Pulsatile Flow Through a Curved Pipe Final

8/8/2019 Pulsatile Flow Through a Curved Pipe Final

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Pulsatile Flow Through A Curved Pipe

VIJAYAALAYAN THIVYATHASAN

00466617

THIS IS MY OWN WORK UNLESS OTHERWISE STATED

June 15, 2010

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Abstract

The main aim o the project is to model blood ow through the human circulatory system. Bymodelling the blood vessel as a slowly curving pipe based on a section o the torus ring, I derive an

unsteady orm o the Dean Equations and then adapt them to my particular model o blood ow.

Apart rom introducing the variable o curvature in a pipe, I will also be introducing Pulsatile Flow

to actor in the efects o the heart as a pumping mechanism.

Contents

1 Introduction 3

1.1 Outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Derivation of the Unsteady Dean Equations 5

2.1 Solutions to the Unsteady Dean Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Analysis o Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.3 Graphical Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Scaling time with 10

18

3.1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.1.1 Looking at a large Womersley parameter: . . . . . . . . . . . . . . . . . . 20

3.2 Analysis o Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.3 Graphical Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4 Conclusion 31

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1 Introduction

In the past Berger et al (1983) have looked into ows in curved pipes, but whilst they have ocusedon the complexity that arises through the diference between various degrees o curvature in compari-son to straight pipes, they have not delved too ar into unsteady, and more specically pulsatile owsthrough curved pipes. Pipe ows play an important part in engineering and also in understanding themechanics o the human blood supply system. This has been researched by Pedley (1980) who coveredquite extensively amongst other arteries, the aorta, which is highly curved. Though Berger et al haveindependently studied the model o a loosely coiled pipe, and unsteady laminar ow in a curved pipe,they have not combined the two like Siggers and Waters [12], who examined unsteady ows in pipes withnite curvature. Siggers et al ocus on the benets o understanding blood ow in the arteries in relationto Atherosclerosis and to this end consider the efects o nite curvature on wall shear stress being asthere is a signicant correlation between these properties and Atherosclerosis.

What we are trying to understand in this article is how the ow varies as a result o the pulsatileow attributed to the heart and we wish to see how it varies rom the models that others have set out.Initially we will allow or the unknown parameters o amplitude and oscillation, and analyse what limi-tations our assumptions bear on our equations. Later on we will make a more rened case dealing withthe Womersley parameter and modelling our ow pending the exibility o our results and whether ornot we will have to constrain our model with urther assumptions.

When we model the pumping mechanism o the heart, we introduce three parameters A, and theDean number K and make the assumption that the pressure G behaves like 1 + A cos(wt) when scaledappropriately. The amplitude A is the unsteady Reynolds number that Siggers et al look into and isassociated with the Womersely number , a dimensionless expression o the pulsatile ow requency inrelation to the viscous efects o the ow and is ormally dened when we rst meet it in our calculations

later on. What we nd is that in the absence o any oscillatory motion, i.e. when our amplitudeA

isequivalent to zero, we end up with the Steady Dean ow. We also make the assumption that as our pipeows along its curvature in the direction, it will eventually settle and so we can simpliy our modelto an axisymmetric one with no dependance, much like the case o two dimensional ow in a straightpipe.

Looking into a simple model o blood ow and bearing in mind that the curvature we look into isslight, we can investigate the ow that arises due to the pumping mechanism o the heart. The curvatureis dened by the ratio o the radius o the pipe to the inner torus radius, say a

b(see gure in the next

section) and as such it is assumed to be much smaller than 1. While this project ocuses on looking atsmall curvature, i we considered a highly curved pipe where a

b 1 this gives rise to a diferent ow. The

efect that curvature has on ow appears to be an increase in stability as the critical Reynolds number,which signies the threshold between stable ow and turbulence, can be higher by a actor o two or more

(Berger et al). (It is possible to conceive o a case where we consider a model in which a straight pipeprecedes our curved pipe, such that the entry conditions o our pipe are altered, but we do not or thetime being)

The Dean number K is also another parameter, which appears in our non-dimensional Navier-Stokesequations and it can be said to characterise the ow. The Dean number is proportional to the centriu-gal orce so by varying the Dean number we shall look at cases o ow, which are driven by diferent

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orces. Berger et al make the point that the importance o the Dean number is in its substitution or the

Reynolds number in the specic case where

a

b 1, which was ound by Dean (1928). The case whichI am currently pursuing looks at the case o a small Dean number, permitting expansion in powers oK and separating equations in terms o leading order. Making approximations as to the relative size othe Dean number is important in characterizing the type o ow we will be looking at. At small Deannumbers we get a ow that is similar to Poiseuille ow but with cross-ow due to the centriugal orce,however slight the curvature. As the Dean number gets larger, we end up with a case where derivedterms in the ow o the curved pipe are less comparable to those ound in the ow along a straight pipe.It is also important to mention that there is an increase in the centriugal orce as K increases, with adevelopment o boundary layers, which can be shown to be thick in the inner bend and thin in the outerbend. At small Dean numbers, our ow attains its maximum velocity at angles +/- 90 degrees [12] andthe vortices are symmetric. The importance o our ratio a

bmust not be understated here either as even

small increments can have efects on the ow resistance that are several times greater but at high Deannumbers there is a discrepancy such that the ow shows signs o higher resistance when a

b

1 (what

Berger et al have classied as loosely coiled) over a highly curved one.

In their analysis o unsteady ow along pipes o slight curvature Siggers et al also deal with the sameparameters that characterise my model o blood ow. Considerations o the Womersley parameter, Deannumber and curvature o the pipe allow them to analyse cases where their dened curvature a

bis uniorm

and the variation o the Womersley parameter at extremely low or high values give rise to varioussolutions. As well as looking into oscillatory pressure gradients, Siggers et al also make use o sinusoidalpressure which removes the steady component o the pressure gradient and gives rise to secondary veloc-ity, allowing or their secondary streaming Reynolds number A which also characterizes the ow. Thevalue o A is important as it parameterizes the steady streaming element along the core o the pipe.Similar to my project which deals with expansions in my dened Dean number K, in their calculationsor the asymptotic solutions to an oscillatory pressure gradient, Siggers et al expand in powers o a

band

or the asymptotic analysis or high requency sinusoidal ow they expand in powers o 1

.

1.1 Outline

The starting point or this project will be to look at the Navier-Stokes equations in cylindrical co-ordinatesor a section o a torus and to non dimensionalise them. We shall assume that the curvature is slight,denoted by the ratio o the pipe radius to the radius o the torus, which we have previously mentioned asab

1. In our simplication o these equations we shall urther assume that as the pipe is only slightlycurved , that we can treat it much like a straight pipe in the sense that over a certain length travelledin the direction, the ow settles and does not change so we are essentially looking at an axisymmetricow. We will then arrive at a parameter K which will characterise the ow, known as the Dean numberand assuming K to be small we shall then expand our velocity and stream unction in powers o K so

that we can analyse the ow. Our uid through the pipe will then need to satisy the no-slip condition onthe boundary o the pipe. Finally we shall graphically compare the ows and how they change throughtime through the two diferent scalings o time we make.

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2 Derivation of the Unsteady Dean Equations

In order to derive our starting equations, we look at ow down a slightly curved pipe and model thisusing polar co-ordinates (r,,z) on a section o a torus ring which is given by (r b)2 + z2 = a2, wherewe have chosen b a and our ow occurs in the direction as shown in the gure below:

Figure 1: Diagram o curved pipe

Our velocity u = (ur, u, uz) ullls the requirements o the axisymmetric Navier-Stokes equation (as wehave decided to obtain solutions which do not have any dependence on ):

1r

r

(rur) +uz

z= 0

Dur

Dt u

2

r

= p

r+

2ur ur

r2

Du

Dt+

uur

r

= G(r,z,t) +

2u u

r2

Duz

Dt= p

z+ 2uz

where G = br

G0(1 + f(t)). In order to simpliy this set o equations we make some assumptions. Giventhat the pipe is slowly curving and that b a with r and z varying over the scale a, we can saely assumethat:

b a so that r = b + ax b and r

1a

1b

1r

Furthermore, to non-dimensionalise our equations, we scale z = az and we take U0 to be a scale o

u. We then use these scales to obtain a suitable scale or ur and uz such that ur uz U0ab

12 ,

and we scale ur with ux. Whilst these scalings are important, what is o most importance is how wewish to scale our time given the signicance o the time driven pressure gradient. Initially we will scale

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t = (ab)12

U0t and later change the scaling depending on any complications that may may arise or in order

to make calculations simpler (Note here that any previous * terms are dimensionless parameters). Uponsubstitution into the axisymmetric Navier-Stokes equations, we can re-arrange the resulting equationsand urther simpliy them with the ollowing denitions or new parameters:

G0a2

U0= 1 and K =

U0a

ab

12

And thus we arrive at the ollowing set o Dean equations:

ux

x+

uz

z= 0

K

Dux

Dt u2

= Kp

r+ 2ux

KDu

Dt= 1 + f(t) + 2u

KDuz

Dt= Kp

z+ 2uz

(where we have dropped the * terms). The derivation o the Dean Equations is shown in greater detailby Mestel [8], [9]. Alongside the Dean number K, we have chosen the pressure gradient 1 + f(t) to beequivalent to 1 + A cos(1t) where we have introduced the non dimensional parameters A and 1, botho which can be dened as we desire in our analysis o the ow. Our velocity here can now be written inthe orm U = (U,V,W ) and i we cross diferentiate to cancel the pressure terms, we can then introduce astream unction o the orm (x, z) such that U = z and W = x. We then have the initial equations:

K(Vt + zVx

xVz) = 1 + A cos(1t) +

2V

K(t + zx xz) = 2 2KV Vzas this is a ow down a pipe o circular cross-section, we can just switch to the (x, z) plane, we can switchto polar co-ordinates (R, ) where x = R sin() and z = R cos(), which then gives us:

K

R(VR RV) = 1 + A cos(1t) KVt + 2V (2.1)

K

R(R R) = 2 Kt 2KV(VR sin + V cos ) (2.2)

Where subscripts denote derivative with respect to the shown variable and where we have dened =

2 . When solving these, we will need to ensure that they ulll the boundary conditions on R = 1

such that:V(1) = 0 (1) = 0 R(1) = 0

We see that there are three parameters in our problem which characterise the ow. Based on the initialassumption that the Dean number K is small, we can expand the above equations in terms o K, andby setting V = V0 + KV1 + K

2V2 + ... and = K1 + K22 + ... we can then match coecients, which

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allow us to then obtain the ollowing equations rom equation (2.1):

O(1) : 0 = 1 + A cos(1t) + 2V0 (2.3)O(K) : 0 = (V0)t + 2V1 (2.4)

O(K2) :1

R(1V0R 1RV0) = (V1)t + 2V2 (2.5)

We can apply a similar process to equation (2.2) where we then obtain the ollowing two equations:

O(K) : 0 = 21 2V0

V0R sin + V0cos

R

(2.6)

O(K2) : 22 = 1t + 2

V0

V1R sin + V1

cos

R

+ V1

V0R sin + V0

cos

R

(2.7)

These equations will be solved or each o the various values o V and and by looking at each term, wecan observe how the non linear interactions and their time averages afect the ow compared to that oa steady ow.

2.1 Solutions to the Unsteady Dean Equations

Looking at our O(1) term we must rst note that as V0 is our value or no curvature with no co-ecientdependence on the Dean number K, then essentially there is no dependence, so we can cancel out theV term and this simplies equation (2.3):

2V0 = (1 + A cos(1t))1

R(RV0R)R = (1 + A cos(1t))(RV0R)R = R(1 + A cos(1t))

RV0R = R2

2(1 + A cos(1t)) + c

V0R = R2

(1 + A cos(1t)) +c

R

V0 = R2

4(1 + A cos(1t)) + c log(R) + d

I we enorce the boundary conditions we nd out that our constants o integration, c = 0 in order orour V0 to be nite at R = 0 and d is easily worked out by placing R = 1, we then get:

V0 =1

4

(1

R2)(1 + A cos(1t)) (2.8)

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I we now look at equation (2.4) we can derive V0t to give the ollowing:

2V1 = 14

(R2 1)(A1) sin(1t)1

R(RV1R)R =

1

4(R2 1)(A1) sin(1t)

(RV1R)R =1

4(R3 R)(A1) sin(1t)

RV1R =1

4

R4

4 R

2

2

(A1) sin(1t) + c

V1R =1

4

R3

4 R

2

(A1) sin(1t) +

c

R

V1 =1

4R

4

16 R2

4 (A1) sin(1t) + c log R + d

Once again i we place boundary conditions, we get that c = 0 and by working out d we get:

V1 =1

64(R4 4R2 + 3)(A1)sin(1t)

=A1

64(R2 1)(R2 3) sin(1t) (2.9)

We must make note o the act that given our 1 here is non-dimensional, and i A1 is large then ourKV1 may also be large and expansion breaks down, but this is a problem that will be dealt with in theollowing section. In order to solve equation (2.5) we need to solve equation (2.6) using what we have soar calculated (note that as V0 has no dependence the nal term V0 disappears):

21 = 2V0V0R sin RR +

1

RR +

1

R2 =

2

4(1 R2)(1 + A cos(1t))

2R4

(1 + A cos(1t))sin

RR +1

RR +

1

R2 =

1

4(R3 R)(1 + A cos(1t))2 sin (2.10)

For all intents and purpose we can treat the t dependent terms as constants as our let hand side onlylooks at and R derivatives. Also o note is how we can separate our 1 term into two unctions, oneor and another or R, giving us

1 =1

4(1 + A cos(1t))

2F(R)G()

and i we use this , then we can instantly come to the conclusion that our G( ) term is sin which allowsus to actor out it out and rewrite our equation (10) in the ollowing orm (note that the nal term onthe LHS changes sign as G is equal to -sin ):

FRR +1

RFR 1

R2F = R3 R

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Finally we have equation (2.7) to be solved, which can be written in the orm (by using equation (2.11)

and the act that = 3

2 ):22 = 1t + 2(V0V1R sin + V1V0R sin )22 =

A1

96(1 + A cos(1t)) sin(1t)

3R 6R3 + 2R5 sin

... +

A1

32(1 + A cos(1t)) sin(1t)

(1 R2)(R3 2R)sin

... +

A1

64(1 + A cos(1t)) sin(1t)

(1 R2)(R3 3R)sin

22 =

A1

192(1 + A cos(1t)) sin(1t)

(15R + 18R3 5R5)sin

Like beore, we can sperate 2 into two sperate unctions o R and such that

22 =

A1

192(1 + A cos(1t)) sin(1t)

F(R)G()

where it is quite clear that our G() is sin(). Ater similar calculations as used to nd 1 we arrive atthe equation

2 =

A1

9216(1 + A cos(1t)) sin(1t)

(R 90R3 + 32R5 5R7)sin (2.14)

where is a variable to be determined by boundary conditions on the stream unction. We now calculate2:

22 = A19216 (1 + A cos(1t)) sin(1t) (R + 90R3 32R5 + 5R7)sin Ater similar calculations we get

2 =

A1

9216(1 + A cos(1t)) sin(1t)

R +

8R3 +

15

4R5 2

3R7 +

5

80R9

sin

which when subjected to our boundary conditions, (where we get that = 12548 and = 46), gives us:

2 =

A1

9216(1 + A cos(1t)) sin(1t)

125

48R 23

4R3 +

15

4R5 2

3R7 +

5

80R9

sin

2 = A1

442368(1 + A cos(1t)) sin(1t)R(125 276R

2 + 180R4 32R6 + 3R8)sin

2 =

A1

442368(1 + A cos(1t)) sin(1t)

R(3R4 26R2 + 125)(R2 1)2 sin (2.15)

2.2 Analysis of Flow

I we wish to see how the introduction o a pulsatile pressure afects our ow it may be convenient to lookat the time averages or each o our values and see how important our parameters are in determining the

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overall nature o the ow. In order to calculate the time average o a variable X, we use the ollowing

ormula X = 1T

t0+T

t0

Xdt

Where t

is our dummy variable and denotes taking the time average our the specied variable. Thisgeneral ormula can be adapted or our own model to give the ollowing equation

X = 12

21

0

Xdt

I we calculate the time average or V0 we get the ollowing

V0 = 12

1

4(1 R2)

21

0

(1 + A cos(1t))dt

=1

2

1

4(1 R2)

t

+A

1sin(1t

)

21

0

=1

4(1 R2)

O note is that this is the solution to the steady Dean equations or V0 which indicates that time has nooverall efect on our ow when we are looking at the leading order term. Let us now look to V1 and itstime average

V1 = 12

A1

64(R2 1)(R2 3)

21

0

sin(1t)dt

=

1

2A1

64 (R2

1)(R2

3) 11 cos(1t

)21

0

= 0

Once again, when compared to the steady solutions to the Dean equations, time appears to have nocontribution to the ow. However, i we look at V2 due to the (1 + A cos(t))

3 term, we should expectsome lasting contribution. Note that the rst term in V2 is just a cos(t) term and so its time averagewill be 0, so we can ocus on the second term.

(1 + A cos(t))3

=

1

2

21

0

(1 + 3A cos(1t) + 3A2 cos2(1t

) + A3 cos3(1t

))dt

=1

2 21

0

(1 + 3A cos(1t) +

3

2A2(cos(21t

) + 1) +

A3

4(cos(31t

) + 3 cos(1t

))dt

We can ignore all the cos(n1t) terms or n = 0, n Z as their time averages are equivalent to zero. Thisleaves us looking at

=1

2

21

0

(1 +3

2A2)dt

= 1 +3

2A2

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I we use this value or calculating our time average or V2 we get the value

V2 = 1737280

(R9 10R7 + 30R5 40R3 + 19R)cos

1 + 32

A2

I we chose our parameter A = 0 then we have the same solution as we do or the steady Dean equationsbut as this is not the case we see that depending on the value we assign to A, the efects o the pulsatileow are now being elt with some orm o efect. These values combined yield the total time average orour ow:

V = 14

(1 R2) K2 1737280

(R9 10R7 + 30R5 40R3 + 19R)cos

1 +3

2A2

+ ...

We can similarly look at our stream unction and note its time average.

1 = 12

14608 R(4 R2)(1 R2)2 sin 21

0(1 + A cos(1t))2dt

=1

2

1

4608R(4 R2)(1 R2)2 sin

21

0

1 + 2A cos(1t

) +

A2

2(cos(21t

) + 1)

dt

=

1

4608R(4 R2)(1 R2)2 sin

1 +

A2

2

As expected, it is clear that depending on our parameter A, unsteadiness has an efect on our ow. Wecan also look at the time average or 2

2 =

A1

442368R(3R4 26R2 + 125)(R2 1)2 sin

21

0

(1 + A cos(1t)) sin(t

)dt

=

A1

442368R(3R4 26R2 + 125)(R2 1)2 sin

2

1

0

sin(1t

) +

A

2sin(21t

)

dt

= 0

Surprisingly, there is no time-averaged contribution rom there 2 term.We can also consider the dimensionless ow rate through the pipe much like Siggers et al (2008) by

considering the equation or each o our Vi and or i = 0, 1, 2 and j = 1, 2. We dene

{X} =20

10

XRdRd

where once again {} perorms the above operation on our specied X variable and i we want to considerthe average then we divide the above ormula by .

{V0} = 14

(1 + A cos(1t))

20

10

(R R3)dRd

=

2(1 + A cos(1t))

R2

2 R

4

4

10

=

8(1 + A cos(1t))

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Similarly

{V1} = A164

sin(1t)

2

0

1

0

(R5 4R3 + 3R)dRd

=A1

48sin(1t)

And

{V2} = 12304

A21 cos(1t)

20

10

(R7 9R5 + 27R5 19R)dRd

... 1737280

(1 + A cos(1t))3

20

10

R2(R8 10R6 + 30R4 40R2 + 19)cos dRd

= 173072 A21 cos(1t)

Using these terms we see that the total dimensionless ow rate through the pipe is:

{V} = 8

(1 + A cos(1t)) + KA1

48sin(1t) K2 17

3072A21 cos(1t) + ...

Looking at the stream unction terms it is quite simple to realise that as both terms have sin , theintegral taken over the whole o 2 averages to zero and so:

{1} = {2} = 0

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2.3 Graphical Interpretation

I we look at the plots or V0 and V1 we see that they are airly similar given their lack o dependenceand as such the only change in the contour plots that they exhibit are the reversal o ow as shown below.Note that the let hand side o contour plot is the inner bend o the tube whilst the right hand side isthe outer bend o the pipe and the white regions denote positive ow:

Figure 2: The contour plot on the let shows V0 at an amplitude o 1 at time t = and we must mention that the contour

graph or V0 does not change as we vary time. The contour graph on the right shows the ow or V1 at time t =32

and o

the same amplitude but at t = 2

the contour graph or V1 is the same as or V0. However, unlike the graphs below, we do

not get a third graph or V1

V2 exhibits quite an interesting display in terms o the ow as we alter the parameters o amplitudeand requency. Not shown below is the act that as time changes and orces the oscillation, we see areversal in the ow much like we did or the V1 contour plots. Also, due to the structure o V2 the valueo 1 acts as a parameter that can be scaled how we wish. The structure o our pressure gradient is1 + A cos t and so we realise that the value o A around the value i 1 is important and so by looking at

amplitude values around 1 and varying time we can see how our ow behaves. By keeping the amplitudeat a constant value, we can change the value o time and note the transormation in the ow.

Figure 3: These contour plots show the change in ow as we maintain the requency 1 = 1 and amplitude A = 1 whilst

changing the time rom 0 to 2

then . It is quite clear that as we progress through time the ow changes direction and i

we increase our value or time urther, we get a repetition in the contour plots as the ow oscillates back and orth. What

we also notice is that at 2

or all these set o graphs as well as or the other set o graphs is that they all exhibit the same

contour plot due to this being the value at which cos is zero and so we have no pulsatile ow but a steady driving gradient.

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The next set o contour plots looks at an amplitude A = 20:

Figure 4: These contour plots show how the ow is slightly diferent at a high amplitude yet also shows how it evolves in

time and that the oscillation is exactly the same

Finally, we wish to see how the oscillation behaves over time when we look at an amplitude A < 1.What we expect and do nd is that as the amplitude heads to zero, the ow loses oscillation andremains more or less constant across time. The transormation and loss o oscillation takes place betweenA = 1 102 and 2 104. Here we look at A = 1 103:

Figure 5: These contour plots show how the ow changes at a really low amplitude and how there is still pulsation even at

really low amplitude values

Each o the contour plots or V2 are at intervals o2 rom 0 to . All the plotted graphs are quite

similar with only slight variation due to the amplitude. There is not much emphasis on the requency 1and or each contour plot we have let the value as 1 as or both V0 and V1 the requency term remainswithin the sin and cos term so it only increases the rate o oscillation. However this is not the case orV2 which has a A

21 value in the rst term and competes with an A

3 value in the second term but stillthe changes this has on the contour plot is by constraining a ratio between 1 and A at which we see

certain contour plots. I we increase 1 we require a higher value o A or our contour plots to remaininvariant in time and essentially give us the same contour plot (as we increase A our contour plots willonly only show the middle graph in Figure 5 as the oscillation about 1 in 1 + A cos t is arbitrary as A isso large). The time averaged equations are also quite helpul to plot as they give us an understandingas to how the ow through the curved pipe looks in general with the efect o time being displayed in asingle contour plot without having to look at multiple graphs and having to compare them. We can plotplot the time averaged equations or the stream unction and non-zero velocity values:

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Figure 6: The contour plot on the let is o V0. The contour plot in the centre is o V2 and the one on the right is o 1

As we plot the values or 1

we must note that as out time dependent term remains within a squaredbracket, there is no oscillation due to a negative pressure gradient. But i we look at the contour plotsor 2 at a xed requency 1 = 1amplitude A = 1 we see there is a change about t = :

Figure 7: The contour plot on the let is at t = 2

and the one on the right is at t = 32

We see that or V0 and 1 there is no oscillation in ow and that or V1 the ow switches between twoxed orms o ow. V2 on the other hand varies with xed amplitudes and oscillates about t =

2 whereas

2 oscillates about t = . These contour plots have been seen beore and are going to be a rame oreerence or comparison with other graphs in the next section.

16


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2.4 Summary

Looking over this section, we have managed to obtain the ollowing important results:

V0 =1

4(1 R2)(1 + A cos(1t))

V1 =A1

64(R2 1)(R2 3) sin(1t)

V2 =1

2304A21 cos(1t)(R

6 9R4 + 27R2 19)

... 1737280

(1 + A cos(1t))3R(R8 10R6 + 30R4 40R2 + 19)cos

V0 =1

4 (1 R2

)V1 = 0V2 = 1

737280(R9 10R7 + 30R5 40R3 + 19R)cos

1 +

3

2A2

{V0} = 8

(1 + A cos(1t))

{V1} = A148

sin(1t)

{V2} = 173072

A21 cos(1t)

1 =1

4608(1 + A cos(1t))

2R(4 R2)(1 R2)2 sin

2 =

A1

442368(1 + A cos(1t)) sin(1t)

R(3R4 26R2 + 125)(R2 1)2 sin

1 =

1

4608R(4 R2)(1 R2)2 sin

1 +

A2

2

2 = 0{1} = 0{2} = 0

What we see rom these set o results is that there is only a weak dependence on time and this canbe attributed to our assumption that KA1 1 and we see rom our time averages that the onlymodication to the ow is the amplitude A. Our expansion starts to break down when A 1 or when1 1 which leads to KA1 not being small and so we thereore need to scale time diferently when wenon-dimensionalise our ow, which is what we do in the next section.

17


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3 Scaling time with 10

Rather than introducing this dimensionless 1 which may be large and contradict the assumptions theanalysis makes where KV1 V0 or K1 1, we can equate our time dependant pressure gradient to1 + cos(0t), where 0 is dimensional, and i we scale time diferently such that in our derivation othe Dean equations we have t = 1

0t, we end up with a slightly diferent set o equations eaturing a

parameter known as the Womersley number

=

a20

The derived set o equations,though more dicult ofer a greater understanding o pulsatile ow andallow or less restrictive assumptions:

KR

(VR RV) = 1 + A cos(t) 2Vt + 2V (3.16)K

R(R R) = 2 2t 2KV (VR sin() + V cos ) (3.17)

Once again, i we match coecients (o K) we end up with the ollowing set o equations rom (3.16) :

O(1) : 2V0t = 1 + A cos t + 2V0 (3.18)O(K) : 2V1t = 2V1 (3.19)

O(K2) :1

R(1V0R 1RV0) = 2V2t + 2V2 (3.20)

and rom equation (3.17), we obtain:

O(K) : 0 = 21 21t 2V0

V0R sin + V0cos

R

(3.21)

O(K2) : 22 = 22t + 2

V0

V1R sin + V1

cos

R

+ V1

V0R sin + V0

cos

R

(3.22)

3.1 Solutions

Looking at equation (3.18) it seems appropriate to take particular solutions and combine them witha general solution in order or us to obtain a V0 that satises the necessary conditions set out by theequation. In this manner we end up with three diferent equations to solve or equation (3.18):

2

V0 2

V0t = 12V0 2V0t = A cos t2V0 2V0t = 0 (3.23)

This gives us a V0 in the orm V0G + V0P1 + V0P2 where we have V0G is the general solution and is aunction o both t and R whilst V0P1 is a particular solution and a unction o time only and V0P2 is onlya unction o R. The particular solutions are simple to calculate as in either case one o the terms on the

18


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let disappear depending on independence o a particular variable and we must also remember that V0 is

independent o . For the particular solutions we get:

V0P1 =A

2sin t (3.24)

V0P2 =1

4(1 R2) (3.25)

The general solution is more complicated to calculate as it is both a unction o t and R. However, i wemodel it in the ollowing ashion we can arrive at a modied Bessel equation.

V0G = Re

F(R)ieit

In which case, when plug this into equation (22), we get;

ReFRR +1

R

FR

i2F ieit = 0

As eit cannot equal zero within a nite time period, then the term in the inner brackets must be equivalentto 0, giving us a special case o the modied Bessel equation o zero order known as Kelvins Function ozero order. As shown by W.W.Bell [1], any equation o the orm:

R2d2F

dR2+ R

dF

dR (ik2R2 + n)F = 0

Has a corresponding solution:

F = C1In(i12 kR) + C2Kn(i

12 kR)

= C1Jn(i32 kR) + C2Yn(i

32 kR)

where Jn(R) is known as the bessel unction o order n, Yn is called Neumanns Bessel unction o the

second kind o order n and C1 and C2 are constants to be determined by our boundary conditions. Inour equation, n = 0 and our k = giving our solution or F:

F(R) = C1J0

i32 R

+ C2Y0

i32 R

To determine our unknown constants C1 and C2 we must use our boundary conditions on V0, but oimportance is that at 0, Y0 has a singularity giving us an innite value at R = 0 so C2 = 0. We thereoreend up with a V0 o the orm:

V0(R, t) =1

4(1 R2) + A

2sin t + Re

C1J0

i32 R

ieit

(3.26)

I we apply the boundary condition, we end up with

0 =A

2 sin t + Re

C1J0

i

3

2

ieit

Now we cannot be too sure as to whether or not our Bessel unction is real or not, so we choose C1appropriately giving us V0 in the nal orm:

V0(R, t) =1

4(1 R2) + A

2sin t +

A

2Re

J0

i32 R

J0

i32 ieit

(3.27)

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which gives

V0R(R, t) = R2

A2

Re

i32

J1

i32 R

J0

i32 ieit

I we now look at equation (3.19) we arrive at the solution

V1 = Re

C3J0

i32 R

ieit

+ (a cos + b sin )Re

C4J1

i32 R

ieit

+ ...

where a and b are constants and we have assumed two possible orms o solutions in the orm o someunction o only t and R whilst the second hal o the solution is a unction o all three variables. I weconsider other unctions o only R, or t we arrive at contradictions. However, similarly, these general

solutions all apart under boundary conditions, orcing our constants C3 and C4 to be set to 0. Lookingat the rst o our vorticity equations (3.21), we once again separate our equations into particular andgeneral solutions.

21 21t = 2V0V0R sin

= sin

1

4(R R3) A

2R sin t AR

2Re

J0

i32 R

J0

i32 ieit

+ ...

... sin

A

2

2(2sin t)Re

i32

J1

i32 R

J0

i32

ieit

+ ...

... sin A

22(1 R2)Re

i 32 J1 i 32 RJ0

i32 ieit

+ ...

... sin A

2

22Re

J0

i32 R

J0

i32 ieit

Re

i32

J1

i32 R

J0

i32 ieit

Taking into account a generalised answer and the rst two particular solutions we get:

1 = Re

C5J1

i32

ieit

+1

4

1

24R5 1

8R3 +

1

16R

sin Re

AiR

4ieit

sin

The remaining our particular solutions pose complex answers which we cant ully analyse nor draw anyuseul conclusions rom so we will have to consider an alternative route.

3.1.1 Looking at a large Womersley parameter: In order to allow our calculations to proceed at a reasonable pace we make the assumption that ourparameter is large, considering the high requency limit. Letting allows us to simpliy our

20


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Bessel unctions to its asymptotic orm. We can also reduce our Laplacian operator to the two possible

orms: 2 2t becomes either RR + 1R

R 2t or RR 2tOwing to the act that as our is large, in order or our equation to balance in orders o magnitude werequire RR to be in balance with our 2t term and we realise that as our R term is o O(1), this leads

to the assumption that 2

R 1

RR

and so we can justiy the neglect o some terms in the 2 operator.Note that in either case we have dropped the 1

R2 term being as it is comparatively the smallest value

in the diferential equation. These assumptions that we have used concerning the Laplacian operator arederived rom the act that oscillatory ow o a viscous uid along a solid wall leads to the ormation oa particular boundary layer known as the Stokes boundary layer and these vorticity oscillations changeexponentially across the small boundary layer they are conned to. Further more, we only use theseassumptions on the terms which are important along the boundary layer. We also have the asymptoticorm o the Bessel unction to consider in order to simpliy our calculations:

J0

i32 R

J0

(i + 1)

2R

e

(i+1)2R

as

J1(i32 R) = 1

i32

d

dr

J0

i32 R

i

12

d

dr

e(i+1)

2R

=i12 (i + 1)

2e(i+1)

2R

as

Where we have used the act that J1(bR) = 1b ddr [J0 (bR)]. As the term (i+1)2 appears oten in thecalculation, we shall assign it the label . Under this assumption we can rewrite V0 as:

V0(R, t) =1

4(1 R2) + A

2sin t +

A

2Re

e(R1)ieit

Using our new assumptions we now have the ollowing set o equations to solve:

1RR 21t = sin

A2

R sin t A2

Re

Re(R1)ieit

+ ...

... +sin

A

2

2(2sin t)Re

e(R1)ieit

+ ...

... +sin

A

22(1 R2)Re

e(R1)ieit

+ ...

... +sin

A

2

22Re

e(R1)ieit

Re

e(R1)ieit

As well as:

21 21t = sin

14

(R3 R)Which we have already solved in the previous chapter and can substitute in directly. Now looking at eachequation separately, we can obtain particular solutions as well as a general solution, all o which can beanalysed with greater ease. Our general solution is o the orm:

1G1 = sin Re

C6ei(R1) + C7e

i(R1))

ieit

21


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However, owing to the act that is large, the term ei(R1) grows exponentially as we head away

rom the surace and towards the centre (as R

1) and so we require C7 = 0. Our particular orm o thesolution is:

1P = sin

1

96

R + 3R3 R5 A

4Re

Reit

+ ...

... +A

2sin Re

2

(1 2)2 +iR

(1 2)

e(R1)ieit

+ ...

... +

A

2

2sin Re

e

it

1 22 +ieit

e(R1)ieit

+ ...

... +A

2sin Re

i

2(1 2) 2R

(1 2)2

i

2(1 2) +1

(1 2)2 +4i2

(1 2)3

R2

e(R1)ieit

+ ...

... +1P6

or an unknown constant. In order to solve the nal particular solution, we need to make use o theact that:

Re[X]Re[Y] 12

Re[(X + X)Y]

or two complex unctions X and Y, where X is the complex conjugate o X. So we now end up havingto solve

1RR 21t =

A

2

2sin Re[(e(R1)ieit e(R1)ieit)e(R1)ieit]

=

A

2

2sin (Re[ie2(R1)ie2it] + Re[e

2(R1)])

which can be solved to give the ollowing particular solution:

1P6 =

A

2

2sin Re

22e2(R1) +

i

2(2 2) e2(R1)e2it

But we must not orget to also include the general solution involving e2it:

1G2 = sin Re

C8e2i(R1) + C9e

2i(R1))

ie2it

and or the same reasons regarding exponential growth as we head away rom the surace, we set C9 = 0.Now we have to use our original denition or 1 in order to solve or 1 whilst amending this denitionin accordance to the assumptions we made at the start o this subsection (note that the value o is

22


23/32

diferent rom the previous section):

1RR = sin Re

C6ei(R1)ieit + C8e

2i(R1)ie2it

+ ...

... +sin

1

96

R + 3R3 R5 A

4Re

Reit

+ ...

... +A

2sin Re

2

(1 2)2 +iR

(1 2)

e(R1)ieit

+ ...

... +

A

2

2sin Re

e

it

1 22 +ieit

e(R1)ieit

+ ...

... +A

2sin Re

i

2(1 2) +1

(1 2)2 +4i2

(1 2)3

2R(1 2)2 +

iR2

2(1 2)

e(R1)ieit

+ ...

... + A

22

sin Re

22 e2(R1) +

i

2(2 2) e2(R1)e2it

Allowing or the various combinations o general solutions that satisy our equation and choosing thosewhich will help satisy our boundary conditions, we arrive at the ollowing set o solutions:

1 = sin Re

C6

i2ei(R1)ieit +

C8

2i2e2i(R1)ie2it

+ ...

... +R sin Re[C12ieit + C13ie

2it] + ...

... +sin

1

96

R +

8R3 +

1

8R5 1

48R7

A

4Re

R3

6eit

+ ...

... +A

2

sin Re2

(1 2

)2

+iR

(1 2

) 2i

(1 2

) 1

2

e(R1)ieit + ...... +

A

2

2sin Re

e

it

1 22 +ieit

2

1

e(R1)ieit

+ ...

... +A

2sin Re

i

2(1 2) +1

(1 2)2 +4i2

(1 2)3

+4 2R

(1 2)2

1

e(R1)ieit

+ ...

.. +A

2sin Re

i(2R2 2(R + 1) + 2)

2(1 2)

1

3e(R1)ieit

+ ...

... +

A

2

2sin Re

44e2(R1) +

i

8(2 2) e2(R1)e2it

(3.28)

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We now need to solve or the unknown constants that are let in the solutions by enorcing our boundary

conditions on 1:

1(1,.t) 0 = sin Re

C6

i2ieit +

C8

2i2ie2it

+ ...

... +sin Re[(C12ieit + C13ie

2it] + ...

... +sin

1

96

+

8+

5

58

A

4Re

1

6eit

+ ...

... +A

2sin Re

2

(1 2)2 +i

(1 2) 2i

(1 2)

1

2ieit

+ ...

... +

A

2

2sin Re

ie

2it

1 22 1

2

1

+ ...

... +

A2

2sin Re

44+ i

8(2 2) e2it

+ ...

... +A

2sin Re

i

2(1 2) +1

(1 2)2 +4i2

(1 2)3

+4 2

(1 2)2

1

ieit

+ ...

... +A

2sin Re

i

2(1 2)

2 4 + 23

ieit

and on R:

1R(1, , t) = sin Re

C6

iieit +

C82i

ie2it

+ ...

... +sin Re[C12ieit

+ C13ie2it

] + ...

... sin

1

96

+

3

8+

23

58

A

4Re

1

2eit

+ ...

... +A

2sin Re

2

2

(1 2)2 +i( 1)(1 2)

1

2ieit

+ ...

... +

A

2

2sin Re

ie

2it

1 22 1

2

+ ...

... +

A

2

2sin Re

2

23+

i

4(2 2) e2it

+ ...

... +A

2sin Re

i

2(1 2

)

+1

(1 2

)2

+4i2

(1 2

)3 +

2(1 )

(1 2

)2 ieit + ...

... +A

2sin Re

i

2(1 2)

( 2)

ieit

24


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By matching coecients, (powers o eit), we end up with the ollowing set o equations:

C6

i2+ C12 = iA

64+

A

21

i(1 + )

3Y+

2 1Y2

+4i

Y3

= D1

C6i

+ C12 = iA24

+A

2

i

2Y (1 + 2)

Y2+

4i2)

Y3

= D2

where Y = (1 + 2) and the values D1 and D2 have been assigned or ease o reerence as they do notcontain any variables in them. We can solve this to get values or both C6 and C12:

C6 =i2

1 i [D1 D1]

C12 =1

1

i D2

iD1

Now looking towards the coecients o e2it we have:

C8 =

A

2

2 2i2

1 2i

1

(1 22) 2 1

8(2 2)

C13 =

A

2

2 1

(1 22)

(1 )

2i

1 2i

1

8(2 2)

2 (1 2)

2i

1 2i

Finally we also get:

= 384A

22

Re 1

3+

(1 2)44

3

2

= 48

A

2

2Re

3

3+

(

2 3)44

+

1

12

I we now solve or (3.20) using the values we have obtained so ar we have an equation or V2:

2V2 2V2t = 1R

(1V0R 1RV0)

V2RR 2V2t = 1R

1V0R

=1

R

R

2+

A

2Re

e(R1)ieit

cos sin

This is quite straight orward to calculate using the methods we have seen beore in this section but it islengthy and as we are more interested in the stream unction which doesnt make use o V2 we can ignorethis equation. I we look at our nal vorticity equation (3.22) we see that due to the act that V1 = 0and V0 = 0 we end up with:

22 22t = 0

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which i we solve or 2 2R2

and by the act that we cannot have exponential growth towards the

centre o the ow, we get :

2RR 22t = 02 = F1() Re

C14e

i(R1)ieit

2 = F1() Re

C14

i2ei(R1) + C16R + C17

ieit

where F1() is unction o that can be decided on depending on any other constraints the system mayhave and can be expected to be proportional to a combination o cos or sin . I we look at our boundaryconditions we are simply solving:

2(1,.t)

0 = F1() ReC14

i2

+ C16 + C17 ieit2R(1,.t) 0 = F1() Re

C14

i+ C16

ieit

what we end up with is that C16 = C14i and that C17 = C14

i1i2

leaving us with a general solution

that satises the boundary conditions and is quite exible.

3.2 Analysis of Flow

Using the derived equations, much like in the rst section, we wish to see how time afects the ow andso we wish to calculate the time averages or V0 and 1:

V0 = 12

14

(1 R2)

2

0

dt

+ A2

2

0

sin tdt

+ A

2Re

J0 i 32 RJ0

i32 2

0

ieit

dt

=1

4(1 R2)

and has ow rate:

{V0} = 8

1 +

A

2sin(t) +

2A

2Re

(e 1 + )eit

With the calculated time average expected and is the same as the previous section. Now or the moreinteresting case o 1 where rather than calculating the time average using the lengthy equations weound, we calculate the time average through our equation (3.21):

21 1t = 2 1 1t = 2 V0V0R sin

26


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and as V0 = g(R) + Re f(R)eit

and the time average o 1t = 0 we can write this as

2 1 = 2

gdg

dR+ Re

f

df

dR

sin

=

1

4(R3 R) + A

2

4Re

2(1 i)

2

e2(R1) e

(1i)2

(R1)

sin

21 = 1 = 14

1

24R5 1

8R3 + L1R

sin + ...

... +A2

4Re

2(1 i)

2

1

2e2(R1) ie

(1i)2

(R1)

sin

1f = 14608

(4R 9R3 + 6R5 R7)sin + ...

... +A2

4Re

2(1 i)

3

2

1

4e2(R1) + e

(1i)2

(R1)

sin

... A2

4Re

(1 i)3

2

3 2i

2

2 5

4

R3

sin + ...

... +A2

4Re

(1 i)3

2

3 2i

2

2 15

4

R

sin

We can then compare this with the time average or 1 using our equations and we get something quitesimilar with diferences arising due to the complex conjugate o :

1s

=

sin

2 A

2

2

Re 13

+(1 2)

44 R3 + 3

3

+(

2 3)

44 R + ...

... +sin 1

4608

R(4 9R2 + 6R4 R6

...

A

2

2sin Re

1

3e(R1)

44e2(R1)

The ow rate o 1 = 0 due to sin .

27


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3.3 Graphical Interpretation

We are interested in the contour plots o the diferent time average or the 1 term o our ow and howthey behave at diferent values o amplitude A as we vary the Womersley parameter . What determineswhether or not is large is how it afects the Bessel unction we have and how it approximates it to anexponential unction. Thereore in our requirement o 1 we can set to around 5 and above. Theratio o the amplitude A to is important due to the arrangement o parameters in the time averageequation and A needs to be larger than , otherwise the purely radial term dominates and we end upwith a ow much similar to that o 1 in the rst section:

Figure 8: The contour plots on the let show the time average or 1f at xed = 8 and changing the amplitude rom

A = 15 in the top graphs to A = 70 in the bottom ones. Whilst the graphs or 1f dont appear to change much we notice

it starts to head rom the centre towards the top and bottom o the pipe. The graph on the right shows the contour plots

or the time average equations o 1s. We look at these two cases as we notice an additional aspect o the ow orming in

the upper and lower regions o the pipe as the amplitude increases in the contour plots or 1s.

The calculated time average or f suggests a deormation in the ow at a lower amplitude and asthis increases the average efect o time leads to an inuence across uid that is ocused very much atthe centre o the ow. The contour plots or 1s acts very much like the 1 seen in the rst section butas the amplitude increase, the efect o time seems to be elt close to the boundary o the pipe. What is

28


29/32

also interesting is that i we lower the value o by hal we end up with a ow that difers greatly rom

our initial 1 and the variation occurs at slight changes in amplitude:

Figure 9: The contour plots on the let show the time average or 1s at xed = 7 at A = 19 and the graph on the right

is A = 20. Unlike 1f which develops into one kind o ow and remains as such, 1s seems to transition b etween three

orms as the amplitude changes. The core also appears to shrink as the additional layers o ow orm.

Figure 10: The contour plot on the let shows the time average or 1s at xed = 5 and an amplitude o A = 8 and the

graph on the right is o1f at amplitude A = 103 just so we can understand how the ow behaves at diferent extremes o

amplitude. I we alter the Womersley parameter to 10, we cannot obtain the above graph or values oA up to and

probably beyond 10150

What we see here is a case o steady streaming, even though there appears to be a problem at thecentre o the ow unlike 1f, but it does show how the ow has settled down over time, or urther worksee Lyne [6]. Unlike the previous section where the chosen scaling restrained the time dependence, whichwas seen in the contour plots, the scaling or this section has allowed or a system which depending on theparameters and A we see a ow whose dependence on time is greater. Furthermore, the time averagesor the last section did not alter with the change in value or amplitude but in this section, the increasein amplitude manages to have a noticeable efect on the ow.

29


30/32

3.4 Summary

Looking over this section, we have managed to obtain the ollowing important results:

V0 =1

4(1 R2) + A

2sin t +

A

2Re

J0

i32 R

J0

i32 ieit

14

(1 R2) + A2

sin t +A

2Re

e(R1)ieit

(in the case o high requency limit)

V1 = 0

V0 = 14

(1 R2)

{V0} = 8

1 + A

2sin(t) + 2A

2Re

(e 1 + )eit

{V1} = V1 = 0

1 = see (3.28)

2 = F1()Re

C14

1

i2ei(R1) 1

iR +

i 1i2

ieit

1f = 14608

(4R 9R3 + 6R5 R7)sin + ...

... + A2

4Re

2(1 i)

3

2

14

e2(R1) + e

(1i)

2 (R1)

sin

... A2

4Re

(1 i)3

2

3 2i

2

2 5

4

R3

sin + ...

... +A2

4Re

(1 i)3

2

3 2i

2

2 15

4

R

sin

1s = sin 2

A

2

2Re

1

3+

(1 2)44

R3 +

3

3+

(

2 3)44

R

+ ...

... +sin 1

4608

R(4 9R2 + 6R4 R6

... A22

sin Re

13 e(R1)

44 e2(R1)

2 = 0{1} = 0{2} = 0

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4 Conclusion

The desire to understand how the ow behaves under a pulsatile driving orce lies in the understandingo blood related diseases, such as high blood pressure or those involving the wall shear stress, much likewhat Siggers et al investigated. The importance o curvature should also be recognised as even with slightcurvature, the centriugal orces introduce cross ow, which is a contributing actor to health concernslinked with blood ow. For example, Peripheral arterial disease is another case which would benet romthe incorporation o pulsatile ow through a curved ow with a narrowing one but or urther compre-hension o this disease, we need to ully understand unsteady ow through a curved pipe.

We have seen how the pulsatile actor within a slightly curved pipe afects the ow and it is interestingto compare the efect o time in diferently scaled models. The next stage would be to look more thor-oughly into steady streaming, see how the ow settles down over time and how it varies with curvature.This would involve urther expansion in powers o the Dean number K so as to include powers o K3

and note the efects that time has on other components o the velocity and the stream unction. We alsoassumed that the Dean number was small and adjusted our model accordingly but an interesting caseto consider would be that o a high Dean number as it exhibits a recognisable boundary layer and theow interaction in that layer is also o importance. A variation in curvature will make the initial modelharder to simpliy but it is also an aspect o the blood ow that needs to be ully understood.. Theconstraints in the model limit how much we can analyse the ow and apply towards real lie scenariosinvolving blood vessels as veins and arteries tend not to have such limited curvature. Another area worthinvestigating is the elasticity associated with the walls o veins and how the deormation in shape efectsthe oscillation o the ow.

The rst hal o this project dealt with a quasi steady ow where we efectively let 0 whereasthe second hal o the project dealt with a high requency limit where and the diference in

the cross sectional pipe ows is due to the steady streaming present. The assumptions made helped tosimpliy the model o the pipe ow yet the calculations were still lengthy and complicated. By lookinginto pulsatile ow through a curved pipe I eel that this project should be able to contribute to an aspecto understanding blood ow. Whilst there are still several areas that still need to be understood whentalking about blood ow, hopeully this project has helped to highlight the importance o both combingtime with pipe curvature. Finally, I would like to thank my project supervisor Doctor Johnathan Mestelor all the invaluable help he has provided in the completion o this project.

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References

[1] BELL, W. W. 1968 Special Functions or Scientists and Engineers. D. Van Nostrad Company Ltd.London. Princeton. New Jersey Toronto Melbourne. Pg 110

[2] BERGER, S.A., TALBOT, L. & Yao, L.S. 1983 Flow in curved pipes. Ann Rev. Fluid Mech. AnnualReviews. Pg 461-512

[3] BOWMAN, F. 1938 Introduction to Bessel Functions, Longmans. Green And Co. Ltd. Pg 41

[4] DAVID, N. Ku. 1977 Blood Flow In Arteries. Ann Rev. Fluid Mech. Pg 399-434

[5] GOLDSMITH, H. L. & SKALAK, R. 1975 Hemodynamics. Annual Reviews. Pg 213-242

[6] LYNE, W. H. 1971 Unsteady viscous ow in a curved pipe. J. Fluid Mech. Pg 45, 1331.

[7] MAZMUNDAR, J. N. 1992 Biouid Mechanics. World Scientic Publishing Co.

[8] MESTEL, J. BioFluids Lecture 15: Steady ow in slightly curved pipes: the Dean equations.http://www2.imperial.ac.uk/ ajm8/BioFluids/lec1015.pd. Imperial College o London

[9] MESTEL, J. BioFluids Lecture 16: Steady ows or various Dean numbers.http://www2.imperial.ac.uk/ ajm8/BioFluids/lec1016.pd . Imperial College o London

[10] PEDLEY, T. J. 1977 Pulmonary Fluid Dynamics. Ann Rev. Fluid Mech. Annual Reviews. Pg 229-274

[11] PEDLEY, T. J. 1980 Fluid Mechanics o Large Blood Vessels. Cambridge University Press

[12] SIGGERS, J. H. & WATERS, S. L. 2008 Unsteady ow in pipes with nite curvature. CambridgeUniversity Press. J. Fluid Mechanics vol.600. Pg 133-165

[13] SKALAK, R., OZKAYA, N. & SKALAK, T. C. 1997 Biouid Mechanics. Ann Rev. Fluid Mech. Pg167-204

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Pulsatile Flow Through a Curved Pipe Final

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