womersley flow
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Louisiana Tech UniversityRuston, LA 71272
Slide 1
Womersley Flow
Steven A. Jones
BIEN 501
Wednesday, April 9, 2008
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 4
Womersley Flow
tAP cos0PP
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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)
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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).
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 15
Reduction to an Ordinary Differential Equation
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.
Louisiana Tech UniversityRuston, LA 71272
Slide 16
Reduction to an Ordinary Differential Equation
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
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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,
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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,
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
Slide 31
Centerline Velocity
Since the true pressure gradient is the real part of the complex pressure gradient, we can write:
AtAdz
dpPhasecos
1
.
Louisiana Tech UniversityRuston, LA 71272
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.
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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:
Louisiana Tech UniversityRuston, LA 71272
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
Louisiana Tech UniversityRuston, LA 71272
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.
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