College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
Solutions to Exercise Eight – External Forced Convection and
Introduction to Internal Flows
1. The components of an electronic
system are located in a 1.5-mlong horizontal duct whose
cross section is 20 cm x 20 cm.
The components in the duct are
not allowed to come into direct
contact with cooling air, and
thus are cooled by air at 30°C
flowing over the duct with a
velocity of 200 m/min. If the
surface temperature of the duct
is not to exceed 65°C, determine
the total power rating of the
electronic devices that can be
mounted into the duct. (Problem
and figure P7.60 from Çengel,
Heat and Mass Transfer.)
This is a flow over a non-circular
cylinder for which the Nusselt
number is given in Table 7-1 on page 414. The Reynolds number is based on the one side of a
square and for air at the mean (film) temperature of (30oC + 65oC)/2 = 47.5oC, we find the
following properties of air from Table A-15: k = 0.02717 W/m▪oC,  = 1.774x10-5 m2/s, and Pr =
0.7235. For D = 20 cm = 0.2 m as the side of the square cross section in the direction of the flow
we find the Reynolds number as..
200 m 1 min
0.2 m
VD VD
min 60 s
Re 


 3.758 x10 4


1.774 x10 5 m 2
s
This is within the range of 5,000 – 100,000 for which the equation for Nu in Table 7-1 i9s valid.
Using thie equation we find
Nu 

hD
 0.102 Re 0.675 Pr 1 3  0.102 3.758 x10 4
k
 0.7235
0.675
13
 112.2
We can now find the heat transfer coefficient from the equation given.
h
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
kNu 0.02717 W 112.2 15.24 W


D
m o C 0.2 m m 2 o C
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise eight solutions
ME 375, L. S. Caretto, Spring 2007
Page 2
We can get the desired answer by using this value of h in our usual equation for convective heat
transfer. The area for heat transfer is the total outside area of the square duct. This is the area
of four sides, each with an area of (0.2 m)(1,5 m) = 0,3 m 2, giving a total surface area for heat
transfer of1.2 m2.
q  hATs  T  



15.24 W
1.2 m 2 65 o C  30 o C  640 W
2 o
m  C
2. Air enters a 25-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a mean
velocity of 7 m/s, and is cooled by the water outside. If the average heat transfer
coefficient is 85 W/m2▪°C and the tube temperature is nearly equal to the water temperature
of 10°C, determine the exit temperature of air and the rate of heat transfer.
We find the following properties of air from Table A-15 at a mean temperature of the air and the
water, (10oC + 50oC)/2 = 30oC:  = 1.164 kg/m3 and cp = 1007 J/kg▪oC. We first find the mass
flow rate of the water.
m  AcVavg  

4
D 2Vavg 
1.164 kg 
0.25 m2 7 m  0.400 kg
3
4
s
s
m
The surface area for heat transfer is the area of the cylinder walls.
As  DL   0.25 m 12 m   9.425 m 2
The temperature at the end of the tube is found from the number of transfer units, NTU, which are
computed below.


85 W
9.425 m 2
2 o
hAs
m

C
NTU 

 1.989
m c p 0.400 kg 1007 J W  s
s
kg  s 1 J
We can now find the temperature at the end of the tube from the following equation.


Tout  Ts  Ts  Tin e  NTU  10 o C  10 o C  50 o C e 1.989  15.47oC
We can now find the log-mean temperature difference
LMT 
Tout  Tin 
T T 
ln  out s 
 Tin  Ts 

15.47o C  50o C
 17.36o C
o
o
 15.47 C  10 C 

ln 
o
o
 50 C  10 C 
With this temperature difference we can find the heat transfer.



85 W
1 kW
Q  hALMT   2 o 9.425 m 2 17.36 o C
 13.9 kW
1000 W
m  C