Louisiana Tech University Ruston, LA 71272 Boundary Layer Theory Steven A. Jones BIEN 501 Friday,...
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Transcript of Louisiana Tech University Ruston, LA 71272 Boundary Layer Theory Steven A. Jones BIEN 501 Friday,...
Louisiana Tech UniversityRuston, LA 71272
Boundary Layer Theory
Steven A. Jones
BIEN 501
Friday, April 11 2008
Louisiana Tech UniversityRuston, LA 71272
Momentum Balance
Learning Objectives:1. State the motivation for curvilinear coordinates.2. State the meanings of terms in the Transport Theorem3. Differentiate between momentum as a property to be transported
and velocity as the transporting agent.4. Show the relationship between the total time derivative in the
Transport Theorem and Newton’s second law.5. Apply the Transport Theorem to a simple case (Poiseuille flow).6. Identify the types of forces in fluid mechanics.7. Explain the need for a shear stress model in fluid mechanics.
The Stress Tensor.Appendix A.5Show components of the stress tensor in Cartesian and cylindrical
coordinates. Vectors and Geometry
Louisiana Tech UniversityRuston, LA 71272
Flow Over a Flat Plate
0U 0U
is small
changes slowy in the direction
changes rapidly in the direction
y
x
x
v
v x
v y
What can we say about this flow?
x
x
y – Boundary layer thickness
Louisiana Tech UniversityRuston, LA 71272
Flow Over a Body
x
yU
U0
Louisiana Tech UniversityRuston, LA 71272
Continuity & Momentum
Use 2-dimensional equations.
2 2
2 2
2 2
2 2
0 continuity
momentum
( momentum)
yx
x x x xx y
y y y yx y
vv
x y
v v v vPv v x
x y x x y
v v v vPv v y
x y y x y
In contrast to previous derivations, we do not say that terms are zero. We say that they are “small.”
Louisiana Tech UniversityRuston, LA 71272
Continuity
Non-dimensionalize so that velocities and lengths are of order 1.
L
UV
y
vV
x
v
L
U
y
Lx
Vy
Ux
o
yx
:conclude
continuity
:directiontheinLengthsticCharacteri
:directiontheinLengthsticCharacteri
:directiontheinVelocitysticCharacteri
:directiontheinVelocitysticCharacteri
0*
*
*
*0
0
Louisiana Tech UniversityRuston, LA 71272
Exercise
If: 0Characteristic Velocity in the direction:
Characteristic Velocity in the direction:
Characteristic Length in the direction:
Characteristic Length in the direction:
o
x U
Uy V
Lx L
y
How do you non-dimensionalize:
2 2
2 2, , , ?x x x x
x y
v v v vv v
x y x y
Louisiana Tech UniversityRuston, LA 71272
x-Momentum
Non-dimensionalize
2 * * 2 * 2 ** 2* *0 0
* * * 2 2 *2 *2
2 * * 2 *** *0
* * * *20
x x x xx y
2x x x
x y
U v v U v vPv v
L x y L x L x y
U v v vPv v
L x y U L x y
P
P
small
21
21
0
Re
LL
U
L
If inertial terms are important
with respect to viscous terms.
Louisiana Tech UniversityRuston, LA 71272
Characteristic Pressure
Non-dimensionalize
20
2
0
ULU
LU
0
P
P 2
important is pressure if 1,orderofis
andsince 21
21
0
Re
LL
U
L
Louisiana Tech UniversityRuston, LA 71272
y-momentum
From y-momentum
* * 2 * 2 *2 2 * 2*0 0 0
2 * * * 2 2 2
y y y yx y
v v v vU U UPv v
L x y y L L x y
2 2
2 2
1y y y yx y
v v v vPv v
x y y x y
2 2 * * ***
* *2 * *
1y y yx y
v v vPv v
y L y x y
Small in comparison to P
x
Louisiana Tech UniversityRuston, LA 71272
Characteristic Pressure
From y-momentum
21constant
2P x U x
0*
y
P
P is not a function of y, so the pressure in the boundary layer is the pressure in the free stream, which can be determined from a potential flow solution.
Take the derivative with respect to x to get something to plug into the x-momentum equation.
dU xdPU x
dx dx
Louisiana Tech UniversityRuston, LA 71272
Final x-momentum, B.C.s
2 * * 2 *** *0
* * * *20
2x x x
x y
U v v vPv v
L x y U L x y
P
2
2x x x
x y
dU xv v vv v U x
x y dx y
0 at 0
at
0 at
x
x
x
v y
v U x y
dvy
dy
Louisiana Tech UniversityRuston, LA 71272
Flow Over a Flat Plate
0U 0U
x
x
x
0
dU xdPU x
dx dx
Velocity at large y does not depend on x.
Louisiana Tech UniversityRuston, LA 71272
Special Case: Flat Plate
2
2x x x
x y
v v vv v
x y y
0 at 0
at
0 at
x
x
x
v y
v U x y
dvy
dy
Use the stream function:
,x yv vy x
Louisiana Tech UniversityRuston, LA 71272
Special Case: Flat Plate2 2 3
2 3y x y x y y
Assume a similarity solution:
0,Uy
x y f f y fx
2 20 0 0 0 0
3 3 3 2
1, ,
2 2 2
U U U U Uf y f f y f
y x x x x y x x x
32 2 3 32
0 02 2 3 3
,U Uf f
y x y x
Louisiana Tech UniversityRuston, LA 71272
Special Case: Flat Plate
2 2 2
2 2
1, ,
2 2 2
f y f f y f
y y x x x y x x
32 2 2 3 3
2 2 2 3,f f
y y y y
2 20 0 0 0 0
3 3 3 2
1, ,
2 2 2
U U U U Uf y f f y f
y x x x x y x x x
3
2 2 3 320 0
2 2 3 3,
U Uf f
y x y x
0Uyx
Louisiana Tech UniversityRuston, LA 71272
Special Case: Flat Plate
32 2 2 2 3
2 2 2 2 3
1
2 2 2
f f y f f f f
y x x y y
2 2 3
2 3y x y x y y
2 2 2
2 2
1, ,
2 2 2
f y f f y f
y y x x x y x x
32 2 2 3 3
2 2 2 3 3,f f
y y y y
Louisiana Tech UniversityRuston, LA 71272
Final Differential Equation
3 2
3 20
2
d f f d f
d d
Or, more simply
12 0f ff
This equation is 3rd order and nonlinear. There is no closed form solution, but it can be solved numerically.
Louisiana Tech UniversityRuston, LA 71272
Final Differential Equation
Remember that f is the stream function, so you must take derivatives to get velocity after you solve for f.
0
1
2
3
4
5
6
7
0.0 0.5 1.0 1.5
Velocity/Uinfinity
x-velocity for flow over a flat plate.
,x y
f fv v
y x
Louisiana Tech UniversityRuston, LA 71272
Integral Momentum Boundary
Instead of seeking an exact solution to the differential equation, we can integrate the equations for continuity and x-momentum.
2
2
2
20 0
x x xx y
x x xx y
dU xv v vv v U x
x y dx y
dU xv v vv v dy U x dy
x y dx y
0
0yx xy
vv vv dy
x y x
Louisiana Tech UniversityRuston, LA 71272
Integral Momentum Boundary
The terms in these equations have specific meanings.
2
20 0 0 0
x x xx y
dU xv v vv dy v dy U x dy dy
x y dx y
2
200
0x xx x
y
v vv vdy
y y y y
Becomes zero because velocity becomes constant far from the boundary.
Wall shear stress
Louisiana Tech UniversityRuston, LA 71272
Integral Momentum Boundary
The terms in these equations have specific meanings.
2
20 0 0 0
x x xx y
dU xv v vv dy v dy U x dy dy
x y dx y
0
xy
vv dy
y
Integrate by parts with , x
y
vu v dv dy
y
00 0
yxy x y xy
vvv dy v v v dy
y y
Louisiana Tech UniversityRuston, LA 71272
Integral Momentum Boundary
vx becomes U far from the boundary, and vy() is obtained from the integrated continuity equation.
0
xx y
vv v U dy
x
00 0
yxy x y xy
vvv dy v v v dy
y y
0 0
0 0 yxy x y x y x
vvv dy v v v v v dy
y y
Both velocities are zero at the wall.
Louisiana Tech UniversityRuston, LA 71272
Integral Momentum Boundary
Because by Continuity:
0 0 0
0 0
yx xy x
x xx
vv vv dy U y v dy
y x y
v vU y v dy
x x
yxvv
x y
Louisiana Tech UniversityRuston, LA 71272
Integral Momentum Boundary
In summary:
0 0 0
x x xy x
v v vv dy U y v dy
y x x
2
20 0
x x xx y
dU xv v vv v dy U x dy
x y dx y
2
20
xw
vdy
y
So:
02 x x
w x
v v Uv U U dy
x x x
Louisiana Tech UniversityRuston, LA 71272
Integral Momentum Boundary
With:
If we knew vx, we would be able to integrate for the wall shear stress. It turns out that the exact form of vx is not as important as one might think, and good results can be obtained with a form that looks reasonable and satisfies the boundary conditions.
02 x x
w x
v v Uv U U dy
x x x