boudarylayer(undisturbed flow encounters the plate)
TRANSCRIPT
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as well. By analogy. we can expect that h(x) and Nu(x) will also decrease withdistance along the plate.
0 0.5 1 1.5 2 2.5 30
0.005
0.01
0.015
0.02
0.025
0.03
u [m/s]
y[m]
x=1mx=2mx=3m
Figure 2: Velocity profiles at three locations along the plate plotted on a sinlgeaxis.
The rate of heat transfer will be highest where the gradients are highest,and clearly this occurs at the leading edge. From our dimensional analysis, weexpect
h cfPrn (3)
since
cf = f(x,Re) (4)
Nu = f(x,Re,Pr) (5)
Example
Consider airflow over a plate of length 1m. Assume transition to turbulenceoccurs at xc = 0.5m, with Rec = 5 105. Given hlam(x) = Clamx0.5 andhturb(x) = Cturbx
0.2, with Clam = 8.845W/(m3/2K) and Cturb = 49.75W/(m
1.8K).
1. Using air at 350 K determine U
2. Develop an expression for the average convection coefficient as a functionof distance from the leading edge, for the laminar region.
3. Develop an expression for the average convection coefficient, hturb(x) as afunction of distance from the leading edge, for the turbulent region.
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4. Plot the local and average coefficients for (0 x L)
Solution
a)
Assumptions: hlam(x) is as given, constant fluid properties.Properties: Air @350K, Table A.4 = 20.92 106m2/s
Rex,c =Uxc
(6)
U =Rex,c
xc=
5 10520.92 1060.5
= 20.9[m/s] (7)
b)
hlam(x) =1
x
x0
hlam(x)dx
=1
xClam(2)x
1/2
= 2Clamx1/2
At any point, the average convection coefficient up to that point is twice thelocal value.
c)
The average convection coefficient in the turbulent region must account for thelaminar portion at the beginning of the plate.
hturb(x) =1
xc
xc0
hlam(x)dx +1
x xc
xxc
hturb(x)dx
=2Clamx
0.5c
xc+
Cturbx xc
x0.8
0.8
x
xc
= 2Clamx1/2
+ 1.25Cturb
x0.2 x
0.8c
x
d)
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0 0.5 1 1.510
20
30
40
50
60
70
80
x [m]
h[W/m2K
Variation of Average and local h
hlamAverage(h
lam)h
turbAverage(h
turb)
Figure 3: Convection coefficients along the plate.
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