users.soe.ucsc.eduhongwang/AMS212B/Notes/...Balance of momentum: U2 L u t + U2 L ()u u = 1 L pˆ + U...

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AMS 212B Perturbation Methods

Lecture 20 Copyright by Hongyun Wang, UCSC

Prandtl’s boundary layer

Navier-Stokes equation:

Conservation of mass:

t

+

u ( ) = 0

Balance of momentum:

u t

+

u ( )

u

= p + µ

u + + µ( )

u ( )

where µ is the first coefficient of viscosity and is the second coefficient of viscosity (dynamic viscosity)

Incompressible flow ( = const = 0)


u = 0


u t

+

u ( )

u = ˆ p +

u

where =µ

0

is the kinematic viscosity and ˆ p =p

0

.

Non-dimensionalization

We use a characteristic length and a characteristic velocity to do the non-dimensionalization.

L: Characteristic length

U: Characteristic velocity (usually the velocity at infinity)

A characteristic time is given by

T =L

U

We consider the non-dimensionalization

t =U

Lt , t =

L

Ut


- 2 -

x =1L

x , x = L x

y =1L

y , y = L y

z =1L

z , z = L z

u =1

Uu , u = U u

= L , =1L

= L2 , =1L2


u = 0


U 2

L

u

t+

U 2

Lu( )u =

1

Lp̂ +

U

L2u

==>

u

t+ u( )u =

p̂

U 2+

1U L

u

Reynolds number:

Re =U L

Normalized pressure

p =p̂

U 2=

p

0 U 2

The momentum equation becomes

u

t+ u( )u = p +

1

Reu

Non-dimensional Navier-Stokes equation

(For simplicity, we use x, t ,

u , to denote

x , t ,

u , )

u = 0

u t

+

u ( )

u = p +1Re

u


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Prandtl’s boundary layer

Consider the two-dimensional steady state flow over a semi-infinite plate. The semi-infinite plate is parallel to the x-axis, starting at x = 0 and extending to x = . The boundary conditions are

At infinity: u , y( ) = u x,( ) = 1, 0( )

On the plate:

u x, 0( ) = 0, 0( ) for x 0

Below we consider the steady state flow in the limit of 1

Re0 Re( ) .

Outer expansion:

The (leading term) outer expansion satisfies

u out( )= 0

u out( )( )

u out( )

= p out( )

u out( ) , y( ) = 1, 0( ) ,

u out( ) x,( ) = 1, 0( )

It is straightforward to verify that the expression below is a solution.

u out( ) x, y( ) = 1, 0( ) , p out( ) x, y( ) = const. = p( )

Assuming the system has a unique solution, the solution given above is the unique solution.

Inner expansion:

For mathematical convenience, we write

u x, y( ) = u x, y( ), v x,y( )( ) .

The equation becomes

ux

+vy

= 0 (E01)

uux

+ vuy

=px

+1Re

2ux2 +

2uy 2

(E02)

uvx

+ vvy

=py

+1Re

2vx2 +

2vy 2

(E03)

Now we use scaling to study the inner expansion.

Let = O1

Re( )

be the thickness of the boundary layer.

We consider the scaling

s = Re( ) y , y =1

Re( )s .

After the scaling, equation (E01) becomes


- 4 -

ux

+ Re( )vs

= 0 (E01B)

Inside the boundary layer, we assume

u = O 1( ) ,u

x= O 1( ) ,

2u

x2= O 1( ) ,

u

s= O 1( ) and

2u

s2= O 1( )

v = 0 both in the outer expansion and on the plate

==> there is no reason to assume v = O 1( ) ,v

x= O 1( ) or

v

s= O 1( )

(Note: v 0 inside the boundary layer is caused by the incompressibility.)

Integrating (E01B) from 0 to s and using v(x, 0) = 0, we get

v x,s( ) =1

Re( )

u x, s ( )

x0

s

d s

For mathematical convenience, we write

v x, s( ) =1

Re( )v* x, s( )

where

v* x, s( ) =u x, s( )

x0

s

ds = O 1( )

Next, we look at (E02). After the scaling, it becomes

uu

x+ v* u

s=

p

x+

1

Re

2u

x2+ Re( )

22u

s2

The two terms on the left side are of the order O(1).

The principle of least degeneracy tells us

1

ReRe( )

22u

s2~ O 1( )

==> Re( )2 1

~1

==> =12

To the leading order, (E02) becomes

uu

x+ v* u

s=

p

x+

2u

s2 (E02B)

Next, we look at (E03). After the scaling, it becomes


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1

Reu

v*

x+

1

Rev* v*

s= Re

ps

+1

Re1

Re

2vx2 + Re

2vs2

==> Reps

= O1

Re

==> ps

= O1

Re

Therefore, to the leading order, (E03) becomes

ps

= 0 (E03B)

Matching to the outer expansion yields

p x,s( ) = const = p( ) (true to the leading order)

Substituting into (E02B), we obtain that u satisfies

uux

+u x, s ( )

x0

s

d s

us

=

2us2 (E02C)

and boundary conditions

u 0, s( ) = 1

u x,( ) = 1 and u x, 0( ) = 0 for x > 0

Self-similar solution (similarity solution):

To show the self-similarity, we consider

˜ u x,s( ) = u x, s( )

It is straightforward to verify that ˜ u x,s( ) satisfies equation (E02C) and associated boundary conditions. Thus, we have

˜ u x,s( ) = u x,s( )

==> u x,s( ) = u x, s( )

This is true for all value of . Selecting =1x

, we get

u x,s( ) = u 1,s

x

That is, the solution depends only on the variable s

x.

Let us introduce function


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f ( ) = u 1, s( )ds0

We have

u 1, s( ) = f ( )

u x,s( ) = u 1,s

x

= f

s

x

u x,s( )

x= f

s

x

s

2x32

u x, s ( )

xd s

0

s

= f s

x

s

2x32

d s

0

s

= s

2xd f

s

x

0

s

= f s

x

s2x

f s

x

12x

d s 0

s

= f s

x

s2x

fs

x

1

2 x

u x,s( )

s= f

s

x

1

x

2u x,s( )

s2 = f s

x

1

x

2

Substituting into equation (E02C) and setting x = 1, we obtain that ƒ( ) satisfies

f s( ) +12

f s( ) f s( ) = 0

and boundary conditions

f 0( ) = 0 , f 0( ) = 0 , f ( ) = 0

Numerical simulations show that the corresponding initial values are

f 0( ) = 0 , f 0( ) = 0 , f 0( ) = 0.3321

Drag on the plate:

Let (x(p), y(p)) be the physical coordinates before the scaling, and (u(p), v(p)) be the physical velocities before the scaling. We have

x =x p( )

L, s = Re

y p( )

L=

U0

vLy p( )


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u p( ) x p( ), y p( )( ) = U 0 u x, s( ) = U0 f s

x

The shear stress at x(p) is given by

0 = µu p( )

y p( )

y p( ) =0

= 0 U0

u

ss=0

s

y p( )

= 0 U0 f 0( )1

x

U0

vL

= 0.3321 0 U0

U0

v x p( )

Drag on the portion of plate from x(p) = 0 to x(p) = X is

Drag X( ) = 2 0.3321 0 U0

U0

v

1

x p( )dx p( )

0

X

= 4 0.3321 0 U0

U0 X

v

This is a good estimate for the drag on a plate of length X.


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Stokes flow

Recall the non-dimensional Navier-Stokes equation

u = 0

u t

+

u ( )

u = p +1Re

u

Below we consider the steady state flow in the limit of Re 0 .

Consider the three-dimensional steady state flow around a sphere. Let R be the radius of the sphere and U0 be the velocity at infinite. The Reynolds number is defined as

Re =U 0R

Note: Here we use R instead of 2R in the Reynolds number so that the non-dimensional radius of sphere is 1 instead of 0.5. After the non-dimensionalization, the velocity at infinite becomes 1. The boundary conditions are

At infinity:

u x, y, z( ) 1, 0, 0( ) as x2

+ y 2+ z 2

On the sphere:

u x, y, z( ) = 0, 0, 0( ) for x2

+ y 2+ z 2

= 1

In the limit of Re 0, the leading term of the non-dimensional Navier-Stokes equation is

u = 0 (E01)

0 = p +1

Re

u (E02)

Question: Why does the p term survive?

Answer: The p term is necessary for satisfying the incompressibility condition.

We select the direction of flow as the z-axis and establish a spherical coordinate system. Let be the polar angle and be the azimuthal angle. The system is symmetric in the -

dimension. So we have

= 0 and

u = ur r,( ), u r,( ), 0( )

(E01) becomes

1r2 r

r 2ur( ) +1

rsinsin u( ) = 0

==> r

r2 sin ur( ) + rsin u( ) = 0 (E01B)

We can define a stream function (r, ) such that


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ur =1

r 2 sinr,( )

u =1

rsin rr,( )

It is straightforward to verify that

u =

rsin

e

and verify the identity

rsinrsin

e

= rsin

1r 2 sin

e r +

1rsin r

e

=

2

r2 +sinr 2

1sin

e (E03)

Using the vector identity

w ( ) =

w ( )

w

and using the incompressibility, we have

u =

u ( )

Applying the curl operator to (E02) and using the vector identity f( ) = 0 , we see that equation (E02) becomes

u ( ) = 0

==>

rsin

e

= 0 (E02B)

Using the vector identity (E03), we can rewrite (E02B) as

2

r2 +sinr 2

1sin

2

r,( ) = 0 (E02C)

The velocity at infinity satisfies

ur r,( ) cos( ) , r

u r,( ) sin( ) , r

Using the velocity at infinity, we get

r,( ) r2 sin cos( ) , r

rr,( ) rsin2 , r


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==> r,( )r 2

2sin2 , r

The velocity on the sphere satisfies

ur 1,( ) = 0 , u 1,( ) = 0

==> 1,( ) = 0 ,r

1,( ) = 0

Let us try a solution of the form

r,( ) = f r( )sin2

Substituting into (E02C) yields

d2

dr2

2r 2

2

f r( ) = 0

This is a fourth order ODE. It is straightforward to verify that it has 4 independent solutions:

f r( ) =1r

, f r( ) = r , f r( ) = r2 , f r( ) = r4

So ƒ(r) has the general form

f r( ) =Ar

+ B r + C r2+ Dr4

==> r,( ) = sin2 Ar

+ B r + C r2+ Dr4

To satisfy the boundary condition at infinity

r,( )r 2

2sin2 , r

we must have D = 0 and C =12

.

To satisfy the boundary condition on the sphere (r = 1),

1,( ) = 0 ,r

1,( ) = 0

we must have

A + B +1

2= 0

A + B + 1= 0

==> A =

1

4

B=3

4

==> r,( ) = sin2 12

r 2+

14r

34

r


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Inside the parentheses, the first term corresponds to the uniform flow, and the second term corresponds to the doublet; together they represent an inviscid flow past a sphere. The third term is called the Stokeslet, representing the viscous correction.

Velocity:

ur = cos 1 +1

2r 3

3

2r

u = sin 11

4r 3

3

4r

Pressure:

To calculate the pressure, we first calculate

u .

u =

u ( ) =

rsin

e

=1

rsin

2

r 2 +sinr2

1sin

e

=sin

r

2

r 2 21r2

12

r 2+

14r

34

r

e

=32

sinr2

e

=1

rsinsin u( )

e r

1r r

r u( )

e

=3r 3 cos

e r +

32r3 sin

e

Using (E02), we obtain

pr

=1

Re3r3 cos

Integrating from r = , we have

p r,( ) = p( ) 1Re

32r2 cos

Stresses and strains:

The non-dimensional strain tensor is

i j{ } =12

u +

u ( )

T

[ ] , i j =12

ui

x j

+uj

xi

The non-dimensional viscous stress tensor (incompressible fluids) is


- 12 -

i j{ } = 21

Re i j{ }

The non-dimensional total stress tensor (incompressible fluids) is

i j{ } = pI + i j{ } = pI + 21

Re i j{ }

We first look at rr.

rr =ur

r= cos

32r 4 +

32r2

On the sphere, r = 1, and we have

rr r= 1 = 0 , rr r= 1 = 0

rr r =1 = p = p( )+

1Re

32

cos

Then we look at r .

r =12

rr

u

r

+

1r

ur

=

34r 4 sin

On the sphere, r = 1, and we have

r r= 1= r r= 1

=2

Re r r =1=

1Re

32

sin

Due to the symmetry of the problem, the total stress on the sphere is parallel to the z-axis (the direction of flow). As a result, we only need to follow the z-component of the stress.

Fz ( ) = rr r =1 cos r r =1sin = p( ) cos +

1Re

32

The non-dimensional total force on the sphere is

F = 2 Fz ( )sin d0

= 2 p( ) cos +1Re

32

sin d

0

=1

Re6

The physical total force on the sphere is

F p( )= F U 0

2R2=

1Re

6 U 02R2

= 6 µU0R

This is the Stokes formula for the total drag on a sphere.


- 13 -

Appendix

Divergence:

Cartesian coordinate system:

u = ux

e x + uy

e y + uz

e z

u =

xux( ) +

yuy( ) +

zuz( )

Cylindrical coordinate system:

u = ur

e r + u

e + uz

e z

u =

1

r rrur( ) +

1

ru( ) +

zuz( )

Spherical coordinate system:

u = ur

e r + u

e + u

e

u =

1

r2 rr2 ur( ) +

1

rsinsin u( ) +

1

rsinu( )

Gradient:


f =fx

e x +

fy

e y +

fz

e z


f =fr

e r +

1r

f e +

fz

e z


f =fr

e r +

1r

f e +

1rsin

f e

Curl operator:


u = ux

e x + uy

e y + uz

e z

u =

uz

y

uy

z

e x +ux

z

uz

x

e y +uy

xux

y

e z



- 14 -

u = ur

e r + u

e + uz

e z

u =

1

r

uz u

z

e r +ur

z

uz

r

e +1

r rr u( )

1

r

ur

e z


u = ur

e r + u

e + u

e

u =

1rsin

sin u( )1

rsin

u

e r

+1

rsin

ur 1

r rru( )

e +

1

r rr u( )

1

r

ur

e

Laplace:


f =

2 fx2 +

2 fy2 +

2 fz 2


f =1r r

rfr

+1r 2

2 f2 +

2 fz2


f =1r 2 r

r 2 fr

+1

r 2 sinsin

f

+1

r 2 sin2

2 f2

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