SUMMER 2022 – 03.08.2022
GMU 351 – HEAT TRANSFER
STEADY STATE HEAT CONDUCTION - PROBLEM SET
SOLUTIONS
1. The inner and outer surfaces of a 5-m x 8-m brick wall of thickness 25 cm and thermal conductivity
0.71 W/m.K are maintained at temperatures of 24 °C and 5 °C, respectively. Determine the rate of
heat transfer through the wall, in W.
qk A
T
L
q  (0.71W m.K) (5m  8m)
(24  5) K
 2158.4W
(0.25m)
2. The inner and outer surfaces of a 0.7-cm-thick 3-m x 3-m window glass in winter are 10 °C and 2
°C, respectively. If the thermal conductivity of the glass is 0.69 W/m.K, determine the amount of heat
loss, in kJ, through the glass over a period of 8 hours. What would your answer be if the glass were 2
cm thick?
qk A
T
L
a) q  (0.69W m.K) (3m  3m)
(10  2) K
 7097.1W
(0.007m)
Amount of heat loss over a period of 8 hours :
7097.1
J 1 kJ
(8  3600 s)  204396 kJ
s 1000 J
b) q  (0.69W m.K) (3m  3m)
(10  2) K
 2484 W
(0.02m)
Amount of heat loss over a period of 8 hours :
3415.5
J 1 kJ
(8  3600 s)  71539 kJ
s 1000 J
3. The inner and outer glasses of a 3-m x 3-m doublepane window are at 10 °C and 2 °C, respectively.
If the 0.5-cm space between the two glasses is filled with still air (k = 0.026 W/m.K), determine the
rate of heat transfer through the window (0.69 W/m.K). The thickness of each glass layer is 0.35 cm.
Thickness of glass layers needs to be provided to solve
air
the problem!
glass
glass
L glass  0.35 cm, k glass  0.69 W m.K
L air  0.5 cm, k air  0.026 W m.K
10 °C
0.5 cm
q
2 °C
0.35 cm
q
T
Rtotal
Rtotal  R1  R2  R3
0.35 cm
R1  R3 
R1
R2
L glass
k glass A

0.0035 m
(0.69 W m.K) (3 m  3 m)
 0.00056 K W
R3
R2 
L air
0.005 m

k air A (0.026 W m.K) (3 m  3 m)
 0.02137 K W
Rtotal  2  0.00056 K W   0.02137 K W  0.02249 K W
q
(10  2) K
 355.7 W
0.02249 K W
4. The roof of an electrically heated home is 7 m long, 9 m wide, and 0.32 m thick, and is made of a
flat layer of concrete whose thermal conductivity is 0.82 W/m.K. The temperatures of the inner and
the outer surfaces of the roof one night are measured to be 19 °C and 4 °C, respectively, for a period
of 12 hours. Determine (a) the rate of heat loss through the roof that night and (b) the cost of that
heat loss to the home owner if the cost of electricity is $0.13/kWh.
qk A
T
L
a) q  (0.82W m.K) (7m  9m)
(19  4) K
 2422W  2.422 kW
(0.32m)
b) The cos t of electricity for a period of 12 hours :
2.422 kW  12 h 
$0.13
 $3.78
kWh
5. Consider a 0.75-m-high and 1.75-m-wide glass window with a thickness of 7 mm and a thermal
conductivity of 0.74 W/m.K. Determine the steady rate of heat transfer through this glass window
and the temperature of its inner surface for a day during which the room is maintained at 20 °C while
the temperature of the outdoors is -5 °C. Take the heat transfer coefficients on the inner and outer
surfaces of the window to be h1 = 8 W/m2.K and h2 = 22 W/m2.K.
q
glass
T
Rtotal
Rtotal  R1  R2  R3
hi = 8 W/m2K
ho = 22 W/m2K
Ti = 20°C
To = - 5 °C
R1 
q
1
1

2
hi A (8 W m .K) (0.75 m  1.75 m)
 0.0952 K W
7 mm
R2 
R1
R2
L glass
k glass A

0.007 m
(0.74 W m.K) (0.75 m  1.75 m)
 0.0072 K W
R3
R3 
1
ho A

1
(22 W m .K) (0.75 m  1.75 m)
2
 0.0346 K W
Rtotal  0.0952  0.0072  0.0346  0.137 K W
q
20  (5) K  182.5 W
0.137 K W
q  q1 
T1 K
 182.5 W
0.0952 K W
T1  17.4 K  17.4 C
T1  20  T  T  20  17.4  2.6 C
6. Consider a 0.85-m-high and 1.7-m-wide double-pane window consisting of two 3-mm-thick layers
of glass (k = 0.76 W/m.K) separated by a 12-mm-wide stagnant air space (k = 0.026 W/m.K).
Determine the steady rate of heat transfer through this double-pane window and the temperature of
its inner surface for a day during which the room is maintained at 25 °C while the temperature of the
outdoors is 6 °C. Take the convection heat transfer coefficients on the inner and outer surfaces of the
window to be h1 = 7 W/m2.K and h2 = 20 W/m2.K.
q
T
Rtotal
Rtotal  R1  R2  R3  R4  R5
R1 
R2 
R3 
1
1

2
hi A (7 W m .K) (0.85 m  1.7 m)
L glass
 0.0989 K W
0.003 m
(0.76 W m.K) (0.85 m  1.7 m)
 0.0027 K W
L air
0.012 m

kair A (0.026 W m.K) (0.85 m  1.7 m)
 0.3194 K W
k glass A

R4  R2
R5 
1
ho A

1
(20 W m .K) (0.85 m  1.7 m)
2
 0.0346 K W
Rtotal  0.0989  0.0027  0.3194  0.0027  0.0346  0.4583 K W
q
(25  6) K
 41.46 W
0.4583 K W
q  q1 
T1 K
 41.46 W
0.0989 K W
T1  4.1K  4.1 C
T1  25  T  T  25  4.1  20.9 C
7. Hot water at Ti=120°C flows through a stainless steel pipe (k = 15 W/m.K) whose inner diameter is
1.8 cm and wall thickness is 0.4 cm. The pipe is to be covered with adequate insulation so that the
temperature of the outer surface of the insulation does not exceed 42 °C when the ambient
temperature is To = 25 °C. Taking the heat transfer coefficients inside and outside the pipe to be hi =
180 W/m2.K and ho = 10 W/m2.K, respectively, determine the thickness of fiberglass insulation (k =
0.038 W/m.K) that needs to be installed on the pipe.
Stainless steel pipe
k = 15 W/mK
Rtotal  R1  R2  R3  R4
R1 
1
1

2
hi Ai (180 W m .K) (2   0.009 m  1 m)
R2 
ln(r2 / r1 )
ln(1.3 / 0.9)

2   kpipe  L 2   15 W m.K  1 m
R3 
ln(r3 / r2 )
ln(r3 / 1.3)

2   kins  L 2   0.038 W m.K  1 m
R3 
ln(r3 / 1.3)
2   0.038 W m.K  1 m
T3 = 42°C
hi = 180 W/m2K
Ti = 120°C
 0.0982 K W
 0.0039 K W
ho = 10 W/m2K
To = 25°C
Fiberglass insulation
k = 0.038 W/mK

0.0159
KW
r3
Rtotal  0.0982  0.0039  4.1883 ln(r3 / 1.3) 
0.0159
KW
r3
R4 
q
1
 4.1883 ln(r3 / 1.3) K W
ho A o

1
(10 W m .K) (2   r3  1 m)
2
(120  25) K

0.0159 
0.0982  0.0039  4.1883 ln(r3 / 1.3)  r
K W
3


Under steady state conditions, q is also equ al to the following :
(42  25) K
q
0.0159
KW
r3
If we equate the two equations and solve for r3 , we get;
r3  0.0253 m  2.53 cm
The insulation thickness, r3  r2, is then;
2.53  1.3  1.23 cm