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advanced videos as we have studied the
Naviers stoke equation we have found
out the naviers stoke equation along X
Direction Y direction and Z direction so
now we have to consider a real life
scenario in which will apply navier
stoke equation and will calculate the
solution of the navier stoke equation
let us solve the exact solution of
navier stoke
let us write down the navier-stokes
equation now it says that minus of dou P
by 2 X now we'll just write down along X
Direction plus mu times mu mu times del
square U is equal is equal to Rho into D
u by DT now plus of course we will
always have the body force per unit mass
of on unit body so this will always be
there and you don't need to consider
this force by solving this Navia stoke
equation because that is not a part of
fluid mechanics which we will we are
considering right now so let us analyze
a real-life problems in which of fluid
is actually flowing in the positive
direction of x axis so it is flowing in
positive direction of x axis now over
your there is some velocity gradient
across Y Direction also due to the
viscosity of the fluid so now we are
only considering you velocity u in in
this entire part so U is a function of X
comma T as well as as well as it is a
function of Y because U is changing with
Y so XY and E so let us write down this
part so that means here V is equals to 0
and W is equals to 0 and secondly we are
considering over here this has an
infinite amount of length so this length
along Z Direction is infinite and the
change in U with respect to said over an
infinite length will be 0 so the change
in Z will be 0 there is only change of V
U along X as well as Y direction so
considering this part let us expand this
part and put put in those values in the
equation where you
we as where of where the W and dou u by
dou Z is equal to 0 so what do we get
over here that is a minus of dou P by
dou X plus mu into dou square u upon dou
X square plus dou square u upon a dou Y
square plus dou square W upon those said
square we are just expanding this part
is equal to let us expand the right hand
side also so that will be Rho times this
entire part so what do we get over here
is Rho into dou u by dou T plus T nu dou
u by dou X plus v dou u by dou y plus a
w dou u by dou Z so this entire part we
have over here so now over here U is
just a function of X of Y and T V is
equal to 0 w is equal to 0 if these two
things are equal to 0 then this part
will tend to 0 W part will tend to 0 and
we know that we have taken into
consideration dou W by dou Z is equal to
0 so if dou W by dou Z is equal to 0
this part will be 0 now if we substitute
in the continuity equation so what is
continuty equation dou u by dou X plus
dou V by dou y plus dou W by dou Z is
equal to 0 we already have these two
terms as 0 that means dou u by dou X is
also equal to 0 that means that means
this part will be also 0 then it is U is
just a function of Y and T which we can
see over here even even over here this
part is 0 so
you're the expert won't be into
consideration it will be just why and so
let us evaluate this what do we get over
here that is minus of dou P by dou X
plus mu times dou square e u upon a dou
Y square is equal to Rho into dou u by
dou T now if the flow is a steady flow
if the flow is a steady flow and we know
that do u by 2 T is equal to 0 that
means steady flow has a three conditions
it is related to velocity viscosity as
well as density but here viscosity
velocity will take into consideration so
that is dou u by dou T will be equals to
0 so this part also tends to 0 if it is
which flow that is a steady flow so what
do we get over here let us find this out
before that we'll just divide by density
because density is not equal to 0 so
that is minus of dou P by dou X into 1
upon density plus mu upon density is
equal to dou square u upon dou Y square
is equal to 0 then we put divided by
density that will put the spot as 0 let
us take this on the other side so we
will get this as keine matic viscosity
into dou square u upon dou Y square is
equals to 1 upon a Rho into dou P by dou
X so that is what they equation we get
we know that kinda matic viscosity is
the ratio of viscosity of the fluid to
its intensity so we have got an equation
of view in terms of this entire part now
let us understand this in badly now over
here this term dou P by dou X will be a
constant why it will be a constant
because this is the change in pressure
along change in pressure along X
direction so this change in pressure
we'll love the wolf fluid to flow from
one end to another end in the pipe or on
the entire horizontal surface so this
part should be constant then only then
only we can able to solve this equation
and over here
pressure pressure is just a function of
X Y Z is a function of X Y Z and Olga
and T also but over here if we
substitute the same thing if we
substitute the same thing this entire
thing along y direction we'll get that
along y direction and z direction we'll
get the three parts will be zero so let
us check this out so let us check this
out from the previous section so we have
seen this in the previous section where
this is the solution of navier-stokes
equation now in our condition we know
that V is equal to zero let us write
down over here that is V equals to 0 W
equals to 0 dou W by dou Z is equal to 0
dou u by dou X is also equals to zero so
all these conditions we have and for
steady flow dou P by dou T and dou W by
dou T is also equal to 0 so all these
conditions we have seen previously so in
this part when we are going to
substitute over here B is 0 entirely and
over your V is also 0 entirely body
pulls we do not consider W is 0 and this
is also 0 that means dou P by dou Y is
equal to 0 and 2 P by dou Z is also
equal to 0 so in this section when we
talk about in this section when we
talked about a pressure which is a
function of XYZ it is no longer a
function of Y because V is equal to 0 it
is no longer a function of Z because W
is equal to 0 so if it is no longer
function of Y & Z then we can write this
equation that is or nu into dou square u
upon dou Y square is equal to 1 upon Rho
DP by DX DP by DX we can replace the
partial derivative by a complete
derivative and this part will be
constant the pressure over the anti
pressure gradient rather is constant
over here now let us solve this equation
so let us multiply to the other side so
what do we get this as dou square god
dou square u upon dou square Y is equals
to 1 upon a Rho into u into dou P by dou
X so we have to integrate this with
respect to Y so if you write this this
is a new upon Rho Rho and Rho will go
away and we can write this equation dou
square y upon those firing upon dou Y
square is 1 upon mu into dou P by dou X
so the density has gone away so now let
us integrate this part with respect
respect to Y if we integrate this part
then what do we get over here doo doo
doo u by 2 x is equals to that is 1 upon
u dou P or not for dou P we have
replaced this P dou P by dou X by DP by
DX so let us write down this as DP by DX
into y plus a constant C again if we
differentiate with respect to Y naught
this is not with respect to X will
differentiate this with respect to Y so
this is so this is equals to we are
differentiating with respect to Y so
this will be equals to u will be equal
to 1 upon mu dou P by dou X into y
square by 2
cy c 1 y plus c 2 so this let us
consider this as C 1 so now we have got
that is U as a function of as a function
of what this is the exact solution of
navier-stokes equation if it is going in
the horizontal direction and all the
conditions also we have applied that is
V is equals to 0 W is equals to 0 dou W
by dou Z dou X is equal to 0 then dou u
by dou X is equal to 0 and dou u by dou
T is equal to 0 this is due to steady
flow this is due to the application of
continuity equation this is an
assumption that though it is infinitely
long along Z direction and this is into
Z and this is do you and the change in
velocity along x direction along the z
direction will be 0 me and W because it
is horizontal going in the horizontal
direction so in short let me write this
all over again you have got you just as
a function of Y that is equals to u is
equal to 1 upon mu dou P by dou X into y
square by 2 plus C 1 y plus C 2 so we
have got the equation of navier-stokes
equation so we have got over here u as a
function of Y it is not a function of X
as well as we have seen all the other
conditions so I think I hope you have
understood how to derive an exact
solution of kneeble stoke equation
considering all the following conditions
first it is a steady flow then we apply
continued equation then we put do u by
dou Z is equals to 0 considering it is
an infinitely length long infinitely
length or long flowing along y direction
and it does not have any gradient along
Z direction and E V and W is equal to 0
Michael
during all these conditions in the
differential equation you get the
solution that is U as a function of Y
but you can see over here it is it is a
quadratic equation which we have
evaluated over here so I hope you have
understood how to calculate the exact
solution of navier-stokes equation thank
you
