Welcome back students. So, in the last lecture
we started with a problem and then I went
back and showed you what different values
of A B and C are. So, solve this problem we
will use the whiteboard.
So, the best way is always to solution to
given first we write what the things are given.
So equation given 1 - M squared phi xx + phi
yy. So drawing the analogy from the general
equation, w write A = 1 - M squared, second,
B is 0 and C is 1. So the first part, we say
when m is less than 1, we have to calculate
the value of b square – 4AC, so B is 0,
so 0 - 4 is 1 - M squared into 1. So - 4 +
4 m squared are 4 M squared – 4.
We have been given M is less than 1 implies
M squared will also be less than 1 which implies
4 M squared is less than 4 so if you bring
4 on this side, it will become 4 M squared
- 4 is less than 0. Therefore, this value
4 M squared - 4 are B squared – 4AC will
be less than 0 which means it is and elliptical
PDE so in the 
second part, when m is greater than 1.
We have found out does not matter what the
value but b squared - 4 AC came out to be
4M Squared - 4, right. So when M is greater
than 1, this means 4 M squared - 4 is going
to be greater than 0, which implies B squared
- 4 AC is greater than 0, this implies it
is hyperbolic PDE. Now the hard part is I
will rub this one here B squared - 4 AC when
M = 1. So B square - 4 AC came out to be 4
M squared - 4 right. So when M = 1 4 M squared
- 4 = 0. This means it is parabolic partial
differential equation.
So, this is how we have solved the problem.
So I will just write the solution here. So,
when M is greater than 1 it was elliptical
PDE M greater than 1 it was hyperbolic 
And for M = 1 it was parabolic PDE. So, time
to move onwards.
Will solve another problem. So, it says that
considers the function phi of x y is e to
the power x + e to the power y and we have
to consider the point x, y as 1, 1 we have
to calculate the exact values of del phi del
x and del phi del y at this point at point
1, 1. Secondly, use the first order of finite
difference with delta x = delta y = 0.1 to
calculate the approximate values of delta
phi by delta x and delta phi by delta y at
1, 1 and calculate the percentage difference
with the exact value. So, this is the application
of all the things that we had done in the
last lecture and this will be a good way to
understand.
So, solution number 3. So, what we have been,
we know that the function phi x, y has been
given as e to the power x + e to the power
y. So, for the first part, we should be able
to calculate del phi del x will be e to the
power x correct and del phi del y is going
to be e to the power y simple partial differential
so we have to calculate del phi del x at 1,
1, it is going to be e to the power 1 are
2.7183. Similarly, del phi by del y at 1,
1 e to the power y that is e to the power
1 2.7183.
So, that was a first question that we have
solved the values of del phi del x at 1, 1
del phi del y at 1, 1. In the second part
we have been given that delta x = delta y
= 0.1 this is what we have to assume let 1,
1 be denoted by i, j. And if we use first
order forward difference del phi del x at
i, j can be written as phi of i, j + 1, j
– phi i, j divided by delta x del phi del
x i, j can be written as phi of phi at what
.1 because delta x is .1, so 1 + delta x so
that is 1.1, 1 – phi at 1, 1 divided by
0.1.
This will be e to the power 1.1 + e to the
power 1 - e to the power 1 + e to the power
1 we just putting in the values and if you
use your calculator, it is going to be 2.8588.
That is del phi del x at 1, 1. So, we are
going to it is everything here and proceed.
So, similarly del phi by del y i, j will be
phi of i, j + 1 – phi of i, j divided by
delta y are the value of the function at 1,
1 .1 – phi at 1, 1 divided by delta y 0.1
are e to the power 1 + e to the power 1 +
1 - e to the power 1 + e to the power 1 divided
by 0.1. This is 2.8588. So, percentage difference
for both cases 
is 2.8588 - 2.7183 divided by 2.7183 and this
percentage difference come out to be if this
5.2%.
So, this is the answer. So, this was the simple
most question and the solution for demonstrating
the percentage difference and the concepts
that we have learned in the last lecture.
So, we proceed from this point now.
To we mentioned about other schemes of may
be machine that is the introduction to finite
volume method, what is finite volume method.
So, the finite volume is the sub domain method
with piecewise definition of field variable
in the neighborhood of chosen control volume,
so, the total solution domain in the Finite
volume is divided into smaller control volume.
So, the total solution to the total domain
is divided into smaller control volumes. Normally
they are rectangular in shape or quadrilateral
inch.
So, nodal points are used with these control
volumes for interpolating the field variable
and usually a single node at the center of
the control volume is used for each control
volume. So, this is the 1 single point in
that control volume that is used and this
is based on the cell balance equation. It
is solved by the integral approach which is
more accurate than the finite difference method
because it is free from truncation error.
So, one thing about the integral approach
is that it is free from the truncation errors.
Cell structure is shown in this figure here,
you see, this is the control volume the dashed
no zone and this at the center point you know,
we calculate the things simple is one method
which is called semi implicit method for pressure
linked equations. So, simple algorithm is
based on finite volume method this is just
one example of finite 
this was just something about the finite volume
method.
So, now going to the things that are very
much relevant to our hydraulic engineering
course that how do we model or do the numerical
modeling of the turbulent flows because most
of the flows in real nature at turbulent for
turbulence is a 3 dimensional unsteady phenomenon.
So, it is a function of time and it also function
of a space. So, a huge amount of information
is needed for the proper description of the
turbulent flows that is a very well established
fact.
Hence, an ideal turbulence method must be
able to capture the features of the turbulent
flow field at a minimum complexity level.
So, ideally the best of the numerical model
that can do this type of turbulent flow modeling
should be able to capture the features of
the turbulent flow with least complexity level.
So, the model should be least complex as well.
Turbulent flows can be modern numerically
to capture the temporal fluctuations the time
average features of the flow field. So, in
the real velocity you know i mean the fluctuations
mean the deviation from the average velocity
and the time average features of the flow
field. So, these because these 2 effect the
turbulence in terms of turbulent stresses
these 2 quantities, you remember we mentioned
about you prime and you bar in our laminar
and turbulent force flows.
So, average flow properties in the turbulent
fluctuations are actually coupled 
and why are the coupled due to nonlinear nature
of the turbulence, this is all the information
that we know from before. And for this particular
reason, we cannot ignore the fluctuations
under any circumstances. So, an ideal turbulent
model should be able to capture all these
fluctuations too.
This turbulent flow is comprised of vortices
with a range of different sizes. Simulations
of all scales of the vortices is not possible,
I mean, the vortices could be very, small
as well. So, the modeling of the smallest
of these vortices might not be possible, because
the reason is we have limitations on the available
computational facilities. The time and the
quantum computing is the reality this might
be an easier thing to do.
But for now, until that point, we still consider
that our computational facilities where we
are able to capture all the scales of what
it says is not yet possible. Further a universal
turbulence model which is valid for all the
turbulence flows is not yet available as well.
So, with this background, we come to one of
the most important techniques to solve these
navier stokes equation and this is called
Reynolds average navier stokes equation as
the name indicates, what we do is, we do the
averaging of the Navier stokes equation Reynolds
because Reynolds was the first scientist to
do it. So, we have already discussed this
term called Reynolds decomposition we have
discuss this and laminar and turbulent flow
module and the Reynolds condition. If you
are able to recall just in just in a case
where you are not I have included that again
as a quick repetition.
So, what is Reynolds decomposition for the
instantaneous velocity is u v w and the pressure
p u can be written as u bar + u prime. And
v is written as v bar + v prime this is the
bar whatever is in the bar denotes the average
value and whatever is with the prime it indicates
the fluctuations. Similarly, w is written
as w bar + w crime and pressure is written
as p bar + p prime.
So, if we say that f dash and d dash or any
to general fluctuating parameters, we also
have dealt this in our previous slide, then
the average of f bar and g bar is going to
be 0 so, f prime whole bar and g prime whole
bar is 0. Also we know that f prime multiplied
by g prime and taking the average of that
quantity is not equal to 0. If you recall
we had Renault shear stress in u prime w prime.
If this was 0 then there would be no shear
stress but this is not equal to 0 there is
a finite shear stress.
The third is the del f prime. So, the del
s of the fluctuation the first differential
and also the second differential and if we
take the whole bar = 0 also if we multiply
both the fluctuations f dash the dash take
the differential and average it will not be
equal to 0. So these are called the Reynolds
condition. So, this we have already seen before.
So, for a velocity field ui we can write ui
prime into uj bar to the whole bar = 0. Also
ui double bar = ui bar. If you take the average
of the average it is the average itself and
the fluctuation if it is average it is 0,
this is Reynolds averaging.
So the continuity equation for the incompressible
flow is this Del uk by del x k = 0, right.
And if we do the we perform the Reynolds averaging
on the above equation, it is going to yield
one equation that is del uk bar by del x k
bar = 0, right? You do not have just uk you
can write uk bar + uk Prime. And so Del, uk
by del x k will be del uk bar by del x k +
del x k uk bar and this will go to 0 So, this
will can be represented as this.
The Navier stokes equation for incompressible
flow can be written as Del ui Del t + Del
ui uj by Del x j = - 1 by rho Del p del xi
+ del xx j to nu Dij this is a standard form
of. So, dij here is a strain rate tensor which
we have seen in the last lecture in our viscous
fluid flow that dij returns half of Del ui
by Del x j + Del uj by Del xi. I think this
is not a problem because we have derived it.
So, if we do the averaging of navier stokes
equation, if we put u = u bar + u prime and
take the whole average of the equation so,
u = u bar + u prime and v = v bar + v prime
and w = w bar + w prime and p = p bar + p
prime and put it into Reynolds equation and
do the whole average of the equation. We are
going to get equation like del ui by del t
whole bar, del ui bar del uj bar So, if you
see this is average term, this is also average
term and this is also average term right.
So, if we use equation to the above equation,
you see you remember that equation to this
one here Del uk bar by del x k = 0 using the
continuity equation it can be written as Del
ui bar by delta t = - 1 by rho del p bar by
del xi + del x j of - tau i j + 2 nu dij.
As if this is equation number 5. So, this
actually is nothing but Reynolds average Navier
stokes equation. There is some things that
we need to find out what is tau ij dij we
already know.
So, in the above equation tau ij is ui prime
multiplied by uj prime this is generally called
Reynolds shear stress and as we have read
about it, is much more than the kinematic
stress and tau ij here is a symmetric tensor
symmetric means, ui prime uj prime = uj prime
ui prime. So, the term rho into when rho multiplied
by tau ij is known as turbulence stress or
Reynolds stress. The objective is to solve
the Reynolds average continuity and momentum
equation to obtain the average flow feel ui
bar and rho.
So the objective of this thing is to solve
the flow for ui bar and p bar we want to find
out the average quantities that is so that
is why we do some turbulence modeling.
So how do we do it we solve the Reynolds stress
equation if we subtract equation to the beginning
equation, second equation from equation 1
we can get Del uk prime divided by del xk
= 0. Similarly, if you subtract equation 4
from equation 3 you see question number 4,
this 1 is 4 and this 1 is 3. So, what we are
doing is we are subtracting the average to
a general equation in both the cases.
We are going to get Del ui prime del t + uj
prime original equation - averaged equation
same is here original equation - average equation
and why do we do this to obtain fluctuation
equations. So, with that this is how we are
we obtain the equations with fluctuation terms
in terms of differential equations.
If expressing uj bar into Del ui bar Del u
uj prime into del ui prime del t + ui in terms
of equation 7. So, if we express that in terms
of equation 7 here, you know and this is called
the Reynolds stress equation where this dij
is called the production of Reynolds stress,
this pi ij is the pressure strain correlation
tensor and epsilon ij is a dissipation where
j ijk is diffusive flux you can try to do
this at your own, but the derivation of this
is outside the scope of the current course,
I have shown you the methods how to get 1
equation from the other but the details is
not in your course curriculum.
And this is Reynolds stress equation, equation
number Pij is given as - tau ik del uj by
del xk - tau jk del ui del xk this is the
expression of the dissipation production and
pressure strain correlation factor epsilon
ij is the dissipation it is expressed like
this to nu del ui prime del xk del uj prime
del xk whole bar.
Where Jik is given in this for this is velocity
fluctuation, this is pressure fluctuation
and this is fluctuation due to the viscosity.
Therefore, JT ijk can also be written as ui
prime uj prime and uk Prime are you see, this
velocity fluctuations is written like this
the pressure fluctuation is written like this
and viscosity fluctuations is written like
this. So, we have seen what all these terms
of we have seen that all what all these term
means, for this Reynolds stress equation.
You see this one here, Pij pi ij a production
of Reynolds stress dissipation diffusive flux
and pressure strain correlation factor. So,
you do not need to remember all the details
here, but the way of doing it is quite an
important and somebody who would be interested
in doing further research or further studies
in M.Tech this will be a complete course in
itself. So, now, how to close this problem
is the important question and that will be
the topic of our next lecture.
And we will also proceed with techniques like
direct numerical simulation and large at simulation
for the modeling of computational fluid dynamics.
So, this will be all for today. Thank you
so much for listening and I will see you in
the next lecture.
