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Hello Friends we have studied the exact
solution of navier stoke equation now we
have to apply that exact solution in
real-life condition so one of the
real-life conditions which we have the
idea of that is of fluid which is
flowing in a horizontal direction so we
have to apply the exact solution of
navier stoke equation on this real life
Real life flow
so let us see this is a real-life
situation where the flow is called as
couette flow model my spelling for quit
anyways so let us solve this part so in
which we have first initially got U as a
function of Y which was solution of
navier-stokes equation so U is equals to
u is equals to that is minus or that is
plus 1 upon 1 upon 2 mu DP by DX into y
square plus y times c1 plus c2 so now
what is upward flow quit flow is a flow
in which we have a horizontal surface
and the fluid is flowing over this
horizontal first surface and there is an
object which is kept on the top surface
of the fluid so we know that over the
your the fluid velocity at Y is equal to
0 u will be equals to 0 because this is
the initial layer which has maximum
viscosity when we consider this is no
slip condition in which the fluid does
not slip over the solid surface we
consider this has no slip condition due
to which it does not slip over the solid
surface and the velocity is zero as it
goes up a or B from this part it will
have go from it will go from minimum
velocity to maximum velocity so at Y is
equals to H this vertical height is at
the top most velocity is equal to
capital u let us consider this as
capital u so we have this differential
equation we have defined the boundary
condition for a simple quick flow we
have to just substitute the boundary
condition in the solution of the
differential equation so now let us
substitute this the first boundary
condition which is which is y is equals
to 0
you is equal to 0 where the hue is equal
to 1 upon 2 mu DP by DX into y square
plus y 1y into c1 plus c2 so if you
substitute this boundary condition what
do we get
the value of C 2 is equal to 0 this part
is 0 this part and this 0 and this part
is 0
all the three terms containing Y and you
are 0 so then your differential equation
becomes that is U is equal to 1 upon 2
mu DP by D X into y squared plus y times
c1 then let us apply the second boundary
condition that is when y is equal to H
you will be equals to u max so this is
the part where u is equal to u max or
maximum velocity so instead of 5 you
have to substitute the value of H so
capital u will be equals to 1 upon 2 mu
DP by D X and instead of by square
we should substitute H square plus h
times into c1 so let us calculate the
value of C 1 so C 1 will be given as
this is U minus 1 upon 2 mu dou P by dou
X into H square is equals to H times C 1
so this will be given as u upon H 1
minus 1 upon 2 nu dou P rather DP by DX
into H is equal to C 1 so we have got
value of C 1 so now our previous
equation was U is equal to 1 upon 2 mu
DP not dou P DP by DX into y square plus
value of C 1 we have to substitute so
this is
I into you upon H minus one upon two new
DP by D X into H into H now let us
evaluate this part so what do we get
over here that is U is equal so let us
take this cotton and is DP by DX 1 upon
2 new from these two terms will take
this as common so this is 1 upon 2 mu DP
by D X into into what we have over here
is y square minus H this is from this
term plus we have over here Y into u
upon H so this is the second term so
this is actually what is the solution of
this equation for a weight flow now if
we consider a simple equate flow simple
quit flow over here for simple quit flow
DP by DX is equal to 0 we do not
consider the effect of pressure gradient
so that time you will get this as u
equals 2 that is the new is is equals to
this part will be 0 over here this part
to DP by DX is e goes to 0 this entire
term will be 0 so that you will be
equals to small you will be equals to Y
times capital u upon H so now if we know
that the value of H is going to remain
constant that is the height of that
height of that entire and fluid above
the solid surface and maximum velocity
at the top surfaces also remain going to
remain constant so you get a Y as a
function however we get U as a function
of Y
and the relationship is linear so that
means what how this is going to change
in the fluid flow so over here Y is
equal to zero at the bottom of surface U
is equal to zero Y is equal to H u will
be equal to capital H during this part
the relation for a simple quit flow is
going to be a linear or linear term and
over here the velocity is linearly
increasing so this is how it will
increase but for oh just ah quit flow
here also Y is equal to 0 U is equal to
0 and Y is equal to H u will be equal to
4 this is capital u and this is capital
u this will be somewhat like this
because over here because in this part
we can see that it is a function of my
sphere hence it is going to rule of like
this so it is going to increase in oh it
is going to increase in a quadratic
curve and this is a linear curve so this
is a quit flow and this is a simple
Kuwait
No I hope you have understood we have
how to derive awkward flow and a simple
quit flow considering the boundary
condition and substituting the boundary
condition in the solution of
navier-stokes equation thank you
