womersley flow

Louisiana Tech UniversityRuston, LA 71272

Slide 1

Womersley Flow

Steven A. Jones

BIEN 501

Wednesday, April 9, 2008


Slide 2

Womersley Flow

Major Learning Objectives:

1. Apply the assumptions of Poiseuille flow to oscillating flow in a pipe.

2. Obtain the momentum equation for this flow.

3. Solve the partial differential equation through complex variables.

4. Relate the flow velocity profile to centerline velocity, flow rate, wall shear stress, and pressure gradient.


Slide 3

Womersley Flow

Minor Learning Objectives:1. Show that the simplifications we arrived at for Poiseuille

flow are consequences of fully developed flow.2. Apply the momentum equation for a case that involves

time and space.3. Use complex analysis in the solution of a partial

differential equation.4. Use complex analysis to recast the solution in the form

of centerline velocity instead of pressure gradient.5. Apply superposition to obtain a solution for an arbitrary

waveform.


Slide 4

Womersley Flow

tAP cos0PP


Slide 5

Womersley Flow

tdzdP cosLaminar pipe flow, but with an oscillating pressure gradient,

Approximation of pulsatile flow in an artery.

0

constant

ddPdrdP

dzdP

trvv

vv

zz

r

,

0

The above “assumptions” can be obtained from the single assumption of “fully developed” flow.


Slide 6

Fully Developed Flow

011

z

vv

rrv

rrz

r

trvv

trvv

trvv

zz

rr

,

,

,

In fully developed pipe flow, all velocity components are assumed to be unchanging along the axial direction, and axially symmetric i.e.:

Examine continuity (for incompressible flow):

0 (no changes in z) 0

rrvr 0 (symmetry)


Slide 7

Fully Developed Flow

rrv f t

0

rrvr

Rr

R

tf

r

tfvr at0

From we conclude

But since we have already ruled out dependence on and z,

It follows that there is no radial velocity, i.e. vr(r,t)=0.

, ,rrv f z t

But so that f(t) must be zero.

Note: It is possible to show from the momentum equation that v must also be zero. We will save that for a day when you really need the extra excitement.


Slide 8

Results from Mass/Momentum

...11

,

0

zeitadz

dp

z

p

trvv

vv

zz

r

offunctionlinearaispressure

Through arguments similar to the above, we find that:

Now look at the z-momentum equation.

2

2

2

2

2

111

dz

vv

rr

vr

rrdz

dp

z

vv

v

r

v

r

vv

t

v

zzz

zz

zzr

z


Slide 9

Kinematic Viscosity

The kinematic viscosity is defined by:

Do not confuse with the velocity v. Yes, they do look very similar, but will not appear with a subscript.

2

2

2

2

2

111

dz

vv

rr

vr

rrdz

dp

z

vv

v

r

v

r

vv

t

v

zzz

zz

zzr

z


Slide 10

Results from Mass/Momentum

tar

vr

rrt

v zz

Or rearranging, and substituting for the pressure gradient:

The removal of the indicated terms yields:

r

vr

rrdz

dp

t

v zz 1

In a typical situation, we would have control over a(t). That is, we can induce a pressure gradient by altering the pressure at one end of the pipe. We will therefore take it as the input to the system (similar to what an electrical engineer might do in testing a linear system).


Slide 11

The Pressure Gradient

tite ti sincos

tizz etrvtrv ,~Re,

tietta Recos

The form we select for the pressure gradient is:

tietta Recos

tiIR AetAAta Recos22

Where the subscript R represents the real part of A. Recall Euler’s rule:

We also define a complex form of the axial velocity:


Slide 12

Complex Velocity

tizevtv ~

It is important to remember the following theorem:

If we have a linear differential equation

Solves the equation, then the real part and the imaginary part of v(t) must satisfy the equation individually.

0tvL

Where L{} is a linear differential operator, and if:


Slide 13

Complex Velocity

titi eBetvL ~~

tBtvL cos

Therefore, instead of solving the equation:

We can solve the equation:

and take the real part of the solution afterwords to obtain the solution to:

tBtvL cos


Slide 14

Reduction to an Ordinary Differential Equation

tar

v

rr

v

t

v zzz

1

2

2

tit

zti

zti

z Aer

ev

rr

ev

t

ev

~1~~2

2

Divide by eit

With these concepts in mind, we substitute for the pressure gradient and for the velocity in the momentum equation.

Through this process, we have removed the time dependence from the equation. This method is similar to a separation of variables method, where vz=T(t)R(r).

Ar

v

rr

vvi zz

z

~1~

~2

2


Slide 15


022

22 yix

dx

dyx

dx

ydx

xryvz and~If we set

The homogeneous equation is transformed into:

222

22 ~

~~Arv

ir

r

vr

r

vr z

zz

If we had merely assumed that vz=T(t)R(r), we would have obtained the same result.


Slide 16


Axyxx

yx

x

yxAxyx

x

yx

x

yx

Axyi

xyx

xx

yx

222

2222

2

22

222

2

2

xryvz and~Arvi

rr

vr

r

vr z

zz 222

22 ~

~~

2

2

2

2

xxxxrr

xxr

x

r

and

r

From/r

vz


Slide 17

Bessel’s Equation

022

22 yix

dx

dyx

dx

ydx

Should be compared to the canonical form of Bessels equation:

0

,0

22

222

22

ypxdx

dyx

dx

dx

ypxx

yx

x

yx lyequivilantor

The above equation:


Slide 18

Bessel’s Equation

notorintegeranispwhetherondependsform

and

xY

pkk

xxJ

p

k

pkk

p0

2

!!

2/1

The solutions to Bessel’s equation are of the form:

In all cases, Yp becomes infinite at r = 0, so it is not

relevant to our case.

When p=0.

kkkmk

k

xkxJ

xxY

k

k

m

Ki

k

kk

o

loglim,00,0

!

21

2log

2

1

1

02

21

0

andforwhere


Slide 19

Bessel’s Equation

.23

01 xiJDy

022

22

yixx

yx

x

yx

Compared to the canonical form of Bessel’s equation:

With the homogeneous equation:

0222

22

ypzz

yz

z

yz

We can deduce that p=0 and the solution is of the form

This form can be verified by substituting z=i3/2x into Bessel’s equation.


Slide 20

Particular Solution

iAxiJDy 23

01

Axyixx

yx

x

yx 22

2

22

is

From direct substitution, it can be verified that the particular solution to:

iAy

We can deduce that the complete solution is of the form:


Slide 21

Ber and Bei Functions

02

2323

0!

21

m

mmm

m

xixiJ

Despite the complication of the complex argument, the Bessel function above has a relatively simple formulation.

The real and imaginary parts of this function are called the Ber and Bei functions, respectively.

02

24

02

4

230

!12

21Bei,

!2

21Ber

,BeiBer

m

mk

m

mk

m

xx

m

xx

xixxiJ where


Slide 22

Boundary Conditions

RiJ

iADiARiJD

230

10123

0

iAriJDy

23

01With we must satisfy the no-slip

condition at r = R.

RiJ

riJ

iAvy z

230

023

1

Therefore,


Slide 23

Particular Solution Equation

R

iJ

R

riJ

i

ARvz

where23

23

0

0

2

2

1~

or

Now back out from x to r and from y to vz.

tiz e

iJ

R

riJ

i

ARv

23

23

0

0

2

2

1Re


Slide 24

Womersley Number

R

Is called either the Womersley number of the alpha parameter. It is a measure of the ratio of the unsteady part of the momentum equation to the viscous part. When is small, the unsteadiness is not important, and the solutions become parabolas (Poiseuille solutions) that vary in magnitude, but not in shape. If is large, however, the shapes of the profiles are not parabolic.

The parameter


Slide 25

Complex Form

is less frightening than it looks. General computational software, such as MatLab has built in functions for the Bessel functions, and it can also handle the complex arithmetic. One can therefore literally program the above expression in MatLab and take the real part to obtain the physical (real) velocity.

The complex form:

tiz e

iJ

R

riJ

i

ARtrv

23

23

0

0

2

2

1Re,


Slide 26

Centerline Velocity

Above is the solution for velocity in terms of the pressure gradient A. What if we have a case where the pressure gradient was not measured, but the centerline velocity was. We can write an expression for velocity in terms of the centerline velocity given in the form:

tktvz cos,0


Slide 27

Centerline Velocity

From the solution for velocity as a function of r and t:

tiz e

iJR

riJ

i

ARtrv

1Re,

230

230

2

2

tiz e

iJR

iJ

i

ARtv

1

0

Re,023

0

230

2

2

We would like to determine the velocity profile in terms of the centerline velocity, uz(0,t). Evaluate the above at r = 0.


Slide 28

Centerline Velocity

Now note that 100 J , so:

titiz e

iJ

iJ

i

ARe

iJ

iJJ

i

ARtv

23

0

230

2

2

230

2300

2

2 1Re

0Re,0

Recall that we can talk about a complex velocity rv~

such that tiervtrv ~Re,

. Then

3 220

23 2

0

10, i t i t

z

J iARv t e e

iJ i

.


Slide 29

Centerline Velocity.

We were not given the complex form of , tktvz cos,0

but we were given its magnitude (k) and phase (), and we know that for any product of complex numbers AB,

BAAB and .PhasePhasePhase BAAB

Therefore, we can obtain the magnitude and phase of the pressure term A from tvz ,0 as:

23

0

230

2

2 1,0~

iJ

iJ

i

ARtvk z

,

23

0

230

2

2

1 iJ

iJ

R

ikA


Slide 30

Centerline Velocity

and

,

23

0

230

2

2 1PhasePhase,0~Phase

iJ

iJRAtv

23

0

230

2

2 1Phase,0~PhasePhase

iJ

iJRtvA


Slide 31

Centerline Velocity

Since the true pressure gradient is the real part of the complex pressure gradient, we can write:

AtAdz

dpPhasecos

1

.


Slide 32

The Stress Tensor

For fluids:

1 1 1

2 2

1 1 1 1 12

2 2

1 1 1

2 2

r r r z

r r r z

r z z z

vv v v vr

r r r r z r

v vv v v vr

r r r r r z r z

vv v v v

z r z r z z

τ

The shear stress has 4 non-zero components.


Slide 33

Shear Stress, Bottom Surface

10 0

2

2 0 0 0

10 0

2

z

z

v

r

v

r

τ

Along the bottom surface, we are concerned only with xy.

zzr

v

r


Slide 34

Wall Shear Stress

axJaxaxJxdx

dp

pp

p1

Similarly, we could recast the solution in the form of wall shear stress. From this simple geometry, the shear stress reduces to:

r

vzrz

To evaluate the derivative, we will need to know the derivatives of the Bessel functions. I.e. we must know that in general:

axaJaxJdx

d10

so that specifically:


Slide 35

Wall Shear Stress

axJaxaxJxdx

dp

pp

p1

It is left to the student to show that:

tirz e

iJ

R

riJ

i

AR

230

231

23Re

and that since:

tie

iJi

iJ

i

ARtQ

23

023

231

2

4 21Re

the flow rate as a function of time is:


Slide 36

Carotid Artery Waveform

v cm/s

Time (s)

A typical example of a carotid artery centerline velocity is shown here. Clearly this waveform is not a pure sine wave. It has a DC component and higher harmonics.

0 1 1 2 2

3 3 4 4

cos cos 2

cos 3 cos 4 ...

clv t A A t A t

A t A t


Slide 37

Superposition

If the centerline velocity is periodic. Then it can be represented as a Fourier series:

0

1cos,0n

nnz tnktv

Because the differential equation is linear, the solution to the above “input” is the same as the sum of solutions to each term. Thus, since we have solved the equation for a single cosine wave, we have solved them for a general case.

womersley flow

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