Welcome back to the video course on fluid
mechanics. Today we will start a new chapter
which is one of the most important in fluid
mechanics, Navier-Stokes Equations and Applications.
We have seen various aspects of fluid mechanics
is the last few lectures, starting from statics,
kinematics, dynamics, full flow and then we
discussed the dimension analysis and now in
this chapter we will concentrate more upon
the viscous flows.
As all of you know viscous flow is very important
since most of the real flows we have to consider
viscous flows and we have to consider the
viscous effect in the derivation of the basic
equations and then we have to get the solutions
according to by considering the viscous effects.
As you can see here in this slide, this is
flow over an airfoil, we can see that say,
since the fluid passing over the airfoil is
viscous then the flow pattern is totally different,
what we considered earlier like when we considered
the added fluid flow over the potential flow
then here if the fluid flow is viscous, the
effect is much more different. The way of
approach is we have to consider viscous flows;
we have to consider the shear stress, all
these we have to consider and then we have
to consider the applications accordingly.
So the objectives of this section of this
chapter on Navier-Stokes equations and applications,
first we will see the viscous flows.
The importance of viscous flow and then how
generally we will be trying to solve the viscous
flows; and then we go to the Navier-Stokes
equations, we will derive the Navier-Stokes
equations and we will discuss briefly the
applications of this Navier-Stokes equations;
and then we will discuss the exact solution
for some of the simple cases; and also we
will be discussing the Plane Poiseuille flow;
flow like Couette flow; stokes flow; and Porous
media flow.
So the objectives of this chapter includes
say the deliberation on the Navier-Stokes
equations, its derivations and the applications;
and exact solutions available and further
studies on this.
Now as I mentioned, viscous flow most of the
real fluid flows are viscous, so that we are
to consider the shear stress. So the dynamics
of fluid flow which we considered earlier
we have derived the Euler equation and the
Bernoulliís equations, there we neglected
the effect of viscosity.
So the derivations of the equations are much
simpler and the solutions we can easily get.
That is what we are seen as far as the Eulerís
equations and the Bernoulliís equations are
considered. And even though these equations
are derived for the simple case of adil fluid
flow or potential flows but still we try to
use these equations with certain simplifications
for the real cases. That is what we have seen
in the lectures when we discussed the Euler
and Bernoulliís equations.
But we can see that, here either we are approximating
that the viscosity is negligible, so that
the shear stress acting on the fluid is not
much and then we derived the equations. But
that weight is an approximation and the accuracy
of the solution will be affected by if you
use Euler equation or Bernoulliís equations,
but when we deal with the real fluids the
viscosity is much more important and the shear
stresses play a major role.
Now when we consider this, the viscous flows,
for all most of the real flows which we deal
in all the aspects of life is actually the
viscosity effect is there, we have to consider
the viscosity.
So the real fluids exhibit viscous effects
and then most of the cases this stick to the
solid surface.
When we put it say in water or in say oil,
when we put the spoon and then take it out
you can see that some fluid stick to the solid
surface, so there is stresses within the body.
So we have to consider this effect. Here also
you can see that in volcano when the lava
is flowing, we can see that it is highly viscous
fluid is flowing and then we have to consider
the viscosity effect and then lot of parameters
change. When we consider the Newtonian fluid,
we can consider most of the time the stresses
vary linearly with respect to strain so we
can derive the Newtonís law of viscosity.
Newtonís law viscosity is very much used
in all the derivations of the viscous flows.
And then of course when we consider say when
a fluid is contained in a container or in
pipe or whatever it is, then the fluid will
be sticking to the solid surface so that we
have to consider the nocil condition. So we
can see that if you consider here say the
fluid is, if you consider the pipe flow, as
far as the pipe flow is concerned, say the
fluid it is coming like this then we can see
that here the pipe is solid surface and then
so here the velocity of this 0 on this surface
and on this surface. So we can see that the
velocity variation is like this, so this is
the velocity. Here due to the mostly condition
on the boundary, so here the pipe flow is
concerned this side and this side or throughout
the periphery of the pipe we can see that
the velocity is 0 and the velocity be maximum.
So viscous through it concerned we have to
consider this nosilic condition as far as
the with respect to when it is contained in
the in a container or in a pipe as we have
seen.
So viscous effect is very important and the
corresponding shear stress we have to consider.
Viscous flows the shear stress are taken into
account and then generally as we have seen
in most of the other cases we consider Newtonís
second law so that we can write the mass of
the particle multiplied by the acceleration
of particle that will be equal to, we can
equate the adequate sum of the forces.
So here if you consider any particular case
for viscous flow is concerned, the various
forces will be net pressure force on particle,
net gravity force on the particle, net stress
force on particle. So if you consider the
viscous flows and then using the Newtonís
second law with respective to a particle we
can write, the mass of the particle multiplied
by the acceleration of the particle that will
be equal to net pressure force on particle
plus net gravity force on particle plus net
stress force on particle. So this stress force
can be shear stress or normal stress for the
pressure force depending upon the case.
This generally for simple cases we consider
like this the Newtonís second law for a particle
and then we try to drive remaining equation
for that particular problem. It is difficult
to drive for each individual case like this
so we can drive a generalized equation. These
generalized equations are called Cauchyís
equations and the Navier-Stokes Equations.
So before discussing the generalized equation
for viscous flows, we will see a particular
case how we are deriving Gavinian equation
with respect to simple case by considering
the Newtonís second law. So let us consider
the flow between two parallel plates. This
earlier also we have discussed but with respect
to the viscous flows now what we have considered
earlier in the laminar flow, now again here
we are discussing when we deal with viscous
flow how we can apply the Newtonís second
law and directly derive the relationship for
velocity for various other parameters.
So here in this slide we can see that we consider
the flow of viscous between two very large
parallel flat plates placed at a distance
2y apart. So we can see here these two plates
this plates at 2y at a distance of 2y between
them and then the flow is taking place from
left to right like this so this is a inlet
of a flow here is the outlet of the flow.
So we consider the axis at the centre of the
of the plate between the plate, so here it
is origin is here X direction is here Y direction
is here and then acceleration due to gravity
we have to be consider.
So the elevation datum is here, as we have
already seen in this case also due to the
nocilic condition the velocity variation will
be like this, the parabolic variations to
maximum here at the center line and minimum
of 0 on the plates. So here what we are trying
to do? Here we consider the viscous flow;
we derive the expression for velocity for
this particular case for laminar flow. Before
considering before deriving the equations
we will be using number of assumptions so
that we can use the Newtonís second law.
Here the assumptions used are the flow is
steady and fluid is incompressible where as
the time dependent variables you can neglect
since flow is steady and the density variation
is not there since flow is fluid is incompressible.
Second assumption is fluid has viscosity it
will not slip at the surface of the plates
that means we consider the no-slip condition.
And as we have seen for this particular case
velocity distribution is parabolic.
In this case the flow is two dimensional.
So we can use the continuity equation which
we have derived earlier in the differential
form, in the continuity equation u and v are
the velocity components in X and Y direction,
for this case we can write del u by del x
plus del u by del y is equal to 0 as in equation
number 1. Due to the special nature of this
particular problem, we can say that streamlines
are straight and parallel to the X direction
as you can see in this figure. Now here this
particular problem is concerned, this velocity
v is equal to 0, so that del v by del x is
equal to 0 and del v by del y is equal to
0. So by using this condition here in the
continuity equation we can write del v by
del x is also equal to 0. That means this
is a parallel flow which we consider here
with respect to this plate so flow between
the parallel plates, so v component is 0,
so we consider only the velocity in the X
direction u. So now our aim is to derive the
Gavinian equation for velocity or other parameters
for this particular problem. As we were using
earlier we can consider a small fluid mass
and then we will consider what are the forces
acting on that and then we will be applying
the Newtonís second law.
Here since velocity does not change with X
velocity field so we got v is equal to 0,
that means we can see here the velocity is
varying with respect to y only, so effectively
this case is a one dimensional flow. Now we
will consider the flow with the velocity profile
unchanging as the fluid moves downstream.
This we can say this particular case the flow
is fully developed flow and then we consider
a small flow particle of size delta X by delta
Y like this and then what are the various
forces active on this particle or this small
volume of the fluid?
So here this particular case is concerned,
we can see that the various forces acting
here are mainly the pressure force, so there
is a pressure variations that is why the flow
is from left to right and then since we consider
the viscous flow, so the shear stress we have
to consider here. Here if you consider the
pressure flow on this plates of this fluids
mass, so here it p is the pressure then this
face we can write p minus del p by del s into
delta x by 2, and on this face this p plus
del p by del s into delta x by 2, and on this
face it is p plus del p by del y into delta
y by 2, and the corresponding other side it
is p minus del p by del y into delta y by
2. So these are the pressure forces acting
on this fluid mass.
And then shear forces are due to the shear
to force we can write correspondingly, we
can write on this face it will be tow the
shear stress tow minus del tow by del x into
delta x by 2 and correspondingly, this side
we can see it is tow plus del tow. So this
is on this face actually here tow minus delta
by delta x delta x by 2, this face and this
face it is tow plus delta by del x into delta
x by 2 and similar way on this face the shear
stress is tow minus delta by del y into delta
y by 2 and here tow plus delta by del y into
delta y by 2.
So for this particular case the forces acting
are the pressure forces on the fluid mass
for the fluid the say delta y by delta is
the mass of fluid which is consider and then
the shear forces for the corresponding shear
stresses on the size of the fluid mass which
we consider. So now here our aim is to drive
the expression for the velocity variation.
Here we are using the Newtonís second law.
In x direction if we consider, so here the
mass into acceleration, so delta Mp if that
is the mass for this particular fluid element
which we consider so delta Mp into ax, ax
is the acceleration in the x direction, so
that is equal to delta Fx,pressure the pressure
component the x direction plus delta Fx,gravity
the gravity force components and delta Fx,shear
the shear force component.
So from the figure the earlier figure so here
we can see that, the pressure component in
x direction we can get us delta Fx,pressure
is equal to minus del p by del x into delta
x into delta y and similarly the gravity force
is concerned so here the shear force as well
as the pressure force and then of course the
gravity force also we have to consider for
this problem. So gravity force is concerned
delta Fx,gravity we can write test rho into
gx into delta x into delta y where gx is the
acceleration due to gravity in the x direction
and then the shear force is concerned we can
write this as in the x direction, we can write
as the difference between tow plus del tow
by del y into delta y by 2 into delta x minus
tow minus delta y by del y in delta y by 2
into delta x, so that we can write as delta
y delta by into delta x into delta y. So this
gives the shear force.
And then the acceleration is concerned this
as far as the fluid element this concerned
acceleration x is equal to del u by del t
plus u into del u by del x plus v into del
u by del y, so here since we consider as steady
state flow del u by del t is 0 and this del
u by del x we are already seen for this parallel
flow it is 0 and then v component v is 0.
So you can see that finally the acceleration
is equal to 0.
Now if you substitute in these equations since
the left hand side here mass into acceleration,
acceleration is already 0, so we can see this
becomes 0 and then we can just equate the
other by some of the forcers, so now the pressure
force we can write as del p by del x into
delta x into delta y; and gravity force rho
into gx into delta x into delta y plus the
shear force delta by del y into delta x by
delta y. So we can add all this and then if
you use the Newtonís second law we can write
minus del p by del x plus rho gx plus delta
by del y is equal to 0. So this is coming
from the Newtonís second law.
Now here we introduce, we assume the z coordinates
pointing opposite to gravity vector so that
we can just approximate this rho gx in this
fashion. Here we can write equation number
2 as del p by del x minus rho g this x direction
with respect to z we can write us del z by
del x so minus rho g del z by del x plus del
tow by del y is equal to 0. This equation
2 is transformed this way and then here for
constant p and g, so the pressure and acceleration
due to gravity we can write equation number
2 as minus del by del x p plus gamma z plus
delta y del y is equal to 0 as in equation
number 3.
So in a very similar way in by direction we
can write us minus del by del y p plus gamma
z plus delta y by del z is equal to 0.
Now here we get an expression with respect
to the shear stress, so we can use Newtonís
law viscosity to convert to get in terms of
velocity. So here now the velocity profile
is not changing in x direction as we are seen,
so hence del tow by del x is equal to 0 that
means in y direction is concerned, this equation
becomes del by del y of p plus gamma z is
equal to 0. So pressure distribution is from
this equation number 4, we get the pressure
distribution is hydrostatic. So now as far
as equation 3 is concerned this we can write
as so now this pressure is varying only with
respect to x so we can write minus d by dx
of p plus gamma z and then plus del tow by
del y is equal to 0 as in equation number
5.
So here if you are assuming the flow is laminar
that we can write the rate of shear deformation
of plate particle, tow is approximately equal
to rate of shear deformation of fluid particle
that means the shear stress is approximately
equal to rate of shear deformations, so that
this phi dash that is equal to half del u
by del x plus del v by del x so here we can
see that this becomes del u by del x is 0,
so that this become phi dash is equal to half
del u by del y, so that we can write tow is
equal to mu du by dy which is also coming
from the Newtonís law of viscosity.
So now directly we can write this from the
Newtonís law of viscosity and substitute
here so that we will get from this equation
if you substitute equation number 5, if you
substitute for the Newtonís law of viscosity
tow is equal to mu du by dy finally we get
minus d by dx plus, p plus gamma z plus mu
del square u by del y square is equal to 0.
So now finally we get the velocity variation
as a function of y so that we can write mu
d square u by dy square is equal to d by dx
of p plus gamma z. So here in this particular
case if you write p tilta is equal to d by
dx of p plus gamma z, so that we can write
d square u by dy square is equal to p tilta
by mu, as in equation number 7.
So we finally get an expression for velocity
in terms of pressure, so we can integrate
twice this expression d square u by dy square,
so that we will get a velocity u is equal
to p tilta by 2 mu y square plus C1 y plus
C2. So integrate twice we get this expression.
So now for this particular case we have a
boundary condition so the flow is between
two parallel plates, on the top as well as
bottom where the fluid touches the plates,
there we can assume the velocity is 0, so
that we will get the boundary condition to
boundary condition. So the boundary conditions
here are y is equal to plus or minus y u is
equal to 0, so we get C1 is equal to 0 and
finally we get C2 is equal to minus p delta
y square by 2 mu.
So finally we get an expression mu is equal
to minus p delta y square by 2 mu into 1 minus
y by y whole square. So here now with respect
to this figure we can see so here we get a
velocity variation is parabolic as now we
have derived. So we have applied the boundary
conditions I am using the Newtonís law of
viscosity to get an expression for these velocity
variations, so that is what we did here.
So here in this particular case we consider
the origin as on the central line of the between
the plates, so we get an expression for u
as minus p tilta y square by 2 mu into 1 minus
small y by y, so this y gives the variation.
So whether between the plates, so the y is
starting from the central line between the
plates. So this is the expression for the
velocity. Now for this particular case we
know that the velocity is maximum at the center
so y is equal to 0 we can get the maximum
velocity umax is equal to minus p tilta y
square by 2 mu.
So this is the maximum velocity and then if
you want to find out the average velocity
V bar is equal to 1 by 2 Y integral minus
Y to Y within the limit u dy that is equal
to we integrate we can write minus 2 by 3
p tilta Y square by 2 mu, so that we can show
that this will be an average velocity will
be two third of the maximum velocity. So now
in this expression we can put back with respect
to the pressure variations so p tilta we can
write as with respect to d by dx of p plus
gamma z so that is equal to u is equal to
minus y square by 2 mu d of p plus gamma z
by dx into 1 minus y by whole square. So this
is the final expression and umax it can be
written as minus Y square by 2 mu of d p plus
gamma z by dx and b bar is two third umax
as we have already shown that is equal to
minus y square 3 mu d of p plus gamma z by
dx.
So like this we can derive various parameters
so here the aim of this deliberation here
is for viscous flow we consider the viscous
effect that means the shear force and the
pressure force and also the gravity force
are considered and for this simple case of
flow between parallel plates which is one
dimensional to parallel flow there we have
derived the equation for velocity variation.
So by here we use the Newtonís second law
to derive the equation. In a very similar
way for various simple cases as demonstrated
earlier in the case when we discuss the laminar
flow problems we can derive various expressions
for simple simplified cases. Now in the very
similar way our aim here is to derive the
fundamental equations for the viscous flow.
Here what we are going to discuss further
is we are deriving the fundamental equations
which we called Navier-Stokes Equations or
the first form of the equation Cauchyís equation
and then say later Navier and Strokes there
modified this equations and called as Navier-Stokes
Equations. So our aim here is to derive the
equations which are generally applicable to
any kinds of problems, applicable to most
of the fluid flow problems.
First, we will derive Cauchyís equation the
Cauchyís equations are applicable for all
kinds of problems compressible and incompressible
or work varieties of fluid flow problems and
then we will discuss especially we consider
the incompressible flow, from the Cauchyís
equations we will derive the Navier-Stokes
Equation. Here as I mentioned as we have already
discussed for the flow between parallel plates
here we will use the Newtonís second law
and the Newtonís law of viscosity by using
the differential approach we will derive the
basic equations.
So first we will derive the Cauchyís equations
and then based upon the Cauchyís equation
we will derive the Navier-Stokes Equations.
So here the fluid flow problems which we generally
consider are three dimensional in nature,
but here to derive the equations we are using
two dimensional flow so that it will be much
easier to understand. The steps involve for
three dimensional flow equations are also
same but here we explain as far as 2-D flow,
two dimensional flow is concerned and then
we will discuss the three dimension equations
but the derivation is for two dimensional
flow.
Now let us consider here we derive the Cauchyís
equation, so let us consider the element within
the fluid as shown here. Here a fluid element
of size delta X delta Y and delta Z we are
considering. The various forces as we are
seen earlier the forces are the pressure force.
If you consider this fluid element delta Fx,pressure
and delta Fx,gravity the gravity force and
delta Fx,stress forces, the stress can be
normal stress or shear stresses.
Now either this face or the with respect to
the X direction or Y direction we have to
consider the pressure force gravity force
and the stress force. If we consider two dimensional
flow with constant velocity if we use Newtonís
second law of motion as we have as discussed
earlier, here for this flow problem we can
write the Newtonís second law mass into acceleration
is equal to the average of sum of the forces.
So here for this fluid element which we consider
the mass is delta Mp and then if acceleration
x direction is ax then we can write delta
Mp into ax is equal to delta Fx,pressure plus
delta Fx,gravity plus Fx,stress. It is written
in this equation number 1.
So in x direction delta Mp ax is equal to
delta Fx,pressure plus delta Fx,gravity plus
delta Fx,stress So in the similar way in y
direction we can write delta Mp into ay acceleration
into y direction that is equal to delta Fy,pressure
delta Fy,gravity plus delta Fx,stress. So
now as far as the mass of the fluid is concerned,
fluid element is concerned, if rho is the
density and here we consider fluid element
as size delta X and delta Y and delta Z size.
So mass is equal to delta Mp is equal to rho
into delta X into delta Y into delta Z and
then acceleration is concerned the local acceleration
and convergent acceleration we consider. So
in two dimensional problem which we consider
here ax is equal to del u by del t plus u
into del u by del x plus v into del u by del
y and a by is equal to del u by del t plus
u into del u by del x plus v into del v by
del y. And then the pressure force here for
this typical problem here we can write the
pressure force here it is explained further
here this fluid element is delta X by delta
Y.
The various pressure and stresses acting on
a rectangular fluid particle are written here,
so here the pressure force is in this direction
p minus del p by del x into delta X by 2,
and the other direction p plus del p by del
x into delta X by 2 and here this direction
the pressure force is p plus del p by del
y into delta y by 2 and here this direction
pressure forces p minus del p by del y into
delta y by 2.
And then if you consider the normal stress
force, so here this face it is sigma x minus
del sigma x by del x into delta x by 2 and
here this phase it is sigma plus del sigma
x into delta x so delta x by 2 and on this
face sigma y minus del sigma y by del x into
del y by 2 and on this face sigma y plus del
sigma y by del x into del y by 2 and similar
way these shear stress components, so here
in this direction it is towyx plus del towyx
by del y into delta y by 2 and on this face
towyx minus delta yx by del y into delta y
by 2 and on this face it is towxy minus del
towxy by del y delta x by 2 and on this face
towx by plus del towxy by del y into delta
x by 2. These are the pressure and stresses
acting on the rectangular fluid element.
Now if we consider the net pressure force
we can write here del Fx,pressure is equal
to del p by minus del p by del x so delta
x into delta y into delta z we consider the
fluid element thickness have delta z and delta
fy in the y direction pressure force is concerned
minus del p by del y into delta x into del
y del z and similarly the gravity force on
the fluid element del Fx,gravity is rho gx
delta x delta y delta z and del Fy,gravity
is rho gy delta x delta y delta z.
So here now the stresses are the normal stresses
and the shear stresses. So depending upon
the cases for the fluid which we considered
definitely shear stresses are there, normal
stress whether depending upon the problem
we consider. So now all these stresses are
shown here. By considering the forces on the
particle other than the pressure, so del Fx,stress
with respect to the normal stress and shear
stress we can write as with respect to the
fluid element here we can write, del sigma
x by del x plus del towyx del y del y into
delta x delta y delta z.
So that is the net stress force in the x direction
and similarly the net stress force in y direction
can make as del sigma y by del y plus del
towxy by del x into delta x delta y delta
z. So now whole the force components the stress
forces regarding normal stresses and the shear
stresses and the pressure forces and the gravity
forces are known and we have already seen
the mass of the fluid element and the acceleration.
So now we can use Newtonís second law to
substitute all these components to get the
final form of the equations.
Now substituting the mass and acceleration
and all of the forces into equation 1 and
2 that means in two dimensions equations we
are writing the equations for two dimension
cases, so equation 1 and 2 and then shrinking
our particle to the point x y here, this figure
shows we consider this equation 1 and 2 and
then we consist particular point in the differential
approach so we can get rho into del u by del
t plus u into del u by del x plus v into del
u by del y is equal to minus del p by del
x plus rho gx plus del sigma x by del x plus
del towx by del y equation number 3. So this
is the final form of the equation and then
in y direction we get rho into del v by del
t plus u into del v by del x plus v into del
v by del y is equal to minus del p by del
y plus rho gy plus del sigma y by del y plus
del towx by del x equation number 4.
So this equation number 3 and 4 are the basic
equations as far as viscous flows are concerned
and these equations are called Cauchyís equations.
So these equations are two dimensional equations,
we got two equations from the Newtonís second
law of motion. So we got two equations. So
with these two equations in two dimensional
case we can supplement with respect to the
continuity equation, the continuity equation
gives del u by del x plus del u by del y is
equal to 0.
Here for two dimensional problem if you inspire
if you check this equation 3, 4 and the continuity
equation we can we have for 2-D problem we
have three equations but we can see we have
seven unknowns. The unknowns here, unknowns
can be u v the velocity component, p the pressure
component, t sigma x and sigma y the normal
stress component and towxy, towyx and towzx
the shear stress component.
So this is a case for two dimensional problem.
So for three dimensional problem in the very
similar way we can write the equations so
there will be three equations the Cauchyís
equations three equations will be there, and
then we can supplement with respect to the
continuity equation for three dimensional
flow. So here the unknowns so if you consider
these three equations of motion derive using
the Newtonís second law and the continuity
equation we define four equations and then
we can see that there will be ten unknowns
u v w with component and the pressure p and
three stress component sigma x sigma y sigma
z and three shear stress components. So three
plus three plus three velocities so nine plus
one ten unknowns for general cases and here
we have four equations.
So depending upon the problem we can supplement
with various other equations like equations
of state and then we can try to solve the
problem. So this Cauchyís equation which
we derived here are the basis equations for
viscous flows, so here this Cauchyís equations
represent a unique balance of body and surface
forces with inertial forces. So here the Cauchyís
equations which we derived here are valid
for compressible and incompressible fluid
flow.
The advantage is depending upon most of the
cases which we discussing this fluid mechanism
course will be incompressible fluid flow but
the Cauchyís equations we can also use for
compressible flow also by considering the
changing the density also and then here say
using the Cauchyís equations only we will
be deriving the Navier-Stokes Equation. So
the vector differential form we can write
this the Cauchyís equations which we derived
here in a vector form or vector differential
form. If you consider P as this capital P
as a stress component then we can write g
minus a into rho plus del dot P is equal to
0, where P is the stress forces; where g is
the acceleration; a is the body force component;
so g is the acceleration due to gravity; and
this P is the stress and this comes from the
various other components. So the vector differential
form of the Cauchyís equation can be written
like this.
As I mention here this equation the Cauchyís
equations are applicable to all fluid flow
problems either compressible incompressible
or any kind. So only thing is that depending
upon the problem we have to change certain
parameters and then we can derive the equations.
So here now for the incompressible fluid flow
problem we are going to use the Cauchyís
equations and then we would derive the Navier-Stokes
Equations. So now, we will see introduction
to Navier-Stokes Equations.
So Navier-Stokes Equations as I mentioned,
for the incompressible viscous flows or even
the viscous flows these are the basic different
equations. So we have already seen the form
for Cauchyís equation and based upon the
Cauchyís equation, this Navier-Stokes Equations
are derived in 1827 by Navier on the basis
of inter-molecular forces and in 1845 in the
nineteen century by Stokes I consider the
normal and tangential stresses as linear functions
of the rate of deformations.
As discussed these are the differential form
of the conservation of linear momentum applicable
in describing the motion of a fluid particle
in any space and time. Here the time component
this compressible viscous time is considered
and then for the variation of any parameters
either the velocity in xyz direction uvw or
the pressure we can utilize this equation.
So that is the beauty of the Navier-Stokes
Equations.
So here we derived the Navier-Stokes Equations
for the Cauchyís equation. We have already
seen there are four 2-D problems as we have
discussed here seven unknowns in times of
u v p sigma x sigma y towxy and towyx and
3-D problems ten unknowns. So what we will
do here for Cauchyís equation we try to represent
this stresses either normal stress or shear
stress in terms of say the velocity components
by utilizing the Newtonís law of viscosity
or the stresses and strain relationship. So
here we try to relate the stresses to velocity
field using the Newtonís law of viscosity
and viscous stresses are proportional to rates
of strain of fluid. So in Cauchyís equations
the shear stresses are proportional to shear
rate of strain.
So that we can write towxy is equal to towyx
is equal to mu del u by del y plus del v by
del x and also if we consider the normal viscous
stresses we can write or we can assume in
this particular case normal viscous stresses
are proportional to the stretching or volumetric
strain rate, so that we can write the normal
stress component in next direction sigma x
is equal to 2 mu del u by del x as in equation
number 7, and sigma y is equal to 2 mu del
v by del y as in equation number 8.
So let us assume for the case we consider
the incompressible fluid flow we assume the
viscosity as constant, so for this case so
this case, this term del sigma x by del x
plus del towyx by del y we can write by using
this equation 6, this equation 7 and 8 we
can write this is equal to mu into del square
u by del x square plus del square u by del
y square plus mu del by del x of del u by
del x plus del v by del y.
So by using the equation 6, 7, 8 these terms
the del sigma x by del x plus del towyx by
del y can be written in this form here the
order of differentiation interchanged for
this last term.
That is why this form is obtained, with respect
to x and y the order of differentiation is
interchanged and then this term is concerned
here we can see that it for two dimensional
problems is del u by del x plus del u by del
y for incompressible viscous flow this term
will be equal to 0. So that finally we get
this del sigma x by del x plus del towyx by
del y is equal to mu into del square u by
del x square plus del square u by del y square.
So by using continuity equation this term
becomes 0 and finally now we can use this
approximation the Cauchyís equation 1 here
rho del u by del t plus u into del u by del
x plus v del u by del y is equal to minus
del p by del x plus rho gx plus del sigma
x by del x plus del towyx by del y. So here
for these terms we substitute from the earlier
this approximation, so finally we get rho
del u by del t plus u into del u by del x
plus v into del u by del y is equal to minus
del p by del x plus rho gx plus mu into del
square u by del y square plus del square u
by del y square equation number 9.
So this is the basic equation of Navier-Stokes
Equation for two dimensions problems in x
direction. So similar way for equation 4 in
the y direction we can write rho into del
v by del t plus u into del v by del x plus
v into del v by del y is equal to minus del
p by del y plus rho gy plus mu del square
v by del x square plus del square v by del
y square as in equation number 10. So for
two dimensional problems this equation 9 and
10 which we derived based upon the Newtonís
second law, these equations are called the
Navier-Stokes Equations for incompressible
viscous flow.
So here we have demonstrated the case for
two dimensions. In a very similar way we can
derive for three dimensional problem and here
I have just written for three dimensional
problem the equations are rho in del u by
del t plus u into del u by del x plus v into
del u by del y plus w into del u by del z
is equal to minus del p by del x plus rho
gx plus mu into del square u by del x square
plus del square u by del y square plus del
square u by del z square this is the equation
in x direction for 3-D problem, three dimensional
problem by considering the velocity components
u v and w. So very similar way in y direction
we can write rho into del v by del t plus
u into del v by del x plus v into del v by
del y plus w into del v by del z is equal
to minus del p by del y plus rho gy plus mu
into del square v by del x square plus del
square v by del y square plus del square v
by del z square equation number 12.
So similar way is z direction rho into del
w by del t plus u into del w by del x plus
v into del w by del y plus w into del w by
del z is equal to minus del p by del z plus
rho gz plus mu into del square w by del x
square plus del square w by del y square plus
del square w by del z square.
So three equations in xyz direction and then
we can use the continuity equation 3-D del
u by del x plus del v by del y plus del w
by del z is equal to 0, so here we have got
four equations and then we have got four unknowns
u, v, w the velocity component and the pressure
component, if the density and the viscosity
are known for the problem. So these are the
Navier-Stokes Equation in 3-D and 2-D also
we have seen, so these are fundamental equations
for incompressible viscous fluid flow.
