College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
April 11 Homework Solutions
7-16
During a cold winter day, wind at
55 km/h is blowing parallel to a
4-m-high and 10-m-long wall of a
house. If the air outside is at 5°C
and the surface temperature of
the wall is 12°C, determine the
rate of heat loss from that wall
by convection. What would your
answer be if the wind velocity
was doubled?
Here we have convection over a
flat surface with a length of 10 m
and a width of 4 m. The mean
temperature is (5oC + 12oC)/2 = 8.5oC. We find the following properties of air at this temperature
from Table A-15: k = 0.02428 W/moC,  = 1.413x10-5 m2/s, and Pr = 0.7340.
We first find the Reynolds number.
55 km 1000 m 1 h
10 m
VL
h
km 3600 s
Re L 

 1.081x10 7
5
2

1.413x10 m
s
This means that the flow becomes turbulent over the plate and we can use the Nusselt number
equation for combined laminar and turbulent flow.
Nu L 




hL
 0.037 Re 0L.8  871 Pr 1 / 3  0.037 1.081x10 7
k

0.8

 871 0.7340
1/ 3
 1.336 x10 4
We then use this Nusselt number to find the heat transfer coefficient and the heat transfer.
h
kNuL 1.336 x10 4 0.02484 W 32.43 W

 2 o
L
10 m
m o C
m  C



32.43 W
1 kW
Q  h ATs  T   2 o
40 m 2 12 o C 5o C
 9.08 kW
1000 W
m C
If the wind velocity were doubled, the Reynolds number would be doubled and we would repeat
the calculations above, starting with this revised Reynolds number..
Nu L 




hL
 0.037 Re 0L.8  871 Pr1 / 3  0.037 2.162 x10 7
k
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348

0.8

 871 0.7340
1/ 3
 2.384 x10 4
Phone: 818.677.6448
Fax: 818.677.7062
April 11 homework solutions
h
ME 375, L. S. Caretto, Spring 2007
Page 2
kNuL 2.384 x10 4 0.02484 W 57.88 W

 2 o
L
10 m
m o C
m  C
57.88 W
1 kW
Q  h ATs  T   2 o 40 m 2 12 o C 5 o C 
 16.21 kW
1000 W
m C
7-32
A transformer that is 10 cm long, 6.2 cm wide,
and 5 cm high is to be cooled by attaching a
10-cm by 6.2-cm wide polished aluminum
heat sink (emissivity = 0.03) to its top surface.
The heat sink has seven fins, which are 5 mm
high , 2 mm thick, and 10 cm long. A fan
blows air at 25°C parallel to the passages
between the fins. The heat sink is to dissipate
12 W of heat and the base temperature of the
heat sink is not to exceed 60°C. Assuming the
fins and the base plate to be nearly
isothermal and the radiation heat transfer to
be negligible, determine the minimum freestream velocity the fan needs to supply to
avoid overheating.
With the simplifying assumptions given in this
problem, we really have a problem of 7 flat plates
with heat transfer from two sides. The spaces at
the base of the fins also form flat plates. So we
find the heat transfer from one flat plate with a
total area of all the fins plus the unifnned surface
at the base of the fins.
We have to work this problem in reverse. We are
essentially given the required heat transfer – the
12W that the transformer has to dissipate – and
data on the fin area, and the temperature difference. We can therefore compute the heat transfer
coefficient and work backwards to find the air velocity.
The flat plates in this case have a length of 10 cm and a width of 0.5 cm. The total area from one
side of one fin is (0.005 m)(0.1 m) = 0.0005 m 2. There are 14 such sides giving an area of 0.007
m2. In addition, the total area at the top of the transformer is (0.1 m)(0.062 m) = 0.0062 m 2. The
bases of the seven fins take up an area of 7(0.1 m)(0.002 m) = 0.0014 m 2. Thus the unfinned
area on the base is the difference in these areas: 0.0062 m 2 – 0.0014 m2. = 0.0048 m2. The total
area for the air is the sum of the fin area and the unfinned area: 0.007 m 2 + 0.0048 m2 = 0.0118
m2. From these area data and the other given data we can compute the heat transfer coefficient
as follows.
h
Q
12 W
29.06 W


2
o
o
ATs  T  0.0118m 60 C  25 C
m 2 o C



We need air properties to continue. The mean temperature is (60oC + 25oC)/2 = 42.5oC. We find
the following properties of air at this temperature from Table A-15: k = 0.02681 W/moC,  =
1.726x10-5 m2/s, and Pr = 0.7248. The Nusselt number based on the common length of 0.1 m for
all the surfaces that we are treating as flat plates, can now be computed as follows
April 11 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 3
29.06 W
0.1 m
hL m 2 o C
Nu L 

 108.4
0.02681 W
k
m o C
Now we have to decide which equation to use for Nu: the laminar or turbulent equation. Since
the length is short we choose the laminar equation for the first attempt.
Nu L  0.664 Re
0.5
L
Pr
13
 Re
0.5
L
Nu L

0.664 Pr1 3
2


108.4
 Re L  
 3.302 x10 4
13 
 0.6640.7248 
Since this is less than the critical Reynolds number of 500,000 we have made a good guess!
Now we can complete our solution by finding the velocity that gives us this Reynolds number.
Re L 
VL

1.726 x10 5 m 2
3.302 x10 4
 Re L
s
 V

 5.70 m/s
L
0.1 m


7-67C In flow across tube banks, why is the Reynolds number based on the maximum velocity
instead of the uniform approach velocity?
The uniform approach velocity is not as important to the flow phenomena across tube banks as
the local velocity between tubes. Thus the local maximum velocity is a better indicator of the heat
transfer effects and is used in the correlation equations.
8-22
Cooling water available at 10°C is used to condense steam at 30°C in the condenser of a
power plant at a rate of 0.15 kg/s by circulating the cooling water through a bank of 5-mlong 1.2-cm-internal-diameter thin copper tubes. Water enters the tubes at a mean velocity
of 4 m/s, and leaves at a temperature of 24°C. The tubes are nearly isothermal at 30°C.
Determine the average heat transfer coefficient between the water and the tubes, and the
number of tubes needed to achieve the indicated heat transfer rate in the condenser.
Here we are using the data given in the problem to determine the heat transfer coefficient that is
responsible for the heat transfer performance represented by those data.
To start, we find the following properties for water at the average temperature of (10oC + 24oC)/2
= 17.5oC from Table A-9:  = 998.7 kg/m3, cp = 4183.8 J/kgoC, and latent heat of vaporization, hfg
= 2431 kJ/kg. The mass flow rate of water in one tube can be found as follows.
m  AcV  

4
D 2V 
998.7 kg 
0.012 m2 4 m  0.4518 kg
3
4
s
s
m
We can apply the first law of thermodynamics to determine the heat addition rate required to
increase the temperature from the inlet temperature of 10oC to the outlet temperature of 24oC in a
single tube.


0.4518 kg 4183.8 J
2.646 x10 4 J
o
o
Q1 tube  m c p Tout  Tin  
24
C

10
C

 2.646 x10 4 W
s
s
kgo C
The log-mean temperature difference, which is used to calculate the heat transfer, is found below.
April 11 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 4
Tout  Tin 
24 o C  10 o C
LMT 

 11.63o C
o
o
 T  Ts 
 24 C  30 C 
 ln  o

ln  out
o
 10 C  30 C 
 Tin  Ts 
The surface area, As, available for heat transfer is the wall area of the cylindrical tube.
As  DL   0.012 m 5 m   0.1885 m 2
We now have all the data required to compute the average heat transfer coefficient.
Q  h ALMT 
1.21x105 W
Q
2.646 x10 4 W
 h


m 2 o C
ALMT  0.1885 m 2 11.63o C



The number of tubes required are found by determining the total heat transfer required to
condense 0.15 kg/s of steam with the latent heat found above: hfg = 2431 kJ/kg. The total heat
transfer required is
0.15 kg 2431 kJ 1000 W  s
Q total  m steamh fg 
 3.6365 x10 5 W
s
kg
1 kJ
Since each tube has a heat transfer of 2.656x104 W, the number of tubes required is simply the
ratio of the two heat transfers.
N tubes 
Q total 3.6365 x105 W

 13.8
Q1 tube 2.646 x10 4 W
Thus, 14 tubes are required .
8-29C Consider fully developed flow in a circular pipe with negligible entrance effects. If the
length of the pipe is doubled, the pressure drop will (a) double, (b) more than double, (c)
less than double, (d) reduce by half, or (e) remain constant. (This problem was not assigned
and was included here when the instructor misread 39 as 29.)
For fully developed flow in a pipe the pressure drop is proportional to the length. Thus if entrance
effects are negligible, the pressure drop in the entire pipe is proportional to length. So, if the pipe
length is doubled, the pressure drop will double.
8-39
Determine the convection heat transfer coefficient for the flow of (a) air and (b) water at a
velocity of 2 m/s in an 8-cm-diameter and 7-m-long tube when the tube is subjected to
uniform heat flux from all surfaces. Use fluid properties at 25oC.
We have to repeat the same calculations for both air and water. Starting with air we find the
following properties for air at 25oC from Table A-15: k = 0.02551 W/moC,  = 1.562x10-5 m2/s,
and Pr = 0.7296. We start by computing the Reynolds number.
2m
0.08 m
s
Re D 

 1.0243x10 4

1.562 x10 5 m 2
s
VD
This corresponds to turbulent flow. Using the rule of thumb that the entry length is 10 diameters
for turbulent flow gives an entry length of 10(0..08 m) = 0.8 m, which is only (0.8 m)/(7 m) =
April 11 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 5
11.4% of the total tube length. For the accuracy of the correlation equations, we can assume that
the entire length is fully developed turbulent flow. (Using the equation that the entry length to
diameter ratio is 1.359Re1/4 given a slight larger entry length of 13.6 diameters.)
We can use the Gnielinski equation for the Nusselt number
Nu 
 f 8Re  1000Pr
12
1  12.7 f 8 Pr 2 3  1
Where we find the friction factor from the first Petukhov equation
2
1
1

 

f 
 
  0.0313
4
 0.790 ln Re  1.64   0.790 ln 1.0243 x10  1.64 
Substituting this friction factor and the Reynolds and Prandtl numbers found above into the
Nusselt number equation gives
Nu 
 f 8Re  1000Pr
0.0313 81.0243x10 4  10000.7296

 31.03
12
12
1  12.7 f 8 Pr 2 3  1
1  12.70.0313 8 0.7296 2 3  1
We can now find the average heat transfer coefficient for air.
h
9.89 W
kNu 31.03 0.02551 W

 2 o
o
m  C
D
0.08 m m  C
Repeating the calculations for water, we start by finding the properties of water at 25 oC from
Table A-9: : k = 0.607 W/moC,  = 0.891x10-3 kg/ms,  = 997 kg/m3, and Pr = 6.14; we first use
properties to find the Reynolds number.
997 kg 2 m
0.08 m
3
VD
s
m
Re D 

 1.7904 x10 5
3

0.891x10 kg
ms
Again, this is turbulent and the entry length of 10 diameters is small enough so that we can
neglect entry effects. We then repeat the calculations of friction factor, Nusselt number, and heat
transfer coefficient as done for air.
2
1
1

 

f 
 
  0.0160
5
 0.790 ln Re  1.64   0.790 ln 1.7904 x10  1.64 
Nu 
 f 8Re  1000Pr  0.0160 81.704 x10 5  10006.14  934.1
12
12
1  12.7 f 8 Pr 2 3  1
1  12.70.0160 8 6.14 2 3  1
h
7087 W
kNu 934.1 0.607 W


o
m 2 o C
D
0.08 m m  C
We see that the heat transfer coefficient in water is over 700 times the heat transfer coefficient for
air. This is due to a combination of effects. The viscosity and density of water give a higher
Reynolds number for the same flow velocity and diameter. This, in combination with a higher
Prandtl number results in a higher Nusselt number. The effect of the larger Nusselt number is
magnified by the higher thermal conductivity of water.
April 11 homework solutions
8-41
ME 375, L. S. Caretto, Spring 2007
Page 6
An 8-m long, uninsulated square duct of cross section 0.2 m by 0.2 m and relative
roughness 10-8 passes through the attic space of a house. Hot air enters the duct at 1 atm
and 80oC at a volume flow rate of 0.15 m3/s. The duct surface is nearly isothermal at 60oC.
Determine the rate of heat loss from the duct to the attic space and the pressure difference
between the inlet and outlet sections of the duct.
If we assume that the air will not cool very much in the duct length we can use the inlet
temperature of 80oC to find the properties of air from Table A-15: k = 0.02953 W/m▪o
2.097x10-5 m2/s, = 0.9994 kg/m 3, cp = 1008 J/kgoC, and Pr = 0.7296. We start by computing
the mean velocity of air from the volume flow rate and the cross sectional area of a2 = (0.2 m)2=
0.04 m2. (Note a denotes the width of one side of the square duct; L denotes the 10 m length of
the duct.)
V 
V 0.15 m 3
1
3.75 m


2
Ac
s
s
0.04 m
Dh 
4 Ac 4a 2

 a  0.2 m
p
4a
We now compute the Reynolds number based on the hydraulic diameter.
3.75 m
0.2 m
VDh
s
Re h 

 35,765

2.097 x10 5 m 2
s
This corresponds to turbulent flow. Using the rule of thumb that the entry length is 10 diameters
for turbulent flow gives an entry length of 10(0.2 m) = 2 m, which is only (2m)/(8 m) = 25% of the
total tube length. For the accuracy of the correlation equations, we can assume that the entire
length is fully developed turbulent flow.
For this problem, the relative roughness,/Dh = (10-8 m)/(0.2 m) = 5x10-8.
We can use the Gnielinski equation for the Nusselt number with the friction factor found from the
Colebrook equation. We are justified in using the Gnielinski formula, rather than a special
formula for roughness, for this duct because the roughness is so small. The Colebrook equation
is shown below.
 D
1
2.52
 2 log 10 

 3.7 Re f
f





Since this is an implicit equation, we can use the Haaland equation to estimate f
1.11
 6.9
 6.9   D 1.11 
 5 x10 8  
1
   6.69 
 1.8 log 10 

 
   1.8 log 10 
f
 35765  3.7  
 Re  3.7  
f  0.0224
Substituting this estimate for f into the Colebrook equation gives our final answer for f.
 D
1
2.52
 2 log 10 


f
 3.7 Re f
8



2.52
  2 log 10  5 x10 
  6.66 
 3.7

35
,
765
0
.
0224



f  0.0226
Before completing the heat transfer calculation, we can use this friction factor to find the pressure
drop.
April 11 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 7
2
p  f
L V 2
Dh 2
0.9994 kg  3.75 m 


8m
s  1 N  s 2 Pa  m 2
m3

 0.0225
 6.34 Pa
0.2 m
2
kg  m 1 N
Substituting the friction factor and the Reynolds and Prandtl numbers found above into the
Gnielinski equation for the Nusselt number gives
Nu 
 f 8Re  1000Pr  0.0226 835765  10000.7296  82.03
12
12
1  12.7 f 8 Pr 2 3  1 1  12.70.0226 8 0.7296 2 3  1
We can now find the average heat transfer coefficient.
h
kNu 82.03 0.02953 W 12.11 W

 2 o
Dh
0.2 m m o C
m  C
Next, we find the exit temperature of the air by computing the number of transfer units, NTU, and
substituting that NTU value into the equation for exponential behavior of the fluid temperature..
12.11 W
4.2 m8 m
hAs h4aL 
m 2 o C
NTU 


 0.5130
m c p
0.9994 kg 0.15 m 3 1008 J W  s
Vc p
s
kg  s 1 J
m3


Tout  Ts  Ts  Tin e  NTU  60 o C  60 o C  80 o C e 0.5130  72.0oC
We can find the heat transfer by using h and the log-mean temperature difference or by using the
first law. Here we use the first law because it is simpler.
0.9994 kg 0.15 m 3 1008 J W  s
Q  m c p Tout  Tin   Vc p Tout  Tin  
72.0 o C  80 o C = –1212 W
3
3
kg  s 1 J
m
m


The negative sign for the heat transfer indicates that the heat is transferred from the fluid to the
wall.