,λ,i )3 dλ 0.4 ∞ ∞ = Ar,3 F31 Eb,1,λ dλ + F32 (qr,o,λ )2 dλ . 0.4
0.4
The fraction of the total blackbody radiation emitted in a wavelength interval λ1 T λ2 T is found from (4.7), i.e., λ1 λ2 Eb,1,λ λdλ − Eb,1,λ dλ 0 Fλ1 T λ2 T = 0 . σSB T 4 Then
∞
Eb,1,λ dλ = (1 − F00.4T )σSB T14 .
0.4
For surface 2, only 5% of the emitted radiation is at wavelengths above 0.4 µm and then ∞ (qr,o,λ )2 dλ = 0.05(qr,o,λ )2 . 0.4
Then (S˙ e,α )3 = Ar,3 [F31 (1 − F00.4T )σSB T14 + F32 (0.05)(qr,o )2 ]. 382
From Table 4.1, we obtain, F00.4T1 = F0160 = 0, as expected. For the geometry of the enclosure, we obtain F13 = F23 = 1. Then, using the reciprocity rule (4.34), we have F 3 1
=
F 3 2
=
Ar,1 F13 =1× Ar,3 Ar,2 F23 =1× Ar,3
0.8(m) = 0.22 3.6(m) 1.0(m) = 0.28. 3.6(m)
Using the numerical values, the radiation absorbed is (S˙ e,α )3
=
3.6(m2 )[0.22 × 5.67 × 10−8 (W/m2 K4 ) × (400)4 (K)4 + 0.28 × 0.05 × 7 × 105 (W/m2 )]
=
36,430 W.
The energy conversion due to radiation emission is found from (4.65), i.e., ∞ ˙ Se,α = −Ar,3 ( r,λ )3 (Eb,λ )3 dλ. 0
Using Figure Pr.4.34(ii), and using the deﬁnition of the fraction of the blackbody emissive power, we have (S˙ e, )3 = −Ar,3 (1 − F00.4T3 )σSB T34 . For T3 = 500 K, from Table 4.3, we have F00.4T3 = 0. Then, (S˙ e, )3
= −3.6(m2 ) × 5.67 × 10−8 (W/m2 K) × (500)4 (K)4 = −12,758 W.
Then, from the energy equation, Qku,3 = (qku Aku )3 = (S˙ e,α )3 + (S˙ e, )3 Solving for qku,3 , we have qku,3 =
36,430(W) − 12,758(W) = 6,576 W/m2 . 3.6(m2 )
(b) The irradiation on surface 1 is given by (qr,i,λ )1 = F13 (qr,o,λ )3 + F12 (qr,o,λ )2 . Since F12 = 0, then from the equation for (qr,o,λ )3 , we have (qr,i,λ )1 = F13 [(ρr,λ )3 (qr,i,λ )3 + ( r,λ )3 (Eb,λ )3 ]. The radiation leaving the surface in the infrared and visible ranges of the spectrum is ∞ (qr,i,λ )1 dλ. Qr,1 (IR + V ) = Ar,1 0.4
Using Figure Pr.4.34(ii), we then have Qr,1 (IR + V )
= Ar,1 F13 σSB T34 = 0.8(m2 ) × 1 × 5.67 × 10−8 (W/m2 K4 ) × (500)4 (K4 ) =
2,835 W.
The radiation heat transfer reaching surface 1 in the ultraviolet range is ∞ (qr,i,λ )1 dλ. Qr,1 (U V ) = Ar,1 0.4
383
Again, using the equation for (qr,i )1 and Figure Pr.4.34(ii), we have ∞ (qr,i,λ )3 dλ. Qr,1 (U V ) = Ar,1 F13 ρr,3 0.4
The fraction reﬂected by surface 3 is
∞
(qr,i,λ )3 dλ = 0.95 × (qr,o )2 F32 .
0.4
Then Qr,1 (U V ) = Ar,1 F13 ρr,3 × 0.95 × (qr,o )2 F32 . From the numerical values, we have Qr,1 (U V )
=
0.8(m2 ) × 1 × 0.95 × 7 × 105 (W/m2 ) × 0.28
=
148,960 W.
(c) For a fraction of IR and V radiation equal to 3 % of the radiation in the UV, we have σSB T3,4 max Qr,1 (IR + V ) = = 0.03 Qr,1 (U V ) Qr,1 (U V ) Solving for T3 , T3,max = [
0.03 × 148,960(W) 1/4 = 529.9 K. ] 5.67 × 10−8 (W/m2 K)
COMMENT: Note that Qku,3 = 2.367 × 104 W, while Qr,1 (U V ) = 1.490 × 105 W. This shows that a large fraction of energy emitted by the lamp arrives on the workpiece.
384
PROBLEM 4.35.DES GIVEN: Consider the eﬃciency of a solar collector with and without a glass cover plate. In a simple model for a solar collector, all the surfaces around the collector, participating in the radiation exchange, are included by deﬁning a single average environment temperature Ta . A solar collector and the various heat ﬂows considered are shown in Figure Pr.4.35. The irradiation from the sun has a magnitude (qr,i )s = 800 W/m2 and the irradiation from the atmosphere (diﬀuse irradiation) is given by (qr,i )a = σSB Ta4 , where Ta = 290 K is the eﬀective atmospheric temperature. The cover plate (made of low iron glass) has a total transmittance for the solar irradiation of τr,2 = 0.79. For the infrared emission, the glass surface can be considered diﬀuse and gray with an emissivity r,2 = 0.9. The absorber plate (coated with the matte black) is opaque, diﬀuse, and gray and has absorptivity αr,1 = 0.95. The glass plate is at a temperature T2 = 310 K and heat losses by surfaceconvection heat transfer occur at the rate of 400 W/m2 . The interior of the solar collector is evacuated and the bottom heat losses by conduction are at a rate of qk,1a = 40 W/m2 . The collector surface is rectangular with dimensions w = 1 m and L = 2 m. For the cases (i) with a glass cover, and (ii) with no glass cover, determine the following. (Use T1 = 340 K.) SKETCH: Figure Pr.4.35 shows the collector with the cover plate. (qr,i)a (W/m2) 2
(qr,i)s (W/m ) DQkuE2a (W)
w
Environment, Ta
L . (Se, )2 (W) ∋
τr,2 (qr,i)a+s (W/m ) 2
Cover Glass Plate T2
Coolant Flow DQuE0 (W)
Insulation Box
Vacuum
DQuEL (W)
Absorber Plate T1
Water Qk,1a (W)
Figure Pr.4.35 A solar collector with a cover glass plate and the associated heat transfer terms.
OBJECTIVE: For the cases (i) with a glass cover, and (ii) with no glass cover, determine the following. (a) Determine the amount of heat transferred to the coolant Qu L0 (W). (b) Determine the thermal eﬃciency η, deﬁned as the heat transferred to the coolant divided by the total irradiation. SOLUTION: (i) With a Glass Cover: (a) From the assumptions above, the integralvolume energy equation (2.9) applied to the collector gives QA,1 = S˙ 1 , where QA,1 = Qu L − Qu 0 + Qku 2a + Qk,1a and S˙ 1 = (S˙ e,α )1 + (S˙ e, )2 . 385
The energy conversion by radiation absorption is due to absorption of solar radiation at the absorber plate (mostly UV and V) and absorption of radiation from the environment at the cover plate (IR and V). Then, noting that only a fraction τr,2 of the solar radiation is transmitted through the cover plate, we have (S˙ e,α )1 = τr,2 αr,1 (qr,i )s Ar,1 + αr,2 (qr,i )a Ar,2 . Using the numerical values and using (4.18), αr,2 = r,2 , Ar,1 = Ar,2 = w × L and (qr,i )a = σSB Ta4 . Then (S˙ e,α )1
=
0.79 × 0.95 × 800(W/m2 ) × 1(m) × 2(m) + 0.9 × 5.67 × 10−8 (W/m2 K4 ) × 2904 (K4 ) × 1(m) × 2(m)
=
(1,200.9 + 721.8)(W) = 1,922.7 W.
The radiation emitted by the absorber plate is reabsorbed internally or at the cover plate. The cover plate emits radiation to the surroundings which is given by (S˙ e, )2 = −Ar,2 r,2 σ SBT24 . Using the values given, = −1(m) × 2(m) × 0.9 × 5.67 × 10−8 (W/m2 K4 ) × 3104 (K4 )
(S˙ e, )2
= −942.5 W. Then, solving for the net convective heat transfer to the ﬂuid, we have Qu L0
≡
Qu L − Qu 0 = (S˙ e,α )1 + (S˙ e, )2 − Qku 2a − Qk,2a
= =
1,922.7(W) − 942.5(W) − [200(W/m2 ) + 40(W/m2 )] × 1(m) × 2(m) 500.2W
(b) The thermal eﬃciency of the collector is η
= =
Qu L − Qu 0 500.2(W) = −8 2 (qr,i )s + (qr,i )a 800(W/m ) + 5.67 × 10 (W/m2 K4 ) × 2904 (K4 ) × 1(m) × 2(m) 0.3122 = 31.22%.
(ii) Without a Glass Cover: (a) The radiation absorbed is (S˙ e,α )1 = Ar,1 αr,1 [(qr,i )s + (qr,i )a ]. From the numerical values, we have (S˙ e,α )1
=
1(m) × 2(m) × 0.95[800(W/m2 ) + 5.67 × 10−8 (W/m2 K4 ) × 2904 (K4 )]
=
2,282(W).
The radiation emission from the absorber plate at T1 is (S˙ e, )1
= −Ar,1 r,1 σSB T14 = −1(m) × 2(m) × 0.95 × 5.67 × 10−8 (W/m2 K4 ) × 3404 (K)4 = −1,439.6 W.
Then the integralvolume energy equation (2.9) gives, where we now have Qku 1a as surface 2 has been removed, Qu L0
= Qu L − Qu 0 = (S˙ e,α )1 + (S˙ e, )1 − Qku 1a − Qk,1a = 2,282(W) − 1,439.6(W) − 1(m) × 2(m) × [200(W/m2 ) + 40(W/m2 )] =
362 W.
(b) The thermal eﬃciency is then η=
Qu,L − Qu,0 362(W) = 22.60%. = (qr,i )s + (qr,i )a 1,602(W) 386
COMMENT: Both the surface convection and conduction heat losses are diﬀerent when a cover plate is used. The surfaceconvection heat loss from the highertemperature absorber plate is larger than that from the cover plate. This will be discussed in Chapter 6. With the cover plate, the temperature of the absorber increases, and the conduction heat losses also increase. The use of an additional cover plate (as in a doubleglazed ﬂat plate solar collector), increases the thermal eﬃciency even more.
387
PROBLEM 4.36.FUN GIVEN: Cirrus clouds are the thin, high clouds (usually above 6 km) in the form of trails or streaks composed of “delicate white ﬁlaments, or tenuous white patches and narrow bands.” Due to the atmospheric conditions at these high altitudes (low temperature and high relative humidity), these clouds contain large amounts of ice crystals. An important eﬀect of these clouds in the atmosphere is the absorption and emission of thermal radiation, which plays an important role in the upper troposphere water and the heat budget. This may signiﬁcantly aﬀect the earth’s climate and the atmospheric circulation. A cirrus uncinus cloud (a hooklike cloud appearing at an altitude between 5 to 15 km, and indicating a slowly approaching storm) has an average thickness L. The extinction coeﬃcient for the cloud is σex = 2.2 × 10−3 1/m. The intensity of the irradiation at the top of the cloud is (qr,i )s = 830 W/m2 , as shown in Figure Pr.4.36. Assume that the surface reﬂectivity of the cloud is zero. Assume that the volumeaveraged density and the speciﬁc heat of the cloud are ρ = 0.8 kg/m3 and cp = 2,000 J/kgK. SKETCH: Figure Pr.4.36 shows the irradiation heating cloud layer. (qr,i)s (W/m2)
Cloud x L
σe,x Earth's Surface
Figure Pr.4.36 Irradiation heating of a cloud layer.
OBJECTIVE: (a) Determine the amount of radiation absorbed by the cloud S˙ e,α /A, for L = 1,000 m. (b) The lengthaveraged cloud temperature T L is deﬁned as 1 T L = L
L
T dx. 0
Determine the time variation of the lengthaveraged cloud temperature dT L /dt for the cloud thickness L = 1,000 m. Neglect the heat transfer by surface convection and the energy conversion due to phase change (i.e., the air is in equilibrium with the cloud). SOLUTION: (a) From Table C.1(d), the local volumetric absorption is S˙ e,σ = (1 − ρr )(qr,i )s σex e−σex x . V Integrating this over length L along the x direction, we have L ˙ L S˙ e,σ Se,σ = dx = (1 − ρr )(qr,i )s σex e−σex x = (1 − ρr )(qr,i )s (1 − e−σex L ). A v 0 0 For zero surface reﬂectivity, ρr = 0, we have S˙ r,α = (qr,i )s (1 − e−σex L ). A 388
Using the numerical values, we have S˙ e,σ A
=
−3 830(W/m2 ) 1 − e−2.2 × 10 (1/m)1,000(m) = 738 W/m2 .
Therefore, 89% of the incident radiation is absorbed by the cloud. (b) For a diﬀerentialvolume at a distance x along the cloud is ∇ · q = −ρcp
dT + s˙ e,σ . dt
From Table C.1(d), the volumetric absorption of irradiation at the depth x is s˙ e,σ =
S˙ e,σ = (1 − ρr )qr,i σx e−σx x = (qr,i )s σex e−σex x . V
Neglecting any other heat transfer from this cloud, we have 0 = −ρcp
dT + (qr,i )s σx e−σx x . dt
Integrating this energy equation over the cloud thickness L and dividing the result by L, we have d 1 L 1 L −σx x T dx + (qr,i )s σx e dx 0 = −ρcp dt L 0 L 0 = −ρcp
d (qr,i )s T L + 1 − e−σx L . dt L
Thus, the time variation of T L is dT L (qr,i )s = 1 − e−σx L . dt ρcp L From the numerical values given, we have dT L dt
=
−3 830(W/m2 ) (1/m) × 1,000(m) −2.2 × 10 × 1−e 0.8(kg/m3 ) × 2,000(J/kgK) × 1,000(m)
=
4.6 × 10−4 K/s = 40 K/day.
COMMENT: Note the relation between the energy equations used in (a) and (b). The lengthaveraged temperature is the temperature used in the integral form of the energy equation, for which the energy conversion in (a) applies. The energy absorption by the cloud is by the water droplets, mostly by the ice particles. This energy aﬀects the growth of the ice particles which in turn aﬀects the air temperature and air circulation within and underneath the cloud. This may have important implications on the upper atmosphere circulation and the earth surface energy budget. The eﬀect of radiation absorption on the growth rate of ice particles is explored in Problem 4.37.
389
PROBLEM 4.37.FUN GIVEN: Consider a cirrus cloud with an average thickness L = 1 km and at an altitude of r = 8 km from the earth’s surface. The top of the cloud is exposed to the deep space and receives solar radiation at the rate (qr,i )s = 1,353 W/m2 . The bottom of the cloud is exposed to the earth’s surface. The extinction coeﬃcient for the cloud is σex = 2.2 × 10−3 1/m. This is shown in Figure Pr.4.37(a). (i) A spherical ice particle at the top of the cloud has a temperature of T1 = −35◦C, and a diameter d = 100 µm. The ice surface is opaque, diﬀuse, and gray with an emissivity r,1 = 1. The particle is moving under the eﬀect of the draft air currents and has no preferred orientation relative to the solar irradiation. The deep sky behaves as a blackbody with a temperature of T3 = 3 K. (ii) Another ice particle at the bottom of the cloud has the same temperature, dimensions, and surface radiation properties, but it is exposed to the earth’s surface. The earth’s surface is opaque, diﬀuse, and gray, has a surfaceaveraged temperature T2 = 297 K, and a total emissivity r,2 = 0.9. SKETCH: Figure Pr.4.37(a) shows a cloud layer with ice particles located (i) at top and (ii) at the bottom of the cloud layer undergoing radiation heat transfer. (qr,i)s (W/m2)
Deep Sky T3 = 3 K r,3 = 1 ∋
T1 = 238 K r,1 = 1 (i) Ice Particle on Top ∋
L = 1 km x r
Cloud
Earth's Surface
(ii) Ice Particle on Bottom T2 = 297 K r,2 = 0.9 ∋
Figure Pr.4.37(a) Iceparticle heat transfer for a particle (i) at the top, and (ii) at the bottom of a cloud layer.
OBJECTIVE: For each of these two particles perform the following analyses. (a) Track the heat transfer vector and show the energy conversions. Note that the particles lose heat by surface convection, and that phase change (frosting or sublimation) also occurs. (b) Draw the thermal circuit diagram for the particles. Assuming that the cloud is optically thick (i.e., there is a signiﬁcant attenuation of the radiation across the cloud) the radiation heat transfer between two positions 1 and 2 in the cloud with emissive powers Eb,1 and Eb,2 is given by Qr,12 =
(Eb,1 − Eb,2 ) . 3σex /4Ar
This resistance can be added in series with the surfacegrayness and viewfactor resistances between two surfaces enclosing the cloud. (c) Determine the net heat transfer for each particle. (d) Neglecting the surfaceconvection heat transfer and assuming a steadystate condition, determine the rate of growth by frosting (or by sublimation) of the ice particles. For the heat of sublimation use ∆hsg = 2.843 × 106 J/kg and for the density of ice use ρs = 913 kg/m3 . (e) From the above results, are these particles expected to grow or decay in size? Would the radiation cooling or heating rate diﬀer for particles of a diﬀerent size? SOLUTION: (a) The heat transfer vector tracking and the energy conversion terms are shown in Figure Pr.4.37(b).
390
Exposed to Deep Sky qr
T3 qu
qr,i
qku . . Ssg+ Se,α
qr qk
 dE dt V Ice Particle T2
qr
x
Exposed to Earth Surface
Figure Pr.4.37(b) Track of the heat ﬂux vector around an ice particle in a cloud layer.
(b) The thermal circuit diagram for heat transfer from a particle at the bottom of the cloud layer is shown in Figure Pr.4.37(c).
Qr,2
. S1
Qr,21 (qr,o)2
Qr,3
Qr,13
(qr,o)1
(qr,o)1
(qr,o)3
Q2
Q3 (Rr, )1
Eb,1 T1 Eb,1 (Rr, )1
(Rr,F)21
(Rr,F)13 (Rr,σex)13
(Rr, )3 Eb,3 T3
T2 Eb,2 (Rr, )2
Figure Pr.4.37(c) Thermal circuit diagram for an ice particle at the bottom of the cloud layer.
(c)(i) For an ice particle at the top of the cloud, the integralvolume energy equation (2.9) energy equation is QA,1 = −(ρcp V )1
dT1 + S˙ 1 . dt
The energy conversion is due to radiation absorption and solidgas phase change (sublimation), i.e., S˙ 1 = S˙ e,α + S˙ sg . The energy conversion due to radiation absorption is S˙ e,α = αr,1 (qr,i )s A1 . The net heat transfer at the particle surface is QA,1 = Qku + Qr,13 + Qr,12 where Qr,13 and Qr,12 are the surfaceradiation heat transfer between particle and sky and particle and earth. The surface radiation heat transfer between the particle and sky is given by Qr,13 =
σSB (T14 − T34 ) (Rr,Σ )13 391
where, 1 − r,1 1 1 − r,3 + + . r,1 A1 F13 A1 r,3 A3
(Rr,Σ )13 =
Considering that A3 A1 and F13 =1 we have, (Rr,Σ )13 =
1 A1 r,1
The surfaceradiation heat transfer between the particle and the earth is aﬀected by the presence of the cloud, which absorbs and scatters radiation. Assuming that the diﬀusion approximation for the radiation ﬂux within the cloud applies (5.64), this surface radiation heat transfer is given by Qr,12 =
σSB (T14 − T24 ) , (Rr,Σ )12
where the overall radiation resistance, and using the resistance for the volumetric absorption of radiation, is given by (Rr,Σ )12 =
1 − r,1 1 1 − r,2 3σex L + + + . r,1 A1 F12 A1 r,2 A2 4A1
Considering that A2 A1 and F12 = 0.5 we have (Rr,Σ )12 =
1 A1
1 r,1
+1+
3σex L 4
.
Therefore, the net radiation heat transfer and absorption for the ice particle at the top of the cloud is Qr,1,t = −αr,1 (qr,i )s A1 +
σSB (T14 − T34 ) σ (T 4 − T24 ) + SB 1 1 1 1 3σex L +1+ A1 r,1 A1 r,1 4
or Qr,1,t σSB (T14 − T34 ) σSB (T14 − T24 ) . = −αr,1 (qr,i )s + + 1 3σex L 1 A1 +1 + r,1 4 r,1 Using the numerical values, we have Qr,1,t A1
5.67 × 10−8 (W/m2 K4 ) × (238.154 − 34 )(K4 ) + 1 5.67 × 10−8 (W/m2 K4 ) × (238.154 − 2974 )(K4 ) 1 3 × 2.2 × 10−3 (1/m) × 1,000(m) + +1 1 4 2 2 = −1,353(W/m ) + 182(W/m ) − 71(W/m2 ) = −1,242 W/m2 .
= −1 × 1,353(W/m2 ) +
The negative sign means that the particle is being heated by radiation. (ii) For the particle at the bottom of the cloud, the radiation absorption is S˙ e,α = αr,1 qr,i A1 . The irradiation is attenuated by volumetric absorption, as discussed in Section 2.3.2(E). The irradiation ﬂux at the bottom of the cloud is given by (2.42), i.e., qr,i = (qr,i )s (1 − ρr,1 )e−σex L . 392
With r,1 = 1 (i.e., ρr,1 = 0), we have qr,i = (qr,i )s e−σex L . Then, as given in Table C.1(d) S˙ e,α = αr,1 (qr,i )s σex e−σex L A1 . The net surfaceradiation heat transfer, in analogy to the particle at the top, is Qr,12
=
Qr,13
=
σSB (T14 − T24 ) 1 A1 r,1 1 A1
σSB (T14 − T34 ) . 1 3σex L +1+ r,1 4
Therefore, the net radiation heat transfer is, σSB (T14 − T24 ) σSB (T14 − T34 ) Qr,1,b = −αr,1 (qr,i )s σex e−σex L + + 1 3σex L 1 A1 +1+ r,1 4 r,1 Using the numerical value, we have Qr,1,b A1
−3 = −1 × 1,353(W/m2 ) × 2.2 × 10−3 (1/m) × e−2.2 × 10 (1/m) × 1,000(m)
5.67 × 10−8 (W/m2 K4 ) × (238.154 − 2974 )(K4 ) 1 5.67 × 10−8 (W/m2 K4 ) × (238.154 − 34 )(K4 ) + 3 × 2.2 × 10−3 (1/m) × 1,000(m) 2+ 4 = (−0.33 − 258.8 + 49.8)( W/m2 ) = −209 W/m2 . +
Note that the particle at the bottom of the cloud is also being heated, but at a much smaller rate. (d) For a steadystate condition, the integralvolume energy equation (2.9) becomes QA1 = S˙ 1 . Neglecting surface convection and using the results calculated in item (b), we have Qr,1 = S˙ sg The energy conversion by sublimation (Table 2.1) is given by S˙ sg = −M˙ sg ∆hsg A1 The rate of sublimation is related to the rate of growth by dV1 M˙ sg = −ρs dt Then 1 dV1 Qr,1 qr,1 = = A1 dt A1 ρs ∆hsg ρs ∆hsg 393
For the particle at the top 1 dV1 A1 dt
=
−1,242(W/m2 ) 913(kg/m3 ) × 2.843 × 106 (J/kg)
= −4.78 × 10−7 m/s = −0.478 µm/s. For the particle at the bottom 1 dV1 A1 dt
=
−209(W/m2 ) 913(kg/m3 ) × 2.843 × 106 (J/kg)
= −8.05 × 10−8 m/s = −0.0805 µm/s. (e) Both particles are expected to decrease in size. The particle at the top would disappear faster than the particle at the bottom. Larger particles would take longer to sublimate due to their larger volume. COMMENT: The surfaceconvection heat transfer from the particle surface inﬂuences the particle growth (or decay) rates.
394
PROBLEM 4.38.DES GIVEN: A ﬂatplate solar collector is modeled as a surface with equivalent total absorptance αr,1 and emittance r,1 , which represent the surface and wavelength average of the absorptivity and emissivity of all the internal and external surfaces (it also accounts for the transmissivity of the cover plate). The solar collector and the model are shown in Figure Pr.4.38. The solar collector receives solar irradiation (qr,i )s = 800 W/m2 and atmospheric irradiation (qr,i )a = σSB Ta4 , where Ta = 290 K is the eﬀective atmospheric temperature (it accounts for the emission and scattering of the radiation by the atmosphere). This radiation is incident on the collector plate surface, which has area L×w = 1×2(m2 ). The collector surface temperature is T1 = 310 K and it loses heat by surfaceconvection heat transfer at a rate qku 1∞ = 400 W/m2 and also by surface radiation emitted to the surroundings (S˙ e, )1 . The bottom of the collector loses heat by conduction at a rate qk,1∞ = 50 W/m2 . The total absorptance of the solar collector is αr,1 = 0.9 and the total emittance is r,1 = 0.3. SKETCH: Figure Pr.4.38 shows the solar collector and the various heat transfer from the collector plate.
(i)
(ii) Solar Irradiation qr,i
(qr,i)a (W/m2)
Qr,1 (W)
Qku (W) τr,2,s = 0.79 r,2 = 0.9
2
(qr,i)s (W/m )
L=2m
∋
Air Flow (Wind)
DQuE0
w=1m
DQuEL
Qk,1 (W)
Water Inlet, DQuE0
DQuEL0 (W)
Water Outlet, DQuEL
Figure Pr.4.38(i) and (ii) A solar collector and the collector plate heat transfer.
OBJECTIVE: (a) Determine the amount of heat transferred to the ﬂuid Qu L0 = Qu L − Qu 0 (W). (b) Determine the thermal eﬃciency η, deﬁned as the ratio of the heat transferred to the ﬂuid to the total irradiation. SOLUTION: (a) The integralvolume energy equation (2.9) applied to the collector gives QA,1 = S˙ 1 , where QA,1 = qku,1∞ Aku + qk,1∞ Ak + Qu L,0 and S˙ 1 = (S˙ e,α )1 + (S˙ e, )1 . The radiation absorbed is (noting that r,1 = αr,1 ) (S˙ e,α )1
= Ar,1 [(αr,1 )s (qr,i )s + (αr,1 )V +IR σSB Ta4 ] = 1(m) × 2(m) × [0.90 × 800(W/m2 ) + 0.3 × 5.67 × 10−8 (W/m2 K4 ) × (290)4 (K)4 ] =
1,680 W. 395
The radiation emitted is (S˙ e, )1
= −Ar,1 (αr,1 )V +IR σSB T14 = −1(m) × 2(m) × 0.3 × 5.67 × 10−8 (W/m2 K4 ) × (310)4 (K)4 = −314.18 W.
Then the convection heat transfer is Qu L0
= (Qu,L − Qu,0 ) = (S˙ e,α )1 + (S˙ e, )1 − qku,1∞ Aku − qk,1∞ Ak = 1,680.62(W) − 314.18(W) − 1(m) × 2(m) × [400(W/m2 ) + 50(W/m2 )] = 466.4 W.
(b) The thermal eﬃciency is deﬁned as η
=
Qu L0 Ar,1 [(qr,i )s + σSB Ta4 ]
=
466.4(W) = 0.1942. 1(m) × 2(m) × [800(W/m ) + 5.67 × 10−8 (W/m2 K4 ) × 2904 (K4 )] 2
The eﬃciency in converting total solar irradiation to sensible heat of the water stream is 19.42%. COMMENT: Note that the eﬀective temperature of the collector T1 is smaller than the water temperature. This temperature is basically an average temperature for the cover plate, which is the emitting surface.
396
PROBLEM 4.39.DES GIVEN: Table Pr.4.39 gives a short list of materials used as selective surface coatings, i.e., coatings that have diﬀerent absorption properties for shorter and longer wavelength ranges. Other materials are listed in Table C.19. For each of the two applications below, choose from Table Pr.4.39 the coating that results in the optimum performance. Table Pr.4.39 Spectral absorptivity and emissivity properties for some selective coatings. Coating αr,λ (0.3 ≤ λ ≤ 3 µm) r,λ (3 ≤ λ ≤ 50 µm) black, chrome electrodeposited copper oxide aluminum hardanodized Teﬂon
0.95 0.87 0.03 0.12
0.15 0.15 0.80 0.85
Figure Pr.4.39(i) shows a solar collector. The absorber plate is exposed directly to the solar and atmospheric irradiation (qr,i )a , as shown in Figure Pr.4.39(i). The intensity of the solar radiation is (qr,i )s = 800 W/m2 . The net radiation emitted by the atmosphere is given by (qr,i )a = σSB Ta4 , where Ta = 290 K is the eﬀective atmospheric temperature (i.e., the “sky” temperature). The absorber plate has a surface L × w = 1 × 2 m2 and it is at a temperature of T1 = 80◦C. The collector is insulated from below and the surfaceconvection heat transfer is neglected. Choose a material from Table Pr.4.39 that would result in the maximum convection heat removal Qu L0 (W) (i.e., maximum heating of the ﬂuid) from the collector and determine the heat ﬂow rate. Figure Pr.4.39(ii) shows a radiative cooler. A satellite in orbit around the earth uses radiation cooling to reject heat [Figure Pr.4.39(ii)]. As the satellite rotates, it is temporarily exposed to the sun. The intensity of the solar irradiation is (qr,i )s = 1,353 W/m2 . The plate has an area L × w = 50 × 50 cm2 and its temperature is T1 = 250 K. The deep sky temperature is Tsky = 3 K. Choose a material that would give the maximum heat transfer Qr,1 (W) and determine this heat ﬂow rate. SKETCH: Figures Pr.4.39(i) and (ii) show the two applications of selective radiation coatings.
(i)
(qr,i)s (W/m2)
(qr,i)a (W/m2)
(ii)
2 (qr,i)s (W/m2) (qr,i)sky (W/m )
(Sr, )1
w
L
DQuE0 (W)
w Q1 (W)
DQuEL (W)
Insulation
Absorber Plate T1 = 80 OC = 353.15 K αr,s , αIR
L (Sr, )1
Radiative Cooler T1 = 250 K αr,s , αIR
Figure Pr.4.39(i) A solar collector. (ii) A radiative cooler. Both use selective radiation absorption.
OBJECTIVE: (a) For the solar collector in Figure Pr.4.39(i), choose a material from Table Pr.4.39 that would result in the maximum convection heat removal Qu L0 (W) (i.e., maximum heating of the ﬂuid) from the collector and determine the heat ﬂow rate. (b) For the radiative cooler in Figure Pr.4.39(ii), choose a material that would give the maximum heat transfer Qr,1 (W) and determine this heat ﬂow rate.
397
SOLUTION: (a) For a ﬂat plate solar collector we need a large absorptivity in the wavelength range characteristic of the solar irradiation and a small emissivity in the wavelength characteristic of radiation emitted at low temperatures. Both the black, chrome electrodeposited and the copperoxide coatings would perform well under these conditions. To calculate the net convection heat transfer Qu L0 , we use the integralvolume energy equation, Qt,1 = S˙ 1 where QA,1
= Qu,L − Qu,0 ≡ Qu L0
and S˙ 1 = (S˙ e,α )1 + (S˙ e, )1 The energy conversion due to the absorption of irradiation is given by (4.64), i.e., ∞ (S˙ e,α )1 = Ar,1 (αr,λ )1 (qr,i,λ )1 dλ. 0
We assume that the surfaces are gray in the wavelength range shown in Table Pr.4.39 and that solar irradiation occurs at the short wavelength and atmospheric irradiation occurs at the long wavelength range. Then we have (using r,λ = αr,λ ), S˙ e,α = Ar,1 [αr,λ (0.3 ≤ λ < 3 µm)(qr,i )s + r,λ (3 ≤ λ < 50 µm)σSB Ta4 ]. Using the values given for black, chrome electrodeposited coating, we have = 1(m) × 2(m)[0.95 × 800(W/m2 ) + 0.15 × 5.67 × 10−8 (W/m2 K4 ) × (290)4 (K)4 ] = 1,640 W.
(S˙ e,α )1
The emitted radiation is emitted in the larger wavelength range, and we have (S˙ e, )1
= −Ar,1 r,λ (3 ≤ λ < 50 µm)σSB T14 = −1(m) × 2(m) × 0.15 × 5.67 × 10−8 (W/m2 K4 ) × (353.15)4 (K)4 = −264 W.
Then the convection heat transfer is Qu L0
= =
(S˙ e,α )1 + (S˙ e, )1 1,640(W) − 264(W) = 1,376 W.
(b) In order to reject the maximum amount of heat possible, the radiative cooler should have a large emissivity in the large wavelength range and a small absorptivity at short wavelength ranges. Both the hardanodized aluminum and the Teﬂon would perform well under these conditions. The net radiation heat transfer is due to absorption and emission of radiation. Then the integralvolume energy equation is QA,1 = (S˙ e,α )1 + (S˙ e, )1 . The radiation absorption is (S˙ e,α )1
4 = Ar,1 [αr,λ (0.3 ≤ λ < 3 µm)(qr,i )s + r,λ (3 ≤ λ < 50 µm)σSB Tsky ]
=
0.5(m) × 0.5(m) × [0.03 × 1,353(W/m2 ) + 0.80 × 5.67 × 10−8 (W/m2 K4 ) × (3)4 (K)4 ]
=
10.15 W.
The radiation emitted is (S˙ e, )1
= −Ar,1 r,λ (3 ≤ λ < 50 µm)σSB T14 = −0.5(m) × 0.5(m) × 0.80 × 5.67 × 10−8 (W/m2 K4 ) × (250)4 (K)4 = −44.30 W. 398
Then Q1 = QA,1
= =
(S˙ e,α )1 + (S˙ e, )1 10.15 − 44.30(W) = −34.15 W.
COMMENT: Note that for (b), the negative QA,1 indicates that heat must be provided to surface A1 . This heat can be provided by sensible heat, i.e., cooling down, of the satellite. Other selective surfaces are listed in Table C.19.
399
PROBLEM 4.40.DES.S GIVEN: The high temperature of the automobile exhaust can be used to promote catalytic reactions and conversions (this is called a closecoupled converter) of some gaseous pollutants (such as unburned hydrocarbons) over catalytic metaloxide surfaces. When the converter is placed close to, but downstream of an internal combustion engine, the exhaust pipe leading to the converter is insulated. One scenario is the addition of a radiation shield around of the outside of the exhaust pipe. This is shown in Figure Pr.4.40(a). The gap between the pipe and the shield contains air, and heat transfer occurs by conduction and by surface radiation. The external surface of the shield exchanges heat with its surroundings by surface radiation and surface convection. The surfaceconvection heat loss of the external surface of the shield is estimated as qku = 4,000 W/m2 . The surroundings behave as a blackbody at T3 = 300 K. The exhaust pipe has an outside diameter D1 = 5 cm, a surface temperature of T1 = 800 K, and its surface is diﬀuse, opaque, and gray with a surface emissivity of r,1 = 0.7. The shield has an inside diameter D2,i = 5.4 cm and an outside diameter D2,o = 5.5 cm and is made of chromium coated carbon steel AISI 1042. Its surface is opaque, diﬀuse, and gray and has a surface emissivity r,2 = 0.1. For the thermal conductivity of air use ka = 0.04 W/mK. SKETCH: Figure Pr.4.40(a) shows the exhaust pipe, the radiation shield, and the surroundings of the shield.
Ambient Surface: T3 = 300 K r,3 = 1 ∋
qku (W/m2)
qr (W/m2)
qr (W/m2)
Air
qk (W/m2)
qu (W/m2)
D1
Exhaust Pipe
D2,i D2,o
Shield 0.1
T1 = 800 K r,1 = 0.7 ∋
r,2 =
∋
Figure Pr.4.40(a) Insulation of an automobile exhaust pipe.
OBJECTIVE: (a) Determine the net heat transfer from the exhaust pipe to the ambient for a L = 1 m long pipe. (b) Comment on the eﬀect of the pipe wall conduction thermal resistance on the total heat transfer from the exhaust pipe. (c) Keeping all the other conditions the same, would the heat transfer from the exhaust pipe increase or decrease with an increase of the inside diameter of the shield (while keeping the thickness constant)? (Suggestion: Plot the variation of Q1 for 5.2 cm ≤ D2,i ≤ 7 cm.) SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.4.40(b) under a steadystate condition and with no energy generation present. Applying the integralvolume energy equation to nodes T1 , T2,i , T2,o and T3 , we have
Q1 = Qk,12 + Qr,12 = Qku + Qr,23 = Q3 . 400
Qr,1 T1 Eb,1
Qr,12
Qr,2,i
(qr,0)1
(Qk)2,i2,o Eb,2,i T2,i
(qr,0)2,i
Qr,23
Qr,2,o (qr,0)2,o
T2,o Eb,2,o
Qr,3 Eb,3 T3
(qr,0)3
Q3
Q1 (Rr, )2,i
(Rk)2,i2,o
(Rr, )3
(Rr,F)12
(Rr,F)23
(Rr, )3
(Rr, )1
Qk,12
Rk,12
Figure Pr.4.40(b) Thermal circuit diagram.
The conduction, surface radiation and surface convection heat transfer rates are given by Qk,12
=
Qr,12
=
Qku
T1 − T2,i Rk,12
σSB (T1 4 − T2,i 4 ) (Rr,Σ )12 = Aku,3 qku
Qr,23
=
(Qk )2,i2,o
=
4 − T3 4 ) σSB (T2,i (Rr,Σ )23 T2,i − T2,o . (Rk )2,i2,o
The conduction thermal resistance through the air gap is Rk,12 =
ln[2.7(cm)/2.5(cm)] ln(R2,i /R1 ) = = 0.3062◦C/W. 2πka L 2π × 0.04(W/mK) × 1(m)
The overall radiation thermal resistance through the air gap is (Rr,Σ )12 =
1 − r,1 1 1 − r,2 + + . Ar,1 r,1 Ar,1 F12 Ar,2 r,2
Using F12 =1 and the areas = πD1 L = π × 5 × 10−2 (m) × 1(m) = 0.157 m2 = πD2 L = π × 5.4 × 10−2 (m) × 1(m) = 0.170 m2 ,
Ar,1 Ar,2 we have (Rr,Σ )12 =
1 − 0.7 1 1 − 0.1 + + = 62.04 1/m2 . 0.157(m2 ) × 0.7 0.157(m2 ) × 1 0.170(m2 ) × 0.1
The conduction thermal resistance through the pipe wall is (using ks for steel from Table C.16), (Rk )2,i2,o =
◦ ln[2.75(cm)/2.7(cm)] ln(R2,o /R2,i ) = = 5.84 × 10−5 C/W 2πks L 2π × 50(W/mK) × 1(m)
The overall radiation resistance from the pipe surface to the surrounding is (Rr,Σ )23 =
1 − r,2 1 1 − r,3 + + Ar,2,o r,2 Ar,2,o F23 Ar,3 r,3
Using F23 =1, Ar,3 Ar,2,o and Ar,2,o = πD2,o L = π × 5.5 × 10−2 (m) × 1(m) = 0.173 m2 , 401
we have (Rr,Σ )23 =
1 − 0.1 1 + = 57.80 1/m2 . 2 0.173(m ) × 0.1 0.173(m2 )
Noting that (Rk )2,i2,o Rk,12 , (Rr,Σ )12 , and (Rr,Σ )23 , we can assume that T2,i = T2,o T2 . Then the energy equation for node T2 becomes QA,2 = Qk,12 + (Qr,Σ )12 − Qku + (Qr,Σ )23 = 0. Using the equation for the heat transfer rates, we have T1 − T 2 σSB (T14 − T24 ) σSB (T24 − T34 ) + − Qku − = 0. Rk,12 (Rr,Σ )12 (Rr,σ )23 This is a fourth order polynomial equation on T2 . Solving for T2 using a solver (such as SOPHT), we obtain T2 = 619.7 K. The heat transfer rate Q1 is then given by Q1 =
T1 − T2 σSB (T14 − T24 ) + = 827.7 W. Rk,12 (Rr,Σ )12
(b) The conduction resistance through the pipe wall, if taken into account, would reduce the heat transfer rate Q1 . However, here, due to the relatively high thermal conductivity of steel and the small thickness of the pipe wall, its eﬀect on Q1 is negligible. (c) Increasing the diameter of the shield D2,i increases the conduction resistance through the air gap, but also decreases the radiation resistance, and increases the surface convection. The combined result of these eﬀects is shown in Figure Pr.4.40(c). Note that there is a minimum in the heat transfer rate Q1 for D2,i = 5.8 cm, for which Q1 = 808 W. 1,200
1,100
Q1 , W
1,000
900
800
700 0.05
Minimum Heat Transfer Rate from Surface 1 0.06
0.07
0.08
0.09
0.10
D2,i , m Figure Pr.4.40(c) Variation of the heat loss with respect to the shielded diameter D2,i .
COMMENT: Here, the dependency of the surfaceconvection heat transfer rate on the external diameter D2,o through qku is neglected. This will be discussed in Chapter 6. Also, with the increase in the gap size, between the shield and the pipe, there is an increase in the importance of the surfaceconvection heat transfer in the gap (when compared to the conduction heat transfer). This thermobuoyant ﬂow and heat transfer is explained in Chapters 6. When the conduction resistance across the pipe wall is not neglected, we obtain T2,i = 619.758 K, T2,o = 619.709 K and Q2,i2,o = 827.708 W. Note the negligible diﬀerence between these results and the results obtained in part (a). 402
PROBLEM 4.41.DES GIVEN: During continuous thermal processing of silicon wafers, to heat the wafers to a desired temperature, the wafers are stacked vertically and moved through an evacuated, cylindrical radiation oven at speed uw . This is shown in Figure Pr.4.41(i). Consider a unit cell formed by two adjacent wafers. This is shown in Figure Pr.4.41(ii). The wafers have a diameter D1 = 100 mm, the distance separating the wafers is l = 30 mm, and the wafers are placed coaxially to the oven wall, which is also cylindrical. The diameter of the ceramic oven is D2 = 300 mm. The oven surface is opaque, diﬀuse, and gray with surface emissivity r,2 = 0.9 and surface temperature T2 = 800 K. The wafers surface is also diﬀuse, opaque, and gray with emissivity r,1 = 0.01. SKETCH: Figure Pr.4.41(a) shows the waferfurnace.
(i) Furnace
(ii) uw
Axis of Symmetry
Wafers
Unit Cell Qr,12
l
D1
∋
D2
Oven Wall T2 = 800 K r,2 = 0.9 ∋
Wafer T1 = 500 K r,1 = 0.01
Figure Pr.4.41(a)(i) Silicon wafers moving and heating in a radiation oven. (ii) The unit cell formed by two adjacent wagers is also shown.
OBJECTIVE: (a) Assuming that the wafers have a uniform temperature T1 = 500 K, determine the net heat transfer by surface radiation between the oven and a wafer Qr,1 . (b) During this process, the wafers enter the oven at an initial temperature lower than 500 K. As they move through the oven, they are heated by surface radiation from the oven walls until they reach a near steadystate temperature. Assuming that the axial variation of temperature (along the thickness) is negligible, quantitatively sketch the radial distribution of the wafer temperature at several elapsed times (as the wafer moves through the oven). (c) Apply a combined integraldiﬀerential length energy equation to a wafer. Use the integral length along the thickness and a diﬀerential length along the radius. Express the surfaceradiation heat transfer in terms of a diﬀerential radiation resistance, which depends on a diﬀerential view factor. SOLUTION: (a) The surface of a wafer exchanges radiation heat transfer with the other wafers facing it and with the oven surface. As the wafer has a uniform temperature T1 , the surface of two adjacent wafers, which are at the same temperature T1 , and the oven form a twosurface enclosure, as shown in Figure Pr.4.41(a)(ii). Then, the net surface radiation heat transfer is given by (4.47), i.e., Qr,12 =
σSB (T14 − T24 ) , (Rr,Σ )12
where the overall radiation resistance is (Rr,Σ )12
=
1 − r,1 1 1 − r,2 + + . Ar,1 r,1 Ar,1 F12 Ar,2 r,2
The view factor F12 is obtained from Figure 4.11(a) and by using the summation rule. From Figure 4.11(a), using R1 = 100(mm)/30(mm) = 3.33, and R2 = 3.33, we obtain F11 = 0.72. Then using 403
the summation rule, we have F12 = 1 − F11 = 1 − 0.72 = 0.28. The areas are 2 × π × (100 × 10−3 )2 (m2 ) = 0.0157 m2 4 = π × 300 × 10−3 (m) × 30 × 10−3 (m) = 0.0283 m2 .
2πD12 4 = πD2 l
=
Ar,1 = Ar,2
Then, the overall radiation resistance becomes (Rr,Σ )12
=
1 − 0.01 1 1 − 0.9 + + 0.01 × 0.0157(m2 ) 0.28 × 0.0157(m2 ) 0.9 × 0.0283(m2 ) 6,302.54(1/m2 ) + 227.36(1/m2 ) + 3.93(1/m2 )
=
6,537 1/m2 .
=
The net heat transfer by surface radiation is then 5.67 × 10−8 (W/m2 K4 )(5004 − 8004 )(K4 ) 6,537(1/m2 ) = −3.008 W.
Qr,12
=
(b) Figure Pr.4.41(b) shows the qualitative, radial temperature distribution of wafer temperature at several elapsed times. Also shown is the expected steadystate temperature distribution. (c) The integraldiﬀerential length analysis for a wafer with thickness w gives (q · sn )dA qr 2π[(r + ∆r)2 − r2 ] −qk,r 2πrw + qk,r+∆r 2π(r + ∆r)w = lim + lim ∆A ∆A→0 ∆A→0 ∆V π[(r + ∆r)2 − r2 ]w π[(r + ∆r)2 − r2 ]w 2qr −qk,r 2r + qk,r+∆r 2(r + ∆r) = lim + . ∆A→0 w 2r∆r + ∆r2 t
= 1
1.0
0.8 Time increasing 0.6 T1 − T1(t = 0) T2 − T1(t = 0) 0.4
0.2 t = 0 s 0 0
0.2
0.4
0.6
0.8
1.0
r/R1
Figure Pr.4.41(b) Qualitative radial distribution of the wafer temperature, at several elapsed times.
Considering that for small ∆r∆r2 2r∆r, we have
(q · sn )dA lim
∆A→0
∆A
∆V
= lim
∆A→0
2qr + w
−rqk,r + rqk,r+∆r + ∆rqr,r+∆r r∆r
404
.
The second term can be rewritten as −rqk,r + rqk,r+∆r + ∆rqr,r+∆r lim = ∆A→0 r∆r
qk,r+∆r − qk,r qk,r+∆r + ∆A→0 ∆r r qk 1 ∂ ∂qk + = (rqk ). = ∂r r r ∂r
lim
Thus, the integraldiﬀerential length energy equation becomes 2qr 1 ∂ ∂T + (rqk ) = −ρcp . w r ∂r ∂t The heat ﬂux by radiation is qr = lim
∆A→0
Qr,12 , ∆Ar
where Qr,12 =
σSB (T 4 − T24 ) (∆Rr,Σ )12
and (∆Rr,Σ )12 =
1 − r,1 1 1 − r,2 + + , ∆Ar r,1 ∆Ar F∆Ar 2 Ar,2 r,2
where F∆Ar 2 is the view factor between the diﬀerential area ∆Ar and surface 2. Then qr
=
=
lim
∆A→0
1 − r,1 r,1
σSB (T 4 − T24 ) 1 − r,2 ∆Ar 1 + + F∆Ar 2 r,2 Ar,2
σSB (T 4 − T24 ) . 1 − r,1 1 + r,1 F∆Ar ,2
The view factor F∆Ar 2 is the view factor from a cylindrical shell (ring), with radial length ∆r, to the oven surface. This is given in reference [9] as a function of geometric parameters. COMMENT: The radiation heat ﬂow rate into the wafer Qr,i is rather small, but the mass of a silicon wafer is not very large. Therefore, speedy heatup is possible. An increase in l will increase Qr,1 .
405
PROBLEM 4.42.FAM GIVEN: A gridded silicon electric heater is used in a microelectromechanical device, as shown in Figure Pr.4.42. The heater has an electrical resistance Re and a voltage ∆ϕ is applied resulting in the Joule heating. For testing purposes, the heater is raised to a steadystate, high temperature (i.e., glowing red). The gridded heater is connected to a substrate through four posts (made of silicon oxide, for low conductivity kp ), resulting in conduction heat loss through four support posts. The substrate is at Ts and has an emissivity r,s . The upper heater surface is exposed to large surface area surroundings at Tsurr . Treat the heater as having a continuous surface (i.e., solid, not gridded), with a uniform temperature T1 . Ts = 400◦C, Tsurr = 25◦C, w = 0.5 mm, a = 0.01 mm, l = 0.01 mm, r,1 = 0.8, r,s = 1, Re = 1,000 ohm, ∆ϕ = 5 V, kp = 2 W/mK. You do not need to use tables or ﬁgures for the view factors. Use (2.28) for S˙ e,J . SKETCH: Figure Pr.4.42(a) shows the heater and the supporting posts.
\\
Asurr >> A1
Silicon T1 , r,1 ∋
\
\ \ \ \ \ \ \ \ \ \ \ Tsurr \
Se,J
()
w
(+) w l
a
a
kp Silicon Oxide ∋
Ts ,
Substrate r,s
Figure Pr.4.42(a) A miniature gridded heater connected to a substrate by four support posts and raised to a glowing temperature.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the heater temperature T1 . SOLUTION: (a) The thermal circuit diagram is shown in Figure 4.42(b). Heat transfer from the heater is by surface radiation to the surroundings and the substrate, and by conduction through the posts. (b) The energy equation, from Figure Pr.4.41(b), is Qr,1surr + Qr,1s + Qk,1s = S˙ e,J . From (2.28), we have ∆ϕ2 . S˙ e,J = Re The view factors between the upper surface and surroundings is unity F1surr = 1. The view factor between the lower surface and the substrate F1s is also assumed unity, using w l in Figure 4.11(b). The surface radiation for unity view factor and Ar,surr A1 , and for r,s = 1, are given by (4.49), i.e., Qr,1surr Qr,1s
4 = Ar,1 r,1 σSB (T14 − Tsurr )
= Ar,1 r,1 σSB (T14 − Ts4 ), Ar,1 = w2 . 406
Tsurr Qr,surr
Eb,surr (Rr,S)surr
Eb,
Se,J =
Eb,
(Rr,S)s Qr,s
Dj2 Re
T Rk,s
Qk,s
Eb, Ts
Ts
Figure Pr.4.42(b) Thermal circuit diagram.
The conduction resistance is found from Table 3.2, i.e., Qk,1s =
T1 − T2 4Ap kp (T1 − Ts ), Ap = a2 . = Rk,1s l
The energy equation becomes 4 w2 r,1 σSB (2T14 − Tsurr − Ts4 ) +
4a2 kp ∆ϕ2 (T1 − Ts ) = l Re
or (5 × 10−4 )2 (m2 ) × 0.8 × 5.67 × 10−8 (W/m2 K4 ) × [2T14 − (298.15)4 (K4 ) − (673.15)4 (K4 )] + 52 (V 2 ) 4 × (10−5 )2 (m2 ) × 2(W/mK) − 673.15)(K) = × (T 1 1,000(ohm) 10−5 (m) or 1.134 × 10−14 × (2T14 − 7.902 × 109 − 2.053 × 1011 ) + 8.000 × 10−5 × (T1 − 673.15) = 0.025. Solving for T1 , we have T1 = 860.5 K. COMMENT: Note that the dull red is identiﬁed as the Draper point in Figure 4.2(a) with T = 798 K. In practice, the heater is a gridded silicon with a polysilicon coating. Also note that since w/l = 50, the F1s , from Figure 4.11(b), is unity.
407
PROBLEM 4.43.FAM GIVEN: A spherical cryogenic (hydrogen) liquid tank has a thin (negligible thickness), doublewall structure with the gap space ﬁlled with air, as shown in Figure Pr.4.43. The air pressure is one atm. r,1 = 0.05, r,2 = 0.05, R1 = 1 m, R2 = 1.01 m, T1 = −240◦C, T2 = −80◦C. Use (3.19) for low and moderate gas pressure gases to determine any pressure dependence of the gas conductivity. Use Table C.22 for the atmospheric pressure properties of air and use T = 150 K. SKETCH: Figure Pr.4.43(a) shows the tank, walls, and the air gap. Vent
T1
Metallic Walls T2
Q1 = Q21 qr
R2
Liquid Hydrogen R1
qk
r,1
Air at Pressure p(atm) r,2
∋
∋
Figure Pr.4.43(a) A cryogenic (liquid hydrogen) tank has a doublewall structure with the space between the thin walls ﬁlled with atmospheric or subatmospheric pressure air.
OBJECTIVE: (a) Draw the thermal circuit diagram for heat ﬂow between the outer and inner walls. (b) Determine the rate of heat transfer to the tank Q21 . (c) Would Q21 change if the air pressure is reduced to 1/10 atm? How about under ideal vacuum (p = 0, k = 0)? SOLUTION: (a) Figure Pr.4.43(b) shows the thermal circuit diagram. The heat ﬂows by conduction and radiation from surface 2 to surface 1.
Qk,21
Rk,21 T1
T2
Q1
Q1 = Q21 Eb,1 Qr,21
Eb,2 Rr,5
Figure Pr.4.43(b) Thermal circuit diagram.
(b) From Figure Pr.4.43(b), we have Q21 =
T2 − T1 Eb,2 − Eb,1 + , Rk,21 Rr,Σ
where from Table 3.3, we have
Rk,21 = Rk,12 =
1 1 − R1 R2
408
/4πk,
and for a twosurface enclosure of the gap space we have from (4.47), 1 − r 1 − r 1 + + . Rr,Σ = Ar r 1 Ar,1 F12 Ar r 2 Here F12 = 1 and Ar,1 = 4πR12 and A2 = 4πR22 . From Table C.22, for air at T = 150 K, we have k = 0.0158 W/mK Then
Rk,21
= =
Rr,Σ
= =
1 1 − 1 1.01
Table C.22
/4π × 0.0158(W/mK)
4.986 × 10−2 K/W 1 − 0.05 1 1 − 0.05 + + 2 2 2 2 4π × (1) (m ) × 0.05 4π × (1) (m ) 4π × (1.01)2 (m2 ) × 0.05 (1.512 + 0.07957 + 1.482)(1/m2 ) = 3.074 1/m2 .
Then Q21
= =
(193.15 − 33.15)(K) 5.67 × 10−8 (W/m2 K4 )[(193.15)4 − (33.15)4 ](K4 ) + 3.074(1/m2 ) 4.986 × 10−2 (K/W) 3,209(W) + 25.65(W) = 3,235 W.
(c) From (3.19), we note that there is no pressure dependence of k for the monatomic gases at low and moderate pressures. This is also true for the diatomic gas mixtures such as air. Then Q21 = 3,235 W p = 0.1 atm. For p = 0 which gives k = 0, we have radiation heat transfer only, and Q21 = 25.65 W
p = 0.
COMMENT: Air is made of oxygen and nitrogen and their condensation temperature (boiling point) at p = 1 atm is given in Table C.4 as Tlg = 90.0 K and Tlg = 197.6 K, respectively. Then it becomes necessary to evacuate the gap space in order to avoid condensate formation which collects in the gap, at the bottom of the tank under gravity. In practice, the outer surface is insulated to lower the surface temperature much below −80◦C. Note the small contribution due to radiation (due to the small emissivities). Also as the gas pressure drops, the possibility of gas molecules colliding with the walls becomes greater than that for the intermolecular collisions. Then the size (gap between the walls) should be included. This is left as an end of Chapter 3 problem and includes the Knudsen number KnL = λm /L as a parameter, where λm is the molecular meanfree path given by (1.19) and L = R2 − R1 .
409
PROBLEM 4.44.FUN GIVEN: In surface radiation through multiple, opaque layer systems, such as the one shown in Figure Pr.4.44, the rate of conduction through the layers can be signiﬁcant. To include the eﬀect of the layer conductivity, and also the layer spacing indicated by porosity, use the approximation that the local radiation heat transfer is determined by the local temperature gradient, i.e., qr,x = −kr
dT T2 − T1 = −kr , dx l1 + l 2
kr = kr (ks , , r , T2 , l2 ),
where kr is the radiant conductivity. SKETCH: Figure Pr.4.44 shows a multiple, ﬁnite thickness l1 and conductivity ks parallel layer (opaque) system and its thermal circuit model.
(i) Multiple, Opaque, FiniteConductivity Layers (Slabs) T1' ,
∋
Solid with Diffuse, Gray, Opaque Surface and FiniteThickness and Conductivity
ks T1
r
T2 ,
∋
Gap: kf = 0
(ii) Thermal Circuit Model Using Radiant Conductivity kr Qr,x
r
a T1
Ar
Rkr
a x
l2
l1
T1'
Rk
dT qr,x =  kr dx
Rr,5
T2
l1 + l2 , Ar = a2 Ar kr
Figure Pr.4.44(i) Surfaceradiation and solid conduction through a system of parallel slabs. (ii) Thermal circuit diagram.
OBJECTIVE: (a) Using the thermal circuit diagram representing Qr,x in Figure Pr.4.44(ii), start from (4.47) and use F1 2 = 1 for l2 a, and then use (4.72) to linearize T14 − T24 and arrive at
qr,x =
4 r σSB T 3 (T1 − T2 ) . 2 − r
(b) Then add Rk and use T1 − T2 to arrive at the radiant conductivity expression kr kr =
1 , 1− (2 − r ) + ks 4 r σSB T 3 l2
=
l2 l1 + l 2
or kr 1 = 4σSB T 3 l2 4σSB T 3 l2 (1 − ) (2 − r ) + ks r or kr r Nr−1 (1 − )−1 = , 3 4σSB T l1 r + Nr−1 (2 − r )
Nr =
4σSB T 3 l1 . ks
SOLUTION: (a) Starting from (4.47) and assuming F1 2 = 1 [from Figure 4.11(b)], we have Eb,1 − Eb,2 1 qr,x = qr,1 2 = 1 1 − r Ar 1 − r + + Ar r Ar Ar r 4 4 4 σSB (T1 − T2 ) r σSB (T1 − T24 ) = . = 2 2 − r −1 r 410
Now, from (4.72), using T14 − T24
=
(T12 + T22 )(T12 − T22 )
= =
(T12 + T22 )(T1 + T2 )(T1 − T2 ) 4T 3 (T1 − T2 ),
where we have deﬁned 4T 3 ≡ (T12 + T22 )(T1 + T2 ) as the average temperature for the case of T1 → T2 → T . Then qr,x =
4 r σSB T 3 (T1 − T2 ) . 2 − r
(b) From Table 3.2, we use the conduction resistance to write, using Figure Pr.4.44(i), qr,x =
1 T 1 − T 1 T 1 − T 1 = . Ar Rk l1 /ks
Now, noting that the conduction and radiation resistance are in series, we have qr,x
T1 − T2 T1 − T2 = l1 2 − r Ar (Rk + Rr,Σ ) + ks 4 r σSB T 3 T2 − T1 1 = − l1 (2 − r )l2 l1 + l 2 + (l1 + l2 )ks 4 r σSB T 3 l2 (l1 + l2 ) T2 − T1 1 l2 = − , = 1− (2 − r ) l1 + l2 l1 + l 2 + ks 4 r σSB T 3 l2 T2 − T1 ≡ −kr l1 + l 2 =
or kr =
1 . 1− (2 − r ) + ks 4 r σSB T 3 l2
or kr 1 = 4σSB T 3 l2 4σSB T 3 l2 (1 − ) (2 − r ) + ks r Now using the conductionradiation number Nr deﬁned by (4.75), we have kr r Nr−1 (1 − )−1 = . 4σSB T 3 l1 r + Nr−1 (2 − r ) Note that using l instead of l2 gives kr 1 = . 3 3 4σSB T l1 4σSB T l1 (1 − ) (2 − r )(1 − ) + ks r COMMENT: Note that for ks → ∞, i.e., an ideally conducting solid, we have kr =
4 r σSB T 3 l2 , (2 − r )
ks → ∞.
This shows an increase in kr as the fraction of high conductivity solid (ks → ∞) increases (i.e., decreases). Note that when r → 0, then there is no heat transfer (inﬁnite radiation resistance), because heat has to be transferred by radiation between surfaces in order to be conducted across the layer (here zero conductivity is assumed for the ﬂuid occupying the space between the surfaces). 411
PROBLEM 4.45.FUN.S GIVEN: The measured eﬀective thermal conductivity of porous solids, such as that for packed zirconium oxide ﬁbers given in Figure 3.13(b), does include the radiation contribution. The theoretical prediction can treat the conduction and radiation heat transfer separately. In a prediction model (derivation for cubic particles is left as an end of the chapter problem), the eﬀective, combined (total) conductivity for a periodic porous solid is given by kkr kr
= kk + kr =
4DσSB T 3
k
= kf
ks kf
r (1 − )1/3 Nr−1 , r + Nr−1 (2 − r )
Nr =
4σSB T 3 D ks
0.280−0.757 log +0.057 log(ks /kf )
Using these, we can compare the predicted and measured results. Here is the porosity, D is the ﬁberdiameter, r is the ﬁber emissivity, kf is the ﬂuid, and ks is the solid conductivity. D = 10 µm, ρ = 1,120 kg/m3 . Use ρ = ρf + (1 − )ρs = (1 − )ρs , for ρf ρs , to determine . For air use kf = 0.0267(W/mK) + 5.786 × 10−5 (W/mK4 ) × (T − 300)(K), and use the only data available for zirconium oxide ks and ρs in Table C.17. For emissivity use r = 0.9 − 5.714 × 10−4 (T − 300)(K) based on Table C.18 for zirconium oxide. OBJECTIVE: (a) Plot the variation of kkr with respect to T for 300 K ≤ T ≤ 1,000 K, for the zirconium oxide and air system. (b) Compare the results with the experimental results of Figure 3.13(b). (c) Is radiation contribution signiﬁcant in this material? SOLUTION: (a) From Table C.17, we have for zirconia ρs = 5,680 kg/m3 ,
ks = 1.675 W/mK
Table C.17 for the porosity.
Using ρ = ρf + (1 − )ρf and noting that ρf (from Table C.22) is much smaller than ρs , we have 1− =
1,120(kg/m3 ) ρ = = 0.1972 ρs 5,680(kg/m3 )
or = 0.8028. Using a solverplotter, the results for kkr versus T is plotted in Figure Pr.4.45. Note the rather linear increase with respect to T . This is due to the assumed linear increase in kf with T . (b) Using the experimental results plotted in Figure 3.13(b), we choose T = 400◦C = 673.15 K and this experimental result is also shown in Figure Pr.4.45. The agreement is rather good. (c) The radiation conductivity kr is not large, kr = 0.0004399 W/mK at T = 1,000 K, due to the small particle diameter D. The lower emissivity at higher temperatures also makes kr small. COMMENT: In Figure 3.13(b), the results for higher and lower ρ do not agree as well with the predictions, however, in all of these data, the role of radiation is not signiﬁcant in the temperature range considered, because D is small.
412
∋
kkr = kk + kr , W/mK
0.20
0.15
Experiment
0.10
0.05 300
= 0.8028
Prediction
400
500
600
700
800
900
1,000
T, K Figure Pr.4.45 Variation of the total thermal conductivity of zirconium oxide air packed bed of ﬁbers, with respect to temperature.
413
PROBLEM 4.46.FAM.S GIVEN: A spherical carbon steel AISI 1010 piece of diameter D = 1 cm, initially at T1 (t = 0) = 1,273 K, is cooled by surface radiation to a completely enclosing cubic oven made of white refractory brick with each side having a length L = 10 cm and a surface temperature T2 = 300 K. Assume that all the surfaces are opaque, diﬀuse, and gray. For the carbon steel sphere use the higher value for the emissivity of oxidized iron, listed in Table C.18. OBJECTIVE: (a) Using a software, plot the variation of the piece temperature with respect to time. (b) Determine the time it takes for the piece to reach T1 = 600 K and compare this result (i.e., the numerical solution) with the one predicted by (4.82) (i.e., the analytic solution). SOLUTION: (a) Figure Pr.4.46 shows the variation in the workpiece temperature with respect to time, along with the associated computer code from SOPHT. (b) For the carbon steel sphere (node T1 ), we have (from Tables C.16 and C.18, and the given geometry), ρ1 = 7,830 kg/m3 , cp,1 = 434 J/kgK, r,1 = 0.89, V1 = πD3 /6 = 5.236 × 10−7 m3 , A1 = πD2 = 3.141 × 10−4 m2 . For the white refractory brick walls (node T2 ), we have (from Table C.18, and the given geometry), r,1 = 0.29 (for T = 1,373 K), A1 = 6(L × L) = 6(0.1 m × 0.1 m) = 0.06 m2 , and using (4.82) gives T2 + T1 (t = 0) T2 + T1 σSB T23 1 −1 T1 −1 T1 (t = 0) + 2tan − ln t= − 2tan , ln Rr,Σ (ρcp V )1 4 T2 − T1 T2 − T1 (t = 0) T2 T2 where T1 (= 0) = 1,273 K, T2 = 300 K, and T1 = T1 (t) = 600 K. The radiation resistance between the sphere and the walls is given by (4.48) as Rr,Σ
=
(Rr, )1 + (Rr,F )12 + (Rr, )2 1 − r 1 1 − r = + , + r A 1 A1 F12 r A 2
where F12 = 1. Substituting into (4.82) and solving for t, we give t = 162 s. COMMENT: The numerical solution for this relatively simple problem, is very accurate. Figure Pr.4.46 gives t = 161.7 s.
414
1,280
T ,K
1,120 960 800 (161.7s, 600K)
640 480 0
40
80
120 t ,s
160
200
// Problem 4.46 // Note that the initial condition on T1 will be speciﬁed in the Solve window // Volumetric Transient Node, 1 T1’ = dT1dt // Must be ﬁrst equation in the set QA1 = rho1*cp1*V1*dT1dt + S1dot // Conservation of energy, W QA1 = Qr1 // Summation of heat transfer leaving node, W rho1 = 7830 // Density of medium, kg/mˆ3 cp1 = 434 // Speciﬁc heat, J/kgK d1 = 0.01 // sphere diameter, m V1 = pi*d1ˆ3/6 // Volume, mˆ3 S1dot = 0 // Energy conversion, W // Surface radiation for node 1 Qr1 = Qr12 // Surface radiation heat transfer, W // Net radiation heat transfer between surface 1 & 2 Qr12 = (Eb1Eb2)/Rrs12 // Radiation heat transfer between surfaces, W Eb1 = sigmaSB*T1ˆ4 // Emissive power of node 1, W/mˆ2 Eb2 = sigmaSB*T2ˆ4 // Emissive power of node 2, W/mˆ2 T2 = 300 // Temperature of node 2, K // Twosurface radiation resistance Rrs12 = (1epsilonr1)/(Ar1*epsilonr1)+1/(Ar1*F12)+(1epsilonr2)/(Ar2*epsilonr2) // equivalent radiation resistance, 1/mˆ2 epsilonr1 = 0.89 // Surface 1 emissivity epsilonr2 = 0.29 // Surface 2 emissivity Ar1 = pi*d1ˆ2 // Surface 1 area, mˆ2 l2 = 0.1 // cube side length, m Ar2 = 6*l2ˆ2 // Surface 2 area, mˆ2 F12 = 1 // View factor sigmaSB = 5.67e8 // StefanBoltzmann constant Figure Pr.4.46 Variation of workpiece temperature with respect to time and the computer code from SOPHT.
415
PROBLEM 4.47.FUN GIVEN: An idealized bed of solid particles is shown in Figure Pr.4.47. The heat is transferred through the bed by surface radiation and the cubic solid particles have a ﬁnite conductivity ks , while conduction through the ﬂuid is neglected (kf = 0). We use the radiant conductivity kr deﬁned through dT T2 − T1 = −kr , dx l1 + l 2 = kr (ks , , r , T, l2 ).
= −kr
qr,x kr
SKETCH: Figure Pr.4.47(a) shows the bed and a simpliﬁed thermal circuit model.
(i) Surface Radiation and Solid Conduction in a Bed of Cubic Particles ∋
l1 Solid with Diffuse, Gray, Opaque Surface and Finitel1 Dimensions and Conductivity
T1' , r T2 ,
T1
ks
∋
Gap: kf = 0
(ii) Thermal Circuit Model Using Radiant Conductivity kr
r
Qr,x a T1
T2 T1'
Rk a l1
Rkr
l2
Rr,Σ
l1 + l2 , Ar = a2 Ar kr
Figure Pr.4.47(a)(i) A bed of cubical particles with surface radiation and solid conduction. (ii) Simpliﬁed thermal circuit diagram.
OBJECTIVE: (a) For the thermal circuit model shown in Figure Pr.4.47(ii), determine the radiation resistance Rr,Σ between two adjacent surfaces 1 and 2. Assume that F1 2 = 1 and use the linearization given in (4.72). (b) Add the conduction resistance using series resistances to arrive at kr 4σSB T l2 (1 − ) 3
2/3
1
=
4σSB T l2 (1 − ) (2 − r ) 1/3 + ks r 3 3 l1 l2 1− = (l1 + l2 )3 (l1 + l2 )3 3
= kr 4σSB T 3 l1
1/3
r Nr−1 (1 − )−1 , r + Nr−1 (2 − r )
=
Nr =
4σSB T 3 l1 . ks
SOLUTION: (a) Starting from (4.47), with F1 2 = 1 and all surfaces having the same emissivity r , we have Qr,1 2
=
1 − r Ar (1 − )
2/3
r
+
Eb,1 − Eb,2 1 Ar (1 − )
Ar (1 − ) (Eb,1 − Eb,2 ) 2(1 − r ) +1 r Ar (1 − )2/3 (Eb,1 − Eb,2 ) 2 −2+1 r Ar (1 − )2/3 σSB (T14 − T24 ) , 2 − r r
2/3
+
1 − r Ar (1 − )2/3 r
2/3
=
=
=
Ar = a × a,
416
=1−
l13 , (l1 + l3 )
(1 − )2/3 =
l12 , (l1 + l2 )3
where (1 − )2/3 is the solid area fraction (as compared to 1 − which is solid volume fraction). Now similar to (4.72), we linearize this for T1 → T2 → T , i.e.,
(T14 − T24 ) 4T 3
=
(T12 + T22 )(T12 − T22 )
= (T12 + T22 )(T1 + T2 )(T1 − T2 ) = 4T 3 (T1 − T2 ) ≡ (T12 + T22 )(T1 + T2 )
where T is an average temperature. Then
Qr,1 2
Ar (1 − )2/3 4σSB T 3 (T1 − T2 ) 2 − r r T 1 − T 2 . 2 − r 1 r 4σSB Ar T 3 (1 − )2/3
=
=
(b) Now adding the conduction heat transfer (Table 3.2) as a series resistance, we have
Qr,12
qr,12
=
=
≡
T1 − T2 1
l1
2 − r r
+ Ar (1 − )2/3 ks Ar (1 − )2/3 4σSB T 3 Qr,12 T1 − T2 = 2 − r l1 1 Ar + r (1 − )2/3 ks (1 − )2/3 4σSB T 3 T2 − T1 . −kr l1 + l 2
Then using l1 = (l1 + l2 )(1 − )1/3 , and l2 = (l1 + l2 ) 1/3 , we have
kr 4σSB T 3 l2 (1 − )2/3
=
1 4σSB T 3 l2 (1 − )1/3 (2 − r ) 1/3 + ks r
or
kr r Nr−1 (1 − )1/3 = , 3 4σSB T l1 r + Nr−1 (2 − r )
417
Nr =
4σSB T 3 l1 . ks
COMMENT: Note that for l1 → 0 ( → 1), we have kr =
4σSB r T 3 l2 . 2 − r
For ks → ∞, we have kr =
4σSB r T 3 l2 (1 − )1/3 . 2 − r
Figure Pr.4.44(b) shows the variation of the dimensionless radiant conductivity 4σSB l1 T 3 /kr with respect to the inverse of conductionradiation Nr−1 = ks /(4σSB T 3 l1 ) for various surface emissivity r . The results are for = 0.478 (corresponding to a squarearray arrangement of touching spherical particles). 0.8
0.7
= 0.476
kr 4ISB l1T 3
0.6
r
=1
0.3
0.8
0.5 0.4
Nr < 0.1, No Significant Solid Temperature Drop
0.8062 (ks )
Nr > 10, Significant Solid Temperature Drop
0.6 0.4
0.2
0.2
0.1 0 102
101
1
0.1
10
0.05
102
ks 1 = 4ISB l1T 3 Nr Figure Pr.4.47(b) Variation of dimensionless radiant conductivity with respect to the inverse of conductionradiation number.
Note that from the above relation for kr , for r = 1, we have kr = (1 − )1/3 , 4σSB T 3 l1 where l2 (1 − )1/3 = l2
l1 l2 = l1 = 1/3 l1 . l1 + l 2 l1 + l 2
Now using = 0.476, we have a value of kr /4σSB T 3 l1 = 0.8062 which matches the value in Figure Pr.4.44(b). From the ﬁgure, note the role of Nr on the radiant conductivity.
418
PROBLEM 4.48.FUN GIVEN: ∗ = σex L > 10, the extinction coeﬃcient σex and the radiant In the limit of a optically thick medium, σex conductivity are used interchangeably and are related through qr,x
dT dx 16 σSB T 3 dT 4 dEb =− . = − 3σex dx 3 σex dx
≡
−kr
Then kr =
16σSB T 3 . 3σex
For a packed bed of cubical particles of ﬁnite conductivity ks , surface emissivity r , and linear dimension l1 , with interparticle spacing l2 , the radiant conductivity can be shown to be approximated by 1
kr =
1 ks (1 − )
1/3
+
(2 − r ) 1/3 4σSB r T 3 l2 (1 − )2/3
or σex =
16σSB T 3 3ks (1 − )1/3
+
4(2 − r ) 1/3 3 r l2 (1 − )2/3
.
OBJECTIVE: (a) Determine σex for a bed of alumina cubical particles at T = 500 K, with r = 0.7 = 0.4, ks = 36 W/mK, and (i) l2 = 3 cm, and (ii) l2 = 3 µm. (b) Repeat (a) for amorphous silica particles, r = 0.45, ks = 1.38 W/mK, keeping other parameters the same. (c) Compare these with the results of Figure 2.13 and comment. SOLUTION: (a) Alumina, (i) l2 = 0.03 m σex
=
16 × 5.67 × 10−8 (W/m2 K4 ) × (500)3 (K3 ) 3 × 36(W/mK)(1 − 0.4)1/3
+
4(2 − 0.7) × 0.41/3 =
3 × 0.7 × 3 × 10−2 (m) × (1 − 0.4)2/3 1.245(1/m) + 85.51(1/m) = 86.75 1/m.
(ii) For l2 = 3 × 10−6 m, we have σex
=
1.245(1/m) + 8.551 × 105 (1/m) = 8.551 × 105 1/m.
(b) Amorphous silica, (i) l2 = 0.03 m σex
= =
16 × 5.67 × 10−8 × (500)3
+
4(2 − 0.45) × 0.41/3
3 × 1.38(1 − 0.4)1/3 3 × 0.45 × 3 × 10−2 (m) × (1 − 0.4)2/3 32.47(1/m) + 158.6(1/m) = 191.0 1/m.
(ii) For l2 = 3 × 10−6 m, we have σex
=
32.47(1/m) + 1.585 × 106 (1/m) = 1.586 × 105 1/m.
(c) In Figure 2.13, for a packed bed of spherical particles (material is not identiﬁed), we have σex of about 20 1/m for a particle diameter of 3 cm and about 2 × 105 1/m for a particle diameter of 3 µm. These are in general agreement with the above results.
419
COMMENT: In the expression used for kr , a onedimensional conduction in series with surface radiation is used. Therefore, the model should be considered an approximation. Note that the conduction contribution decreases as the particle spacing decreases. Also note that from the expression for radiant conductivity, if we deﬁne the phonon meanfree path as λph =
1 2 r l2 (1 − )2/3 = , σex 4(2 − r ) 1/3
then λph decreases with decreasing r (smaller surface emission) and increasing (there is less surface to emit radiation).
420
PROBLEM 4.49.FAM GIVEN: Spherical, pure, roughpolish aluminum particles of diameter D1 and emissivity r,1 are heated by surface radiation while traveling through an alumina ceramic tube kept at a high temperature T2 . The tube has an inner diameter D2 , a length l, and an emissivity r,2 . This is shown in Figure Pr.4.49(a). A particle arrives at the entrance to the tube with an initial, uniform temperature T1 (t = 0), and exits the tube with a ﬁnal, uniform temperature T1 (t = tf ). Assume that, throughout the time of travel, the fraction of radiative heat transfer between the particle and the open ends of the tube is negligible (i.e., the view factor F1ends = 0). D1 = 10 µm, D2 = 3 mm, l = 5 cm, T1 (t = 0) = 20◦C, T1 (t = tf ) = Tsl (aluminum, Table C.16), T2 = 1,283 K. Evaluate the emissivities from Table C.18 and the properties of aluminum at T = 300 K (Table C.16). SKETCH: Figure Pr.4.49(a) shows a particle ﬂowing in a tube while being heated by surface radiation from the tube wall. Inlet D1
x
Aluminum Particle, T1(t), r,1, ks,1 ∋
2
Ar,1 = πD1
up
π
3 V1 = _ 6 D1
l
Alumina Tube, T2, r,2 ∋
Ar,2 = πD2l
D2 Exit
Figure Pr.4.49(a) Particles heated in a tube by surface radiation.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the speed up = l/tf at which the particle must move through the tube in order to exit at the melting temperature T1 (t = tf ) = Tsl . (c) Approximate the internal conduction resistance to be (D1 /2)/(Ar,1 ks,1 ), where ks,1 is the thermal conductivity of the solid aluminum, and determine if the assumption of uniform temperature within the particle is valid. SOLUTION: (a) To determine the particle speed up = l/tf , we must determine the time tf for the particle to be heated to the melting temperature of aluminum T1 (t = tf ) = Tsl = 933 K (Table C.16). Since the sphere is very small and consists of aluminum which has a high thermal conductivity, we will initially assume it to behave as a lumpedcapacitance thermal mass. The thermal circuit diagram is shown in Figure Pr.4.49(b). Qr,12 T1(t) E b,1
(qr,o )1 (Rr, )1
(qr,o )2
(Rr,F)12
Eb,2 T2 (Rr, )2
(HcV)1 dT1 dt
Figure Pr.4.49(b) Thermal circuit diagram.
421
The sphere is a lumped system with a single resistive radiation heat transfer. Applying conservation of energy to node T1 , we have Q A
∂T1 ∂t ∂T1 = −(ρcV )1 ∂t = −(ρcV )1
=
Qr,12
=
Eb,1 − Eb,2 (Rr,Σ )12 σSB (T14 − T24 ) (Rr, )1 + (Rr,F )12 + (Rr, )2
=
= −(ρcV )1
∂T1 . ∂t
The solutions to this diﬀerential equation is given as T2 + T1 (t = tf ) σSB T23 1 − ln T2 + T1 (t = 0) + 2tan−1 T1 (t = tf ) − 2tan−1 T1 (t = 0) tf = . ln T2 − T1 (t = 0) (Rr,Σ )12 (ρcV )1 4 T2 − T1 (t = tf ) T2 T2 The volume and areas relevant to the problem are V1
= πD13 /6 = π × (10 × 10−6 )3 (m3 )/6 = 5.236 × 10−16 m3 ,
Ar,1 = A1
= πD12 = π × (10 × 10−6 )2 (m2 ) = 3.142 × 10−10 m2 ,
Ar,2 = A2
= πD2 l = π × 0.003(m) × 0.05(m) = 4.712 × 10−4 m2 ,
From Table C.16 (T1 = 300 K), ρ1 = 2,702 kg/s and c = 903 J/kgK. The thermal capacitance of the particle is then (ρcV )1
=
2,702(kg/m ) × 903(J/kgK) × 5.236 × 10−16 (m3 )
=
1.278 × 10−9 J/K.
3
From Table C.18, the emissivity of the rough polish aluminum is r,1 = 0.18 and of the nylon is r,2 = 0.78. The grayness resistances are then (Rr, )1
1 − r,1 1 − 0.18 = r,1 Ar,1 0.18 × 3.142 × 10−10 (m2 )
=
2
1.450 × 1010 1/m , 1 − r,2 1 − 0.78 = r,2 Ar,2 0.78 × 4.712 × 10−4 (m2 )
= (Rr, )2
=
2
=
598.58 1/m .
From the summation rule for the view factors from surface 1, n
Fij = F11 + F12 + F1ends = 1.
j=1
Since F11 = 0 and F1ends ≈ 0, then F12 ≈ 1. Then the view factor resistance between surfaces 1 and 2 is (Rr,F )12
=
1 1 = Ar,1 F12 3.142 × 10−10 (m2 ) × 1
=
3.183 × 109 1/m ,
2
and the total radiative resistance is (Rr,Σ )12
=
(Rr, )1 + (Rr,F )12 + (Rr, )2
=
1.450 × 1010 (1/m ) + 3.183 × 109 (1/m ) + 598.58 (1/m )
=
1.768 × 1010 1/m .
2
2
2
422
2
Upon substitution into our thermal conservation of energy equation T2 + T1 (t = 0) T2 + T1 (t = tf ) 1 σSB T23 − ln tf = ln (Rr,Σ )12 (ρcV )1 4 T2 − T1 (t = tf ) T2 − T1 (t = 0) −1 T1 (t = tf ) −1 T1 (t = 0) +2tan − 2tan , T2 T2 we can solve for tf [noting T1 (t = 0) = (20 + 273.15)(K) = 293.15 K] as 1,283(K) + 293(K) 1,283(K) + 933(K) 5.67 × 10−8 (W/m2 K4 ) × (1,283)3 (K3 ) 1 − ln × t = ln f 2 1,283(K) − 293(K) + 1,283(K) − 933(K) 4 1.768 × 1010 (1/m ) × 1.278 × 10−9 (J/K) 933(K) 293(K) −1 −1 2 × tan − 2 × tan 1,283(K) 1,283(K) 5.30(1/s) × tf = 0.25 × (1.846 − 0.465 + 1.257 − 0.449) tf
=
0.103 s.
The velocity of the particle through the tube must then be such that it travels the entire length of the tube in tf seconds, i.e., up
= l/tf = 0.05(m)/0.103(s) = 0.48 m/s.
(b) The assumption of uniform temperature distribution can be validated by considering the conductionradiation number Nr , given by (4.74) where Nr =
Rk,i 3 ), Rr,Σ /(4σSB Tm
where Rk,i is the internal conduction resistance and Tm is the radiation mean temperature. From Table C.16 for aluminum, ks,1 = 237 W/mK. Then Rk,i
= =
3 Tm
= =
D1 /2 10 × 10−6 (m)/2 = Ar,1 ks,1 3.142 × 10−10 (m2 ) × 237(W/mK) 67.15 K/W, (T22 + T12 )(T2 + T1 ) = 0.25 × [(1,2832 + 9332 )(1,283 + 933)](K3 ) 4 1.394 × 109 K3 .
Upon substitution into Nr , we have Nr
= =
Rk,i 3 , Rr,Σ /4σSB Tm 67.15(K/W) 1.768 × 10 (1/m )/[4 × 5.67 × 10−8 (W/m2 K4 ) × 1.394 × 109 (K3 )] 10
2
= 1.2 × 10−6 .
Here, Nr < 0.1, therefore the assumption of uniform temperature within the particle is valid. COMMENT: Surfaceconvection heating will also occur during the ﬂight and should be included.
423
PROBLEM 4.50.FUN GIVEN: A highly insulated thermos depicted in Figure Pr.4.50 has ﬁve layers of insulation shields on the outside. The wall has two glass layers separated by an evacuated space [Figure Pr.4.50(a)(i)], or a cork board [Figure Pr.4.50(a)(ii)]. T1 = 90◦C, T3 = 20◦C, l = 1 mm, L = 7 mm, r,s = 0.04, r,2 = 0.9, r,3 = 1. SKETCH: Figure Pr.4.50(a) shows the thermos with ﬁve radiation shields with and without a cork board between the glass layers.
(i) Without Cork
Radiation Shields r,s = 0.04
Thermos Wall with Vacuum
∋
Hot Fluid T1 = 90oC
q1
Ak = A r = A
∋
(ii) With Cork
Thermos Wall with Cork
Hot Fluid T1 = 90oC
Surrounding Surface T3 = 20oC r,3 = 1 ∋
Glass Plates r,2 = 0.9
q1
l = 1 mm
l = 1 mm L = 7 mm
Figure Pr.4.50(a) Surface radiation from a thermos having radiation shields (i) With a cork board wall. (ii) Without a cork board wall.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the heat transfer per unit area associated with each of these designs. Comment on the preference of the cork board or the vacuum. Also, assume that all the surfaces are diﬀuse and gray. The glass is assumed opaque to radiation. SOLUTION: (i) Thermos wall with vacuum: (a) The heat transfer is by conduction and radiation. Figure Pr.4.50(b) shows the thermal circuit diagram. Note that r,i = r,o = r,2 . (b) Using Figure Pr.4.50(b), we have Qk,1i
=
Qr,io
=
Qk,o2
=
Qr,23
=
T1 − Ti Rk,1i Eb,i − Eb,o (Rr,Σ )io To − T2 Rk,o2 Eb,2 − Eb,3 . (Rr,Σ )23 424
Qr,i0
(Rr, )i
Qr,0
(Rr,F)i0
(Rr, )0
(Rr, )2
(Rr,F)2s
Q1
Surroundings
SurfaceGrayness Resistance
Qr,s
Qr,s Eb,s Ts Eb,s
(qr,0)s
(qr,0)2
T2 Eb,2 Rk,02
SurfaceGrayness Resistance
Qr,2s
Qr,2
Qk,02 Eb,0 T0
(qr,0)0
Rk,1i
(qr,0)i
SurfaceGrayness Resistance
Qr,i Ti Eb,i
Radiation Shield
Conduction Resistance
(Rr, )s
Qk,1i T1
Glass Plate SurfaceGrayness Resistance
SurfaceGrayness Resistance
Qr,s3 (qr,0)s
(Rr, )s
Conduction Resistance
Vacuum
SurfaceGrayness Resistance
Qr,3 (qr,0)3
(Rr,F)s3
Eb,3 T3 (Rr, )3
Glass Plate
For n plane, parallel radiation shields with the same surface properties, this block is repeated n times.
Figure Pr.4.50(b) Thermal circuit diagram without cork board.
From Figure Pr.4.50(b), we have −Q1 = Qk,1i = Qr,io = Qk,o2 = Qr,23 = Q3 . The blackbody emissive power is Eb,i = σSB Ti4 . All the thermal resistances can be calculated from the data given. Then, the equations above form a system of 4 equations and 4 unknowns. The unknowns are Ti , To , T2 , and Q1 . The equations are nonlinear because of the T 4 terms. To solve for the resistances, the thermal conductivity of the glass plate is needed. From Table C.17, we have kg = 0.76 W/mK. The resistances are Ak Rk,1i
=
Ar (Rr,Σ )io
= = =
Ak Rk,o2
=
Ar (Rr,Σ )23
= = =
l 0.001(m) 2 = 0.0013 K/(W/m ), Ak = Ar = A = kg 0.76(W/mK) Ar (Rr, )i + Ar (Rr,F )io + Ar (Rr, )o 1 − r,i 1 1 − r,o + + r,i Fio r,o 1 − 0.9 1 − 0.9 +1+ = 1.22 0.9 0.9 l 0.001(m) 2 = 0.0013 K/(W/m ) = kg 0.76(W/mK) Ar (Rr, )2 + Ar (Rr,F )2s1 + 5 2 (Rr, )s + (Rr,F )s1s2 + Ar (Rr, )3 1 − r,s 1 − r,2 1 1 1 − r,3 + +5 2 + + r,2 F 2 3 r,s Fs1s2 r,3 1 − 0.04 1 − 0.9 1−1 +1+5 2 = 246.11. +1 + 0.9 0.04 1
The energy equations are then written as −
Q1 Eb,i − Eb,o Eb,2 − Eb,3 T1 − Ti To − T2 = = = = . 2 2 A 1.22 246.11 0.0013[K/(W/m )] 0.0013[K/(W/m )]
The thermal resistances between surfaces 1 and i, surfaces i and o, and surfaces o and 2, are small compared to the resistance between surfaces 2 and 3. This allows us to use the approximation T2 To Ti T1 . Using this approximation (another reason to adopt this is discussed in COMMENT), we have −
Q1 Eb,1 − Eb,3 . A 246.11 425
Q3
Using the numerical values, we have Q1 − A
2 5.67 × 10−8 (W/m K4 ) 363.154 (K)4 − 293.154 (K)4 246.11 2 2.305 W/m .
=
(ii) Thermos wall with cork: (a) The thermal circuit for the system with the cork is shown in Figure Pr.4.50(c).
Qk,i0 T0
Ti Rk,1i
Qk,02
Rk,i0
SurfaceGrayness Resistance
(qr,0)2
T2 Eb,2 Rk,02
(Rr, )s
Q1
SurfaceGrayness Resistance
Qr,s
Eb,s Ts Eb,s
(qr,0)s
(Rr,F)2s
(Rr, )2
Surroundings
SurfaceGrayness Resistance
Qr,s
Qr,2s
Qr,2
Qk,1i
Conduction SurfaceGrayness Resistance Resistance
Conduction Resistance
Radiation Shield
Qr,s3 (qr,0)s
(Rr, )s
Conduction Resistance
T1
Glass Plate
Cork
Qr,3 (qr,0)3
(Rr,F)s3
For n plane, parallel radiation shields with the same surface properties, this block is repeated n times.
Eb,3 T3 (Rr, )3
Glass Plate
Q3
Figure Pr.4.50(c) Thermal circuit diagram with cork board.
(b) The heat transfer within the space between the glass plates, instead of being by radiation, is by conduction through the cork. Between surfaces 1 and 2, conduction is the only heat transfer mode. This allows us to write the heat ﬂux between surfaces 1 and 2 as Qk,12 =
T1 − T2 , (Rk,Σ )12
where l L l + + . kg kc kg
Ak (Rk,Σ )12 =
From Table C.17, for cork board, we have kc = 0.043 W/mK. Thus Ak (Rk,Σ )12
=
0.007(m) 0.001(m) 0.001(m) + + 0.76(W/mK) 0.043(W/mK) 0.76(W/mK)
=
0.165 K/(W/m ).
2
The heat ﬂux can then be written as, noting that Ak = Ar = A, −
Q1 Eb,2 − Eb,3 T1 − T2 = = 2 A 246.11 0.165[K/(W/m )]
The conduction resistance is again much smaller than the radiation resistance. This allows us to assume that T2 T1 . Using this approximation, the heat ﬂux can be calculated from −
Q1 A
Eb,1 − Eb,3 2 = 2.305 W/m . 246.11
COMMENT: Using a solver (to solve the nonlinear system of equations) results in the following values: (i) Ti = 363.147 K, (ii) Ti = 363.147 K,
To = 362.889 K, To = 362.774 K,
T2 = 362.886 K, T2 = 362.771 K,
− Q1 /A = 2.294 W, − Q1 /A = 2.290 W.
Note that the temperatures Ti to T2 are nearly equal to T1 . Also, the calculated heat ﬂuxes are very close to those found using the approximations. Due to the existence of the ﬁve radiation shields, the use of vacuum 426
or cork between the glass plates has little eﬀect on the heat loss. The relative magnitude of the conduction and radiation resistances can be compared by using the conductionradiation number Nr deﬁned in (4.75). For the radiation between surfaces 2 and 3 with T1 = T2 , the average temperature is Tm
1/3 1/3 T22 + T32 (T2 + T3 ) 363.152 + 293.152 (363.15 + 293.15) = = = 329.4 K. 4 4
Comparing the conduction resistance between surfaces o and 2 with the radiation resistance between 2 and 3, we have from (4.74) Nr
=
3 3 Rk,o2 Ak Rk,02 4σSB Tm 4σSB Tm = (Rr,Σ )23 Ar (Rr,Σ )23
=
4 × 5.67 × 10−8 (W/m K4 ) × 329.43 (K)3 × 0.0013[K/(W/m )] = 4.28 × 10−5 . 246.11 2
2
As Nr 1, the conduction resistance can be neglected. Because of the series arrangement for the resistances, the system is radiationresistance dominated.
427
PROBLEM 4.51.FAM.S GIVEN: A person with a surface temperature T1 = 31◦C is standing in a very large room (Ar,2 Ar,1 ) and is losing heat by surface radiation to the surrounding room surfaces, which are at T2 = 20◦C [Figure Pr.4.51(a)]. Model the person as a cylinder with diameter D = 0.4 m and length L = 1.7 m placed in the center of the room, as shown in Figure Pr.4.51(a). Neglect surfaceconvection heat transfer and the heat transfer from the ends of the cylinder. Assume that all the surfaces are opaque, diﬀuse, and gray. Assume negligible contact resistance between the clothing and the body. SKETCH: Figure Pr.4.51(a) shows the person losing heat by surface radiation, to the surrounding walls, (i) with no clothing, and (ii) with a layer of clothing.
Physical Model
T1 = 31oC
Room Temperature T2 = 20oC Qr,12
An Approximation (ii) With Clothing Room Temperature T2 = 20oC L = 1.7 m
Qr,12
Clothing
L = 1.7 m
Qr,s2 Ts ,
T1 = 31oC r,1 = 0.9 ∋
D = 0.4 m
∋
(i) No Clothing
r,s
= 0.7
T1 = 31oC l = 1 cm D = 0.4 m
Figure Pr.4.51(a) A physical model and approximation for surfaceradiation heat exchange between a person and his or her surrounding surfaces, with (i) no clothing and (ii) a layer of clothing.
OBJECTIVE: For a steadystate condition, (a) draw the thermal circuit and (b) determine the rate of heat loss for the case of (i) no clothing covering a body with a surface emissivity r,1 = 0.9, and (ii) for the case of added clothing of thickness l = 1 cm with a conductivity k = 0.1 W/mK and a surface emissivity r,s = 0.7. Comment on the eﬀect of the clothing, for the given temperature diﬀerence. SOLUTION: (i) No Clothing: (a) For no clothing, the thermal circuit is shown in Figure Pr.4.51(b).
428
Qr,1
Qr,12
T1 Eb,1
Qr,2
(qr,0)1
Eb,2 T2
(qr,0)2
Q1
Q2 (Rr,F)12
(Rr, )2
(Rr, )1
Figure Pr.4.51(b) Thermal circuit diagram for no clothing included.
(b) From Figure Pr.4.51(b), the heat transfer rate from the body to the walls is given by Qr,12 =
Eb,1 − Eb,2 . (Rr,ε )1 + (Rr,F )12 + (Rr,ε )2
The radiation thermal resistances are (Rr,ε )1 =
1 − r,1 1 − 0.9 2 = 0.05201 1/m = Ar,1 r,1 [π × 0.4(m) × 1.7(m)] × 0.9
(Rr,F )12 =
1 1 2 = 0.4681 1/m = Ar,1 F12 [π × 0.4(m) × 1.7(m)] × 1 (Rr,ε )2 (Rr,ε )1 .
Here F12 = 1, because the cylinder is surrounded by the room walls. Also Ar,2 Ar,1 (note that no assumption is made about r,2 ). This allows us to neglect the wall surfacegrayness resistance. Solving for Qr,12 , we have Qr,12 =
5.67 × 10−8 (W/m2 K4 ) × [304.154 (K4 ) − 293.154 (K4 )] 2
2
0.05201(1/m ) + 0.4681(1/m )
= 127.8 W.
(ii) With Clothing: (a) By covering the body with clothing, a conduction thermal resistance is created in the path of the heat transfer. Figure Pr.4.51(c) shows the thermal circuit. Clothing
Walls
Clothing Surface
Conduction SurfaceGrayness ViewFactor SurfaceGrayness Resistance Resistance Resistance Resistance
Qr,s
Qk,1s Ts Eb,s
T1
Qr,s2 (qr,0)s
Qr,2 Eb,2 T2
(qr,0)2
Q1
Q2 (Rr, )s
(Rr,F)s2
(Rr, )2
Rk,1s
Figure Pr.4.51(c) Thermal circuit diagram for clothing included.
(b) Applying the energy equation to the Ts node, we have Qk,1s = Qr,s2 . The conduction and radiation heat transfer rates are Qk,1s =
Qr,s2 =
T1 − Ts Rk,1s
Eb,s − Eb,2 . (Rr,ε )s + (Rr,F )s2 + (Rr,ε )2 429
The thermal resistances are Rk,1s =
ln [(0.2 + 0.01)(m)/0.2(m)] ln (R2 /R1 ) = = 0.0457 ◦C/W 2πkL 2 × π × 0.1(W/mK) × 1.7(m)
(Rr,ε )s =
1 − r,s 1 − 0.7 2 = 0.191 1/m = Ar,s r,s [π × (0.4 + 0.02)(m) × 1.7(m)] × 0.7
(Rr,F )s2 =
1 1 2 = 0.446 1/m = Ar,s Fs2 [π × (0.4 + 0.02)(m) × 1.7(m)]1 (Rr,ε )2 (Rr,ε )s .
Then Qr,12 =
σSB Ts4 − T24 σSB Ts4 − T24 T1 − Ts ◦ ( C/W) = = 2 2 2 , 0.0456 0.191(1/m ) + 0.446(1/m ) 0.637(1/m )
where T1 = 304.15 K, T2 = 293.15 K, and σSB = 5.67 × 10−8 W/m2 K4 . This is an implicit equation for Ts . The solution can be obtained iteratively. First, we rewrite this as an algebraic equation in Ts , Ts = T1 − 4.059 × 10−9 (K−3 ) × (Ts4 − T24 ) and using T1 and T2 , we have Ts = 304.15(K) − 4.059 × 10−9 (K−3 ) × (Ts4 − 7.385 × 109 )(K4 ). Using the method of successive substitutions, the equation above is rearranged as Tsnew = 304.15(K) − 4.059 × 10−9 (K−3 ) × [(Tsold )4 − 7.385 × 109 (K4 )] Table Pr.4.51 presents the results for three iterations. Table Pr.4.51 Results obtained for three iterations. Tsold , K
Tsnew , K
300 301.10 300.70
301.25 300.70 300.94
After about 10 iterations, the solution converges to Ts = 300.87 K. The heat transfer rate can then be calculated from = Qk,1s =
(304.15(K) − 300.87(K) = 71.93 W. 0.0456(◦C/W)
Alternatively, we can use a solver (such a SOPHT). COMMENT: The clothing has the shape of a cylindrical shell and requires the appropriate equation for the conduction thermal resistance. If the equation for a slab is used instead, the conduction thermal resistance is Rk,12 =
0.01(m) l = = 0.0457 ◦C/W Ak,ave k [π × ( 0.42+0.40 )(m) × 1.7(m)] × 0.1(W/mK) 2
This is a good approximation for this problem, because l/R = 0.01/0.2 = 0.05 1. 430
To determine whether the conduction thermal resistance could be neglected when compared to the radiation thermal resistance, the conductionradiation number Nr could be used. The linearized Nr is given by (4.75), i.e., Nr =
3 Rk,1s 4σSB Tm , Rr,Σ
where the linearized average temperature Tm is Tm =
1/3 Ts2 + T22 (Ts + T2 ) . 4
The highest value for Ts is achieved when the conduction resistance is negligible and it is equal to T1 . Using Ts = T1 = 304.15 K, we ﬁnd that Tm = 298.7 K and Nr is Nr =
4 × 5.67 × 10−8 (W/m2 K4 ) × (298.7)3 (K3 ) × 0.0456(◦C/W) 2
0.637(1/m )
= 0.433.
This indicates that the conduction thermal resistance is 43.3% of an equivalent (linearized) radiation thermal resistance and therefore, cannot be neglected (the assumption that Rk,1s Rr,Σ does not apply). Note that using the deﬁnition of Nr and the calculated temperature we obtain, Nr =
T1 − Ts 304.15(K) − 300.87(K) = 0.425. = Ts − T2 300.87(K) − 293.15(K)
The small diﬀerence is due to the linearization of the diﬀerence (Eb,s − Eb,2 ) used in the deﬁnition of the radiation thermal resistance used in the equation for Nr .
431
PROBLEM 4.52.FAM GIVEN: A thin ﬁlm is heated with irradiation from a laser source with intensity qr,i = 106 W/m2 , as shown in Figure Pr.4.52(a). The heat losses from the ﬁlm are by surface emission and by conduction through the substrate. Assume that the ﬁlm can be treated as having a uniform temperature T1 (t) and that the conduction resistance through the substrate can be treated as constant. SKETCH: Figure Pr.4.52(a) shows the radiation heating of a thin ﬁlm with heat loss by substrate conduction. qr,i = 106 W/m2 (ρcp)1 = 106 J/m3K αr = 1 r=0
T1(t) T1(t = 0) = 20 C
L1 = 10 µm
∋
L2 = 5 mm
Ts,2 = 20 C
k = 1.3 W/mK
Figure Pr.4.52(a) Laser irradiation heating of a thin ﬁlm on a substrate.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) For an initial temperature T1 (t = 0) = 20◦C, determine the time needed to raise the temperature of the ﬁlm T1 to 500◦C. SOLUTION: (a) Figure Pr.4.52(b) shows the thermal circuit for the problem. Note that the thin ﬁlm is lumped into a single node and the thick ﬁlm is modeled as a conduction resistance constant with time. T1
Qk,12
(Sr,=)1
Rk,12
T2 Q2
Figure Pr.4.52(b) Thermal circuit diagram.
(b) The energy equation is applied to the thin ﬁlm to determine the time needed to raise the ﬁlm temperature to 500◦C. The integralvolume energy equation is (4.76) QA = − (ρcp V )1
dT1 + S˙ 1 . dt
From Figure Pr.4.52(b), we notice that QA has only a conduction component. The energy convection terms are due to radiation absorption with αr = 1 and radiation emission with r = 0. Then from (4.66) we have T1 − T2 dT1 + αr qr,i Ar . = − (ρcp V )1 Rk,12 dt 432
The conduction resistance is given by Rk,12 =
L2 5 × 10−3 (m) 3.85 × 10−3 [◦C/(W/m2 )] = = . ks Ak 1.3 (W/mK) Ak Ak
The thermal capacitance is (ρcp V )1 = 106 (J/m ◦C ) × 10 × 10−6 (m) Ak = 10(J/m ◦C)Ak . 3
2
The energy conversion term is 2 S˙ 1 = αr qr,i Ar = (1) × 106 (W/m )Ar .
The solution to this integralvolume energy equation is given in Section 3.5.2, i.e., T1 − Ts,2 − a1 τ1 t = −τ1 ln , T1 (t = 0) − Ts,2 − a1 τ1 where 3.85 × 10−3 [◦C/(W/m2 )] 10(J/m2 ◦C)Ak = 3.85 × 10−2 s Ak
τ1
=
(ρcp V )1 Rk,12 =
a1
=
2 S˙ 1 106 (W/m )Ar = = 105 1/s, (ρcp V )1 10(J/m2 ◦C)Ak
and Ar = Ak has been used. Then
500 (◦C) − 20 (◦C) − 105 (1/s) × 3.85 × 10−2 (s) t = −3.85 × 10 (s) ln 20 (◦C) − 20 (◦C) − 105 (1/s) × 3.85 × 10−2 (s) = 0.0051 s = 5 ms. −2
COMMENT: The assumption of constant substrate resistance is probably not a valid assumption for small elapsed times. In this case, there is a penetration of the transient conduction front and the equivalent thermal resistance changes with time.
433
PROBLEM 4.53.DES GIVEN: A pipeline carrying cryogenic liquid nitrogen is to be insulated. Two scenarios, shown in Figure Pr.4.53, are considered. The ﬁrst one [Figure Pr.4.53(i)] consists of placing the pipe (tube) concentrically inside a larger diameter casing and ﬁlling the space with microspheres insulation material. The microspheres have an eﬀective thermal conductivity of k = 0.03 W/mK. The tube has an outside diameter D1 = 2 cm and the casing has an inside diameter D2 = 10 cm. Another scenario [Figure Pr.4.53(ii)] consists of placing a thin polished metal foil between the tube and the casing, thus forming a cylindrical shell with diameter D3 = 6 cm, and then evacuating the spacings. Both the tube and casing have an emissivity r,1 = r,2 = 0.4 and the thin foil has an emissivity r,3 = 0.05. The tube is carrying liquid nitrogen and has a surface temperature T1 = 77.3 K and the casing has a surface temperature T2 = 297 K. SKETCH: Figure Pr.4.53 shows the tube insulation. (i)
L
Microspheres Insulation Tube
D1
Nitrogen
T2 = 297 K
Casing
D2
T1 = 77.3 K Qr,12
(ii)
L Shield (Thin Metallic Foil) r,2 = 0.05
Vacuum
∋
D1
Nitrogen Tube
T2 = 297 K r,2 = 0.4
D2
T1 = 77.3 K r,1 = 0.4 ∋
∋
Casing
D3
Figure Pr.4.53(i) and (ii) Two scenarios for insulation of a cryogenic ﬂuid tube.
OBJECTIVE: (a) Determine the net heat transfer to liquid nitrogen for the two scenarios using a tube length L = 1 m. (b) How thick should the microsphere insulation be to allow the same heat transfer as that for the evacuated, radiation shield spacing? SOLUTION: (a) (i) For the microsphere insulation, the conduction thermal resistance, from Table 3.1, is LRk,12
=
ln[5(cm)/1(cm)] ln (R2 /R1 ) = = 8.54 K/(W/m). 2πk 2π(0.03)(W/mK)
Then, the conduction heat transfer is, Qk,12 L
=
T1 − T2 (77.3 − 297)(K) = −25.7 W/m. = Rk,12 8.54(K/W)
(ii) With one radiation shield placed between surfaces 1 and 2, the overall radiation thermal resistance is 1 − r,3 1 − r,1 1 1 − r,2 1 + +2 + . (Rr,Σ )12 = + Ar,1 r,1 Ar,1 F13 Ar,3 r,3 Ar,3 F32 Ar2 r,2 For F13 = F32 = 1 and using the values given, L(Rr,Σ )12
=
23.87 + 15.92 + 2 × 100.80 + 5.31 + 4.77 = 251.47 1/m. 434
Then, the net radiation heat transfer is 4
σSB (W/m2 K4 ) × (77.34 − 2974 )(K) Qr,12 = = −1.743 W/m. L 251.47(1/m2 ) (b) Equating the expression for the conduction heat transfer to the result for the net radiation heat transfer, we have −1.743(W/m) =
(77.3 − 297)(K) . LRk,12
Solving for LRk,12 we obtain LRk,12 = 126.0 K/(W/m). From the expression for Rk,12 and solving for R2 , we obtain ﬁnally R2 = 2.042 × 108 m. COMMENT: Note how eﬀective the radiation shield and vacuum are in reducing the heat transfer from the pipe. One assumption used is that conduction and surface convection heat transfer through the evacuated gap are negligible.
435
PROBLEM 4.54.FAM GIVEN: Automatic ﬁre sprinklers, shown in Figure Pr.4.54(a), are individually heat activated, and tied into a network of piping ﬁlled with pressurized water. When the heat ﬂow from a ﬁre raises the sprinkler temperature to its activation temperature Tm = 165◦F, a lead alloy solder link will melt, and the preexisting stress in the frame and spring washer will eject the link and retainer from the frame, allowing the water to ﬂow. An AISI 410 stainless steel sprinkler having a mass Ms = 0.12 kg and an initial temperature T1 (t = 0) = 72◦F is used to extinguish a ﬁre having a temperature T∞ = 1,200◦F and an area Ar,∞ much greater than the area of the sprinkler Ar,1 = 0.003 m2 . Assume that the dominant source of heat transfer is radiation and that the lumped capacitance analysis is valid. SKETCH: Figure Pr.4.54(a) shows the ﬁre sprinkler actuated by heat transfer and raised to a threshold temperature.
MeltingActivated Sprinkler
Fusible Alloy Spring Valve
Water Supply
Figure Pr.4.54(a) A ﬁre extinguisher actuated by rise in temperature caused by surfaceradiation heating.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the elapsed time t needed to raise the sprinkler temperature to the actuation temperature. SOLUTION: (a) The thermal circuit diagram is shown in Figure 4.54(b).
dT1 dt Q1
T1(t)
Qr,1 Eb,1(t)
(ρcpV )1
Rr,Σ
Eb,
r,1
Figure Pr.4.54(b) Thermal circuit diagram.
(b) We use the transient lumpedcapacitance analysis and from (4.80), we have 3 T∞ + T1 σSB T∞ 1 − ln T∞ + T1 (t = 0) + 2 tan−1 T1 − 2 tan−1 T1 (t = 0) t= ln T∞ − T1 (t = 0) Rr, (ρcp V )1 4 T∞ − T1 T∞ T∞ 436
where Rr, (Rr, )1 (Rr,F )1∞ (Rr, )2 Rr,
=
(Rr, )1 + (Rr,F )1∞ + (Rr, )2 1 − r 1 − 0.13 = = = 2,230.8 1/m3 Ar r 1 0.003(m2 ) × 0.13 1 1 = = = 333.33 1/m3 , F1∞ = 1 Ar,1 F1∞ 0.003(m2 ) 1 − r = 0, Ar,o Ar,1 Ar r 2 =
2,564.13 1/m3 .
The properties are (AISI 410 stainless steel, Table C.16), ρ = 7,770 kg/m3 , k = 25 W/mK, and cp = 460 J/kgK and (AISI 410 stainless steel, Table C.19), r,1 = 0.13. The volume V1 , is ρ=
M , V
V =
M 0.12(kg) = = 1.544 × 10−5 m3 . ρ 7,770(kg/m3 )
Using (4.80) with T∞ = 1,200◦F = 422 K, T1 (t = 0) = 72◦F = 295.4 K, and Tm = 165◦F = 347 K, we have 922 + 347 5.67 × 10−8 (W/m2 K4 ) × (922)3 (K3 ) 1 − ln 922 + 295.4 + 2 tan−1 347 t = ln −5 2 4 922 − 347 922 − 295.4 922 2,564.13(1/m ) × (7,770 × 460 × 1.544 × 10 )(J/K) 295.4 −2 tan−1 922 1 3.14 × 10−4 t = [0.7916 − 0.664 + 0.7199 − 0.62] 4 3.14 × 19−4 t = 0.05687 s t = 181.1 s 3 min. COMMENT: The time needed to start the sprinkler is rather high, t = 23 min. The heat transfer by surface convection reduces t. In order to obtain a more accurate prediction, the thermobuoyant ﬂow surface convection should be included.
437
PROBLEM 4.55.FUN GIVEN: Heat transfer by conduction and surface radiation in a packed bed of particles with the void space occupied by a gas is approximated using a unitcell model. Figure Pr.4.55(i) shows a rendering of the cross section of a packed bed of monosized spherical particles with diameter D and surface emissivity r . A twodimensional, periodic structure with a square unitcell model is used. The cell has a linear dimension l, with the gas and solid phases distributed to allow for an interparticle contact and also for the presence of the pore space, as shown in Figure Pr.4.55(ii). The thermal circuit model for this unit cell is shown in Figure Pr.4.55(iii). The surface radiation is approximated by an optically thick medium treatment. This allows for a volumetric presentation of radiation (this is discussed in Section 5.4.6). This uses the concept of radiant conductivity kr . One of the models for kr is 4σSB T 3 D 4 r σSB T 3 D . kr = = 2 2 − r −1 r SKETCH: Figure Pr.4.55 shows the cross section of the packed bed of spheres, the unitcell model, the thermal circuit model for the unit cell. The radiant conductivity is combined with the gas conductivity in Rkr,f .
(i) Physical Model qkr CrossSection of a Packed Bed of Spherical Particles ∋
Porosity
ks kf D Gas Solid r
∋
(iii) Thermal Circuit Model for Unit Cell
(ii) TwoDimensional UnitCell Model
qkr Th w DRk,sE1
Solid
Gas
qkr
(Rkr,f)1 (Rk,s)2
l/2
(Rkr,f)3
∋
r
a1l/2
(Rk,s)1 a1l/4
(1 − a1)(1 − )l
a1l
(Rkr,f)2 (Rkr,f)2
(1 − a1) l ∋
l
(Rk,s)3
∋
l Tc
qkr
Figure Pr.4.55(i) Physical model of a packed bed of spherical particle with the pore space ﬁlled with a gas. (ii) A simpliﬁed, twodimensional unitcell model. (iii) Thermal circuit model for the unit cell.
OBJECTIVE: Using the geometric parameters shown in Figure Pr.4.55(ii), show that the total thermal conductivity for the thermal circuit model of Figure Pr.4.55(iii) is kkr
=
qkr = (1 − a1 )(1 − r )ks + (Th − Tc )l a1 + (1 − a1 ) (kf + kr ). 1 1 1 + + (kf + kr ) + ks 4(kf + kr ) 4ks 438
Here we have combined the surface radiation with the gas conduction such that in Figure Pr.4.55(iii), Rkr,f uses kf + kr as the conductivity. SOLUTION: The eﬀective thermal conductivity is deﬁned as qkr ≡
(Th − Tc )l Th − Tc = , Akr = lw, Akr Rkr k
where w is the depth of the unit cell. The heat ﬂows through the various resistances shown in Figure Pr.4.55(iii). Combining these, we have 1 1 = + Rkr (Rk,s )1
1 1 1 (Rkr,f )1
+
1
+ (Rk,s )3 .
+ (Rkr,f )2 + (Rkr,f )3
(Rk,s )2
The six resistances are determined using Table 3.1 for the slab resistance along with the geometrical parameters of Figure Pr.4.45(ii). Then, we have (Rk,s )1
=
(Rkr,f )1
=
(Rk,s )2
=
(Rk,f )2
=
(Rk,s )3
=
(Rk,f )2
=
l (1 − a1 )(1 − )lwks l/2 a1 l w(kf + kr ) 2 l/2 a1 l wks 2 l/4 a1 lw(kf + kr ) l/4 a1 lwks l . (1 − a1 ) lw(kf + kr )
Combining these, we have kkr =
1 = (1 − ar )(1 − r )ks + wRkr
1 + (1 − a1 ) (kf + kr ). 1 1 1 + + a1 (kf + kr ) + ks 4a1 (kf + kr ) 4a1 ks
which is the desired expression. COMMENT: To verify the results, take the case of a1 = 1 and = 0. Then by setting kf +kr = ks , we will have kkr = ks , as expected. Also for the case of = 0 and a1 = 0, we have kkr = ks , as expected. Note that kr is given in terms of the surface emissivity r . In Section 5.4.6 we will give another expression for kr .
439
PROBLEM 4.56.FAM.S GIVEN: The rangetop electrical heater has an electrical conductor that carries a current and produces Joule heating S˙ e,J /L(W/m). This conductor is covered by an electrical insulator. This is shown in Figure Pr.4.56(a). In electrical insulator should be a good thermal conductor, in order to a avoid large temperature drop across it. It should also have good wear properties, therefore various ceramics (especially, oxide ceramics) are used. Here we consider alumina (Al2 O3 ). Consider a heater with a circular cross section, as shown Figure Pr.4.56(a). Neglect the conduction resistance between the electrical conductor (central cylinder) and the electrical insulator (cylindrical shell). Assume a steadystate surface radiation heat transfer only (from the heater surface). Ri = 1 mm, S˙ e,J /L = 5 × 103 W/m, T2 = 30◦C, r,1 = 0.76. SKETCH: Figure Pr.4.56(a) shows the heater and its insulation shell.
RangeTop Electrical Heater Surroundings
T2 , Ar,2 >> Ar,1
Ro Electrical Insulator, k1 2Ri T1 , r,1 , Ar,1 ∋
Length, L Rk,c = 0
Electrical Conductor, TH Se,J /L (W/m)
Figure Pr.4.56(a) A rangetop electrical heater with a cylindrical heating element made of an inner electrical conductor and an outer electrical insulator.
OBJECTIVE: (a) Draw the thermal circuit diagram for the heater. (b) Plot the heater temperature TH with respect to the outer radius Ro , for 2 ≤ Ro ≤ 20 mm, and the conditions given below. 3 )], where Tm is deﬁned (c) Plot TH with respect to the conductionradiation number Nr = Rk,H−1 /[Rr,Σ /(4σSB Tm by (4.73). SOLUTION: (a) Figure Pr.4.56(b) shows the thermal circuit diagram. The heat ﬂow per unit length S˙ e,J /L encounters conduction and a surfaceradiation resistances.
Qr,H1
Qr,12
Se,J Ar,1
TH
Rk,H1 T1
Eb,1
Rr,5
Eb,2 T2
Figure Pr.4.56(b) Thermal circuit diagram.
440
(b) The energy equation for the heating surface Ar,1 , is written using Figure 4.56(b), i.e., QA,1 = Qr,12 = S˙ e,J , or and since the same ﬂows through the electrical insulator, we have TH − T1 Rk,H 1
TH − T1 = S˙ e,J ln(Ro /Ri ) 2πk1 L Eb,1 − Eb,2 = 2πRo L r,1 σSB (T14 − T24 ) = S˙ e,J , Rr,Σ
=
=
where we have used Table 3.2 for Rk,H 1 and (4.49) for Rr,Σ , with (Rr, )2 → 0 for Ar,2 Ar,1 and F12 = 1. Then solving for T1 and TH , we have T1 TH
=
T24
S˙ e,J /L + 2πRo r,1 σSB
1/4
(S˙ e,J /L) ln(Ro /Ri ) 2πk1 1/4 ˙ e,J /L S (S˙ e,J /L) ln(Ro /Ri ) = T24 + + . 2πRo r,1 σSB 2πk1 = T1 +
The thermal conductivity is found at T = 1,300 K in Table C.14, as k1 = 6.0 W/mK
Table C.14.
We also have S˙ e,J /L = 5 × 103 W/m, T2 = (273.15 + 30) K = 303.15 K, and Ri = 1 mm. Using these numerical values, we have TH
(303.15)4 (K4 ) +
1/4 5 × 103 (W/K) 5 × 103 (W/m) × ln(Ro /0.001) + −8 2 4 2π × 6(W/mK) 2πRo × 0.76 × 5.67 × 10 (W/m K ) 1/4 Ro 1.846 × 1010 + 1.326 × 102 (K) ln = 8.446 × 109 + . Ro 0.001
=
Figure Pr.4.56(c) shows the variation of TH with respect to Ro . Note that as Ro increases, the surface area Ar,1 increases proportional to Ro , while the conduction resistance increases as ln(Ro ). This results in a continuous decrease of TH as Ro increases. 2,000
TH , K
1,800 1,600 1,400 1,200 1,000 0.002
0.004
0.006
0.01
0.02
Ro , m Figure Pr.4.56(c) Variation of the heater temperature with respect to insulator outer radius.
441
(c) From (4.74), we have
Nr
=
Rk,H 1 3 = Rr,Σ /(4σSB Tm )
=
3 4σSB Tm
ln(Ro /Ri ) 2πk1 L 1 3 2πRo L r,1 (4σSB Tm )
Ro r,1 ln(Ro /Ri ). k1
From (4.73), we have Tm =
(T12 + T22 )(T1 + T2 ) 4
1/3 .
Using the numerical values, we have
Nr
−8
= 4 × 5.67 × 10
3 (W/m2 K4 )Tm
= 2.873 × 10−8 Ro ln T1
T24 +
=
S˙ e,J /L 2πRo r,1 σSB
8.446 × 109 +
=
Ro 0.001 1/4
Ro 0.001 6(W/mK)
Ro × 0.76 × ln
×
3 Tm
1.846 × 1010 Ro
1/4 .
Figure Pr.4.56(d) shows the variation of Nr with respect to Ro . As Ro → Ri , the conduction resistance (and therefore, Nr ) decreases and TH → T1 . 0.6
Nr
0.4
0.2
0.1 0.08 0.06 0.002
0.004
0.006
0.01
0.02
Ro , m Figure Pr.4.56(d) Variation of conductionradiation number with respect to insulator outer radius.
COMMENT: Note that increasing the dielectric layer thickness Ro − Ri , decreases the heater temperature TH by reducing the overall resistance. Also note that even for Nr < 0.1, there is still a conduction resistance and this inﬂuences TH .
442
PROBLEM 4.57.FAM GIVEN: Ice is formed in a water layer as it ﬂows over a cooled surface at temperature Tc . Assume that the surface of the water is at the saturation temperature Tls and that the heat transfer across the water layer (thickness L) is by steadystate conduction only. The top of the water layer is exposed to the roomtemperature surroundings at temperature T∞ , as shown in Figure Pr.4.57(a). Assume that water and the surrounding surfaces are opaque, diﬀuse, and gray (this is a reasonable assumption for water in the near infrared range which is applicable in this problem). Tls = 0◦C, Tc = −10◦C, T∞ = 300 K, L = 2 mm, r,l = 1. Assume that the ice is being formed at the top surface of the water layer. Evaluate the water properties at T = 280 K (Tables C.4, and Table C.13). SKETCH: Figure Pr.4.57(a) shows ice formation by conduction through the water layer. There is also surface radiation between the water surface and the surrounding surfaces.
T
g
Sls = − mls ∆hls Al Ice Formation
Ar, >> Ar,l
Ak,l = Ar,l = Al Tls, r,l ∋
L
Liquid Water (and Ice)
Tc
Figure Pr.4.57(a) Ice is formed in a thin water layer cooled from below.
OBJECTIVE: (a) Draw the thermal circuit diagram for the water surface. (b) Determine the rate of ice formation per unit area m ˙ ls = M˙ ls /Al . SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.4.57(b). Heat transfer from the surroundings to the surface of the water is by surface radiation and heat transfer across the water layer is by conduction. (b) The energy equation, from Figure Pr.4.57(b), is Qr,l∞ + Qk,lc = S˙ ls . The view factor between the surface of the water and surroundings is unity, Fl∞ = 1. The surface radiation for the unity view factor, for Ar,l Ar,∞ , and r,1 = 1, is given by (4.49), i.e., Qr,l∞
4 = Ar,l σSB (Tls4 − T∞ ).
The conduction resistance is found from Table 3.2, and when used in Qk,lc , gives Qk,lc =
Tls − Tc Ak,l kw (Tls − Tc ). = Rk,lc L
Then, the energy equation becomes 4 Ar,l σSB (Tls4 − T∞ )+
Ak,l kw (Tls − Tc ) = S˙ ls . L 443
T Eb, Qr,l
(Rr,5)l
Sls = − mls ∆hls Al
Eb,l Tls
Qk,1c
Rk,lc
Tc
Figure Pr.4.57(b) Thermal circuit diagram.
Since S˙ ls /Al = −m ˙ ls ∆hls , the energy equation becomes 4 )+ Ar,l σSB (Tls4 − T∞
Ak,l kw (Tls − Tc ) = −m ˙ ls Al ∆hls , L
From Tables C.4 and C.13, we have kw = 0.582 W/mK, and ∆hls = −∆hsl = −333.6 × 103 J/kg. Using the numerical values given, we have Ar,l × 5.67 × 10−8 (W/m2 K4 ) × (273.154 − 3004 )(K4 ) +
(273.15 − 263.15)(K) = 2 × 10−3 (m) Ak,l × 0.582(W/mK)
−m ˙ ls Al × (−333.6 × 103 )(J/kg), −143.6Ar,l + 2,910Ak,l = −m ˙ ls Al × (−333.6 × 103 )(J/kg), or m ˙ ls = 8.284 × 10−3 kg/m2 s = 8.284 g/m2 s. where we have used Al =Ar,l =Ak,l . COMMENT: Note that surface radiation is not negligible. In practice the ice is formed adjacent to the cooled surface and heat is conducted through the ice. Also note that the heat gained by radiation is approximately 5% of the heat removed by conduction.
444
Chapter 5
Convection: Unbounded Fluid Streams
PROBLEM 5.1.FAM GIVEN: In order to protect exhaust line walls from exposure to hightemperature exhaust gases, these walls are covered by a sacriﬁcial layer. This is shown in Figure Pr.5.1(a). Upon exposure to high temperature exhaust gas and a rise in temperature, this sacriﬁcial layer undergoes a pyrolytic thermal degradation, produces pyrolytic gases, and becomes porous. The pyrolytic gas ﬂows toward the heated surface, thus providing for transpiration cooling and prevention of the large heat load Qs from reaching the wall. Treat the pyrolytic gas as air at T = 600 K and assume a steadystate gas ﬂow that is uniform through the layer. Assume that the area for conductionconvection is πDl and use the planar presentation of the resistance as given by (5.14). Tf,1 = 300 K, Tf,2 = 900 K, k = 0.5 W/mK, uf = 50 cm/s, D = 80 cm, l = 1 m, L = 1.5 cm. SKETCH: Figure Pr.5.1(a) shows the sacriﬁcing layer lining and the heat transfer to the layer. Exhaust Gas Qu k
Gas Evolved by Pyrolysis of Sacrificing Layer
L D Tf,2 Tf,1
uf
Sacrificing Layer (a FiberFiller Composite)
Qs
Qs
l
Q1 FlueGas Stream
x
Control Surface Sr,c Combustion Chamber
Figure Pr.5.1(a) Exhaust line walls covered by sacriﬁcial layers.
OBJECTIVE: (a) Draw the thermal circuit diagram and show the energy equation for surface node Tf,1 . (b) For the conditions given above, determine the rate of heat ﬂowing into the wall. SOLUTION: (a) The thermal circuit diagram for node Tf,1 is shown in Figure Pr.5.1(b). Q(x=0) Tf,1
Tf,2
Q1 (Rk,u)12
Figure Pr.5.1(b) Thermal circuit diagram.
(b) The heat ﬂow rate Q1 is determined from the energy equation for node 1, which is found by examining Figure Pr.5.1(b) as QA = Q1 + Q(x = 0) = 0. Then from (5.23), we have Ak,u k PeL (Tf,1 − Tf,2 ) L ePeL − 1 k uf L , α = = πDl, PeL = . α (ρcp )f Q1 = −
Ak,u
446
From Table C.22, for air at T = 600 K, we have ρf = 0.589 kg/m3 cp,f = 1038 J/kgK
Table C.22 Table C.22.
Then α
= =
PeL
=
0.5(W/mK) 0.589(kg/m3 ) × 1038(J/kgK) 8.178 × 10−4 m2 /s. 0.5(m/s) × 0.015(m) = 9.171. 8.178 × 10−4 (m2 /s)
Using the numerical values, Q(x = 0) is Q1
= −
9.171 π × 0.8(m) × 1(m) × 8(W/mK) × 9.171 (300 − 900)(K) = 767.3 W. 0.015(m) e −1
COMMENT: Note that as PeL increase, less heat ﬂows into the substrate. Materials which produce signiﬁcant pyrolytic gases, as a result of thermal degradation, are used.
447
PROBLEM 5.2.FAM GIVEN: The axial conductionconvection resistance may be large, when compared to other heat transfer resistances. In ﬂow through a tube, as shown in Figure Pr.5.2, the axial conductionconvection resistance Rk,u is compared to the average convection resistance Ru L (this will be discussed in Chapter 7). The average velocity of the ﬂuid ﬂowing in the tube is uf and the average ﬂuid inlet and outlet temperatures are Tf,1 and Tf,2 . The average convection resistance is given (for the case of small N T U , to be discussed in Chapter 7) by Ru L =
1 . 3.66πLkf
uf = 0.2 m/s, D = 5 cm, and L = 30 cm. Evaluate the properties for air at T = 350 K from Table C.22, and for engine oil at T = 350 K from Table C.23. SKETCH: Figure Pr.5.2 shows the two resistances in a tube ﬂow and heat transfer. Ak,u = FD2/4 Ts Average Convection Resistance
DRuEL
uf D Tf,1
Rk,u Axial ConductionConvection Resistance
Fluid Stream
Tf,2
L
Figure Pr.5.2 Comparison of lateral (surfaceconvective) and axial (conductionconvective) resistances.
OBJECTIVE: (a) For the conditions given above determine the ratio of Rk,u /Rku when the ﬂuid is air. (b) Determine the ratio of Rk,u /Rku when the ﬂuid is engine oil. SOLUTION: Using the deﬁnition of Rk,u given in Table 5.1, and using Ak,u = πD2 /4, the ratio of the two resistances is ePeL − 1 Ak,u kf PeL ePeL 1 3.66πLkf L (πD2 /4)kf ePeL − 1 1 PeL ePeL 3.66πLkf L
Rk,u Ru L
=
=
=
3.66 × 4
L2 ePeL − 1 , D2 PeL ePeL
where from (5.9), PeL =
uf L . αf
From Tables C.22 and C.23 at T = 350 K air : αf = 2.944 × 10−5 m2 /s −8
engine oil : αf = 7.74 × 10
2
m /s 448
Table C.22 Table C.23.
Then for (a), 0.2(m/s) × 0.30(m) = 2,038. 2.944 × 10−5 (m2 /s)
=
PeL
This will be a very large argument for the exponential function. By noting that for PeL 1, ePeL − 1 1, ePeL the ratio of the resistances can be expressed as Rk,u L2 1 = 3.66 × 4 2 . Ru L D PeL And so, for (a), Rk,u Ru L
= =
(0.30)2 (m2 ) 1 (0.05)2 (m2 ) 2,038 0.2586 for air.
3.66 × 4 ×
For (b), PeL
=
0.2(m/s) × 0.30(m) = 8.065 × 105 7.44 × 10−8 (m2 /s)
Rk,u Ru L
=
3.66 × 4 ×
=
6.535 × 10−3 for engine oil.
(0.30)2 (m2 ) 1 (0.05)2 (m2 ) 8.065 × 105
COMMENT: Since we are comparing two resistances, in addition to PeL , the ratio L/D is also important. This is because the resistance Ru L decreases with an increase in L, while the resistance Rk,u increases with an increase in L. Therefore, a large PeL is needed to make the axial conductionconvection resistance negligible, unless L/D is small. Here the surfaceconvection resistance dominates for the oil. But for air at this speed, the axial conductionconvection resistance is relatively signiﬁcant.
449
PROBLEM 5.3.FAM GIVEN: Transpiration surface cooling refers to ﬂowing a ﬂuid through a permeable solid toward the surface to intercept and remove a large amount of heat ﬂowing to the surface. This imposed heat input Qs is called the heat load. The ﬂowing ﬂuid opposes the axial conduction heat transfer and results in a lower surface temperature, compared to that of conduction heat transfer only (i.e., no permeation). This is shown in Figure Pr.5.3(a). Air is made to ﬂow through a porous ceramic slab to protect a medium (a substrate) beneath the ceramic. Qs = −103 W, uf = 10 cm/s, Tf,1 = 20◦ C, k = 0.5 W/mK, w = l = 20 cm, L = 5 cm. Evaluate the air properties at T = 500 K. SKETCH: Figure Pr.5.3(a) shows air ﬂowing toward the surface and intercepting the imposed heat load. Porous Wall
Substrate
Ak,u uf
Transpiration Air Stream l
T
 Qs (Heat Load)
Tf,2 Tf (x) Tf,1 0
L w x
L
Figure Pr.5.3(a) Transpiration surface cooling.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) For the conditions given below, determine the surface temperature Tf,2 . (c) For comparison, determine Tf,2 using uf = 0, i.e., (5.15). SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.5.3(b).
(Qk,u)12 Tf,1
Tf,2
Qs
Figure Pr.5.3(b) Thermal circuit diagram.
(b) The energy equation for node Tf,2 , shown in Figure Pr.5.3(b), is
where from (5.13)
(Qk,u )12
QA = −(Qk,u )12 + Qs = 0, Tf,1 − Tf,2 = (Rk,u )12
(Rk,u )12
=
PeL
=
L(ePeL − 1) Ak,u kPeL ePeL uf L k , αf = , αf (ρcp )f 450
Ak,u = lw.
Solving the energy equation for Tf,2 , we have Tf,2
= Tf,1 − Qs (Rk,u )12 = Tf,1 − Qs
L ePeL − 1 . lwk PeL ePeL
From Table C.22, for air at T = 500 K, we have ρf = 0.706 kg/m3 and cp,f = 1017 J/kgK. Then α
=
PeL
=
k 0.5(W/mK) = = 6.964 × 10−4 m2 /s (ρcp )f 0.706(kg/m3 ) × 1017(J/kgK) 0.1(m/s) × 0.05(m) = 7.180. 6.964 × 10−4 (m2 /s)
The temperature of surface 2 is then Tf,2
= =
e7.180 − 1 0.05(m) × 0.2(m) × 0.2(m) × 0.5(W/mK) 7.180 × e7.180 293.15(K) + 347.9(K) = 641.1 K = 367.9◦C.
293.15(K) − (−103 )(W) ×
(c) For the case of uf = 0, we have from (5.15), (Rk,u )12 = Rk,12 =
L , Ak,u k
and Tf,2 = Tf,1 − Qs
L , lwk
or Tf,2
= =
0.05(m) 0.2(m) × 0.2(m) × 0.5(W/mK) 293.15(K) + 2,500(K) = 2,793 K = 2,520◦C.
293.15(K) − (−103 )(W) ×
COMMENT: Note that by providing a cold air ﬂow, the surface temperature is reduced signiﬁcantly. The air ﬂow removes the heat by convection and prevents a large fraction of the heat load from entering the substrate. The heat ﬂow rate into the substrate is given by (5.23).
451
PROBLEM 5.4.FUN GIVEN: The onedimensional, axial conductionconvection thermal resistance is given by (5.14), i.e., Rk,u =
ePeL − 1 . Ak,u kf PeL ePeL L
OBJECTIVE: Show that in the limit as PeL → 0, this resistance becomes the conduction thermal resistance for a slab (Table 3.2). SOLUTION: We begin by taking the limit of (5.14) for PeL → 0, i.e., ePeL − 1 L = PeL →0 Ak,u kf PeL ePeL Ak,u kf
lim Rk,u = lim
PeL →0
L
ePeL − 1 . PeL →0 PeL ePeL lim
Applying the L’Hopital rule, we have ePeL − 1 ePeL 1 = lim = lim = 1. PeL →0 PeL ePeL PeL →0 ePeL + PeL ePeL PeL →0 1 + PeL lim
Therefore, lim Rk,u =
PeL →0
L . Ak,u kf
This is the conduction resistance for a slab given in Table 3.2. The area for conduction is the same as the area for conduction and convection. COMMENT: The P´eclet number is a ratio of the ﬂuid axial conduction and convection resistances, as given by (5.9), i.e., PeL =
Rk,f . Ru,f
When the convection resistance becomes very large, the heat transfer occurs primarily by conduction, being controlled by the conduction thermal resistance given above. Similarly, as the convection resistance become very small (PeL → ∞), the primary transport will be by convection.
452
PROBLEM 5.5.FAM GIVEN: In a space shuttle, a permeable Oring is used as a thermal barrier and in order to optimize its function, the permeation of combustion ﬂue gas allows for gradual pressure equalization around it. This Oring is shown in Figure Pr.5.5. The braided carbon ﬁber Oring has an average porosity . The mass ﬂow rate through the Oring is M˙ f . Assume an ideal, square crosssectional area L × L. The length of the Oring is l = 1 m. = 0.5, L = 0.7 cm, M˙ f = 3.25 g/s, Tf,1 = 1,700◦C, Tf,2 = 100◦C. For the gas use the properties of air at T = 900◦C. Use thermal conductivity of carbon at T = 900◦C. Use (3.28) to determine the eﬀective thermal conductivity k. SKETCH: Figure Pr.5.5 shows the permeable Oring.
Permeable, Braided ORing Phenolic
Gas Mf Stream
Braided Carbon ORing, Idealized as Square CrossSection, L x L
Tf,1
Tf,2 x
CarbonFiber Strands (12,000 per CrossSection)
Figure Pr.5.5 A permeable Oring, made of braided carbon ﬁber, is used as a thermal barrier and gradual pressure equalizer.
OBJECTIVE: Determine the rate of heat transfer (Qk,u )12 = Qx=L . SOLUTION: The heat transfer rate (Qk,u )12 is given by (5.13), i.e., (Qk,u )12
=
Tf,1 − Tf,2 (Rk,u )12
=
(Tf,1 − Tf,2 )Ak,u k
PeL ePeL , ePeL − 1
where we have used the eﬀective conductivity k for the carbonﬁber and air composite. Since we are given the porosity, we use (3.28) for k, i.e., k = kf
ks kf
0.280−0.757 log −0.057 log(ks /kf ) .
The P´eclet number is given by (5.9), i.e., PeL =
k uf L , α = , α (ρcp )f
where M˙ f = Ak,u ρf uf
or
uf =
M˙ f , Au ρf
Ak,u = Au = lL.
Interpolating from Table C.14, we have, for carbon at T = 900 K, ks
= 2.435W/mK 453
Table C.14.
Interpolating from Table C.22, we have, for air at T = 900K, ρf = 0.392 kg/m3 cp,f = 1,111 J/kgK
Table C.22 Table C.22
kf = 0.0625 W/mK
Table C.22.
Then, ks kf
=
2.435(W/mK) = 38.96 0.0625(W/mK)
k
=
0.0625(W/mK) × (38.96)0.280−0.757×log(0.5)−0.057 log(38.96)
=
0.0625(W/mK) × 4.609 = 0.2881 W/mK.
Also, uf
= =
3.25 × 10−3 kg/s 1(m) × 0.007(m) × 0.392(kg/m3 ) 1.184 m/s
Then PeL
= =
uf L(ρcp )f uf L = α k 1.184(m/s) × 0.007(m) × 0.392 × 1,111(J/m3 K) = 12.53. 0.2884(W/mK)
For the heat ﬂow rate we have, for Au = lL, (Qk,u )12
=
(1,700 − 100)(◦C) × 1(m) × 0.007(m) ×
12.53 × e12.53 = 140.3 W. e12.53 − 1
COMMENT: Although intended as a thermal barrier, here PeL is large enough to cause a large heat ﬂow. The dimensionless temperature distribution along the ﬂow direction is given in Figure 5.3 and shows the strong inﬂuence of convection.
454
PROBLEM 5.6.FUN GIVEN: The temperature distribution in a ﬂuid stream with axial conduction and convection and subject to prescribed temperatures Tf,1 and Tf,2 at locations x = 0 and x = L, respectively, is given by (5.12). OBJECTIVE: Starting from the dimensionless, onedimensional steadystate diﬀerentialvolume energy equation (5.7), and by using (5.10) and (5.11), derive (5.12). SOLUTION: Equation (5.7) is a dimensionless energy equation and is a secondorder, ordinary diﬀerential equation with the boundary conditions given by (5.10) and (5.11), i.e., d2 Tf∗ dx∗2
dTf∗ dx∗ ∗ ∗ Tf (x = 0) − PeL
Tf∗ (x∗
= 0 = 1
= L) = 0
The ﬁrst integration gives an exponential solution dTf∗ ∗ = a1 ePeL x . dx∗ Integrating this again, gives Tf∗ =
a1 PeL x∗ e + a2 . PeL
Using the boundary conditions, we have a1 + a2 or a1 = (1 − a2 )PeL PeL ePeL ePeL 0 = a1 + a2 or a2 = −a1 . PeL PeL 1 =
Eliminating a2 ,
a1
=
a1
=
a1
=
ePeL 1 + a1 PeL
PeL
PeL + a1 ePeL PeL , 1 − ePeL
so that, a2 = −
ePeL . 1 − ePeL
Using these expressions for a1 and a2 , we have Tf ∗
= = = = =
PeL ePeL x∗ ePeL − PeL Pe 1−e 1 − ePeL L ePeL x∗ − ePeL 1 − ePeL ePeL − ePeL x∗ ePeL − 1 ePeL − 1 − (ePeL x∗ − 1) ePeL − 1 ePeL x∗ − 1 . 1 − PeL e −1 455
COMMENT: Note that for large PeL , we have ePeL 1 and ePeL x∗ 1. Then ∗
Tf ∗ = 1 − ePeL x
−PeL
∗
= 1 − ePeL (x
−1)
, ∗
which has an exponential behavior, as shown in Figure 5.3. For small values of PeL , we expect ePeL and ePeL x as ePeL = 1 + PeL + .... . Then Tf∗ = 1 −
PeL x∗ = 1 − x∗ , PeL
which is the expected linear behavior, as shown in Figure 5.3.
456
PROBLEM 5.7.FUN GIVEN: Impermeable, extended surfaces (ﬁns) are used to assist in surfaceconvection heat transfer by providing an extra surface area. By allowing ﬂow through the ﬁns (e.g., in boiling heat transfer, the surface tension is used to draw the liquid through the ﬁns and this liquid evaporates on the surface), the heat ﬂow rate at the base of the ﬁn can increase substantially. To demonstrate this, consider the permeable ﬁns shown in Figure Pr.5.7(a). Assume that the ﬂuid stream starts from the ﬁn top and leaves very close to the base (x = 0). Then a unidirectional ﬂow with velocity uf can be assumed (this imply that the ﬂuid stream continues to ﬂow through the base). Here water is allowed to ﬂow through ﬁns made of sintered metallic particles. L = 2 mm, R = 0.5 mm, Tf,1 = 70C, Tf,2 = 80◦C, k = 20 W/mK. Evaluate the water properties at T = 350 K. SKETCH: Figure Pr.5.7(a) shows a simpliﬁed model for the permeable ﬁns attached to a surfaces. Fluid Stream Entrance uf,2
Tf,1 < Tf,2
Permeable (Porous) Fin, k
R g
Ideally Insulated and Impermeable Fluid Stream Exit
Tip
x
Tf,2
L
(qk,u)12
Base
Substrate
Assuming Fluid Stream Continues
Se,J
Figure Pr.5.7(a) A permeable ﬁn is used to direct a ﬂuid stream toward the base. A simple model is also used.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the heat ﬂow rates through each ﬁn (Qk,u )12 and (qk,u )12 , for (ii) uf = 0.1 m/s, and uf = 0. SOLUTION: (a) Figure Pr.5.7(b) shows the thermal circuit diagram. The only heat transfer to be determined is (Qk,u )12 . Tf,1 Qk,u
Rk,u
12
12
Tf,2 Se,J
Figure Pr.5.7(b) Thermal circuit diagram.
(b) From Table 5.1, we have (Qk,u )12
= =
Tf,1 − Tf,2 (Rk,u )12 Tf,1 − Tf,2 , L ePeL −1 Aku k PeL ePeL
PeL =
Here we have Ak,u = πR2 . 457
uf L , α
α =
k . (ρcp )f
From Table C.23, for water at T = 350 K, we have = 975.7 kg/m3 = 4,194 J/kgK
ρf cp,f
Table C.23 Table C.23.
(i) Using the numerical values, we have, for uf = 1 m/s, α
=
20(W/mK) (975.7)(kg/3 ) × 4,194(J/kgK)
=
4.887 × 10−6 m2 /s
PeL
=
(Qk,u )12
=
= (qk,u )12
=
0.1(m/s) × 2 × 10−3 (m) = 40.92 4.887 × 10−6 (m2 /s) (70 − 80)(◦C) −3 2 × 10 (m) e40.92 − 1 π(5 × 10−4 )2 (m2 ) × 20(W/mK) 40.92 × e40.92 −10(◦C) = −3.214 W 3.111(◦C/W) (Qk,u )12 3.214(W) =− Ak,u π(5 × 10−4 )2 (m2 )
= −4.092 × 106 W/m2 . (ii) For uf = 0, we have from (5.15) (Rk,u )12
= Rk,12 = = =
L Aku k
2 × 10−3 (m) π × (5 × 10−4 )2 (m2 ) × 20(W/mK) 127.3◦C/W.
Then (Qk,u )12
=
−10(◦C) = −7.854 × 10−2 W 127.3(◦C/W)
(qk,u )12
=
(Qk,u )12 −7.855 × 10−2 W = Ak,u π(5 × 10−4 )2 (m2 )
= −1.000 × 105 W/m2 . Comparing the results for (i) and (ii), the heat transfer rate at the base is signiﬁcantly increased (by a factor equal to PeL = 40.92). This is due to interception (and removal by convection) of the heat ﬂow by the opposing ﬂuid ﬂow (as shown by the temperature distribution of Figure 5.3). COMMENT: Note that we have assumed that locally the ﬂuid and solid have the same temperature Tf . This is the condition of negligible surfaceconvection heat transfer resistance between the ﬂuid and solid. This resistance will be discussed in Chapter 7. Also, we have neglected the eﬀect of added conductivity (this is called thermal dispersion) due to the nonuniformity of the velocity in the pores.
458
PROBLEM 5.8.FUN GIVEN: Capillary pumping (or wicking) refers to ﬂow of liquid through and toward the porous solids by the force of surface tension (an intermolecular force imbalance at the liquidgas interface). In capillary pumped evaporators, heat is also provided to the porous solid surface such that the liquid is completely evaporated on the surface. Figure Pr.5.8(a) shows three capillarypumped evaporators, distinguished by the relative direction of the heat and liquid ﬂows. These are used in heat pipes (to be discussed in Example 8.1) and in enhanced, surface evaporations. Figure Pr.5.8(b) renders a counter heatwater capillary evaporator, where the liquid ﬂow rate M˙ l ﬂowing through the wick(distributed as attached, permeable cylinders) is evaporated at location L1 . Assume that the liquid is at Tf,1 = Tf,2 = Tlg at x = L1 , such that (Qk,u )12 = 0. The temperature at T (x = L1 ) is the saturation temperature, so the heat for evaporation is provided by conduction in the region adjacent to the surface, i.e., L1 ≤ x ≤ L1 + L2 . Then Qk,23 is determined from Table 3.2. This simple thermal circuit model is also shown in Figure Pr.5.8(b). R = 0.5 mm, L2 = 150 µm, Tlg = 100◦C, T3 = 105◦C, k = 10 W/mK. Determine saturated water properties from table C.27, at T = 373.15 K. SKETCH: Figure Pr.5.8(a) shows three diﬀerent capillarypumped evaporators and Figure Pr.5.8(b) shows the counter heatliquid ﬂow evaporator considered.
Capillary Pumped Evaporators (i) Counter HeatLiquid Flow Liquid
(iii) CounterCross HeatLiquid Flow
Wick (Wet Porous Solid)
Vapor g
(ii) Cross HeatLiquid Flow
qk Vapor
g
Slg
Slg
Slg
Liquid Substrate
qk
Vapor
qk
Liquid
Figure Pr.5.8(a) Various capillary pumped evaporators, based on the relative direction of the heat and liquid ﬂows.
(i) Counter HeatLiquid Flow Capillary Evaporator Water Flow Ml , Tf,1 = Tf,2
Ml
Wick (Wet, Porous Solid)
Slg
T3  qk,23
Rk,23
L1 L2
Tf,2 = Tlg
 Qk,23
x
Vapor Surface Evaporation Slg , Tf,2 = Tlg
(ii) Thermal Circuit Diagram
g
T3
Se,J
Figure Pr.5.8(b) A counter heatliquid ﬂow capillary evaporator with a simple conduction heat transfer model in the region adjacent to surface, i.e., L1 ≤ x ≤ L1 + L2 .
OBJECTIVE: (a) Determine the liquid mass ﬂow rates M˙ l and m ˙ l = M˙ l /Ak , for the given conditions. (b) Comment on the practical limit for the reduction of L2 .
459
SOLUTION: (a) Since in the simple heat transfer model the heat supplied to the evaporation region is by conduction only, we have, from Figure Pr.5.8(b), Qk,23 = S˙ lg . From Tables 2.1, for S˙ lg , and 3.2, for Qk,23 , we have Tf,2 − T3 Rk,23
Tlg − T3 Tlg − T3 = L2 /(Ak k) L2 /(πR2 k) = −M˙ l ∆hlg .
=
From Table C.27, at T = 373.15 K, we have ∆hlg = 2.257 × 106 J/kg
Table C.27.
Then using the numerical values, we have M˙ l
= −
= − = m ˙l
=
Tlg − T3 πR2 k(Tlg − T3 ) = L2 ∆hlg L2 ∆hlg πR2 k π × (5 × 10−4 )2 (m2 ) × 10(W/m◦C)(100 − 105)(◦C) 1.5 × 10−4 (m) × 2.257 × 106 (J/kg)
1.160 × 10−7 kg/s M˙ l M˙ l k(Tlg − T3 ) = =− = 0.1477 kg/sm2 . 2 Ak L2 ∆hlg πR
(b) Reducing L2 is limited by the fabrication technique. If sintered particles are used to make the wick, L2 is limited to the diameter of a single particle. COMMENT: One advantage of the distributed wick stack region is that it allows for the passage of vapor in the areas between the stacks. This avoids the passage of both phases through the wick (the counter ﬂow of the liquid and vapor) and allows for a larger M˙ l (for a given driving capillary pressure). The liquid ﬂow will be ultimately limited by the formation of a vapor blanket on top of the stacks.
460
PROBLEM 5.9.FAM GIVEN: In order to protect a substrate from high temperatures resulting from intense irradiation, evaporation transpiration cooling is used. This is shown in Figure Pr.5.9(a). Liquid water is supplied under a porous layer and this liquid is evaporated by the heat reaching the liquid surface, which is at temperature Tlg = Tf,1 . The heat ﬂow to the liquid surface is only a fraction of the prescribed irradiation, because the watervapor ﬂow intercepts and carries away a fraction of this heat by convection. ρf = 0.596 kg/m3 , cp,f = 2,029 J/kgK, ∆hlg = 2.257 × 106 J/kg (water at 100◦C), k = 15 W/mK, L = 40 cm, w = 15 cm, L = 1.5 cm, Tlg = Tf,1 = 100◦C, αr,2 =0.9, qr,i = 105 W/m2 , ρl = 958 kg/m3 . Note that from conservation of mass across the liquid surface, ρf uf = ρl ul . SKETCH: Figure Pr.5.9(a) shows the evaporation transpiration cooling to protect a substrate from high temperatures resulting from intense irradiation.
qr,i Heat Returned by Convection
(1  αr,2) qr,i
Ak,u
l
Porous Layer Tf,2 Permeable, Sintered Metal Particles
L
Tlg = Tf,1 Makeup Water Stream
Ak,u αr,2 qr,i Heat Used for Evaporation
Water Vapor Stream
x k
Slg = Mlg Dhlg
Q1 = 0 Substrate
w
Figure Pr.5.9(a) Evaporation transpiration cooling.
OBJECTIVE: (a) For the condition given in Figure Pr.5.9(a), draw the thermal circuit diagram. (b) Determine M˙ lg and Tf,2 . SOLUTIONS: (a) The thermal circuit diagram is shown in Figure Pr.5.9(b).
(b) Thermal Circuit Model Ak,u αr qr,i Tf,2
(Qk,u)21 Qu,21 = Ak,u (ρcp u)f (Tf,2  Tf,1)
Mlg Q (x = 0) Tlg = Tf,1 Q1 = 0
x Slg = Mlg ,hlg
Figure Pr.5.9(b) Thermal circuit diagram.
461
(b) From Figure Pr.5.9(b), for node Tf,2 , we have QA,2 = (Qk,u )21 = Ak,u αr,2 q1,i , where from (5.13) and (5.14) Ak,u k PeL ePeL (Tf,2 − Tf,1 ). L ePeL − 1
(Qk,u )21 = and for node Tf,1 , we have
QA,1 = −Q1
= S˙ lg = −M˙ lg ∆hlg ,
where from (5.23), Q1 = PeL =
uf L , α
ul =
M˙ lg , Ak,u ρl
α =
Ak,u k PeL (Tf,1 − Tf,2 ) L ePeL − 1
k , (ρcp )f
Ak,u = lw,
ρl ul = ρf uf .
We have also used the continuity equation to equate the liquid and vapor from rates at the liquid surface. We need to solve these two energy equations simultaneously for Tf,2 and uf (or M˙ lg ). Since the relation for uf is nonlinear, a numerical solution is needed and a solver can be used. Then thermal diﬀusivity is α
=
15(W/mK) 0.596(kg/m3 ) × 2029(J/kgK)
=
1.240 × 10−2 m2 /s.
The two energy equations and the deﬁnition of PeL give PeL 0.4(m) × 0.15(m) × 15(W/mK) PeL ePeL (Tf,2 − 100)(◦C) 0.015(m) ePeL − 1 0.4(m) × 0.15(m) × 15(W/mK) PeL (100 − Tf,2 )(◦C) 0.015(m) ePeL − 1
=
uf (m/s) × 0.015(m) 1.240 × 10−2 (m2 /s)
=
0.4(m) × 0.15(m) × 0.9 × 105 (W/m2 )
=
−0.4(m) × 0.15(m) × 0.596(kg/m3 ) × uf × 2.257 × 106 (J/kg).
The results are uf
=
0.06207 m/s
PeL Tf,2
= =
0.07508 186.7◦C.
For M˙ lg , we have M˙ lg
= Aku ρf uf =
0.4(m) × 0.15(m) × 0.596(kg/m3 ) × 0.06207(m/s) = ×10−3 kg/s = 2.220 g/s.
COMMENT: Note that since PeL is small, nearly all the absorbed irradiation energy reaches the evaporation surface. Due to the large heat of evaporation, a small velocity is found for qr,i = 105 W/m2 . For qr,i = 106 W/m2 , we would have uf = 0.4083 m/s. Also, the P´eclet number is small due to the large k.
462
PROBLEM 5.10.FUN GIVEN: Anesthetic drugs are supplied as liquid and are evaporated, mixed with gases (such as oxygen), and heated in portable vaporizer units for delivery to patients. This is shown in Figure Pr.5.10(a). The drugs, such as enﬂuorane, isoﬂuorane, etc., have thermophysical properties similar to that of refrigerant R134a (Table C.28). The drug is sprayed into the vaporizer. The heating of the gas mixture (drug and oxygen) is by surface convection and here it is prescribed by Qku L . This heat in turn is provided by Joule heating from a heater wrapped around the tube. Assume that droplets evaporate completely. Tf,1 = 15◦C, M˙ l = 2 × 10−5 kg/s, M˙ O2 = 2 × 10−4 kg/s, Qku L = −6 W. Use the speciﬁc heat capacity of oxygen (at T = 300 K, Table C.22) for the mixture, and use ∆hlg from Table C.28, at p = 1 atm. SKETCH: Figure Pr.5.10(a) shows the vaporizer. The oxygen (gas) and drug (liquid) streams enter and a gas mixture exits. The mixture is heated by surface convection. Liquid Nozzle (Injector)
Surface Convection  Qku L Droplets Warm Gas Mixture Stream, Tf,2
Oxygen Stream MO2 , TO2 = Tf,1
To Patient
Liquid Anesthetic Drug Stream Ml , Tl = Tf,1
L Slg , Evaporation
Control Volume, V Control Surface, A
Figure Pr.5.10(a) A liquid anesthetic drug is evaporated, mixed with oxygen, and heated in a vaporizer. The drug is sprayed into the vaporizer.
OBJECTIVE: (a) Draw the thermal circuit diagram for the control volume V . (b) Determine the exit ﬂuid stream temperature Tf,2 , for the given conditions. SOLUTION: (a) Figure Pr.5.10(b) shows the thermal circuit diagram.  Qku
L
Tf,1 Mf
Tf,2 Qu,1
Slg
Qu,2
Figure Pr.5.10(b) Thermal circuit diagram.
(b) From continuity, we have M˙ f = M˙ O2 + M˙ l . From Figure Pr.5.10(b), and from the energy equation, (5.17), we have QA = Qku L + Qu,2 − Qu,1 = S˙ lg = Qku L + M˙ f cp,f (Tf,2 − Tf,1 ) = S˙ lg . Solving for Tf,2 , we have Tf,2
S˙ lg − Qku L M˙ f cp,f −M˙ l ∆hlg − Qku L = Tf,1 + . (M˙ O + M˙ l )cp,f = Tf,1 +
2
463
From Table C.28, we have, for p = 1 atm = 0.1013 MPa, ∆hlg = 2.172 × 105 J/kg
Table C.28.
From Table C.22, for oxygen at T = 300 K, we have cp,f = 920 J/kgK Table C.22. Using the numerical values, we have Tf,2
= = =
−2 × 10−5 (kg/s) × 2.172 × 105 (J/kg) − [−6(W)] (2 × 10−4 + 2 × 10−5 )(kg/s) × 920(J/kg◦C) ◦ 15( C) + 8.182(◦C) 23.18◦C.
15(◦C) +
COMMENT: Since the vaporizer unit is portable, the surface convection heating Qku L , which is provided by Joule heating needs to be minimized and for an ideally insulated tube this would give S˙ e,J = 5 W. Note that we did not addressed the sensible heat required to heat the droplet from Tf,1 to Tlg (Tlg = 249.2 K). This heat is not signiﬁcant because M˙ O2 is much larger than M˙ l . In Problem 8.2, a more complete analysisdescription is made. In Section 6.9, we will discuss the heat and mass transfer resistances Rku and RDu , which inﬂuence the droplet evaporation rate.
464
PROBLEM 5.11.FUN GIVEN: In surface evaporation from permeable membranes, the heat for evaporation is partly provided by the ambient gas (by surface convection) and partly by the liquid reservoir (through the conductionconvection heat transfer through the membrane). This is shown in Figure Pr.5.11. The ambient gas may contain species other than the vapor produced by the evaporation. These other species are called the inert or noncondensables and provide a resistance to the vapor mass transfer. This will be discussed in Section 6.9 and here we do not address the mass transfer resistance and assume that the gas is made of the vapor only. Consider using superheated steam to evaporate water from a permeable membrane. We assume that the gas is moving and has a farﬁeld temperature Tf,∞ and that there is a surfaceconvection heat transfer resistance Rku,2∞ between the surface and the gas stream. This is also shown in Figure Pr.5.11. The surface temperature Tf,2 is equal to the saturation temperature Tlg (pg ). Also, since (Rk,u )12 depends on M˙ l , the liquid mass ﬂow is determined such that it simultaneously satisﬁes (Qk,u )12 and S˙ lg . Tf,1 = 100◦C, Tf,∞ = 110◦C, Tlg = 95◦C, Ak,a = Aku = 1 m2 , Rku,2∞ = 0.25 K/W, k = 1 W/mK, L = 1 cm. Determine the water properties at T = 373.15 K, from Table C.27. SKETCH: Figure Pr.5.11 shows the permeable membrane with the water ﬂowing through it and evaporating on the steammembrane interface. Permeable Membrane Liquid (Water)
Ak,u = Aku
k
Ml
Ml
Gas Stream Tf, , uf,
Ml
L
Tlg = Tf,2 Tf,
Tf,1
Slg (Rku)12 Rku,2 (Qk,u)12  Qku,2
Figure Pr.5.11 Water supplied through a permeable membrane is evaporated on the gasside interface. The thermal circuit diagram is also shown.
OBJECTIVE: Determine M˙ l for the given conditions. SOLUTION: From Figure Pr.5.11, the energy equation is QA = −(Qk,u )12 + Qku,2∞ = S˙ lg or −
Tf,2 − Tf,∞ Tf,1 − Tf,2 + = −M˙ l ∆hlg . (Rk,u )12 (Rku )12
From (5.14), we have (Rk,u )12
=
ePeL − 1 L , Ak,u k PeL ePeL
PeL
=
uf L , α
α = 465
k , (ρcp )f
uf =
M˙ l . ρf Ak,u
From Table C.27, we have ρf = 958 kg/m3 cp,f = 4,217 J/kgK
Table C.27 Table C.27
∆hlg = 2.257 × 10 J/kg 6
Table C.27.
Then using the numerical values, we have α
=
PeL
=
1(W/mK) = 2.475 × 10−7 m2 /s 958(kg/m3 ) × 4,217(J/kg) M˙ l × 10−2 (m) = 42.17M˙ l (s/kg) 958(kg/m3 ) × 1(m2 ) × 2.475 × 10−7 (m2 /s) ˙
(Rk,u )12
=
e42.17Ml (s/kg) − 1 10−2 (m) . 2 1(m ) × 1(W/mK) 42.17M˙ l (s/kg) × e42.17M˙ l (s/kg)
The energy equation becomes −
(100 − 95)(◦C) (95 − 110)(◦C) = −M˙ l (s/kg) × 2.257 × 106 (J/kg). + (Rk,u )12 0.25(◦C/W)
Solving for M˙ l , we have M˙ l = 0.5476 kg/s. COMMENT: This liquid ﬂow rate corresponds to PeL = 23.09 and uf = 0.5716 mm/s, which are relatively large, This is because nearly all of the heat is arriving from the liquid side. The surfaceconvection resistance Rku,2∞ corresponds to a laminar ﬂow parallel to the interface. Means of reducing this resistance will be discussed in Chapter 6 (i.e., using perpendicular ﬂow or turbulent ﬂow, etc.)
466
PROBLEM 5.12.FAM GIVEN: Consider an adiabatic methaneair ﬂame in a packedbed of spherical alumina particles of diameter D = 1 mm and a bed porosity = 0.4. Assume the average temperature of the bed to be T = 1,300 K. The inlet conditions are Tf,1 = 289 K and p1 = 1 atm and the mixture is stoichiometric. OBJECTIVE: (a) Determine the eﬀective thermal conductivity for the bed. Use the radiant conductivity correlation for spheres given by a3 ks −1 a2 kr = 4DσSB T Fr = 4DσSB T a1 r tan + a4 r 4DσSB T 3 3
3
correlation for radiant conductivity for packed bed of particle with 0.4 ≤ ≤ 0.6,
where Fr is the radiant exchange factor, a1 = 0.5756, a2 = 1.5353, a3 = 0.8011, a4 = 0.1843. (b) Determine the eﬀective radiant conductivity. (c) Using the sum of these conductivities, at the above average temperature, determine the adiabatic ﬂame speed. Use the results of Example 5.4, as needed. SOLUTION: (a) The eﬀective thermal conductivity for a bed of spherical particles in random arrangement can be estimated from (3.28), k = kf
ks kf
0.280−0.757 log()−0.057 log(ks /kf ) .
For alumina at T = 1300 K, interpolating from Table C.14 gives ks = 6 W/mK. For air at T = 1300 K, interpolating from Table C.22 gives kf = 0.0791 W/mK. For a porosity = 0.4, the eﬀective thermal conductivity is
k
6(W/mK) = 0.0791(W/mK) 0.0791(W/mK) = 0.6158 W/mK.
6(W/mK) 0.280−0.757 log(0.4)−0.057 log[ 0.0791(W/mK) ]
(b) The radiant thermal conductivity using the diﬀusion approximation can be estimated from the given relation a3 a2 kf kr = 4DσSB T 3 a1 r tan−1 + a 4 . r 4DσSB T 3 For D = 1 mm and using r = 0.78 obtained from Table C.18 (a1 , a2 , a3 , and a4 are constants) we have kr
=
4 × 0.001(m) × 5.67x10−8 (W/m K4 ) × (1300)3 (K3 ) 0.8011 1.5353 0.0791(W/mK) 0.5756 × 0.78 × tan−1 × + 0.1843 2 0.78 4 × 0.001(m) × 5.67x10−8 (W/m K4 ) × (1300)3 (K3 )
=
0.19 W/mK.
2
(c) The ﬂame speed, assuming a zerothorder reaction, and constant properties is given by (5.55),
uf,1
2
Rg Tf,2 2kf −∆Ea = ar exp( ) ρf,1 ρF,1 cp,f Rg Tf,2 ∆Ea (Tf,2 − Tf,1 )
1/2 .
The inlet conditions are Tf,1 = 289 K, p = 1 atm, and the mixture is stoichiometric. These are the conditions for Example 5.4 and the ﬂame speed obtained there is uf,1 = 0.4109 m/s. Using the expression for uf,1 from above, 467
and noting from Section 5.4.6 that for a porous medium k = k + kr , the ratio of the ﬂame speeds with and without the porous medium is (uf,1 )packed bed (uf,1 )plain medium
=
k + kr kf
1/2 .
Therefore, (uf,1 )packed bed
=
0.4109(m/s)
0.63(W/mK) + 0.19(W/mK) 0.0824(W/mK)
1/2 = 1.30 m/s.
COMMENT: An increase of the ﬂame speed (and burning rate) is achieved by increasing the medium conductivity. The presence of the highconductivity solid and the occurrence of the reaction in the gas phase only results in a local temperature diﬀerence between the solid and the gas. This leads to high local gas temperatures (a phenomenon called the local superadiabatic ﬂame temperature).
468
PROBLEM 5.13.FAM GIVEN: To achieve the same ﬂame speed that was obtained using the porous medium in Problem Pr.5.12, turbulent ﬂow may be used. The turbulent intensity aﬀects the ﬂame speed. The laminar ﬂame speed is uf,1 = 0.4109 m/s and the packedbed ﬂame speed is 1.30 m/s. OBJECTIVE: Using the same adiabatic, stoichiometric methaneair ﬂame, determine the needed turbulent intensity T u to achieve the same ﬂame speed. SOLUTION: The turbulent intensity is deﬁned by (5.70) as Tu =
uf,1 2 u2f,1
.
The ratio of the turbulent ﬂame speed uf,1 to the laminar ﬂame speed uf,1 is correlated to the mean square of the velocity ﬂuctuation u2 f,1 by (5.71), i.e., uf,1 2 uf,1 =1+ 2 . uf,1 uf,1 Here the laminar ﬂame speed is that obtained in Example 5.4, i.e., uf,1 = 0.4109 m/s and the packedbed ﬂame speed of Problem 5.12 is uf,1 = uf,1 = 1.30 m/s. If a turbulent ﬂame speed equal to the ﬂame speed obtained within the packed bed (from Problem 5.12) is desired, the required mean square of the velocity ﬂuctuation is 1.30(m/s) uf,1 2 2 2 2 − 1 = 0.3653 (m/s)2 . uf,1 = uf,1 − 1 = (0.4109) (m/s) uf,1 0.4109(m/s) For this mean square of the velocity ﬂuctuation, the turbulent intensity is Tu =
uf,1 2 u2f,1
=
0.3653(m/s)2 = 0.2162. (1.30)2 (m/s)2
COMMENT: The correlation (5.71) applies to low turbulent intensity and is an approximation.
469
PROBLEM 5.14.FUN GIVEN: In a premixed fueloxidant stream, as the fuel is oxidized, its density ρF decreases and this decrease inﬂuences the combustion and chemical kinetics. Consider a ﬁrstorder chemicalkinetic model for reaction of methane and oxygen given by n˙ r,F = −ar ρF e−∆Ea /Rg Tf . Start with the fuelspecies conservation equation (B.51). Assume onedimensional, incompressible ﬂow, and negligible mass diﬀusion. Use a constant, average temperature Tf = (Tf,0 + Tf,L )/2 to represent the average temperature over length L. The ﬁnal expression is −∆Ea /Rg Tf
ρF,L = ρF,0 e(−ar L/uf )e
.
OBJECTIVE: Derive the relation for the fuel density as it undergoes reaction over a length L with a velocity uf , inlet density ρF,0 , and temperature Tf,0 . SOLUTION: From (B.51), for a steadystate condition, onedimensional ﬂow, and negligible diﬀusion, we have the following species F (fuel) conservation equation, ∆Ea − d ρF uf = n˙ r,F = n˙ r,F = −ar ρF e Rg Tf . dx Assuming an incompressible ﬂow and using (B.50), the above equation becomes ∆Ea − dρF R g Tf . = −ar ρF e uf dx After rearranging the above and integrating over the length L, we have,
ρF,L
ρF,0
dρF = ρF
L
∆Ea −ar − Rg Tf e dx uf
∆Ea −ar L − Rg Tf ρF,L = e . ln ρF,0 uf Solving for ρF,L , the fuel density at location L, we have, −∆Ea /Rg Tf
ρF,L = ρF,0 e(−ar L/uf )e
.
COMMENT: This chemicalkinetic model can be compared with (2.19), with aF = 1 and aO = 0. This model is used in combustion in porous media. Note that the gas temperature used is the average over the length L and is considered constant. The above expression is a useful approximation for well mixed gas.
470
PROBLEM 5.15.FAM GIVEN: To estimate the ﬂame speed of the atmospheric (p1 = 1 atm) gasoline (assume it is octane) reaction with air (premixed), assume that the chemical kinetic model for this reaction can be approximated as being zeroth order and represented by a preexponential factor ar and an activation energy ∆Ea given in Table 5.3 for zerothorder reaction of methane and air. The reaction is represented by C8 H18 + 12.5O2 + 47.0N2 → 8CO2 + 9H2 O + 47.0N2 . Use Figure 5.9 for the adiabatic ﬂame temperature, Table 5.2 for the heat of reaction, and evaluate the properties of air at the average ﬂame temperature Tf δ = (Tf,1 + Tf,2 )/2. Determine the density ρf,1 from the idealgas law using Tf,1 = 16◦C. OBJECTIVE: Determine (a) the Zel’dovich number Ze, and (b) the ﬂame speed uf,1 , for a premixed gasolineair ﬂame. SOLUTION: (a)The Zel’dovich number is given by (5.53), Ze =
∆Ea (Tf,2 − Tf,1 ) . 2 Rg Tf,2
The adiabatic ﬂow temperature Tf,2 is found from Figure 5.9 (for octane), Tf,2 =
2,310◦C = 2,583.15 K
Tf,1 =
16◦C = 289.15K.
The chemical kinetic constants are assumed to be those given in Table 5.3 for methane oxidization (for a zerothorder reaction), ar ∆Ea
= =
1.3 × 108 kg/m3 s Table 5.3 2.1 × 108 J/kmole Table 5.3.
The Zel’dovich number is then Ze
2.1 × 108 (J/kmole)(2,583.15 − 289.15)(K) . 8,314(J/kmoleK)(2,583.15)2 (K)2 8.684.
= =
As Ze> 5, we can use the high activation energy approximation. Equation (5.55) can then be used for uf,1 . (b) The ﬂame speed is given by (5.55), 1/2 ∆Ea − a 2k f r e Rg Tf,2 , = ρf,1 cp,f ρF,1 Ze
uf,1
where kf , ρf,1 , and cp,f need to be speciﬁed. The average speciﬁc heat capacity is given by (5.35), cp,f
=
−∆hr,F (ρF,1 /ρf,1 ) . Tf,2 − Tf,1
The heat of reaction for octane is found in Table 5.2, ∆hr,F = −48.37 × 106 J/kg
Table 5.2.
For the given chemical reaction, ρF,1 ρf,1
=
νC8 H18 MC8 H18 = 0.06234. νC8 H18 MC8 H18 + νO2 MO2 + νN2 MN2 471
The molecular masses are found in Table C.4. The speciﬁc heat capacity is then cp,f
= =
−48.37 × 106 ( J/kg) × 0.06234 (2,310 − 16)(K) 1,315 J/kgK.
The average ﬂame temperature is Tf δ =
Tf,1 + Tf,2 = 1,436 K. 2
From Table C.22, at T = 1,436 K, we have kf = 0.08447 W/mK
Table C.22.
The gas density at Tf,1 and p1 = 1 atm is found from (3.18), i.e., ρf,1
= =
p 1.013 × 105 (Pa) × 30.26(kg/kmole) = Rg /M1 Tf,1 8.314 × 103 (J/kmoleK) × 289.15(K) 1.275 kg/m3 ,
where M1
= = =
νC8 H18 MC8 H18 + νO2 MO2 + νN2 MN2 νC8 H18 + νO2 + νN2 [1 × (8 × 12.011 + 18 × 1.006) + 12.5 × 2 × 15.99 × 5.999 + 47 × 2 × 14.007](kg/kmole) 1 + 12.5 + 47 30.26 kg/kmole.
The adiabatic ﬂame speed is then 8 3 2 × 0.08447 × 1.3 × 10 (kg/m s) exp − uf,1
2.1 × 108 (J/kgmole) 8.314 × 103 (J/kg moleK) × 2,583.15(K) 1.275(kg/m3 ) × 1,315(J/kgK) × 0.06234 × 1.275(kg/m2 ) × 8.684
=
=
1.037 m/s = 103.7 cm/s.
1/2
COMMENT: The chemical kinetic model used here is not a realistic representation of the gasolineair reaction. The measured adiabatic ﬂame speed at one atm pressure is given in table C.21(a) as uf,1 = 0.38 m/s. This is much lower than the predicted value. The use of a more realistic kinetic model and temperature dependent properties results in the need for a numerical solution of the energy equation (5.27) and the species conservation equation (5.29). This is commonly done to predict the ﬂame speed.
472
PROBLEM 5.16.FUN GIVEN: In order to achieve higher ﬂame temperatures, pure oxygen is used instead of air in burning hydrocarbons. Consider stoichiometric methaneoxygen premixed combustion. Tf,1 = 16◦C, cp,f = 3,800 J/kgK. Use the chemical kinetic constants for the zerothorder reaction given in Table 5.3. Use the thermal conductivity of air at the average gas temperature Tf δ = (Tf,1 + Tf,2 )/2 for the mixture. OBJECTIVE: (a) Determine the adiabatic ﬂame temperature Tf,2 . (b) Determine the laminar, adiabatic ﬂame speed uf,1 . (c) Compare this ﬂame speed with the laminar, adiabatic ﬂame speed of methaneair in Example 5.4 (i.e., uf,1 = 0.4109 m/s) SOLUTION: (a) The adiabatic ﬂame temperature is given by (5.35), i.e., Tf,2 = Tf,1 −
∆hr,F cp,f
ρF ρf
. 1
We need to determine the fuel mass fraction (ρF /ρf ). This is found from the stoichiometric reaction CH4 + 2O2 → 2CO2 + 2H2 O. Then ρF,1 ρf,1
νCH4 MCH4 νCH4 MCH4 + νO2 MO2 1 × (12.011 + 1.008 × 4) 1 × (12.011 + 1.008 × 4) + 2 × 2 × 15.999 16.04 = 0.2004. 80.04
= = =
Then from Table 5.2, we have −∆hr,F = 5.553 × 107 J/kg and Tf,2
=
−5.553 × 107 (J/kg) × 0.2004 3,800(J/kgK) 16(◦C) + 2,928(◦C)
=
2,944◦C.
=
16(◦C) −
(b) The laminar, adiabatic ﬂame speed is given by (5.55), i.e., 1/2 ∆Ea − a 2k f r e Rg Tf,2 . = ρf cp,f ρF,1 Ze
uf,1
The chemicalkinetic constants are given in Table 5.3. For methane, with a zerothorder reaction, ar = 1.3 × 108 kg/m3 and ∆Ea = 2.10 × 108 J/kmole. The Zel’dovich number, Ze, is Ze
= = =
∆Ea (Tf,2 − Tf,1 ) 2 Rg Tf,2 2.1 × 108 (J/kmole) × 2,928(K) 8.314 × 103 (J/kmoleK) × (2,944 + 273.15)2 (K2 ) 7.146. 473
As Ze> 5, (5.55) can be used to ﬁnd uf,1 . The density ρf,1 is determined from the idealgas relation (3.18), i.e., ρf,1 =
p1 , (Rg /M1 )Tf,1
where M1
= = =
νCH4 MCH4 + νO2 MO2 νCH4 + νO2 [1 × (12.011 + 1.008 × 4) + 2 × 2 × 15.999](kg/kmole) 1+2 80.04 = 26.68 kg/kmole. 3
Then ρf,1
=
=
1.013 × 105 (Pa) 8.314 × 103 (J/kmoleK) × (289.15)(K) 26.68(kg/kmole) 1.124 kg/m3 .
The air thermal conductivity is found from Table C.22 at the average ﬂame temperature Tf δ
(16 + 2,944)(◦C) + 273.15(◦C) 2 = 1,753 K
=
From Table C.22, for air at T = 1,753 K, we have: kf = 0.09520 W/mK. Then 2 × 0.09520(W/mK) × 1.3 × 108 (kg/m3 s) × uf,1 = 1.124(kg/m3 ) × 3,800(J/kgK) × 1.124 × 0.2004(kg/m3 ) × 7.146 1/2 2.10 × 108 (J/kg) exp − 8.314 × 103 (J/kgK) × (2,944 + 273.15)(K) = [3.600 × 104 (m2 /s2 ) × 3.893 × 10−4 ]1/2 = 3.744 m/s. (c) Comparing with the results of Example 5.4, with Tf,2 = 1,918◦C, the adiabatic ﬂame temperature is much higher for the methane burning in pure oxygen. The predicted laminar adiabatic ﬂame speed is also much higher in pure oxygen, compared to the value of 0.4109 m/s which is predicted in air. COMMENT: The measured laminar ﬂame speed is uf,1 = 6.919 m/s. The diﬀerence is due to the constant thermal conductivity used in the predictions, and due to the simpliﬁed chemical kinetic model used.
474
PROBLEM 5.17.FAM.S GIVEN: Surfaceradiation drying of wet pulp in paper production uses permeable ceramic foams for both combustion and surface emission. This is shown Figure Pr.5.17(a). The premixed, gaseous fuelair ﬂows into the foam, and after an initial ignition, it undergoes steady combustion. The ﬂue gas heats the foam and leaves while the foam radiates to the load (wet paper). The steady combustion requires a mixture ﬂow rate that in turn is determined by the heat transfer. This ﬂow rate, or speciﬁcally the velocity uf,1 , can be several times the laminar, adiabatic ﬂame speed given by (5.55). Consider stoichiometric methaneair combustion with the mixture arriving at Tf,1 and at 1 atm pressure. Assume that F2p = 1 and both the radiantburner surface and the wet paper are blackbody surfaces. uf,1 = 0.2 m/s, Tf,1 = 16◦C, a = 25 cm, w = 60 cm, Tp = 60◦C, cp,f = 1,611 J/kgK, ρf,1 = 1.164 kg/m3 , (ρF /ρf )1 = 0.05519. SKETCH: Figure Pr.5.17(a) shows the permeable ceramic foam with surface radiation to a wetpulp sheet being dried.
Paper Drying System Pulp, Tp a w
Qu,2
Qr,2p
Qu,1
Tf,2 = Ts Sr,c Occurring at Top of Heater
Premixed MethaneAir Stream
Figure Pr.5.17(a) A surfaceradiation burner used for drying a wetpulp sheet.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Assuming local thermal equilibrium between the gas and the permeable solid (i.e., the ceramic foam), determine the radiantsurface temperature Ts = Tf,2 and the radiation heat transfer rate Qr,2p . (c) What is the radiant heat transfer eﬃciency of this heater? SOLUTION: (a) Figure Pr.5.17(b) shows the thermal circuit diagram. In addition to the surfaceradiation heat transfer, the convection heat ﬂow Qu,2 is partly exchanged with the wetpulp sheet by surface convection Qku (discussed in Chapter 6). (b) From Figure Pr.5.17(b), the integralsurface energy equation becomes Qu,2 + Qr,2 − Qu,1 = S˙ r,c , where Qu,2 − Qu,1 Qr,2 Au ˙ Sr,c
= ρf,1 cp,f uf,1 Au (Tf,2 − Tf,1 ) Eb,2 − Eb,p 4 = Qr,2p = = Ar,2 σSB (Tf,2 − Tp4 ), as r,2 = r,p = 1 1 Ar,2 F2p = Ar,2 = aw = −ρF,1 uf,1 Au ∆hr,F [from (5.34)]. 475
Tf,1 Qu,1 (Sr,c)12 Q2 = 0
Control Surface, A2 Ts = Tf,2 Eb,f,2
Qu,2
(Rr,5)2p
Qu
Qr,2 = Qr,2p
Qku
Eb,p Tp
Figure Pr.5.17(b) Thermal circuit diagram.
Combining these, we will have the equation to solve for Tf,2 , i.e.,
4 − Tp4 ) = − ρf,1 cp,f uf,1 (Tf,2 − Tf,1 ) + σSB (Tf,2
ρF,1 ρf,1 uf,1 ∆hr,F . ρf,1
Then we can solve for Qr,2p . Alternatively, we can solve the above equations simultaneously using a software (such as SOPHT). From Table 5.2, we have ∆hr,F = −5.553 × 107 J/kg. Then, using the numerical values , we have 1.164(kg/m3 ) × 1,611(J/kgK) × 0.2(m/s) × [Tf,2 − 289.15(K)] + 5.67 × 10−8 (W/m2 K4 ) × 4 [Tf,2 − 333.154 (K4 )] = −1.164(kg/m3 ) × 0.05519 × 0.2(m/s) × (−5.553 × 107 )(J/kg) 4 Qr,2p = 0.25(m) × 0.6(m) × 5.670 × 10−8 (W/m2 K4 ) × [Tf,2 − 333.154 (K4 )].
The solutions are
Tf,2 Qr,2p
= 1,476 K. = 40,259 W.
(c) The surfaceradiation eﬃciency η is deﬁned as
η
=
Qr,2p = 37.62%. −ρF,1 uf,1 Au ∆hr,F
COMMENT: Figure Pr.5.17(c) shows the variation of η with respect to uf,1 . Note that η decreases as uf,1 decreases. It is possible to avoid the low eﬃciency associated with the high velocities. This can be done by using a distributed fuel supply, instead of the premixed fuelair considered here, along with an impermeable radiation surface. 476
0.5 0.4
D
0.3 0.2 0.1 0 0
1
2
3
4
5
uf,1 , m/s Figure Pr.5.17(c) Variation of surfaceradiation eﬃciency with respect to the ﬂame speed uf,1 .
477
PROBLEM 5.18.FUN GIVEN: In burning gaseous fuel in a tube or in a porous medium, it is possible to locally create gas temperatures above the adiabatic ﬂame temperature. This is done by conduction of heat through the bounding solid (tube or solid matrix) from the high temperature region to the lower temperature region. This is called heat recirculation and the process of combustion with this local increase in the gas temperature is called the superadiabatic combustion. This is rendered for a premixed fuel (methane) and oxidant (air) in Figure Pr.5.18(a). In this idealized rendering, three diﬀerent regions are identiﬁed, namely the gaspreheat, combustion, and solidheating regions. The heat recirculation begins as surface convection from the ﬂue gas to the solid, then it is conducted along the solid (ﬂowing opposite to the gas ﬂow), and is ﬁnally returned to the gas (premixed fueloxidant) by surface convection. The idealized gas temperature distribution (for the three regions) is also shown in the ﬁgure and can be represented by the internodal energy conservation equation and the temperatures labeled Tf,1 to Tf,4 . The surface convection out of the solidheating region is given per unit gas ﬂow crosssectional area, i.e., Qku /Au . Qku /Au = 5 × 104 W/m2 , Tf,1 = 20◦C. Assume negligible heat loss in the combustion region (i.e., an adiabatic combustion region). Use the heat of combustion of methane ∆hr,F and the stoichiometric mass fraction of the fuel from Table C.21(a), and a constant speciﬁc heat capacity corresponding to air at T = 1,500 K. SKETCH: Figure Pr.5.18(a) shows the ﬂow and reaction, and the anticipated gas temperature distribution along the tube.
(i) Local Superadiabatic Combustion Heat Circulation Through Axial Tube Conduction Tube Wall
Tf,1 Premixed MethaneAir Stream
Tf,2
Sr,c
Tf,3
Tf,4
(Mcp)f Qku
Au
Gas Preheat Region
Combustion Region
SolidHeating Region
Heat Recirculation  Qku =  Qku,34
(ii) Ideal Axial Temperature Distribution Tf Tf,4 Tf,3
,Texcess
Tf,2 Tf,1
,Tcombustion x
Figure Pr.5.18(a) Heat recirculation and local superadiabatic temperature in combustion in a tube.
OBJECTIVE: (a) Draw the thermal circuit diagram. Determine (b) Tf,2 , (c) Tf,3 , (d) Tf,4 , and (e) the excess temperature ∆Texcess = Tf,4 − Tf,3 for stoichiometric, premixed methaneair combustion. (f) Comment on the eﬀect of using temperaturedependent speciﬁc heat capacity on the predicted excess temperature. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.5.18(b). The surfaceconvection heat transfer out of the solidheating region control volumes is equal in magnitude and opposite in sign to that entering the preheat region. 478
Combustion Region Control Volume Preheat Region SolidHeating Region Control Volume Control Volume Fluid Stream T S T T Tf,4 r,c f,2 f,3 f,1 (Mcp)f
Qu,1 Qku Solid
Qku Qk,s Heat Recirculation
Figure Pr.5.18(b) Thermal circuit diagram.
(b) There are three regions and we write the energy equation (5.33) for each of them. The control volumes are shown in Figure Pr.5.18(b). The energy equations are −Qu,1 + Qu,2 − Qku = 0 −Qu,2 + Qu,3 = S˙ r,c
gaspreheat region
−Qu,3 + Qu,4 + Qku = 0
solidheating region.
combustion region
Now we use a constant and uniform cp,f , similar to that used in (5.34), and the energy equations then become Qku Au
=
(ρcp )f (Tf,3 − Tf,2 )
=
S˙ r,c Au
Qku Au
=
0.
(ρcp )f (Tf,2 − Tf,1 ) −
(ρcp )f (Tf,4 − Tf,3 ) + Next, from (5.34), we replace S˙ r,c by
S˙ r,c = Au (ρF uf )2 ∆hr,F . Then as in (5.38), we have Tf,3 = Tf,2 −
∆hr,F (ρF /ρf )2 . cp,f
The density and the speciﬁc heat capacity of air at T = 1,500 K is found from Table C.22, i.e., ρf = 0.235 kg/m3 cp,f = 1,202 J/kgK
Table C.22 Table C.22.
From Table C.21(a), for the methaneair stoichiometric combustion, we have ∆hr,F = −5.553 × 107 J/kg ρF = 0.0552 ρf 2
Table C.21(a) Table C.21(a)
For the gaspreheat region, we have Tf,2
= Tf,1 +
Qku /Au . (ρcp )f
The second term on the right side will appear in all the energy equations, and we therefore calculate its value, 2
5 × 104 W/m Qku /Au = 177.0◦C. = 3 (ρcp )f 0.235(kg/m ) × 1,202(J/kgK) 479
The expression for Tf,2 then becomes Tf,2 = 20(◦C) + 177.0(◦C) = 197.0◦C. (c) In the combustion region, we have Tf,3
= =
−5.553 × 107 (J/kg) × 0.0552 1,202(J/kgK) 191.0(◦C) + 2,550(◦C) = 2,747◦C.
197.0(◦C) −
(d) In the solidheating region, we have Tf,4
Qku /Au (ρcp )f ◦ 2,747( C) − 177.0(◦C) = 2,569◦C.
= Tf,3 − =
(e) The excess temperature is Tf,4 − Tf,3
=
Qku /Au = 177.0◦C. (ρcp )f
(f) As shown in Example 5.4, the ﬂuegas speciﬁc heat capacity is larger than that of air, because it contains CO2 and H2 O. The speciﬁc heat capacity of methane is also strongly temperature dependent [Figure 3.1(b)]. In principle, we should use a nonuniform mixture speciﬁc heat capacity. Using a larger mixture speciﬁc heat capacity results in a smaller excess and adiabatic ﬂame temperatures. COMMENT: Note that due to the smaller cp,f , the adiabatic ﬂame temperature is larger than that found in Example 5.4. In chapter 6, we will determine Qku from the average gas and solidsurface temperatures. Surfaceradiation heat transfer among the various regions can also be signiﬁcant.
480
PROBLEM 5.19.FAM GIVEN: Consider nonadiabatic (i.e., lateral heat losses Qloss not being negligible), onedimensional ﬂow and reaction of premixed methaneair gaseous mixture. There is a heat loss, per unit ﬂow crosssectional area, Qloss /Au = qloss = 105 W/m2 . The mixture is stoichiometric and the initial temperature and pressure are Tf,1 = 298 K and p1 = 1 atm. Use the Table C.21(a) data for adiabatic, stoichiometric ﬂame (ρF /ρf , ∆hr,F , and uf,1 ), and for the average speciﬁc heat use cp,f = 1,611 J/kgK. OBJECTIVE: Determine the ﬁnal temperature Tf,2 . SOLUTION: The integralvolume energy equation applied to a control volume including the ﬂame, under steadystate conditions, reduces to QA = S˙ r,c . The energy generation occurs by conversion from chemicalbond to thermal energy S˙ r,c . The net heat ﬂux leaving the control surface QA has contributions of convection heat transfer in and out of the ﬂame and heat loss. The energy equation is (5.34), i.e., − (ρf cp,f uf Tf )1 Au + (ρf cp,f uf Tf )2 Au + Qloss = −n˙ r,F ∆hr,F . Dividing the equation by Au and noting that n˙ r,F = (ρF uf )1 Au we have − (ρf cp,f uf Tf )1 + (ρf cp,f uf Tf )2 + qloss = − (ρF uf )1 ∆hr,F , where qloss = Qloss /Au and Au is the ﬂow crosssection area. From the conservation of mass equation (continuity) for a steadystate, uniform ﬂow, we have ρf uf = (ρf uf )1 = (ρf uf )2 = constant, and the energy equation is ﬁnally written as − (cp,f Tf )1 + (cp,f Tf )2 +
qloss =− (ρf uf )1
ρF ρf
∆hr,F . 1
For the stoichiometric adiabatic reaction between methane and air, from Table C.21(a) we obtain ∆hr,F = −55.53 MJ/kgF , ρF /ρf = 0.0552, and uf,1 = 0.338 m/s. The temperatureaverage speciﬁc heat is assumed to be cp,f = 1,611 J/kgK. The temperature and pressure at the inlet are Tf,1 = 298 K and p = 1 atm = 101.3 kPa. The density of the gas mixture at the inlet, assuming idealgas behavior and calculating the molecular weight for the gas mixture from the stoichiometric reaction (see Example 5.2), is given by ρf,1 =
p Rg Tf,1 Mf
=
1.013 × 103 (Pa) 3 = 1.130 kg/m . 8,314(J/kmoleK) 298(K) 27.63(kmole)
Solving the energy equation for Tf,2 we have ρF ∆hr,F qloss Tf,2 = Tf,1 − − ρf 1 cp,f (ρf uf )1 cp,f = =
2 −55.53 × 106 (J/kgF ) 105 (W/m ) − 3 1,611(J/kgK) 1.130(kg/m ) × 0.338(m/s) × 1,611(J/kgK) 298(K) + 1,903(K) − 162.5(K) = 2,039 K.
298(K) − 0.0552(kgF /kgf )
COMMENT: The increase in the heat loss will eventually cause the extinguishment of the ﬂame. Note that the adiabatic ﬂame temperature (Qloss = 0) is Tf,2 = 2,187 K and that this heat loss results in a reduction of 162.5 K. The heat loss also inﬂuences the ﬂame speed uf,1 . 481
PROBLEM 5.20.FAM.S GIVEN: The adiabatic ﬂame temperature and speed are referred to the condition of no lateral heat losses (Qloss = 0) in combustion of a unidirectional premixed fueloxidant stream. The presence of such losses or gains decreases the ﬂame temperature Tf,2 , given by (5.35), and also the ﬂame speed given by (5.55). This can continue until the ﬂame temperature decreases below a threshold temperature required to sustain ignition and combustion. Tf,1 = 16◦C, ρf,1 = 1.164 kg/m3 , (ρF /ρf )1 = 0.05519, cp,f = 1,611 J/kgK, kf = 0.07939 W/mK, Au = 10−2 2 m , 0 ≤ Qloss ≤ 800 W. Use Tables 5.2 and 5.3 for ∆hr,F , ar , and ∆Ea . OBJECTIVE: (a) Consider a laminar, stoichiometric premixed methane air combustion. For the conditions given above, plot the ﬂame temperature Tf,2 and the ﬂame uf,1 speed for the given range of Qloss . (b) Comment on the quenching of the ﬂame as Qloss increases and Tf,2 decreases. SOLUTION: (a) The ﬂame temperature Tf,2 is given by (5.35), and ∆hr,F is given in Table 5.2, i.e., ∆hr,F ρF Qloss Tf,2 = Tf,1 − − cp,f ρf 1 Au ρf,1 cp,f uf,1 Qloss (W) −5.553 × 107 (J/kg) × 0.05519 − −2 2 1,611(J/kgK) 10 (m ) × 1.164(kg/m3 ) × 1,611(J/kgK) × uf,1 (m/s) Qloss = 289.15(K) + 1,902(K) − 5.333 × 10−2 (mK/sW) × . uf,1
= 16(◦C) −
The ﬂame speed is given by (5.55), and ar and ∆Ea are given in Table 5.3, i.e., 1/2 ∆Ea − 2kf ar e Rg Tf,2 uf,1 = ρf cp,f ρF,1 Ze Ze
=
∆Ea (Tf,2 − Tf,1 ) 2 Rg Tf,1
=
2.10 × 108 (J/kmole) × (Tf,2 − 289.15(K)) 2 8.314 × 103 (J/kmoleK) × Tf,2
=
2.526 × 104 (K) ×
uf,1
=
(Tf,2 − 289.15(K)) 2 Tf,2
2 × 0.07939(W/mK) × 1.3 × 10 (kg/m s) 1.164(kg/m3 ) × 1,611(J/kgK) × 0.05519 × 1.164(kg/m3 ) × Ze 8
3
1/2 2.10 × 108 (J/kmole) e 8,314(J/kmoleK) × Tf,2 −
1/2 2.526 × 104 (K) 1.713 × 10 (m /s ) Tf,2 e = . Ze
5
2
2
−
The solution to these three equations is fully deﬁned by the speciﬁcation of Qloss and Tf,2 . Thus, plots of Tf,2 and uf,1 as functions of Qloss can be made. These are shown in Figure Pr.5.20. While the decrease in Tf,2 is not very noticeable, uf,1 decreases signiﬁcantly with increase in Qloss . This is because of the Tf,2 proportionality through Ze and the exponential relation dependence through the activation term. (b) If we continue to increase Qloss beyond 800 W, the ﬂame may quench. COMMENT: If we continue to increase Qloss , at Qloss 880 W, the ﬂame speed would tend to zero and no solution will be found. This can be deﬁned as the theoretical quenching limit. 482
uf,1 x 104, m/s, Tf,2 , K
4200 uf,1 x 104, Flame Speed
3600 3000 2400
Tf,2 , Flame Temperature
Quenching
1800 0 Adiabatic
160
320
480
640
800
Qloss , W
Figure Pr.5.20 Variation of ﬂame temperature and speed with respect to the heat loss.
483
PROBLEM 5.21.FUN GIVEN: A surfaceradiation burner, which uses distributed, direct fuel supply, is shown in Figure Pr.5.219a). The radiation is from the impermeable surface having a uniform temperature Ts and facing the radiationload surface at temperature TL . The oxidant stream (air) is at mass ﬂow rate M˙ O and temperature Tf,1 . The fuel stream (methane) is divided into three smaller streams with each ﬂow rate designated as M˙ F,i , i = 1, 2, 3. The product stream M˙ P leaves at the exit port at temperature Tf,4 . Here we assume that Tf,4 = Tf,3 = Tf,2 = Ts . This is a design requirement that in practice is obtained by using more than three fuel stream ports and by taking the pressure drop in the fuel membrane and the combustion chamber into the account. r,s = 0.9, r,L = 1, FsL = 1, TL = 700 K, Tf,1 = 300 K, M˙ O = 0.013 kg/s, M˙ F,1 = 3 × 10−4 kg/s, M˙ F,2 = M˙ F,3 = 2 × 10−4 kg/s. Use an integralvolume energy equation for each of the three segments. Assume a constant speciﬁc heat capacity cp,f = 1,600 J/kgK. SKETCH: Figure Pr.5.21(a) shows the burner, the fuel and oxidant ports, and the heat transfer load surface. Distributed, Direct Fuel Supply, Tf,1 MF,3
MF,1
MF,2
Tf,4 = Ts
L/3
Product Stream MP = MF,4 Tf,3 = Ts Permeable Fuel Supply Membrane
Sr,c
Sr,c
w
Sr,c
PowderFilled Combustion Chamber
L
TL ,
RadiationLoad Surface
Tf,2 = Ts
r,s
Ts ,
Impermeable Radiation Surface
Oxidant Stream MO , Tf,1
r,L
Radiation Load, Qr,sL
Figure Pr.5.21(a) A distributed, direct fuel supply surfaceradiation burner. The radiation surface is impermeable.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the surface temperature Ts . , (c) Determine the burner eﬃciency η = Qr,sL /S˙ r,c , S˙ r,c = −∆hr,F i M˙ F,i . SOLUTION: (a) Figure Pr.5.21(b) shows the thermal circuit diagram. The three segments are connected by having the stream leaving one segment enter the next segment downstream, after fuel is added. The mass ﬂow rate of the product M˙ P,i increases as more fuel is added.
MF,3 Exit (Sr,c)3 Qu,4
MF,2 (Sr,c)2
Qu,3' TF,3 Qu,3 Qr,3L
MF,1 (Sr,c)1
Qu,2' TF,2 Qu,2
Qr,2L
Entrance
Qu,1' TF,1 Qu,1
Qr,1L
Figure Pr.5.21(b) Thermal circuit diagram.
484
MO
(b) The energy equations for the three segments are written according to (5.33) as Qu,2 − Qu,1 + Qr,1L = (S˙ r,c )1 Qu,3 − Qu,2 + Qr,2L = (S˙ r,c )2 Qu,4 − Qu,3 + Qr,3L = (S˙ r,c )3 . Now we note that the temperature for the oxidant and the fuel entering the burner is the same and equals Tf,1 . By adding the three equations, we have Qu,4 − Qu,1 − (Qu,2 − Qu,2 ) − (Qu,3 − Qu,3 ) +
3
Qr,iL
i=1
=
3
(S˙ r,c )i
i=1
= −∆hr,F
3
M˙ F,i .
i=1
Now using (5.34) and (4.49), for a twosurface enclosure with FsL = 1 and r,L = 1, we have 3 3 3
˙ ˙ ˙ ˙ MO2 + Mf,i cp,f Ts − MO2 + Mf,i cp,f Tf,1 + r,s wLσSB (Ts4 − TL4 ) = −∆hr,F M˙ f,i , i=1
i=1
i=1
where we have used Tf,4 = Ts and for the exit product mass ﬂow rate we have used the sum of the oxidant and the total fuel mass ﬂow rate. From Table C.21(a), we have for methane, ∆hr,F = −5.553 × 107 J/kg
Table C.21(a).
Using the numerical values, we then have (1.3 × 10−2 + 7 × 10−4 )(kg/s) × 1,600(J/kgK) × (Ts − 300)(K) + 0.9 × 0.5(m) × 1(m) × 5.67 × 10−8 (W/m2 K4 )(Ts4 − 7004 )(K4 ) = −(−5.553 × 107 )(J/kg) × 7 × 10−4 (kg/s). Solving for Ts , we get Ts = 1,040 K. (c) With the given expression for η, η
=
Qr,sL S˙ r,c
=
r,s wLσSB (Ts4 − TL4 ) 3
M˙ F,i −∆hr,F i=1
= =
2.367 × 104 (W) 3.887 × 104 (W) 0.6090 = 60.90%.
COMMENT: This eﬃciency is rather low and can be increased by preheating the air and the fuel using heat exchangers. The burner can be made from ceramics such as zirconia and the thermal stress/strain during the cyclic use should be addressed. Also, by using pure oxygen instead of air, we can increase the surface temperature and the burner eﬃciency.
485
PROBLEM 5.22.FAM GIVEN: A watercooled thermal plasma generator, used for spray coating and shown in Figure Pr.5.22(a), is to produce an argon gas plasma stream with an average exit temperature of Tf,2 = 5,000 K. The Joule heating is by direct current (dc) and uses Je = 200 A, and ∆ϕ = 150 V provided by a power supply. The heat transfer to the water coolant is Qku and other heat losses are given by Qloss . Assume S˙ ij = 0. Qku = 5 kW, Qloss = 3 kW, Tf,1 = 300 K. Use a constant speciﬁc heat capacity for the ionized argon and use a value equal to ﬁve times that for argon at T = 1,500 K. SKETCH: Figure Pr.5.22(a) shows the plasma torch.
Spray Coating Using Thermal Plasma to Melt Entrainned Particles Argon Gas Cathode () Mf , Tf,1
Anode (+)
Coolant (Water) Qloss
Se,J
Particle Stream
Qku
Plasma Stream
Thermal Plasma (Argon Gas)
Substrate
Particle Coating
us
Figure Pr.5.22(a) Generation of an argon plasma stream using the Joule heating.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the argon gas ﬂow rate M˙ f . SOLUTION: (a) Figure Pr.5.22(b) shows the control volumes and the thermal circuit diagram.
Thermal Circuit Diagram: Internodal Energy Conversion Mf , Tf,1
Control Volume, V Control Area, A
Tf,1
Qu,1
Se,J = JeDj Qku
Qloss Qku
Qloss
Qu,2 Tf,2
Se,J = Je,j Mf , Tf,2
Figure Pr.5.22(b) Thermal circuit diagram.
(b) From Figure Pr.5.22(b), or from (5.73), we have the energy equation QA
= Qu,1 + Qu,2 + Qku + Qloss = S˙ ij + S˙ e,J = −M˙ f cp,f (Tf,1 − Tf,2 ) + Qku + Qloss = Je ∆ϕ. 486
The speciﬁc heat capacity of argon at T = 1,500 K is given in Table C.22, and we use a value ﬁve times this, i.e., cp,f = 5 × 520(J/kgK) = 2,600 J/kgK Table C.22. Then solving for M˙ f , we have Je ∆ϕ − Qku − Qloss M˙ f = . cp,f (Tf,2 − Tf,1 ) Using the numerical values, we have M˙ f
=
200(A) × 150(V) − 5,000(W) − 3,000(W) 2,600(J/kgK) × (5,000 − 300)(K)
=
1.800 × 10−3 kg/s = 1.800 g/s.
COMMENT: The gas speciﬁc heat capacity is a strong function of temperature. Upon gas dissociation and ionization at high temperatures, it increases further.
487
PROBLEM 5.23.FAM GIVEN: An acetyleneoxygen torch has 70% by weight of its stoichiometric oxygen provided by a pressurized tank and this is called the primary oxygen. Due to the fast chemical reaction of C2 H2 and O2 and the fast diﬀusion of O2 , the remaining 30% of the oxygen is provided by entraining the ambient air and this is called the secondary oxygen. These are shown in Figure Pr.5.23(a). Along with the entrained oxygen, nitrogen is also entrained and this inert gas tends to lower the ﬁnal temperature Tf,2 . The products of combustion (i.e., the ﬂue gas) ﬂows over a surface to be welded. Then a fraction of the sensible heat is transferred to the surface by surfaceconvection heat transfer Qku , and the remainder ﬂows with the gas as Qu,2 . Tf,1 = 20◦C, p1 = 1 atm, Qku = 0. Use a constant and uniform speciﬁc heat of cp,f = 3,800 J/kgK. SKETCH: An acetyleneoxygen torch, with primary (pure oxygen) and secondary oxygen (mixed with nitrogen) supplies, is shown in Figure Pr.5.23(a). Acetylene Fuel MC2H2
Primary Oxygen MO2
Torch Tf,1 Qu
Plasma Stream
Entrained (Secondary) Oxygen MO2 , MN2
Sr,c
Intramedium Convection
Qu Flame
Mf
Tf,2
Qu,2
Surface Convection, Qku
Figure Pr.5.23(a) An acetyleneoxygen torch used for welding of a surface. The oxygen is provided by a tank and by entraining air.
OBJECTIVE: (a) Draw the thermal circuit diagram for this combustion and heat transfer. (b) For the conditions given above, determine the ﬂue gas temperature Tf,2 . SOLUTION: (a) Figure Pr.5.23(b) shows the thermal circuit diagram with the internodal energy conversion and surfaceconvection heat transfer connected to node Tf,2 .
Qu,1 Tf,1
Sr,c Q u,2 MO2 Mf MN 2 MC2H2
Q2 Qku Tf,2
Figure Pr.5.23(b) Thermal circuit diagram.
(b) We need to determine ρF /ρf,1 , in order to solve for Tf,2 from (5.35), i.e., Tf,2 = Tf,1 −
∆hr,F ρF,1 , cp,f ρf,1 488
where we have assumed a constant cp,f . To determine ρF,1 /ρf,1 , we note that the two stoichiometric reactions are, 5 C2 H2 + O2 → 2CO2 + H2 O 2 5 C2 H2 + O2 + 9.40N2 → 2CO2 + H2 O + 9.40N2 . 2 Then, similar to Example 5.4, we have νC2 H2 MC2 H2 νC2 H2 MC2 H2 ρF,1 = 0.70 × + 0.30 × . ρf,1 νC2 H2 MC2 H2 + νO2 MO2 νC2 H2 MC2 H2 + νO2 MO2 + νN2 MN2 Using the numerical values, we have ρF,1 ρf,1
1 × (12.011 × 2 + 1.008 × 2) + 1 × (12.011 × 2 + 1.008 × 2) + 2.5 × 2 × 15.999 1 × (12.011 × 2 + 1.008 × 2) 0.30 × 1 × (12.011 × 2 + 1.008 × 2) + 2.5 × 2 × 15.999 + 9.40 × 2 × 14.007 18.227 7.8114 + = 106.03 369.30 = 0.17190 + 0.021152 = 0.19305. =
0.70 ×
With ∆hr,F = −4.826 × 107 J/kg from Table 5.2, [also listed in Table C.21(a)], we have Tf,2
= =
−4.826 × 107 (J/kg) × 0.19305 3,800(J/kgK) 20(◦C) + 2,452(◦C) = 2,472◦C. 20(◦C) −
COMMENT: From Figure 5.9, using pure oxygen gives an adiabatic ﬂame temperature of about 3,100◦C. Therefore, the nitrogen dilution should be avoided in order to achieve a higher adiabatic temperature. Note that high Catom content and low Hatom content results in a high (ρF /ρf )1 . This makes acetylene a good fuel for achieving high adiabatic ﬂame temperatures.
489
PROBLEM 5.24.FAM GIVEN: The acetyleneoxygen, ﬂamecutting torch is used with low carbon and low alloy irons. A simple mixer that requires pressurized oxygen and acetylene is shown in Figure Pr.5.24(a). In addition to the heat provided by the reaction C2 H2 + 2.5O2 → 2CO2 + H2 O, the excess oxygen provided by the torch reacts with the iron and releases heat. One of the reactions is 3Fe + 2O2 → Fe3 O4 . This reaction is highly exothermic, i.e., ∆hr,F = −6.692 × 106 J/kg of Fe. Figure Pr.5.24(a) shows this iron oxidation. The rate of this reaction is controlled by the speed of the torch moving on the cutting surface, and with other variables. Assume that this reaction will add an extra energy conversion such that we can approximate the contribution for this iron oxidation by adding 30% to the heat of reaction of C2 H2 . We also model the excess oxygen by adding 40%, by weight, to the stoichiometric oxygen needed to burn the acetylene. Tf,1 = 20◦C, cp,f = 3,800 J/kgK. SKETCH: Figure Pr.5.24(a) shows the torch and the reactingeroding workpiece.
AcetyleneOxygen Cutting Torch with Extra Heat Generated by Oxidation of Iron Workpiece O2 , For Acetylene and Iron C2H2 , Acetylene Tf,1 Qu,1
(Sr,c)12
C2H2 +
5 2 O2
2CO2 + H2O
Tf,2
Qu,2
Qku
Qku Q32
T3 (Sr,c)3
Q3 3Fe + 2O2
Fe3O4
Figure Pr.5.24(a) A cutting torch with reacting workpiece.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the combustion product gas temperature Tf,2 . SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.5.24(b). The oxidization of Fe is shown as (S˙ r,c )3 and a fraction of this heat generation rate is transferred to the gas. The gas in turn heats the solid in the regions away from where (S˙ r,c )3 is generated, thus providing the preheating necessary for the iron oxidation. (b) The combustion product gas temperature Tf,2 is given by (5.35), i.e., Tf,2 = Tf,1 −
∆hr,F ρF,1 , cp,f ρf,1 490
Mf Tf,1 Qu,1
(Sr,c)12 Qu,2
Tf,2
Q32 T3 Q3
Qku (Sr,c)3
Figure Pr.5.24(b) Thermal circuit diagram.
where ∆hr,F is that given in Table 5.2 for C2 H2 , times 1.3. To determine (ρF /ρf )1 , we begin from the stoichiometric C2 H2 oxidation, C2 H2 + 2.5O2 → 2CO2 + H2 O. Next we use the approximation given for the extra oxygen supplied for the iron oxidation, i.e., C2 H2 + 1.4 × 2.5O2 → 2CO2 + H2 O + 0.4 × 2.5O2 . Then the density ratio (ρF /ρf )1 becomes ρF,1 ρf,1
= = =
νC2 H2 MC2 H2 νC2 H2 MC2 H2 + νO2 MO2 1 × (12.011 × 2 + 1.008 × 2) 1 × (12.011 × 2 + 1.008 × 2) + 1.4 × 2.5 × 2 × 15.999 26.04 = 0.1886. 138.0
From Table 5.2, we have C2 H2 : ∆hr,F = −4.826 × 107 J/kg
Table 5.2.
Then Tf,2
= =
1.3 × (−4.826 × 107 )(J/kg) × 0.1886 3,800(J/kgK) 20(◦C) + 3,114(◦C) = 3,134◦C.
20(◦C) −
COMMENT: From Figure 5.9, we note that the adiabatic ﬂame temperature for the stoichiometric acetylene oxidation in pure oxygen is nearly 3,100◦C. Thus the addition of 30% to the heat of combustion is nearly compensated by the excess oxygen that must be heated to the ﬁnal temperature Tf,2 .
491
PROBLEM 5.25.FUN GIVEN: The combustion product gas from a ﬁreplace chamber enters its chimney at a ﬂow rate M˙ f and a temperature Tf,1 . This is shown in Figure Pr.5.25. Assume that the surfaceconvection heat transfer to the chimney wall is negligible. Ts = 120◦C, Tf,1 = 600◦C, M˙ f = 2.5 × 10−3 kg/s, R = 15 cm, L = 5 m, r,f = 0.1 (for large soot concentration). Use cp,f for air at T = 600 K. Use r,s = αr,s for ﬁreclay brick. SKETCH: Figure Pr.5.25 shows the chimney with the combustion product gas stream passing through it. Air Stream R
Tf,2 Chimney Sr,= + Sr,
Ts , =r,s
L
Tf,1 Fireplace Mf , FlueGas Stream Burning Wood Logs
Figure Pr.5.25 A chimney bounding the combustion product gas stream exiting a ﬁreplace chamber.
OBJECTIVE: (a) Show that the surface radiation emitted by the chimney wall is negligible compared to that emitted by the combustion product gas stream. (b) Determine the combustion product gas stream exit temperature Tf,2 . SOLUTION: (a) The condition for neglecting the surface emission is given in (5.81), i.e., Ts4 4 1. Tf,1 Here we have (120 + 273.15)4 (K4 ) = 0.0410 1. (600 + 273.15)4 (K4 ) (b) Using (5.81), we have 1 3 Tf,2
=
1 3 Tf,1
+
Ar r,f αr,s σSB . (M˙ cp )f
From Table C.22, for air at T = 600 K, we have cp,f = 1,038 J/kgK
Table C.22.
From Table C.18, we have for ﬁreclay brick r,s = αr,s = 0.75 492
Table C.18.
Also, Ar = 2πRL = 2 × π × 0.15(m) × 5(m) = 4.713 m2 . Then 1 3 Tf,2
Tf,2
=
1 4.713(m3 ) × 0.1 × 0.75 × 5.67 × 10−8 (W/m2 K4 ) + 3 3 (873.15) (K ) 2.5 × 10−3 (kg/s) × 1,038(J/kgK)
=
1.502 × 10−9 (1/K3 ) + 7.723 × 10−9 (1/K3 )
=
476.8 K = 203.7◦C.
COMMENT: Inclusion of the surfaceconvection heat transfer will further decrease Tf,2 and will be discussed in Chapter 7. Here the surface emission from the wall was neglected. When included, Tf,2 will increase by a small amount.
493
Chapter 6
Convection: SemiBounded Fluid Streams
PROBLEM 6.1.FUN GIVEN: Surfaceconvection heat transfer refers to heat transfer across the boundary separating a ﬂuid stream and a condensedphase (generally solid) volume, as rendered in Figure Pr.6.1(a) for a stationary solid. Assume a uniform solid surface temperature Ts and a farﬁeld temperature Tf,∞ = Ts . SKETCH: Figure Pr.6.1(a) renders the general surfaceconvection of a semibounded ﬂuid stream. SemiBounded Fluid Stream uf, , Tf,
y
y,vf sn
x,uf
Fluid Ts
Tf,
Tf
L Solid
Ts (Uniform) us = 0
qk,s
Figure Pr.6.1(a) A rendering of a semibounded ﬂuid stream passing over a solid surface with Tf,∞ = Ts .
OBJECTIVE: (a) Draw the heat ﬂux vector tracking starting from qk,s and ending with qu away from the surface. At the surface use qk,f = qku and in the thermal boundary layer show both conduction and convection. Neglect radiation heat transfer. (b) Quantitatively draw the ﬂuid temperature distribution Tf (y) at the location shown in Figure Pr.6.1(a). (c) Show the thermal boundary layer thickness δα on the same graph. Show the viscous boundary layer thickness for Pr > 1 and Pr < 1. (d) Draw the thermal circuit diagram for the solid surface and write the expression for the average heat transfer rate Qku L . (e) What is the average surfaceconvection resistance Rku L , if Ts is the same as Tf,∞ , i.e., for the solid surface temperature to be made equal to the ﬂuid stream farﬁeld temperature? (f) If Tf,∞ = Ts what should Rku L be for there to be no surfaceconvection heat transfer (ideal insulation) SOLUTION: (a) Figure Pr.6.1(b) shows the heat ﬂux vector tracking. Note that as in Figures 6.3 and 6.7, at the surface qk,f = qku , i.e., surfaceconvection heat transfer is the ﬂuid conduction heat ﬂux, since uf = 0 on the surface.
uf, , Tf, qu qu qk qk,f = qku sn Ts (Uniform)
Thermal Boundary Layer, da dn (Pr > 1) y dn (Pr < 1) qu
Tf,
Ts L
Tf
Tf, Qku
L
Rku
L
Ts
qk,s
Figure Pr.6.1(b) Various features of the surfaceconvection heat transfer.
(b) The temperature distribution Tf (y) is also shown in Figure Pr.6.1(b), starting from Tf = Ts at y = 0 and having Tf = Tf,∞ , far away from the surface. Using the concept of the thermal boundarylayer thickness, the farﬁeld conditions are applied at y = δα . (c) The thermal boundary layer thickness δα is shown in Figure Pr.6.1(b) and this is where Tf reaches Tf,∞ to within a small diﬀerence given by (6.20).
496
For Pr > 1, from (6.48), we have δν > δα , and for Pr < 1 we have δν < δα . These are shown in Figure Pr.6.1(b). (d) The thermal circuit diagram is shown in Figure Pr.6.1(b). From this ﬁgure, or from (6.49), we have Qku L =
Ts − Tf,∞ . Rku L
(e) If Ts = Tf,∞ , there will be no surface convection heat transfer, and therefore there will be no thermal boundary layer. If there is no thermal boundary layer, then Rku L = 0. This shows that the surface temperature approaches the ﬂuid farﬁeld temperature as Rku L → 0. This is a method for controlling (maintaining) the surface temperature. (f) For an ideally insulated surface, we have Qku L = 0 for Rku L → ∞ and this would require a large resistance (i.e., vacuum) for a ﬁnite diﬀerence between Ts and Tf,∞ . Similarly Qku L = 0 when Ts = Tf,∞ . COMMENT: Heat transfer between a semibounded ﬂuid passing over a solid surface requires heat transfer by ﬂuid conduction across the interface. This heat transfer is greatly inﬂuenced by the ﬂuid motion, and other ﬂuid properties, which in turn inﬂuence the gradient of temperature ∇Tf , and directly depends on the ﬂuid conductivity kf .
497
PROBLEM 6.2.FUN GIVEN: A surface, treated as a semiinﬁnite plate and shown in Figure Pr.6.2, is to be heated with a forced, parallel ﬂow. The ﬂuids of choice are (i) mercury, (ii) ethylene glycol (antifreeze), and (iii) air. Ts = 10◦C, Tf,∞ = 30◦C, uf,∞ = 0.2 m/s, L = 0.2 m. Evaluate the properties at T = 300 K, from Tables C.22 and C.23. SKETCH: Figure Pr.6.2 shows a forced, parallel ﬂow over a semiinﬁnite plate. The thermal boundarylayer thickness is also shown. Ts = 10oC Parallel Flow uf, = 0.2 m/s Tf, = 30oC
δα(L)
qu x qku(x) L = 0.2 m
Figure Pr.6.2 A semibounded ﬂuid stream exchanging heat with its semiinﬁnite plate bounding surface.
OBJECTIVE: For the tailing edge of the plate x = L, do the following: (a) Determine the local rate of heat transfer per unit area qku (W/m2 ). (b) Determine the thermal boundarylayer thickness δα (mm). Use the Nusselt number relation for Pr = 0. (c) For mercury, also use the relation for Nusselt number for a zero viscosity (i.e., Pr = 0) and compare the results with that obtained from the nonzero viscosity relations. SOLUTION: (a) The Reynolds number is given by (6.45), i.e., ReL =
uf,∞ L . νf
For ReL < 5 × 105 , the ﬂow regime is laminar. The local Nusselt number at x = L for laminar, parallel ﬂow over a ﬂat plate is given by (6.44), i.e., 1/2
NuL (x = L) = 0.332ReL Pr1/3 . The local surfaceconvection heat ﬂux, from (6.44), is qku =
NuL kf (Ts − Tf,∞ ). L
(b) The thermal boundarylayer thickness for laminar ﬂow at x = L is given by (6.48) δα (x = L) =
5L
1
1/2 ReL
Pr1/3
Table Pr.6.2 shows the thermophysical properties at T = 300 K for the three ﬂuids and the numerical results obtained for δα and qku . (c) With the zero viscosity (or Prandtl number) assumption, the Nusselt number is given by (6.30) and δα is given by (6.21), i.e., NuL (x = L) =
PeL π
1/2 ,
δα (x = L) = 3.6 498
αf L uf,∞
1/2 ,
for Pr = 0,
Table Pr.6.2 Properties (from Tables C.22 and C.23) for the three ﬂuids and the numerical results. Fluid νf , kf , Pr ReL δα (x = L), NuL (x = L) qku , m2 /s W/mK mm W/m2 mercury ethylene glycol air
0.112 × 10−6 18.09 × 10−6 15.66 × 10−6
8.86 0.2515 0.0267
0.0240 193 0.69
3.571 × 105 2,211 2,554
5.801 3.680 22.39
57.23 90.22 14.83
−50,703 −2,269 39.59
where from (5.9), the Peclet number is PeL =
uf,∞ L . αf
For mercury, from Table C.23 at T = 300 K, αf = 4.70 × 10−6 m2 /s. Then PeL
=
8,511
NuL (x = L)
=
52.05
qku δα
= −46,115 W/m =
2
7.805 mm.
COMMENT: Liquid metal ﬂow makes for very eﬀective surfaceconvection heat transfer. Also, liquids are more eﬀective than gases in surfaceconvection heat transfer. Finally, treating mercury as a Pr = 0 ﬂuid results in a qku which is within 10(hydrodynamic) boundary layer formed on the plate does not inﬂuence the surfaceconvection heat transfer when Pr is very small.
499
PROBLEM 6.3.FAM GIVEN: The top surface of a microprocessor chip, which is modeled as a semiinﬁnite plate, is to be cooled by forced, parallel ﬂow of (i) air, or (ii) liquid Refrigerant12. The idealized surface is shown in Figure Pr.6.3. The eﬀect of the surfaces present upstream of the chip can be neglected. Evaluate the properties at T = 300 K. SKETCH: Figure Pr.6.3 shows the surface of a microprocessor chip subjected to parallel ﬂow. The thermal boundarylayer thickness is also shown. Ts = 50 oC
δα(L)
Parallel Air or Refrigerant Flow uf, = 0.5 m/s Tf, = 20 oC Qku
L
w = 0.04 m
L = 0.2 m
Figure Pr.6.3 Surface of a microprocessor is cooled by a semibounded ﬂuid stream.
OBJECTIVE: (a) Determine the surfaceconvection heat transfer rate Qku L (W). (b) Determine the thermal boundarylayer thickness at the tail edge of the chip δα (mm). SOLUTION: (a) The Reynolds number is deﬁned in (6.45) as ReL =
uf,∞ L . νf
For ReL < 5 × 105 , the ﬂow regime is laminar. The averaged Nusselt number (averaged over L) for laminar, parallel ﬂow over a semiinﬁnite ﬂat plate is given by (6.51), i.e., 1/2
NuL = 0.664ReL Pr1/3 . For the turbulent regime (ReL > 5 × 105 ), the averaged Nusselt number (averaged over L) for parallel ﬂow over a semiinﬁnite ﬂat plate is given by (6.67), i.e., 4/5
NuL = (0.037ReL − 871)Pr1/3 . (b) The thermal boundarylayer thickness for laminar ﬂow, at x = L, is given by (6.48), i.e., δα (x = L) = 5
1 L 1/2 (ReL ) Pr1/3
and for turbulent ﬂow δα is given by (6.66), i.e., δα (x = L) = 0.37
1 L . 1/5 (ReL ) Pr1/3
From (6.149), the average surfaceconvection resistance is Rku L =
L , Aku kf NuL
where the surface area is Aku = wL. The averaged surfaceconvection heat transfer from the plate is given by (6.49), i.e., Qku L =
(Ts − Tf,∞ ) . Rku L 500
The thermophysical properties are evaluated at T = 300 K from Tables C.22 and C.23. Table Pr.6.3 lists the thermophysical properties for the ﬂuids and the numerical results obtained for Qku L and δα (x = L), Fluid
air R12
Table Pr.6.3 Thermophysical properties for the ﬂuids and numerical results. νf , kf , Pr ReL δα (x = L), NuL Rku L , m2 /s W/mK mm K/W 15.66 × 10−6 0.195 × 10−6
0.0267 0.072
0.69 3.5
6,386 (laminar) 5.128 × 105 (turbulent)
14.16 3.515
46.89 754.9
19.97 0.4599
Qku L , W 1.502 65.23
COMMENT: Air has a larger boundarylayer thickness than liquid R12. The Nusselt number and the heat transfer rate are larger for liquid R12.
501
PROBLEM 6.4.FUN GIVEN: As discussed in Section 6.2.5, the stream function ψ, for a twodimensional, laminar ﬂuid ﬂow (uf , vf ), expressed in the Cartesian coordinate (x, y), is deﬁned through uf ≡
∂ψ , ∂y
vf ≡ −
∂ψ . ∂x
OBJECTIVE: Show that this stream function satisﬁes the continuity equation (6.37). SOLUTION: The continuity equation for laminar incompressible ﬂow, in two dimensions, using the Cartesian coordinates, is given by (6.37), i.e., ∂vf ∂uf + = 0. ∂x ∂y Substituting the above, we have ∂2ψ ∂2ψ − = 0. ∂x∂y ∂y∂x Thus, the deﬁnition of ψ given above automatically satisﬁes the continuity equation (6.37). COMMENT: With no need to include the continuity equation in the analysis, the momentum equation is used to determine ψ. This is investigated in the next problem.
502
PROBLEM 6.5.FUN GIVEN: As discussed in Section 6.2.5, the twodimensional, (x, y), (uf , vf ) , laminar steady viscous, boundarylayer momentum equation (6.36) can be reduced to an ordinary diﬀerential equation using a dimensionless similarity variable 1/2 uf,∞ η≡y νf x and a dimensionless stream function ψ∗ ≡
ψ (νf uf,∞ x)1/2
, uf ≡
∂ψ ∂ψ , vf ≡ − . ∂y ∂x
OBJECTIVE: (a) Show that the momentum equation (6.36) reduces to 2
2 ∗ d3 ψ ∗ ∗d ψ + ψ = 0. dη 3 dη 2
This is called the Blasius equation. (b) Show that energy equation (6.35) reduces to d2 Tf∗ dη 2
∗
dTf 1 = 0, + Prψ ∗ 2 dη
Tf∗ =
Tf − Tf,∞ . Ts − Tf,∞
SOLUTION: (a) We start with (6.36), written as uf
∂uf ∂uf ∂ 2 uf + vf − νf = 0. ∂x ∂y ∂y 2
Using the stream function ψ, the dimensionless stream function ψ ∗ and the similarity variable η, we transform uf and vf into ψ ∗ , x and η. We start with dψ ∂η dψ ∗ ∂ψ = = (νf uf,∞ x)1/2 ∂y dη ∂y dη dψ ∗ = uf,∞ dη
uf ≡
uf,∞ νf x
1/2
or uf dψ ∗ . = uf,∞ dη Also, ∗ ∂ 1 νf uf,∞ 1/2 ∗ ∂ψ 1/2 ∗ 1/2 ∂ψ vf ≡ − = − [(νf uf,∞ x) ψ ] = − (νf uf,∞ x) + ( ) ψ ∂x ∂x ∂x 2 x 1 νf uf,∞ 1/2 dψ ∗ = − ψ∗ η 2 x dη or vf
ν u 1/2 f f,∞ x
1 = 2
dψ ∗ ∗ −ψ . η dη
503
Next these velocity components are diﬀerentiated with respect to x and y, and we have uf,∞ d2 ψ ∗ η 2x dη 2 1/2 2 ∗ uf,∞ d ψ = uf,∞ νf x dη 2 u2f,∞ d3 ψ ∗ = . νf x dη 3
∂uf ∂x
= −
∂uf ∂y ∂ 2 uf ∂y 2
Substituting these into the above momentum equation, we have ∂ψ ∗ −uf,∞ ∂η
uf,∞ d2 ψ ∗ η 2x dη 2
1/2 2 ∗ u2f,∞ d3 ψ ∗ uf,∞ d ψ 1 νf uf,∞ 1/2 dψ ∗ ∗ − ψ uf,∞ − ν =0 + η f 2 x dη νf x νf x dη 3 dη 2
or uf,∞ x
or 2
d3 ψ ∗ 1 d2 ψ ∗ − − ψ∗ 2 dη 2 dη 3
=0
2 ∗ d3 ψ ∗ ∗d ψ + ψ = 0. dη 3 dη 2
(b) Starting from (6.35), we have uf
∂Tf ∂Tf ∂ 2 Tf + vf − αf = 0. ∂x ∂y ∂y 2
We already have the expression for uf and vf from part (a), then ∂Tf ∂x ∂Tf ∂y ∂ 2 Tf ∂y 2
1/2 dTf ∂η y uf,∞ η dTf dTf =− = dη ∂x 2 νf x3 dη 2x dη 1/2 uf,∞ dTf = νf x dη =
=
uf,∞ d2 Tf . νf x dη 2
Substituting these into the energy equation, and using Tf∗ , we have 1/2 1/2 2 ∗ dTf∗ dTf∗ uf,∞ y uf,∞ 1 νf uf,∞ 1/2 dψ ∗ uf,∞ d Tf dψ ∗ ∗ uf,∞ − + − ψ − α =0 η f dη 2 νf x3 dη 2 x dη νf x dη νf x dη 2 −
∗ 2 ∗ uf,∞ d Tf 1 uf,∞ ∗ dTf ψ − αf =0 2 x dη νf x dη 2
d2 Tf∗ dη 2 COMMENT: Also note that outside the boundary layer, i.e., when uf = uf,∞ , we have uf dψ ∗ = 1, = uf,∞ dη
dψ ∗ = dη
for
uf = uf,∞ .
Then outside the boundary layer, we have vf
ν u 1/2 f f,∞ x
dψ ∗ − ψ∗ η dη
=
1 2
=
1 (η − ψ ∗ ) 2
504
∗
dTf 1 = 0. + Prψ ∗ 2 dη
This would tend to a constant as η becomes large. This constant is vf ν u 1/2 = 0.86054 f f,∞ x Note that the momentum equation written in terms of velocity (u, v) is second order in y, while the use of the stream function results in a thirdorder diﬀerential equation.
505
PROBLEM 6.6.FUN GIVEN: The thirdorder, ordinary Blasius diﬀerential equation 2
2 ∗ d3 ψ ∗ ∗d ψ + ψ = 0, dη 3 dη 2
subject to surface and farﬁeld mechanical conditions dψ ∗ = ψ∗ = 0 dη dψ ∗ = 1. dη
at η = 0 : for η → ∞ :
Note that with an initialvalue problem solver, such as SOPHT, the second derivative of ψ ∗ at η = 0 must be guessed. This guess is adjusted till dψ ∗ /dη becomes unity for large η. Hint: d2 ψ ∗ /dη 2 (η = 0) is between 0.3 to 0.4. OBJECTIVE: Use a solver to integrate the dimensionless transformed boundarylayer momentum equation. Plot ψ ∗ , dψ ∗ /dη = uf /uf,∞ , and d2 ψ ∗ /dη 2 , with respect to η. SOLUTION: The solver we choose is an initialvalue solver, such as SOPHT, where the initial values (i.e., at η = 0) for ψ ∗ , dψ ∗ /dη, and d2 ψ ∗ /dη 2 must be provided for this thirdorder, ordinary diﬀerential equation. Therefore, in place of the condition for η → ∞, we choose guess :
d2 ψ ∗ = constant at dη 2
η=0
such that
dψ ∗ =1 dη
for η → ∞.
Note that we can write the Blausius equation as a set of ﬁrstorder diﬀerential equations, i.e, g z f
1 = − fg 2 = g = z,
here g = d3 ψ ∗ /dη 3 , z = d2 ψ ∗ /dη 2 , and f = dψ/dη. The initial conditions are f (η = 0)
=
z(η = 0) g(η = 0)
= =
0 0.332
after iterating to get z(η → ∞) = 1.
The results are plotted in Figure Pr.6.6. The results show that the streamwise velocity uf /uf,∞ increases and reaches a value of unity at η 5. The stream function ψ ∗ increase monotonically, while d2 ψ ∗ /dη 2 decrease and vanishes at η 5. COMMENT: The results are sensitive to the initial choice for d2 ψ ∗ /dη 2 at η = 0. Because of the similarity between the momentum and energy equations, (6.35) and (6.36), this derivative is the same as the temperature derivative and therefore 0.332 is also the constant appearing in the solution (6.44) for the surface ﬂuid conduction heat transfer rate (i.e., surface convection). 506
2.0
2 * d O* O*, dD , ddDO2
1.6
;*
O
d * uf uf, = dD = 1
1.2
uf uf, = 0.99
0.8
d
O
d2
0.4
*
O
*
dD
Boundary Layer
dD2
FarField
0 0
1.2
2.4
D
3.6
4.8
D=5
Edge of Boundary Layer
Figure Pr.6.6 Variations of ψ ∗ , dψ ∗ /dη, and d2 ψ ∗ /dη 2 with respect to η.
507
6
PROBLEM 6.7.FAM GIVEN: During part of the year, the automobile windshield window is kept at a temperature signiﬁcantly diﬀerent than that of the ambient air. Assuming that the ﬂow and heat transfer over the windshield can be approximated as those for parallel ﬂow over a semiinﬁnite, ﬂat plate, examine the role of the automobile speed on the surfaceconvection heat transfer from the window. These are shown in Figure Pr.6.7. The ambient air is at −10◦C and the window surface is at 10◦C. The window is 1 m long along the ﬂow direction and is 2.5 m wide. Use the average temperature between the air and the window surface to evaluate the thermophysical properties of the air. SKETCH: Figures Pr.6.7(i) and (ii) show an automobile windshield window and its idealization as a semiinﬁnite plate.
(i) Physical Model
Qku
L
Parallel Air Flow
uf,
(ii) An Approximation of Heat Transfer from Windshield Parallel Air Flow Tf, = 10oC (1) uf, = 2 km/hr (2) uf, = 20 km/hr (3) uf, = 80 km/hr
Ts = 10oC
Qku
L
w = 2.5 m
L=1m
Figure Pr.6.7 (i) Fluid ﬂow and heat transfer over an automobile windshield window. (ii) Its idealization as parallel ﬂow over a semiinﬁnite plate.
OBJECTIVE: (a) To the end of automobile speed on the surfaceconvection heat transfer from the window, determine the average Nusselt number NuL . (b) Determine the average surfaceconvection thermal resistance Aku Rku L [◦C/(Wm2 )]. (c) Determine the surfaceaveraged rate of surfaceconvection heat transfer Qku L (W). Consider automobile speeds of 2, 20, and 80 km/hr. Comment on the eﬀects of the ﬂowregime transition and speed on the surfaceconvection heat transfer. SOLUTION: (a) The Reynolds number is given by (6.45), i.e., ReL =
uf,∞ L . νf
For ReL < 5 × 105 the ﬂow regime is laminar. The average Nusselt number (averaged over L) for laminar, parallel ﬂow over a ﬂat plate is given by (6.51), i.e., 1/2
NuL = 0.664ReL Pr1/3 . 508
For ReL > 5 × 105 the ﬂow regime is turbulent and the averaged Nusselt number is given by (6.67), i.e., 4/5
NuL = (0.037ReL − 871)Pr1/3 . (b) The average surfaceconvection thermal resistance is calculated from (6.49), i.e., Aku Rku L =
L . kf NuL
(c) The averaged surfaceconvection heat transfer is then obtained from (6.49) as Qku L = Aku
(Ts − Tf,∞ ) . Aku Rku L
The thermophysical properties of air are obtained from Table C.22. For the average temperature Tδ = (Ts + Tf,∞ )/2 = 273.15 K, we have kf = 0.0251 W/mK, νf = 13.33 × 10−6 m2 /s, and Pr = 0.69. Table Pr.6.7 lists the numerical results obtained for the three vehicle diﬀerent speeds. Table Pr.6.7 Numerical results obtained for the three diﬀerent speeds. uf,∞ , ReL ﬂow regime NuL Aku Rku L , Qku L , ◦ C/(W/m2 ) W m/s 0.5556 5.556 22.22
0.4168 × 105 4.168 × 105 16.67 × 105
laminar laminar turbulent
119.8 378.8 2,335
0.3326 0.1052 0.0171
150.3 475.4 2,930
COMMENT: As the vehicle speed increases, the ﬂow regime changes from laminar to turbulent. The Nusselt number for the turbulent regime is larger than that for the laminar regime. This eﬀect, associated with the increase in the Reynolds number, causes the total heat transfer to increase by more than one order of magnitude, when the vehicle speed is changed only by a factor of four.
509
PROBLEM 6.8.FAM.S GIVEN: A square ﬂat surface with side dimension L = 40 cm is at Ts = 120◦C. It is cooled by a parallel air ﬂow with farﬁeld velocity uf,∞ and farﬁeld temperature Tf,∞ = 20◦C. OBJECTIVE: (a) Use a solver (such as SOPHT) to plot the variation of the averaged surfaceconvection heat transfer rate Qku L (W) with respect to uf,∞ (m/s) from zero up to the sonic velocity. Use (3.20) to ﬁnd the sonic velocity. (b) Determine the air velocity needed to obtain qku L = 1,200 W/m2 . SOLUTION: (a) The average surfaceconvection heat transfer rate is given by (6.49) as Qku L =
Ts − Tf,∞ , Rku L
where the average surfaceconvection resistance is also given by (6.49) as Rku L =
L . Aku kf NuL
For ReL < 5 × 105 the ﬂow regime is laminar. The averaged Nusselt number (averaged over L) for laminar, parallel ﬂow over a ﬂat plate is given by (6.51) and in Table 6.3 as 1/2
NuL = 0.664ReL Pr1/3 . For the turbulent regime, the averaged Nusselt number (averaged over L) for parallel ﬂow over a ﬂat plate is given by (6.67) and in Table 6.3 as 4/5
NuL = (0.037ReL − 871)Pr1/3 . The Reynolds number is given by (6.45) as ReL =
uf,∞ L . νf
The ﬂuid properties are evaluated at the ﬁlm temperature, given by (Ts + Tf,∞ )/2 = 70◦C. From Table C.22, interpolation gives νf = 19.66 × 10−6 m2 /s, kf = 0.0295 W/mK and Pr = 0.69. For the given conditions, the transition from laminar to turbulent ﬂow then occurs at uf,∞ = 24.60 m/s. The speed of sound, assuming that air behaves as an ideal gas, is given by (3.20), i.e., 1/2 cp Ru 1/2 as = Tf,∞ = (kRTf,∞ ) cv Mg Where k for air is 1.4, R = 287J/kgK and Tf,∞ is in Kelvin. This gives as = [1.4 × 287(J/kgK) × 293.15(K)]1/2 = 343 m/s. A plot of the surfaceconvection heat transfer rate as a function of the air speed is shown in Figure Pr.6.8. (b) From the numerical data, for qku = Qku L /Aku = 1200 W/m2 , we obtain uf,∞ = 3.78 m/s. For an analytic answer, the plot obtained could be used to identify that the desired heat ﬂux lies in the laminar ﬂow regime, and then the appropriate equation for Qku L could be solved for uf,∞ . COMMENT: Note the sudden rise in the rate of increase of the surfaceconvection heat transfer rate as the turbulent regime is entered. It is important to note that we have neglected any kind of transition region. In a real ﬂow, the transition from laminar to turbulent ﬂow would take place over a range of the Reynolds number. As the sonic speed is reached, the compressibility of the gas should be included in the Nul correlation.
510
12 x 10 3 10 x 10 3
DQ E
ku L
,W
8 x 10 3 Turbulent 6 x 10 3 ReL,t = 5 x 10 5 (uf, = 24.60 m/s)
4 x 10 3 2 x 10 3
as = 343 m/s
Laminar 0
50
100
150
200
250
300
350
uf, , m/s Figure Pr.6.8 Variation of the surfaceconvection heat transfer rate with respect to the farﬁeld velocity.
511
PROBLEM 6.9.FAM GIVEN: On a clear night, a water layer formed on a paved road can freeze due to radiation heat losses to the sky. The water and the pavement are at the freezing temperature T1 = 0◦C. The water surface behaves as a blackbody and radiates to the deep sky at an apparent temperature of Tsky = 250 K. The ambient air ﬂows parallel and over the water layer at a speed uf,∞ = 9 m/s and temperature Tf,∞ , which is greater than the water temperature. Assume that the surface convection is modeled using a surface that has a length L = 2 m along the ﬂow and a width w = 1 m (not shown in the ﬁgure) perpendicular to the ﬂow. These are shown in Figure Pr.6.9(a). Neglect the heat transfer to the pavement and evaluate the air properties at T = 273.15 K (Table C.22). SKETCH: Figure Pr.6.9(a) shows the thin water layer exposed to a warm air stream and a cold radiation sink.
Tsky Air
uf, Tf,
r,1
T1 = 273.15 K
L
Figure Pr.6.9(a) Radiation cooling of a thin water ﬁlm and its surfaceconvection heating.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the maximum ambient temperature below which freezing of the water layer occurs. (c) When a given amount of salt is added to the water or ice, the freezing temperature drops by 10◦C. If the water surface is now at 10◦C, will freezing occur? Use the property values found for (b), the given uf,∞ and Ts ky, and the freestream air temperature found in (b). SOLUTION: (a) To ﬁnd the maximum Tf,∞ above which melting will occur, we will assume that the water layer is at a uniform, lumped temperature. The thermal circuit is then shown in Figure Pr.6.9(b).
Tsky
Tf,
Eb,sky Rku
Qku
L
L
Qr,1
Rr,Σ Eb,1
T1
Ar1=Aku,1
Figure Pr.6.9(b) Thermal circuit diagram.
512
(b) From Figure Pr.6.9(b), by applying the conservation of energy principle to node T1 for steadystate conditions, we have Q A
=
Qku L + Qr,1
=
0.
We then evaluate each of the heat transfer terms as (i) Heat Transfer By Surface Convection: From Table C.22 for air at T = 273.15 K, kf = 0.0251 W/mK, νf = 13.33 × 10−6 m2 /s, and Pr = 0.69. For parallel ﬂow over a ﬂat plate, we have ReL =
uf,∞ L 9(m/s) × 2(m) = = 1.350 × 106 , νf 13.33 × 10−6 (m2 /s)
turbulent ﬂow regime.
From Table 6.3, for combined laminarturbulent ﬂow, NuL
=
Rku L
=
Qku L
=
4/5
(0.037ReL − 871)Pr1/3 = [0.037(1.350 × 106 )4/5 − 871](0.69)1/3 = 1,854 L 2(m) = 0.0215◦C/W = Aku NuL kf 2(m) × 1(m) × 1,854 × 0.0251(W/mK) T1 − Tf,∞ . Rku L
(ii) Heat Transfer By Surface Radiation: The water surface and the night sky are assumed to behave as black bodies, r,sky = r,1 = 1, and the view factor from the water to the sky is F1sky = 1. Then 1 − r 1 − r 1 1 1 2 Rr,Σ = + + =0+ +0= = 0.5 1/m r Ar 1 Ar,1 F1sky r Ar sky Ar,1 F1sky 2(m2 ) Qr,1
=
4 ) σSB (T14 − Tsky Eb,1 − Eb,sky = . Rr,Σ Rr,Σ
Then from Figure Pr.6.9(b), the energy equation for node T1 is (for no net heat transfer or for maximum Tf,∞ ) T1 − Tf,∞ Eb,1 − Eb,sky + = 0. Rku L Rr,Σ Solving for Tf,∞ Tf,∞
= T1 + Rku L
4 σSB (T14 − Tsky ) Rr,Σ
=
273.15 + (0.0215)(◦C/W)
=
277.20 K = 4.05◦C.
5.67 × 10−8 (W/m2 K) × (273.154 − 2504 )(K4 ) 2
0.5(1/m )
(c) For Tsl = −10◦C = 263.15 K, we repeat the above determinations of Qku L and Qr,1 . Qku L
=
Qr,1
=
(263.15 − 277.20)(K) = −653.5 W, 0.0215 −8 5.67 × 10 (W/m2 K4 ) × (263.154 − 2504 )(K4 ) = 100.8 W. 0.5(1/m2 )
Since  Qku L > Qr,w , the ice will melt. COMMENT: It is possible to cool a body below the ambient temperature using radiation heat transfer.
513
PROBLEM 6.10.DES GIVEN: A square ﬂat plate, with dimensions a × a, is being heated by a thermal plasma (for a coating process) on one of its sides. To prevent meltdown and assist in the coating process, the other side is cooled by impinging air jets. This is shown in Figure Pr.6.10. In the design of the jet cooling, a single, largediameter nozzle [Figure Pr.6.10(i)], or nine smaller diameter nozzles [Figure Pr.6.10(ii)] are to be used. a = 30 cm, Ts = 400◦C, Tf,∞ = 20◦C. Single nozzle: D = 3 cm, , Ln = 6 cm, L = 15 cm, uf A = 1 m/s. Multiple nozzles: D = 1 cm, Ln = 2 cm, L = 5 cm, uf A = 1 m/s. Use the average temperature between the air and the surface to evaluate the properties of the air. SKETCH: Figure Pr.6.10 shoes a single and a nine jet arrangement for cooling of a ﬂat surface. (i) BackSurface Cooling with a Single Impinging Jet Thermal Plasma
Qku
(ii) BackSurface Cooling with Multiple Impinging Jets Thermal Plasma
L
Qku
L
Square Plate Aku Ln = 6 cm
Back Surface Temperture Ts = 400oC
Ln = 2 cm D = 1 cm
Exit Conditions uf = 1 m/s , Tf, = 20oC
Nozzle
Square Plate
Aku
L = 5 cm Exit Conditions uf = 1 m/s , Tf, = 20oC
Surface Temperture Ts = 400oC
a = 30 cm
D = 3 cm L = 15 cm L = 15 cm a = 30 cm
Figure Pr.6.10 (i) A single impinging jet used for surface cooling. (ii) Multiple impinging jets.
OBJECTIVE: For each design, do the following: (a) Determine the average Nusselt number NuL . (b) Determine the average surfaceconvection thermal resistance Aku Rku L [◦C/(W/m2 )]. (c) Determine the rate of surfaceconvection heat transfer Qku L (W). SOLUTION: (i) Single Nozzle: (a) The average Nusselt number for cooling with a single, round nozzle is given by (6.71), i.e., 1/2
1/2 NuL = 2ReD Pr0.42 (1 + 0.005Re0.55 D )
1 − 1.1D/L , 1 + 0.1(Ln /D − 6)D/L
where the Reynolds number is based on the nozzle diameter and is given by (6.68), i.e., ReD =
uf A D . νf
Equation (6.71) is valid for L/D > 2.5. For the nozzle given, L/D = 15/3 = 5, thus satisfying this constraint The properties for air at Tδ = (400 + 20)/2 = 210◦C = 483 K are, from Table C.22, kf = 0.0384 W/mK, νf = 35.29 × 10−6 m2 /s, and Pr = 0.69. The area for surface convection is Aku = a2 . Then Reynolds number becomes ReD =
uf A D 1(m/s) × 0.03(m) = 850.1. = νf 35.29 × 10−6 (m2 /s) 514
The average Nusselt number, from (6.71) becomes NuL = 46.43. (b) The average surfaceconvection thermal resistance is obtained from (6.49), i.e., L 0.15(m) 2 = 8.413 × 10−2 ◦C/(W/m ). = kf NuL 0.0384(W/mK) × 46.43
Aku Rku L =
(c) The averaged surfaceconvection heat transfer is obtained from (6.49) as Qku a = Aku
(Ts − Tf,∞ ) 400(◦C) − 20(◦C) = 406.5 W. = (0.3)2 (m)2 × 2 Aku Rku L 8.413 × 10−2 [◦C/(W/m )]
(ii) Square Array of Multiple Nozzles: (a) The average Nusselt number for cooling with a square array of round nozzles is given by (6.72), i.e., 6 −0.05 2L /D 1 − 2.2(1 − )1/2 L n 2/3 1/2 NuL = ReD Pr0.42 (1 − ) . 1+ 0.6 D 1 + 0.2(Ln /D − 6)(1 − )1/2 (1−)1/2 This is valid for L/D > 1.25. For the nozzles, L/D = 5/1 = 5, thus satisfying this constraint. The Reynolds number becomes ReD =
uf,∞ D 1(m/s)0.01(m) = 283.4. = νf 35.29 × 10−6 (m2 /s)
The void fraction deﬁned by (6.73) is =1−
πD2 = 0.9921. 16L2
The average Nusselt number becomes NuL = 28.35. (b) The average surfaceconvection thermal resistance becomes Aku Rku L =
L 0.05(m) 2 = 4.593 × 10−2 ◦C/(W/m ). = kf NuL 0.0384(W/mK) × 28.35
(c) The averaged surfaceconvection heat transfer can then be obtained from Qku a = Aku
(Ts − Tf,∞ ) 400(◦C) − 20(◦C) = 744.6 W. = (0.3)2 (m)2 × 2 Aku Rku L 4.593 × 10−2 [◦C/(W/m )]
COMMENT: Note that under these conditions the single nozzle removes slightly more heat from the surface. However, the multiple nozzles results in a more uniform surface cooling.
515
PROBLEM 6.11.FUN GIVEN: Permanent damage occurs to the pulp of a tooth initially at T (t = 0) = 37◦C, when it reaches a temperature Tp = 41◦C. Therefore, to prevent nerve damage, a water coolant must be constantly applied during many standard tooth drilling operations. In one such operation, a drill, having a frequency f = 150 Hz, a burr diameter Db = 1.2 mm, a tooth contact area Ac = 1.5 × 10−7 m2 , and a coeﬃcient of friction between the drill burr and the tooth µF = 0.4, is used to remove an unwanted part of the tooth. The contact force between the drill burr and the tooth is F = 0.05 N. During the contact time, heat is generated by surface friction heating. In order to keep the nerves below their threshold temperature, the tooth surface must be maintained at Ts = 45◦C by an impinging jet that removes 80% of the generated heat. The distance between the jet and the surface Ln is adjustable. Use the dimensions shown in Figure Pr.6.11(a)(ii). Tf,∞ = 20◦C, uf = 0.02 m/s, D = 1.5 mm, L = 4 mm, µF = 0.4 Pas, f = 150 1/s, ∆ui = 2πf Rb , pc = Fc /Ac , Fc = 0.05 N, Ac = 1.5 × 10−7 m2 , Db = 2Rb = 1.2 mm. Use the same surface area for heat generation and for surface convection (so surface area Aku will not appear in the ﬁnal expression used to determine Ln ). Determine the water properties at T = 293 K. SKETCH: Figure Pr.6.11(a) shows the waterjet cooling of a tooth during drilling.
(i) Anatomy of a Tooth Dentin Pulp
(ii) Physical Model of Friction Heated and Jet Cooled Enamel Surface
Enamel (Crown) . Frictional Heating, Sm,F CementoEnamel Junction
Ts = 45oC Tp
Gingival Crevice (Gumline)
Gingival Sulcus (Space Between Gum and Tooth)
Jet Exit Conditions: Tf, = 20oC uf = 0.02 m/s L = 4 mm
Gingiva (Gum)
Cementum
Dentin and Enamel
Bone
Ligament Root Canal
Sm,F D = 1.5 mm L = 4 mm
Pulp Blood Vessels and Nerves
Ln
Figure Pr.6.11(a)(i)Cooling of tooth during drilling. (ii) Physical model of friction heated and jet cooled enamel surface.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Write the surface energy equation for the tooth surface. (c) Determine the location Ln of the jet that must be used in order to properly cool the tooth. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.6.11(b).
Qk,sp
Sm,F
Qku
L
Tf,
Tp Rk,sp
Ts
Rku
L
Figure Pr.6.11(b) Thermal circuit diagram.
516
(b) From Figure Pr.6.11(b), we have QA = Qku = Aku qku =
S˙ i = 0.8S˙ m,F .
i
(c) The heat generated by surface friction heating, from (2.53), is S˙ m,F /Aku = µF pc ∆ui ,
∆ui = 2πf Rb = 2πf
Db 2
where pc
=
Fc 0.05(N) = = 3.333 × 105 N/m2 Ac 1.5 × 10−7 (m2 )
∆ui
=
2πf Rb = 2πf
=
0.5655 m/s.
1.2 × 10−3 Db = 2π × 150(1/s) × (m) 2 2
Then, S˙ m,F /Aku
=
0.4 × 3,333 × 106 (N/m2 ) × 0.5655(m/s)
=
75,391 W/m2 .
From the energy equation, after dropping Aku , we have qku L
S˙ m,F = 0.8 × 75,400(W/m2 ) Aku 60,313 W/m2 .
0.8 ×
= =
Next, we use the Nusselt number for qku L using (6.49), i.e., kf (Ts − Tf,∞ ) L kf qku L = NuL (Ts − Tf,∞ ). L
Qku L = Aku NuL
Properties (water, T = 293 K, Table C.23): interpolated values; kf = 0.595 W/mK, Pr = 7.528, νf = 106.7×10−8 m2 /2. Using Ts = 45◦C, Tf,∞ = 20◦C, L = 4 mm, we have 60,313(W/mK) = NuL × NuL
=
0.595(W/mK) × (45 − 20)(K) 0.004(m)
16.22.
The Nusselt number relation is found from Table 6.3, i.e.,
NuL = 2Re
1/2
0.42
Pr
(1 + 0.005Re
0.55 1/2
)
D 1 − 1.1 L D Ln −6 1 + 0.1 D L
where, ReD
=
uf A D 0.02(m/s) × 0.0015(m) = = 28.12. νf 106.7 × 10−8 (m2 /s) 517
Then NuL
=
= = =
2 × (28.12)1/2 × (7.528)0.42 × [1 + 0.005 × (28.12)0.55 ]1/2 × 1.5(cm) 1 − 1.1 × 4(cm) Ln 0.0015(m) −6 × 1 + 0.1 × 0.0015(m) 0.004(m) 0.5875 25.14 × 1 + (666.7Ln − 6) × 0.03750) 0.5875 25.14 × 25Ln + 0.775 14.77 . 25Ln + 0.775
Solving for Ln in the above relation, we have 16.22
=
405.5Ln + 12.57
=
14.77 25Ln + 0.775 14.77
Ln
=
0.005425 m = 0.5425 cm.
COMMENT: The presence of the drill in the impinging jet area is neglected. This is a reasonable nozzle to surface distance.
518
PROBLEM 6.12.FAM GIVEN: Heatactivated, dry thermoplastic adhesive ﬁlms are used for joining surfaces. The adhesive ﬁlm can be heated by rollers, hot air, radiofrequency and microwaves, or ultrasonics. Consider a ﬂat fabric substrate to be coated with a polyester adhesive ﬁlm with the ﬁlm, heated by a hot air jet, as shown in Figure Pr.6.12(a). The ﬁlm is initially at T1 (t = 0). The thermal set temperature is Tsl . Assume that the surfaceconvection heat transfer results in the rise in the ﬁlm temperature with no other heat transfer. Ln = 4 cm, L = 10 cm, D = 1 cm, uf = 1 m/s, l = 0.2 mm, Tf,∞ = 200◦C, Tsl = 120◦C. Determine the air properties at T = 350 K. For polyester, use Table C.17 and the properties of polystyrene. SKETCH: Figure Pr.6.12(a) shows the thin plastic ﬁlm heated by a hotair jet. Hot Air Jet Thin Adhesive Film, Thermally Activated T1(t)
D = 2 cm Ln 2L
Tf, , uf 2L
l Substrate
 Qku,1
Figure Pr.6.12(a) A heatactivated adhesive ﬁlm heated by a hot air jet.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the elapsed time needed to reach Tsl , for the conditions given above. SOLUTION: (a) Figure Pr.6.12(b) shows the thermal circuit diagram. The only heat transfer is assumed to be Qku,1∞ , i.e., no heat losses are allowed. Qku,1 T1(t)
Q1 = 0
 (HcpV)1 dT1 dt
Figure Pr.6.12(b) Thermal circuit diagram.
(b) Assuming a uniform temperature for the ﬁlm, the transient temperature is given by (6.156), i.e., T1 (t = 0) = Tf,∞ + [T1 (t = 0) − Tf,∞ ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ), where a1 =
S˙ 1 − Q1 = 0, (ρcp V )1
τ1 = (ρcp V )1 Rku L .
The surfaceconvection resistance is given by (6.49) as Rku L
=
Aku
=
L Aku N uL kf 2L × 2L 519
and NuL is given in Table 6.3 as 1/2
NuL
=
1/2 2ReD Pr0.42 (1 + 0.005Re0.55 D )
ReD
=
uf D . νf
1 − 1.1D/L , 1 + 0.1(Ln /D − 6)D/L
From Table C.22, for air at T = 350 K, we have νf = 2.030 × 10−5 m2 /s kf = 0.0300 W/mK
Table C.22
Pr = 0.69
Table C.22.
Table C.22
Then ReD
=
NuL
=
=
1(m/s) × 0.02(m) = 985.2 2.030 × 10−5 (m2 /s) 2 × (985.2)1/2 × (0.69)0.42 (1 + 0.005 × (985.2)0.55 )1/2 × 1 − 1.1 × 0.02(m)/0.1(m) , 1 + 0.1{[0.04(m)/0.02(m)] − 6} × [0.02(m)/0.10(m)] 0.78 2 × 26.86 × 1.105 × 1.006 × = 50.33. 0.92
Then Rku L
V1
0.10(m) (0.20)2 (m2 ) × 50.33 × 0.030(W/mK) = 1.656 K/W = 2L × 2L × l = (2 × 0.10)2 × 2 × 10−4 (m3 ) = 8.0 × 10−6 m3 . =
For polystyrene, from Table C.17, we have ρ1 = 1,050 kg/m3 cp = 1,800 J/kgK
Table C.17 Table C.17.
Then τ1
= =
1,050(kg/m3 ) × 1,800(J/kgK) × 8 × 10−6 (m3 ) × 1.656(K/W) 25.04 s.
We can solve the temperature T1 (t) expression for t, since a1 = 0, and we have, with T1 (t) = Tsl T1 (t) − Tf,∞ t = τ1 ln T1 (t = 0) − Tf,∞ (120 − 200)(◦C) = −25.04(s) × ln = 20.31 s. (20 − 200)(◦C) COMMENT: Note that we have neglected the heat losses from the ﬁlm to the substrate by conduction and to the surroundings by surface radiation. Inclusion of these would be increase the elapsed time. Note that we need slightly less than one time constant (τ1 ) to reach the desired temperature.
520
PROBLEM 6.13.DES GIVEN: A pure aluminum plate is to be rapidly cooled from T1 (t = 0) = 40◦C to T1 (t) = 20◦C. The plate has a length L = 12 cm and thickness w = 0.2 cm. The plate is to be cooled using water by placing it at a distance Ln = 10 cm from a faucet with diameter D = 2 cm. The water leaves the faucet at a temperature Tf,∞ = 5◦C and velocity uA = 1.1 m/s. There are two options for the placement of the plate with respect to the water ﬂow. The plate can be placed vertically, so the water ﬂows parallel and on both sides of the plate. Then the water layer will have a thickness l = 2 mm on each side of the plate [shown in Figure Pr.6.13(a)(i)]. Alternately, it can be placed horizontally with the water ﬂowing as a jet impingement [shown in Figure Pr.6.13(a)(ii)]. Assume that the results for impinging jets can be used here, even though the jet ﬂuid (water) is not the same as the ambient (air) ﬂuid. Use the water as the only ﬂuid present. Assume a uniform plate temperature. Estimate the parallel, farﬁeld velocity uf,∞ using the mass ﬂow rate out of the faucet. This approximate ﬂux is assumed uniform over the rectangular ﬂow cross section (l × 2L) and is assumed to be ﬂowing over a square surface (2L × 2L). SKETCH: Figure Pr.6.13(a) shows the plate to be cooled by the water from a faucet. Two diﬀerent plate orientations are considered.
(i) Parallel Configuration Faucet (Nozzle)
(ii) Perpendicular Configuration
D = 2 cm
D = 2 cm
Tf, , Duf EA
Tf, , Duf EA Ln = 10 cm
Water Flow Vertically Placed Plate
Water Jet
L = 12 cm
w = 0.2 cm Horizontally L = 12 cm Placed Plate
uf, w = 0.2 cm Water Layer Thickness l = 0.2 cm
Figure Pr.6.13(a) Cooling of an aluminum plate under a faucet. (i) Parallel conﬁguration. (ii) Perpendicular conﬁguration.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine which of the orientations gives the shorter cooling time. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.6.13(b).
Q1
T1
Qku
 (HcpV)1 dT1
Tf,
L
Rku
L
dt
Figure Pr.6.13(b) Thermal circuit diagram.
521
(b) (i) Parallel Flow: The mass ﬂow rate of water out of the faucet is 2
πD M˙ f = ρf uf Au = ρf uf . 4 Properties: (water at T = 280 K, Table C.23): kf = 0.5675 W/mK, cp = 4,204 J/kgK, νf = 152 × 10−8 m2 /s, Pr = 11.36; (pure aluminum at T = 300 K, Table C.16): ρs = 2,702 kg/m3 , cp,s = 903 J/kgK, ks = 237 W/mK. This ﬂow is approximately divided on both side of the plate, i.e., M˙ f ≡ ρf 2 × (l × 2L) × uf,∞ . Then, equating the two expressions for M˙ f gives, uf,∞
= uf
πD2 /4 2(l × 2L)
=
1.1(m/s) ×
=
0.36 m/s.
π(0.02)2 (m2 )/4 2 × (0.002)(m) × 2 × (0.12)(m)
we determine the elapsed time needed to cool the plate, using (6.156), i.e., T1 (t) − Tf,∞
=
τ1
=
[T1 (t = 0) − Tf,∞ ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ) S˙ 1 − Q1 (ρcp V )1 Rku L , a1 = . (ρcp V )1
Since there is no energy conversion and all other forms of heat transfer are assumed negligible, both S˙ 1 and Q1 equal to zero. Then a1 = 0, or, T1 (t) = Tf,∞ = [T1 (t = 0) − Tf,∞ ]e−t/τ1 , τ1 = (ρcp V )1 Rku L . The Reynolds number is uf,∞ 2L , 2L = 24 cm = 0.24 m νf 0.36(m/s) × (0.24)(m) = 56,842 < ReL,t = 5 × 105 . ReL = 152 × 10−8 (m2 /s) ReL =
The ﬂow remains laminar over the plate. From (6.49), we have Rku L =
2L , Aku NuL kf
where Aku = 2L2 = 2(0.12)2 (m2 ) = 0.0288 m2 , since there are two sides to the plate. From Table 6.3, we have NuL
=
1/2
0.664ReL Pr1/3 = 0.664(56,842)1/2 (11.36)1/3 = 355.9.
Then Rku L =
0.24(m) = 0.04126 K/W. 0.0288(m2 ) × 355.9 × 0.5675(W/mK) 522
Next, V1 V1 τ1
= L2 w,
w = 0.2 cm = 0.002 m
= =
(0.12) (m2 ) × (0.002)(m) = 2.88 × 10−5 m3 (ρcp V )1 Rku L = [2,702(kg/m2 ) × 903(J/kgK) × 2.88 × 10−5 (m3 )] × 0.04126(K/W)
=
2.899 s.
2
Then
t
= −τ1 ln
T1 (t) − Tf,∞ 20 − 5 = −2.899(s) × ln = 2.456 s. T1 (t = 0) − Tf,∞ 40 − 5
(ii) Perpendicular Flow: Again, we have T1 (t) − Tf,∞
=
[T1 (t = 0) − Tf,∞ ]e−t/τ1 ,
τ1 = (ρcp V )1 Rku L
The nozzle Reynolds number is ReD
= =
uf D , D = 2 cm = 0.02 m νf 1.1(m/s) × (0.02)(m) = 14,474. 152 × 10−8 (m2 /s)
Also, L D
=
Rku L
=
Aku
12(cm) =6 2(cm) L Aku NuL kf
= L2 = (0.12)2 (m2 ) = 0.0144 m2 .
From Table 6.3, we have NuL
1.1D 1− 1/2 L + 0.005Re0.55 D ) D Ln −6 1 + 0.1 D L
=
1/2 2ReD Pr0.42 (1
=
2 × (14,474)1/2 × (11.36)0.42 × [1 + 0.005(14,474)0.55 ]1/2 × 1.1 × 0.02(m) 1− 0.12(m) 0.1(m) 0.02(m) −6 × 1 + 0.1 × 0.02(m) 0.12(m) 0.8167 = 778.6. 667.7 × 1.404 × 0.9833
= Then Rku L =
L 0.12 = = 0.01189 K/W. Aku Nukf 0.0144(m2 ) × (778.6) × 0.5675(W/mK)
Next τ1
=
(ρcp V )1 Rku L
=
(2702(kg/m ) × 903(J/kgK) × 2.88 × 10−5 ) × 0.01189(K/W) = 1.325 s. 3
Then
t
= −τ1 ln
T1 (t) − Tf,∞ 20 − 5 = −1.325(s) × ln = 1.123 s. T1 (t = 0) − Tf,∞ 40 − 5 523
This cooling time is less than that for the parallel arrangement, therefore the perpendicular orientation should be used for rapid cooling. COMMENT: In order to verify the lumped capacitance assumption, we must show that the Biot number is much less than one (for both ﬂows). The Biot number is deﬁned by (6.128) as Biw =
Rk . Rku L
For aluminum, from Table C.14, at T = 300 K, we have ks = 237 W/mK. (i) Parallel Flow: Biw
= =
w/Ak ks Rku L 0.002(m)/[0.0144(m2 ) × 237(W/mK)] = 0.0142. 0.04126(K/W)
(ii) Perpendicular Flow: The conduction resistance is the same, and therefore, Biw
=
5.86 × 10−4 (K/W) = 0.0493. 0.001149(K/W)
In both cases, the Biot number is much less than one and the lumped capacitance assumption is therefore valid.
524
PROBLEM 6.14.FAM GIVEN: A bottle containing a cold beverage is awaiting consumption. During this period, the bottle can be placed vertically or horizontally, as shown in Figure Pr.6.14. Assume that the bottle can be treated as a cylinder of diameter D and length L. We wish to compare the surfaceconvection heat transfer to the bottle when it is (i) standing vertically or (ii) placed horizontally. For the vertical position, the surfaceconvection heat transfer is approximated using the results of the vertical plate, provided that the boundarylayer thickness δα is much less than the bottle diameter D. D = 10 cm, L = 25 cm, Ts = 4◦C, Tf,∞ = 25◦C. Neglect the end areas. Use the average temperature between the air and the surface to evaluate the thermophysical properties of the air. SKETCH: Figure Pr.6.14 shows two positions of a beverage bottle.
Surrounding Air Tf, = 25 oC
L = 25 cm
g
(i) Vertically Arranged
Cola
Co
la
Bottle Ts = 4 oC < Tf,
(ii) Horizontally Arranged
D = 10 cm
Figure Pr.6.14 Thermobuoyant ﬂow and heat transfer from beverage bottles. (i) Standing vertically. (ii) Placed horizontally
OBJECTIVE: (a) Determine the average Nusselt numbers NuL and NuD . (b) Determine the average surfaceconvection thermal resistances Aku Rku L [◦C/(W/m2 )] and Aku Rku D [◦C/(W/m2 )]. (c) Determine the rates of surfaceconvection heat transfer Qku L (W) and Qku D (W). SOLUTION: (i) Vertical Position: (a) The Rayleigh number is given by (6.88) as
RaL =
gβ(Ts − Tf,∞ )L3 . νf αf
The properties for air at Tδ = (Ts + Tf,∞ )/2 = 288 K are obtained from Table C.22: kf = 0.026 W/mK, νf = 14.60 × 10−6 m2 /s, Pr = 0.69, and from (6.77) we have βf = 1/Tave = 3.472 × 10−3 1/K. With νf αf = νf2 /Pr, the Rayleigh number becomes
RaL =
9.81(m2 /s) × 3.472 × 10−3 (1/K) × (25 − 4)(K) × (0.25)3 (m)3 = 3.618 × 107 . (14.60 × 10−6 )2 (m2 /s)2 /(0.69) 525
Since RaL < 109 , from (6.91) the ﬂow is laminar. For thermobuoyant ﬂow over a vertical ﬂat plate, the average Nusselt number is obtained from (6.92) as a1
=
0.503 4 = 0.5131 3 1 + ( 0.492 )9/16 4/9 Pr
NuL,l
=
2.8
ln 1 +
= 41.18
2.8 1/4 a1 RaL
NuL,t
=
0.13Pr0.22 1/3 0.81 0.42 RaL = 33.88 (1 + 0.61Pr )
NuD
=
[(NuL,l )6 + (NuL,t )6 ]1/6 = 43.08.
(b) The average surfaceconvection thermal resistance is found from (6.49) as Aku Rku L =
L 0.25(m) 2 = 2.232 × 10−1 ◦C/(W/m ). = kf NuL 0.026(W/mK) × 43.08
(c) The surfaceaveraged surfaceconvection heat transfer is found from (6.49) as Qku L = Aku
Ts − Tf,∞ 4(◦C) − 25(◦C) = −7.390 W. = π × 0.1(m) × 0.25(m) × 2 Aku Rku L 2.232 × 10−1 [◦C/(W/m )]
(ii) Horizontal Position: (a) For the horizontal cylinder, the Rayleigh number is found from Table 6.4, i.e., RaL =
gβ(Ts − Tf,∞ )D3 9.81(m2 /s) × 3.472 × 10−3 (1/K) × [25(◦C) − 4(◦C)](0.1)3 (m)3 = 2.315 × 106 . = νf αf (14.60 × 10−6 )2 (m2 /s)2 /(0.69)
A correlation for the average Nusselt number for a horizontal cylinder is given in Table 6.4. Using the values given a1
=
0.503 4 = 0.5131 3 1 + ( 0.492 )9/16 4/9 Pr
NuD,l
=
ln 1 +
1.6
= 16.24
1.6 1/4 0.772a1 RaD
NuD,t
=
0.13Pr0.22 1/3 Ra = 13.55 (1 + 0.61P r0.81 )0.42 D
NuD
=
[(NuD,l )3.3 + (NuD,t )3.3 ]1/3.3 = 18.55.
(b) The average surfaceconvection thermal resistance is Aku Rku D =
D 0.10(m) 2 = 2.073 × 10−1 ◦C/(W/m ). = kf NuD 0.026(W/mK) × 18.55
(c) The averaged surfaceconvection heat transfer is Qku D = Aku
Ts − Tf,∞ 4(◦C) − 25(◦C) = −7.956 W. = π × 0.1(m) × 0.25(m) × 2 Aku Rku D 2.073 × 10−1 [◦C/(W/m )]
COMMENT: For the vertical plate, since RaL < 109 , the ﬂow regime is laminar. For the laminar thermobuoyant ﬂow over a vertical ﬂat plate, the average Nusselt number could also be determined from using the similar relation (6.89), i.e., NuL =
0.503 4 1/4 Ra = 39.79 3 1 + ( 0.492 )9/16 4/9 L Pr
526
and the average surfaceconvection thermal resistance and heat transfer rate are Aku Rku L =
Qku L = Aku
L 0.25(m) 2 = 2.416 × 10−1 ◦C/(W/m ) = kf NuL 0.026(W/mK) × 39.79
Ts − Tf,∞ 4(◦C) − 25(◦C) = −6.826 W. = π × 0.1(m) × 0.25(m) × 2 Aku Rku L 2.42 × 10−1 [◦C/(W/m )]
These are considered close to the values obtained using the combined laminarturbulent correlation. Note also that the horizontal position results in a slightly larger heat ﬂow. Also, note that determining δα (L) from (6.90) we have
δα (L)
= =
20 3.93L Pr + 21 0.01574m,
1/4
(GrL Pr)−1/2 1/2
where GrL =RaL /Pr. Then δα (L) 0.01574(m) = = 0.1574, D 0.10(m) which satisﬁes the needed constraint that δα (L) D.
527
PROBLEM 6.15.FAM GIVEN: The ﬁreplace can provide heat to the room through surface convection and surface radiation from that portion of the ﬁreplace wall heated by the combustion products exiting through a chimney behind the wall. This heated area is marked on the ﬁreplace wall in Figure Pr.6.15(a). Assume this portion of the wall (including the ﬁreplace) is maintained at a steady, uniform temperature Ts . The surface convection is by a thermobuoyant ﬂow that can be modeled as the ﬂow adjacent to a heated vertical plate with length L. The surface radiation exchange is between this heated portion of the wall and the remaining surfaces in the room. Assume that all the remaining wall surfaces are at a steady uniform temperature Tw . Ts = 32◦C, Tf,∞ = 20◦C, Tw = 20◦C, r,s = 0.8, r,w = 0.8, w = 3 m, L = 4 m, a = 6 m. Determine the air properties at 300 K (Table C.22). SKETCH: Figure Pr.6.15(a) shows the surface convection and radiation from a portion of the ﬁreplace wall to the rest of the room.
w
Thermobuoyant Flow g
a
r,s
Ts
Qku
L
Qr Fireplace a
a
r,w , Tw For All Nonshaded Areas Tf, uf, = 0
Figure Pr.6.15(a) Surface convection and radiation from a heated wall.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the surfaceconvection heat transfer rate. (c) Determine the surfaceradiation heat transfer rate. (d) Assume the ﬁre provides an energy conversion rate due to combustion of S˙ r,c = 19,500 W. (This would correspond to a large 5 kg log of wood burning at a constant rate to total consumption in one hour.) Determine the eﬃciency of the ﬁreplace as a roomheating system (eﬃciency is deﬁned as the ratio of the total surface heat transfer rate to the rate of energy conversion S˙ r,c ). SOLUTION: The thermal properties for air are evaluated at T = 300 K from Table C.22 and are kf = 0.0267 W/mK, νf = 15.66 × 10−6 m2 /s, αf = 22.57 × 10−6 m2 /s, Pr = 0.69, and βf = 1/T = 1/300 1/K. (a) The thermal circuit diagram is given in Figure Pr.6.15(b). (b) The area for surfaceconvection is Aku = L × w = 4(m) × 3(m) = 12 m2 . Then for surface convection from (6.49), we have Qku L =
Ts − T∞ Rku L
where from (6.49) Rku L =
1 . Aku NuL kf /L 528
Ts
Qku
L
Rku
L
Tf,
Qs (Rr,F)sw
Eb,s
(qr,o)s
(Rr, )w
(Rr, )s
Eb,w Tw
(qr,o)w
Qr,s
Qr,sw
Qr,w
Figure Pr.6.15(b) Thermal circuit diagram.
The ﬂuid ﬂow for surfaceconvection is modeled as a thermobuoyant ﬂow over a ﬂat vertical plate. The NuL is given in Table 6.4 as NuL = [(NuL,l )6 + (NuL,t )6 ]1/6 , where NuL,l
2.8
=
2.8
ln 1 +
1/4
a1 RaL NuL,t
0.13Pr0.2 1/3 Ra . (1 + 0.61Pr0.81 )0.42 L
=
The RaL and a1 are deﬁned in Table 6.5 as 2
RaL
= =
a1
= =
gβf (Ts − T∞ )L3 9.81(m/s ) × 1/300(1/K) × (32 − 20)(K) × (4)3 (m) = νf αf 15.66 × 10−6 (m2 /s) × 22.57 × 10−6 (m2 /s) 7.105 × 1010 0.503 4 3 [1 + (0.492/Pr)9/16 ]4/9 0.5131.
Then upon substitution, the NuL,l and NuL,t are NuL,l = 266.31 NuL,t = 424.33, Then NuL then is NuL = [(266.31)6 + (424.33)6 ]1/6 = 428.55. Then Rku L is Rku L
= =
1 12(m2 ) × 428.55 × 0.0267(W/m◦C)/4(m) 0.0291◦C/W.
And ﬁnally, Qku L is Qku L
= =
(32 − 20)(◦C) 0.0291(◦C/W) 411.9 W. 529
3
(c) The heated surface and the rest of the room constitute an enclosure and the surfaces are gray and diﬀuse. Then we apply the radiation enclosure analysis. The wall area is Aw = 5 × a2 = 5 × (62 )(m2 ) = 180 m2 . Then for surface radiation exchange in a twosurface enclosure, we have from (4.47), Qr,sw =
Eb,s − Eb,w σ(Ts4 − Tw4 ) = . (RΣ,r )sw (Rr, )s + (Rr,F )sw + (Rr, )w
Solving for the resistances, we have 1 − r,s 1 − 0.8 1 = = 0.0208 2 As r,s 12(m2 ) × 0.8 m Fsw = 1 by inspection 1 1 1 (Rr,F )sw = = = 0.0833 2 2 As Fsw 12(m ) × 1 m 1 − r,w 1 − 0.8 1 (Rr, )w = = = 0.001389 2 2 Aw r,w 180(m ) × 0.8 m (Rr, )s =
RΣ,r = (Rr, )s + (Rr,F )sw + (Rr, )w = (0.0208 + 0.0833 + 0.001389)
1 1 . 2 = 0.1055 m m2
Therefore, Qr,sw =
5.67 × 10−8 (W/m2 K4 ) × [305.154 − 293.154 ](K4 ) = 691.0 W. 1 0.1055( 2 ) m
(d) The eﬃciency η is deﬁned as Qku L + Qr,sw QA = ˙ Sr,c S˙ r,c 1,102.9(W) 411.9(W) + 691.0(W) = = 0.05656 = 5.656%. = 19,500(W) 19,500(W)
η=
COMMENT: Note that the surfaceradiation heat transfer is larger than surface convection. Higher eﬃciencies are possible by heating a larger portion of the wall, allowing for convection directly into the room, and forcing an air stream around the ﬁreplace.
530
PROBLEM 6.16.FUN GIVEN: As discussed in Section 6.5.1, the twodimensional (x, y), (uf , vf ), laminar viscous, thermobuoyant boundary layer (for vertical, uniform surface temperature plate) momentum equation (6.80) can be reduced to an ordinary diﬀerential equation using a dimensionless similarity variable y η= x
Grx 4
1/4 ,
and a dimensionless stream function ψ∗ =
4νf
∂ψ ∂ψ gβf (Ts − Tf,∞ )x3 . 1/4 , uf = ∂y , vf = − ∂x , Grx = νf2 Grx 4 ψ
OBJECTIVE: (a) Show that the momentum equation (6.80) reduces to 2 ∗ d3 ψ ∗ ∗d ψ + 3ψ −2 dη 3 dη 2
dψ ∗ dη
2
+ Tf∗ = 0 Tf∗ =
Tf − T∞ . Ts − Tf,∞
(b) Show that the energy equation (6.79) reduces to d2 Tf∗ dη 2
+ 3Prψ ∗
dTf∗ νf = 0 Pr = . dη αf
SOLUTION: We start from (6.80), written as uf
∂uf ∂uf ∂ 2 uf + vf − νf − gβf (Tf − Tf,∞ ) = 0. ∂x ∂y ∂y 2
Using the stream function ψ, the dimensionless stream function ψ ∗ and the similarity variables η, we transform uf and vf into ψ ∗ , x and η. We start with uf ≡
∂ψ ∂y
= =
1/2 1/4 Grx dψ dη dψ ∗ 1 Grx = 4νf dη dy dη x 4 4 1/2 ∗ dψ 4νf Grx x 4 dη
or
4νf x
dψ ∗ 1/2 = dη . Grx 4
uf
Also, vf
1/4 Grx ∂ ∂ψ = − 4νf = − ψ∗ ∂x ∂x 4 1/4 νf Grx dψ ∗ − 3ψ ∗ . = η x 4 dη 531
Next, these velocity components are diﬀerentiated, with respect to x and y, and we have 1/2 2 ∗ 1/2 ∂uf Grx Grx d ψ dψ ∗ 1 η = 4νf − 2 + 2 ∂x 4 4 dη 4x dη 2 2x 3/4 2 ∗ 4νf Grx d ψ ∂uf = 2 ∂y 4 x dη 2 2 3 ∗ ∂ uf 4νf Grx d ψ = . 2 ∂y x2 4 dη 3 Substituting these in the above momentum equation and using Tf∗ =
Tf − Tf,∞ , Ts − Tf,∞
and after rearranging the terms, we have 2 ∗ d3 ψ ∗ ∗d ψ + 3ψ −2 dη 3 dη 2
dψ ∗ dη
2
+ Tf∗ = 0.
(b) We start from (6.79), rewritten as uf
∂Tf ∂Tf ∂ 2 Tf + vf − αf = 0. ∂x ∂y ∂y 2
We have ∂Tf ∂x
=
∂Tf ∂y
=
∂ 2 Tf2 ∂y 2
=
∗ ∂Tf∗ ∂η Ts − Tf,∞ dTf =− η ∂η ∂x 4x dη 1/4 ∗ ∂Tf ∂η dTf∗ Ts − Tf,∞ Grx (Ts − Tf ∞ ) = ∂η ∂y x 4 dη 1/4 1/2 2 ∗ ∗ dTf d Tf Ts − Tf,∞ Grx ∂ 1 Grx = (Ts − Tf ∞ ) . 2 ∂y x 4 dη 4 x dη 2
(Ts − Tf ∞ )
Substituting into the above energy equation, using the velocity results from part (a), and after rearranging, we have d2 Tf∗ dη 2
+ 3Prψ ∗
dTf∗ = 0. dη
COMMENT: Note that for Pr → 0 (liquid metals), the temperature distribution will be linear in η (because the second derivative in zero). Also, note that from the above relation for ∂Tf /∂y, we have the surface heat ﬂux as −k
1/4 dTf∗ ∂Tf 1 Grx = q = . ku ∂y y=0 x 4 dη η=0
532
PROBLEM 6.17.FUN GIVEN: The dimensionless, transformed coupled boundarylayer momentum and energy equations for thermobuoyant ﬂow, are ∗ 2 2 ∗ dψ d3 ψ ∗ ∗d ψ + 3ψ − 2 + Tf∗ = 0 dη dη 3 dη 2 dTf∗ d2 Tf∗ ∗ = 0, + 3Prψ dη dη 2 subject to the surface and farﬁeld thermal and mechanical conditions dψ ∗ = ψ ∗ = 0, Tf∗ = 1 dη dψ ∗ = 0, Tf∗ = 0. η→∞: dη
at
η=0:
for
Use Pr = 0.72 and plot ψ ∗ , dψ ∗ /dη, d2 ψ ∗ /dη 2 , Tf∗ , and dTf∗ /dη, with respect to η. Note that with an initialvalue problem solver such as SOPHT, the second derivative of ψ ∗ and ﬁrst derivative of Tf∗ at η = 0 must be guessed. The guesses are adjusted till d2 ψ ∗ /dη 2 becomes zero for large η. Use d2 ψ ∗ /dη 2 (η = 0) = 0.6760 and dTf∗ /dη ∗ (η = 0) = −0.5064. OBJECTIVE: Using a solver, integrate these coupled equations SOLUTION: The solver we choose is an initialvaluesolver, such as SOPHT, where the initial values (i.e., at η = 0) for ψ ∗ , dψ ∗ /dη, d2 ψ ∗ /dη 2 , Tf∗ , and dTf∗ /dη must be provided for these coupled thirdand secondorder, ordinary diﬀerential equations. Note that using a set of arbitrary notations we can write these as ﬁve ﬁrstorder, ordinary diﬀerential equations. These are g i h
= −3f g + 2z 2 − h = −3Prf i = i
z f
= g = z.
The variations of ψ ∗ , dψ ∗ /dη, d2 ψ ∗ /dη 2 , Tf∗ , and dTf∗ /dη, with respect to η, are plotted in Figure Pr.6.17. The results show that for η = 5.66, the xdirection velocity represented by dψ ∗ /dη will have a magnitude 1/100 of its peak (or maximum value). This is designated as the edge of the boundary layer. COMMENT: Note that results are a strong function of Pr. In general, the derivatives are guessed until Tf∗ = dψ ∗ /dη = 0 far from the surface (large η). Also note that from (6.90) we have δα L
GrL 4
1/4 = 3.804
for Pr = 0.72,
while the numerical results for Tf∗ = 0.01 show that this is 4.4176. This is because (6.90) is an approximation to results over a large range of Pr. 533
1.0
Tf*(D = 0) = 1
Pr = 0.72
dT * d O* d 2O* O*, dD , dD2 , Tf , dDf
0.8 Tf*
0.6
O*
0.4 0.2
D = 3.724 [from (6.90)] Tf* = 0.01 at D = 4.4176
d O* dD
0 d 2O* dD2
0.2 0.4
dTf* dD
BoundaryLayer Region
0.6 0
1
2
3
D=
4
y GrL L 4
5
6
7
8
1/4
Figure Pr.6.17 Variation of the dimensionless, velocity and temperature variables with respect to the similarity variable.
534
PROBLEM 6.18.FAM GIVEN: An aluminum ﬂat sheet, released from hot pressing, is to be cooled by surface convection in an otherwise quiescent air, as shown in Figure Pr.6.18. The sheet can be placed vertically [Figure Pr.6.18(a)(i)] or horizontally [Figure Pr.6.18(a)(ii)]. Both sides of the sheet undergo heat transfer and in treating the horizontal arrangement, treat the lower surface using the Nusselt number relations listed in Table 6.5 for the top surface. w = L = 0.4 m, Tf,∞ = 25◦C, Ts = 430◦C. Use air properties at Tf δ = (Ts + Tf,∞ )/2. SKETCH: Figure Pr.6.18(a) shows the two arrangements. (i) Vertical Arrangement
(ii) Horizontal Arrangement
Aluminum Sheet
L
w L
g
Ts > Tf,
Ts > Tf,
g
Quiescent Air Tf, , uf, = 0
w
Figure Pr.6.18(a) A sheet of aluminum is cooled in an otherwise quiescent air. (i) Vertical placement. (ii) Horizontal placement.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the surfaceconvection heat transfer rate Qku L for the two arrangements and for the conditions given above. SOLUTION: (a) Figure Pr.6.18(b) shows the thermal circuit diagram.
Qku
L
Tf,
Ts Rku
L
Figure Pr.6.18(b) Thermal circuit diagram.
(b) (i) For the vertical placement, from Table 6.5, we have NuL
=
NuL,l
=
NuL,t
=
a1
=
RaL
=
[(NuL,l )6 + (NuL,t )6 ]1/6 2.8 2.8 ln 1 + a1 Ra1/4 0.13Pr0.22 Ra1/3 (1 + 0.61Pr0.81 )0.42 0.503 4 9/16 4/9 3 0.492 1+ Pr gβf (Ts − Tf,∞ )L3 . νf αf 535
From Table C.22, for air at Tf δ
(430 + 25)(◦C) Ts + Tf,∞ = + 273.15(K) 2 2 500.7 K,
= =
the properties are kf = 0.0395 W/mK
Table C.22
νf = 3.730 × 10 m /s 5
−5
αf = 5.418 × 10 Pr = 0.69
2
Table C.22
2
m /s
Table C.22 Table C.22,
and treating air as an ideal gas, from (6.77), = T1 = 1 = 1.997 × 10−3 1/K Tf δ f
βf Then, RaL
= =
a1
=
NuL,l
=
9.81(m2 /s) × 1.997 × 10−3 (1/K) × (430 − 25)(K) × (0.4)3 (m3 ) (3.730 × 10−5 )(m2 /s) × 5.418 × 10−5 (m2 /s) 2.512 × 108 0.503 4 = 0.5131 9/16 4/9 3 0.492 1+ 0.69 2.8 ln[1 +
= 65.99
2.8
] 0.5131 × (2.512 × 108 )1/4
NuL,t
=
0.13 × (0.69)0.22 × (2.512 × 108 )1/3 = 64.64 [1 + 0.61 × (0.69)0.81 ]0.42
NuL
=
[(65.99)6 + (64.64)6 ]1/6 = 73.33.
From (6.124), and noting that Aku = 2wL, we have Qku L
= Aku NuL
kf (Ts − Tf,∞ ) L
=
2 × (0.4)2 (m2 ) × 73.33 ×
=
938.5 W.
0.0395(W/mK) × (430 − 25)(K) 0.4(m)
(ii) For the horizontal placement, from Table 6.5, we have NuL
=
NuL,l
=
[(NuL,l )10 + (NuL,t )10 ]1/10 1.4 1.4 ln 1 + 1/4 0.835a1 RaL
NuL,t
=
0.14RaL ,
1/3
where a1 is the same as in (i) and for the deﬁnition given for L in Table 6.5, we have L = =
L Aku Lw = = Pku 2(L + w) 4 0.4(m) = 0.1 m. 4 536
for each side
Then using the results of (i), we have RaL
=
2.512 × 108 ×
NuL,l
=
ln 1 +
1 = 3.925 × 106 (4)3 1.4 1.4
0.835 × 0.5131 × (3.925 × 106 )1/4
=
19.76
NuL,t
=
0.14 × (3.925 × 106 )1/3 = 22.08
NuL
= =
Qku L
=
[(19.76)10 + (22.08)10 ]1/10 22.72 0.0395(W/mK) × 405K 2 × (0.4)2 (m2 ) × 22.72 × 0.10(m) 1,163 W.
=
COMMENT: Note that although for the of horizontal placement NuL is smaller, due to the smaller length used in the scaling of NuL , the heat transfer rate is larger for this arrangement. The correlations valid for both laminar and turbulent ﬂows trend to add uncertainties, compared to the correlations valid only for a given range of RaL [16]. For example using (6.89), for the vertical arrangement and for the RaL < 109 , we have NuL
= = =
4 1/3 a1 RaL 3 4 × 0.5131 × (2.512 × 108 )1/4 3 86.13.
This is to be compared to NuL = 64.63 in which we used in (b)(i).
537
PROBLEM 6.19.FAM GIVEN: Water, initially at T = 12◦C, is boiled in a portable heater at one atm pressure, i.e., it has its temperature raised from 12◦C to 100◦C. The heater has a circular, nickel surface with D = 5 cm and is placed at the bottom of the water, as shown in Figure Pr.6.19. The amount of water is 2 kg (which is equivalent to 8 cups) and the water is to be boiled in 6 min. The properties for water are given in Table C.23. SKETCH: Figure Pr.6.19 shows boiling from the bottom surface of an electrical water heater.
g
Boiler Water m = 2 kg
Heater Surface Ts
DQkuED
Tf, = Tlg Se,J
D = 5 cm
Figure Pr.6.19 An electric water heater using boiling surfaceconvection heat transfer.
OBJECTIVE: (a) Determine the timeaveraged (constant with time) electrical power needed S˙ e,J (W) assuming no heat losses. 2 (b) Determine the critical heat ﬂux qku,CHF (W/m ) for this ﬂuid and then comment on whether the required electrical power per unit area is greater or smaller than this critical heat ﬂux. Note that the surfaceconvection heat transfer rate (or the electrical power) per unit area should be less than the critical heat ﬂux; otherwise, the heater will burn out. (c) Determine the required surface temperature Ts , assuming nucleate boiling. Here, assume that the eﬀect of the liquid subcooling on the surfaceconvection heat transfer rate is negligible. When the subcooling is not negligible (i.e., the water is at a much lower temperature than the saturation temperature Tlg ), the larger temperature gradient between the surface and the liquid and the collapse of the bubbles away from the surface, will increase the rate of heat transfer. (d) Determine the average surfaceconvection thermal resistance Aku Rku L [◦C/(W/m2 )] and the average Nusselt number NuL . SOLUTION: (a) The integralvolume energy equation (2.9) applied to a control volume containing the water, assuming constant properties and a uniform temperature (i.e., a lumpedcapacitance analysis), is dT . dt The heat losses to the ambient are neglected. One component of these heat losses is the energy leaving the water surface in the form of vapor. It is assumed that most of the vapor formed at the heater surface recondenses as the bubbles rise in the liquid. Then, we have QA = −ρcp V
dT . dt Integrating from Ti = 12◦C at t = 0 to Tf = 100◦C at t = 6 min gives −Qku D = −ρcp V
Qku D = ρcp V
Tf − Ti . t
538
For water at Tave = (Tf + Ti )/2 = 329 K from Table C.23, cp = 4,183 J/kgK. Also, ρV = M = 2 kg. Then Qku D = 2(kg) × 4,183(J/kgK)
100(◦C) − 12(◦C) = 2045 W. 360(s)
The integralvolume energy equation applied to the electrical heater gives Qku D = S˙ e,J or
S˙ e,J = 2,045 W.
(b) The critical heat ﬂux for pool boiling from a horizontal surface is given by (6.100) qku,CHF × ρg ∆hlg
ρ2g gσ∆ρlg
1/4 = 0.13.
The properties for saturated water and steam at a pressure of 1 atm are given in Table C.26, ρl = 958.3 kg/m3 , ρg = 0.596 kg/m3 , σ = 0.05891 N/m, ∆hlg = 2.257 × 106 J/kg, µl = 277.53 × 10−6 m2 /s, Prl = 1.73, kl = 0.6790 W/mK, and cp,l = 4,216 J/kgK. Solving for qku,CHF we have 3
0.13ρg ∆hlg 0.13 × 0.596(kg/m ) × 2.257 × 106 (J/kg) 2 6 = qku,CHF = ! "1/4 = 1.098 × 10 W/m . 1/4 2 3 2 2 ρg gσ∆ρlg
(0.596) (kg/m ) 9.81(m2 /s)×0.05891(N/m)×957.7(kg/m3 )
The surfaceconvection heat transfer rate provided by the heater, at the critical heat ﬂux condition, is Qku,CHF = Aku qku,CHF =
π 2 × (0.05)2 (m)2 × 1.098 × 106 (W/m ) = 2,156 W. 4
The critical heat ﬂux is slightly larger than the required heat transfer rate. Therefore, for the desired heating rate, the electrical heater will operate at low surface temperatures, characteristic of nucleate boiling. (c) The total surfaceconvection heat rate is given by (6.98) Qku D = Aku
Ts − Tlg . Aku Rku D
The average surfaceconvection resistance for pool nucleate boiling from a horizontal surface is determined from (6.98), i.e., Aku Rku D = a3s
∆h2lg µl c3p,l (Ts − Tlg )2
σ g∆ρlg
1/2 Prnl =
Aku (Ts − Tlg ) . Qku D
For a nickel surface, from Table 6.2, we have as = 0.006. For water, from (6.98), we have n = 3. Solving the above for Ts − Tlg gives (Ts − Tlg ) = 3
a3s ∆h2lg Qku D Prnl Aku µl c3p,l
×
σ g∆ρlg
1/2 ,
and subbing in numerical values gives
Ts − Tlg
6.84 × 10−4 (K3 m2 /W) × 2,045(W) = π 2 2 4 (0.05) (m)
or, for Tlg = 100◦C, we have Ts = 108.9◦C. 539
1/3 = 8.93K,
Table Pr.6.19 Summary of Nusselt numbers and average surfaceconvection thermal resistances in order of decreasing surfaceconvection thermal resistance. Fluid Flow Arrangement with NuL or NuD Aku Rku L or Aku Rku D , ◦ C/(W/m2 ) or without phase change (6.7): laminar, parallel ﬂow over a ﬂat plate: uf,∞ = 0.5556 m/s (6.14): laminar, thermobuoyant ﬂow around a vertical cylinder (6.14): laminar, thermobuoyant ﬂow around a horizontal cylinder (6.7): laminar, parallel ﬂow over a ﬂat plate: uf,∞ = 5.556 m/s (6.10): perpendicular ﬂow with a single, round impinging jet (6.10): perpendicular ﬂow with an array of 9 round impinging jets (6.10): turbulent, parallel ﬂow over a ﬂat plate: uf,∞ = 22.22 m/s (6.19): nucleate, poolboiling on a horizontal ﬂat surface
119.8
3.326×10−1
43.08
2.232×10−1
18.55
2.073×10−1
378.8
1.052×10−1
46.43
8.413×10−2
28.35
4.953×10−2
2,335
1.711×10−2
8,587
8.575×10−6
(d) Using this Ts , the average surfaceconvection thermal resistance is Aku Rku D = Aku
Ts − Tlg 2 = 8.75 × 10−6 ◦C/(W/m ). Qku D
The average Nusselt number is determined from (6.99), i.e., NuD =
D = 8,587. kl Aku Rku D
COMMENT: Problems 6.7, 6.10, 6.14 and 6.19 present applications of four diﬀerent ﬂuid ﬂow arrangements for surfaceconvection heat transfer. The choice of a certain process for convection heating or cooling of a surface depends initially on the desired rate of cooling or heating. Table Pr.6.19 summarizes the results obtained for the Nusselt number and average surfaceconvection thermal resistance in order of decreasing surfaceconvection thermal resistance. The low speed laminar, parallel ﬂow and the thermobuoyant ﬂows have the largest surfaceconvection thermal resistances. As a consequence, they provide the lowest surfaceconvection cooling or heating power. However, the ﬂuid propelling costs are minimal (or zero) making them attractive in situations where the desired heating/cooling powers can be achieved by an increase in the available surfaceconvection area. Residential applications such as climate control and cooling of the condenser of refrigerators are typical examples. Highspeed, parallel, laminar ﬂows and perpendicular ﬂows provide higher surfaceconvection heat transfer at a cost of ﬂuid propelling power. For low conductance substrates, multiple jets provide a more uniform rate of heat transfer over the surface. Finally, heat transfer with phase change gives the highest heat transfer rates. Modiﬁcations of the surface can increase the rates of heat transfer in the nucleate boiling regime (i.e., heat transfer enhancement). The choice of a surfaceconvection heating/cooling mechanism depends on additional constraints, for example, available space, cost of ﬂuid propelling power, weight, availability of ﬂuids (e.g, situations were the necessary amount of a liquid is not available), whether liquids can be used (e.g., when the system has to be kept dry), continuous versus intermittent operation, reliability of the system (e.g., when the system has to operate by itself for long periods of time, or in a remote location), and environmental concerns related to chemical or thermal pollution (e.g., gaspollutant emissions, hotgas discharges in the atmosphere, and hotwater discharge into lakes and rivers).
540
PROBLEM 6.20.FAM GIVEN: Steam is produced by using the ﬂue gas from a burner to heat a pool of water, as shown in Figure Pr.6.20(a). The water and the ﬂue gas are separated by a plate. On the ﬂuegas side (modeled as air), the measurements show that the ﬂuegas, farﬁeld temperature is Tf,∞ = 977◦C and ﬂows parallel to the surface at uf,∞ = 2 m/s, while the ﬂuegas side surface of the plate is at Ts,2 = 110◦C. The heat ﬂows through the plate (having a length L = 0.5 m and a width w) into the water (water is at the saturated temperature Tlg = 100◦C and is undergoing nucleate boiling). Evaluate the ﬂuegas properties as those of air at the ﬂuegas ﬁlm temperature (i.e., at the average temperature between the ﬂuegas side surface temperature of the plate and the ﬂuegas, farﬁeld temperature). For water, use the saturation liquidvapor properties given in Table C.26. SKETCH: Figure Pr.6.20(a) shows the surface separating the ﬂow from the boiling water. Water Pool, Tlg = 100 oC
g
Nucleate Boiling SurfaceConvection Heat Transfer
Ts,1 Boiler Base Plate Ts,2 = 110 oC
L = 0.5 m Air uf, = 2 m/s Tf, = 977 oC
Parallel Flow SurfaceConvection Heat Transfer
Figure Pr.6.20(a) A solid surface separating ﬂue gas and water with a large diﬀerence between the farﬁeld temperatures causing the water to boil and produce steam.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the surface temperature of the plate on the water side Ts,1 . For the nucleate boiling Nusselt number correlation, use as = 0.013. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.6.20(b). (b) The energy equations for nodes Ts,2 and Ts,1 gives Qku L,2 = Qk,21 = Qku L,1 where Qku L,2 is the surfaceconvection heat transfer rate on the parallelﬂow, gas side, Qk,21 is the conduction heat transfer rate through the plate, and Qku L,1 is the surfaceconvection heat transfer rate on the nucleateboiling water side. Note that no information is provided about the material or thickness of the base plate. However, the above equation states that, under steadystate, the two surfaceconvection heat transfer rates are the same. We use this to determine Ts,1 . (i) For the parallel ﬂow, the gas side, the surfaceconvection heat transfer rate is given by (6.49) as Qku L,2 =
Tf,∞ − Ts,2 . Rku L,2
The average surfaceconvection thermal resistance can be obtained from the conditions given. The properties for air at Tδ = (977+110)/2 = 544◦C = 817 K are obtained from Table C.22: kf = 0.0574 W/mK, νf = 84.16×10−6 541
Qu,lg Tlg
DQkuEL,1
Nucleate Boiling
DRkuEL,1 SurfaceConvection Heat Transfer
Ts,1 Qk,21
Rk,21
Conduction Heat Transfer Through the Base Plate
Ts,2 Parallel Flow
DQkuEL,2
DRkuEL,2 SurfaceConvection Heat Transfer
Tf, Qu, Figure Pr.6.20(b) Thermal circuit diagram.
m2 /s, and Pr= 0.70. The Reynolds number is given by (6.45), i.e., ReL =
uf,∞ L 2(m/s) × 0.5(m) = = 11,882. νf 84.16 × 10−6
For ReL = 11,882 < ReL,t = 5 × 105 the ﬂow is in the laminar regime. For the laminar regime the average Nusselt number is given in Table 6.3 as 1/2
NuL,2 = 0.664ReL Pr1/3 = 0.664(11882)1/2 (0.70)1/3 = 64.27. The average surfaceconvection thermal resistance is given by (6.49), i.e., Aku Rku L,2 =
L 0.5(m) = 0.1355◦C/(W/m2 ). = kf NuL,2 0.057(W/mK) × 64.27
(ii) For the nucleate boiling, the liquid side, the surfaceconvection heat transfer rate is also given by (6.49) as Qku L,1 =
Ts,1 − Tlg . Rku L,1
The properties for water at Tlg = 100◦C are found from Table C.26: kl = 0.679 W/mK, µl = 277.53 × 10−6 Pa.s, ρl = 958.3 kg/m3 , ρg = 0.596 kg/m3 , cp,l = 4,220 J/kgK, Prl = 1.73, σ = 0.05891 N/m, and ∆hlg = 2.257 × 106 J/kg. The average Nusselt number is given Table 6.6 as 1/2 L µl c3p,l (Ts,1 − Tlg )2 g∆ρlg NuL,1 = Pr−n l , kl a3s ∆h2lg σ where n = 3 for water. The average surfaceconvection thermal resistance is given by (6.49) as 1/2 a3s ∆h2lg σ L = Prn . Aku Rku L,1 = kl NuL,1 µl c3p,l (Ts,1 − Tlg )2 g∆ρlg Note that Aku Rku L,1 cannot be determined, because Ts,1 is not known. All the other properties are known and substituting their values we obtain Aku Rku L,1 =
0.006957 K3 /(W/m2 ). (Ts,1 − Tlg )2 542
From the energy equations, we get an algebraic equation in Ts,1 , Tf,∞ − Ts,2 Ts,1 − Tlg = 0.006957 . 0.1355 (Ts,1 −Tlg )2 substituting for Tf,∞ and Ts,2 and rearranging the righthand side 44.51 = (Ts,1 − Tlg )3 . Solving the equation above for Ts,1 , Ts,1 = Tlg + (44.51)1/3 = 100 + 3.54 = 103.5◦C. COMMENT: Note that the diﬀerence in the surface and farﬁeld temperatures for the ﬂue gas side is Tf,∞ − Ts,2 = 867◦C, while for the boiling water side it is Ts,1 − Tlg = 3.5◦C. The temperature diﬀerence Ts,1 − Tlg is called the surface superheat.
543
PROBLEM 6.21.FAM GIVEN: A glass sheet is vertically suspended above a pan of boiling water and the water condensing over the sheet and raises its temperature. This is shown in Figure Pr.6.21(a). Filmwise condensation and uniform sheet temperature Ts are assumed. Note that the condensate is formed on both sides of the sheet. Also assume a steadystate heat transfer. l = 1 mm, L = 15 cm, w = 15 cm, Tlg = 100◦C, Ts = 40◦C. Use the saturated water properties at Tlg . SKETCH: Figure Pr.6.21(a) shows the glass sheet and the surrounding water vapor. l
Suspended Glass Sheet Saturated Water Vapor
w
Filmwise Condensation (Water)
L Uniform Ts Qku
L
. Ml
Water Vapor, Tlg
Boiling Water
qk
Figure Pr.6.21(a) A glass sheet is vertically suspended in a watervapor ambient and the heat released by condensation raises the sheet temperature.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the heat transfer rate Qku L , for the conditions given above, for each side. (c) Determine the condensate ﬂow rate M˙ l = −M˙ lg , for each side. (d) Is this a laminar ﬁlm condensate ﬂow? SOLUTION: (a) Figure Pr.6.21(b) shows the thermal circuit diagram. Qku
L
Ts Tlg Slg =  Mlg ,hlg
Rku
L
Figure Pr.6.21(b) Thermal circuit diagram.
(b) From (6.49) and Table 6.6, for ﬁlmwise condensation on vertical surfaces, we have Qku L NuL
kl = Aku NuL (Ts − Tlg ) L 1/4 g∆ρlg ∆hlg L3 = 0.9428 . kl νl (Tlg − Ts ) 544
From Table C.27, we have for saturated water, ρl ρg
= =
958 kg/m3 0.596 kg/m3
∆hlg kl
= =
2.257 × 106 J/kg 0.680 W/mK
νl
=
µl 279 × 10−6 (Pas) = = 2.91 × 10−7 m2 /s ρl 958(kg/m3 )
Then using the numerical values, we have
NuL
= =
9.81(m/s2 ) × (958 − 0.596)(kg/m3 ) × 2.257 × 106 (J/kg) × (0.15)3 (m3 ) 0.9428 × 0.680(W/mK) × 2.91 × 10−7 (m2 s) × (100 − 40)(K) 1,480.
1/4
Then Qku L
=
0.15(m) × 0.15(m) × 1,480 ×
=
9058 W.
0.680(W/mK) × (40 − 100)(K) 0.15(m)
(c) Using the energy equation for the control volume shown in Figure Pr.6.21(b), we have Qku L = S˙ lg = −M˙ lg ∆hlg or M˙ lg
Qku L ∆hlg −9,058(W) = − 2.257 × 106 (J/kg) = −
=
4.005 × 10−3 kg/s
=
4.013 g/s
(d) Using (6.114), we have 4qku L L µg ∆hlg
= =
4 × 9,058(W) × 0.15(m) (0.15) (m ) × 271 × 10−6 (Pas) × 2.257 × 106 (J/kg) 394.9 < 1,800. 2
2
The ﬁlm condensate ﬂow is in the laminar regime, and the choice of NuL was correct. COMMENT: Note that we have assumed a steady state heat transfer, where in practice Ts increases and eventually reaches Tlg .
545
PROBLEM 6.22.FAM GIVEN: To boil water by electrical resistance heating would require a large electrical power per unit area of the heater surface. For a heater having a surface area for surface convection Aku , this power from (2.28) is ∆ϕ2 S˙ e,J = , Re
Aku = πDl,
where ∆ϕ is the applied voltage, Re is the electrical resistance, and D and l are the diameter and length of the heater surface. Consider the waterboiler shown in Figure Pr.6.22(a). Using Figure 6.20(b), assume a surface superheat Ts − Tlg = 10◦C is needed for a signiﬁcant nucleate boiling. Then use the nucleate boiling correlation of Table 6.6. as = 0.01, D = 0.5 cm, l = 12 cm, Tlg = 100◦C. Use saturated water properties at T = Tlg . SKETCH: Figure Pr.6.22(a) shows the Joule heater in the water boiler.
Water, Tl, = Tlg Heater Length, l
Heater Se,J =
,j2
Aku
D
Re
+
Electrical Dj  Power Line
Figure Pr.6.22(a) A Joule heater is used to boil water.
OBJECTIVE: (a) Draw the thermal circuit diagram for the heater. (b) Determine the surfaceconvection heat transfer rate Qku D , for the conditions given above. (c) For an electrical resistance of Re = 20 ohm, what should be the applied voltage ∆ϕ, and the electrical current Je ? SOLUTION: (a) Figure Pr.6.22(b) shows the thermal circuit diagram for the heater. The energy equation is QA = Qku L = S˙ e,J . Qku
L
Ts Tlg
,j2 Se,J = R
Rku
L
e
Figure Pr.6.22(b) Thermal circuit diagram.
(b) From Table 6.6, we have for the nucleate boiling regime and using (6.49), Qku L
kl (Ts − Tlg ) L 1/2 3 3 g∆ρlg 1 µl cp,l (Ts − Tlg ) = Aku 3 Pr−3 l , ∆hlg σ as
= Aku NuL
where we have used n = 3 for water and L drops out of the relation. 546
Table C.26 list the water properties at Tlg = 373.15 K, i.e., µl cp,l
g∆ρlg σ
= =
277.5 × 10−6 Pas 4, 220 J/kgK
∆hlg σ
= 2,257 × 106 J/kg = 0.05891 N/m
∆ρlg PrL 1/2
= ρl − ρg = (958.3 − 0.596)(kg/m3 ) = 957.7 kg/m3 = 1.73 1/2 9.81(m/s) × (957.7)(kg/m3 ) = = 399.3 1/m 0.05891(N/m)
Qku L
= =
π × 0.005(m) × 0.12(m) × 2.775 × 10−4 (Pas) × (4,220)3 (J/kg)3 × 103 (K)3 × 399.3(1/m) (0.01)3 × (2.257 × 106 )2 (J/kg)2 × (1.73)3 595.1 W.
(c) The required electrical potential is ∆ϕ
=
(S˙ e,J Re )1/2
=
(Qku L Re )1/2
= =
[595.1(W) × 20(ohm)]1/2 109.1 V.
From (2.32), we have Je
=
∆ϕ 109.1 V = 5.455 A. = Re 20(ohm)
COMMENT: Boiling water requires a large electric power per unit area. For this reason, stored electric power, such as in batteries, can not be used. Also, note that from (2.32), we have Re =
ρe l , Ae
where Ae is the electrical conductor crosssection area. To produce a large electrical resistance, a small Ae is used (see Example 2.10).
547
PROBLEM 6.23.FAM GIVEN: To reduce the air conditioning load, the roof of a commercial building is cooled by a water spray. The roof is divided into segments with each having a dedicated sprinkler, as shown in Figure Pr.6.23(a). Assume that the impingingdroplet ﬁlm evaporation relation of Table 6.6 can be used here. ˙ d /ρl,∞ = 10−3 m/s. L = 4 m, Tf,∞ = 30◦C, Ts = 210◦C, D = 100 µm, ud = 2.5 m/s, m Evaluate the water properties at T = 373 K. SKETCH: Figure Pr.6.23(a) shows the water sprinkler and the roof panel. Roof Panel to be Cooled, Ts L L
Water Droplet Spray, Tl,
Remotely Controlled Valve
Water Sprinkler and Supply Pipe
Figure Pr.6.23(a) Waterspray cooling of a roof panel.
OBJECTIVE: (a) Draw the thermal circuit diagram for the panel surface. (b) Using the conditions given above, determine the rate of surfaceconvection heat transfer Qku L from the roof panel. SOLUTION: (a) Figure Pr.6.23(b) shows the thermal circuit diagram Ts Qku
Rku
L
L
Tl,
Figure Pr.6.23(b) Thermal circuit diagram.
(b) The rate of surfaceconvection heat transfer is given by (6.49) as Qku L =
kl Ts − Tl,∞ = Aku (Ts − Tl,∞ )NuL , Rku L L
Aku = L2 .
From Table 6.5, we have Qku L
=
Aku ρl,∞ ∆hlg,∞
m ˙ d ρl,∞
m ˙ d /ρl,∞ ηd 1 − + (m ˙ d /ρl )◦
Aku × 1,720(Ts − Tl,∞ )0.912 D−1.004 ud −0.764
m ˙d ρl,∞
=
5 × 10−3 m/s
o
ηd
=
3.68 × 104 (Ts − Tl,∞ )1.691 D−0.062 . ρl,∞ ∆hlg,∞ 548
(m ˙ d /ρl,∞ )2 , (m ˙ d /ρl )◦
From Tables C.4, and C.23, we have for water at T = 373 K ∆hlg = 2.256 × 106 J/kg (cp,l )∞ = 4,218 J/kgK
Table C.4 Table C.23
3
ρl,∞ = 960.2 kg/m
Table C.23,
Then ∆hlg,∞
= = =
(cp,l )∞ (Tlg − Tl,∞ ) + ∆hlg 4,218(J/kgK) × (100 − 30)(K) + 2.256 × 106 (J/kg) 2.551 × 106 J/kg
Next, ηd
= =
3.68 × 104 3
960.2(kg/m ) × 2.251 × 106 (J/kg) 0.1732.
× (210 − 30)1.691 (K)1.691 × (1 × 10−4 )−0.062 (m)−0.062
For Qku L , we have Qku L
=
3
(4)2 (m)2 × 960.5(kg/m ) × 2.541 × 106 (J/kg) × 10−3 (m/s) −3 10 (m/s) × 0.1732 × 1 − + (4)2 (m)2 × 1,720 × (210 − 30)0.912 (K)0.912 × 5 × 10−3 (m/s) (10−4 )−1.004 (m)−1.004 (2.5)−0.746 (m/s)−0.746
=
5.432 × 106 (W) + 3.286 × 106 (W)
=
8.718 × 106 W.
(10−3 )2 (m/s)2 5 × 10−3 (m/s)
COMMENT: This cooling rate is held only for a short time to reduce the roof panel temperature to a low, safe and desirable value. Note that water spray cooling is economical and when no highhumidity damage is expected, this cooling method can be eﬀective.
549
PROBLEM 6.24.FAM GIVEN: In using water evaporation in surfaceconvection heat transfer, compare pool boiling by saturated water (Tl,∞ = Tlg ), as shown in Figure 6.20(b), and droplet impingement by subcooled water droplets (Tl,∞ < Tlg ) as shown in Figure 6.26. For pool boiling, the peak in Qku L is given by the critical heat ﬂux, i.e., (6.100), and the minimum is given by (6.101). For impinging droplets, the peak shown in Figure 6.26 is nearly independent of the droplet mass ﬂux, in the high mass ﬂux regime, and the minimum is approximately correlated by (6.116). The correlations are also listed in Table 6.6. Pool boiling: Tlg = 100◦C, Ts = 300◦C. Impinging droplets: m ˙ d = 1.43 kg/m2 s, ud = 3.21 m/s, D = 480 µm, Tl,∞ = 20◦C, Tlg = 100◦C, ◦ Ts = 300 C, evaluate properties at 310 K. Note that not all these conditions are used in every case considered. SKETCH: Figure Pr.6.24 shows the two surface cooling methods using water evaporation.
(i) Pool Boiling: VaporFilm Regime
(ii) Impinging Droplets: VaporFilm Regime
Liquid Tl,
Tlg Vapor Ts
D
Vapor Film
Substrate qku
Droplet
Ts
L
Vapor Vapor Film
Substrate qku
L
Figure Pr.6.24 Selection of a water evaporation surface cooling method, between (i) pool boiling and (ii) impingement droplets.
OBJECTIVE: Select between water pool boiling [Figure Pr.6.24(i)] and droplet impingement [Figure Pr.6.24(ii)], by comparing (i) peak, and (ii) minimum, surfaceconvection heat ﬂux qku L . SOLUTION: (a) Peak (or critical) heat ﬂux: (i) For pool boiling, from Table 6.6, we have qku L = qku,CHF = 0.13ρg ∆hlg
σ∆ρlg ρ2g
1/4 .
From Table C.26, for water at T = 373.15 K, we have ρl = 958.3 kg/m3 ρg = 0.596 kg/m3
Table C.26 Table C.26
∆hlg = 2.257 × 10 J/kg Table C.26 σ = 0.05891 N/m Table C.26. 6
Then, qku L
= qku,CHF = 0.13 × 0.596(kg/m3 ) × 2.257 × 106 (J/kg) × 1/4 0.05891(N/m) × 9.81(m/s2 ) × (958.3 − 0.596)(kg/m3 ) (0.596)2 (kg/m3 )2 =
1.099 × 106 W/m2 .
This is close to the value shown in Figure 6.20(b). 550
(ii) For impinging droplets, reading from Figure 6.26 for m ˙ d = 1.43 kg/m2 s, we have qku L = qku,peak = 2 × 106 W/m2
Figure 6.26.
Thus the impinging droplets gives a higher qku L . (b) Minimum heat ﬂux: (i) For pool boiling, from Table 6.6, we have qku L
= qku,min = 0.09ρg ∆hlg
σg∆ρlg ρl + ρg
1/4 .
=
0.09 × 0.596(kg/m3 ) × 2.257 × 106 (J/kg) × 1/4 0.0591(N/m) × 9.81(m/s2 ) × (958.3 − 0.596)(kg/m3 ) (958.3 + 0.596)2 (kg/m3 )2
= =
(1.211 × 105 × 0.1567)(W/m2 ) 1.896 × 104 W/m2 .
(ii) For impinging droplets, we have from Table 6.6, qku L
=
m ˙d m ˙ d /ρl,∞ ρl,∞ ∆hlg,∞ ηd 1 − + ρl,∞ (m ˙ d /ρl )◦
1,720(Ts − Tl,∞ )0.912 D−1.004 ud −0.764
(m ˙ d /ρl,∞ )2 . (m ˙ d /ρl )◦
From Table C.23, at T = 310 K, we have ρl,∞ cp,l
= =
995.3 kg/m3 Table C.23 4,178J/kgK Table C.23.
Then from Table 6.6, we have ∆hlg,∞
ηd
= =
(cp,l )∞ (Ts − Tl,∞ ) + ∆hlg 4,178(J/kgK) × (100 − 20)(K) + 2.257 × 106 (J/kg)
=
2.591 × 106 J/kg.
= = =
3.68 × 104 (Ts − Tl,∞ )1.691 D−0.062 ρl,∞ ∆hlg,∞ 3.68 × 104 3
995.3(kg/m ) × 2.591 × 106 (J/kg) 0.3150.
(300 − 20)1.691 (K)1.691 (4.80 × 10−4 )−0.062 (m)−0.062
Then qku L
=
3
995.3(kg/m ) × 2.591 × 106 (J/kg) × 1.43(kg/m2 s) 1.43(kg/m2 s) + 1,720(300 − 20)0.912 (K)0.912 × × 0.3150 × 1 − 995.3(kg/m3 ) 995.3(kg/m3 ) × 5 × 10−3 (m/s) (4.8 × 10−4 )−1.004 (m)−1.004 (3.21)−0.746 (m/s)−0.746
= =
(1.43)2 (kg/m2 s)2 (995.3)2 (kg/m3 ) × 5 × 10−3 (m/s)
(8.317 × 105 + 1.090 × 105 )(W/m2 ) 9.407 × 105 W/m2 .
This is much larger than that for the vaporﬁlm pool boiling. COMMENT: Here droplet impingement gives higher heat ﬂuxes. Note that the correlation used for vaporﬁlm regime of impingement droplets uses a temperature dependence that is much stronger than that found in the experimental results given in Figure 6.26. 551
PROBLEM 6.25.FUN GIVEN: A person caught in a cold cross wind chooses to curl up (crouching as compared to standing up) to reduce the surfaceconvection heat transfer from his clothed body. Figure Pr.6.25(a) shows two idealized geometries for the person while crouching [Figure Pr.6.25(a)(i)] and while standing up [Figure Pr.6.25(a)(ii)]. Ds = 50 cm, Dc = 35 cm, Lc = 170 cm, T1 = 12◦C, Tf,∞ = −4◦C, uf,∞ = 5 m/s, ki = 0.1 W/mK. Use air properties (Table C.22) at T = 300 K. SKETCH: Figure Pr.6.25(a) shows the two positions. (i) Crouching
(ii) Standing Ds T1
T1
Lc Tf, uf,
Tf, uf, Dc
Figure Pr.6.25(a) Two positions by a person in a cold cross ﬂow of air.
(i) Crouching position. (ii) Standing position. OBJECTIVE: (a) Draw the thermal circuit diagram and determine the heat transfer rate for the idealized spherical geometry. (b) Draw the thermal circuit diagram and determine the heat transfer rate for the idealized cylindrical geometry. Neglect the heat transfer from the ends of the cylinder. (c) Additional insulation (with thermal conductivity ki ) is to be worn by the standing position to reduce the surfaceconvection heat transfer to that equal to the crouching position. Draw the thermal circuit diagram and determine the necessary insulation thickness L to make the two surface convection heat transfer rates equal. Assume that T1 and the surfaceconvection resistance for the cylinder will remain the same as in part (b). (d) What is the outsidesurface temperature T2 of the added insulation? SOLUTION: From Table C.22, the properties of air at T = 300 K are ρf = 1.177 kg/m3 , νf = 15.66 × 10−6 m2 /s, Pr= 0.69, and kf = 0.0267 W/mK. (a) Sphere: The thermal circuit diagram is shown in Figure Pr.6.25(b). T1
Rku
D,s
Qku
D,s
Tf,
Figure Pr.6.25(b) Thermal circuit diagram.
The average surfaceconvection heat transfer rate from the sphere is then Qku D,s =
T1 − Tf,∞ , Rku D,s
where Rku D,s =
Ds . Aku,s NuD,s kf 552
The surface area for convection is the area of the sphere, or Aku,s
= πDs2 = π × 0.52 (m2 ) = 0.7854 m2 .
The Nusselt number for the sphere, from Table 6.4, is 1/2
2/3
NuD,s = 2 + [0.4ReD,s + 0.06ReD,s ]Pr0.4 . The Reynolds number is found as ReD,s
=
uf,∞ Ds 5(m/s) × 0.5(m) = = 1.596 × 105 . νf 15.66 × 10−6 (m2 /s)
There was no Reynolds number restrictions given for the applicability of the Nusselt number correlation, so we will assume the correlation is valid. Then the Nusselt number becomes NuD,s
=
2 + (0.4(1.596 × 105 )1/2 + 0.06(1.596 × 105 )2/3 )0.690.4 = 291.9,
and the surfaceconvection heat transfer resistance becomes Rku D,s
= =
0.5(m) 0.7854(m ) × 291.9 × 0.0267(w/mK) 0.08167◦C/W. 2
The surface convection heat transfer from the sphere is then Qku D,s
12◦C − (−4◦C) = 195.9 W. 0.08167◦C/W
=
(b) Cylinder: The thermal circuit diagram is shown in Figure Pr.6.25(c). The average surfaceconvection heat T1
Rku
D,c
Qku
D,c
Tf,
Figure Pr.6.25(c) Thermal circuit diagram.
transfer rate from the cylinder (neglecting the heat transfer from the ends) is then Qku D,c =
T1 − Tf,∞ , Rku D,c
where Rku D,c =
Dc . Aku,c NuD,c kf
The surface area for convection is the area of the cylinder not including the end areas, or Aku,c
= πDc Lc = π × 0.35(m) × 1.70(m) = 1.869 m2 .
The Nusselt number for a cylinder in cross ﬂow, from Table 6.3, is Reynolds number dependent and has the form 2 NuD,c = a1 ReaD,c Pr1/3 .
The Reynolds number is found as ReD,c
=
uf,∞ Dc 5(m/s) × 0.35(m) = = 1.1175 × 105 . νf 15.66 × 10−6 (m2 /s) 553
For this Reynolds number, the constants are given in Table 6.3 as a1 = 0.027 and a2 = 0.805. Then the Nusselt number becomes NuD,c
0.027 × (1.1175 × 105 )0.805 × (0.69)1/3 = 276.4,
=
and the surfaceconvection heat transfer resistance becomes Rku D,c
=
0.35(m) = 0.02538◦C/W. 1.869(m2 ) × 276.4 × 0.0267(w/mK)
The surface convection heat transfer from the sphere is then Qku D,c
=
12◦C − (−4◦C) = 630.5 W. 0.0254◦C/W
Which is signiﬁcantly higher than that for the sphere. (c) Added insulation: The thermal circuit diagram is shown in Figure Pr.6.25(d). Insulation of thermal conT1
T2
Rk,i
Qku
D,c =
Rku
Qku
D,c
Tf,
D,s
Figure Pr.6.25(d) Thermal circuit diagram.
ductivity ki = 0.1 W/mK and thickness L (to be determined) is added to the cylinder such that the average surfaceconvection heat transfer rate from the cylinder is equal to that from the sphere. The surfaceconvection resistance from the cylinder Rku D,c and the temperature T1 are assumed the same as in Part (b). Therefore, Qku D,c =
T1 − Tf,∞ T1 − Tf,∞ = Qku D,s = . Rk,i + Rku D,c Rku D,s
Given that Qku D,c = Qku D,s , we then have Rk,i + Rku D,c = Rku D,s or Rk,i
= Rku D,s − Rku D,c = 0.08167◦C/W − 0.02538◦C/W = 0.05629◦C/W.
For the cylindrical system, the conduction resistance in the radial direction across the insulation thickness L = R2 − R1 is Rk,i =
ln(R2 /R1 ) 2πLc ki
or R2 /R1
=
exp(Rk,i 2πLc ki )
=
exp[0.05629(◦C/W) × 2 × π × 1.7(m) × 0.1(W/mK)] = 1.062.
The inner radius is the uninsulated radius of the cylinder and is R1
= Dc /2 = 0.35(m)/2 = 0.175 m,
R2
= R1 × 1.062
and then
=
0.175(m) × 1.062 = 0.1858 m. 554
The thickness of insulation is then L = R2 − R1 = 0.186(m) − 0.175(m) =
0.0108 m = 1.08 cm.
(d) The outside surface temperature of the insulation can be found by applying the energy equation between node T2 and T1 in Figure Pr.6.33(d) as Qku D,s =
T1 − T2 . Rk,i
Then T2
= T1 − Qku D,s × Rk,i = 12(◦C) − 195.9(W) × 0.05629(◦C/W) = 0.97◦C.
COMMENT: Note that the two Nusselt numbers are nearly the same, but the heat transfer rates are diﬀerent to the surface area.
555
PROBLEM 6.26.FUN GIVEN: A thermocouple is placed in an air stream to measure the stream temperature, as shown in Figure Pr.6.26(a). The steadystate temperature of the thermocouple bead of diameter D is determined through its surfaceconvection (as a sphere in a semibounded ﬂuid stream) and surfaceradiation heat transfer rates. uf,∞ = 2 m/s, Tf,∞ = 600◦C, Tw = 400◦C, r,w = 0.9, r,s = 0.8, D = 1 mm. Neglect the heat transfer to and from the wires and treat the surfaceconvection heat transfer to the thermocouple bead as a semibounded ﬂuid ﬂow over a sphere. Assume the tube length L is large (i.e, L → ∞). Evaluate the ﬂuid properties at T = 350 K (Table C.22). Numerical hint: The thermocouple bead temperature should be much closer to the air stream temperature than to the tube surface temperature. For iterations, start with a a guess of T = 820 K. OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the thermocouple bead temperature Ts . (c) Comment on the diﬀerence between Ts and Tf,∞ . How can the diﬀerence (measurement error) be reduced? SKETCH: Figure Pr.6.26(a) shows the thermocouple placed in a tube to measure the ﬂuid stream temperature at a location. Thermocouple bead, T2 Air Flow uf, Tf, Thermocouple Bead
T1 D
Thermocouple Wires
Figure Pr.6.26(a) A thermocouple used for measuring a ﬂuid stream temperature.
SOLUTION: From Table C.22, the properties of air at T = 350 K are νf = 20.30 × 10−6 m2 /s, Pr= 0.69, and kf = 0.030 W/mK. (a) The thermal circuit diagram is shown in Figure Pr.6.26(b). Tf,
Rku
Qku
T2
(Rr,Σ)21 Qr
T1
Figure Pr.6.26(b) Thermal circuit diagram.
(b) The conservation of energy applied to the spherical thermocouple bead gives Qr + Qku 4 − T1 ) T2 − Tf,∞ + (Rr,Σ )21 Rku
σSB (T24
556
=
=
0.
The radiation resistances are found as (Rr,Σ )21 = (Rr, )2 + (Rr,F )21 + (Rr, )1 . Since L → ∞, then we can assume A1 → ∞. Also, since L is very large compared to the tube diameter, we can neglect the tube ends on the radiation to the thermocouple bead. The view factor from the bead to the tube can then be assumed F12 ≈ 1. The area of the bead is A2 = πD2 = π × (0.001)2 = 3.142 × 10−6 m2 . Thus we have (Rr,Σ )21
= = =
1 − r,2 1 1 − r,1 + + A2 r,2 A2 F21 A1 r,1 1 − 0.8 1 + +0 3.142 × 10−6 (m2 ) × 0.8 3.142 × 10−6 (m2 ) × (1) 7.957 × 104 m−2 + 3.183 × 105 m−2 = 3.978 × 105 m−2 .
For the surfaceconvection resistance, we have Rku =
1 , A2 NuD kf /D
where the NuD is found from Table 6.4 as 1/2
2/3
NuD = 2 + (0.4ReD + 0.06ReD )Pr0.4 . with the ReD determined as ReD =
uf,∞ D 2(m/s) × 0.001(m) = = 98.52. νf 20.30 × 10−6 (m2 /s)
The Nusselt number is then NuD
=
2 + [0.4(98.52)1/2 + 0.06(98.52)2/3 ]0.690.4 = 6.526,
and the surfaceconvection resistance is Rku
=
1 = 1626◦C/W. 3.142 × 10−6 (m2 ) × 6.526 × 0.030(W/mK)/0.001(m)
The energy balance cannot be solved explicitly for T2 , therefore we must use a solver or iterate. We can rearrange to facilitate iteration as new
T2
Rku σSB (old T24 − T14 ) (Rr,Σ )21 1626(◦C/W) × 5.67 × 10−8 (W/m2 K4 ) × (old T24 − 2.053 × 1011 )(K4 ) 873.15(K) − 3.978 × 105 (m−2 )
= Tf,∞ − = =
873.15(K) − 2.3176 × 10−10 (K−3 ) × (old T24 − 2.053 × 1011 )(K4 ).
To iterate, we guess an initial T2 = old T2 and calculate new T2 . Then the new T2 of the previous iteration becomes the old T2 of the next, and so on, until new T2 − old T2  < criterion (i.e., the same). Faster convergence can be achieved averaging the old and new values of the previous iteration to become the old value of the next iteration. Table Pr.6.26 shows the values at the iteration steps, using averaged guesses and a criterion of 0.1 K. The initial T2 is taken to be closer to the ﬂuid stream temperature Tf,∞ at old T2 = 820 K. Table Pr.6.26 Results of Numerical Iteration. iteration old T2 , K new T2 , K average diﬀerence 1 2 3 4
820 817.97 817.43 817.34
815.95 816.98 817.25 817.30
557
817.97 817.43 817.34 817.32
4.05 0.99 0.18 0.04
Therefore, the bead temperature is T2 817.3 K. (c) The temperature diﬀerence T2 −Tf,∞ = 55.8 K is rather large and is due to the heat loss from the thermocouple bead. The heat loss is by radiation and conduction along the wires. At high temperatures, radiation heat transfer can become signiﬁcant compared to the surfaceconvection heat transfer. To reduce the measurement error, the radiation losses from the bead must be reduced. This can be accomplished by placing a thin, cylindrical radiation shield around the bead that would be heated by the ﬂuid stream to a temperature nearer to the thermocouple bead temperature and that would not greatly disturb the ﬂuid ﬂow around the bead. The shield increases the net sum radiation resistance between the bead and the tube wall. The measurement error can also be reduced by decreasing the bead diameter to reduce the area for radiation heat transfer. COMMENT: Note that errors due to the bead and wire surfaceradiation heat transfer and conduction along the wire (where placed along a nonuniform temperature ﬁeld) should be reduced for accurate measurements.
558
PROBLEM 6.27.FAM GIVEN: In order to prevent the ﬂame from blowoﬀ by a cross wind, a lighter is desired with a ﬂame anchor (i.e., ﬂame holder) in the from of a winding wire placed in the airfuel stream undergoing combustion. This is shown in Figure Pr.6.27(a)(i). The wire retains (through its heat storage) a high temperature and will maintain the ﬂame around it, despite a large, intermittent cross ﬂow. Assume that the combustion of the nbutane in air is complete before the gas stream at temperature Tf,∞ and velocity uf,∞ reach the ﬂame holder. Treat the ﬂame holder as a long cylinder with steadystate, surfaceconvection heating and surfaceradiation cooling. The simpliﬁed heat transfer model is also shown, in Figure Pr.6.27(a)(ii). Tf,∞ = 1,300◦C, D = 0.3 mm, Tsurr = 30◦C, r,s = 0.8. Use the adiabatic, laminar ﬂame speed, for the stoichiometric nbutane in air, from Table C.21(a), for the farﬁeld ﬂuid velocity uf,∞ . You do not need to use tables or graphs for the view factors. Assume the properties of the combustion products are those of air at T = 900 K. SKETCH: Figure Pr.6.27(a) shows the ﬂame holder and a simpliﬁed heat transfer model for the wire.
(i) Physical Model
(ii) Heat Transfer Model \ \ \\
\
Wire Flame Holder
\ \
Tsurr
Qr,ssurr
Wire Ts , r,s
D
Qku
D,s
Tf, uf,
Lighter
Figure Pr.6.27(a)(i) A windingwire, ﬂame holder used in a lighter. (ii) A simpliﬁed heat transfer model for the wire.
OBJECTIVE: (a) Draw the thermal circuit for the ﬂame holder. (b) Determine the ﬂame holder temperature Ts for the given conditions. SOLUTION: (a) Figure Pr.6.27(b) shows the thermal circuit diagrams. The heat is added by surface convection to the wire and is removed by radiation to the surroundings. The surface area for surface radiation and convection are the same i.e., Ar,s = Aku,s = As . (b) From Figure Pr.6.27(b), the energy equation is QA,s = Qku D,s∞ + Qr,ssurr = 0. The surface convection is given by (6.124) as Qku D,s∞
=
Ts − Tf,∞ (Rku )D
= Aku NuD
kf (Ts − Tf,∞ ), D
where NuD is found from Table 6.3 for cross, forced ﬂow over a cylinder, i.e., NuD = a1 ReaD2 Pr1/3 , 559
ReD =
uf,∞ D , νf
uf, Qku
D,s
Tf, Rku
D
Ar,s = Aku,s = A Ts Eb,s Qr,ssurr
Rr,5 Eb,surr Tsurr
Figure Pr.6.27(b) Thermal circuit diagram.
and a1 and a2 depend on the magnitude of ReD . From Table C.21(a), for nbutaneair, we have uf,∞ = uf,1 = 0.379 m/s
Table C.21(a).
From Table C.22, for air at T = 900 K, we have kf = 0.0625 W/mK
Table C.22
νf = 9.860 × 10−5 m2 /s Pr = 0.7
Table C.22 Table C.22.
Then ReD
=
a1
=
0.379(m/s) × 3 × 10−4 (m) = 1.153 9.860 × 10−5 (m2 /s) 0.683, a2 = 0.466 Table 6.3
NuD
=
0.683 × (1.153)0.466 × (0.7)1/3 = 0.6481.
The surface radiation for Asurr As and Fssurr = 1, is given by (4.49), i.e., 4 Qr,ssurr = Ar,s r,s σSB (Ts4 − Tsurr ).
Then the energy equation becomes As NuD
kf 4 (Ts − Tf,∞ ) + As r,s σSB (Ts4 − Tsurr )=0 D
or NuD
kf 4 (Ts − Tf,∞ ) + r,s σSB (Ts4 − Tsurr ) = 0. D
Using the numerical values, we have 0.6481 ×
0.0625(W/mK) × (Ts − 1,573)(m) + 0.8 × 5.67 × 10−8 (W/m2 K) × [Ts4 − (303.15)4 (K4 )] = 0 3 × 10−4 (m)
which gives an algebraic equation in Ts , 1.350 × 102 × [Ts − 1,573(K)] + 4.536 × 10−8 × (Ts4 − 8.446 × 109 ) = 0. Solving for Ts , we have Ts = 1,094 K. COMMENT: The wire can be made of tungsten to be resistive to oxidation at such high temperatures. Note that NuD is less than unity. This is because as ReD → 0, NuD for long cylinders tends to zero (unlike that for spheres for which NuD → 2 in the conduction limit). This is due to the available conduction area as R2 → ∞ in (3.61), as compared to (3.64). 560
PROBLEM 6.28.FUN GIVEN: Consider measuring the temperature Tf,∞ of an air stream using a thermocouple. A thermocouple is a junction made of two dissimilar materials (generally metals, as discussed in Section 2.3.2), as shown in Figure Pr.6.28. The wires are electrically (and if needed, thermally) insulated. The wire may not be in thermal equilibrium with the stream. This can be due to the nonuniformity of temperature along the wire. Then the temperature of the thermocouple bead (i.e., its tip) Ts,L may not be close enough to Tf,∞ , for the required accuracy. Consider an air stream with a farﬁeld temperature Tf,∞ = 27◦C and a farﬁeld velocity uf,∞ = 5 m/s. Assume that the bare (not insulated) end of the wire is at temperature Ts,o = 15◦C. Consider one of the thermocouple wires made of copper, having a diameter D = 0.2 mm and a barewire length L = 5 mm. Evaluate the properties of air at 300 K. SKETCH: Figure Pr.6.28 shows the thermocouple junction and the idealized model for heat transfer from one of its wires.
(ii) A Thermocouple for Measurement of Tf,
(i) Idealized Thermocouple Wire Model
Probe Ts,0
Ts,L
Tf, , uf ,
Ts,0
L = 5 mm Tf, = 27 C uf , = 5 m/s
Thermocouple Bead Ts,L Dissimilar Thermocouple Wires
D = 0.2 mm
Insulation Wrapping
Individual Wire Insulation Wrapping
Figure Pr.6.28 A thermocouple junction used for temperature measurement in an air stream. (i) Idealized thermocouple wire model. (ii) Thermocouple for measurement of Tf,∞ .
OBJECTIVE: Using the extended surface analysis, determine the expected uncertainty Tf,∞ − Ts,L . SOLUTION: The properties of air are determined from Table C.22 for T = 300 K and are kf = 0.0267W/mK, νf = 15.66 × 10−6 m2 /s, Pr = 0.69. For pure copper from Table C.14 and T = 300 K, we have ks = 401 W/mK. We will use the extended surface (ﬁn) results for the temperature distribution along the wire, i.e., from (6.143) we have, Ts (x) − Tf,∞ cosh[m(Lc − x)] . = Ts,o − Tf,∞ cosh(mLc ) We are interested in the end location along the thermocouple wire which is at the end or at x = Lc (i.e., corrected length). Then we have Ts (L) − Tf,∞ Ts,0 − Tf,∞ Ts (Lc ) − Tf,∞ Ts,0 − Tf,∞ Ts (Lc ) − Tf,∞
= = =
cosh[m(Lc − Lc )] cosh(mLc ) cosh(0) cosh(mLc ) Ts,0 − Tf,∞ . cosh(mLc )
From (6.141), we have Lc
= L+
Lc
=
D 4
0.005(m) +
2 × 10−4 (m) = 0.00505 m. 4 561
For m, from (6.144) we have m=
Pku NuD kf Ak ks D
1/2 ,
Pku = πD,
Ak =
πD2 . 4
For NuD , from Table 6.3 we have NuD = a1 ReaD2 Pr1/3 , where ReD =
uf,∞ D 5(m/s) × (2 × 10−4 )(m) = = 63.86. νf 15.66 × 10−6 (m2 /s)
Then from Table 6.4, we have a1 NuD
m
= =
0.683, a2 = 0.466 0.683(63.8)0.466 (0.69)1/3
=
4.187
1/2
π × 2 × 10−4 (m) × 4.187 × 0.0267(W/mK) = π(2 × 10−4 )2 (m2 ) −4 × 401(W/mK) × 2 × 10 4 7.3 × 10−5 1/2 = ( ) 1/m = 167.0 1/m. 2.519 × 10−4
The expected uncertainty is Tf,∞ − Ts,L
=
27(◦C) − 15(◦C) = 8.71◦C. cosh[167.0(1/m) × 0.00505(m)]
COMMENT: The uncertainty can be large for large Tf,∞ − Ts,o and for small L. Here, the error is 8.7◦C. In order to reduce this error, we can for example, triple the length and the error would be reduced to 1.9 ◦C. With care, Tf,∞ − Ts,o can also be reduced by placing the thermocouple wire along nearly isothermal paths in the ﬂow ﬁeld. Also note 1/2 that mLc = BiD = 0.8434, or BiD = 0.7112. This also shows that there is a signiﬁcant temperature variation within the wire.
562
PROBLEM 6.29.FUN GIVEN: Shapememory actuation devices are capable of recovering a particular shape upon heating above their transformation temperature. These alloys can be made of nickel and titanium and display two types of material properties. When at a temperature below their transformation temperature Tt , they display the properties of martensite and, when above this temperature they display the properties of austenite. The NiTi alloy shown in Figure Pr.6.29(a) is shaped as a spring and deforms from a compressed state [Figure Pr.6.29(a)(i)] to an extended state [Figure Pr.6.29(a)(ii)] when heated above its transformation temperature. This spring is being tested for its suitability for use in the closing of heating ducts within a desired elapsed time. In order to close the duct, the spring must extend a lightweightbeam induced closing mechanism within 20 s. The air ﬂow within the heating duct has a velocity uf,∞ and temperature Tf,∞ , near the spring. The spring has a length L, an outer diameter D2 , an inner diameter D1 , and an initial temperature T1 (t = 0). Assume that the entire spring is at a uniform temperature T1 (t) and the dominant surface heat transfer is by surface convection. Martensite: ρ = 6,450 kg/m3 , k = 8.6 W/m◦C, cp = 837.36 J/kg◦C, T1 (t = 0) = 21◦C, Tt = 50◦C, L = 4 cm, D2 = 0.5 cm, D1 = 0.4 cm, uf,∞ = 5 m/s, Tf,∞ = 77◦C. Evaluate the properties of air at 350 K. SKETCH: Figure Pr.6.29(a) shows the thermally actuated shapememory spring.
(i) Martensite T1 < Tt
uf, Tf,
ρcp , k
D1 D2
L T1(t), T1(t = 0) uf, Tf,
(ii) Austenite T1 > Tt
ρcp , k
D1 D2
L T1(t), T1(t = 0)
Figure Pr.6.29(a) Springs made of shapememory alloy and used for thermal actuation.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Use the properties of the lowtemperature form of NiTi listed above and determine if the spring will activate during the time allowed. SOLUTION: Properties (air, T = 350 K, Table C.22): kf = 0.0300 W/mK, ρf = 1.012 kg/m3 , cp,f = 1,007 J/kgK, νf = 20.3 × 10−6 m2 /s, αf = 29.44 × 10−6 m2 /s, Pr = 0.69. (a) The thermal circuit diagram is shown in Figure Pr.6.29(b).
563
T1(t) Initial Temperature: T1(t = 0)
 (ρcpV)1dT1
Rku
D
Qku
D
Tf, uf,
dt
Figure Pr.6.29(b) Thermal circuit diagram.
(b) Assuming surfaceconvection heat transfer with transient, lumped capacitance treatment of the substrate, from (6.156) we have T1 (t) − Tf,∞
=
τ1
=
[T1 (t = 0) − Tf,∞ ]e(−t/τ1 ) + a1 τ1 [1 − exp(−t/τ1 ) ] (S˙ − Q) (ρcp V )1 (Rku )D a1 = . (ρcp V )1
Since there is no energy conversion in the spring a1 = 0. Then, we have T1 (t) − Tf,∞ = [T1 (t = 0) − Tf,∞ ] exp−t/τ1 The volume is V1
= π×
D22 − D12 × L = 2.827 × 10−7 m3 . 4
To determine τ1 , the surfaceconvection resistance Rku D is needed. Form (6.124), we have Rku D =
D2 , Aku NuD kf
where Aku
= πD2 L = π × 0.005(m) × 0.04(m) = 6.283 × 10−4 m2 .
The Nusselt number is found for a cylinder in cross ﬂow as given in Table 6.4, i.e., NuD = a1 ReaD2 Pr1/3
Table 6.4,
where ReD
= = =
uf,∞ D2 νf 5(m/s) × 0.005(m) 20.3 × 10−6 (m2 /s) 1,231.5.
From Table 6.3, a1 a2
= =
0.683 0.466.
Then NuD Rku D
=
0.683 × (1,231.5)0.466 × (0.69)1/3
=
16.63
=
0.005(m) = 15.95 K/W. 6.28 × 10−4 (m2 ) × 16.63 × 0.0300(W/mK) 564
The time constant τ1 is τ1
= =
6,450(kg/m3 ) × 837.36(J/kgK) × 2.827 × 10−7 (m2 ) × 15.95(K/W) 24.36 s.
From (6.156), we solve for the time needed for the spring to reach T1 = 323 K, i.e., (323.15 − 350.15)(K) =
(294.15 − 350.15)(K)e(−t/24.36(s))
−27(K) = −56e[−t/24.36(s)] 0.48214 = e[−t/24.26(s)] ln(0.48214) = −t/24.36(s) t
=
17.77 s.
This is less than 20 s and the device performs as required. COMMENT: This problem assumes that the temperature of the spring is uniform. To verify this, the Biot number given by (6.128) should be less than 0.1, i.e., BiD =
Rk < 0.1, Rku D
where Rk
=
ln(D2 /D1 ) = 1.032 K/W. 2πks L
Using this BiD = 0.06473. Then the assumption of a uniform spring temperature is valid.
565
PROBLEM 6.30.FAM GIVEN: A steel cylindrical rod is to be cooled by surface convection using an air stream, as shown in Figure Pr.6.30. The rod can be placed perpendicular [Figure Pr.6.30(i)] or parallel [Figure Pr.6.30(ii)] to the stream. The Nusselt number for the parallel ﬂow can be determined by assuming a ﬂat surface. This is justiﬁable when the viscous boundarylayer thickness δν is smaller than D. D = 1.5 cm, L = 40 cm, uf,∞ = 4 m/s, Tf,∞ = 25◦C, Ts = 430◦C. Determine the air properties at Tf δ = (Ts + Tf,∞ )/2. SKETCH: Figure Pr.6.30(a) shows the two arrangement.
(i) Cross (Perpendicular) Flow uf, Tf,
(ii) Parallel Flow
D
L
L
D
uf, Tf,
Figure Pr.6.30 A steel rod is cooled in an air stream with the choice of placing it perpendicular or parallel to the stream. (i) Cross (perpendicular) ﬂow. (ii) Parallel ﬂow.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the heat transfer rates Qku D and Qku L , for the conditions given above (i.e., δα < D). (c) Is neglecting the surface curvature and using a ﬂat surface, for the parallel ﬂow, justiﬁable? SOLUTION: (a) Figure Pr.6.30(b) shows the thermal circuit diagram, where for the cross ﬂow the resistance is shown as Qku D and for the parallel ﬂow as Qku L . Qku
D
or Qku
L
Tf,
Ts Rku
D
or Rku
L
Figure Pr.6.30(b) Thermal circuit diagram.
(b) The surfaceconvection heat transfer is given by (6.124), i.e., Qku D or
L
= Aku NuD or
L
kf . NuD or L (Ts − Tf,∞ )
(i) The Nusselt number for the cross ﬂow is given in Table 6.4 as NuD = a1 Rea2 Pr1/3 , where constants a1 and a2 depend on ReD , ReD
=
uf,∞ D . νf
From Table C.22, for air at temperature Tf δ
= =
(430 + 25)(◦C) Ts + Tf,∞ = + 273.15 2 2 500.7 K, 566
the properties are kf = 0.0395 W/mK
Table C.22
νf = 3.730 × 10 m /s Pr = 0.69
Table C.22
5
2
Table C.22.
Then ReD
=
a1 NuD
= =
Qku D
= =
4(m/s) × 0.015(m) = 1,608.6 3.730 × 10−5 (m2 /s) 0.683, a2 = 0.466 Table 6.4 0.683 × (1608.6)0.466 × (0.69)1/3 = 18.83 0.0395(W/mK) × (430 − 25)(K) π × (0.015)(m) × (0.4)(m) × 18.83 × 0.015(m) 378.6 W.
(ii) The Nusselt number for the parallel ﬂow us given in Table 6.3 and depends on the Reynolds number ReL ReL =
uf,∞ L 4(m/s) × 0.4(m) = = 4.290 × 104 < ReL,t = 105 , νf 3.730 × 10−5 (m2 /s)
laminar ﬂow regime.
Then from Table 6.3, we have NuL
=
1/2
0.664ReL Pr1/3
0.664 × (4.290 × 104 )1/2 × (0.69)1/3 = 121.5 0.0395(W/mK) × 405K = π × 0.015(m) × 0.4(m) × 121.5 × 0.40(m) = 91.61 W, =
Qku L
which is 24% of the result for cross ﬂow. (c) For laminar ﬂow, the viscous boundarylayer thickness at the tailing edge is given by (6.48), i.e., δν
= = =
νf L 1/2 ) uf,∞ 1/2 3.730 × 10−5 (m2 /s) × 0.4(m) 5× 4(m/s) 0.009657 m. 5(
Then δν 0.009657(m) = = 0.6438 < 1. D 0.015(m) COMMENT: The criterion δα /D < 1 is not a rigorously derived condition. However, for the conditions given, the cross ﬂow results in a large heat transfer rate. The cross ﬂow results in a relatively large qku in the front stagnation region. Also, at higher Reynolds numbers due to ﬂow separation and ﬂow vortices and turbulence, qku is large in the rear section of the cylinder.
567
PROBLEM 6.31.FUN GIVEN: An automobile brake rotor is idealized as a solid disc, as shown in Figure Pr.6.31(a). In a laboratory test the rotor is friction heated at a rate S˙ m,F , under steadystate heat transfer and its assumed uniform temperature becomes Ts . Assume that the heat transfer is by surface convection only and that the ﬂuid (air) motion is only due to the rotation (rotationinduced motion) (Table 6.4). ω = 130 rad/s, R = 30 cm, Tf,∞ = 20◦C, S˙ m,F = 2 × 104 W. Determine the air properties at T = 400 K. SKETCH: Figure Pr.6.31(a) shows the rotating disc. Rotating Disc Air, Tf, uf, = 0 (RotationInduced Motion) Ts
Aku r
M
R
Sm,F
Figure Pr.6.31(a) A rotating disc is heated by friction and cooled by surface convection.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the radial location rtr that the ﬂow regime changes from laminar to turbulent (Rer,tr = 2.4 × 105 ). (c) Integrate the local surface convection over the entire surface area (two sides, neglect the edge). (d) Determine the rotor temperature Ts . SOLUTION: (a) Figure Pr.6.31(b) shows the steadystate thermal circuit diagram. The surface averaged surface convection Qku R is used. Qku
R
Ts
Tf,
Rku
R
Sm,F
Figure Pr.6.31(b) Thermal circuit diagram.
(b) From Table C.22, for air at T = 400 K, we have kf
=
0.0331 W/mK
νf = 2.550 × 10−5 m2 /s Pr = 0.69. From Table 6.4, we have Rer,tr =
2 ωrtr = 2.4 × 105 νf
or rtr
= =
2.4 × 105 × νf ω
1/2 =
2.4 × 105 × 2.550 × 10−5 (m2 /s) 130(rad/s)
2.170 × 10−1 m = 21.70 cm. 568
1/2
(c) Then the inner portion, 0 ≤ r ≤ 21.70 cm is subject to laminar ﬂow regime, and the outer portion, 21.70 cm ≤ r ≤ 35 cm, is subject to turbulent ﬂow regime. Then from (6.50), (6.49) and Table 6.4, we have Qku L Ts − Tf,∞
R Nur kf Nur kf dr + dr 2πr r r 0 rtr 2 1/2 ωr 2 0.8 rtr 0.585 R ωr νf 2πkf 0.021 Pr1/3 dr dr + 2πkf 0.6 0.95 ν f rtr 0 + 1/3 Pr Pr 0.8 1/2 ω 1 2 rtr 1 2.6 R 2πkf × 0.585 ω r 0 + 2πkf × 0.021 × r rtr Pr1/3 0.6 0.95 νf 2 νf 2.6 + 1/3 Pr Pr 1/2 130(rad/s) 2π × 0.0331(W/mK) × 0.585 1 × × (0.2170)2 (m2 ) + −5 2 0.95 0.6 2 2.55 × 10 (m /s) + 0.69 (0.69)1/3 0.8 130(rad/s) 2π × 0.0331(W/mK) × 0.021 × × (0.69)1/3 × 2.550 × 10−5 (m2 /s) 1 [(0.35)2.6 − (0.2170)2.6 ](m)2.6 2.6 (3.33 + 16.00)(W/K) 19.33 W/K = 19.33 W/◦C.
=
=
=
=
= =
rtr
2πr
(d) From Figure Pr.6.31(b), the energy equation is QA
= Qku R = S˙ m,F =
19.33(W/◦C) × (Ts − Tf,∞ ) = S˙ m,F
or Ts
= Tf,∞ +
S˙ m,F 42.22(W/◦C)
=
20(◦C) +
=
1.055◦C.
2 × 104 (W) 19.33(W/◦C)
COMMENT: The ﬂow regime transition is similar to that occurring for the forced and thermobuoyant parallel ﬂows. The material used for the disc should be chosen so as to withstand the resulting surface temperature which is very high.
569
PROBLEM 6.32.FAM GIVEN: A person remaining in a very cold ambient will eventually experience a drop in body temperature (i.e., experience hypothermia). This occurs when the body no longer converts suﬃcient chemicalbond energy to thermal energy to balance the heat losses. Consider an initial uniform temperature of T1 (t = 0) = 31◦C and a constant energy conversion of S˙ r,c = 400 W. The body may be treated as a cylinder made of water with a diameter of 40 cm and a length of 1.7 m, as shown in Figure Pr.6.32(ii). Assume that the lumpedcapacitance analysis is applicable. Evaluate the properties at the average temperature between the initial temperature and the farﬁeld ﬂuid temperature. SKETCH: Figure Pr.6.32 shows a person undergoing surfaceconvection heat losses.
(i) Physical Model
(ii) An Approximation Ambient Air Tf, , uf,
Tf, uf,
DQkuED
Q L = 1.7 m
sr,c = 400 W Water T1 = 31oC
D = 40 cm
Figure Pr.6.32 (i) Surfaceconvection heat transfer from a person. (ii) Its geometric presentation.
OBJECTIVE: Determine the elapsed time for a drop in the body temperature of ∆T1 = 10◦C. (a) Consider the ambient to be air with a temperature of Tf,∞ = −10◦C, blowing at uf,∞ = 30 km/hr across the body (i.e., in cross ﬂow). (b) Consider the ambient to be water with a temperature of Tf,∞ = 0◦C with a thermobuoyant motion (uf,∞ = 0 km/hr) in the water along the length of the cylinder. For the thermobuoyant motion, use the results for a vertical plate and assume that the body temperature is the timeaveraged body temperature (i.e., an average between the initial and the ﬁnal temperature). This results in a constant surfaceconvection resistance. SOLUTION: (a) Air With Forced, Cross Flow: This is a transient problem in which a lumpedcapacitance analysis is to be used. The integralvolume energy equation (2.9) applied to the body gives dT1 + S˙ r,c . dt The net heat transfer at the control surface (wrapped around the body) is due only to surface convection. Thus, we write T1 − Tf,∞ QA = Qku D = . Rku D QA,1 = −(ρcp V )1
Note that both the resistance and the heat transfer are averaged over D, because the air is in cross ﬂow. The energy conversion is from chemical bond to thermal energy, S˙ r,c . Substituting this into the energy equation we have T1 − Tf,∞ dT1 + S˙ r,c = −ρ1 cp,1 V1 Rku D dt The solution to this is given by (6.156), and as in Example 6.15, solving for t we have T1 (t) − Tf,∞ − a1 τ1 t = −τ1 ln , T1 (t = 0) − Tf,∞ − a1 τ1 570
where τs a1
= ρ1 cp,1 V1 Rku D S˙ r = . ρ1 cp,1 V1
The average surfaceconvection resistance is obtained from the Nusselt number using (6.49). The properties for air, at Tδ = [(31+21)/2−10]/2 = 8.0◦C = 281.15 K, are found from Table C.22, as kf = 0.0256 W/mK, νf = 14.01x10−6 m2 /s, Pr = 0.69. The Reynolds number based on diameter is given in Table 6.4 as ReD =
uf,∞ D 8.333(m/s) × 0.4(m) = 2.379 × 105 . = νf 14.01 × 10−6 (m2 /s)
From Table 6.4, the correlation for NuD for cross ﬂow over a circular cylinder is found with a1 = 0.027 and a2 = 0.805. The Nusselt number is NuD = a1 ReaD2 Pr1/3 = 0.027(2.379 × 105 )0.805 × (0.69)1/3 = 507.74. The average surfaceconvection resistance is given by (6.49), i.e., Rku D =
D 0.4(m) = 0.0144◦C/W. = kf NuD Aku 0.0256(W/mK) × 507.74 × π × 0.4(m) × 1.7(m)
The properties for water at Tδ = (31 + 21)/2 = 26◦C = 299.15 K, are found from Table C.23 as ρl = 997.8 kg/m3 and cp,l = 4,182 J/kgK. Thus, τ1 and a1 are 3
τ1 = 997.8(kg/m ) × 4,182(J/kgK) ×
a1 =
π × (0.4)2 (m)2 × 1.7(m) × 0.0145(◦C/W) = 12,837 s 4
400(W) 3
997.8(kg/m ) × 4,182(J/kgK) ×
π×(0.4)2 (m)2 4
× 1.7(m)
= 4.487 × 10−4 ◦C/s.
Solving for t, we have t = −12,837(s) × ln
21(◦C) − [−10(◦C)] − 4.487 × 10−4 (◦C/s) × 12,837(s) = 4,284 s = 71.41 min. 31(◦C) − [−10(◦C)] − 4.487 × 10−4 (◦C/s) × 12,837( s)
(b) Water With Thermobuoyant Flow: The above energy equation and transient analysis remain the same. The average Nusselt number for the thermobuoyant ﬂow over a vertical ﬂat plate is given in Table 6.5. The properties for water, at Tδ = [(31 + 21)/2 + 0]/2 = 286.15 K, are obtained from Table C.23 as kf = 0.581 W/mK, νf = 1.28 × 10−6 m2 /s, αf = 0.139 × 10−6 m2 /s, Pr = 9.31, and βf = 0.00018 1/K. The timeaveraged body temperature is T 1 = 26◦C. The Rayleigh number is deﬁned in Table 6.4 as gβf (T 1 − Tf,∞ )L3 9.81(m/s ) × 0.00018(1/K) × [26(◦C) − 0(◦C)] × (1.7)3 (m)3 = 1.27 × 1012 . = νf αf 1.28 × 10−6 (m2 /s) × 0.139 × 10−6 (m2 /s) 2
RaL =
From Table 6.5, a1
=
0.6205
NuL,l NuL,t
= =
659.8 1,198.2
NuL
=
1,203.7.
The average surfaceconvection resistance is Rku L =
L 1.7(m) = 0.001137◦C/W. = kf NuL Aku 0.581(W/mK) × 1,203.7 × π × 0.4(m) × 1.7(m) 571
The properties for the body remain the same. Then, τ1 is 3
τ1 = 997.8(kg/m ) × 4182(J/kgK) ×
π × (0.4)2 (m)2 × 1.7(m) × 0.001137(◦C/W) = 1,013.5 s 4
and a1 remains the same, i.e., a1 = 4.487 × 10−4 ◦C/s. Solving for t gives
21(◦C) − 0(◦C) − 4.487 × 10−4 (◦C/s) × 1,013.5(s) t = −1,013.5(s) × ln = 401.9 s = 6.7 min. 31(◦C) − 0(◦C) − 4.487 × 10−4 (◦C/s) × 1,013.5(s) COMMENT: Although the motion in the water is due to thermobuoyancy and of a smaller velocity (as compared to the forced ﬂow), due to the larger thermal conductivity of water, the surfaceconvection heat transfer is larger (i.e., the surfaceconvection resistance is smaller) for water.
572
PROBLEM 6.33.FAM GIVEN: A methaneair mixture ﬂows inside a tube where it is completely reacted generating a heating rate of S˙ r,c = 104 W. This heat is removed from the tube by a cross ﬂow of air, as shown in Figure Pr.6.33(a). Evaluate the properties of air at T = 300 K. SKETCH: Figure Pr.6.33(a) shows energy conversion by combustion in a tube and heat removal by surface convection from the tube. Air in Cross Flow Tf, = 20 oC uf, = 5 m/s
Ts Methane and Air Flow
r,s
= 0.7
Sr,c = 104 W
D = 15 cm
L=1m Surrounding Surface Tf, = 20 oC r=1
Figure Pr.6.33(a) Heat removal from a combustion tube.
OBJECTIVE: (a) Draw the thermal circuit diagram and determine the tube surface temperature with no surface radiation. (b) Repeat part (a) with surface radiation included. SOLUTION: (a) Neglecting radiation, the thermal circuit is shown in Figure Pr.6.33(b). Tf,
Qku
Rku
D
Ts
D
. Sr,c Qs
Figure Pr.6.33(b) Thermal circuit diagram for the case of no surface radiation heat transfer.
Applying the integralvolume energy equation (2.9), with Qs = 0, to the tube we have QA = −(ρcp V )s
dTs + S˙ r,c , dt
573
where the lumped capacitance is assumed for into node Ts . As the only heat loss occurs by surface radiation and the energy conversion is by combustion, for this steadystate problem, the energy equation becomes Ts − Tf,∞ = S˙ r,c . Rku D For air, from Table C.22, at T = 300 K, we have νf = 15.66 × 10−6 m2 /s, kf = 0.0267 W/mK, Pr = 0.69. The Reynolds number is uf,∞ D 5(m/s) × 0.15(m) = 47,893. = ReD = νf 15.66 × 10−6 (m2 /s) The Nusselt number for crossﬂow over a circular cylinder, from Table 6.3, with a1 = 0.027 and a2 = 0.805, is NuD = 0.027Re0.805 Pr1/3 = 0.027 × (47,893)0.805 × (0.69)1/3 = 139.7. D The average surfaceconvection resistance is given by (6.124), i.e., Rku D =
D 0.15(m) = 0.08532◦C/W. = Aku NuD kf π × 0.15(m) × 1(m) × 139.7 × 0.0267(W/mK)
Then from the energy equation, we have Ts = Tf,∞ + Rku D S˙ r,c = 20(◦C) + 0.08532(◦C/W) × 104 (W) = 873.2◦C = 1,146K. (b) With the inclusion of surface radiation, the thermal circuit is shown in Figure Pr.6.33(c). Tf,
Eb, (Rr,F) (qr,o)
Qku
Rku
D
D
Qr,s
(Rr,F)s (qr,o)s (Rr, )s
. Sr,c
Eb,s
Ts
Qs
Figure Pr.6.33(c) Thermal circuit diagram for the case with surfaceradiation heat transfer.
The energy equation then becomes
Ts − Tf,∞ Eb,s − Eb,∞ + = S˙ r,c Rku D (RΣ,r )s∞
The thermal resistance for radiation is given by (4.47), i.e., 1 − r 1 − r 1 (RΣ,r )s∞ = + + r Ar ∞ Ar,s Fs∞ r Ar s 1 − 0.7 1 + = 0+ π × 0.15(m) × 0.1(m) × 1 (0.7) × π × 0.15(m) × 1(m) =
3.032
2
1/m . 574
The surfaceconvection resistance and the energy conversion term remain the same. Then the energy equation becomes Ts (K) − 293.15(K) σSB [Ts4 (K4 ) − (293.15)4 (K4 )] + = 104 W. 2 0.08532(◦C/W) 3.032(1/m ) Using a solver (such as SOPHT), the results is Ts = 722.6 K COMMENT: Note the signiﬁcant drop in Ts caused by the surfaceradiation heat transfer. The assumption of a uniform tube temperature may not be valid if L/D is very large.
575
PROBLEM 6.34.FUN GIVEN: Consider surfaceconvection heat transfer from a sphere of radius R and initial temperature Ts (t = 0) as rendered in Figure Pr.6.34(a). The time dependence of the uniform sphere temperature, with surface convection, is given by (6.156) and is valid for BiR < 0.1. Also for BiR > 10 the surfaceconvection resistance becomes negligible and then the constant surface temperature, distributed transient temperature given in Figure 3.33(b)(ii) becomes valid. In the Biot number regime 10 < BiR < 0.1, numerical or series, closedform solutions are used. In an existing series solution, when the elapsed time is suﬃciently large (i.e., FoR > 0.2) such that the penetration distance has reached and passed the center of the spheres, a single term from this series solution may be used to obtain Ts = Ts (r, t). This solution for the center of the sphere, i.e., r = 0, is Ts∗ (r = 0, t) =
2 Ts (r = 0, t) − Tf,∞ = a1 e−a2 FoR , Ts (t = 0) − Tf,∞
FoR =
tαs > 0.2, R2
where the constants a1 and a2 depend on BiR and are listed for some values of BiR in Table Pr.6.34. From (6.128), we have BiR =
Rk,s NuD kf /D . = Rku D ks /R
Table Pr.6.34 The constants in the oneterm solution.
Lumped
Constant Surface Temperature
BiR
a1
a2
(3BiR )1/2
0 0.01 0.10 1.0 10 100 ∞
1.000 1.003 1.030 1.273 1.943 1.999 2.000
0 0.1730 0.5423 1.571 2.836 3.110 3.142 = π
0 0.1732 0.5477 1.414 4.472 14.14 ∞
OBJECTIVE: (a) Show that (6.156) can be written as Ts∗ (t) =
Ts (t) − Tf,∞ = e−3FoR BiR , Ts (t = 0) − Tf,∞
BiR < 0.1.
(b) Plot Ts∗ (t) with respect to FoR , for 0.01 ≤ FoR ≤ 1, and for BiR = 0.01, 0.1, 1, 10, and 100. (c) On the above graph, mark the center temperature Ts (r = 0, t) for FoR = 0.2 and 1.0, and for the Biot numbers listed in part (b). (d) For FoR = 0.2 and 1.0, also mark the results found from Figure 3.33(b)(ii), noting that this corresponds to BiR → ∞. (e) Comment on the regime of a signiﬁcant diﬀerence among the results of the lumpedcapacitance treatment, the distributedcapacitance treatment with BiR → ∞, and the singleterm solution for distributed capacitance with ﬁnite BiR . SKETCH: Figure Pr.6.34(a) shows a sphere placed in a ﬂuid stream with surface convection, and timedependent temperature. OBJECTIVE: (a) Show that (6.156) can be written as Ts∗ (t) =
Ts (t) − Tf,∞ = e−3FoR BiR Ts (t = 0) − Tf,∞
BiR < 0.1.
(b) Plot Ts∗ (t) with respect to FoR for 0.01 ≤ FoR < 1, and for BiR = 0.01, 0.1, 1, 10 and 100. 576
r
uf, Tf,
R
Ts(r, t), or Ts(t = 0)
Figure Pr.6.34(a) A sphere of initial temperature Ts (t = 0) is placed in a ﬂuid stream with Tf,∞ , and uf,∞ .
(c) On the above graph, mark the center temperature Ts∗ (r = 0, t) for FoR = 0.2 and 1.0, and for the Biot numbers listed in part (b). (d) For FoR = 0.2 and 1.0, also mark the results found from Figure 3.33(b)(ii), noting that this graph corresponds to BiR → ∞. (e) Comment on the regime a signiﬁcant diﬀerence among the three solutions. SOLUTION: (a) Starting from (6.156), we write for a1 = 0, and charging t to FoR using FoR = tαs /R2 , we have Ts∗ (t) =
Ts (t) − Tf,∞ = exp(−FoR R2 /αs τ ). Ts (t = 0) − Tf,∞
Next, using the deﬁnition of τ given by (6.157), we have FoR R2 αs τ
=
FoR R2 αs (ρcp )s V Rku D
Noting that Rku D = D/(Aku NuD kf ), (αρcp )s = ks , V = πD3 /6 and that Aku = 4πR2 , FoR R2 kf = 3FoR × NuD . αs τ ks From Example 6.15, we have BiD = NuD
kf . 4ks
BiR = NuD
kf , ks
As Bix ∝ x2 , BiR = 4BiD , so that
and we can then write FoR R2 = 3FoR BiR . αs τ Using this, we have Ts∗ (t) = e−3FoR BiR . (b) Figure Pr.6.34(b) shows the variation of Ts∗ (t) with respect to FoR for several values of BiR . For large BiR , Ts (t) quickly (i.e., small FoR ) becomes equal to Tf,∞ , i.e., Ts∗ (t = 0) → 0. For very small BiR , the sphere temperature does not change unless FoR is very large. (c) The center temperature found from the oneterm solution is marked (with closed circles) on Figure Pr.6.34(b). For BiR < 0.1, the two results are identical. This can be also noticed from Table Pr.6.34, where a22 = 3BiR , for BiR ≤ 0.1. For larger BiR , there is a diﬀerent between the two, especially at FoR = 0.2. As FoR increases beyond 1, the diﬀerence decreases again, regardless of the value of BiR .
577
Single Term Distributed Solution Valid
Ts*(t) =
Ts(t)  Tf, T (r = 0, t)  Tf, , Ts*(r = 0, t) = s Ts(t = 0)  Tf, Ts(t = 0)  Tf,
Ts*(r = 0, t) 1.0
0.01 0.1 Lumped Capacitance Valid
BiR = 1 BiR = 1
0.8
Region of Significant Difference 0.2 FoR 1 10 BiR 0.1
Lumped Capacitance Approximation 0.6
Ts*(t) 10 10
0.4
Figure 3.33(b)(ii) BiR , FoR = 0.2
100
0.2779 0.2
100
Figure 3.33(b)(ii) BiR , FoR = 1
1
Constant Surface Temperature Valid Figure 3.33(b)(ii)
10 and 000
0 0.01
0.02
0.04
0.08
0.2
FoR =
0.6
1.0
tas R2
Figure Pr.6.34(b) Variation of dimensionless temperatures with respect to dimensionless time. The dots represent Ts∗ (r = 0, t).
(d) From Figure 3.33(a)(ii), for FoR = 0.2, we have Ts∗ (r = 0, t) = 1 − 0.72 = 0.28 (the exact value is 0.2779 and is found by using the singleterm solution for BiR → ∞). For FoR = 1, we have Ts∗ (r = 0, t) = 1 − 1 = 0. These are also marked in Figure Pr.6.34(b). (e) In the regime marked by 0.2 ≤ FoR ≤ 1,
10 ≤ BiR ≤ 0,
the singleterm solution gives more accurate results. COMMENT: When this region is encountered, the lumpedcapacitance and constant surfacetemperature approximations are not valid. But in practice, most problems fall outside this region and satisfy the requirements for the approximation lumped capacitance or constant surfacetemperature solutions. The graphical Heisler results give the results for FoR ≥ 0.2 and an arbitrary BiR for spheres, long cylinders, and ﬁnite slabs and can be found in Chapter 6, reference 19.
578
PROBLEM 6.35.FAM GIVEN: In a rapid solidiﬁcationcoating process, a liquid metal is atomized and sprayed onto a substrate. The atomization is by gas injection into a spray nozzle containing the liquidmetal stream. The injected gas is small compared to the gas (assume air) entrained by the droplet spray stream. This entrained gas quickly cools the droplets such that at the time of impingement on the substrate the droplets contain a threshold amount of liquid that allows for them to adhere to each other and to the substrate surface. This is shown in Figure Pr.6.35(a), where a plastic balloon is coated with a tin layer and since the droplets are signiﬁcantly cooled by surface convection, the balloon is unharmed. Assume that each droplet is independently exposed to a semibounded air stream. T1 (t = 0) = 330◦C, Tf,∞ = 40◦C, uf,∞ (relative velocity) = 5 m/s, D = 50 µm. The Nusselt number can be determined (Table 6.4) using the relative velocity and the properties of tin are given in Tables C.5 and C.16. Determine the air properties at T = 400 K. SKETCH: Figure Pr.6.35(a) shows the ﬂight of droplets. Liquid Metal Droplet Stream Deposit Relative Velocity D uf,
Injection Gas Stream
Pattern (Balloon)
T1(t = 0) Entrained Air Tf,
Figure Pr.6.35(a) A plastic balloon is spray coated with tin droplets solidifying on its surface.
OBJECTIVE: (a) Draw the thermal circuit diagram for a tin droplet cooled from the initial temperature T1 (t = 0) to its solidiﬁcation temperature Tsl . Assume a uniform temperature T1 (t). (b) By neglecting any motion within the droplet, determine if a uniform droplet temperature can be assumed; use Rk,s = D/4Aku ks . (c) Determine the time of ﬂight t, for the given conditions. SOLUTION: (a) Figure Pr.6.35(b) shows the thermal circuit diagram for the droplet cooling. The sensible heat (and not the phase change) is included. Qku Tf,
D
T1(t) = 0 Rku
D
 (HcpV)1 dT1 dt
Figure Pr.6.35(b) Thermal circuit diagram.
(b) From (6.128) and(6.124), we have BiD =
Rk,1 D/4Aku ks kf = = NuD . Rku D D/Aku NuD kf 4ks
From Tables C.16, we have for tin: Tsl = 505 K, ρs = 7,310 kg/m3 , cp,s = 227 J/kgK, ks = 66.6 W/mK. From Table C.22, we have for air at T = 400 K: kf = 0.0331 W/mK, νf = 2.55 × 10−5 m2 /s, Pr = 0.69. 579
Then from (6.124) we have ReD =
uf,∞ D 5(m/s) × 5 × 10−5 (m) = = 9.804. νf 2.55 × 10−5 (m2 /s)
From Table 6.4, we have NuD
1/2
2/3
=
2 + (0.4ReD + 0.06ReD )Pr0.4
=
2 + [0.4(9.804)1/2 + 0.06(9.804)2/3 ](0.69)0.4 = 3.317.
Then BiD = 3.317 ×
0.0331(W/mK) = 4.121 × 10−4 < 0.1 4 × 66.6(W/mK)
and the variation of temperature within droplet is therefore negligible. (c) From (6.156), with Q1 = S˙ 1 = 0, a1 = 0, and we have T1 (t) − Tf,∞ = e−t/τ , T1 (t = 0) − Tf,∞ or
t τ1
t
T1 (t) − Tf,∞ = −τ1 ln T1 (t = 0) − Tf,∞
D Aku NuD kf
=
(ρcp V )1 Rku D = (ρcp V )1
=
(ρcp )1
=
7,310(kg/m3 ) × 227(J/kgK) ×
=
6.297 × 10−3 s
VD 1 D2 1 = (ρcp )1 Aku NuD kf 6 NuD kf
= −6.297 × 10−3 (s) × ln =
2.602 × 10−3 s
=
2.602 ms.
1 (5 × 10−5 )2 (m2 ) × 6 3.317 × 0.0331(W/mK)
505(K) − (273.15 + 40)(K) (330 − 40)(K)
COMMENT: Note that we have assumed that each droplet is independently exposed to a semiinﬁnite air stream. In practice, the cloud of droplets heat the stream and also reduces NuD through modifying the ﬂuid ﬂow and temperature distribution around each droplet. This results in a larger resistance and a larger time constant, requiring larger elapsed times to reach the desired temperature.
580
PROBLEM 6.36.FAM GIVEN: In order to enhance the surfaceconvection heat transfer rate, ﬁns (i.e., extended surfaces) are added to a planar surface. This is shown in Figure Pr.6.36. The surface has a square geometry with dimensions a = 30 cm and w = 30 cm and is at Ts,o = 80◦C. The ambient is air with a farﬁeld velocity of uf,∞ = 1.5 m/s and a temperature of Tf,∞ = 20◦C ﬂowing parallel to the surface. There are N = 20 rectangular ﬁns made of pure aluminum and each is l = 2 mm thick and L = 50 mm long. Assume that the Nusselt number is constant and evaluate the properties at the average temperature between the plate temperature and the farﬁeld ﬂuid temperature. SKETCH: Figure Pr.6.36 shows the ﬁns attached to the heat transfer surface. Base, Ts,o = 80 oC
l = 2 mm
Parallel Air Flow o Tf, = 20 C uf, = 1.5 m/s
Rectangular Fin (20)
a = 30 cm
w = 30 cm L = 50 mm
Figure Pr.6.36 An extended surface with parallel, forced ﬂow.
OBJECTIVE: (a) Determine the rate of heat transfer for the plate without the ﬁns. (b) Determine the rate of heat transfer for the plate with the ﬁns. Treat the ﬂow over the ﬁns as parallel along the width w, thus neglecting the eﬀect of the base and the neighboring ﬁns on the ﬂow and heat transfer. SOLUTION: (a) Without Fins: The air is in parallel ﬂow over the plate surface (along w). For the properties for air, at Tδ = (80 + 20)/2◦C= 323.15 K, from Table C.22, we have: kf = 0.0283 W/mK, νf = 17.73 × 10−6 m2 /s, Pr = 0.69. The Reynolds number, based on the length w, is given by (6.45) as Rew =
uf,∞ w 1.5(m/s) × 0.3(m) = 25,381 < Rew,t = 105 = νf 17.73 × 10−6 (m2 /s)
laminar ﬂow.
From Table 6.3 for parallel ﬂow with Rew < 105 , the Nusselt number is 1/3 Nuw = 0.664Re1/2 = 0.664 × (25,295)0.5 × (0.69)1/3 = 93.48. w Pr
The average surfaceconvection resistance is Rku w =
w 0.3(m) = 1.26◦C/W. = kf Nuw Aku 0.0283(W/mK) × 92.48 × 0.3(m) × 0.3(m)
The surfaceconvection heat transfer rate is given by (6.49) as Qku w =
Ts − Tf,∞ 80(◦C) − 20(◦C) = 47.62 W. = Rku w 1.26(◦C/W) 581
(b) With Fins: The air is again in parallel ﬂow over the ﬁns. The ﬁns surface is considered a ﬂat semiinﬁnite (along the direction of the ﬂow) surface. Then, the Nusselt number remains the same as the one used above. From Table C.14, for aluminum, ks = 238 W/mK. The geometric parameters used in the ﬁn eﬃciency, given by (6.147), are Lc = L + l/2 = 0.05(m) + 0.002(m)/2 = 0.051 m Pku,f = 2w + 2l = 0.604 m Ak = wl = 0.3(m) × 0.002(m) = 6.00 × 10−4 m2 Aku,f = Pku Lc = 0.604(m) × 0.051(m) = 0.0308 m2 Ab = aw − Nf Ak = (0.3)2 (m)2 − (20)6 × 10−4 (m2 ) = 0.078 m2 . The ﬁn parameter is given by (6.144), i.e., m=
Nu kf
Pku ww Ak ks
1/2
=
93.48×0.0283(W/mK) 0.3(m) 10−4 (m2 ) × 238(W/mK)
0.604(m) × 6×
1/2
= 6.107 1/m.
Then the eﬃciency is ηf =
tanh(mLc ) tanh[6.107(m×)0.051(m)] = 0.9689. = mLc 6.107(m) × 0.051(m)
From (6.152) and (6.153), the average surfaceconvection resistances for the bare surface and ﬁn are Rku w,b =
Rku w,f =
w 0.3(m) = 1.454◦C/W = kf Nuw Ab 0.0283(W/mK) × 93.48 × 0.078(m2 )
w 0.3(m) = 0.190◦C/W. = kf Nuw Aku,f ηf Nf 0.0283(W/mK) × 93.48 × 0.0308(m2 ) × 0.97 × 20
From (6.151), the overall thermal resistance is 1 RΣ RΣ
= =
1 1 + Rku w,b Rku w,f 0.1680◦C/W.
The surfaceconvection heat transfer rate is Qku w =
Ts − Tf,∞ 80(◦C) − 20(◦C) = 357.1 W. = RΣ 0.1680(◦C/W)
COMMENT: The use of the ﬁns has increased the surfaceconvection heat transfer from the plate by a factor of 7.5.
582
PROBLEM 6.37.FAM GIVEN: An automobile discbrake converts mechanical energy (kinetic energy) to thermal energy. This thermal energy is stored in the disc and is transferred to the ambient by surface convection and surface radiation and is also transferred to other mechanical components by conduction (e.g., the wheel, axle, suspension, etc). The rate of energy conversion decreases with time due to the decrease in the automobile speed. Here, assume that it is constant and is (S˙ m,F )o = 6 × 104 W. Assume also that the heat loss occurs primarily by surfaceconvection heat transfer from the disc surface. The disc is made of carbon steel AISI 4130 (Table C.16) and its initial temperature is T1 (t = 0) = Tf,∞ = 27◦C. The disc surfaceconvection heat transfer is from the two sides of disc of diameter D = 35 cm, as shown in Figure Pr.6.37(a), and the disc thickness is l = 1.5 cm. The Nusselt number is approximated as that for parallel ﬂow over a plate of length D and determined at the initial velocity. The average automobile velocity is uf,∞ = 40 km/hr and the ambient air is at Tf,∞ . Evaluate the air properties at Tf,∞ . SKETCH: Figure Pr.6.37(a) shows the physical and an approximation models of the disc brake.
(i) Physical Model
(ii) An Approximation Energy Conversion at Brake PadRotor Interface (Sm,F)o
Air Flow Over Disc uf, , Tf,
Caliper l D
T1(t = 0) = Tf,
Rotor (Disc)
Figure Pr.6.37(a) An automobile brake cooled by parallel ﬂow. (i) Physical model. (ii) Approximate model.
OBJECTIVE: (a) Assuming that the lumpedcapacitance analysis is applicable, determine the temperature of the disc after 4 s [T1 (t = 4 s)]. (b) Using this temperature [i.e., T1 (t = 4 s)] as the initial temperature and setting the heat generation term equal to zero (i.e., the brake is released), determine the time it takes for the disc temperature to drop to t1 = 320 K. (c) Evaluate the Biot number and comment on the validity of the lumpedcapacitance assumption. For the Biot number, the conduction resistance is based on the disc thickness l, while the surface convection resistance is based on the disc diameter D. SOLUTION: (a) This is a transient problem, with surface heating due to friction, and cooling by surface convection. The corresponding thermal circuit is shown in Figure Pr.6.37(b). Using a lumpedcapacitance analysis, the integralvolume energy equation (2.9) becomes .QA = −(ρcp V )1
dT1 + (S˙ m,F )o . dt
For surfaceconvection, heat transfer only, we have QA = Qku D =
dT1 T1 − Tf,∞ + S˙ m,F = −(ρcp V )1 Rku D dt
The solution for this equation is given by (6.156), i.e., T1 (t) − Tf,∞ = [T1 (t = 0) − Tf,∞ ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ), 583
. (Sm,F)o
Qku
D
Tf,
T1
Q1
Q2 Rku
D
 (ρcpV)1dT1 dt
Figure Pr.6.37(b) Thermal circuit diagram.
where
S˙ m,F (ρcp V )1 To determine the surfaceconvection resistance, the Nusselt number is needed. The properties for air, from Table C.22, evaluated at 300 K, are νf = 15.66 × 10−6 m2 /s, kf = 0.0267 W/mK, and Pr = 0.69. The Reynolds number for parallel ﬂow over a ﬂat plate with length D is given in Table 6.4 as τ1 = (ρcp V )1 Rku D ,
ReD =
a1 =
uf,∞ D (40/3.6)(m/s) × 0.35(m) = 248,333 < ReD,t = 5 × 105 = νf 15.66 × 10−6 (m2 /s)
This Reynolds number is still in the laminar regime. The average Nusselt number, from Table 6.4, is given by 1/2
NuD = 0.664ReD Pr1/3 = 0.664 × (248,333)1/2 × (0.69)1/3 = 292.4. Taking into account both sides of the plate, the average surface convection resistance is then given by (6.124) as Rku D =
D 0.35(m) = NuD kf Aku 292.4 × 0.0267(W/mK) × 2 ×
π×(0.35)2 (m2 ) 4
= 0.233◦C/W.
For carbon steel AISI 4130, from Table C.16, we have ρ = 7,840 kg/m3 , cp = 460 J/kgK, and ks = 43 W/mK. Then the parameters τ1 (time constant) and a1 are τ1
a1
3
=
(ρcp V )1 Rku D = 7,840(kg/m ) × 460(J/kgK) ×
=
1212.6s (S˙ m,F )o 6 × 104 = 3 (ρcp V )1 7,840(kg/m ) × 460(J/kgK) ×
=
π × (0.35)2 (m2 ) × 0.015(m) × 0.233(◦C/W) 4
π(0.35)2 (m2 ) 4
× 0.015(m)
= 11.53◦C/s
Now the plate temperature after t = 4 s, for T1 (t = 0) = Tf,∞ = 300 K, is T1 (t = 4 s) = 300.15(K) + 11.53(◦C/s)] × 1,212.6(s)[1 − e−4(s)/1212.6(s) ] = 346.2 K (b) Once the break is released, the energy conversion due to friction stops. In this case, a1 = 0. The time constant is still the same, because neither the disc properties nor the surfaceconvection resistance have changed. Then the time to cool the disc down to T1 (t) = 320 K is T1 (t) − Tf,∞ 320(K) − 300.15(K) t = −τ1 ln = −1,212.6(s) × ln = 1020 s = 17.0 min T1 (t = 4 s) − Tf,∞ 346.2(K) − 300.15(K) (c) The Biot number is given by the ratio of the conduction thermal resistance (through half the plate thickness) and the surfaceconvection resistance, i.e., Ak Rk , Bil = Aku Rku D where Ak = Aku = πD2 /4. Then we have Bil =
1.744 × 10−4 [◦C/(W/m2 )] (l/2)/ks = = 3.89 × 10−3 . D/(NuD kf ) 4.483 × 10−2 [◦C/(W/m2 )]
As Bil 1, the lumped capacitance analysis can be used. COMMENT: Note that compared to the t = 4 s heatup period, the cooldown period is very long. In Problem 6.39, the determination of (S˙ m,F )o is described. 584
PROBLEM 6.38.FAM GIVEN: In a portable, phasechange hand warmer, titanium bromide (TiBr4 , Table C.5) liquid is contained in a plastic cover (i.e., encapsulated) and upon solidiﬁcation at the freezing temperature Tsl (Table C.5) releases heat. A capsule, which has a thin, rectangular shape and has a crosssectional area Ak = 0.04 m2 , as shown in Figure Pr.6.38(a), is placed inside the pocket of a spectator watching an outdoor sport. The capsule has a planar surface area of Ak = 0.04 m2 . The pocket has a thick insulation layer on the outside of thickness Lo = 2 cm (toward the ambient air) and a thinner insulation layer on the inside of thickness Li = 0.4 cm (toward the body). The eﬀective conductivity for both layers is k = 0.08 W/mK. The body temperature is Tb = 32◦C. The outside layer is exposed to surface convection with a wind blowing as a cross ﬂow over the body (diameter D) at a speed uf,∞ = 2 m/s and temperature Tf,∞ = 2◦C. For the surfaceconvection heat transfer, use cross ﬂow over a cylinder of diameter D = 0.4 m. Assume that the heat ﬂow is steady and that the temperatures are constant. Treat the conduction heat transfer as planar and through a crosssectional surface area Ak . Determine the air properties at T = 300 K from Table C.22. For the surfaceconvection heat transfer, assume a cross ﬂow over a cylinder of diameter D = 0.4 m. Assume that the heat ﬂow is steady and that the temperatures are constant. Use planar geometry for the conduction resistances. SKETCH: Figure Pr.6.38(a) shows the phasechange material sandwiched between two insulation layers. The outside insulation layer is exposed to the surface convection. D/2 Li
Lo Air
Body
uf, Tf,
Tb k
k Ak Tsl , Ssl PhaseChange Material Capsule
Figure Pr.6.38(a) Simpliﬁed, physical model for heat transfer from a hand warmer.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine all the thermal resistances that the heat ﬂow from the capsule encounters. Use planar geometry for the conduction resistances. (c) Determine the heat rate toward the body and toward the ambient air. (d) Determine the total energy conversion rate S˙ sl (W). SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.6.38(b). . Ssl Rk,slb Tb
Qk,slb
Rk,sls Tsl
Rku
D
Qku
D
Tf,
Ts
Qk,sls
Figure Pr.6.38(b) Thermal circuit diagram.
585
(b) The thermal resistances are as follows. (i) Internal conduction thermal resistance: Rk,slb =
Li 0.04(m) = = 1.25◦C/W. kak 0.08(W/mK) × 0.04(m2 )
(ii) External conduction thermal resistance Rk,sls =
Lo 0.02(m) = = 6.25◦C/W. kak 0.08(W/mK) × 0.04(m2 )
(iii) Surface convection thermal resistance The properties for air at T = 300 K, from Table C.22: kf = 0.0267 W/mK, νf = 15.66 × 10−6 m2 /s, Pr = 0.69. For air in cross ﬂow, the Reynolds number is given by (6.124) as ReD =
uf,∞ D 2(m/s) × 0.4(m) = = 51,086. νf 15.66 × 10−6 (m2 /s)
From Table 6.4, with a1 = 0.027 and a2 = 0.805, the Nusselt Number is NuD = 0.027Re0.805 Pr1/3 = 0.027 × (51,086)0.805 × (0.69)1/3 = 147.2. D The area for convection relevant to the problem is the area over the pocket heater, or Aku = Ak . Therefore, the surface convection thermal resistance is Rku D =
D 0.4(m) = = 2.54◦C/W. 2 Aku NuD kf (0.04)(m ) × 147.2 × 0.0267(W/mK)
(c) The fraction of heat ﬂowing toward the body is (from Table C.5, for TiBr4 , Tsl = 312.2 K = 39.2◦C) Qk,slb =
Tsl − Tb (39.2 − 32)(◦C) = 5.76 W. = Rk,slb 1.25(◦C/W)
(d) The fraction of heat ﬂowing toward the ambient air is RΣ,sl∞ = Rk,sls + Rku D = (6.25 + 2.54)(◦C/w) = 8.79◦C/W.
Qk,sl∞ =
Tsl − Tf,∞ (39.2 − 2)(◦C) = 4.23 W. = RΣ,sl∞ 8.79(◦C/W)
(e) The energy conversion rate is then found from applying the integralvolume energy equation of energy to the Tsl node, i.e., S˙ sl = Qk,slb + Qk,sl∞ = (5.76 + 4.23)(W) = 9.99 W. COMMENT: Note that although a thicker insulation layer was allowed on the outside, due to the low ambient air temperature a signiﬁcant portion of the heat generated is still lost to the ambient air.
586
PROBLEM 6.39.FAM.S GIVEN: To analyze the heat transfer aspects of the automobile rearwindow defroster, the window and the very thin resistive heating wires can be divided into identical segments. Each segment consists of an individual wire and an a × L × l volume of glass aﬀected by this individual wire/heater. Each segment has a uniform, transient temperature T1 (t). This is shown in Figure Pr.6.41. In the absence of any surface phase change (such as ice or snow melting, or water mist evaporating), the Joule heating results in a temperature rise from the initial temperature T1 (t = 0), and in a surface heat loss to the surroundings. The surface heat loss to the surroundings is represented by a resistance Rt . The surrounding farﬁeld temperature is T∞ . T1 (t = 0) = −15◦C, T∞ = −15◦C, l = 3 mm, a = 2 cm, L = 1.5 m, S˙ e,J = 15 W, Rt = 2◦C/W. Determine the glass plate properties from Table C.17. SKETCH: Figure Pr.6.39(a) shows a disc brake and its air ﬂow and heat transfer characteristics. Automobile DiscBrake: Physical Model Rotor Angular Velocity, ω
Brake Fluid
Brake Pad
Axle ∆ui
Rotor (Disc) Tr(t), r,r
Qk,ra 0
Qk,rw 0
Caliper Sm,F Energy Conversion at Brake PadRotor Interface
r
u=0
Rotor Angular Velocity, ω
Aku
Air Flow into Disc
To Wheel
Opposite Side of Rotor Ts,
Air Flow over Disc uf, , Tf,
Aku
Surrounding Surface
Figure Pr.6.39(a) An automobile disc brake showing the air ﬂow around it.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Show that the lumped capacitance approximation is valid using l for the conduction resistance. (c) Assuming no surface phase change occurs, determine the steadystate temperature of the glass. (d) Still assuming no surface phase change occurs, determine the glass temperature after an elapsed time t = 5 min. SOLUTION: (a) The circuit diagram is shown in Figure Pr.6.39(b). The heat loss to the pad is negligible because of the small conductivity of the pad material (k ≡ 0.6 W/mK) which can be used as organic compound. The heat transfer to wheel and axle is also neglected.
587
(b) Thermal Circuit Model for Automobile Disc Brake Tp
Negligible Heat Loss to Pad Qρck(t) = 0
SemiBounded Air Flow
Rρck(t)
Rotor, Tr Surrounding Solid Surface (Rr,Σ)k Ts,
Rku,r
Qr,r
Tf, Qku,r
dT  (ρcpV)r r + Sm,F dt
Rk,rw Negligible Heat Loss SemiBounded Air Flow to Axle (Thermal Qk,rw 0 Wheel, Tw Circuit Not Shown) Q k,ra 0 Rku,w (Rr,Σ)w Ts, Tf, Surrounding Solid Surface dT  (ρcpV)w w dt
Figure Pr.6.39(b) Thermal circuit diagram.
(b) The integralvolume energy equation (6.155) is applicable, i.e., Qku,r + Qr,r = −(ρcp V )t
dTr + S˙ m,F (t). dt
Here S˙ m,F (t) is given and is zero for t > τ . The surface convection is given by (6.124) as Qku,r =
Tr − Tf,∞ Rku D
Rku D =
,
D . Aku,r NuD kf
For NuD , we use parallel ﬂow with L = D, i.e., we use Table 6.3, and we need to determine the magnitude of ReD . ReD =
uf,∞ D . νf
From Table C.22, for air we have, νf = 1.566 × 10−5 m2 /s kf = 0.0267 W/mK
Table C.22
Pr = 0.69
Table C.22.
Table C.2
Then ReD =
22.22(m/s) × 0.35(m) = 4.966 × 105 ≤ ReD,t = 5 × 105 . 1.566 × 10−5 (m2 /s)
From Table 6.3, we use the laminarregime correlation, i.e., NuD
1/2
=
0.664ReD Pr1/3
= =
0.664 × (4.966 × 105 )1/2 × (0.69)1/3 413.5.
The surface areas for the surfaceconvection and surfaceradiation heat transfer are the two sides of the disc, i.e., Aku,r = Ar,r
=
2(πD2 /4)
=
0.1924 m2
588
(c) Rotor Temperature 320
Tr , K
314
Slow Decay
308 Surface ConvectionRadiation Cooling 302 Friction Energy Conversion and Energy Storage 296 Tr(t = 0) = 293.15, Initial Temperature 290 20
0 4
40
60
80
100
t, s Figure Pr.6.39(c) Time variation of the disc (rotor) temperature during and after brake.
The surfaceradiation heat transfer for Ar,∞ Ar,r and r,0 = 1.0, is given by (4.49), i.e., Qr,r
4 = Av,r r,r σSB (Tr4 − Ts,∞ )
=
0.1923(m2 ) × 0.4 × 5.67 × 10−8 (W/m2 K4 ) × [Tr4 (t) − (293.15)(K)].
From Table C.16 for carbon steel AISI 1010, we have, ρr
=
7830 kg/m2
cp,r kr
= =
434 J/kgK 64 W/mK .
Also Vr = πD2 l/4 = 1.443 × 10−3 m3 . Then S˙ m,F
= =
(22.22)2 (m/s)2 0.65 t(s) × 1500(kg) × 1− 2 4(s) 4s t(s) 6.017 × 104 (W) 1 − . 4(s)
(c) Using a software (such as SOPHT), Figure Pr.6.15(c) shows the time variation of Tr for 0 < t < 100 s. Note that during the friction heating, there is a very rapid increase in Tr and during this period, the surface convectionradiation heat transfer is not signiﬁcant. Using this assumption (Qku D Qr,r 0) the energy equation can be integrated to ﬁnd u2 0.65 t2 (ρcp V )t Ma a t − Tr (t) = Tr (t = 0) + (for t τ ). 2 τ 2τ which increases monotonically for t ≤ τ . From Figure Pr.6.15(c), note that even after 96 s of elapse time, Tr is still high. COMMENT: In order to examine the validity of the lumpedcapacitance approximation for the rotor we need to show that the Biot Number is very small (i.e., less than 0.1). From (6.130), we have BiD =
l kf Rk,l = NuD Rku D D ks
0.015(m) 0.0267(W/mK) × 0.35(m) 64(W/mK)
=
413.5 ×
=
7.393 × 10−3 < 0.1.
Note that we have used the thickness of the disk as the length for conduction. 589
PROBLEM 6.40.FAM GIVEN: A microprocessor chip generates Joule heating and needs to be cooled below a damage threshold temperature of 90◦C. The heat transfer is by surface convection from its top surface and by conduction through the printedcircuitboard substrate from its bottom surface. The surface convection from the top surface is due to air ﬂow from a fan that provides a parallel ﬂow with a velocity of uf,∞ . The conduction from the bottom surface is due to a temperature drop across the substrate of Tp − Ts . The substrate is fabricated from a phenolic composite and has a thermal conductivity of ks . Neglect the contact resistance between the processor and the substrate. Assume that the energy conversion occurs uniformly within the microprocessor chip. Neglect the edge heat losses. Assume the processor is at a uniform temperature Tp . S˙ e,J = 35 W, Tf,∞ = 25◦C, uf,∞ = 0.5 m/s, w = 7 cm, as = 1.5 mm, ks = 0.3 W/m2 K, L = 3.5 cm, l = 1 mm, Nf = 16. Evaluate the properties of aluminum at T = 300 K. Evaluate the properties of air at T = 300 K. SKETCH: Figures Pr.6.40(a) and (b) show the microprocessor cooled by surface convection.
(a) Pentium Pro Microprocessor Microprocessor
x
ixx
qu qku qk Microprocessor at Uniform Temperature, Tp
qk qk
Substrate (Printed Circuit Board)
qku
Ts
qu
(b) Two Different SurfaceConvection Designs
Tf, uf,
Tf, uf,
l w
Microprocessor
Aluminum Fins
Tp Se,J
L
as
w
w
Substrate, ks Ts
Ts
(i) No Fins
Se,J
No Contact Resistance Microprocessor, Tp
(ii) With Fins
Figure Pr.6.40 Surfaceconvection cooling of a microprocessor. (a) Physical Model. (b) Two diﬀerent
surfaceconvection designs (i) without, and (ii) with back ﬁns. OBJECTIVE: (a) Draw the thermal circuit diagram. 590
(b) Determine Tp for the case with no ﬁns and with Tp − Ts = 10◦C. (c) Determine Tp for the case with aluminum ﬁns and with Tp − Ts = 1◦C. (d) Comment on the diﬀerence between the two cases with respect to the damage threshold temperature. SOLUTION: (a) The thermal circuit diagram for both cases is shown in Figure Pr.6.40(b). . Se,J Rk,ps Ts
Rku
w
Qku
w
Tf,
Tp
Qk,ps
Figure Pr.6.40(b) Thermal circuit diagram.
Here Tp and S˙ e,J are uniform within the volume of the processor chip. Therefore, we will consider the processor as lumped and model it with a single node Tp . From this node, we have surfaceconvection heat transfer from the top to the air, and conduction heat transfer from the bottom through a substrate. The conduction causes a temperature drop across the substrate of Tp − Ts , which is given. For node Tp , for steady state conditions, we have from Figure Pr.6.40(b). Q A = Qk,ps + Qku w Tp − Ts Tp − Tf,∞ + Rk,ps Rku w
= S˙ e,J = S˙ e,J ,
where Rk,ps
=
Rku w
=
as 0.0015(m) as = = = 1.02◦C/W ks A ks (w × w) 0.3(W/mK) × (0.07)2 (m)2 w . Aku Nuw kf
For the case with the ﬁns, the ﬂuid ﬂow between the ﬁns is assumed to be a parallel ﬂow over a ﬂat plate. Then, since w and uf,∞ are the same for both cases, we can use the same Nuw in parts (b) and (c). From Table C.22 for air, at T = 300 K, we have νf = 15.66 × 10−6 m2 /s, kf = 0.0267 W/mK and Pr = 0.69. Then Rew is Rew =
uf,∞ w 0.5(m/s) × 0.07(m) = = 2235 < 5 × 105 , laminar ﬂow. νf 15.66 × 10−6 (m2 /s)
From Table 6.3, for parallel laminar ﬂow over a ﬂat plate, we have Nuw
1/3 = 0.664Re1/2 w Pr
Nuw
= 0.664 × (2235)1/2 × (0.69)1/3 = 27.74.
Then Nuw
0.0267(W/mK) kf = 27.74 × = 10.58 W/m2 K. w 0.07(m)
The integralvolume of energy equation (2.9) for node Tp , then becomes Tp − Ts Tp − T∞ + Rk,ps Rku,p∞ ◦ Tp − 25 C Tp − Ts + 1.02(◦C/W) 1/[Aku × 10.58(W/m2 ◦C)] 591
= S˙ e,J =
35 W.
(b) Case With No Fins, Tp − Ts = 10◦C For this case, Aku is the top surface of the plate, i.e., Aku = w × w = 0.0049 m2 . The energy equation then becomes Tp − 25(◦C) 10(◦C) + 1.02(◦C/W) 1/[0.0049(m2 ) × 10.58(W/m2 K)]
=
35 W.
Solving for Tp , we have Tp = 511◦C. (c) Case With Fins, Tp − Ts = 1◦C For this case, Aku is the eﬀective area of the ﬁns and the base area (over which surface convection occurs). Then Aku = (Ab + Nf Aku,f ηf ), where Ab = A − Nf Ak
= w × w − Nf × (w × l) = 0.0049(m2 ) − 16 × [0.07(m) × 0.001(m)] = 0.0049(m2 ) − 16 × [7 × 10−5 (m2 )] = 0.00378 m2
and Aku,f
= Pku,f × Lc = 2(w + l) × (L + l/2) = {2 × [0.07(m) + 0.001(m)]} × [0.035(m) + 0.001(m)/2] = 0.142(m) × 0.0355(m) = 0.005041 m2 .
To ﬁnd ηf we must ﬁrst ﬁnd the ﬁn parameter m. The ﬁns are fabricated from aluminum. From Table C.16, at T = 300 K, ksl = 237 W/mK. Then, m = =
Pku,f Nuw ksl Ak
kf w
1/2
0.142(m) × 10.58(W/m2 K) 237(W/mK) × 7 × 10−5 (m2 )
1/2
= 9.516 m−1 .
Then, from (6.147) and (6.149), we have ηf
=
tanh(0.3378) tanh(mLc ) tanh[9.516(m−1 ) × 0.0355(m)] = = 0.964. = −1 mLc 9.516(m ) × 0.0355(m) 0.3378
Aku = Ab + Af ηf
= Ab + Nf Aku,f ηf =
[0.00378(m2 ) + 16 × 0.005041(m2 ) × 0.964] = 0.0815 m2 .
The energy equation becomes Tp − Ts Tp − 25(◦C) + ◦ 1.02( C/W) 1/[Aku × 10.58(W/m2 K)] Tp − 25(◦C) 1(◦C) + ◦ 1.02( C/W) 1/[0.0815(m2 ) × 10.58(W/m2 K)]
=
35 W
=
35 W.
Solving for Tp , we obtain Tp = 64.5◦C. (d) For the case with the ﬁns, the calculated temperature is well above the damage threshold of Tp,max = 90◦C. With the ﬁns, there is a dramatic drop in the temperature to below this damage threshold temperature. This is due to the increased surface area, allowing for a much increased amount of surfaceconvection heat transfer to 592
the ﬂuid. Fins, or some equivalent heat transfer enhancement mechanism, are required for safe operation of this processor chip. COMMENT: Note that the ﬁn eﬀectiveness is Γf =
0.00378 + 0.08066 Ab + Af ηf = = 17.23. A 0.0049
This shows a very eﬀective ﬁn attachment.
593
PROBLEM 6.41.FAM GIVEN: To analyze the heat transfer aspects of the automobile rearwindow defroster, the window and the very thin resistive heating wires can be divided into identical segments. Each segment consists of an individual wire and an a × L × l volume of glass aﬀected by this individual heater. Each segment has a uniform, transient temperature T1 (t). This is shown in Figure Pr.6.41(a). In the absence of any surface phase change, the Joule heating results in a temperature rise, from the initial temperature T1 (t = 0) = −15◦C, and a surface heat loss to the surroundings. The surface heat loss to the surroundings is represented by a resistance Rt . The surrounding farﬁeld temperature is T∞ . T1 (t = 0) = −15◦C, T∞ = −15◦C, l = 3 mm, a = 2 cm, L = 1.5 m, S˙ e,J = 15 W, Rt = 2◦C/W. Determine the glass plate properties from Table C.17. SKETCH: Figure Pr.6.41(a) shows a unit cell on a glass window, where a thin resistive heater heats the glass volume around it. Very Thin Resistive Heater
Surroundings T , Rt
Glass
L l a
Se,J T1 (t) One Segment
Qt
Figure Pr.6.41(a) Thinﬁlm electric heaters on a glass surface.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Show that the lumpedcapacitance approximation using l for the conduction resistance. Assuming no surface phase change occurs, determine (c) the steadystate temperature of the glass, and (d) the glass temperature after an elapsed time t = 5 min. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.6.41(b). . Se,J Rt
Q1 = 0 T1(t)
Qt,1
T
Figure Pr.6.41(b) Thermal circuit diagram.
(b) To show validity of the lumped capacitance assumption, we must show Bi 1, or Bi < 0.1. For glass from Table C.17, at T = 293 K, we have ρ = 2710 kg/m3 , cp = 837 J/kgK, k = 0.76 W/mK, α = 0.34 × 10−6 m2 /s, and V = a × l × L = 9 × 10−5 m3 .
594
Then, Bil
=
Rk
=
Bil
=
Rk,l Rk 0.003(m) l l = = 0.1316◦C/W = kAk kaL (0.76)(W/mK) × (0.02)(m) × (1.5)(m) 0.1316 = 0.0658 1 lumped assumption is valid. 2
(c) The lumpedcapacitance analysis, with a single external resistance heat transfer, results in (6.156), i.e., T1 (t) = T∞ + [T1 (t = 0) − T∞ ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ), where τ1
=
a1
=
(ρcp V )Rt = [(2710)(kg/m3 )(837)(J/kgK)(9 × 10−5 )(m3 )] × (2)(◦C/W) = 408.29 s S˙ 1 − Q1 (15 − 0)(W) = = 0.0735◦C/s. 3 (ρcp V ) (2710)(kg/m )(837)(J/kgK)(9 × 10−5 )(m3 )
Then T1 (t) T1 (t)
= −15(◦C) + [(−15(◦C) + 15(◦C)] × e−t/408.29(s) + 0.0735(◦C/s) × 408.29(s)) × [1 − e−t/408.29(s) ] = −15(◦C) + 0.0 + 30(◦C)[1 − e−t/408.29(s) ].
As t → ∞, then e−t/τ1 → 0 and T1 (t → ∞) = −15(◦C) + 30(◦C)(1 − 0) = 15◦C = 288.15K. (d) At t = 5 min = 300 s, we have T1 (t = 300 s) = −15(◦C) + 30(◦C)[1 − e−300(s)/408.29(s) ] = 0.61◦C = 273.76K. COMMENT: Note that this heating rate is able to raise the glass temperature above 0◦C in 5 min. For faster response a higher heating rate is needed.
595
PROBLEM 6.42.FAM GIVEN: In particle spray surface coating using impingingmelting particles, prior to impingement the particles are mixed with a high temperature gas as they ﬂow through a nozzle. The time of ﬂight t (or similarly the nozzletosurface distance) is chosen such that upon arrival at the surface the particles are heated (i.e., their temperature is raised) close to their melting temperature. This is shown in Figure Pr.6.42(a). The relative velocity of the particlegas, which is used in the determination of the Nusselt number, is ∆up . Consider lead particles of diameter D ﬂown in an air stream of Tf,∞ . Assume that the particles are heated from the initial temperature of T1 (t = 0) to the melting temperature Tsl with surfaceconvection heat transfer only (neglect radiation heat transfer). T1 (t = 0) = 20◦C, Tf,∞ = 1,500 K, D = 200 µm, ∆up = 50 m/s. Determine the air properties at T = 1,500 K (Table C.22), and the lead properties at T = 300 K (Table C.16). SKETCH: Figure Pr.6.42(a) shows the solid particles entrained in hot gas then after surfaceconvection heating arriving at the substrate for deposition.
Gas o
T1(t = 0) = 20 C
T1(t) = Tsl
Particles Spray Used for Coating Tf, uf,= ∆up
Moving Surface
Figure Pr.6.42(a) A particle spray coating surfacecoating process using
impingingmelting particles.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the Biot number BiD , based on the particle diameter D. Can the particles be treated as lumped capacitance? (c) Determine the time of ﬂight t needed to reach the melting temperature Tsl . SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.6.42(b).
DQkuED
T1
Tf,
dT1 − (ρcV)1 dt
DRkuED
Figure Pr.6.42(b) Thermal circuit diagram.
(b) From Table C.16 for lead at 300 K, we have ρs = 11,340 kg/m3 , cp,s = 129 J/kgK, ks = 35.3 W/mK, αs = 24.1 × 10−6 m2 /s, and Tsl = 601 K. From Table C.22 for air at 1500 K, we have ρf = 0.235 kg/m3 , cp,f = 1202 J/kgK, kf = 0.0870 W/mK, νf = 229 × 10−6 m2 /s, and Pr = 0.7. 596
The Biot number based on D is deﬁned as Rk,s Rk,u D kf kf = . NuD = NuD 4ks D 4ks
BiD
=
To determine the NuD , we ﬁrst must determine the ReD , given as ReD
∆up D 50(m/s) × (200 × 10−6 )(m) = = 43.67. νf 229 × 10−6 (m2 /s)
=
Then from Table 6.4 for a sphere we have NuD
1/2
2/3
=
2 + (0.4ReD + 0.06ReD )Pr0.4
=
2 + [0.4(43.67)1/2 + 0.06(43.67)2/3 ](0.7)0.4 = 4.937.
Then substituting in to the above, the Biot number is BiD
kf 4ks 0.0870(W/mK) = 3.042 × 10−3 . (4.937) 4 × 35.3(W/mK)
= NuD =
(c) Since the BiD 1, we can analyze the lead droplets as lumped capacitance systems. Applying conservation of energy around the droplet gives Qku =
dT1 Tf,∞ − T1 . = −(ρcp V )1 Rku D dt
The droplet is a lumped system with a single resistive heat transfer. The solution to this is given by (6.156) as T1 (t) = Tf,∞ + [T1 (t = 0) − Tf,∞ ]e−t/τ1 + a1 τ1 (1 − e−t/τ1 ) where τ1 = (ρcp V )1 Rku D ,
a1 =
S˙ 1 − Q1 . (ρcp V )1
For this case, S˙ 1 = 0 and Q1 = 0, therefore a1 = 0. Then τ1
= = = = =
(ρcp V )1 Rku D D (ρcp V )1 Aku NuD kf 2 V1 D (ρcp )1 D (ρcp )1 = NuD kf Aku NuD kf 6 11,340(kg/m3 ) × 129(J/kgK) × (200 × 10−6 )2 (m2 ) 4.937 × 0.0870(W/mK) × 6 0.02270 s.
Now, solving for t, for T1 = Tsl = 601 K, we have Tsl
= Tf,∞ + [T1 (t = 0) − Tf,∞ ]e−t/τ1
Tsl − Tf,∞ T1 (t = 0) − Tf,∞
= e−t/τ1
t
= −τ1 ln
Tsl − Tf,∞ T1 (t = 0) − Tf,∞ (601 − 1,500)(K) = −0.02270(s) × ln = 6.685 × 10−3 s. (293.15 − 1,500)(K)
COMMENT: Note that here about onethird of a time constant is needed to reach the desired particle temperature. 597
PROBLEM 6.43.FAM GIVEN: A rectangular (square cross section) metal workpiece undergoing grinding, shown in Figure Pr.6.43(a), heats up and it is determined that a surfaceconvection cooling is needed. The fraction of the energy converted by friction heating S˙ m,F , that results in this heating of the workpiece, is a1 . This energy is then removed from the top of the workpiece by surface convection. A single, round impinging air jet is used. Assume steadystate heat transfer and a uniform workpiece temperature Ts . S˙ m,F = 3,000 W, a1 = 0.7, Tf,∞ = 35◦C, uf = 30 m/s, D = 1.5 cm, L = 15 cm, Ln = 5 cm. Evaluate properties of air at T = 300 K. SKETCH: Figure Pr.6.43(a) shows the workpiece and surface convection cooling. Jet Exit Conditions: Tf, , uf
D Ln
Aku = 2L x 2L Grinding Belt
Workpiece, Ts L
l Sm,F
Figure Pr.6.43(a) Grinding of a metal workpiece.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the workpiece temperature Ts . (c) What should the ratio of the workpiece thickness l to its conductivity ks be for the uniform temperature assumption to be valid? SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.6.43(b). DQkuEL
Ts
Tf,
Qs . DRkuEL Sm,F Figure Pr.6.43(b) Thermal circuit diagram.
(b) Applying the conservation of energy equation to the boundary node Ts at the interface of the workpiece and the grinder belt, and noting steadystate, we have QA = Qku L + Qs = S˙ m,F . It is given that the fraction of energy conversion by friction heating S˙ m,F that results in heating of the workpiece is a1 . This is the same energy that must be removed for steadystate conditions to exist. So we then have Qku L Ts − Tf,∞ Rku L
= a1 S˙ m,F = a1 S˙ m,F , 598
where Rku L =
1 Aku NuL
kf L
.
Solving for NuL , we note that we have a single, round nozzle, impinging jet. Therefore, from Table 6.3, NuL =
1/2 2ReD Pr0.42 (1
D 1 − 1.1 1/2 L . + 0.005Re0.55 D ) D Ln −6 1 + 0.1 D L
From Table C.22 at T = 300 K, νf = 15.66 × 10−6 m2 /s, kf = 0.0267 W/mK, and Pr = 0.69. Then ReD
=
NuL
= =
uf D 30(m/s) × 0.015(m) = νf 15.66 × 10−6 (m2 /s) 2.874 × 104 2 × (2.874 × 104 )1/2 × (0.69)0.42 × [1 + 0.005(2.874 × 104 )0.55 ]1/2 0.015(m) 1 − 1.1 × 0.15(m) × 0.015(m) 0.05(m) 1 + 0.1 −6 0.015(m) 0.15(m) 412.3.
=
Then solving for Rku L , we have Aku Rku L
=
2L × 2L = 4L2 = 4 × (0.15)2 (m)
=
0.09 m2
=
2
1 0.0267(W/mK) 0.09(m ) × 412.3 × 0.15(m)
= 0.1514◦C/W.
2
From the conservation of energy equation, Ts is Ts
= Tf,∞ + a1 Rku L S˙ m,F = 35(◦C) + (0.7)[0.1514(◦C/W)][3000(W)] 352.9◦C = 626.1 K.
=
(c) The lumped assumption is valid when Bi < 0.1. From (6.130) and noting for conduction across the thickness of the workpiece that Ak = Aku , we have BiL
=
Rk,s Rk Ak = = Rku L Rku L Aku
l ks
< 0.1
l ks
1 NuL kf /L
−1 < 0.1.
Solving for l/ks we have L 0.15(m) , = 0.1 × NuL kf 412.3 × 0.0267(W/mK)
or l ks
< 1.36 × 10−3 ◦C/(W/m2 ).
COMMENT: This l/ks can be easily achieved for metals (ks > 10 W/mK). 599
PROBLEM 6.44.FUN GIVEN: A microprocessor with the Joule heating S˙ e,J is cooled by surface convection for one of its surfaces. An oﬀtheshelf surface attachment is added to this surface and has a total of Nf squarecrosssectional aluminum pin ﬁns attached to it, as shown in Figure Pr.6.44. Air is blown over the ﬁns and we assume that the Nusselt number can be approximated using the farﬁeld air velocity uf,∞ and a cross ﬂow over each squarecrosssectional cylinder ﬁn (i.e., the Nusselt number is not aﬀected by the presence of the neighboring ﬁns). This is only a rough approximation. Tf,∞ = 35◦C, uf,∞ = 2 m/s, S˙ e,J = 50 W, D = 2 mm, a = 5 cm, Nf = 121, L = 2 cm. Evaluate the air and aluminum properties at T = 300 K. Assume that the NuD correlation of Table 6.3 is varied. SKETCH: Figure Pr.6.44(a) shows the extended surface. Tf, uf,
Fins (Square Cross Section) D
L
D
Ts
Microprocessor
Fastener
Se,J
a
a
Figure Pr.6.44(a) A microprocessor with the Joule heating and a surfaceconvection cooling. There is an attached extended surface for reduction of the microprocessor temperature.
OBJECTIVE: (a) Draw the thermal circuit diagram. (b) Determine the ﬁn eﬃciency. (c) Determine the steadystate surface temperature Ts . (d) Determine the ﬁn eﬀectiveness. SOLUTION: (a) The thermal circuit diagram is shown in Figure Pr.6.44(b). The steadystate, uniform surface temperature is Ts . Qku,s Tf,
Qs = 0 Se,J
Ts
Rku,s
uf,
Figure Pr.6.44(b) Thermal circuit model.
(b) The ﬁn eﬃciency is given by (6.147) as ηf m
tanh(mLc ) mLc 1/2 Pku,f NuD kf = . Ak ks D =
600
We use Table 6.3 for NuD , where D is the side length for the square cross section cylinder. Here D , Ak = D2 , 4 where we used (6.141) and a similarity to circular pin ﬁns. From Table C.14, for aluminum, ks = 237 W/mK. From Table C.22, for air at T = 300 K, we have Pku,f = 4D,
Lc = L +
air: νf = 1.566 × 10−5 m2 /s
Table C.22
kf = 0.0267 W/mK Pr = 0.69
Table C.22 Table C.22.
Then ReD
=
uf,∞ D 2(m/s) × 2 × 10−3 (m) = = 255.4 νf 1.566 × 10−5 (m2 /s)
This is outside the range of ReD given in Table C.6.3, however, for lack of an alternative we will use the available results, i.e., NuD
= a1 Rea2 Pr1/3 = =
0.102Re0.675 Pr1/3 D 0.102 × (255.4)0.675 × (0.69)1/3
=
3.800.
Then
m
=
4NuD kf D2 ks
1/2
=
4 × 3.800 0.0267(W/mK) −3 2 2 (2 × 10 ) (m ) 237(W/mK)
=
20.69(1/m) × 0.0205(m) = 0.4241.
1/2 = 20.69 1/m
Lc = 0.02(m) + 0.002/4(m) = 0.0205 m mLc Next, interpolating from Table 6.6, we have ηf =
0.3998 tanh(mLc ) tanh(0.4241) = = 0.9426. = mLc 0.4241 0.4241
(c) The energy equation for the microprocessor volume is written, using Figure Pr.6.28(b), as Qku,s∞ = S˙ e,J . From (6.149), we have Qku,s∞ = (Ab + Af ηf )NuD
kf (Ts − Tf,∞ ), D
where we have used NuD for the base and the ﬁn surfaces. Here Ab
= a2 − Nf D2
Af
= (0.05)2 (m2 ) − 121 × (0.002)2 (m2 ) = 2.016 × 10−3 m2 = Nf × 4DLc =
4 × 121 × 0.002(m) × (0.0205)(m) = 1.984 × 10−2 m2 .
Then S˙ e,J = Qku,s∞ = (2.016 × 10−3 + 1.984 × 10−2 × 0.9426)(m2 )× 0.0267(W/mK) × (Ts − Tf,∞ ) 0.002(m) 50(W) = 1.051(W/◦C)[Ts − 35(◦C)] Ts = 82.57◦C 3.800 ×
601
(d) The ﬁn eﬀectiveness is deﬁned in Section 6.8.2 as Γf
= =
Ab + Af ηf A (2.016 × 10−3 + 1.984 × 10−2 × 0.9426)(m2 ) = 8.287. (0.05)2 (m2 )
COMMENT: A ﬁn eﬀectiveness of Γf = 8.287 is high enough to allow for maintaining the microprocessor at a temperature below the damage threshold (which is around 100◦C). Also note that we have used a NuD correlation that is a only an approximation for the collection of the ﬁns used here. A more accurate value of tanh(0.4241) = 0.4004 can be obtained from most pocket calculators.
602
PROBLEM 6.45.FUN GIVEN: Desiccants (such as silica gel) are porous solids that adsorb moisture (water vapor) on their large interstitial surface areas. The adsorption of vapor on the surface results in formation of an adsorbed water layer. This is similar to condensation and results in liberation of energy. The heat of adsorption, similar to the heat of condensation, is negative and is substantial. Therefore, during adsorption the desiccant heats up. The heat of adsorption for some porous solids is given in Table C.5(b). Consider a desiccant in the form of pellets and as an idealization consider a spherical pellet of diameter D in a mistair stream with farﬁeld conditions Tf,∞ and uf,∞ . Assume that the released energy is constant. S˙ 1 = S˙ ad = ∆had ρad V /to , D = 5 mm, ρad = 200 kg/m3 , T1 (t = 0) = 10◦C, Tf,∞ = 10◦C, (ρcp )1 = 106 J/m3 K, uf,∞ = 3 cm/s, ∆had = 3.2 × 106 J/kg, to = 1 hr. Evaluate properties of air at T = 300 K. SKETCH: Figure Pr.6.45(a) shows the desiccant pellet (porous zeolite) in cross, moistair ﬂow. Desiccant (Porous Zeolite)
Air D
WaterVapor
Tf, uf,
T1(t)
Had
T(t = 0)
Sad Heat Release Due to WaterVapor Adsorption
Figure Pr.6.45(a) A desiccant pellet in a cross, moist air ﬂow.
OBJECTIVE: (a) Draw the thermal circuit diagram for the pellet. (b) Determine the pellet temperature after an elapsed time to . SOLUTION: (a) Figure Pr.6.45(b) shows the thermal circuit diagram.
 (HcpV )1 . . S1 = Sad
dT1 dt
Rku
D
Qku
D
Tf,
T1(t)
Figure Pr.6.45(b) Thermal circuit diagram.
(b)The temperature of the pellet is given by (6.156), i.e., T1 (t) τ1
= Tf,∞ + [T1 (t = 0) − Tf,∞ ]e−t/τ1 + a1 τ1 (1 − et/τ1 ) S˙ 1 − Q1 πD3 . = (ρcp V )1 Rku D , a1 = , V1 = (ρcp V )1 6
The average surfaceconvection resistance Rku D is found from (6.124), i.e., Rku D =
D , Aku NuD kf 603
Aku = πD2 .
The Nusselt number is found from Table 6.4, i.e., 1/2
2/3
NuD = 2 + (0.4ReD + 0.06ReD )Pr0.4 , ReD =
uf,∞ D . νf
From Table C.22, at T = 300K, we have for air kf = 0.0267 W/mK −5
νf = 1.566 × 10
Table C.22 2
m /s
Table C.22
Pr = 0.69. Then ReD
=
0.03(m/s) × 5 × 10−3 (m) = 9.5785 1.566 × 10−5 (m2 /s)
NuD
=
2 + [0.4(9.5785)1/2 + 0.06(9.5785)2/3 ](0.69)0.4 = 3.300
Rku D
=
τ1 S˙ 1 a1 T1 (t = to = 1 hr)
5 × 10−3 (m) = 722.4◦C/W π × (5 × 10−3 )2 (m2 ) × 3.300 × 0.0267(W/mK) π = 106 (J/m3 K) × (5 × 10−3 )3 (m3 ) × 722.4(◦C/W) = 47.28 s 6 6 (J/kg) × 200(kg/m3 ) × π × (5 × 10−3 )3 /6(m3 ) 3.2 × 10 = 1.164 × 10−2 W = S˙ ad = 3,600(s) =
1.164 × 10−2 (W) = 0.1778 × 10−2 ◦C/s 106 (J/m3 K) × π × (5 × 10−3 )3 /6(m3 )
= 10(◦C) + 0.1778(◦C/s) × 47.28(s) × [1 − e−3,600(s)/47.28(s) ] = 18.41◦C.
COMMENT: Since the vapor is slow in diﬀusing into the porous pellet, the energy release rate is rather low. Also since the time constant τ1 , is much less than the elapsed time of interest, the above T1 is the steadystate temperature during the heat release period to .
604
PROBLEM 6.46.FUN.S GIVEN: Consider the concept of the critical radius discussed in Example 6.13. An electricalcurrent conducting wire is electrically insulated using a Teﬂon layer wrapping, as shown in Figure Pr.6.46(a). Air ﬂows over the wire insulation and removes the Joule heating. The thermal circuit diagram is also shown. L = 1 m, R1 = 3 mm, uf,∞ = 0.5 m/s. Evaluate the air properties at T = 300 K. Thermal conductivity of Teﬂon is given in Table C.17. SKETCH: Figure Pr.6.46(a) shows the insulated wire and the thermal circuit diagram.
(i) Physical Model uf, Tf,
(ii) Thermal Circuit Model
L
Qku
Qk,12
D,2
Qu
Se,J Tw
R2
Rk,12
T2
Rku
D,2
Tf,
R1
T2 Tw
Electrical Conductor
Se,J
Electrical Insulator (Teflon)
Figure Pr.6.46(a)(i) An electricalcurrent carrying wire is electrically insulated with a Teﬂon layer wrapping. (ii) Thermal circuit diagram.
OBJECTIVE: (a) Plot the variation of RΣ = Rk,12 + Rku D,2 with respect to R2 , for R1 ≤ R2 ≤ 3R1 . (b) Determine R2 = Rc (where RΣ is minimum). (c) Show the contributions due to Rk,12 and Rku D,2 at R2 = Rc . (d) Also determine Rc from the expression given in Example 6.13, i.e., Rc =
a2 ks a2 −1 2 aR
1/a2 .
SOLUTION: (a) The total resistance to the heat ﬂow is given in Example 6.13 as RΣ =
ln(R2 /R1 ) 1 + . 2πLks πLNuD,2 kf
From Table C.17, we have for Teﬂon ks = 0.26 W/mK
Table C.17.
From Table C.22 for air, at T = 300 K, we have kf = 0.0267 W/mK −5
νf = 1.566 × 10 Pr = 0.69
Table C.22 2
m /s
Table C.22 Table C.22.
605
The Reynolds number ReD,2 and NuD,2 are given in Table 6.3 as
ReD,2
= = =
NuD,2
uf,∞ 2R2 νf 0.5(m/s) × 2 × R2 1.566 × 10−5 (m2 /s) 6.386 × 104 (1/m) × R2
2 = a1 ReaD,2 Pr1/3 .
Since R2 ≥ R1 , we begin from R2 = R1 . Then ReD,2
=
6.386 × 104 (1/m) × 3 × 10−3 (m)
=
191.6.
For R2 = 3R1 , ReD,2 = 574.7. From Table 6.3, we have for 191.6 < ReD,2 < 574.7, a1 = 0.683
a2 = 0.466
Table 6.3.
Figure Pr.6.46(b) shows the variation of RΣ with respect to R2 for R1 ≤ R2 ≤ 3R1 . The numerical values are obtained and plotted using a solver (such as SOPHT).
(b) Minimum in Total Resistance 1.75
RΣ , K/W
1.73 1.71 1.69 1.67
Rc = 5.259 mm
1.65 3
4
5
6
7
8
9
R2 , mm
R1
Figure Pr.6.46(b) Variation of RΣ with respect to R2 for R1 ≤ R2 ≤ 3R1 .
(b) The minimum in RΣ occurs at R2 = Rc = 5.259 mm. (c) The value of the two resistances at R2 = Rc are Rk,12
=
0.3436◦C/W
Rku D,2
=
1.314◦C/W.
Here the surfaceconvection resistance is much larger.
606
(d) The result of Example 6.13 for Rc for the above values for a1 , a2 , etc, is
Rc aR
1 a2 a2 ks a2 −1 2 aR a2 uf,∞ = kf a1 a2 Pr1/3 νf =
=
0.0267(W/mK) × 0.683 ×
(0.5)0.466 (m/s)0.466 × (0.69)1/3 (1.566 × 10−5 )0.466 (m2 /s)0.466
2.024(W/mK)/m0.466 1 0.466 0.466 × 0.26(W/mK) = 0.466−1 0.466 2 × 2.027(W/mK)/m =
Rc
= =
(0.08655)2.146 (m) 0.005259 m
=
5.259 mm.
As expected, this is equal to the numerical/graphical result of (b). COMMENT: From Figure Pr.6.46(b), note that RΣ is rather independent of R2 near Rc (nearly ﬂat). Therefore a range of R2 can be used with a nearly equal RΣ .
607
PROBLEM 6.47.FUN GIVEN: In designing ﬁns, from (6.149) we note that a combination of high ﬁn surface area Af and high ﬁn eﬃciency ηf are desirable. Therefore, while high ηf (ηf → 1) is desirable, ηf decreases as Af increases. From (6.149) the case of ηf → 1 corresponds to Biw → 0. Note that tanh(z) =
ez − e−z ez + e−z sinh(z) z2 , sinh(z) = , cosh(z) = , ex = 1 + z + + ... . cosh(z) 2 2 2
OBJECTIVE: Show that in the limit of mLc → 0, the ﬁn eﬃciency tends to unity. SOLUTION: We begin with tanh(z)
=
sinh(z)
=
sinh(z) , cosh(z) ez + e−z ez − e−z , cosh(z) = . 2 2
Then tanh(z) =
ez − e−z . ez + e−z
Next we expand ez using a Taylor series as ez
=
e−z
=
z3 z2 + + .... 2! 3! z3 z2 − ..... 1−z+ 2! 3! 1+z+
Using these, we have z3 z2 + .. − 1 + z − 2! 3! z3 z2 + .. + 1 − z + 1+z+ 2! 3! 2z 3 2z + + .... 3! . 2z 2 + .... 2+ 2! 1+z+
tanh(z) =
=
z2 z3 + − ... 2! 3! 2 3 z z − + ... 2! 3!
We are interested in the ﬁn eﬃciency, given by (6.147), ηf =
tanh(z) , z = mLc = Bi1/2 w . z
and the limit of z = Bi1/2 w → 0. Then with tanh(z) from above, tanh(z) lim (ηf ) = lim z→0 z→0 z 2z 2 2 + 3! + ... = lim