WEBVTT
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so in this lecture we will look at flow of
blood in a channel or flow of fluid in a channel
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as we know that in our cardiovascular system
there are number of channels of different
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diameters so it is important to understand
flow in a channel problem so in this lecture
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what we will look at is fully developed flow
of a newtonian fluid and fully developed flow
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of a casson fluid which can represent or which
can model the rheological behaviour or the
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viscosity or the viscous behaviour of the
blood in a channel so you must have studied
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in your basic fluid mechanics course that
the flow of fluid can be modelled by the
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conservation equations and if we want to model
the full three dimensional flow of a fluid
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we can represent these conservation equations
in the differential form and these are the
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conservation of mass and momentum if the fluid
properties are dependent on the temperature
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then a energy equation also needs to be solved
but since in our case the flow is isothermal
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and the fluid is incompressible that is blood
is incompressible so the properties are independent
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of temperature so we just need to solve two
conservation equation conservation of mass
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and conservation of momentum so let us look
at the general form of these conservation
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equations first
the mass conservation equation is written
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in this form the first term in this equation
is rate of increase of mass per unit volume
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the first term and the second term is the
mass that is being convicted or need net rate
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of mass addition per unit volume by convection
so if you consider a control volume which
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might be a part of the channel or a channel
so the first part represents you must see
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here that these equations are in terms of
per unit volume so that is why it is rate
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of increase of mass per unit volume which
will be zero in our case because the density
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of the blood is a constant and the second
term it represents the net rate of mass addition
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that is provided by the convection so that
is flow coming in and the flow going out from
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the boundaries of the domain so that is mass
conservation equation and for an incompressible
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fluid which has a constant density this equation
this term becomes so the first term becomes
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zero
so this equation can be given by divergence
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of velocity vector is zero the second conservation
that we consider is momentum conservation
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and in the momentum conservation the general
momentum conservation equation looks like
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this the first term represents the rate of
increase of momentum per unit volume so again
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this is in terms of per unit volume rho
v is the momentum per unit volume so rate
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of increase of momentum per unit volume first
term the second term is rate of momentum addition
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by convection per unit volume here the
first term on the right hand side is the rate
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of momentum addition by molecular transport
per unit volume and the last term is body
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forces of the gravitational force or any other
force so this can be gravitational force or
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any other body force
this term includes the molecular transport
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per unit volume where p is the thermodynamic
pressure which is normal stress and t is the
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tau tau is the viscous stresses it can have
it will have normal as well as shear stress
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components so for a fully developed pipe flow
let us consider a circular cylinder a cylinder
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which has circular cross section and flow
has happening in the axial direction which
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we represent a z direction here and the radial
coordinate is r and angular coordinate is
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theta and because there is no flow in the
angular direction so that velocity will be
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zero so let us try to understand the assumptions
that will be involved in it the flow is incompressible
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so that means the density is constant t z
steady state so that means the del by del
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t term will be zero things do not change
with time we do not consider any body forces
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including gravity in this analysis and we
also neglect the entrance and exit effects
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which means the flow is fully developed so
what fully developed flow means that with
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increase in distance or with distance the
velocity profile do not change so if i plot
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velocity profile at z is equal to z one and
if i plot a velocity profile at z is equal
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to z two it remains unchanged so what does
it essentially means that z velocity at z
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one and velocity at z two they are same so
if we take a first approximation to the
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gradient then this will be zero so that means
del v over del v z over del z is equal to
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zero for a fully developed flow ok
there is no tangential flow that means
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there is no flow in the angular direction
and there is no radial flow as you can see
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from here that if there is flow in the radial
direction then the velocity profile in the
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z direction will not remain same so v theta
tangential flow is zero so that means v theta
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is equal to zero there is no radial flow so
that means v r is equal to zero and the gradients
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in theta direction are also zero so that means
v z or z component of velocity or the axial
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component of velocity is independent of theta
so v theta is zero and del by del theta is
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also zero for the tangent the no tangential
flow ok so let us look at the conservation
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equation again in the expanded form in cylindrical
coordinate so what we see here is the continuity
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and three components of momentum equation
r theta and z coordinates and let us try to
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see different terms in these equations because
flow is a the steady state so all the transient
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term will go to zero v r and v theta r zero
so all the term having v r and v theta will
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go to zero
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we neglect body forces now the shear stress
is a function of velocity gradient
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so if v z is non zero this term gives us that
del v z over del z is equal to zero ok from
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here we can see that all the stresses except
tau or tau r z will be zero because tau r
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z have v z over v r all other stresses will
be zero in this case so tau r r is zero tau
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theta r is zero tau theta theta is zero tau
r theta is zero this is zero this is zero
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this is also zero tau z z is zero because
it will have v z over del z so this we have
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seen then this is also zero so this component
is also zero tau theta z is zero because del
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by del theta is zero so only two components
remaining again this will be again zero because
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tau z over del z and the shear stress will
not vary because the flow is fully developed
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so shear stress will not vary in the axial
direction so this is also zero
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so if we write down the non zero terms in
all the equations we will end up this with
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this this we have already said that if the
flow is fully developed then del v z over
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del z is equal to zero so that means v z is
not a function of z so from our two assumptions
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that says that v z is a function of r only
because it was not a function of theta and
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it is not a function of z from these two equations
we see that p is a not function is not a func[tion]
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a is constant with respect to r it does not
change along the radial coordinate and p does
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not change along the theta coordinate so that
again means that p is a function of p z so
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this partial derivative we can change it to
tangents total derivative you can see
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here in this equation this has been rearranged
the the second term has been brought towards
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left hand side and we see that one over r
del over del r r tau r z is equal to minus
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d p by d z so let us try to integrate this
and if you integrate this what we will end
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up with r tau r z is equal to minus d p by
d z del by del r of r tau r z is minus d p
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by d z into r so that means r tau r z is equal
to minus d p by d z into r square by two or
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tau r z is equal to minus d p by d z r by
two so this is a general equation and we get
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relationship between the shear stress and
pressure gradient and the radius of the channel
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or radial coordinate ok
so this relationship we can also look at a
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force balance in a channel so let us see if
you have a fluid element which has a length
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of d z and the pressure at the two ends is
p and p plus d p the radius of this fluid
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element is cylindrical fluid element the radius
is r and if sorry this is p plus d p and
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we write the force balance that this will
be acting in this direction in the element
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so minus d p into pi r square that is equal
to tau into two pi r d z so pi pi cancels
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out r cancels out and tau is equal to minus
d p by d z r by two and this is an important
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relationship between the shear stress and
the radial coordinate and this is independent
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of the constitutive equation of the fluid
it is true for a newtonian as well as non
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newtonian fluid so let us look at now for
the fully developed flow of a newtonian fluid
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and for a newtonian fluid the shear stress
is proportional to the shear rate so that
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tau is equal to minus mu del v z by v r so
we substitute this here we will end up with
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minus mu del v z by del r is equal to minus
d p by d z into r by two and if we bring mu
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on the other side then it will be one over
mu d p by d z r by two so let us integrate
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it and we will get v z is equal to one over
mu d p by d z r square by four plus c where
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c is a constant so this has already been integrated
for us so we here end up with this equation
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for v z
so having obtained the velocity profile in
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a fully developed pipe flow for a newtonian
fluid let us now we can also derive the
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flow rate in the channels so flow rate using
this profile we can say the volumetric flow
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rate which is q is equal to integral zero
to r v z into two pi r which is the area of
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a differential element that we will take on
a cross section into d r ok so after the integration
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we will obtain this relationship between flow
rate and the pressure gradient this relationship
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is known as hagen poiseuille law which is
which says that for constant pressure gradient
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q is proportional to r to the power four for
given pressure gradient for given d p by d
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z q is proportional to r to the power four
or for given q delta p is proportional to
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or delta p y l for a for that matter is
proportional to r to the power four because
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r is a because l for a channel if it is
a constant then we can say the pressure drop
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is proportional to r to the power four for
given flow rate now this has very interesting
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implications in blood flow we can show using
this relationship that delta q over q is proportional
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to four delta r over r that means a very small
change only one percent change in flow rate
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will cause four percent change sorry one
percent change in the radius will cause four
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percent change in the flow rate
similarly if the flow rate is constant then
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very small change in the radius will cause
significant change in the pressure drop so
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this is for given pressure drop whereas this
is for given of a constant flow rate so this
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is an effective way in the cardiovascular
system to change the blood pressure so the
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hypertension is generally caused by narrowing
of the blood vessels blood vessels become
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narrow and the pressure increases significantly
similarly the muscle tension or the pressure
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hypertension can be reduced by smoothing
the muscles ok so hagen poiseuille flow has
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a number of interesting implications in the
cardiovascular system now having looked at
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the flow in a of flow of a newtonian fluid
in a circular channel now let us consider
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the case of a non newtonian fluid so a we
will consider the flow of a casson fluid now
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why casson fluid because the first and foremost
reason is because blood is a non newtonian
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fluid which follows the casson model or
which has shown that casson model can be
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used to model the flow of blood accurately
ok the another reason is because as you can
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see in this relationship casson model includes
a yield stress behaviour as well as a power
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law kind of behaviour so by understanding
how we can model flow of a casson fluid we
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can also use the analysis or some parts of
the analysis to model of the fluid which show
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yield stress behaviour for example bingham
plastic fluid or we can model the flow of
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a power law fluid using the similar methodology
as we will use for casson fluid so the fluid
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shows a yield stress and as we have seen in
the previous slide that for all the fluids
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tau is equal to pressure gradient into r by
two that means tau is proportional to the
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shear stress
so if i plot this on a x y graph let us
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say on the y axis i show radius as shown in
this figure and tau so from this relationship
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tau will be zero at the centre and then it
will increase and become maximum at the wall
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so let us call this value as tau wall ok
now yield stress means that below yield stress
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there will be no flow or the shear rate will
be zero if there is no flow so if because
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the as we can see from this plot that the
shear stress is maximum at the wall so if
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the value of the yield stress is more than
the wall shear stress then there will be no
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flow in the channel even if some shear stress
less than the yield stress is being applied
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so now if the yield stress is more than stress
on the wall so let us draw this plot again
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where we plotted yield the stress verses
r curve if tau y which is yield stress is
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less than tau wall then it will fall somewhere
in between let us say that at tau is equal
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to tau y at this point where the shear stress
is tau y let us say the radius we call it
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r c or the core radius why do we call it core
we will be looking at it in a minute if
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the radius that means that below r c or when
the radius is less than r c in that region
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the stress is less than so we can have two
regions this is core and in this region tau
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is less than tau y whereas in this region
tau is greater than tau y
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so when the tau is greater than tau y that
means when the radius is between r c and r
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then fluid will follow root tau is equal to
root tau y plus root m gamma dot m gam[ma]-
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where gamma dot is the shear rate now if the
radius is less than r then the shear rate
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will be zero however that does not mean that
there will be no flow at the wall so let us
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draw this picture of the channel again and
this is r c so we will have a velocity profile
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something like this in the region where r
is greater than r c now at this point at the
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interface or at r is equal to r c from this
equation we will be able to find out what
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is the velocity and the velocity in this region
in the core region will be the same so the
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fluid will have two kind of regions one is
the core region where the fluid move like
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a rigid body move like a plug or a piston
and there will be gradients in the near wall
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annular region so we can analyse the flow
field for a casson fluid by considering these
28:58.169 --> 29:07.990
two regions separately a similar approach
is required while considering the flow of
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any non newtonian fluid which shows the yield
stress behaviour for example a bingham plastic
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fluid or a herschel burkley fluid ok so let
us consider the region where r is greater
29:26.530 --> 29:35.310
than r c it will follow this law so if we
rearrange this we will obtain this relationship
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that is work out the step in between so
we will have root m gamma dot is equal to
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root of tau minus root of tau y and if we
square on both the sides then we will get
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m gamma dot is equal to tau plus tau y minus
two root tau tau y ok
30:16.410 --> 30:27.710
so we have obtain a relationship between the
shear rate and the shear stress we have shear
30:27.710 --> 30:35.890
stress in two terms on the right hand side
and one is the shear stress only in the
30:35.890 --> 30:45.120
other term has root of tau there ok so now
because our objective is to find out the velocity
30:45.120 --> 30:55.110
profile so we will substitute the shear stress
as a function of radius which also include
30:55.110 --> 31:04.180
d p by d z in for all these cases d p by d
z is a constant so the pressure is we are
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varying linearly or the pressure gradient
is a constant so if we substitute these here
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what we obtain is this we can substitute that
shear rate is equal to minus del v z over
31:20.059 --> 31:28.420
del r is equal to one over m as it is in place
of tau we substitute minus d p by d z r by
31:28.420 --> 31:35.970
two plus tau y we will not touch it as of
now two again we substitute the value of shear
31:35.970 --> 31:50.370
stress as minus d p by d z r by two
so let us now integrate this equation because
31:50.370 --> 32:02.140
the right hand side have only r or constants
because yield stress and the pressure gradient
32:02.140 --> 32:11.000
both are constant in independent of the radius
so we can integrate this to obtain v z is
32:11.000 --> 32:34.070
equal to minus one over m within bracket minus
d p by d z r square by four plus tau y r minus
32:34.070 --> 32:59.270
two root of tau y minus d p by d z now one
over root two r to the power three by two
32:59.270 --> 33:14.100
divided by three by two so if we so if we
write this down this is what we get and
33:14.100 --> 33:22.230
then integration constant now we apply
a no slip boundary condition on the wall so
33:22.230 --> 33:30.070
that we can find out this integration constant
c so we can substitute and get the velocity
33:30.070 --> 33:41.280
profile so we can write that v z at r is equal
to capital r is equal to zero because of the
33:41.280 --> 33:47.100
no slip boundary condition and we substitute
r is equal to capital r in the equation so
33:47.100 --> 34:02.549
minus d p by d z capital r square by four
plus tau y capital r minus root of two tau
34:02.549 --> 34:17.409
y minus d p by d z capital r to the power
three by two plus integral constant so if
34:17.409 --> 34:26.520
we say this equation as a and this equation
as b or we can do to eliminate c is a minus
34:26.520 --> 34:35.019
b if we do that v z minus zero is v z minus
one over m and now we can subtract the terms
34:35.019 --> 34:42.909
one by one so minus d p by d z minus d p by
d z multiplied by r square minus capital r
34:42.909 --> 34:55.259
square by four plus tau y r minus capital
r minus this all root two tau y minus d p
34:55.259 --> 35:05.759
by g d z r power three by two what we have
missed here is this is divided by three by
35:05.759 --> 35:14.079
two here so this will be two by three and
this all inside
35:14.079 --> 35:25.559
now we look at this further this equations
so v z is equal to the same equation and now
35:25.559 --> 35:38.859
we can substitute this in terms of r c
tau y is minus d p by d z r c by two so we
35:38.859 --> 35:46.229
substitute and obtain this in terms of r c
the radius of the core so we will get velocity
35:46.229 --> 36:07.420
profile v z is equal to minus one over m minus
d p by d z and this will be r square minus
36:07.420 --> 36:22.349
capital r square by four plus minus d p by
d z into r c by two multiplied by r minus
36:22.349 --> 36:40.170
capital r minus two by three root two minus
d p by d z we have two d p by d z so its square
36:40.170 --> 36:54.119
into r c by two here and this is multiplied
by r power three by two minus capital r power
36:54.119 --> 37:05.690
three by two so if we take one by four minus
d p by d z out from this bracket minus one
37:05.690 --> 37:13.400
by four m minus d p by d z what we are left
with is minus r square minus capital r square
37:13.400 --> 37:19.430
here and this will be because this is force
so we can multiply by two and two here so
37:19.430 --> 37:31.019
that will be two r c r minus capital r and
again if we multiply this by this two will
37:31.019 --> 37:39.319
cancel out and you may multiply this by four
by four so we have eight by three and this
37:39.319 --> 37:45.799
four will go out and d p by d z has gone out
already what we are left inside the bracket
37:45.799 --> 37:51.249
is root r c r to the power three two minus
capital r by three by two so that is a velocity
37:51.249 --> 38:02.700
profile in terms of the m constant d p by
d z and r c and r
38:02.700 --> 38:14.650
so having obtained this velocity profile now
remember that the velocities that we obtained
38:14.650 --> 38:22.589
r is a general velocity profile and this is
only in the annular region so if you want
38:22.589 --> 38:29.299
to obtain or if you want to obtain a flow
rate we need to obtain the velocity in the
38:29.299 --> 38:39.259
entire region so at this point where r is
equal to r c by continuity of velocity we
38:39.259 --> 38:45.180
can obtain the velocity of the solid core
if we substitute in the velocity profile obtained
38:45.180 --> 38:54.339
r is equal to r c that was the primary region
why we were interested to substitute tau y
38:54.339 --> 39:01.650
in terms of r c because we will know tau y
anyway but we have substituted this in terms
39:01.650 --> 39:09.559
of r c so that this equation we can also find
out from this equation we also find out the
39:09.559 --> 39:19.140
velocity of the core so after substituting
r is equal to r c small r is equal to r
39:19.140 --> 39:30.319
c that is the velocity profile ok now to
obtain flow rate we need to integrate it zero
39:30.319 --> 39:38.269
to r v z two pi r d r and because the velocity
profile is different in the core region and
39:38.269 --> 39:44.479
in the annular region so we can divide into
two parts first we can integrate zero to r
39:44.479 --> 39:56.049
c v z c two pi r d r and then we can integrate
r c two r sorry this is v z two pi r d r so
39:56.049 --> 40:02.650
after substitution we will get after the
integration and the substitution we will get
40:02.650 --> 40:09.670
this flow rate which is pi r to the power
four by eight m minus d p by d z into this
40:09.670 --> 40:19.959
entire sector where this factor the s i z
i in this factor is equal to two tau y y r
40:19.959 --> 40:37.339
divided by minus d p by d z this you might
notice this is a yield stress and this is
40:37.339 --> 40:51.349
the pressure gradient ok
so in this lecture what we have looked
40:51.349 --> 41:04.410
at is flow fully developed flow in a channel
for the laminar flow of a newtonian fluid
41:04.410 --> 41:13.029
and of a casson fluid in the casson fluid
there are two things that we should remember
41:13.029 --> 41:22.739
that if the there is an yield stress then
we need to think about that the fluid in the
41:22.739 --> 41:29.440
channel will have two region in one region
where the shear rate will be varying across
41:29.440 --> 41:38.739
the cross section whereas in the centre the
fluid will flow as a core now the another
41:38.739 --> 41:46.749
lesson for power law of fluids is that we
can rearrange the equation as gamma dot is
41:46.749 --> 41:58.160
equal to shear rate to the power alpha where
alpha can be any number and once we have rearrange
41:58.160 --> 42:10.479
this what we need to know is we can substitute
tau is equal to minus d p by d z r by two
42:10.479 --> 42:17.269
and when we substitute this for constant pressure
gradient we can get the velocity gradient
42:17.269 --> 46:11.390
in terms of radius and then we integrate it
to obtain the velocity itself ok
46:11.390 --> 46:21.109
and now our task is to find this constant
in the integration and this can be found by
46:21.109 --> 46:27.209
the use of a boundary condition the no slip
boundary condition at the channel wall which
46:27.209 --> 46:39.949
essentially say that when a liquid is in contact
with a solid wall then there is no slip between
46:39.949 --> 46:53.309
the fluid molecule that is in contact with
the wall and the wall so v fluid at the wall
46:53.309 --> 47:01.849
is equal to v wall whatever the velocity of
the solid wall will be the velocity of the
47:01.849 --> 47:07.200
fluid that is in contact with the wall and
in this case because the pipe is a stationary
47:07.200 --> 47:14.989
so this is zero so we will use this boundary
condition that at r is equal to capital r
47:14.989 --> 47:20.959
where r is the channel radius the velocity
v z is zero
47:20.959 --> 47:29.489
so if we substitute that then we will end
up with v z is equal to zero at one over mu
47:29.489 --> 47:41.440
d p by d z capital r square by four plus c
and if we subtract then we will get one over
47:41.440 --> 47:48.489
mu minus d p by d z capital r square minus
small r square so you might notice that if
47:48.489 --> 47:59.420
you substitute from this equation let
us say this is equation one and equation two
47:59.420 --> 48:10.329
and if you subtract two minus one you will
get this ok so what we have done here we have
48:10.329 --> 48:20.099
obtained the velocity profile for the newtonian
fluid in a channel now if i want to obtain
48:20.099 --> 48:32.199
this in terms of v maximum then i can
say that at v z at r is equal to zero that
48:32.199 --> 48:46.230
will be v maximum right so v maximum is equal
to one over mu one over four mu minus d p
48:46.230 --> 49:06.170
by d z so we can write v z is equal to v max
into capital r square minus small sorry this
49:06.170 --> 49:17.769
is r square so if we divide v by v max
and we will get this is equal to divided by
49:17.769 --> 49:29.119
r square now v z is equal to so what we will
get v z is equal to v max into one minus small
49:29.119 --> 49:41.039
r square by capital r square ok so this is
a parabolic velocity profile in a channel
49:41.039 --> 49:48.839
which is also known as hagen poiseuille flow
you can also calculate the flow rate and average
49:48.839 --> 49:57.529
velocity for this pipe flow now we will
look at the flow of a casson fluid for a casson
49:57.529 --> 50:03.640
fluid you might remember from the previous
lecture that the constitutive equation that
50:03.640 --> 50:11.729
is the relationship between shear stress and
a strain rate is given by this formula that
50:11.729 --> 50:19.209
is this square root of shear stress is equal
to square root of yield stress plus square
50:19.209 --> 50:27.939
root of m gamma dot where gamma dot is shear
rate and m is a consistency index
50:27.939 --> 50:40.729
so again for this fluid also the relationship
between tau and r is a still valid and this
50:40.729 --> 50:49.780
shows that tau is proportional to r by two
for this fluid casson fluid shows yield stress
50:49.780 --> 51:04.079
behaviour that means if the applied stress
tau is less than tau y that means flow is
51:04.079 --> 51:16.170
there is no gradient right so from this what
we can see or what we can observe that tau
51:16.170 --> 51:29.589
in a channel will be maximum at the wall at
r is equal to capital r if stress shear stress
51:29.589 --> 51:42.359
at the wall is less than the yield stress
that means there is no flow in channel however
51:42.359 --> 51:54.079
if shear stress is greater than the yield
stress then it will follow casson fluid flow
51:54.079 --> 52:09.020
or it will follow the casson fluid model so
from this equation it is clear that the shear
52:09.020 --> 52:24.670
stress will in a channel the shear stress
will be zero at the centre and it will vary
52:24.670 --> 52:38.670
to a value tau w here now somewhere in between
this will be equal to tau y let us say this
52:38.670 --> 53:04.660
radius is r c so that means that in the channel
the condition will be such that there is
53:04.660 --> 53:15.349
yield stress the stress is less than yield
stress in the centre so there will be no gradients
53:15.349 --> 53:26.869
or zero gradients there whereas near the wall
above r c the there will be gradients so the
53:26.869 --> 53:35.569
core this core region will flow as a rigid
body and the gradients will be present in
53:35.569 --> 53:49.630
the annular kind of region there so for
r c is rigid body like movement will be
53:49.630 --> 53:54.809
in the core and the casson equation will be
valid in this region ok
53:54.809 --> 54:01.489
let us now look at this equation so again
write down this casson equation and let us
54:01.489 --> 54:08.759
rearrange this so if we want to write down
m gamma dot this will be equal to root of
54:08.759 --> 54:24.549
tau minus root of tau y square that means
m gamma dot is equal to tau plus tau y minus
54:24.549 --> 54:37.989
two root tau tau y ok now we substitute
tau is equal to minus d p by d z r by two
54:37.989 --> 54:46.959
in this and what we will end up with with
this equation that shear rate minus d v z
54:46.959 --> 54:53.049
by d v r which is equal to is equal to one
over m tau is replaced by this value minus
54:53.049 --> 54:59.989
d p by d z r by two plus tau y which is yield
stress minus two root of tau y into again
54:59.989 --> 55:08.990
we have substituted tau here ok further
let us move on to take this equation and we
55:08.990 --> 55:19.529
integrate and get the velocity from this
so if we take minus sign on the right hand
55:19.529 --> 55:26.880
side it becomes v z is equal to minus one
over m the first term minus d p by d z r square
55:26.880 --> 55:35.880
by four the second term tau y is a constant
so tau y r minus two root of tau y minus d
55:35.880 --> 55:51.779
p by d z is a constant two by two is become
a root two here and there is r so this is
55:51.779 --> 56:04.450
r to the power three by two so that will be
divided by three by two ok
56:04.450 --> 56:10.749
so now we apply the there is again an integration
constant here so now we apply no slip boundary
56:10.749 --> 56:17.859
condition at the wall and we can eliminate
c so if we do that then we will end up with
56:17.859 --> 56:35.369
v z is equal to minus one over m minus d p
by d z r square so in this place now we are
56:35.369 --> 56:48.499
writing it for the walls so capital r square
by four plus tau y capital r minus two root
56:48.499 --> 57:09.269
two by three root of tau y minus d p by d
z r to the power three by two plus c and we
57:09.269 --> 57:24.989
subtract equation two from equation one
so this will be zero at r is equal to capital
57:24.989 --> 57:37.279
r so we get from one minus two v z is equal
to minus one over m minus d p by d z r square
57:37.279 --> 57:45.499
minus capital r square by four tau y r minus
capital r and this term sorry for this mistake
57:45.499 --> 57:53.309
so two two by three tau y minus d p by d z
r to the power three by two minus capital
57:53.309 --> 58:05.390
r to the power three by two and now bring
this here again and in this that is noticed
58:05.390 --> 58:18.929
that tau is proportional to or tau is equal
to minus d p by d z at r by two and if we
58:18.929 --> 58:30.160
do that at r is equal to r c where tau is
equal to tau y so we substitute this in this
58:30.160 --> 58:38.490
equation then we get this value that tau y
is equal to minus d p by d z r c by two and
58:38.490 --> 58:49.979
we get this velocity profile so after substituting
we will get v z is equal to minus one over
58:49.979 --> 59:02.910
m we can take minus d p by d z out of this
bracket
59:02.910 --> 59:12.890
four also can be taken out so this will be
small r square minus capital r square plus
59:12.890 --> 59:22.650
tau y we replace by minus d p by d z so minus
d p by d z has gone out there is a two there
59:22.650 --> 59:31.390
and we have taken four out so we will multiply
by two here two r c small r minus capital
59:31.390 --> 59:40.859
r minus tau y is equal to minus d p by d z
so minus d p by d z has gone out and within
59:40.859 --> 59:45.430
bracket what remains is r c two and two is
cancel out so