Faculty of Engineering
Mechanical Engineering Department
ME 233 Heat Transfer (2)
Solution Sheet [0]
Natural Convection
1- A power transistor mounted on the wall dissipates 0.18 W. The surface temperature of the transistor is to be
determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 Any heat transfer
from the base surface is disregarded. 4 The local atmospheric pressure is 1 atm. 5 Air properties are evaluated at
100C.
Properties The properties of air at 1 atm and the given film temperature of
100C are (Table A-15)
Power
transistor, 0.18 W
D = 0.4 cm
 = 0.1
k  0.03095 W/m. C
  2.306 10 5 m 2 /s
Pr  0.7111
1
1


 0.00268 K -1
Tf
(100  273 ) K
Air
35C
Analysis The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh
number and thus the Nusselt number depends on the surface temperature which is unknown. We start the solution
process by “guessing” the surface temperature to be 165C for the evaluation of h. This is the surface temperature
that will give a film temperature of 100C. We will check the accuracy of this guess later and repeat the calculations
if necessary.
The transistor loses heat through its cylindrical surface as well as its top surface. For convenience, we take
the heat transfer coefficient at the top surface of the transistor to be the same as that of its side surface. (The
alternative is to treat the top surface as a vertical plate, but this will double the amount of calculations without
providing much improvement in accuracy since the area of the top surface is much smaller and it is circular in shape
instead of being rectangular). The characteristic length in this case is the outer diameter of the transistor,
Lc  D  0.004 m. Then,
Ra 
g (Ts  T ) D 3
2
Pr 
(9.81 m/s 2 )(0.00268 K -1 )(165  35 K )(0.004 m) 3

0.387 Ra 1 / 6

Nu  0.6 

1  0.559 / Pr 9 / 16


(2.306  10 5 m 2 /s) 2
2


0.387 (292 .6)1 / 6



0
.
6



8 / 27


1  0.559 / 0.7111 9 / 16




(0.7111 )  292 .6
2


 2.039
8 / 27 



k
0.03095 W/m. C
Nu 
(2.039 )  15 .78 W/m 2 .C
D
0.004 m
As  DL  D 2 / 4   (0.004 m )( 0.0045 m )   (0.004 m) 2 / 4  0.0000691 m 2
h
and
Q  hAs (Ts  T )  As  (Ts 4  Tsurr 4 )
0.18 W  (15 .8 W/m 2 .C)( 0.0000691 m 2 )(Ts  35 ) C

 (0.1)( 0.0000691 m 2 )(5.67  10 8 ) (Ts  273 ) 4  (25  273 K ) 4


 Ts  187 C
which is relatively close to the assumed value of 165C. To improve the accuracy of the result, we repeat the
Rayleigh number calculation at new surface temperature of 187C and determine the surface temperature to be
Ts = 183C
Discussion W evaluated the air properties again at 100C when repeating the calculation at the new surface
temperature. It can be shown that the effect of this on the calculated surface temperature is less than 1C.
2- Water is boiling in a pan that is placed on top of a stove. The rate of heat loss from the cylindrical side surface of
the pan by natural convection and radiation and the ratio of heat lost from the side surfaces of the pan to that by the
evaporation of water are to be determined.
Vapor
2 kg/h
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with
constant properties. 3 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of (Ts+T)/2 =
(98+25)/2 = 61.5C are (Table A-15)
k  0.02819 W/m. C
Pan
Ts = 98C
Air
T = 25C
  1.910  10 5 m 2 /s
Water
100C
Pr  0.7198
1
1


 0.00299 K -1
Tf
(61 .5  273 )K
Analysis (a) The characteristic length in this case is the height of the pan, Lc  L  0.12 m. Then,
Ra 
g (Ts  T ) L3
2
Pr 
(9.81 m/s 2 )(0.00299 K -1 )(98  25 K )(0.12 m) 3
(1.910  10 5 m 2 /s) 2
(0.7198 )  7.299 10 6
We can treat this vertical cylinder as a vertical plate since
35 L
Gr
1/ 4

35(0.12 )
(7.299 10 / 0.7198 )
6
1/ 4
 0.07443 < 0.25
and thus D 
35 L
Gr 1/ 4
Therefore,
2
2











6 1/ 6 
1/ 6
0.387 (7.299  10 )
0.387 Ra




Nu  0.825 
 0.825 
 28 .60
8 / 27 
8 / 27 
  0.492  9 / 16 
  0.492  9 / 16 




1  

1  







  Pr 

  0.7198 





k
0.02819 W/m. C
Nu 
(28 .60 )  6.720 W/m 2 .C
L
0.12 m
As  DL   (0.25 m )( 0.12 m )  0.09425 m 2
h
and
Q  hAs (Ts  T )  (6.720 W/m2 .C)(0.09425 m 2 )(98  25)C  46.2 W
(b) The radiation heat loss from the pan is
Q
 A  (T 4  T 4 )
rad
s
s
surr


 (0.95 )( 0.09425 m 2 )(5.67  10 8 W/m 2 .K 4 ) (98  273 K ) 4  (25  273 K ) 4  56.1 W
 = 0.95
(c) The heat loss by the evaporation of water is
Q  m h fg  (2 / 3600 kg/s )( 2257 kJ/kg )  1.254 kW  1254 W
Then the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water then becomes
f 
46 .2  56 .1
 0.082  8.2%
1254
3- Water is boiling in a pan that is placed on top of a stove. The rate of heat loss from the cylindrical side surface of
the pan by natural convection and radiation and the ratio of heat lost from the side surfaces of the pan to that by the
evaporation of water are to be determined.
Vapor
2 kg/h
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with
constant properties. 3 The local atmospheric pressure is 1 atm.
Properties The properties of air at 1 atm and the film temperature of (Ts+T)/2 =
(98+25)/2 = 61.5C are (Table A-15)
k  0.02819 W/m. C
  1.910  10
5
Pan
Ts = 98C
Air
T = 25C
2
m /s
Pr  0.7198
1
1


 0.00299 K -1
Tf
(61 .5  273 )K
Water
100C
 = 0.1
Analysis (a) The characteristic length in this case is the height of the pan, Lc  L  0.12 m. Then,
Ra 
g (Ts  T ) L3

2
Pr 
(9.81 m/s 2 )(0.00299 K -1 )(98  25 K )(0.12 m) 3
(1.910  10
5
2
m /s)
2
(0.7198 )  7.299 10 6
We can treat this vertical cylinder as a vertical plate since
35 L
Gr
1/ 4

35(0.12 )
(7.299 10 / 0.7198 )
6
1/ 4
 0.07443 < 0.25
and thus D 
35 L
Gr 1/ 4
Therefore,
2
2











6 1/ 6 
1/ 6
0.387 (7.299  10 )
0.387 Ra




Nu  0.825 
 0.825 
 28 .60
8 / 27 
8 / 27 
9
/
16
9
/
16
  0.492 

  0.492 





1  

1  







  Pr 

  0.7198 





k
0.02819 W/m. C
Nu 
(28 .60 )  6.720 W/m 2 .C
L
0.12 m
As  DL   (0.25 m )( 0.12 m )  0.09425 m 2
h
and
Q  hAs (Ts  T )  (6.720 W/m2 .C)(0.09425 m 2 )(98  25)C  46.2 W
(b) The radiation heat loss from the pan is
Q
 A  (T 4  T 4 )
rad
s
s
surr


 (0.10 )( 0.09425 m 2 )(5.67  10 8 W/m 2 .K 4 ) (98  273 K ) 4  (25  273 K ) 4  5.9 W
(c) The heat loss by the evaporation of water is
Q  m h fg  (2 / 3600 kg/s )( 2257 kJ/kg )  1.254 kW  1254 W
Then the ratio of the heat lost from the side surfaces of the pan to that by the evaporation of water then becomes
f 
46 .2  5.9
 0.042  4.2%
1254
9-34 A glass window is considered. The convection heat transfer coefficient on the inner side of the window, the
rate of total heat transfer through the window, and the combined natural convection and radiation heat transfer
coefficient on the outer surface of the window are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with
constant properties. 3 The local atmospheric pressure is 1 atm.
Room
T = 25C
Properties The properties of air at 1 atm and the film temperature of (Ts+T)/2 =
(5+25)/2 = 15C are (Table A-15)
L = 1.2 m
Glass
Ts = 5C
 = 0.9
Q
k  0.02476 W/m. C
  1.471  10 5 m 2 /s
Outdoors
-5C
Pr  0.7323
1
1


 0.003472 K -1
Tf
(15  273 )K
Analysis (a) The characteristic length in this case is the height of the window, Lc  L  1.2 m. Then,
Ra 
g (T  Ts ) L3c
2
Pr 
(9.81 m/s 2 )(0.003472 K -1 )( 25  5 K )(1.2 m) 3
(1.471 10 5 m 2 /s) 2
2
(0.7323 )  3.986 10 9
2











9 1/ 6 
1/ 6
0.387 (3.986 10 )
0.387 Ra




Nu  0.825 
 0.825 
 189 .7
8 / 27 
8 / 27 
  0.492  9 / 16 
  0.492  9 / 16 




1  

1  







  Pr 

  0.7323 





k
0.02476 W/m. C
Nu 
(189 .7)  3.915 W/m2 .C
L
1.2 m
As  (1.2 m)(2 m)  2.4 m 2
h
(b) The sum of the natural convection and radiation heat transfer from the room to the window is
Q
 hA (T  T )  (3.915 W/m2 .C)(2.4 m 2 )(25  5)C  187.9 W
convection
s

s
Q radiation  As  (Tsurr 4  Ts 4 )
 (0.9)( 2.4 m 2 )(5.67  10 8 W/m 2 .K 4 )[( 25  273 K ) 4  (5  273 K ) 4 ]  234 .3 W
Q total  Q convection  Q radiation  187.9  234.3  422.2 W
(c) The outer surface temperature of the window can be determined from
kA
Q t
(346 W)(0.006 m)
Q total  s (Ts,i  Ts,o ) 
 Ts,o  Ts,i  total  5C 
 3.65 C
t
kAs
(0.78 W/m. C)( 2.4 m 2 )
Then the combined natural convection and radiation heat transfer coefficient on the outer window surface becomes
or
Q total  hcombined As (Ts ,o  T,o )
Q total
346 W
hcombined 

 20.35 W/m2 .C
As (Ts ,o  T,o ) (2.4 m 2 )[ 3.65  (5)]C
 and thus the thermal resistance R of a layer is proportional to the temperature drop across that
Note that T  QR
layer. Therefore, the fraction of thermal resistance of the glass is equal to the ratio of the temperature drop across the
glass to the overall temperature difference,
Rglass
Tglass
5  3.65


 0.045 (or 4.5%)
R total TR total 25  (5)
which is low. Thus it is reasonable to neglect the thermal resistance of the glass.
4- The equilibrium temperature of a light glass bulb in a room is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant properties. 3 The local
atmospheric pressure is 1 atm. 4 The light bulb is approximated as an 8-cm-diameter sphere.
Properties The solution of this problem requires a trial-and-error approach since the determination of the Rayleigh
number and thus the Nusselt number depends on the surface temperature which is unknown. We start the solution
process by “guessing” the surface temperature to be 170C for the evaluation of the properties and h. We will check
the accuracy of this guess later and repeat the calculations if necessary. The properties of air at 1 atm and the
anticipated film temperature of (Ts+T)/2 = (170+25)/2 = 97.5C are (Table A-15)
k  0.03077 W/m. C
Lamp
60 W
 = 0.9
Air
T = 25C
  2.279 10 5 m 2 /s
Pr  0.7116
1
1


 0.002699 K -1
Tf
(97 .5  273 )K
D = 8 cm
Analysis The characteristic length in this case is Lc = D = 0.08 m. Then,
Ra 
g (Ts  T ) D 3
2
Pr 
(9.81 m/s 2 )( 0.002699 K -1 )(170  25 K )( 0.08 m ) 3
(2.279  10 5 m 2 /s) 2
(0.7116 )
Light,
6W
 2.694  10 6
Nu  2 
0.589 Ra 1 / 4
1  0.469 / Pr  
9 / 16 4 / 9
 2
0.589 (2.694 10 6 )1 / 4
1  0.469 / 0.7116  
9 / 16 4 / 9
 20 .42
Then
k
0.03077 W/m. C
Nu 
(20 .42 )  7.854 W/m 2 .C
D
0.08 m
As  D 2   (0.08 m) 2  0.02011 m 2
h
Considering both natural convection and radiation, the total rate of heat loss can be written as
Q  hA (T  T )  A  (T 4  T 4 )
s
s

s
s
surr
2
(0.90  60 ) W  (7.854 W/m .C)( 0.02011 m )(Ts  25 )C
2
 (0.9)( 0.02011 m 2 )(5.67  10 8 W/m 2 .K 4 )[(Ts  273 ) 4  (25  273 K ) 4 ]
Its solution is
Ts = 169.4C
which is sufficiently close to the value assumed in the evaluation of properties and h. Therefore, there is no need to
repeat calculations.