Quanitification of BL Effects in Engineering Utilitites… P M V Subbarao Professor Mechanical...
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Transcript of Quanitification of BL Effects in Engineering Utilitites… P M V Subbarao Professor Mechanical...
Quanitification of BL Effects in Engineering Utilitites…
P M V SubbaraoProfessor
Mechanical Engineering Department
I I T Delhi
Engineering Parameters for Flat Plate Boundary Layer Solutions
ODE for Flat Plate Boundary Layer
x
UyFxU
,)(
xxx
U
y
2
1,
)()(2
1 FF
x
U
x
)(2
12
yFx
U
yx
)(FU
y
)(2
2
F
x
UU
y
)(3
3
F
x
UU
y
3
3
2
22
yyxyxy
The Blasius Equation
• The Blasius equation with the above boundary conditions exhibits a boundary value problem.
• However, one boundary is unknown, though boundary condition is known.
• However, using an iterative method, it can be converted into an initial value problem.
• Assuming a certain initial value for F=0, Blasius equation can be solved using Runge-Kutta or Predictor- Corrector methods
Blasius Solution, Laminar Boundary Layer
0
1
2
3
4
5
6
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
f, f', f''
f()
f'()
f''() F
F
F
FFF ,,
F F' F''0 0 0 0.332060.1 0.00166 0.03321 0.332050.2 0.00664 0.06641 0.331990.3 0.01494 0.0996 0.331810.4 0.02656 0.13277 0.331470.5 0.04149 0.16589 0.330910.6 0.05974 0.19894 0.330080.7 0.08128 0.23189 0.328920.8 0.10611 0.26471 0.327390.9 0.13421 0.29736 0.325441 0.16557 0.32978 0.323011.1 0.20016 0.36194 0.320071.2 0.23795 0.39378 0.316591.3 0.27891 0.42524 0.312531.4 0.32298 0.45627 0.307871.5 0.37014 0.48679 0.302581.6 0.42032 0.51676 0.296671.7 0.47347 0.54611 0.290111.8 0.52952 0.57476 0.282931.9 0.5884 0.60267 0.275142 0.65003 0.62977 0.26675
Runge’s Numerical Reults
2.2 0.7812 0.68132 0.248352.3 0.85056 0.70566 0.238432.4 0.9223 0.72899 0.228092.5 0.99632 0.75127 0.217412.6 1.07251 0.77246 0.206462.7 1.15077 0.79255 0.195292.8 1.23099 0.81152 0.184012.9 1.31304 0.82935 0.172673 1.39682 0.84605 0.161363.1 1.48221 0.86162 0.150163.2 1.56911 0.87609 0.139133.3 1.65739 0.88946 0.128353.4 1.74696 0.90177 0.117883.5 1.83771 0.91305 0.107773.6 1.92954 0.92334 0.098093.7 2.02235 0.93268 0.088863.8 2.11604 0.94112 0.080133.9 2.21054 0.94872 0.07191
F F' F''
4 2.30576 0.95552 0.064234.1 2.40162 0.96159 0.05714.2 2.49806 0.96696 0.050524.3 2.595 0.97171 0.044484.4 2.69238 0.97588 0.038974.5 2.79015 0.97952 0.033984.6 2.88827 0.98269 0.029484.7 2.98668 0.98543 0.025464.8 3.08534 0.98779 0.021874.9 3.18422 0.98982 0.01875 3.2833 0.99155 0.015915.1 3.38253 0.99301 0.013475.2 3.48189 0.99425 0.011345.3 3.58137 0.99529 0.009515.4 3.68094 0.99616 0.007935.5 3.7806 0.99688 0.00658
F F' F''
5.6 3.88032 0.99748 0.005435.7 3.98009 0.99798 0.004465.8 4.07991 0.99838 0.003655.9 4.17976 0.99871 0.002976 4.27965 0.99898 0.00246.1 4.37956 0.99919 0.001936.2 4.47949 0.99937 0.001556.3 4.57943 0.99951 0.001246.4 4.67939 0.99962 0.000986.5 4.77935 0.9997 0.000776.6 4.87933 0.99977 0.000616.7 4.97931 0.99983 0.000486.8 5.07929 0.99987 0.000376.9 5.17928 0.9999 0.000297 5.27927 0.99993 0.00022
F F' F''
7.1 5.37927 0.99995 0.000177.2 5.47926 0.99996 0.000137.3 5.57926 0.99997 9.8E-057.4 5.67926 0.99998 7.4E-057.5 5.77925 0.99999 5.5E-057.6 5.87925 0.99999 4.1E-057.7 5.97925 1 3.1E-057.8 6.07925 1 2.3E-057.9 6.17925 1 1.7E-058 6.27925 1 1.2E-058.1 6.37925 1 8.9E-068.2 6.47925 1 6.5E-068.3 6.57925 1 4.7E-068.4 6.67925 1 3.4E-068.5 6.77925 1 2.4E-068.6 6.87925 1 1.7E-068.7 6.97925 1 1.2E-068.8 7.07925 1 8.5E-07
F F' F''
8.9 7.17926 1.00001 5.9E-079 7.27926 1.00001 4.1E-079.1 7.37926 1.00001 2.9E-079.2 7.47926 1.00001 2E-079.3 7.57926 1.00001 1.4E-079.4 7.67926 1.00001 9.3E-089.5 7.77926 1.00001 6.3E-089.6 7.87926 1.00001 4.3E-089.7 7.97926 1.00001 2.9E-089.8 8.07926 1.00001 1.9E-089.9 8.17926 1.00001 1.3E-0810 8.27926 1.00001 8.5E-09
F F' F''
Recall that the nominal boundary thickness is defined such that u = 0.99 U when y = . By interpolating on the table, it is seen that u/U = F’ = 0.99 when = 4.91.Since u = 0.99 U when = 4.91 and = y[U/(x)]1/2, it follows that the relation for nominal boundary layer thickness is
NOMINAL BOUNDARY LAYER THICKNESS
91.4x
U
91.4C,U
xC
2/1
Let the flat plate have length L and width b out of the page:
DRAG FORCE ON THE FLAT PLATE
L
b
00
yy
o y
u
y
u
L
ooD dybdAF0
)(2
2
F
x
UU
yy
u
The shear stress o (drag force per unit area) acting on one side of the plate is given as
Since the flow is assumed to be uniform out of the page, the total drag force FD acting on the plate is given as
The term u/y = 2/y2 is given from as
The shear stress o(x) on the flat plate is then given as
)0(Fx
UUo
2/12
)(332.0 xo
URe
But from the table F (0) = 0.332, so that boundary shear stress is given as
Thus the boundary shear stress varies as x-1/2. A sample case is illustrated on the next slide for the case U = 10 m/s, = 1x10-6 m2/s, L = 10 m and = 1000 kg/m3 (water).
F F' F''0 0 0 0.332060.1 0.00166 0.03321 0.332050.2 0.00664 0.06641 0.331990.3 0.01494 0.0996 0.331810.4 0.02656 0.13277 0.331470.5 0.04149 0.16589 0.33091
)0(Fx
UU
Boundary Shear Stress
0
0.0001
0.0002
0.0003
0 0.02 0.04 0.06 0.08 0.1
x (m)
o (
Pa)
Sample distribution of shear stress o(x) on a flat plate:
Variation of Local Shear Stress along the Length
x
UU332.0o
Note that o = at x = 0.Does this mean that the drag force FD is also infinite?
U = 0.04 m/sL = 0.1 m = 1.5x10-5 m2/s = 1.2 kg/m3
(air)
The drag force converges to a finite value!
L
oD dxbF0
bLU
Fc D
D 2
UL
cD ReRe ,)(664.0 2/1
DRAG FORCE ON THE FLAT PLATE
2/1664.0 bLUUFD
L
dxxUbU0
2/1332.0
Boundary Layer Thicknesses
• The application of integral balances to the control volume shown in above figure delivers more engineering properties namely,
• the boundary layer displacement thickness
• the boundary layer momentum deficiency thickness and
• the energy deficiency thickness .
Selection of a CV for Engineering BL Analysis
• The control volume chosen for the analysis is enclosed by dashed lines.
• This selection is not arbitrary but very clever.
• Since velocity distributions are known only at the inlet and exit, it is imperative that the other two sides on the control volume be streamlines, where no mass or momentum crosses.
Impermeable Streamlines of A BL
• The lower side should be the wall itself; hence the drag force will be exposed.
• The upper side should be a streamline outside the shear layer, so that the viscous drag is zero along this line.
The Displacement Thickness
• Conservation of mass is applied to this Engineering CV @SSSF:
0. CS
Adv
000
HY
Udyudy
Assuming incompressible flow, this relation simplifies to
000
HY
Udyudy
UHudyY
0
UHdyUuUY
0
UHdyUuUYY
0
HYUdyuUY
0
*
0
UdyuUY
Y
dyU
u
0
* 1
* is the engineering formal definition of the boundary-layer displacement Thickness and holds true for any incompressible flow.
Momentum Thickness as Related to Flat-Plate Drag
Conservation of Integral x momentum
CS
x AdvuDF
.
HY
UdyUudyuD00
Y
dyuHUD0
22 Y
dyuUuUuHUD0
22
Y
dyuUuD0
Y
dyU
u
U
u
U
D
02
1
Physical Interpretation of Momentum and displacement thicknesses
Their ratio of displacement thickness to momentum thickness is called the shape factor, is often used in boundary-layer analyses:
1*
Boundary layer development along a wedge
We know that the velocity distribution outside the boundary layer is a simple power function
mCxxU
ODE for Wedge BL Flows
Introducing the same relationship for the stream function as
x
UyFxU
,)(
012
1 2
FmFFm
F
This is the so-called Falkner-Skan equation which describes a flow past a wedge.
Types of Boundary Layers
01
2
2
y
u
x
p
y
uv
x
uu
x
p
y
u
y
0
2
2
If the external pressure gradient, , then,
at the wall and hence is at a maximum there and falls away steadily.
Favourable Pressure Gradient
If the external pressure gradient, , then, 0
x
p
00
2
2
yy
uat the wall
Must decrease faster as y increasesy
u
Adverse Pressure Gradient
If the external pressure gradient, , then, 0
x
p
00
2
2
yy
uat the wall
Must Increase initially and then decrease as y increasesy
u