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Applied BioFluid Mechanics - Lee Waite and Jerry Fine

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202 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, second-order, 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 zero-order 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 zero-order 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 zero-order Bessel function of the first kind, Y0 is a zero-order 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 no-slip 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 Three-dimensional 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 three-dimensional 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