WEBVTT
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in this lecture we will be looking at the
pulsatile flow as we know that the flow in
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the cardiovascular system is pulsatile we
have in on an average in a healthy human
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being about that the heart beats about seventy
two times per minute so the flow process
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or the circulatory process repeats itself
about ones per second the frequency is about
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ones per second so the flow is not steady
as is assumed in the poiseuille flow for example
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rather it the the pressure gradient or
the pressure that drives the flow in the circulatory
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system it changes almost every second and
it is periodic in nature so the flow is pulsatile
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and periodic so in order to understand or
in order to study the flow in the cardiovascular
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system it is important that we understand
and we know what is the flow in a the pulsatile
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flow in a rigid channel for example so
that is what we are going to study in this
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lecture ok now what we see in this graph
here is the pressure gradient d p by d x verses
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time this is a typical pressures verses
time diagram in the circulatory system so
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the pressure goes through certain changes
the it it peaks first the the pressure
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gradient is highest first and then there
is a dip and then again a bit increase and
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then again and then the process repeat itself
so the pressure gradient is pulsatile the
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pressure is pulsatile in nature it is periodic
in nature it its time dependent it is not
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steady and same goes for the pressure gradient
so we need to study that because the flow
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is driven by the pressure gradient in the
circulatory system and the pressure and the
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pressure gradient they are time dependent
they are pulsatile they are periodic so the
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associated flow is also going to be pulsatile
it is going to be periodic in nature and it
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will be time dependent so while this graph
is periodic in nature it is of course time
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dependent
but by just looking at the graph we cannot
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guess the function or time dependent function
that the pressure follows so here we might
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recall from our under graduate classes that
any periodic function say f t which has a
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time period of t it can be represented as
a sum of one constant term and a fundamental
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of period t and its harmonics so a periodic
function can be represented as a sum of f
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t can be represented as a constant so let
us say this constant is a zero plus some
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cosine terms let us say a one cos omega t
plus a two cos two omega t plus so on plus
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b one sin omega t plus b two sin two omega
t plus so on where omega is the frequency
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then the unit is radian per second it can
be two pi over t where t is the time period
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so now this any signal which is periodic
and time dependent that can be represented
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as the sum of a constant term plus sin
and cosine terms and these constants a zero
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a one a two b one b two they can be evaluated
by different methods so one of the methods
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is where for example one calculates that a
zero is equal to one over t integral zero
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to t f t d t and similar terms for a one
and b one there you will have two by t zero
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to t f t sin omega t d t and so on so forth
ok
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so the point i am trying to make here is
that a periodic signal which is arbitrary
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can be decomposed into sinusoidal or and sin
and cosine terms now this sin and cosine terms
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the the coefficients of these sin and cosine
terms can be obtained by using fourier
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transform ok now as you might remember
that sin and cosine terms can be also represented
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as a complex exponential terms so you can
write e to the power i theta is equal to cos
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theta plus i sin theta so for differentiation
and integration purposes it is easier to represent
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the fourier series in terms of exponential
term you can write e to the power i theta
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is equal to cos theta and i sin theta and
so let us say that the pressure gradient del
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p over del z can be represented as sum of
a n which is a coefficient in to exponential
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term e to the power i n omega t where n is
an integer which value will change from zero
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to infinity at n is equal to zero the first
term will be constant and e to the power i
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n omega t so a n will be a n minus b n
if you want to relate it from the previous
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terms ok so the pressure gradient can be
represented as the sum of different harmonics
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harmonics are n is equal to one two three
four for a can be represented as a sum
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of the harmonics so our target or our goal
here now is to understand or to obtain a relationship
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for pulsatile flow where the pressure gradient
is say exponential and the del p by del
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z is del p by del z constant and into a
function of time which is exponential function
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ok
so we will start with the conservation
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equations that we have been doing as of until
now the mass conservation which is also called
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continuity momentum r theta and z so you see
they are all in cylindrical coordinates and
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the stresses the equation has been written
in terms of stresses we will assume the flow
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in a cylindrical tube and flow is in the axial
direction so the flow is going to be axisymmetric
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and when flow is axisymmetric that means v
theta is equal to zero and the terms the
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gradient of the variables in the theta direction
will be zero so this will end up that we will
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not have any term in the theta coordinate
and the respective or the v terms containing
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v theta or del by del theta or the gradient
in the theta direction will be eliminated
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in the equations r and z directions also so
we will end up this set of equations for the
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axisymmetric continuity r and z and so
this continuity equation we have only two
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terms del v by del z and the radial
term and similarly and r and z directions
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now we assume at this point that flow is fully
developed so if the flow is fully developed
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then we might recall that del v z by del z
that is the gradient of the axial velocity
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in the axial direction is zero or the velocity
profile does not change along the axial direction
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so if del v by del z is equal to zero then
from here we can see that v r is also going
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to be zero or if we assume v r is equal
to zero so if v r is equal to zero then this
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term is going to be zero and del v by del
z is equal to zero so we have two conclusions
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from here that the flow is fully developed
and v r is equal to zero so we are going to
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neglect the terms remember that v r is equal
to zero so this is zero this term is also
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zero v z is there but v r is equal to zero
so this term is also zero now because this
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term has v r so this term will go to zero
and this term will also go to zero so effectively
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what we are left with by is minus one over
rho del p by del r is equal to zero where
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rho is non zero so we will have del p by del
r is equal to zero in this the z momentum
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equation we have del v by del z del t so
it will be non zero but this term will be
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gone because v r is equal to zero again the
flow is fully developed so this term is zero
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this term will be there but again del v
by del z del v z by del z is equal to zero
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so this term will also be zero so now we
will have our momentum equation in the
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r direction which is del p by del r is equal
to zero so that simply tell us that p is independent
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of r p is not a function of r and p is anyway
not a function of theta p is a function of
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time and p is a function of the axial direction
so p is a function of time and p is a function
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of z and for the z direction momentum equation
we have del v z over del t is equal to minus
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one over rho del p by del z nu one over r
del by del r r v del r del v z by del r so
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there are three terms this is transient term
this is pressure gradient term and the third
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term is the viscous term
so effectively now we are left with the reject
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term let us non dimensionalise the terms in
this v z momentum equation and the scale
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for non dimensionalisation let us take
a velocities scale v to non dimensionalise
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velocity the both the length or the distance
terms r star and z star r non dimensionalise
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by the channel radius so r by a z by a p star
pressure will be non dimensionalise by
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the dynamic pressure rho v square and t is
non dimensionalise by the time step or or
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say frequency so it is omega t now we substitute
this in the equation then what we will
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get is del v z star and it will be multiplied
by v divided by del t star and it will be
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omega will come from here is equal to minus
one over rho del p star so because of p
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star you will have rho v square term there
divided by del z star so for z star it will
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be multiplied by a plus nu lets collect all
the terms of a here so we will have one
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over r star and there will be one a for r
star del by del r star and this will be a
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square into there will be one a for r in the
numerator and in the another a for r in the
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denominator so we will have both of them cancelling
out so we will simply have r star del by del
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r star now um for v z we will have v z star
and one v here
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let us now eliminate some of the terms so
we can make this one so we can multiply
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by by a square over nu v if we do that then
we will have first term as omega v will cancel
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out omega a square by nu into del v z star
over del t star is equal to so when we multiply
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this by nu rho and rho will cancel out a square
so there will be only one a because one a
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will cancel out and only one v divided by
nu so you might have recognised this non dimensional
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number by now and this is multiplied by
there is a minus sign here del p star by del
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z star plus nu square a a square by nu v and
this will be one so we will have simply one
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over r star del over del t star r star del
v z star over del r star ok so we can see
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that there are two non dimensional group whereas
this is r e or reynolds number and this group
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is a new dimensionalise group which is known
as womersley number ok
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so womersley number if we look at it is represented
by alpha square this alpha square is equal
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to omega a square by nu as we have seen in
the non dimensionalisation of the differential
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equation and if we rearrange it then it terms
out that it is a omega into a by nu so a omega
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is a velocity scale and a is the length scale
and nu is the kinematic viscosity so it
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resembles a reynolds number and this velocity
scale is the oscillatory velocity so it is
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the ratio of inertial forces and viscous forces
but this inertial force is the oscillatory
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inertial force and the viscous force so this
is not flow this is force you can also rearrange
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it alpha is equal to a over ome[ga]- nu
over omega sorry this should be nu over
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omega and this is you might see that nu
over omega the units there nu is meter square
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per second and omega is also one over second
so second second over will cancel out and
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nu over omega is equal to meter square so
this nu over omega power point five is
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going to be meter and this is oscillatory
boundary layer thickness so alpha is also
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a ratio of the channel radius and the boundary
layer thickness so that says that at low womersley
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number the viscous effects are going to dominate
and the boundary layer is going to be thick
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whereas at large womersley number the at large
womersley number the boundary layer will be
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thin and alpha will be large so when alpha
is large boundary layer is thin viscous effects
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are negligible ok we will come back to this
so anyway after non dimensionalisation if
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we go back to our equation this equation has
been written again in this form alpha square
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del v z square over del t star del v z
star over del t star is equal to minus r e
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reynolds number del p star by del z star plus
one over r star del over del r star r star
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del v z star over del r star one might be
worrying about the that the reynolds number
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is generally dependent on or or defined in
terms of the radius so one can do that and
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then there will be a two term coming in
in picture there so as as i was talking about
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that if alpha square is small then in that
case the the viscous term will be significant
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but when the alpha is small and it is less
than one then alpha square will be further
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small and this term will go away that means
at low reynolds number it is not this term
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is zero so you will have that the pressure
gradient term it the pressure gradient term
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and the viscous term they will balance each
other whereas if the inertial force is high
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or the womersley number is large in that case
the viscous force will be negligible or the
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viscous term sorry the viscous term will
go away you can take this on this side and
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then one can see that the viscous term will
go away and the viscous term goes away then
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the transient inertial term or the oscillatory
inertial term that will be balanced by
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the pressure or pressure gradient will be
balance by the inertial force or the transient
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inertial force so ok
now our objective is to understand or to find
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out a relationship between the pressure gradient
and flow or or say axial velocity because
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v theta is equal to zero and v r is equal
to zero so the only direction in which velocity
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is non zero is the axial direction so then
we can integrate it to find out the flow rate
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so we our objective is to find out that
what is the flow rate the relationship
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between flow rate and the pressure gradient
so let us assume a harmonic pressure gradient
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that del p by del z is equal to del p cap
by del z e to the power i omega t and we
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because the pressure gradient is harmonic
so the force coming from this let us assume
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the solution is v z is equal to v z cap i
omega t so our objective is to find out this
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v z in terms of the pressure gradient ok
so first we will look at the two asymptotic
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cases the first one where the womersley
number is small so as we discussed just
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few minutes back that at low womersley number
when alpha is small that means the inertial
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term is negligible
so when we neglect the inertial term what
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we will end up with that the pressure gradient
is balanced by the viscous term so this term
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will go away and we will have del over del
r of r del v z over del r is equal to one
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over rho into nu which will be mu the dynamic
viscosity into del p over del z we have i
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think missed one r here so one over r when
it goes here it becomes r here
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now we substitute that del p by del z is equal
to del p cap over del z e to the power i omega
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t and similarly for v z so we can also substitute
v z is equal to v z cap e to the power i omega
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t we substitute both of this in this equation
then we will get del over del r r v z cap
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over del r because e to the power the exponential
term does not depend on the radius so we can
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take this out of the differentiation equal
to r over mu del p cap over del z e to the
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power i omega t now because the exponential
terms they can cancel out so we will just
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write this and integrate it and what we
will get is r del v z over del r is equal
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to we have miss a this is correct r del v
z over del r is equal to r square over two
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mu del p cap over del z plus a integration
constant let us say c one now we take this
30:07.550 --> 30:16.220
r on the other side so let us delete this
and this will go away and we will have
30:16.220 --> 30:46.160
c one by r now this c one has to be zero as
del v z over del r is finite at r is equal
30:46.160 --> 30:54.520
to zero so if we substitute that then c one
is going to be zero so this term will go away
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and then we further integrate it so we will
get v z is equal to r square over four mu
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del p cap over del z plus c two so v z is
equal to r square by four mu del p by del
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z plus c two and now we use the boundary
conditions so we had v z cap is equal to r
31:42.220 --> 31:57.280
square over four mu del p cap over del z plus
c two we need to find c two and the boundary
31:57.280 --> 32:13.670
condition no slip boundary condition on wall
which says that at r is equal to a which is
32:13.670 --> 32:29.120
wall v z cap is going to be zero
so that means this is zero and a square by
32:29.120 --> 32:45.880
four mu del p cap over del z plus c two if
we subtract these the this equation let
32:45.880 --> 32:58.250
us say equation b and equation a and if we
do b minus a then we will end up with v z
32:58.250 --> 33:10.160
is equal to minus because it will be minus
so we will do b minus a b v z is equal to
33:10.160 --> 33:31.420
minus one over four mu del p cap over del
z and a square minus r square so if we substitute
33:31.420 --> 33:45.460
this in v z if remember we had v z is equal
to v z cap e to the power i omega t so we
33:45.460 --> 34:07.630
will have v z is equal to minus one over four
mu del p over del z a square minus r square
34:07.630 --> 34:20.070
into e to the power i omega t now you might
see because of this negative sign you
34:20.070 --> 34:29.190
can say that v this term and this term
they make the pressure gradient so because
34:29.190 --> 34:39.909
of this minus sign v z and del p by del z
or the flow velocity and the pressure gradient
34:39.909 --> 34:46.970
they are out of phase one eighty degree out
of phase so they are completely opposite this
34:46.970 --> 34:56.450
is the relationship that we just derived and
if we plot del p by del z which will be a
34:56.450 --> 35:08.390
say sinusoidal function that is point
five t by t it is minimum and then it is
35:08.390 --> 35:17.110
say zero at this now the so it is at d
p by d z is equal to zero it is zero and then
35:17.110 --> 35:25.410
it is maximum at t by t so it is going to
be sinusoidal function anyway so at the
35:25.410 --> 35:33.660
minimum when d p by d z is minimum it is negative
minus one then at that point you will see
35:33.660 --> 35:40.090
that the velocity is in the forward direction
and parabolic velocity profile when obtains
35:40.090 --> 35:48.590
when d p by d z is maximum then the velocity
is in the negative direction and the entire
35:48.590 --> 35:55.460
velocity is negative at at every point whereas
when d p by d z is equal to zero then the
35:55.460 --> 36:01.200
velocity is also zero at point seven five
we can see and same point two five so they
36:01.200 --> 36:09.370
are the the velocity and d p by d z they are
one eighty degree out of phase ok one can
36:09.370 --> 36:20.220
also write this as v z is equal to minus one
over four mu a square minus r square del p
36:20.220 --> 36:31.690
by del z so del p by del z and v z they are
out of phase ok let me not confuse you
36:31.690 --> 36:39.020
from this sign ok so we have looked at
asymptotic solution at small womersley number
36:39.020 --> 36:44.700
which suggest that its small womersley number
the expression is very much similar to what
36:44.700 --> 36:55.450
we have for a poiseuille flow but the velocity
and the pressure gradient they are out of
36:55.450 --> 37:03.360
phase now let us look at large alpha or when
womersley number is large then as we suggested
37:03.360 --> 37:10.040
that as at large womersley number the inertial
forces will be dominant and the viscous forces
37:10.040 --> 37:16.220
can be neglected so this viscous force can
be neglected as we followed earlier let us
37:16.220 --> 37:31.610
substitute v z and del p by del z here
so we know that v z is equal to v z cap
37:31.610 --> 37:47.100
i omega t that gives me that del v z over
del t is equal to sorry this is v z cap
37:47.100 --> 38:00.420
e to the power i omega t and del v z over
del t will be v z cap i omega exponential
38:00.420 --> 38:06.260
i omega t
so let us substitute this here and what we
38:06.260 --> 38:19.780
will get is i omega v z cap exponential i
omega t is equal to minus one over rho del
38:19.780 --> 38:31.210
p cap over del z exponential i omega t now
i omega t i omega t will cancel out and what
38:31.210 --> 38:48.870
we will have is v z cap is equal to minus
one over i rho omega del p cap over del z
38:48.870 --> 39:01.150
we can change minus one to i square so
we will get i over rho omega del p cap over
39:01.150 --> 39:15.310
del z so what we see from here that in this
case velocity is plug flow that is there is
39:15.310 --> 39:25.080
no gradient of velocity in the radial direction
the velocity is uniform and which is understandable
39:25.080 --> 39:31.090
because the velocity profile comes because
of the viscous term and we have neglected
39:31.090 --> 39:43.170
the viscous term so the velocity profile is
plug flow ok and as you can see from this
39:43.170 --> 39:54.210
that this term has i so the velocity v z will
be ninety degree out of phase or ninety degree
39:54.210 --> 40:01.970
phase difference between the pressure gradient
and the flow velocity
40:01.970 --> 40:09.010
so as you can see from this graph here the
similar graph for the pressure gradient as
40:09.010 --> 40:18.950
we saw for small alpha and for velocity
at large womersley number we can see the velocity
40:18.950 --> 40:27.770
profile is parabolic and wherever there
is velocity is maximum the velocity sorry
40:27.770 --> 40:36.250
wherever the pressure gradient is maximum
on those pressure maximum or minimum the velocities
40:36.250 --> 40:46.460
are zero there whereas when the velocity when
the pressure gradients are zero when the pressure
40:46.460 --> 40:56.551
gradient is zero at those time instants velocity
have magnitude as maximum and the velocity
40:56.551 --> 41:05.220
profile is plug flow that means the velocity
is uniform everywhere in the channel cross
41:05.220 --> 41:14.040
section ok so now having looked at the
solution at two limits two asymptotic limits
41:14.040 --> 41:21.550
when the womersley number is small and when
the womersley number is a large let us look
41:21.550 --> 41:30.370
at the solution and try to find out if we
can find the solution for the entire range
41:30.370 --> 41:37.720
of womersley number so we again come back
to the v z momentum equation that we have
41:37.720 --> 41:46.080
obtained after reducing the navier stokes
equations to the axisymmetric ones and then
41:46.080 --> 41:53.610
neglecting the terms based on that v r is
equal to zero and del v z over del z is equal
41:53.610 --> 42:00.780
to zero that is flow is fully developed so
then we have got this equation
42:00.780 --> 42:07.250
now in this equation as we have been doing
earlier for the two cases two asymptotic
42:07.250 --> 42:15.760
cases let us now substitute v z and del p
by del z so if we do that again we will have
42:15.760 --> 42:33.690
del v z over del t is equal to i omega
v z cap exponential i omega t that is equal
42:33.690 --> 42:54.580
to minus one over rho del p cap over del z
e to the power i omega t plus nu one over
42:54.580 --> 43:10.090
r and let us break this into two terms
so what we will get is into r del two v
43:10.090 --> 43:21.860
z over del r two and this r and r will cancel
out so we will just get rid of both of
43:21.860 --> 43:32.180
them plus we will have one over r and the
differentiation of r is one so we will have
43:32.180 --> 43:48.610
del v z over del r into e to the power i omega
t now all the exponential terms in the three
43:48.610 --> 43:57.590
equations are same so they can be cancelled
out and let us rearrange this to as
43:57.590 --> 44:04.950
a second order differential equations so we
can write this as this will be v z cap
44:04.950 --> 44:37.540
so we can write this as nu del two v z cap
over del r two plus nu over r del v z cap
44:37.540 --> 44:57.390
over del r minus i omega v z cap is equal
to so because we have brought this on that
44:57.390 --> 45:11.220
side and then it will be equal to one over
rho del p cap over del z ok so this is the
45:11.220 --> 45:17.420
equation that we got in the previous
slide
45:17.420 --> 45:31.810
so now let us rearrange it to this equation
so first we need to divide by nu and
45:31.810 --> 45:52.960
v z cap is a function of r only so we can
have this as total derivative so we can write
45:52.960 --> 46:22.520
this as d two v z cap over d r two plus one
over r d v z cap over d r this because
46:22.520 --> 46:28.960
we have been writing this as a function of
r so this is also a function of r so d v z
46:28.960 --> 46:38.210
over d r plus if we want to change this i
to i cube so minus one is equal to i square
46:38.210 --> 47:01.740
so this will be i cube omega by nu v z cap
is equal to one over rho into nu or we can
47:01.740 --> 47:13.860
also write this as mu del p cap over del z
so we have obtained this relationship now
47:13.860 --> 47:29.170
this is if you look at this is a linear
o d e or terms have power one only and
47:29.170 --> 47:45.900
this is second order and non homogeneous
so we have to go back if we want to find out
47:45.900 --> 47:51.820
the solution of this equation we have to go
back to our under graduate mathematics where
47:51.820 --> 47:58.910
we learnt how to solve non homogeneous second
order linear ordinary differential equation
47:58.910 --> 48:07.420
so if we go back to this you might have heard
or you might have read about bessel equations
48:07.420 --> 48:12.670
if you have done a heat transfer course where
you have looked at some of the complex
48:12.670 --> 48:20.140
equations there also you might encounter bessels
of functions otherwise you would have definitely
48:20.140 --> 48:27.720
done these in your under graduate mathematics
so if the ordinary differential equation
48:27.720 --> 48:34.000
a homogeneous differential equation if it
looks like this x square d two y by d x two
48:34.000 --> 48:40.160
plus x d y by d x plus x square minus n square
y is equal to zero and if we divide by this
48:40.160 --> 48:53.330
by x square and take n is equal to zero so
we will end up with d two y by d x two
48:53.330 --> 49:11.020
plus one over x d y by d x plus y is equal
to zero this is a homogeneous equation and
49:11.020 --> 49:18.170
when the equation is non homogeneous this
term in place of zero it will become non zero
49:18.170 --> 49:25.810
so now let us compare the form of this equation
with the equation that we have derived
49:25.810 --> 49:32.870
so d two y by d x two d two v z by d r two
one over x d y by d x one over r d v z over
49:32.870 --> 49:44.119
del r or d v z over d r plus i cube omega
by v v z is equal to zero if we take the homogeneous
49:44.119 --> 49:55.060
part of our equation so if we want to have
the similarity in the two equations then
49:55.060 --> 50:04.520
we will need to change our variables a bit
so let us do that and if we assume or another
50:04.520 --> 50:12.530
variable let us call this s and if we assume
that s square is equal to i cube omega by
50:12.530 --> 50:22.600
nu into r square so then we can write this
equation as let us divide the entire equation
50:22.600 --> 50:48.061
by i cube omega by nu so we will have one
over i cube omega by nu d two v z over d r
50:48.061 --> 51:18.190
two plus one over i cube omega by nu into
r d v z cap over d r plus v z cap is equal
51:18.190 --> 51:25.980
to zero now we have seen that s square is
equal to we have assumed that s square is
51:25.980 --> 51:36.001
equal to i cube omega nu into r square so
from that let us just make this look
51:36.001 --> 51:53.310
better so this is d r two now we can have
this d two v z cap by d s two plus one over
51:53.310 --> 52:02.990
s there are two s here also one here and one
in the differentiation so one over s d v z
52:02.990 --> 52:17.000
cap over d s plus v z cap is equal to zero
so now we have both the equations the bessel
52:17.000 --> 52:24.930
differential equation for n is equal to zero
and the equation that we have obtained for
52:24.930 --> 52:33.990
intermediate womersley number they look
very similar so the next thing is to look
52:33.990 --> 52:46.520
at what is the solution of this ordinary
differential equation and if we look at the
52:46.520 --> 52:53.760
books this is the form of the equation
for n is equal to zero the x y differential
52:53.760 --> 53:02.550
equation will have the solution y is equal
to a j naught x plus b y naught x we have
53:02.550 --> 53:09.280
we have a and b they are constants whereas
j naught x and y naught x are called bessel
53:09.280 --> 53:15.800
functions and j naught x is the bessels function
of first kind and y naught x is the bessel
53:15.800 --> 53:23.190
function of the second kind so if we look
at the nature of the graphs or the values
53:23.190 --> 53:33.010
of this j naught x and y naught x then
we with respect to x then we see that at x
53:33.010 --> 53:43.630
is equal to zero j naught x or the bessel
function of first kind is the value is one
53:43.630 --> 54:01.780
whereas it is indefinite at x is equal to
zero that means y naught at zero is indefinite
54:01.780 --> 54:15.619
whereas our velocity at s is equal to zero
that means at r is equal to zero will be finite
54:15.619 --> 54:24.740
right so let us write down the solution
based on this the solution for v z cap
54:24.740 --> 54:40.510
and what we will have by copying from this
a j naught s plus b y naught s now at as we
54:40.510 --> 54:50.690
have just seen that y naught zero is indefinite
so b is equal to zero so that this term goes
54:50.690 --> 55:00.990
away so we have that v z is finite at r is
equal to zero so our solution is v z is equal
55:00.990 --> 55:14.720
to or v z cap is equal to a j naught s ok
now we have this because our solution
55:14.720 --> 55:22.440
our equation is non homogeneous equation
so what we have obtained is the complementary
55:22.440 --> 55:28.130
solution and this complementary solution is
a v z complementary is equal to a j naught
55:28.130 --> 55:34.580
s we can also find the particular solution
and i will live it to you to find how to
55:34.580 --> 55:39.490
find out the particular solution and this
will turn out it will turn out that v z cap
55:39.490 --> 55:47.170
now the particular solution will be i by
rho rho omega del p by del z which is the
55:47.170 --> 55:56.230
the last term we had
so the total solution will be or the complete
55:56.230 --> 56:03.750
solution for v z will be because it is a linear
equations so both the solution can be super
56:03.750 --> 56:18.320
imposed so we will have v z is equal to complementary
solution which is a j naught s plus v z particular
56:18.320 --> 56:29.690
solution so that is what the complete solution
is now what we need to do that a is unknown
56:29.690 --> 56:36.590
in this remember what was the particular solution
the particular solution we had is v z p is
56:36.590 --> 56:43.870
equal to i by rho omega del p by del z so
we will use this later on and if we use
56:43.870 --> 56:57.860
the no slip boundary condition to find a that
means v z is equal to zero at a j naught s
56:57.860 --> 57:13.350
we had s square is equal to i cube omega by
nu
57:13.350 --> 57:24.950
r square that is what our definition of s
was so at r is equal to a the value of s will
57:24.950 --> 57:53.270
be i cube omega by nu a square plus v z p
cap now from this we can also remember
57:53.270 --> 58:01.680
that alpha square which is womersley number
is equal to a square omega by nu so this is
58:01.680 --> 58:11.120
i to the power three by two omega power one
by two and nu power one by two or we can say
58:11.120 --> 58:20.940
that alpha by a is equal to root of omega
by nu so we substitute this then we will find
58:20.940 --> 58:43.300
a is equal to minus v z p divided by j naught
i to the power three by two this will be sorry
58:43.300 --> 58:49.880
this will be a only not a square so we will
have alpha over a and a a will cancel out
58:49.880 --> 59:04.910
so i to the power three by two a so we will
have v z is equal to v z particular we can
59:04.910 --> 59:17.020
take this out one minus now we substitute
the value of a here so that will be j naught
59:17.020 --> 59:26.910
and the s if you replace s then it will be
i to the power three by two in place of root
59:26.910 --> 59:37.600
omega by nu we can write alpha over a and
r here for s and this is divided by j naught
59:37.600 --> 59:50.990
i to the power three by two alpha so this
is the value of v z cap
59:50.990 --> 59:58.370
and the relationship that we can find out
now what we have found is v z cap and the
59:58.370 --> 1:00:11.200
v z will be v z cap into i omega t so this
is the relationship between the velocity and
1:00:11.200 --> 1:00:19.140
the pressure gradient for pulsatile flow fully
developed flow in a rigid tube we can also
1:00:19.140 --> 1:00:26.620
find out the asymptotic solution by replacing
alpha is equal to for very small alpha
1:00:26.620 --> 1:00:36.590
and for very large alpha so in summary today
what we have looked at is that using fourier
1:00:36.590 --> 1:00:49.020
series any periodic signal it can be decomposed
into sum of sin or cosine terms or what we
1:00:49.020 --> 1:00:55.840
call harmonics and we will try to take or
we will try to do one example for this case
1:00:55.840 --> 1:01:02.230
then we have encoun[tered]- encountered
our new non dimensional number which is very
1:01:02.230 --> 1:01:09.250
important with respect to or in cardiovascular
fluid mechanics which is known as womersley
1:01:09.250 --> 1:01:20.310
number it is the ratio of oscillatory inertial
flow or the oscillatory inertial force
1:01:20.310 --> 1:01:29.250
and the viscous force and so it is kind
of oscillatory reynolds number and then we
1:01:29.250 --> 1:01:36.310
have obtained the solution for pulsatile flow
for a harmonic or for a sinusoidal function
1:01:36.310 --> 1:01:44.380
or for an exponential complex function for
fully developed flow assuming that we obtained
1:01:44.380 --> 1:01:52.440
or we saw that at low small womersley number
at small womersley number or low womersley
1:01:52.440 --> 1:02:00.720
number the expression of the velocity profile
is very similar to what we obtain for a
1:02:00.720 --> 1:02:06.270
poiseuille flow
but the but it is time dependent and it
1:02:06.270 --> 1:02:14.580
is one eighty degree the velocity is one eighty
degree out of phase with the pressure gradient
1:02:14.580 --> 1:02:20.910
whereas at large wave womersley number
the flow profile is same everywhere in the
1:02:20.910 --> 1:02:27.150
cross section that is no effect of viscosity
flow profile look like a plug flow it is
1:02:27.150 --> 1:02:32.880
uniform everywhere in the cross section and
it is at ninety degree out of phase with
1:02:32.880 --> 1:02:38.200
the womersley number so wherever the pressure
gradient we have it is ninety degree out of
1:02:38.200 --> 1:02:45.530
phase with the pressure gradient so wherever
the pressure gradient is maxima or minima
1:02:45.530 --> 1:02:52.450
then because it is sinusoidal function
so when the pressure gradient is at is at
1:02:52.450 --> 1:03:02.970
maxima or minima the velocity is zero and
if it is zero then if the pressure gradient
1:03:02.970 --> 1:03:11.430
is zero then the velocity is maxima or minima
we will also try to look at some of the applications
1:03:11.430 --> 1:03:20.119
and try to have a field of numbers and some
examples of this this womersley solution
1:03:20.119 --> 1:03:43.600
in the context of cardiovascular fluid mechanics
thank you