ME 407 – Heat
Transfer – Homework Answers 
Week 1 Answers
1-22a. Q = 0.12 KWh
1-22b. q = 400 W/m2
1-43. T2 = 74.3 oF
1-68. Q = 78, 624 kJ, if thickness of glass is one
centimeter Q=39312 kJ
1-115. Qstillair = 336W, Qwindyair =
1120 W, Fwindchill = 32.7 oC
Week 2 Answers
2-42. a) Steady b) 1D c)heat generation d)thermal
conductivity is variable
2-62. derivation
2-73. Qwall=6031W
Week 3 Answers
3-50. Qtotal = 450 W
3-65.
Q = 93.9W, DTpipe=0.095oC, DTinsulation=290oC
3-82. T1 = 39.4oC, rcr= 8.3mm
3-70. Q = 95.58 btu/hr %error = 0.035%
Week 4 Answers
3-130. Qtotalfin = 17407W, efin= 7.10
3-132 a)T2 = 174.7oC b)Qfinned= 219W hfin = 0.88, c)equivalent
length = 16.7 cm
3-137 Q = 111Btu/h
Week 5 Answers
4-15. t = 38.5s
4-16. a) T = 167oF b)Qtotal = 984.7
Btu/min
Week 6 Answers
4-36. T0=227 oF
4-47. T(r0,t) = -5.2oC Frozen Oranges
4-61. a)T(0,0,t) = 76.9oC b)T(L,0,t) = 75.4oC
c)Q = 185.0 KJ
Week 7 Answers
5-21. finite
difference equations
5-26. b)T1
= 63.92oC, T2 = 47.84oC T3 = 31.75 oC, T4 =
15.67oC C)Qwall=322 W
5-48. T1=T3=T7=T9=185oC, T2=T4=T5=T6=T8=190oC
Week 8 Answers
5-72. Finite difference equations
5-85. T(10 minutes) = 68.9oC, Tsteadystate
found using computer methods after 540 minutes
Week 9 Answers
6-7 a)Nu = hL/k b,c) Nu = hD/k d) Dh=2ab/(a+b)
6-19. Fd= 81.3N, Q = 20,498W
6-29. Ts = 77.9 oF
Week 10 Answers
6-49. Qconv= 450W
6-51. Ts = 132pC
6-77. Te=75.6oC, Q = 946.2W
6-79. a) Te = 9.7oC b) Q = 33964W c) Wpump=
4368W
Week 11 Answers
7-24b. Ts = 42 oC
7-30. Ts = 140 oF cost = $566.80
7-62. Vinfinity = 11 m/s
Week 12 Answers
9-21. T
= 2700K
9-32. a)
e = 0.155 a=0.155, r=0.845 b) e = 0.246 a=0.246, r=0.754
Week 13 Answers
9-49. qnet = 144 W/m2
9-58. F13 =F31 = 0.24, F23
= F32 = 0.05
9-74. a) Q31 = 3,344,970 Btu/hr b)Q12
= 76,420 Btu/hr Q1 = 3,268,540 Btu/hr
9-80. Q21 = -Q12 = 127,355 Btu/hr
9-93. Q12,1shield = 1857W/m2 Q12, noshield =12,035W/m2
Week 14 Answers
10-19. R = 4055W/m2-oC
10-46. Ai = 0.49 m2
10.76. Tc,out = 70.4 oC, Th, out = 77.4 oC
10.77. A = 52.4 m2
10.82. Q = 31350W,
NTU = 0.438 A = 0.482 m2