Solutions to HW#4
7-16 Wind is blowing parallel to the wall of a house. The rate of heat loss from that wall
is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr
= 5105. 3 Radiation effects are negligible. 4 Air is an ideal gas with constant properties.
Properties The properties of air at 1 atm and the film
temperature of (Ts + T)/2 = (12+5)/2 = 8.5C are
Air
(Table A-15)
V = 55 km/h
k  0.02428 W/m. C
T = 5C
  1.413  10 -5 m 2 /s
Ts = 12C
Pr  0.7340
Analysis Air flows parallel to the 10 m side:
The Reynolds number in this case is
Re L 
L
V L [(55  1000 / 3600 )m/s](10 m)

 1.081  10 7
5
2

1.413  10 m /s
which is greater than the critical Reynolds number. Thus we have combined laminar and
turbulent flow. Using the proper relation for Nusselt number, heat transfer coefficient and
then heat transfer rate are determined to be
hL
 (0.037 Re L 0.8  871) Pr 1 / 3  [0.037 (1.081  10 7 ) 0.8  871]( 0.7340 )1/ 3  1.336  10 4
k
k
0.02428 W/m. C
h  Nu 
(1.336  10 4 )  32 .43 W/m 2 .C
L
10 m
Nu 
As  wL  (4 m)(10 m) = 40 m 2
Q  hAs (T  Ts )  (32 .43 W/m 2 .C)(40 m 2 )(12 - 5) C  9081 W  9.08 kW
If the wind velocity is doubled:
Re L 
V L [(110  1000 / 3600 )m/s](10 m)

 2.163  10 7

1.413  10 5 m 2 /s
which is greater than the critical Reynolds number. Thus we have combined laminar and
turbulent flow. Using the proper relation for Nusselt number, the average heat transfer
coefficient and the heat transfer rate are determined to be
hL
 (0.037 Re L 0.8  871) Pr 1 / 3  [0.037 (2.163  10 7 ) 0.8  871]( 0.7340 )1 / 3  2.384  10 4
k
k
0.02428 W/m. C
h  Nu 
(2.384  10 4 )  57 .88 W/m 2 .C
L
10 m
Nu 
As  wL  (10 m)(4 m) = 40 m 2
Q  hAs (T  Ts )  (57 .88 W/m 2 .C)(40 m 2 )(12 - 5) C  16 ,206 W  16.21 kW
7-31 Air is blown over an aluminum plate mounted on an array of power transistors. The
number of transistors that can be placed on this plate is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr
= 5105. 3 Radiation effects are negligible 4 Heat transfer from the backside of the plate
is negligible. 5 Air is an ideal gas with constant properties. 6 The local atmospheric
pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of (Ts + T)/2 =
(65+35)/2 = 50C are (Table A-15)
k  0.02735 W/m. C
  1.798  10 -5 m 2 /s
Pr  0.7228
Note that the atmospheric pressure will only affect the
kinematic viscosity. The atmospheric pressure in atm is
1 atm
P  (83.4 kPa)
 0.823 atm
101.325 kPa
Air
V = 4 m/s
T = 35C
The kinematic viscosity at this atmospheric pressure will be
  (1.79810 5 m 2 /s ) / 0.823  2.184 10 5 m 2 /s
Transistors
Ts = 65C
L = 25 cm
Analysis The Reynolds number is
Re L 
V L
(4 m/s)(0.25 m)

 4.579  10 4

2.184  10 5 m 2 /s
which is less than the critical Reynolds number ( 5  105 ). Thus the flow is laminar. Using
the proper relation in laminar flow for Nusselt number, the average heat transfer
coefficient and the heat transfer rate are determined to be
hL
 0.664 Re L 0.5 Pr 1 / 3  0.664 (4.579  10 4 ) 0.5 (0.7228 )1 / 3  127 .5
k
k
0.02735 W/m. C
h  Nu 
(127 .5)  13 .95 W/m 2 .C
L
0.25 m
Nu 
As  wL  (0.25 m)(0.25 m) = 0.0625 m 2
Q conv  hAs (T  Ts )  (13 .95 W/m 2 .C)(0.0625 m 2 )(65 - 35) C = 26.2 W
Considering that each transistor dissipates 3 W of power, the number of transistors that
can be placed on this plate becomes
n
26 .2 W
 4.4 
 4
6W
This result is conservative since the transistors will cause the flow to be turbulent, and the
rate of heat transfer to be higher.
7-48 An aircraft is cruising at 900 km/h. A heating system keeps the wings above
freezing temperatures. The average convection heat transfer coefficient on the wing
surface and the average rate of heat transfer per unit surface area are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air
is an ideal gas with constant properties. 4 The wing is approximated as a cylinder of
elliptical cross section whose minor axis is 30 cm.
Properties The properties of air at 1 atm and the film temperature of (Ts + T)/2 = (055.4)/2 = -27.7C are (Table A-15)
k  0.02152 W/m. C
  1.106  10 -5 m 2 /s
18.8 kPa
V = 900 km/h
Note that the atmospheric pressure will only affect the kinematic viscosity. TheT = -55.4C
Pr  0.7422
atmospheric pressure in atm unit is
1 atm
P  (18.8 kPa)
 01855
.
atm
101.325 kPa
The kinematic viscosity at this atmospheric pressure is
  (1.106 105 m 2 /s)/0.1855  5.961105 m 2 /s
Analysis The Reynolds number is
Re 
V D (900  100 0/3600) m/s (0.3 m)

 1.258  10 6
5
2

5.961  10 m /s
The Nusselt number relation for a cylinder of elliptical cross-section is limited to Re <
15,000, and the relation below is not really applicable in this case. However, this relation
is all we have for elliptical shapes, and we will use it with the understanding that the
results may not be accurate.
Nu 
hD
 0.248 Re 0.612 Pr 1/ 3  0.248 (1.258  10 6 ) 0.612 (0.724 )1/ 3  1204
k
The average heat transfer coefficient on the wing surface is
h
k
0.02152 W/m. C
Nu 
(1204 )  86.39 W/m2 .C
D
0.3 m
Then the average rate of heat transfer per unit surface area becomes
q  h(Ts  T )  (86.39 W/m2 .C)[0 - (-55.4)]C  4786 W/m 2
7-53 A steam pipe is exposed to light winds in the atmosphere. The amount of heat loss from the steam
during a certain period and the money the facility will save a year as a result of insulating the steam pipes
are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The plant
operates every day of the year for 10 h. 4 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of
(Ts + T)/2 = (75+5)/2 = 40C are (Table A-15)
k  0.02662 W/m. C
  1.702  10 -5 m 2 /s
Pr  0.7255
Analysis The Reynolds number is
V D (10  1000/3600) m/s (0.1 m)
Re   
 1.632  10 4

1.702  10 5 m 2 /s
The Nusselt number corresponding this Reynolds number is determined to be
Wind
V = 10 km/h
T = 5C
Steam pipe
Ts = 75C
D = 10 cm
 = 0.8
hD
0.62 Re 0.5 Pr 1 / 3
Nu 
 0.3 
1/ 4
k
1  (0.4 / Pr) 2 / 3


  Re  5 / 8 
1  
 
  282 ,000  
4/5

0.62 (1.632  10 4 ) 0.5 (0.7255 )1 / 3   1.632  10 4
 0.3 
1
1/ 4
  282 ,000
1  (0.4 / 0.7255 ) 2 / 3







5/8 
4/5



 71 .19
The heat transfer coefficient is
k
0.02662 W/m. C
h  Nu 
(71.19)  18.95 W/m2 .C
D
0.1 m
The rate of heat loss by convection is
As  DL   (0.1 m)(12 m)  3.77 m 2
Q  hA (T  T )  (18.95 W/m2 .C)(3.77m 2 )(75- 5)C = 5001W
s
s

For an average surrounding temperature of 0  C , the rate of heat loss by radiation and the total rate of heat
loss are
Q rad  As  (Ts 4  Tsurr 4 )


 (0.8)(3.77 m 2 )(5.67  10 -8 W/m 2 .K 4 ) (75  273 K ) 4  (0  273 K ) 4  1558 W
Q
 Q
 Q  5001 1588  6559 W
total
conv
rad
If the average surrounding temperature is 20  C , the rate of heat loss by radiation and the total rate of
heat loss become
Q
 A  (T 4  T 4 )
rad
s
s
surr
2

 (0.8)(3.77 m )(5.67  10 -8 W/m 2 .K 4 ) (75  273 K ) 4  (20  273 K ) 4
 1807 W

Qtotal  Q conv  Q rad  5001 1807  6808 W

which is 6808 - 6559 = 249 W more than the value for a surrounding temperature of 0C. This corresponds
to
Q
249 W
%change  difference  100 
 100  3.8% (increase)

6559 W
Qtotal,0C
If the average surrounding temperature is 25C, the rate of heat loss by radiation and the total rate of heat
loss become
Q rad  As  (Ts 4  Tsurr 4 )

4
 (0.8)(3.77 m 2 )(5.67  10 -8 W/m 2 .K 4 ) (75  273 K ) 4  (25  273 K ) 4

 1159 W
Q total  Q conv  Q rad  5001 1159  6160 W
which is 6559 - 6160 = 399 W less than the value for a surrounding temperature of 0C. This corresponds
to
Q
399 W
%change  difference  100 
 100  6.1% (decrease)

6559
W
Qtotal,0C
Therefore, the effect of the temperature variations of the surrounding surfaces on the total heat transfer is
less than 6%.
7-60 Air flows over a spherical tank containing iced water. The rate of heat transfer to the tank and the rate
at which ice melts are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas
with constant properties. 4 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm pressure and the free stream temperature of 25C are (Table A-15)
k  0.02551 W/m. C
Air
V = 7 m/s
T =25C
  1.562  10 -5 m 2 /s
   1.849  10 5 kg/m.s
Iced water
0C
 s , @ 0C  1.729  10 5 kg/m.s
D = 1.8 m
Pr  0.7296
Analysis The Reynolds number is
V D
(7 m/s)(1.8 m)
Re   
 806 ,658

1.562  10 5 m 2 /s
The proper relation for Nusselt number corresponding to this Reynolds number is



hD
Nu 
 2  0.4 Re 0.5  0.06 Re 2 / 3 Pr 0.4  
k
 s

 2  0.4(806 ,658 )
0.5
 0.06 (806 ,658 )
1/ 4



2/3
(0.7296 )
0. 4 
 1.849
1/ 4
 10 5 
 1.729  10 5 


 790 .1
The heat transfer coefficient is
k
0.02551 W/m. C
h  Nu 
(790 .1)  11 .20 W/m 2 .C
D
1.8 m
Then the rate of heat transfer is determined to be
As  D 2   (1.8 m) 2 = 10.18 m 2
Q  hAs (Ts  T )  (11 .20 W/m 2 .C)(10.18 m 2 )( 25  0)C  2850 W
The rate at which ice melts is
h 
 (333.7 kJ/kg)
  0.00854 kg/s  0.512kg/min
Q  m
  2.850 kW = m
 m
fg
7-70 Air is cooled by an evaporating refrigerator. The refrigeration capacity and the pressure drop across
the tube bank are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to the
temperature of refrigerant.
Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air
properties at the assumed mean temperature of -5C (will be checked later) and 1 atm (Table A-15):
k = 0.02326 W/m-K
 = 1.316 kg/m3
Cp =1.006 kJ/kg-K
Pr = 0.7375
 = 1.70510-5 kg/m-s
Prs = Pr@ Ts = 0.7408
Also, the density of air at the inlet temperature of 0C (for use in the mass flow rate calculation at the inlet)
is i = 1.292 kg/m3.
Analysis It is given that D = 0.008 m, SL = ST = 0.015 m, and V = 4 m/s.
Then the maximum velocity and the Reynolds number
based on the maximum velocity become
Vmax
ST
0.015

V
(4 m/s)  8.571 m/s
ST  D
0.015  0.008
Re D 
V=4 m/s
Ti=0C
Ts=-20C
SL
ST
Vmax D (1.316 kg/m 3 )(8.571 m/s)(0.008 m)

 5294

1.705  10 5 kg/m  s
The average Nusselt number is determined using
the proper relation from Table 7-2 to be
Nu D  0.27 Re 0D.63 Pr 0.36 (Pr/ Prs ) 0.25
D
 0.27 (5294 ) 0.63 (0.7375 ) 0.36 (0.7375 / 0.7408 ) 0.25  53 .61
Since NL > 16. the average Nusselt number and heat transfer coefficient for
all the tubes in the tube bank become
Nu D, NL  FNu D  53.61
h
Nu D, N L k
D

53 .61(0.02326 W/m  C)
 155 .8 W/m 2  C
0.008 m
The total number of tubes is N = NL NT = 3015 = 450. The heat transfer
surface area and the mass flow rate of air (evaluated at the inlet) are
As  NDL  300(0.008 m)(0.4m)  4.524 m 2
 m
 i  iV( NT ST L)  (1.292 kg/m3 )(4 m/s)(15)(0.015 m)(0.4m)  0.4651kg/s
m
Then the fluid exit temperature, the log mean temperature difference, and
the rate of heat transfer (refrigeration capacity) become
 Ah
Te  Ts  (Ts  Ti ) exp   s
 m C p

Tln 
2
2



  20  (20  0) exp   (4.524 m )(155 .8 W/m  C)   15 .57 C
 (0.4651 kg/s)(1006 J/kg  C) 




(Ts  Ti )  (Ts  Te )
(20  0)   20  (15 .57 )

 10 .33 C
ln[(Ts  Ti ) /(Ts  Te )] ln[( 20  0) /(20  15 .57 )]
Q  hAs Tln  (155.8 W/m2  C)(4.524m 2 )(10.33C)  7285 W
For this square in-line tube bank, the friction coefficient corresponding to
ReD = 5294 and SL/D = 1.5/0.8 = 1.875 is, from Fig. 7-27a, f = 0.27. Also,
 = 1 for the square arrangements. Then the pressure drop across the tube
bank becomes
P  N L f
2
Vmax
(1.316 kg/m 3 )(8.571 m/s) 2
 30 (0.27 )(1)
2
2
 1N

 1 kg  m/s 2


  391.6 Pa


Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (0 -15.6)/2 = -7.8C, which is fairly close
to the assumed value of -5C. Therefore, there is no need to repeat calculations.
7-71 Air is cooled by an evaporating refrigerator. The refrigeration capacity and the pressure drop across
the tube bank are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of the tubes is equal to the
temperature of refrigerant.
Properties The exit temperature of air, and thus the mean temperature, is not known. We evaluate the air
properties at the assumed mean temperature of -5C (will be checked later) and 1 atm (Table A-15):
k = 0.02326 W/m-K
 = 1.316 kg/m3
Cp =1.006 kJ/kg-K
Pr = 0.7375
 = 1.70510-5 kg/m-s
Prs = Pr@ Ts = 0.7408
Also, the density of air at the inlet temperature of 0C (for use in the mass flow rate calculation at the inlet)
is i = 1.292 kg/m3.
Analysis It is given that D = 0.008 m, SL = ST = 0.015 m, and V = 4 m/s.
Then the maximum velocity and the Reynolds number
V=4 m/s
based on the maximum velocity become
Vmax 
Re D
ST
0.015
V
(4 m/s)  8.571 m/s
ST  D
0.015  0.008
SL
Ts=-20C
Ti=0C
ST
V D (1.316 kg/m 3 )(8.571 m/s)(0.008 m)
 max 
 5294

1.705  10 5 kg/m  s
The average Nusselt number is determined using
the proper relation from Table 7-2 to be
D
Nu D  0.35(ST / S L ) 0.2 Re 0D.6 Pr 0.36 (Pr/ Prs ) 0.25
 0.35(0.015 / 0.015 ) 0.2 (5294 ) 0.6 (0.7375 ) 0.36 (0.7375 / 0.7408 ) 0.25  53 .73
Since NL > 16. the average Nusselt number and heat transfer coefficient for
all the tubes in the tube bank become
Nu D, NL  FNu D  53.73
h
Nu D, N L k
D

53 .73(0.02326 W/m  C)
 156 .2 W/m 2  C
0.008 m
The total number of tubes is N = NL NT = 3015 = 450. The heat transfer
surface area and the mass flow rate of air (evaluated at the inlet) are
As  NDL  300(0.008 m)(0.4m)  4.524 m 2
Then the fluid exit temperature, the log mean temperature difference, and
the rate of heat transfer (refrigeration capacity) become
 Ah
Te  Ts  (Ts  Ti ) exp   s
 m C p

Tln 
2
2



  20  (20  0) exp   (4.524 m )(156 .2 W/m  C)   15 .58 C



 (0.4651 kg/s)(1006 J/kg  C) 

(Ts  Ti )  (Ts  Te )
(20  0)   20  (15 .58)

 10 .32 C
ln[(Ts  Ti ) /(Ts  Te )] ln[( 20  0) /(20  15 .58)]
Q  hAs Tln  (156.2 W/m2  C)(4.524m 2 )(10.32C)  7294 W
For this staggered arrangement tube bank, the friction coefficient
corresponding to ReD = 5294 and SL/D = 1.5/0.8 = 1.875 is, from Fig. 727b, f = 0.44. Also,  = 1 for the square arrangements. Then the pressure
drop across the tube bank becomes
P  N L f
2
Vmax
(1.316 kg/m 3 )(8.571 m/s) 2
 30 (0.44 )(1)
2
2
 1N

 1 kg  m/s 2


  638.2 Pa


Discussion The arithmetic mean fluid temperature is (Ti + Te)/2 = (0 -15.6)/2 = -7.8C, which is fairly close
to the assumed value of -5C. Therefore, there is no need to repeat calculations.