Applied BioFluid Mechanics  Lee Waite and Jerry Fine
.pdf202 Chapter Seven
then we need to find the first and second derivatives of u with respect to r and plug it into the equation to check for equality. We should also recognize explicitly that dPdx 5 ''Px since the pressure under Poiseuille flow conditions is only a function of x and does not vary with time.
'u 
5 
2 
dP r 



(7.33) 

'r 







4m dx 















'2u 
5 

2 dP 




(7.34) 

'r2 








4m dx 


a4m dx rbb 







'x 
5 maa4m dx b 1 r 
(7.35) 

'P 


2 
dP 
1 

2 dP 

The equality holds true showing that the Poiseuille flow parabolic velocity profile is one simple solution of the Navier–Stokes equations.
7.7 Pulsatile Flow in Rigid
Tubes—Womersley Solution
Consider the following problem. Given a pressure waveform and the geometry of the artery and some fluid characteristics, estimate the flow rate in terms of volume per time. This may be a pulsatile pressure waveform as is seen in the relatively larger arteries in humans. The blood flow rate is a measure of perfusion, or simply said, how much oxygen may be provided to the downstream, or distal tissues. Therefore, a mathematical model of flow rate under pulsatile conditions can be a very useful device.
To estimate flow from a pulsatile driving pressure in rigid tubes, we will begin by assuming a Newtonian fluid, uniform, laminar, axially symmetric, pipe flow. This is similar to the Poiseuille flow problem, but now we are considering pulsatile flow rather than steady.
We will need to have a driving function for the pressure. Recall from Eq. (7.20) that
'P 
` 

5 Re c anejvntd 
(7.20) 

'x 
n50 

Therefore, for each harmonic n, we can write each component of the driving pressure as a complex exponential by using Eq. (7.36).
'P 
` 

5 ane jvnt 
(7.36) 
'x 
n 
Pulsatile Flow in Large Arteries 
203 
For each component of the driving pressure from 0 to n, it is also possible to write the Navier–Stokes equation given in Eq. (7.30). The use of the subscript “n” emphasizes that we are solving for one generic “nth” component and then, at the end, assembling the total solution as a sum of components, and, finally, taking the real part.
'sud 
'sud 
'sud 
vu 'sud 







r 't 
1 r au 'x 
1 vr 'r 1 

'u 
b 






r 









'2sud 

'2sud 
1 'sud 

1 '2sud 
(7.30) 



'P 


b 


5 rgx 2 
'x 
1 ma 'x2 
1 'r2 
1 r 'r 
1 

'u2 


r2 
For this pulsatile flow case we have assumed steady, uniform, laminar, axially symmetric pipe flow. Since the flow is axially symmetric, there is no swirling flow and no velocity in the radial or transverse directions and therefore no change in velocity in either the radial or transverse
direction, so 'u 5 0 and v 5 0 and v 5 0. Since the flow is uniform,
' r u
u 'u
there is no change in velocity u in the x (axial) direction, so 'x 5 0. The flow is also horizontal; therefore, Eq. (7.30) simplifies to the complex equation:
a e jvnt 
'2su 
d 
'su d 

'su 
d 

n 
5 n 
n 

1 n 
n 
2 
n 

(7.37) 
r 
'r2 

'r 
't 





r 



This is a linear, secondorder, partial differential equation (PDE) with a driving function. One possible solution to the PDE is:
un 5 fnsrd e jvnt 
(7.38) 
5 

f is not time dependent
In order to try this solution in Eq. (7.37), we will need to have the derivative of u with respect to time t, the derivative of u with respect to r, and also the second derivative of u with respect to r.

dun 
5 
dfnsrd 
e jvnt 
(7.39) 


dr 
dr 








d2un 

d2fnsrd 
jvnt 
(7.40) 


dr2 
5 
dr2 

e 







dun 
5 jvnfnsrd e jvnt 
(7.41) 


dt 








204 Chapter Seven
Now by using our proposed solution to the differential equation, it is possible to rewrite Eq. (7.37) in the following manner so that an exponential appears in each term.
a 
ejvnt 
d2f 
n 
srd 
e jvnt 1 n 
df 
n 
srd 



n 
5 n 




e jvnt 2 jvnfnsrde jvnt 
(7.42) 


r 
dr2 
dr 




r 


Now it is possible to divide Eq. (7.42) by ejvnt to remove the time dependence. The result would be:
an 

d2 fnsrd 
n dfnsrd 



5 n 

1 
r 

2 jvnfnsrd 
(7.43) 
r 





dr2 
dr 

This allows us to treat time dependence and spatial dependence separately. Also we can replace n with m r and j with j3 to obtain Eq. (7.44).
an 

d2 fnsrd 
1 dfnsrd 
j3vnfnsrd 
(7.44) 


5 

1 
r 

1 


m 


n 



dr2 
dr 


Now we have an ordinary differential equation instead of a partial differential equation and one that is not time dependent! Equation (7.44) looks a little bit like a zeroorder Bessel differential equation. The homogeneous differential equation for the case without a driving function would be:
d2 f 
srd 
1 
df srd 

j3vnf 
srd 


n 

1 
n 
1 
n 

5 0 
(7.45) 



r 

n 


dr2 
dr 










This equation matches a template. A zeroorder Bessel differential equation has the form:
d2g 
1 dg 

2 
g 5 0 
(7.46) 


1 s 

1 l 


ds2 
ds 

The solution to Eq. (7.46) is a Bessel function of order zero and complex arguments, which is well known and arises in problems connected with the distribution of current in conductors of finite size. One homogenous solution is:
fnsrd 5 C1 J0slrd 1 C2 Y0slrd 
(7.47) 
where C1 and C2 are constants, J0 is a zeroorder Bessel function of the first kind, Y0 is a zeroorder Bessel function of the second kind, l is defined in Eq. (7.48), and J0 is given by Eq. (7.49). It can easily be shown, if you know Bessel functions, that in this case C2 0, and we
Pulsatile Flow in Large Arteries 
205 
need only to be concerned with the first term, the one having J0. By comparing our equation with its Bessel equation template, we can write
l2 5 

j3vn 













(7.48) 


n 















Josxd 5 
1 2 


x2 

1 

x4 

2 

x6 

1 

x8 

2 c. (7.49) 



2 
2 
2 
4 
2 
2 
6 
2 
2 
8 
2 


2 
s1!d 


s2!d 


s3!d 


s4!d 


Note that l2 is a constant for a given kinematic viscosity, fundamental frequency, and harmonic n.
The Bessel equation in the Womersley solution is inhomogeneous because of the pressure gradient term. We need to find a find a particular solution to the differential equation, so we propose an easy one.
Let 
fnsrdparticular 5 C3 
We can then set the derivatives of fn(r) with respect to r to zero and solve for the constant C3.

j3vn 

an 





(7.50) 



C3 
5 











m 







n 










C3 
5 
an m 
1 
5 
2an 
(7.51) 











m r 
3 
jrvn 






j vn 



The total solution becomes: 












fnsrd 5 C1Joslrd 2 
an 
(7.52) 



jrvn 












According to the noslip condition, velocity at the wall should be zero. That location is also defined by r R, since r 0 corresponds to the centerline of the vessel. In order to solve for the constant C1, use the boundary condition of u 0 at r R.
0 5 C1JoslRd 2 
an 
(7.53) 

jrvn 

C1 
5 
an 

(7.54) 

jrvnJoslRd 




The total solution to the differential equation, for harmonic n, now becomes:
fnsrd 5 
an 
c 
Joslrd 
2 1d 
(7.55) 
jrvn 
JoslRd 
206 Chapter Seven
This is still the complex form that is time independent. If we take the pressure gradient as the real part of anejvnt and substitute u from Eq. (7.38), the corresponding velocity as a function of r and t becomes:

an 
Joslrd 



un 5 Re c 

e 

2 1 fe jvnt d 
(7.56) 
jrvn 
JoslRd 
Recall that this solution applies to the results of each harmonic. Now to find the velocity as a function of radius r and time t for the entire driving pressure, add together steady flow result u0 to the results from all harmonics.
` 

usr,td 5 u0srd 1 un sr,td 
(7.57) 
n51 

Figure 7.5 shows an example plot of the total velocity u(r) for two different r values in a mathematical model of an artery under pulsatile driving pressure.
At this point we turn to a quantity that is a bit more important than the velocity. It is the flow rate passing through a given cross section of the blood vessel. To find this flow rate one needs only to integrate the velocity function just found multiplied by the differential area, over the entire cross section. The differential area, a simple annulus, may be written as a function of r, that is, 2prdr, so the flow term becomes
R 

Qstd 5 3 usr,td ? 2pr dr 
(7.58) 
0 

Figure 7.5 Graph of axial velocity as a function of radius and time, where r is the radius variable and R is the radius of the artery.
Pulsatile Flow in Large Arteries 
207 
We can perform this integral after looking up integral identities for Bessel functions in a mathematical reference. The one we need to use is
3x J0sxddx 5 x J1sxd
The symbol J1 denotes a Bessel function of the first kind and of the first order. After a number of steps of calculus style derivations which need not be explained here, we reach a solution for the flow rate produced by harmonic n of the pressure gradient.
Qn 5 sspR2ddRe c 
an 
e 
2 J1slRd 
2 1 f ejvnt d 
(7.59) 

jvnr 
lR J0slRd 

These must be summed and added to 



Q0 5 a0 
8m 

(7.60) 





pR4 








which is the average flow rate produced by the constant term in the pressure gradient’s Fourier series. Finally, we have
N 
(7.61) 
Qstd 5 Q0 1 Qnstd 

n51 

Womersley published his solution in 1955 when it wasn’t so easy to make up a computer model of the flow. Bessel functions were not easily accessible then as they are now in all scientific software. So he integrated the velocity terms, solved for flow, and published the following equation for the flow component resulting from each of the driving pressure gradient harmonics:
Qn 5 
pR4 
Mn a 
M10 
bn sin svnt 1 fn 1 e10nd 
m 
a2 
where the pressure gradient` associated with each harmonic is:
'P
'x n 5 Mn cos svnt 1 fnd
The magnitude of the driving pressure is given by:
Mn 5 2A2n 1 B2n
(7.62)
(7.63)
(7.64)
208 Chapter Seven 

or alternatively, recall that an An Bn j 

Mn 5 an 
(7.65) 
The angle of the phase shift is given by: 

fn 5 arguments2and 
(7.66) 
The total flow Q is computed with exactly the same kind of sum just
described. 


M10 




Womersley compiled the values of the constants for 
and for e 
10 
as a 

a2 

function of the parameter.As approaches zero, 
M10 




2 
approaches 1/8 and 
a
e10 approaches 90 and the solution becomes Poiseuille’s law. The reader
may wish to use this alternative to the formula for flow rate involving Bessel functions since some spreadsheets do not provide Bessel functions with complex arguments.
For the sake of completeness, the calculation of M10 and 10 will now
be briefly described. For each harmonic, start by finding 

M0 5 ZJ0saj3>2dZ 
and 
M1 5 ZJ1saj3>2dZ 
u0 5 argumentsJ0saj3>2dd 
and 
u1 5 argumentsJ1saj3>2dd 
(The number of the harmonic enters through .) Next let
d10 5 3 p 2 u1 1 u0
4
and
k5 aM0
2M1
The quantities of interest may now be found.

5 1 
2 



M10 
ssin sd10dd2 1 sk 2 cos sd10dd2 
(7.67) 


k 









sin sd10d 



e10 5 tan21a 

b 
(7.68) 


k 2 cos sd10d 
Recall from Chap. 1 that a is the Womersley number, or alpha parameter, which is a ratio of transient to viscous forces, and is defined by:
vr 
(7.69) 
a 5r Å m 

Pulsatile Flow in Large Arteries 
209 
6
4
Q
2
0 

0 
5 


r 
10 
1 t 
15 
20
25 2
Figure 7.6 Threedimensional plot that shows a flow waveform plotted in terms of time and vessel radius for a pulsatile flow condition. The curve was generated using Womersley’s version of the Navier–Stokes equation solution. The flow Q is shown in cubic millimeters per second, the time t is shown in seconds, the radius r in millimeters.
Figure 7.6 shows a velocity waveform plotted in terms of time and vessel radius for a pulsatile flow condition in the uterine artery of a cow.
Example problem—Womersley solution A pressure gradient is modeled with a simple function,
dPstd 5 2796 2 1250 sin s2ptd 1 531 cos s4ptd dx
Time t is measured in seconds, and the pressure gradient has units of N/m3. A plot of this function, Fig. 7.7, shows that the period T0 is 1.0 s and that v 2p rad/s.
The blood vessel has the following set of parameters.

Quantity 
Symbol 
Value 
Units 


Dynamic viscosity 
m 
0.0035 
Ns/m2 


Density 
r 
1060 
kg/m3 


Kinematic viscosity 

3.302 10 6 
m2/s 


Vessel radius 
R 
0.0025 
m 

























210 Chapter Seven
Pressure gradient driving the flow
500
dP/dx (Pa/m)
0
–500
–1000
–1500
–2000
–2500
–3000
0 
0.1 
0.2 
0.3 
0.4 
0.5 
0.6 
0.7 
0.8 
0.9 
1 





Time (s) 





Figure 7.7 Womersley solution example problem: pressure gradient.
Calculate the Womersley solution for pulsatile flow. In addition to the velocity through the pipe cross section at any time, calculate the flow rate through the vessel as a function of time. Present the results with plots.
Solution:
1.We organize the input data. The pressure gradient is already in Fourier series form. It may also be written as
dPstd 5 Res2796 1 1250je2pjt 1 531e4pjtd
dx
so that a0 796, a1 1250 j, and a2 531.
2.Using the Eq. (1.11) which describes the velocity profile in Poiseuille flow, we write down the steady state (DC analog) solution.
u0 
5 
1 
a0sr2 2 R2d 5256840r2 1 0.35525 




4m 
3. We compute the parameters needed for Womersley which depend on the
vessel geometry and fluid properties. We utilize Eq. (7.48) and write 


Å 



ln 5 
j3vn 




n 











Pulsatile Flow in Large Arteries 
211 
Pieces of the solution are displayed in the following table.
Term 
n 1 
n 2 

ln 
975.4 975.4 j 
1379 1379 j 

an 

0.1877 
0.03983 j 

jrvn 




J0slRd 
1.0758 2.259 j 
5.754 0.6115 j 





4.We calculate and add the pulsatile terms for the velocity solution. Using Eq. (7.56),
u1 5 Re[ss20.03225 2 0.06772jdJ0ss975.4 2 975.4jdrd
2 0.1877de6.283jt]
u2 5 Re[ss20.0007274 1 0.006844jdJ0ss1379 2 1379jdrd
1 0.03983de12.566jt]
Hence, our solution may be constructed as
usr,td 5 u0 1 u1 1 u2
The infinity sign (`) over the sum in Eq. (7.57) indicates that in the most general case a converging series with an infinity of terms would be needed to represent the arbitrary pressure gradient exactly. In this case, with a simple representation of the pressure gradient using three terms, our solution terminates quickly with just three parts. Two plots illustrate the results. Figure 7.8 shows the threedimensional portrayal of axial velocity as a function of time and radial position, and Fig. 7.9 shows the velocity versus time for center and midradius.
5.We calculate the flow rate as a function of time by integrating over the cross section, as indicated in Eq. (7.58). As an alternative we could use the formula in Eq. (7.59)
2 
dd Re c 
an 
e 
2 J1slRd 
2 1 f e 
jvnt 
d 
Qn 5 sspR 




jvnr 
lR J0slRd 

To assist a reader who desires to implement this formula, we tabulate the inner term in braces for each harmonic.

e 
2 J1slRd 
2 1 f 

n 
lRJ0slRd 







10.5783 0.3292 j
20.7088 0.2467 j