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Von Karman Integral Method (BSL). 1. 2. PRANDTL BOUNDARY LAYER EQUATIONS for steady flow are. Continuity. N-S (approx). If we solve these, we can get V x , (and hence d) . Alternative: We can integrate this equation and obtain an equation in d and shear stress t. - PowerPoint PPT Presentation
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IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0yxVV
x y
2
2
yx xx y
VV VdPV V
x y dx y
Von Karman Integral Method (BSL)
PRANDTL BOUNDARY LAYER EQUATIONS for steady flow are
Continuity
N-S (approx)
1
2
If we solve these, we can get Vx, (and hence .
Alternative: We can integrate this equation and obtain an equation in and shear stress
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Von Karman Integral Method (BSL)
If we assume a rough velocity profile (for the boundary layer), we can get a fairly accurate relationship
Integration is ‘tolerant’ of changes in shape
For all the above 3 curves, the integration (area under the curve) will provide the same result (more or less), even though the shapes are very different
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0yxVV
x y
2
2
yx xx y
VV VdPV V
x y dx y
Von Karman Integral Method (BSL)
Prandtl equations for steady flow are
Continuity
N-S (approx)
What is Vy?
0
yx
y
VV dy
x
1
2
VdPV
dx x
Pressure gradient (approx)
3a
3b
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
20
yx x x x
x
V V V VVV dy V
x x y x y
Substitute (3a) and (3b) in (2)
Von Karman Integral Method (BSL)
4
0 0 0 00
yx x x x
x
V V V VVV dy dy dy V dy
x y x x y
Integrate (4) with respect to y, from 0 to infinity
5
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0x
y
V
y
0
0
x
y
V
y
0
yxV
dyx
F
xdV
dx
F y
xG V
xdV
dy
G y
Integration by Parts. Let
Von Karman Integral Method (BSL)
Eqn. 5: On the RHS
0x
y
V
y
Eqn 5: On the LHS, for the marked part
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Von Karman Integral Method (BSL)
0 0
x xx
V VV dy V dy
x x
0 00
yxV Vx
dy dy F dGy x
0
0
F G G dF
This is for the marked region in LHS of Eqn 5
0
xVF dy
x
xG V V
0
00 0xV
F dyx
0 0 0xG V
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Von Karman Integral Method (BSL)
000 0
x x xx x
V V V VV V dy V dy V dy
x x x x
1. To equation (6), add and subtract0
x
VV dy
x
Note : and areindepent ofV
V yx
Substituting in equation (5)
6
To write equation (6) in a more meaningful form:
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2. Note 2
2x x
x
V VV
x x
xx x
VVV V V V
x x x
0 00
0
0 0
2 xx
x
xx
V VV dy V
VV dy
x
V VV dy V dy
x x
dyx x
7
Von Karman Integral Method (BSL)
3. Also
... and multiply both sides by -1
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Combining the above two
2x x x xV V V V V V
x x
2 x xx x
V VVV V V
x x x
Von Karman Integral Method (BSL)
2 x xx x x x
V VVV V V V V V
x x x x
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0
0 0
x x x
VV V V dy V V dy
x x
First term is momentum thickness
Second term is displacement thickness
(Note: The density term is ‘extra’ here)
Note : and areindepent ofV
V yx
Von Karman Integral Method (BSL)
Equation (7) becomes
Note: Integral method is not only applied to Boundary Layer. It can be applied for other problems also.
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example
Assume velocity profile
It has to satisfy B.C.0 0xV at y
xV V at y 0xV
at yy
For zero pressure gradient
2
20
0x
y
V P
y x
33 1
2 2x
y yV V
For example, use
Von Karman Integral Method (BSL)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
sinxV a by
2b
2b sin
2x
yV a
cos 0y
by
Von Karman Integral Method (BSL)
Or for example, use
What condition should we impose on a and b?
What is the velocity gradient at y= ?
0x
y
V
y
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
xV V at y
sin2xV a a
Von Karman Integral Method (BSL)
What is the velocity at y= ?
a V
Check for other two Boundary Conditions
For zero pressure gradient
OK
sin2x
yV V
OK
0 0xV at y
2
20
0x
y
V P
y x
No slip condition
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
0 0
( ) sin 1 sin2 2x x
y yV V V V dy
2 2 1
2V
Von Karman equation gives
2 2 1
2 2
VV
x
2 2( )
(4 )d dx
V
0 2x
y
VV
y
0V
x
Now, to substitute in the von Karman Eqn, find shear stress
Also
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2 22 2
(4 )
x
V x
2
(4 )x xV
2
(4 R) exx
4.7
Rexx
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Calculation for comes out ok
Calculation for Cf also comes out ok
0
212
fxV
C
0
0
2
1
12
fL
L
dx
CL
V
Even if velocity profile is not accurate, prediction is tolerable
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Now numerical method are more common
( . ) x
s V
F V n dA V dV
01. . .1.2
x x xx x x x x x x
P PF P P P x
2 2
0 0
.x x x x x top
s
V dy V dy V m
0t
Conservation of mass
Von Karman Method (3W&R)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0t
Conservation of mass
.V n dA dVt
00
s
0 0
. . 0x x xx x topV dy V dy m
0 0
. .x x xtop x xm V dy V dy
Von Karman Method
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Substitute , rearrange and divide by x
2
0 0
o x x
dP d dV dy V V dy
dx dx dx
0VdP
Vdx x
Outside B.L.
2dP d dV V V
dx dx dx
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0
0 0
x x x
d dV V V V V V dy
dx dx
If is constV
If we assume 2 3xV a by cy dy
0
0
x x
dV V V dy
dx
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0a
3
2b V
0c
32
Vd
0 0xV at y
xV V at y
0xVat y
y
2
20
0x
y
V P
y x
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0
2
0.6461 Re2
fx
x
CV
1.272
RefL
L
C
4.64
Rexx
140
13
dxd
V
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