College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
Solutions to Exercise Seven – Introduction to Convection
1. During air cooling of oranges, grapefruit, and
tangelos, the heat transfer coefficient for combined
convection, radiation, and evaporation for air
velocities of 0.11 < V < 0.33 m/s is determined
experimentally and is expressed as h = 5.05 kairRe1/3/D,
where the diameter D (in meters) is the characteristic
length in the Reynolds and Nusselt numbers. (E.g. Re
= VD/.) The constant 5.05 is dimensionless; the
units for h and k are W/m2·°C and W/m·°C. Oranges
are cooled by refrigerated air at 5°C and 1 atm at a
velocity of 0.3 m/s. Determine (a) the initial rate of
heat transfer from a 7-cm-diameter orange initially at 15°C with a thermal conductivity of
0.50 W/m·°C, (b) the value of the initial temperature gradient inside the orange at the
surface, and (c) the value of the Nusselt number. (Problem and figure P6.11 from Çengel,
Heat and Mass Transfer.)
(a) To determine the heat transfer we first have to find the heat transfer coefficient from the
equation given. We can find the properties of air at the mean (film) temperature of (5 oC
15oC)/2 = 10oC from Table A-15: k = 0.02349 W/m·oC and  = 1.426x10-5 m2/s. We can
compute the Reynolds from the given data on V and D and the kinematic viscosity just found.
0.3 m
0.07 m 
VD VD
s
Re 


 1473


1.426 x10  5 m 2
s
We can now find the heat transfer coefficient from the equation given.
5.05k air Re
h
D
13
5.05

0.02349 W
m C
0.07 m
o
14731 3

20.02 W
m 2 o C
When the orange is at its initial temperature of 15oC, the initial convective heat transfer is
found as follows.


20.02 W
Q  hAT  T   hD 2 T  T   2 o 0.07 m2 15o C 5o C  3.08 W
m  C
(b) Since this is also the conduction heat heat leaving the surface of the potato we can find the
temperature gradient there by the Fourier Law.
dT
Q  kA
dr

dT
q
3.08 W


0.50 W
dr
kA
0.07 m 2
o
m C
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
dT
400 oC

dr
m
Phone: 818.677.6448
Fax: 818.677.7062
Exercise seven solutions
ME 375, L. S. Caretto, Spring 2007
Page 2
(c) We are told that the characteristic length is the diameter so the Nusselt number is Nu = hD/k
for this problem.
0.07 m 
hD m 2 o C

0.02349 W
k
m o C
20.02 W
Nu 
Nu = 57.5
2. A long 8-cm-diameter steam pipe whose external surface temperature is 90°C passes
through some open area that is not protected against the winds. Determine the rate of
heat loss from the pipe per unit of its length when the air is at 1 atm pressure and 7°C and
the wind is blowing across the pipe at a velocity of 50 km/h. The average heat transfer
coefficient for flow over a cylinder is found from the following equation where the
characteristic distance in the Reynolds and Nusselt numbers is the cylinder diameter, D.
58
0.62 Re 0.5 Pr1 3   Re  
1  
Nu  0.3 
 
14
  282,000  
1  0.4 Pr 2 3

45

We first have to find the properties for air at the mean temperature of (90oC + 7oC)/2 = 48.5oC
from Table A-15: k = 0.02724 W/m·K, n = 1.784x10-5 m2/s, and Pr = 0.7232.
The Reynolds number is
Re 
VD VD




50 km 1 h 1000 m
0.08 m
h 3600 s km
1.426 x10
s
5
m
2
 6.228 x10 4
From the equation given for the heat Nusselt number, we find
hD
0.62 Re 0.5 Pr1 3
Nu 
 0.3 
14
k
1  0.4 Pr 2 3
0.3 




  Re  5 8 
1  
 
  282,000  
0.72321 3 1   6.228 x10 4 
1  0.4 0.72322 3 1 4   282,000 
0.62 6.228 x10 4
0.5
45
5 8 4 5



 159.1
We can compute the heat transfer coefficient from the Nusselt number.
0.02724 W
k Nu
h

D
159.1
54.17 W
m o C
 2 o
0.08 m
m  C
The heat transfer is given by the usual convection equation, which we can solve for the heat
transfer per unit length.
 hAT  T   hDL T  T  
Q
hDT  T 
L
Exercise seven solutions
ME 375, L. S. Caretto, Spring 2007

Q
54.17 W
 hDT  T   2 o 0.08 m 90o C  7 o C
L
m  C
Page 3
Q 1130 W

L
m

3. Repeat problem 2 using water instead of air as the flowing fluid.
We first have to find the properties for air at the mean temperature of (90oC + 7oC)/2 = 48.5oC
from Table A-9: k = 0.641 W/m·K,  = 5.682x10-7 m2/s, and Pr = 3.782.
The Reynolds number is
Re 
VD VD




50 km 1 h 1000 m
0.08 m 
h 3600 s km
5.682 x10
s
7
m
2
 1.956 x10 6
From the equation given for the heat Nusselt number, we find
hD
0.62 Re 0.5 Pr1 3
Nu 
 0.3 
14
k
1  0.4 Pr 2 3
0.3 


0.62 1.956 x10

6 0.5

3.782
13
1  0.4 3.782 
2314
  Re  5 8 
1  
 
  282,000  
 
6 5 8 
1   1.956 x10  
  282,000  


45
45
 1104
We can compute the heat transfer coefficient from the Nusselt number.
1104
o
k Nu
8848 W
m

C
h

 2 o
D
0.08 m
m  C
0.641 W
Substitute the results for this problem into the equation for heat transfer per unit length from
problem 2.

Q
8848 W
 hDT  T   2 o 0.08 m 90o C  7 o C
L
m  C

Q 184,564 W

L
m
For the same flow conditions, the convection heat transfer with water is much higher than that
with air.