H.T through flate plate with reynolds.ppt

heat transfer in flat plate

2. Flow and Heat Transfer
over Flat Plates

3. Flow and Heat Transfer over Flat Plates
In this section, for laminar and turbulent flow conditions heat
transfer rate from/ to flat plates is determine.
Drag force exerted by fluid is determine.
Surfaces that are slightly contoured such as turbine blades can
also be approximated as flat plates with reasonable accuracy.
Consider laminar flow of a fluid over a flat plate, as shown in Fig.
The x-coordinate is measured along the plate in the direction of
the flow, y-coordinate is measured from the surface in the
normal direction.

The fluid approaches the plate in the x-direction with a uniform
upstream velocity.

6. Flow and Heat Transfer over Flat Plates
The friction and heat transfer coefficients for flat plate can be
determined theoretically by solving:
Conservation of mass
Momentum
Energy equations approximately
They can also be determined by experimentally and
Expressed by empirical correlations:
Nusselt Number
Reynolds Number
Prandtl Number
nm
eLCR
k
hL
Nu Pr

7. Flow and Heat Transfer over Flat Plates
In thermal boundary layer fluid temperature varies from surface to the
outer edge of the thermal boundary layer.
Fluid properties also varies with temperature
In order to account the variation of properties with temperature properly
Fluid properties evaluated at film temperature
(arithmetic average)
2
TT
T
s
f



8. For laminar Flow over a Flat Plate

9. For laminar Flow over a Flat Plate
Local Friction Co-efficient
Average Friction Co-efficient
 Local Nusselt Number
Average Nusselt Number
5.0
2/1, Re664.0
Re
664.0 

x
x
xfC
5.0
2/1
328.1
328.1 

eL
eL
f R
R
C
3/1
PrRe332.0
2/1
x
x
x
k
xh
Nu 
3/1
PrRe664.0
2/1
L
k
hL
Nu 

10. For Turbulent Flow over a Flat Plate

11. For Turbulent Flow over a Flat Plate
Local Friction Co-efficient
Average Friction Co-efficient
 Local Nusselt Number
Average Nusselt Number
5/1,
Re
0592.0
x
xf
C
5/1
Re
074.0
L
fC
3/1
PrRe0296.0
5/4
x
x
x
k
xh
Nu 
3/1
PrRe037.0
5/4
L
k
hL
Nu 
75
10Re105 
x

12. EXAMPLE : Flow of Hot Oil over a Flat Plate
Engine oil at 60°C flows over the upper surface of a 5-m-long flat
plate whose temperature is 20°C with a velocity of 2 m/s. Determine
the total drag force and the rate of heat transfer per unit width of the
entire plate.
SOLUTION:
Assumptions: 1. The flow is steady and incompressible.
2. The critical Reynolds number is Re
cr
= 5x10
5
.
Properties The properties of engine oil at the film temperature of
from (Table A–10).
smvCmWk
mkg
o
/10242,./144.0
2870Pr,/876
26
3



 
  CT
TTT
o
f
sf
402/2060
2/




13. 4
6
1013.4Re
10242
52
Re






L
L
v
Lu
This value is less than the critical Reynolds number. Thus we
have laminar flow over the entire plate,
Therefore, average friction coefficient is determine
 
00653.0
1013.4328.1
328.1
328.1
5.0
4
5.0
2/1





f
f
eL
eL
f
C
C
R
R
C
)(
sTThAQ 

 3/1
PrRe664.0
5.0
L
k
hL
Nu 
2


V
ACF
fD


14. NF
V
ACF
D
fD
2.57
2
2876
)15(00653.0
2




Drag force acting on the plate per unit width becomes
Similarly, the Nusselt number is determined using the laminar flow
relations for a flat plate,
1918
)2870()1013.4(664.0
PrRe664.0
3/15.04
3/15.0



Nu
Nu
k
hL
Nu
L

15. Then, convective heat transfer co-efficient (h) is determine by
And heat transfer rate through the plate is given by
CmWh
h
Nu
L
k
h
o2
/2.55
1918
5
144.0



WQ
Q
TThAQ
s
11040
)2060)(15(2.55
)(







Discussion: Note that heat transfer is always from the higher-temperature medium
to the lower-temperature one. In this case, it is from the oil to the plate. The heat
transfer rate is per m width of the plate.
The heat transfer for the entire plate can be obtained by multiplying the value
obtained by the actual width of the plate.

16. Combined laminar and
Turbulent Flow over Flat
Plate

17. Combine Laminar and Turbulent Flow
In some cases, a flat plate is
sufficiently long for the flow to
become turbulent.
but not long enough to disregard the
laminar flow region.
In such cases, the average friction
co-efficient and Nusselt number over
the entire plate is determined by
performing the integration of
governing equations.

18. Combine Laminar and Turbulent
Flow
Once the average friction and heat transfer coefficients are determined
Heat transfer rate and the drag force can easily be determine from:
2
2


V
ACF
fD

3/15/4
5/1
Pr)871Re037.0(
Re
1742
Re
074.0


L
LL
f
k
hL
Nu
C )10Re105(
75

L
)(

 TThAQ
s


19. Example:
The local atmospheric pressure of a hilly place (elevation 1610 m),
is 83.4 kPa. Air at this pressure and 20°C flows with a velocity of 8
m/s over a 1.5 m x 6 m flat plate whose temperature is 140°C as
shown in Fig. Determine the rate of heat transfer from the plate, if
the air flows parallel to the
(a)6-m-long side and
(b) the 1.5-m side.
Assumptions:1. Steady operating conditions exist.
2. The critical Reynolds number is Re
cr = 5 x10
5
.
3. Radiation effects are negligible.
4. Air is an ideal gas.

20. Properties: The properties k, , Cp, and Pr of ideal gases are
independent of pressure, while the properties are inversely
proportional to density and thus pressure.

The properties of air at the film temperature and 1 atm pressure
Taken from property (Table A–15)
  
atm
smv
CmWk
CT
CTTT
o
o
f
o
sf
1@
/10097.2
7154.0Pr
/02953.0
80
802/201402/
25







22. smv
v
atmPatmvv
/1050.2
823.0/1006.2
)(/1@
25
5





The atmospheric pressure of hilly place is
P = (83.4 kPa)/(101.325 kPa/atm)=0.823 atm.
Then the kinematic viscosity of air at that place becomes
Analysis:
(a)When air flows along the long side of 6-m and
the Reynolds Number at the end of the plate becomes,
Which is greater than the critical Reynolds Number.
6
5
10884.1Re
1050.2
68
Re






L
L
v
LV

23. Thus, we have combined laminar and turbulent flow, and the average
Nusselt number for the entire plate is determined to be
Then the convective coefficient
2727
706.0)871)10884.1(037.0(
3/15/46


Nu
k
hL
Nu
2
02
9
65.1
/5.13
)2727(
6
0297.0
mA
LwA
CmWh
Nu
L
k
h




3/15/4
Pr)871Re037.0( 
L
k
hL
Nu

24. Now, heat transfer rate
Note that if we disregarded the laminar region and assumed
turbulent flow over the entire plate, we would get Nu = 3466 from
turbulent Eq,
which is 29 % higher than the value calculated above.
(b) When air flow is along the short side, we have L =1.5 m, and the
Reynolds number at the end of the plate becomes
WQ
Q
TThAQ
s
14300
)20140(95.13
)(







5
5
1071.1Re
10548.2
5.18
Re






L
L
v
LV

25. which is less than the critical Reynolds number. Thus we have
laminar flow over the entire plate, and the average Nusselt number
is determine from
And heat transfer rate
CmWh
Nu
L
k
h
Nu
k
hL
Nu
02
3/15.05
/12.8
408
5.1
0297.0
408
706.0)108.4(664.0




WQ
Q
TThAQ
s
8670
)20140(912.8
)(







which is considerably less than the heat transfer rate determined in case (a).

26. Discussion :
Note that the direction of fluid flow can have a significant effect on
convection heat transfer to or from a surface. In this case, we can
increase the heat transfer rate by 65 percent by simply blowing the
air along the long side of the rectangular plate instead of the short
side.

27. Problem
Air at 70F and 14.7psi flow over a plate at 20ft/sec. The plate is
2.5 ft long and maintained at 130 F. Calculate heat transfer from
The plate of 1ft edge.
At 100
o
F the temperature , variables obtained from table as:
=0.071
=0.0459 lb/hr ft.
Pr.=0.706
k= 0.01576
Cp=0.2406 Btu/lb
o
F
U=20ft/sec
F100
2
13070
T
o
m 


U=20
T
=70
o
F
T
s
=130
o
F

28. 3600
04590
522007080
R
e
.
.. 

This value of Reynolds number is lower than the 5x10
5
Therefore, the flow is laminar .
k
hL
Nu
*


L
kNu
h
5
e 10552R .

UL
R
e

We know that

29. hrBtu14147Q
7013015298090Q
ThAQ
/.
)(..






 
FhrftBtu98090h
52
0157607060105526640
h
o2
3
1
2
1
5
/.
.
.....



Now, substitute the values in equation (*)

3
1
2
1
eL
R6640Nu Pr..
For laminar flow Nusselt Number

30. Problem: Air flows over a plate at constant velocity of 100ft/sec
and ambient conditions 3 psi and 70
o
F. The plate is heated to
constant Temperature of 150
o
F. What is the total heat transfer
from leading to a point 1-ft from leading edge.
Data
T

=70
o
F
T
s
=150
o
F
P=3psi
U=100 ft/sec
At 110
o
F the temperature , variables obtained from table as:
=0.06965
=0.04656 lb/hr ft.
Pr.=0.7045
k= 0.015905
F110
2
15070
T
o
m



U=100
T
=70
o
F
T
s=150
o
F

31. 3600
046560
1100069650UL
R
e
.
. 



Cp=0.24055 Btu/lb
o
F
U=100ft/sec
We know that
5
e 103855R .
This value of Reynolds number is greater than the 5x10
5
Therefore, the flow is turbulent .
For turbulent flow Nusselt
Num:
Now, substitute the values and re-arrange the equation

3
1
5
4
eL
R0370
k
hL
Nu Pr..

32.   
FhrftBtu1320h
1
0159050704501038550370
h
o2
3
1
5
4
5
/.
.....



hrBtu1610Q
7015011320Q
ThAQ
/
)(.





