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Worked Out Examples
(Thermal B.L.)
Example 1 (Convection Coefficient):
Air at a free stream temperature of T = 20 C is in parallel flow over a flat plate of length L
= 5 m and temperature Ts = 90 C. However, obstacles placed in the flow intensify mixing
with increasing distance x from the leading edge, and the spatial variation of temperatures
measured in the boundary layer is correlated by an expression of the form T(C) = 20 +
70e(- 600 x y), where x and y are in meters. Determine and plot the manner in which the local
convection h varies with x. Evaluate the average convection coefficient h for the plate.
1. Statement of the Problem
a) Given
 Free stream air temperature T = 20 C
 Plate length L = 5 m
 Plate surface temperature Ts = 90 C
 Correlated measured temperature in the boundary layer: T(C) = 20 + 70e(- 600 x y),
where x and y are in meters
b) Find
 Determine and plot the manner in which the local convection h varies with x.
 Evaluate the average convection coefficient h for the plate.
2. System Diagram
T = 20C
T x, y   20  70  e600xy 
y
Ts = 90C
L=5m
x
3. Assumptions
 Steady state condition
 Uniform free stream air temperature T = 20 C = constant
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Uniform surface temperature Ts = 90 C = constant
Constant thermal conductivity


4. Governing Equations
 Newton's Law of Cooling

qs  hTs  T 
One Dimensional Fourier's Law
T
y
q   k f
On the plate surface y = 0  q s  k f

T
y
y 0
Average Convection Coefficient Definition
h
1
As

As
h  dAs
For the special case of flow over a flat plate, h varies with the distance x from the leading
edge. Thus,
h
1 L
h  dx
L 0
5. Detailed Solution
Local Convection Coefficient, h
qs  hTs  T  … Newton's law of cooling
q s  k f
T
y
… One-dimensional Fourier's law on the plate surface
y 0
Thus,
q s  hTs  T   k f
Therefore,
h



1
T
  k f 
Ts  T
y
kf
Ts  T
kf
Ts  T
kf
Ts  T

y 0


20  70  e 600xy
y


y 0
 0  70   600 x   e 600xy
 70   600 x   1

y 0
T
y
y 0
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42000  k f  x
 h x  
Ts  T
Taking the average of the free stream and surface temperatures:
T 
20  90  55C
2
 kf = 0.02837 W/mK
After plugging numbers into the expression obtained above, it becomes:
hx  17.02  x W/m2K
Using MatLab, the variation of local convection coefficient can be plotted as:
Variation of Local Convection Coefficient
90
80
70
h (W/m2.K)
60
50
h(x)
40
30
20
10
0
0
0.5
1
1.5
2
2.5
x (m)
3
Average Convection Coefficient, h
The average coefficient over the range 0  x  5 m is
h

1 L
h  dx
L 0
1 L
17.02  x  dx
L 0
L
1
x2 
 17.02  
L
2 0

1
L2
17.02 
L
2
 8.51  L



3.5
4
4.5
5
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 h  42.6 W/m2K
6. Critical Assessment
Because the local convection coefficient is a function of x, the average of the convection
coefficient must be obtained by integrating the function over the whole range of the flat plate.
Example 2 (Velocity and Temperature Profiles):
In flow over a surface, velocity and temperature profiles are of the forms
u(y) = Ay + By2 - Cy3 and
T(y) = D + Ey + Fy2 - Gy3
where the coefficients A through G are constants. Obtain expressions for the friction
coefficient Cf and the convection coefficient h in terms of U , T , and appropriate profile
coefficients and fluid properties.
1. Statement of the Problem
a) Given
 Velocity and temperature profiles
u(y) = Ay + By2 - Cy3 and
T(y) = D + Ey + Fy2 - Gy3
where the coefficients A through G are constants.
b) Find
 Expression for the friction coefficient Cf
 Expression for the convection coefficient h
Both expressions must be in terms of U , T , and appropriate profile coefficients and
fluid properties.
2. System Diagram
U , T
Velocity Profile, u(y)
Temperature Profile, T(y)
Velocity B.L.
Thermal B.L.
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3. Assumptions
 Steady state condition
 Constant air properties
 Uniform U , T = constant
4. Governing Equations

Friction Coefficient Definition
Cf 

s
1
U 2
2
Shear Stress Definition
 
On the surface,  s  
u
y
y 0

Newton's Law of Cooling

One Dimensional Fourier's Law
qs  hTs  T 
q   k f
On the plate surface y = 0  q s  k f
5. Detailed Solution
Friction Coefficient, Cf
u
y
T
y
T
y
y 0
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s  

u
y
y 0


Ay  By 2  Cy 3
y

   A  2 By  3Cy 2


Therefore,
s
1
U 2
2

y 0

y 0
   A  2 B0   3C 0 
A
Cf 
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2

A
1
U 2
2
Convection Coefficient, h
q s  hTs  T   k f
T
y
y 0
Thus,
h
1
T
  k f 
Ts  T
y
y 0
Here, Ts = T(y = 0) = D + E (0) + F (0)2 - G (0)3 = D
h 



1

  k f  
D  Ey  Fy 2  Gy 3
D  T
y
kf
D  T
kf
D  T

 0  E  2 Fy  3Gy 2


y 0

y 0
 0  E  2 F 0  3G 0
2

Finally,
h
kf E
D  T
6. Critical Assessment
It is important to recognize (or know) that for both cases, the friction coefficient and
convection coefficient, an analysis must be done on the surface, which implies y = 0 m.
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Example 3 (Viscous dissipation and Heat Transfer Rate):
A shaft with a diameter of 100 mm rotates at 9000 rpm in a journal bearing that is 70 mm
long. A uniform lubricant gap of 1 mm separates the shaft and the bearing. The lubricant
properties are  = 0.03 Ns/m2 and k = 0.15 W/mK, while the bearing material has a thermal
conductivity of kb = 45 W/mK.
y (mm)
Bearing, kb
Tb
1
Bearing, kb
0
Lubricant
x
Shaft
Ts
Lubricant
Shaft
100 mm
diameter
200 mm
Water-cooled surface,
Twc = 30C
(a) Determine the viscous dissipation,  (W/m3), in the lubricant.
(b) Determine the rate of heat transfer (W) from the lubricant, assuming that no heat is lost
through the shaft.
(c) If the bearing housing is water-cooled, such that the outer surface of the bearing is
maintained at 30 C, determine the temperature of the bearing and shaft, Tb and Ts.
1. Statement of the Problem
a) Given
 Di = 0.1 m
  = 9000 rpm = 942.5 rad/s
 L = 0.07 m
 a = 0.001 m (gap)
  = 0.03 Ns/m2
 k = 0.15 W/mK
 kb = 45 W/mK
 Twc = 30 C
 Do = 0.2 m
b) Find
 Viscous dissipation in the lubricant,  (W/m3)
 Rate of heat transfer (W) from the lubricant, assuming no heat loss through the shaft
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
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Temperatures of the bearing and shaft, Tb and Ts
2. System Diagram
Bearing (kb)
a
Shaft (Di, )
Bearing
y
(mm)
1
0
Tb
Lubricant
x
Lubricant (, k)
Shaft
Ts
Water-cooled surface (Twc)
D
o
3.





Assumptions
Steady state condition
Constant fluid properties (, , and k's)
Fully developed flow in the gap (u/x = 0)
Infinite width [L/a = (0.07 m) / (0.001 m) = 70, so this is a reasonable assumption]
p/x = 0 (flow is symmetric in the actual bearing at no load)
4. Governing Equations
 2-D Dissipation Function
2
 u  2  v  2  2  u v  2
 u v 
      2         
 x   y   3  x y 
 y x 

Velocity Distribution in Couette Flow (flow in two infinite parallel plates, but one plate
moving with constant speed)
u( y) 

U
1  p  2
y
  y  ay 
a
2  x 
Heat Diffusion Equation in Cylindrical Coordinates
1   T  1   T    T 
T
 kr
 2
k
  k
  q  c p
r r  r  r     z  z 
t

Fourier's Law in Cylindrical Coordinates


  T  1 T  T 
q   kT  k  er
 e
 ez

r 
z 
 r

2-D Energy Equation
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 T
T    T    T 
  k
    q 
v
  k
y  x  x  y  y 
 x
c p  u
5. Detailed Solution
Viscous dissipation in the lubricant
Assume v  0 in the gap, and the fully developed flow (assumed) implies
u
 0 . Thus,
x
2
2
 u  2  v  2  2  u v  2
 u 
 u v 
      2              
 y 
 x   y   3  x y 
 y x 
Because p/x = 0 (assumed), the velocity distribution becomes:
u( y) 
U
U
1  p  2
y
  y  ay   u ( y )  y
a
a
2  x 
Therefore, the viscous dissipation is
2
 u 
  U
        
 y 
 y  a
2

U 
y     

a
2
where U is the tangential velocity of the shaft, and it is
U  Ri   
Di

2
Finally, the viscous dissipation is
 D  
U 
        i 
a
 2a 
2
2
After plugging in values into this expression above,
 = 6.662  107 W/m3
Rate of heat transfer (W) from the lubricant
The heat transfer rate from the lubricant volume  through the bearing is
qL =   =   ( Di  a  L)
 qL = (6.662  107 W/m3)[()  (0.1 m)  (0.001 m)  (0.07 m)] = 1465 W
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where L = 0.07 m is the length of the bearing normal to the page.
Temperatures of the bearing and shaft, Tb and Ts
First, let us find out the bearing temperature (Tb), which requires considering heat transfer
between Tb and Twc. See the diagram below.
Direction of Heat Transfer
Do /2
r
Di /2
Ts Tb
Twc
Assume that the direction of heat transfer is in only r direction. Then Fourier's law becomes:
q r   k
T
T
or q r   kA
r
r
In our case,
q r  k b  2rL  
T
… (1)
r
Heat diffusion equation is (assuming no heat generation in the bearing)
1   T 
 kr
0
r r  r 
In our case, because kb = constant,
1  
T 
1   T 
  T 
 kb r
  0  kb
r
0 
r
0
r r 
r 
r r  r 
r  r 
Boundary conditions for this differential equation are:
T = Tb @ r = Di /2
T = Twc @ r = Do /2
Let us solve the differential equation with the boundary conditions,
T
  T 
T C1
 C1 
 T (r )  C1 ln( r )  C2

r
0  r
r
r  r 
r
r
The first boundary condition: Tb = C1 ln(Di/2) + C2
The second boundary condition: Twc = C1 ln(Do/2) + C2
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After rearranging, the temperature distribution becomes:
T (r ) 
 r 
Tb  Twc
  Twc
ln 
ln Di Do   Do 2 
Substituting this temperature distribution into equation (1),
q r  k b  2rL  

 r 
T  Twc 1
  Tb  Twc
  Twc   k b  2rL   b
ln 




r  ln Di Do   Do 2 
ln
D
D
r
i
o

Therefore,
qr 
 Tb  Twc 
k b  2L   Tb  Twc 
 qL
ln Di Do 
q L  ln Do Di 
1465W   ln 0.2m 0.1m  81.3C
 30C  
2Lk b
2 0.07m45W / m  K 
Finally, let us find out the shaft temperature (Ts).
y
Tb
k&
a
U
x
Ts
The 2-D energy equation may be simplified for the prescribed conditions (see assumptions)
and further assuming v  0 and q  0 , it follows that
 u 
 T    T    T 
    
c p  u    k
   k
 x  x  x  y  y 
 y 
2
However, because the top and bottom plates are at uniform temperatures, the temperature
field must also be fully developed, in which case (T/x) = 0. For constant thermal
conductivity the appropriate form of the energy equation is then
 u 
 2T
0  k 2    
y
 y 
2
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The desired temperature distribution may be obtained by solving this equation. Rearranging
and substituting for the velocity distribution,
2
 du 
d 2T
U 
k 2          
dy
a
 dy 
2
Integrating twice, we obtain
T ( y)  
 U 
2
2
  y  C1 y  C 2
2k  a 
The boundary conditions are, at y = 0, the surface is adiabatic
dT
dy
 0  C3 = 0
y 0
and at y = a, the temperature is that of the bearing, Tb
T (a)  Tb  
 U 

2
2
2
  a  0  C 2  C 2  Tb  U
2k
2k  a 
Hence, the temperature distribution is
T ( y )  Tb 


y2 
  Di 
U 2 1  2   Tb 
  
2k
2k  2

 a 
2

y2 
1  2 
 a 
and the temperature at the shaft, y = 0, is


0.03N  s / m 2  0.1m

942.5rad / s 
Ts  T (0)  Tb       81.3C  

2k  2
2  0.15W / m  K   2


  Di
2
2
 Ts  303.4C
6. Critical Assessment
We have dealt with both heat conduction and convection situation on this problem. Make sure
you understand the difference between them and how to apply an appropriate equation for a
particular case.
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Example 4 (Use of Similarity Rules and Correlation Parameters):
An industrial process involves evaporation of a thin water film from a contoured surface by
heating it from below and forcing air across it. Laboratory measurements for this surface
have provided the following heat transfer correlation:
Nu L  0.43  Re 0L.58  Pr 0.4
The air flowing over the surface has a temperature of 290 K, a velocity of 10 m/s, and is
completely dry ( = 0). The surface has a length of 1 m and a surface area of 1 m2. Just
enough energy is supplied to maintain its steady-state temperature at 310 K.
(a) Determine the heat transfer coefficient and the rate at which the surface loses heat by
convection.
(b) Determine the mass transfer coefficient and the evaporation rate (kg/h) of the water on
the surface.
(c) Determine the rate at which heat must be supplied to the surface for these conditions.
1. Statement of the Problem
a) Given
Heat transfer correlation equation: Nu L  0.43  Re 0L.58  Pr 0.4
Forcing air properties:
 T = 290 K (temperature)
 U = 10 m/s (velocity)
  = 0 (completely dry)
 Surface dimensions and property:
 L = 1 m (length)
 As = 1 m2 (area)
 Ts = 310 K (temperature)
b) Find
 Heat transfer coefficient
 Rate at which the surface loses heat by convection
 Mass transfer coefficient
 Evaporation rate (kg/h) of the water on the surface
 Rate at which heat must be supplied to the surface for these conditions


2. System Diagram
Heat transfer correlation: Nu L  0.43  Re 0L.58  Pr 0.4
Air
T
U

Surface
Ts
L
As
Thin water film
Ghosh - 550
3.



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Assumptions
Steady state condition
Constant properties
Heat-mass analogy applies:
Heat Transfer
2/5/2016
Mass Transfer
Nu  f1 Re L , Pr 

Sh  f 2 Re L , Sc 
T  Ts
Correlation requires properties evaluated at Tmean  
 300 K
2
4. Governing Equations
 V  L


Reynolds Number: Re L 

Prandtl Number: Pr 

Schmidt Number: Sc 

Average Nusselt Number: Nu 

Average Sherwood Number: Sh 

Newton's Law of Cooling: q conv  h  As  Ts  T 

  hm  As   A,s   A,
Convection Mass Transfer Equation: m

First Law of Thermodynamics (for steady flow process): E in  E out  0



D AB
hL
kf
hm L
D AB

5. Detailed Solution
Properties:
Air (at Tmean = 300 K, 1 atm)
  = 15.89  10-6 m2/s
 kf = 0.0263 W/mK
 Pr = 0.707
Air-water mixture (at Tmean = 300K, 1 atm)
 DAB = 0.26  10-4 m2/s
Saturated water (at Ts = 310 K)
 A, sat = 1/vg = 1/22.93 m3/kg = 0.04361 kg/m3

Ghosh - 550

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2/5/2016
hfg = 2414 kJ/kg
Heat transfer coefficient
First of all, evaluate ReL at Tmean to characterize the flow
Re L 
UL


10m / s   1m
15.89 10
6
2
m /s

 6.293  10 5
and substituting into the prescribed correlation for this surface, find
Nu L  0.43  Re L
h 
0.53

 Pr 0.4  0.43  6.293  10 5
Nu L  k f
L


0.58
 0.707 
0.4
 864.1 
hL
kf
864.1  0.0263W / m  K   22.72 W/m2K
1m 
Rate at which the surface loses heat by convection

 
qconv  h  As  Ts  T   22.71W / m 2  K  1m 2  310 K   290 K   454.2W
Mass transfer coefficient
Using the heat-mass analogy,
Heat: Nu L  0.43  Re 0L.58  Pr 0.4
Mass: Sh L  0.43  Re 0L.58  Sc 0.4
where
Sc 

D AB
15.89  10 6 m 2 / s

 0.6112
0.26  10 4 m 2 / s
Substituting numerical values, and find
Sh L  0.43  Re L
hm 
0.53

 Sc 0.4  0.43  6.293  10 5

0.58
 0.6112
0.4
 815.2 
Sh L  D AB 815.2  0.26  10 4 m 2 / s 

 2.12  10  2 m / s
L
1m
Evaporation rate (kg/h) of the water on the surface
The evaporation rate, with A,s = A,sat (Ts), is
hm L
D AB
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2/5/2016
   

 
  hm  As   A,s   A,   2.12  10 2 m / s  1m 2  0.04361kg / m 3  0kg / m 3
m

  9.243  10 4 kg / s  3.327kg / h
m
Rate at which heat must be supplied to the surface for these conditions
Air
qconv
qevap
qin
Applying the first law of thermodynamics,
E in  E out  0
qin  q conv  q evap  0
where qin is the heat supplied to sustain the losses by convention and evaporation.
qin  q conv  q evap
qin  h  As  Ts  T   m h fg
qin  454.2W   9.243  10  4 kg / s   2414  10 3 J / kg 
qin  454.2W   2231.3W 
qin  2685W
6. Critical Assessment
 Heat-mass analogy has been applied in this problem. Note that convection mass transfer
can be analyzed like convection heat transfer. Equations are very similar to each other.
 Notice that the heat loss from the surface by evaporation is nearly 5 times that due to
convection.
The End