WEBVTT
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in this lecture we are going to continue
pulsatile flow in the previous lecture we
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discussed about pulsatile flow in rigid tubes
so we suggested there that as the flow
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is pulsatile in the cardiovascular system
and any pulsatile the flow is pulsatile
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as well as periodic so as the flow is periodic
and from the fourier series any periodic function
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can be decomposed in a infinite number
of sinusoidal functions or sinusoidal harmonics
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in infinite series of a constant term plus
sine and cosine terms and we need to find
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out the coefficients of this sine and cosine
terms and the constant term for which there
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are standard methods using fourier transform
so the objective then reduce to to understand
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the pulsatile flow when the pressure gradient
is sinusoidal in nature and we consider the
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rigid tube for a fully developed flow we derived
the solution or the relationship between velocity
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and pressure gradient for a sinusoidal pressure
gradient and we did it in terms of when the
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time dependence is exponential in terms of
complex numbers ok so in this lecture we are
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going to continue where we left in the previous
lecture and discuss few more characteristics
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of pulsatile flow in rigid tubes the solution
that we obtained we called it womersley
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solution after doctor womersley who looked
at the pulsatile flow in rigid tubes and
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derived a number of relationships for it
so the relationship what we see here is for
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we said that the velocity v z will be a function
of v z cap which is a function of r and the
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time dependent per term e to the power i omega
t now the v z cap and similarly the relationship
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between the the the pressure gradient the
time dependence of pressure gradient was del
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p by del z is equal to del p cap over del
z e to the power i omega t now this equation
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gives the relationship between the v z cap
and del p over del z for all values of womersley
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number so just to recap womersley number is
equal to alpha square is equal to a square
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omega over nu and it is the ratio of transitional
or transient inertial force and the viscous
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force so in this equation what we see
is these are functions of these j zero
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and j zero is bessel function of first kind
so we need to a recap or we need to just remind
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ourselves what bessel function is what is
the series representation of bessel function
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how does the graph of a bessel function of
first kind look like and then some properties
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which we might require to obtain the
shear stress or to obtain the flow arte in
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pulsatile flows from the womersley solution
so let us look at the bessel function of first
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kind and the series representation is that
you have j n x where n is for any number
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n can be any harmonic and k is equal to zero
to infinity minus one to the power k so it
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is a infinite series and k is a representing
each term minus one or to the power minus
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one to the power k factorial k of factorial
and plus k x by two to the power two k plus
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n so what we are concerned is or we are we
are concerned about is j zero so we will write
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that j zero x is equal to sigma k is equal
to zero to infinity minus one to the power
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k and factorial k will be square x by two
exponent two k because n is equal to zero
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now this is j zero x so if we want to
find out say value of j zero at zero or let
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us just expand it first a bit let us write
few terms so we will see that j zero x is
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equal to for k is equal to zero this is one
this will also be one everything is one minus
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one to the power one so that is one so one
over and k is one so one into x by two square
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plus when k is equal to two we will get term
one divided by two square that means four
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x by two four and so on so we can see that
the value of j zero at x is equal to zero
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will be one as we can see from this graph
and then the graph goes into the negative
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direction and then it oscillates in the magnitude
of this oscillation we can see that it comes
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down ok
so two important properties of bessel functions
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we can obtain these properties by looking
at the properties of the bessel functions
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these two equations so j minus n x is equal
to minus one exponent or minus one to
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the power n j n x and the other property between
relationship between j n x the derivative
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of j n x and j n x and j n minus one x so
let us take if n is equal to zero then what
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we will have is x j zero dash x is equal to
this term is going to be this term is going
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to be zero because n is equal to zero so we
will have x j minus one x but from this relationship
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we have that j n j minus one x is equal to
minus j one x so we will have x minus x j
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one x so this basically give us the relationship
between we can cancel out x from all and we
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have that j zero dash x or the derivative
of j zero with respect to x is equal to minus
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j one x do we might require this later on
now for n is equal to one if we substitute
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one here in the first equation then we will
get x del j one x divided by del x is equal
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to minus j one x plus x j zero x now we do
a bit of recifal of terms so we can put
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a plus here and this becomes equal so from
this let us look at this this is a derivative
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of x j one x that is equal to x j zero x so
you can say the derivative of x x the second
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term is differentiated j one dash x plus
the differentiation of x is one and then j
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one x so we might want to say this explicitly
here that j dash x means del j n x over del
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x so we have obtained two important relationships
here which we might require to use further
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ok
so we have obtained a relationship between
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the velocity and pressure gradient we have
obtained the velocity profile but often when
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we are looking at flow in the pipes we need
to obtain the characteristics of the average
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velocity or the average velocity or flow rate
so let us obtain the flow rate the flow rate
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is q is equal to integral zero to a where
a is the radius of the channel two pi r d
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r multiplied by v z ok and you might we have
the velocity profile here so if we substitute
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that velocity profile then we will require
to integrate it the terms which are independent
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of r they can come out of the integral sign
so we will have i over rho omega del p over
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del z p cap and the exponential term time
dependence e to the power i omega t plus we
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also have two pi here all of this is independent
of r so it can come out of the integral sign
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then we have integral zero to a one minus
j zero i to the power three by two alpha r
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over a divided by j zero i to the power three
by two alpha into d r remember that i have
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included the r term here now if we write this
down again in integrate q is equal to two
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pi i over rho omega del p cap over del z exponential
i omega t the first term when we integrate
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multiply with r so r the integration will
be r square by two and when we put the limits
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we will end up with a square by two minus
integral of limits from zero to a j zero i
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to the power three by two alpha r over a divided
by j zero i to the power three by two alpha
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d r
now i the again there is a r here you might
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notice that this term is independent of r
so it can so also be taken out of the integral
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sign so now the objective is to find out this
integral so let us try to find out this integral
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and we will see that integral zero to r so
we can take the denominator term outside because
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it is independent of r one over j zero i to
the power three by two alpha now what we can
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do we can assume because this if you remember
what we had assumed earlier that i to the
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power three by two alpha over a into r is
equal to s so we can do a change of variable
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here if we do that then we also need to do
the change of variable for d or differential
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terms so d s is equal to i to the power three
by two alpha d r over a now if we substitute
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in place of r so we have s over i is to the
power three by two alpha over a which is for
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r now j zero and this entire term is s and
again d r is d s over i to the power three
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by two alpha and this becomes a square so
we can say that we can get rid of all this
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here and just a square here now all this is
independent of s so we can bring this here
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and this will be a square over i cube alpha
square j zero i is to the power three by two
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alpha we need to also work out the limits
so integral zero to r is equal to it should
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be small a because we are using a for the
radius so i to the power three by two alpha
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a over a so that will be one these are the
limits integral zero to i to the power three
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by two alpha s j zero s d s
now we had just seen the relationship in the
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previous slides that x j zero x is equal
to d by d x of x j one x so what we are going
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to do when we integrate it this equation we
will get this a square i cube alpha square
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j zero i to the power three by two alpha and
s j one s at s is equal to i is to the power
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three by two alpha so let us substitute this
and what we are going to get is a square divided
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by i raised to the power three alpha square
j naught i raised to the power three by two
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alpha and for s if we substitute i raised
to the power three by two alpha j one i raised
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to the power three by two alpha and we will
get a square divided by i raised to the power
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three by two alpha because this will cancel
out and the ratio of j one of i raised
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to the power three by two alpha divided by
j zero i raised to the power three by two
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alpha remember this was an integral in a bigger
expression
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so if we substitute that expression we are
going to get the flow rate to be i pi a square
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which is the cross sectional area of the channel
divided by rho omega this is the pressure
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gradient and this term inside the bracket
so as you might notice that this is a complex
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term and we need to obtain its magnitude and
phase angle so if rep[resent]- we represent
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this term as x one plus i x two and remember
that e to the power i omega t is equal to
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cos omega t plus i sin omega t and we have
another i here so we can substitute that i
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here and from that we can reduce this into
one so we can have multiply x one i x two
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into this so we can get what is the magnitude
of q and that will be pi a square by rho omega
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del p over del z root of x one square plus
x two square and if i is the angle phase angle
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then we will get tan phi is equal to x one
over x two so this is the magnitude of the
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flow rate and this is the phase angle so where
phi is the phase angle between the pressure
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gradient and the flow rate
you might recall from yesterdays analysis
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that at small reynolds number the flow rate
is same as in a poiseuille flow and the
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phase angle or the the between the phase angle
between the two is one eighty degree whereas
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for large reynolds number the phase angle
is about ninety degree between the two so
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what we have looked at based on the velocity
profile till now we have obtained q or
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the flow rate now we will try to obtain tau
which is the wall shear stress tau w and the
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formula for this is equal to minus mu del
v z over del r at r is equal to a so again
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we have the velocity profile here from this
velocity profile we want to obtain wall shear
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stress which is the derivative of this
so to obtain wall shear stress we will
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differentiate the velocity term so it will
be minus mu multiplied by i over rho omega
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del p over del z exponential i omega t which
is the time dependent pressure gradient the
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first term in this bracket because it is a
constant it is not dependent on r so that
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will be zero and there will be a minus sign
so that minus and minus they will make plus
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so we can just get rid of this sign here
and then one over j naught i to the power
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three by two alpha this is also independent
of r and what we are left with is d by d r
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of j naught i to the power three by two alpha
r over a and we had derived earlier that the
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derivative of j naught x is equal to minus
j one x so the derivative of j naught is minus
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j one x so if we can substitute that here
and tau w is equal to mu i over rho omega
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del p over del z we can write this in terms
of capital e to the power i omega t one over
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j naught i to the power three by two alpha
now that the derivative of this will be minus
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j one i to the power three by two alpha r
over a and i to the power three by two alpha
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over a so the wall shear stress is we have
alpha over a so finally when we do the
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algebraic calculations the final relationship
that we will obtain is minus a over i to the
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power three by two alpha del p over del z
we will write this in terms of exponential
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term so del p over del z e to the power i
omega t j one remember that this will be at
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r is equal to a so this a and a will cancel
out so that will be sorry j one i raised to
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the power three by two alpha over j zero i
to the power three by two alpha um this is
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the final value what we have just written
and this del p by del z is written in terms
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of time dependent it is another important
relationship is viscous impedance which
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is the ratio of del p over del z which
is the pressure gradient and the flow rate
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so the viscous impedance is equal to pressure
gradient and the and the ratio of flow rate
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so if we look at this relationship we can
easily find out that z l is equal to del p
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over del z which is this term so this is q
and what we want to do is del p over del z
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divided by q so that will be rho omega by
i pi a square into one over one minus two
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over i to the power three by two alpha j one
i raised to the power three by two alpha divided
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by j naught i raised to the power three by
two alpha so this is the viscous impedance
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term and again we will have the magnitude
of j l here z l here and that will be and
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there will be a theta term which will be
the phase difference between the pressure
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gradient and flow rate and this is quite
handy in calculations of the work that is
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done in the system for the flow or for
the pulsatile flow to happen
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so in the previous class or in the previous
look lecture we looked at what is womersley
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number and womersley number is the ratio of
transient inertial force and the viscous forces
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we also said in the previous class that womersley
number which is so it is ratio of alpha
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square is equal to a square omega over nu
or it can be represented as a over nu over
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omega raised to the power one by two so the
velocity of the flow oscillates but is always
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zero on the wall even though when the flow
is pulsatile there is always no slip boundary
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condition at the wall that means for a viscous
flow the velocity near the wall is going to
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be zero and there are going to be velocity
gradients near the wall so where the viscous
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force dominate whereas in the layer just next
to the wall so if you have the channel here
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and after some distance from the wall where
you have gradient and then beyond that
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distance let us say that this distance is
delta where viscous force is dominate and
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after this the velocity is oscillating and
where the transient inertial forces are dominating
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and thats where you see the more change of
transient effects ok so as we move from
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this distance towards the wall the viscous
force will be dominating and as and as move
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away from here the transient force becomes
more important so at this distance what we
32:18.960 --> 32:27.059
call the transient boundary layer the two
forces are equally important so what is the
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transient inertial force
32:40.549 --> 32:55.570
that is we can say that it is equal to
rho del u by del t so a magnitude will be
32:55.570 --> 33:08.570
rho u divided by say t si proportional
to one over omega so one over omega so this
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force will be rho u omega
and the viscous force will be
33:25.749 --> 33:44.070
mu u by delta square because it is it comes
from mu del two u by del r two so if we substitute
33:44.070 --> 33:56.580
that a from a from dimensional analysis and
at delta we will see that rho u omega is of
33:56.580 --> 34:07.679
the same order of magnitude as rho mu u by
del square so that gives us u and u cancel
34:07.679 --> 34:19.730
out and they gives that delta square is equal
to mu over rho omega or nu over omega so from
34:19.730 --> 34:29.419
this we get the relationship that delta is
equal to nu over omega to the power one by
34:29.419 --> 34:36.310
two so what we are trying to say here that
the thickness of this oscillating boundary
34:36.310 --> 34:45.770
layer can be obtained by the comparison of
the transient inertial force and the viscous
34:45.770 --> 34:53.280
force at the boundary layer thickness or
at the at the just the extend where the
34:53.280 --> 35:01.040
boundary layer is so from this we can obtain
the boundary layer and we can show that alpha
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is the ratio of a which is the cube radius
and transient boundary layer thickness now
35:10.980 --> 35:19.040
we will look at some of the typical values
of alpha square or womersley number that we
35:19.040 --> 35:30.130
encounter in general so we know that alpha
is equal to a alpha square is equal to a square
35:30.130 --> 35:38.380
omega over nu so let us find out what are
typical a omega and nu as the radius of the
35:38.380 --> 35:47.640
channel vary in a system so we will have different
radii but omega and nu are going to be same
35:47.640 --> 35:57.350
for one system so the typical value of omega
we know that seventy two the heart beats
35:57.350 --> 36:07.010
seventy two times per minute so from that
we can calculate omega is equal to two pi
36:07.010 --> 36:20.190
over time period and time period is for
one b at it is sixty divided by seventy two
36:20.190 --> 36:30.450
so that will come out one forty four pi divided
by sixty and that number will be about seven
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point five five ok
next we will calculate the kinematic viscosity
36:40.930 --> 36:50.400
which will be omega sorry mu over rho and
the typical value of the blood viscosity is
36:50.400 --> 36:59.390
about three point five centipoise so if we
change that centipoise into s i units three
36:59.390 --> 37:10.710
point five into ten to the power minus three
k g per meter per second divided by rho which
37:10.710 --> 37:26.620
we will take about ten to the power three
k g per meter sorry this is k g per
37:26.620 --> 37:35.500
meter cube and so that will be of the order
of three point five into ten to the power
37:35.500 --> 37:53.730
minus six meter square per second now we
will look at for say aorta the radius is the
37:53.730 --> 38:00.610
diameter of aorta is about two point five
centimetre so radius is about two point five
38:00.610 --> 38:16.100
divided by two into ten to the power minus
two meters so alpha is equal to root of omega
38:16.100 --> 38:29.130
over nu into a so let us work out this number
a root of omega is seven point five five divided
38:29.130 --> 38:42.570
by nu which is about three point five into
ten to the power minus six so what we will
38:42.570 --> 38:51.780
have roughly root two this number is will
be more than two but for the simplicity
38:51.780 --> 38:59.370
sake let us say that this is root two into
ten to the power six that the square
38:59.370 --> 39:12.190
root of ten to the power six will be ten to
the power three into a so a is in meters
39:12.190 --> 39:24.400
for this to be true so for aorta this will
be one point four which is about root two
39:24.400 --> 39:33.810
value into ten to the power three into one
point two five that is two point five divided
39:33.810 --> 39:42.270
by two into ten to the power minus two so
what we will get is this number is about seventeen
39:42.270 --> 39:53.000
or eighteen so the order of magnitude of alpha
in aorta is about eighteen and because omega
39:53.000 --> 39:58.690
and nu are going to became in the cardiovascular
system so that is the about the maximum value
39:58.690 --> 40:02.110
of alpha that will be there in the circulatory
system
40:02.110 --> 40:07.860
if we look at the capillary system which will
be the say minimum size of the channel so
40:07.860 --> 40:18.550
in the capillaries let us say alpha capillary
and this will be one point four into two to
40:18.550 --> 40:31.170
the power three and a capillary the radius
of capillary will be about say five microns
40:31.170 --> 40:39.360
so five into ten to the power minus six meters
into one point four into ten to the power
40:39.360 --> 40:45.950
three into five into ten to the power minus
six so that will come out about seven into
40:45.950 --> 40:51.390
ten to the power minus three or roughly we
can say that about ten to the power minus
40:51.390 --> 40:57.410
two so this womersley number is going to
be very very small that means the viscous
40:57.410 --> 41:02.350
forces are dominating in this case and the
transient inertial force is neglected so that
41:02.350 --> 41:12.910
is why the transient forces are not
very important in the a small channels ok
41:12.910 --> 41:18.940
so now let us look at some of the velocity
profiles in the channels so these velocity
41:18.940 --> 41:26.150
profiles have been plotted for four different
ha values of alpha three point three four
41:26.150 --> 41:34.300
four point seven two five point seven eight
six point sex seven and in these four cases
41:34.300 --> 41:41.390
the velocity profiles have been plotted for
different values of omega t ranging from
41:41.390 --> 41:47.440
this the first profile here is for omega t
is equal to zero then omega t fifteen omega
41:47.440 --> 41:52.920
t thirty and so on so forth up to one eighty
and you might notice that after one eighty
41:52.920 --> 42:00.310
the profile has become inverted in all the
cases the profile has become inverted so one
42:00.310 --> 42:06.800
eighty to three sixty the profiles will be
same but inverted that one eighty plus
42:06.800 --> 42:11.740
five ok
another thing that you should notice
42:11.740 --> 42:20.030
here that at low values the profile is parabolic
there are large gradient in the centre but
42:20.030 --> 42:26.490
as the reynolds number or as the womersley
number increases the flow profile becomes
42:26.490 --> 42:33.790
flatter in these cases here so all these
have been plotted for pressure gradient when
42:33.790 --> 42:42.850
the pressure gradient vary as del p over del
z cos omega t ok so you can see or you can
42:42.850 --> 42:48.030
easily make out the thickness of the boundary
layer that the thickness of boundary layer
42:48.030 --> 42:57.000
reduces as alpha increase ok boundary layer
is where there is a change in the velocity
42:57.000 --> 43:04.180
gradient near the wall and we can easily make
out in all these cases decrease ok so another
43:04.180 --> 43:14.000
question that we might come across that often
sometimes or many students have this
43:14.000 --> 43:26.740
perception that if the flow is transient or
if the flow is time dependent then is it
43:26.740 --> 43:34.830
laminar or not or being time dependent does
it directly mean that the flow is turbulent
43:34.830 --> 43:43.110
in another misconception what people think
is that if there is recirculation in the
43:43.110 --> 43:51.540
flow that means if you have some kind of
vertices in the flow then people tend to think
43:51.540 --> 43:57.990
that the flow is turbulent in nature but these
are two different thing the turbulence
43:57.990 --> 44:05.220
have eddies of different length scales starting
from very small scale what we call kolmogorov
44:05.220 --> 44:11.530
length scale to the largest possible length
scale in the system
44:11.530 --> 44:19.650
however just the presence of eddies does not
mean that the flow is turbulent so the
44:19.650 --> 44:27.290
question comes how one can define or how
one can think about or how one can determine
44:27.290 --> 44:34.850
that the flow is turbulent so the turbulence
is defined by random fluctuations in the flow
44:34.850 --> 44:45.820
velocity so the key here is the fluctuations
that are random they are they dont follow
44:45.820 --> 44:51.300
any pattern the the fluctuations in the velocity
and consequently in the pressure they are
44:51.300 --> 44:59.020
they are random fluctuations in the flow velocity
so one cannot determine with precise absolute
44:59.020 --> 45:05.090
precision it is not possible to determine
the velocity field in turbulent flows however
45:05.090 --> 45:13.900
there is an order in this randomness also
and the statistical feature of the turbulent
45:13.900 --> 45:21.580
flow can be well defined ok so the question
that we are trying to look at here is that
45:21.580 --> 45:29.040
is the pulsatile flow that we have is turbulent
or it is laminar or something in between so
45:29.040 --> 45:36.540
let us look at this experiment first this
is the first famous reynolds experiment
45:36.540 --> 45:46.520
image and it has been taken from funks book
so when the flow is laminar at low velocities
45:46.520 --> 45:54.130
the dye that was injected at the inlet it
moves just in that line whereas when the flow
45:54.130 --> 46:01.030
become turbulent or from there is transition
the dye is start to defuse in the entire channel
46:01.030 --> 46:08.180
and this is fully turbulent flow where
you see that dye is defused but what i want
46:08.180 --> 46:14.740
you to notice here that even in the turbulent
flow the flow does not become turbulent just
46:14.740 --> 46:20.050
at the entrance
it takes some times for the eddies to develop
46:20.050 --> 46:26.800
for the eddies to grow and for the flow to
become turbulent and this has a consequence
46:26.800 --> 46:37.150
in the pulsatile flow so this image is
velocity with respect to time for a pulsatile
46:37.150 --> 46:45.950
flow so as an this dotted line is the this
line corresponds to the reynolds number and
46:45.950 --> 46:52.720
this reynolds number is based on the velocity
scale which which is the average velocity
46:52.720 --> 47:00.580
in the channel cross section so this dotted
line corresponds to reynolds number two thousand
47:00.580 --> 47:09.560
three hundred which is considered as the critical
reynolds number at which in a smooth pipe
47:09.560 --> 47:19.560
the steady laminar flow transits or the
laminar flow starts becoming turbulent or
47:19.560 --> 47:27.470
the transition starts happening so you might
see here from this that as the velocity
47:27.470 --> 47:37.030
grow and the flow remains laminar for sufficiently
higher value of reynolds number before it
47:37.030 --> 47:44.170
becomes turbulent that means before the fluctuations
in the velocity become random so what does
47:44.170 --> 47:51.420
this suggest that in an accelerating flow
where the velocity of the flow is growing
47:51.420 --> 48:01.990
because it takes some time for the fluctuations
or the eddies to grow or turbulence to set
48:01.990 --> 48:10.560
in the accelerating flow is more stable than
the steady flow at the corresponding reynolds
48:10.560 --> 48:18.980
number so the transition reynolds number for
an accelerating flow where the flow velocity
48:18.980 --> 48:25.650
is increasing is higher
similarly if you look at this end where the
48:25.650 --> 48:33.510
flow is decelerating flow velocity decreases
so because the turbulent eddies they have
48:33.510 --> 48:39.510
to die down and they do not die down just
then and they are it take some time for the
48:39.510 --> 48:48.580
turbulent eddies to down the die down so the
flow will half well below through reynolds
48:48.580 --> 48:56.020
number of two thousand three hundred the flow
remains turbulent and then slowly die down
48:56.020 --> 49:05.030
ok so this is one important conclusion
in terms of turbulence in pulsatile flows
49:05.030 --> 49:12.950
so in summary what we have looked at in
including both the lectures in pulsatile
49:12.950 --> 49:20.390
flow that one of the critical reynolds number
or the critical parameter which is important
49:20.390 --> 49:29.030
for the pulsatile flow in rigid tubes is
womersley parameter or you can also call it
49:29.030 --> 49:34.930
a transient reynolds number or the square
of womersley number can be called transient
49:34.930 --> 49:45.840
reynolds number so this is ratio of transient
inertial force and viscous force then the
49:45.840 --> 49:58.350
womersley solution so we looked at the
flow of a periodic flow time dependent
49:58.350 --> 50:05.750
periodic flow in a rigid channel and obtain
the solution and from that we also obtain
50:05.750 --> 50:13.360
the flow rate and pressure gradient
and it is a function of it because it is
50:13.360 --> 50:19.100
a function of it has womersley solution bessels
function so we also looked at the bessel function
50:19.100 --> 50:28.290
of first kind and its properties and we
briefly discussed about the pulsatile flow
50:28.290 --> 50:38.360
and turbulence that is the flow always turbulent
when the flow is pulsatile or not ok so
50:38.360 --> 50:59.190
with that we will end this lecture
thank you