ENSC 388 Week # 10, Tutorial # 9–Heat Transfer Coefficient
Problem 1: Air flows over the top and bottom surfaces of a thin, square
plate. The flow regime and the total heat transfer rate are to be determined
and the average gradients of the velocity and temperature at the surface are
to be estimated.
Air
T∞ = 10˚C
u∞ = 60 m/s
Ts = 54˚C
L=0.5 m
Solution
Step 1: Write out what you are required to solve for (this is so you don’t
forget to answer everything the question is asking for)
Find:
a)
b)
c)
d)
Flow regime.
Total Heat Transfer Coefficient.
Average gradients of the velocity at the surface.
Average gradients of the temperature at the surface.
Step 2: Read properties from the corresponding table
The properties of air at the film temperature (Table A-2)
𝑇𝑓 =
M. Bahrami
𝑇𝑠 + 𝑇∞
= 32 [℃]
2
ENSC 388
Tutorial # 9
1
ρ = 1.156 [
𝑚3
𝑘𝑔
]
𝑘𝐽
𝑐𝑝 = 1.007 [
]
𝑘𝑔℃
υ = 1.627 × 105 [
𝑊
𝑚2
𝑠
]
k = 0.02603 [ ]
𝑚𝐾
Pr = 0.7276
Step 3: State your assumptions (you may have to add to your list of
assumptions as you proceed in the problem)
Assumptions:
1) Steady operating conditions exist.
2) The critical Reynolds number is Recr = 5105.
3) Radiation effects are negligible.
Step 4: Calculations
(a) The Reynolds number is
Re L 
VL


(60 m/s)(0.5 m)
 1.844  10 6
5
2
1.627  10 m /s
which is greater than the critical Reynolds number. Thus we have turbulent
flow at the end of the plate.
(b) We use modified Reynolds analogy to determine the heat transfer
coefficient and the rate of heat transfer
s 
Cf 
s
0.5V
Cf
2
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2
F
1.5 N
N

3 2
2
A 2(0.5 m)
m 
3 N/m 2

3
0.5(1.156 kg/m )(60 m/s)
 St Pr 2 / 3 
2
 1.442 10 3
Nu L
Nu L
Pr 2 / 3 
Re L Pr
Re L Pr 1 / 3
ENSC 388
Tutorial # 9
2
(1.442 10 3 )
Nu  Re L Pr
 (1.844 10 )(0.7276 )
 1196
2
2
k
0.02603 W/m. C
 W 
h  Nu 
(1196)  62.26  2 
L
0.5 m
 m C 
Q  hAs (Ts  T )  (62.26 W/m 2 C)[2  (0.5 m) 2 ](54  10)C = 1370 [W]
1/ 3
Cf
6
1/ 3
(c) Assuming a uniform distribution of heat transfer and drag parameters
over the plate, the average gradients of the velocity and temperature at the
surface are determined to be
s
u
u
3 N/m 2
s  




 1.60 10 5 [s -1 ]
3
5
2
y 0
y 0  (1.156 kg/m )(1.627 10 m /s )
k
h
T
y
0
Ts  T
M. Bahrami


T
y

0
 h(Ts  T ) (62.26 W/m 2  C)(54  10)C
 C 

 1.05 105  
k
0.02603 W/m  C
m
ENSC 388
Tutorial # 9
3
Problem 2: The wind is blowing across the wire of a transmission line. The
surface temperature of the wire is to be determined.
Wind
𝒖∞ = 𝟒𝟎 𝒌𝒎/𝒉
Transmission
Wire
𝑻∞ = 𝟏𝟎 ℃
𝑻𝒔
𝑫 = 𝟎. 𝟔 𝒄𝒎
Solution
Step 1: Write out what you are required to solve for (this is so you don’t
forget to answer everything the question is asking for)
Find:
a) Surface temperature of the wire.
Step 2: Read properties from the corresponding table
We assume the film temperature to be 10C. The properties of air at the film
temperature (Table A-2)
ρ = 1.246 [
𝑚3
𝑘𝑔
]
υ = 1.426× 10−5 [
𝑊
𝑚2
𝑠
]
k = 0.02439 [ ]
𝑚𝐾
Pr = 0.7336
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ENSC 388
Tutorial # 9
4
Step 3: State your assumptions (you may have to add to your list of
assumptions as you proceed in the problem)
Assumptions:
1)
2)
3)
4)
Steady operating conditions exist.
Radiation effects are negligible.
Air is an ideal gas with constant properties.
The local atmospheric pressure is 1 atm.
Step 4: Calculations
The Reynolds number is
Re 
u D


(40 1000/3600) m/s (0.006 m)  4675
1.426 10 5 m 2 /s
The Nusselt number corresponding to this Reynolds number is determined to
be
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
5/8
0.62 (4675 ) 0.5 (0.7336 )1 / 3   4675  
1  
 0.3 
 
2 / 3 1/ 4 
282
,
000
 
1  0.4 / 0.7336 
 

4/5

 36 .0
The heat transfer coefficient is
h
k
0.02439 W/m. C
 W 
Nu 
(36.0)  146.3  2 
D
0.006 m
 m C 
The rate of heat generated in the electrical transmission lines per meter
length is
W  Q  I 2 R  (50 A) 2 (0.002 Ohm) = 5.0 [W]
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ENSC 388
Tutorial # 9
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The entire heat generated in electrical transmission line has to be transferred
to the ambient air. The surface temperature of the wire then becomes
As  DL   (0.006 m)(1 m) = 0.01885 [m 2 ]
Q
5W
Q  hAs (Ts  T ) 
 Ts  T 
 10C +
 11.8 [C]
2
hAs
(146.3 W/m .C)(0.01885 m 2 )
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ENSC 388
Tutorial # 9
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