ASEN 3113 Thermodynamics and Heat Transfer
Homework #5
Assigned: October 26, 2006
Due: November 2, 2006 (class time)
(Total points noted in each section; must clearly show equations with values and units, drawings,
assumptions, etc.)
1. (20 points)
Heat is transferred between two large parallel plates a distance L = 3 cm apart, at constant
temperatures T1= 320K and T2. The surfaces can be assume to be black (ε = 1). The rate of heat
conduction between the plates is 160 W. Determine a) The temperature T2 and the total rate of
heat transfer per unit area if the gap between the plates is filled with air, b) the total rate of heat
transfer if the gap is evacuated, c) the total rate of heat transfer if the gap is filled with fiber glass
insulation, and d) the thermal conductivity of an insulator to achieve a heat conductivity is 2.0W.
(Disregard any natural convection currents)
Soln:
a) The temperature at T2:
Q L
T  T2
(160W )(0.03m)
Q cond  kA 1
 T2  T1  cond  320 
 104.75K
L
kA
(0.0223W / mC )(1m 2 )
Q rad  A(T14  T24 )  1(5.67 x10 8 W / m 2 K 4 )(1m 2 )[(320 K ) 4  104.75K ) 4  587.72W
Q total  Q cond  Q rad  160W  587.72W  742.7W
b) When the air is evacuated there is only radiative heat transfer, therefore:
Q total  Q rad  587.72W
c) In this case there will be conduction heat transfer through the fiberglass only.
d)
T  T2
(320  104.75) K
Q total  Q cond  kA 1
 (0.046W / mC )(1m 2 )
 330.05W
L
0.03
Q cond L
T  T2
Q cond  kA 1
k 
 0.000279W / mC
L
A(T1  T2 )
2. (20 points)
A refrigerator has dimensions of 6 ft x 4 ft x 3 ft with walls 1 in thick. It consumes 600 W of
power when operating and has a COP of 2.5. The motor remains on for five minutes and off for
15 minutes per cycle. The average temperatures of the inner and outer surfaces are 50 F and
respectively. Determine the average thermal conductivity of the refrigerator walls. What is the
annual cost of operating this refrigerator if the price of electricity is $0.06/kWh.
Soln:
Cross section of the refrigerator is,
Atotal  2[(1.8  1.2)  (1.8  0.8)  (1.2  0.8)]  9.12m 2
The refrigerator has a COP pf 2.5 therefore:
Q  600  2.5  1500W
The refrigerator operates a quarter of the time, therefore;
Q ave  Q / 4  375W
The thermal conductivity of the refrigerator walls is:
T
Q L
Q  kA
k 
 0.112W / mC
L
AT
The number of hours in a year that this refrigerator operates is
Δt= = 365 x 24/4 = 2190 h
The total amount of electricity consumed per year:
Q  Q t  (0.6kW )(2190h)  1314kWh  cos t  (1314kWh)($0.06 / kWh)  $78.84
3. (20 points, 15 points for equations, 2/3 points for correct answer (a)/(b))
Consider a 1.5 m high and 2 m wide glass window whose thickness is 6 mm and which has a
thermal conductivity of k = 0.78 W/(m* ºC). Determine (a)the steady rate of heat transfer
through this glass window and (b) the temperature of its inner surface for a day during which the
room is maintained at 24 ºC while the temperature of the outdoors is -5 ºC. Take the convection
heat transfer coefficients on the inner and outer surfaces of the window to be h1 = 10 W/(m2* ºC)
and h2 = 25 W/(m2* ºC), and disregard any heat transfer by radiation.
Solution:
A  1.5m   2m   3.0m 2
Rinside  Rconv,1 
1

h1 A 10 W

1
m 2 C
3.0m 
2
 0.0333 C
W
L
0.006m

 0.00256 C
W
k1 A 0.78W
3.0m 2
mC
1
1
Routside  Rconv, 2 

 0.0133 C
2
W
h2 A 25W 2
3.0m
m C
R glass 






RTotal  Rconv,1  R glass  Rconv, 2  0.0333  0.00256  0.0133C
RTotal  0.0492 C
W
W
The steady rate of heat transfer through the window glass is
T  T 2 24   5C
Q  1

 589W
RTotal
0.0492 C
W
The inner surface temperature of the window glass can be determined from
T T
Q  1 1  T1  T1  Q Rconv,1  24C  589W  0.0333 C
 4.38C
W
Rconv,1


4. (20 points, 15 points for equations, 5 points for correct answer)
Problem 8-80 in Cengal, See Figure P8-80
A 4-m high and 6-m wide wall consists of long 18 cm X 30 cm cross-section horizontal bricks
[k = 0.72 W/(m* ºC)] separated by 3-cm-thick plaster layers [k = 0.22 W/(m* ºC)]. There are
also 2-cm-thick plaster layers on each side of the wall, and 2 cm thick rigid foam
[k = 0.026 W/(m* ºC)] on the inner side of the wall. The indoor and the outdoor temperatures
are 22 ºC and -4 ºC, and the convection heat transfer coefficients on the inner and the outer sides
are h1 = 10 W/(m2* ºC) and h2 = 20 W/(m2* ºC), respectively. Assume one-dimensional heat
transfer and disregarding radiation, determine the rate of heat transfer through the wall.
Solution
Rinside  Rconv,1 
1

h1 A 10 W
1



 0.303 C
W
0.33m
m 2 C
L
0.02m
R1  R foam 

 2.33 C
2
W
kA 0.026 W
0.33m
mC
L
0.02m
R2  R6  R plaster _ side 

 0.303 C
2
W
kA 0.22 W
0.30m
mC
L
0.18m
R3  R5  R plaster _ center 

 54.55 C
2
W
kA 0.22 W
0.015m
mC
L
0.18m
R4  Rbrick 

 0.833 C
2
W
kA 0.72W
0.30m
mC
1
1
Routside  Rconv, 2 

 0.152 C
2
W
W
h2 A 20
0.33m
m 2 C
1
1
1
1
1
1
1






 0.808 C
W
Rmid R3 R4 R5 54.55 0.833 54.55

2














RTotal  Rinside  R1  2 R2  Rmid  Routside  0.303  2.33  0.606  0.808  0.152 C
RTotal  4.20 C
W
W
The steady rate of heat transfer through the wall per 0.33 m2 is
T  T 2 22   4C
Q  1

 6.19W
RTotal
1.20 C
W
The steady rate of heat transfer through the entire wall becomes
4  6m 2  450W
Q Total  6.19W 
0.33m 2
5. (20 points, 15 points for equations, 2/3 points for correct answer (a)/(b))
A 6-m internal diameter spherical tank made of 1.5 cm-thick stainless steel [k = 15 W/(m* ºC)] is
used to store iced water at 0 ºC. The tank is located in a room whose temperature is 20 ºC. The
walls of the room are also at 20 ºC. The outer surface of the tank is black (emissivity, ε=1), and
heat transfer between the outer surface of the tank and the surroundings is by natural convection
and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the
tank are 80 W/(m2* ºC) and 10 W/(m2* ºC), respectively. Determine (a) the rate of heat transfer
to the iced water in the tank and (b) the amount of ice at 0 ºC that melts during a 24-h period.
The heat of fusion of water at atmospheric pressure is hif = 333.7 kJ/kg.
Solution
The inner and outer surface areas of the sphere are:
2
2
Ainner  Dinner
  6m   113.1m 2
2
Aouter  Douter
  6.03m   114.2m 2
2
We assume the outer surface temperature T2 to be 4 ºC after comparing convection heat transfer
coefficients at the inner and outer surfaces of the tank. With this assumption, the radiation heat
transfer coefficient can be determined from:


2
T2  Tsurr 
hrad   T22  Tsurr

hrad  1 5.67  10 8 W
hrad  5.25W
273  4K   273  20K  273  20K   273  4K 
m K
2
2
2
4
m2 K
The individual thermal resistances are
1
1
Rconv,inside 

 0.000110 C
2
W
hinside A 80 W 2
113.1m
m C
r r
3.015  3m
R1  Rsphere  2 1 
 0.0000088 C
W
4kr1 r2 4 15W

3.015m 3.0m 
mC
1
1
Routside 

 0.000876 C
W
houtside A 10 W 2
114.2m 2
m C
1
1
Rrad 

 0.00167 C
2
W
hrad A 5.25W 2
114.2m
m C
1
1
1
1
1




 0.000575 C
W
Reqv Rconv,outside Rrad 0.000876 0.00167











RTotal  Rconv,inside  R1  Reqv  0.000110  0.0000088  0.000575C
RTotal  0.000694 C
W
W
(a) Then the steady rate of heat transfer to the iced water becomes
T  T 2
20  0C  28,818W
Q  1

RTotal
0.000694 C
W
(b) The total amount of heat transfer during a 24 hr period and the amount of ice which will melt
during this period are

mice

24  3600s   2,489,900kJ
s
Q 2,489,900kJ


 7461kg
hif
333.7 kJ kg
Q  Q t  28.818 kJ