Good morning and welcome you all to this session
on fluid machines. In the last class we discussed
the axial flow and radial flow machines depending
on the direction of flow. And next we discussed
impulse and reaction machines. Now today we
will discuss mainly the principle of similarity
applied to fluid machines. But before that
we will just see how the efficiencies of fluid
machines are defined. As I have already told
and you have seen that fluid machine is a
device which converts stored mechanical energy,
stored energy in the fluid into mechanical
energy and vice versa.
The efficiency accordingly of a fluid machine
is defined as the ratio of output energy divided
by input energy. Now output energy maybe mechanical
energy delivered by the machine or the energy
stored in the fluid depending upon whether
it is turbine or pump. Similarly the input
energy will be mechanical energy, sorry, stored
energy in the fluid or the mechanical energy
depending upon it is turbine or pump. And
in this connection, two efficiencies are defined,
one is hydraulic efficiency concerning the
energy transfer between the fluid and the
rotor and another one is overall efficiency
concerning with the energy transfer between
the fluid and the shaft.
Now the difference in energy at the rotor
and at the output shaft is the energy absorbed
by glands, bearings and other couplings due
to friction. Now, if we see that in detail,
let us write this, there are two efficiencies,
one is hydraulic efficiency, one is hydraulic
efficiency which is given by Eta H which deals
with the energy transferred between the fluid
and the rotor. Another is overall efficiency,
overall efficiency, overall efficiency Eta
O which is concerned with the energy transfer
between the fluid and the output shaft.
Now let us define separately for turbines.
Hydraulic efficiency is defined as the output
energy of the turbines, what is this, this
is the mechanical energy delivered by the
rotor, that is mechanical energy delivered
by the rotor, mechanical energy delivered
by the rotor, mechanical energy delivered
by the rotor divided by, that is in the denominator
is energy, that is the stored energy of fluid
at inlet to rotor or the machine, fluid machine.
This energy of fluid at inlet is the available
energy, stored energy from which the conversion
takes place in the rotor and rotor delivers
mechanical energy in the form of its rotation.
Now here this energy available to the rotor
at the inlet to the rotor and machine is same
since the Stator is fixed member, so does
not move, so there is no mechanical energy
transferred. So therefore if we neglect the
heat loss from the machine and the viscous
dissipation, the mechanical energy at inlet
to the machine and mechanical energy at inlet
to the rotor remains same, this is because
stator remains fixed.
Now if you think of not mechanical energy,
the total energy inclusive of intermolecular
energy also which we call as thermal energy,
then if we neglect the heat transfer loss
from the machine or no heat transfer between
the machine and the surrounding, so by the
conservation of energy, energy at the inlet
to the stator or the inlet to the machine
will remain as inlet to the rotor because
stator does not perform any mechanical work.
So therefore we write energy of fluid at inlet
to rotor or machine.
The overall efficiency is now, the denominator
remains same, the input energy, energy of
fluid, same thing, energy of fluid at inlet
to machine or rotor, whatever you write and
this thing is the mechanical energy, mechanical
energy in output shaft at coupling, in output
shaft at coupling. And the difference between
the two is the friction in glands, bearings
and couplings. And if we now divide this overall
efficiency by the hydraulic efficiency, overall
efficiency if you divide it, this thing is
cancelled, energy of fluid at inlet, what
we get, we get mechanical energy, mechanical
energy at shaft, in shaft, in output shaft.
I am not writing the entire sentence and here
the mechanical energy at rotor delivered by
the rotor, I am not writing at rotor. That
means this becomes overall efficiency divided
by the hydraulic efficiency, mechanical energy
in shaft output shaft divided by the mechanical
energy delivered by the rotor. That means
mechanical energy at rotor. And this is always
less than this because of the mechanical friction
in glands, bearings and couplings. So therefore
this ratio is always less than one and this
is defined as mechanical efficiency.
This is Eta M, mechanical efficiency. And
hence we can write the overall efficiency
is hydraulic efficiency into mechanical efficiency.
Okay. Now next is, similar is the case of
pumps or compressors. In case of pumps and
compressors, in case of pumps and compressors,
we can define similar way the hydraulic efficiency,
here output is the mechanical energy gained
by the fluid, that is mechanical energy, mechanical
energy 
gained by the fluid, gained by the fluid at
final discharge, means discharge from the
machines, at final discharge.
And the denominator is mechanical energy supplied
by the rotor. Mechanical energy supplied to
the rotor or by the rotor, supplied to the
rotor. Out of which how much energy is gained
by the fluid at final discharge, that is the
hydraulic efficiency. That takes care of the
losses in the fluid flow, similar is the case
of turbine. So this is the definition of hydraulic
efficiency of pumps and compressors. Now here
mechanical energy gained means this mechanical
energy at the final discharge relative to
its mechanical energy at inlet because fluid
at inlet has some mechanical energy, that
is why mechanical energy gained by the fluid
at final discharge.
And in the similar way, following the similar
line of thought as we discussed in case of
turbines, the overall efficiency in this case
the output will remain same, mechanical energy
gained by the fluid, by the fluid at discharge,
final discharge or simply at discharge, now
here mechanical energy supply to the rotor,
here input will be mechanical energy supplied
to the shaft. That means here the final, initial
input point is shaft, like the turbine, the
final output point is shaft.
So therefore overall efficiency deals with
this energy supply to the shaft and the energy
gained that final discharge and in the similar
way, hydraulic, overall efficiency divided
by the hydraulic efficiency is the mechanical
efficiency which is the ratio of what, ratio
of, if you divide this by this is the mechanical
energy supplied to the rotor divided by mechanical
energy supply to shaft. That means mechanical
energy to rotor and I am writing in short,
mechanical energy to shaft, input shaft.
And this is always less than one because energy
supplied to shaft is not going to the rotor
because of the frictions in those elements
glands, bearings and couplings, so rotor receives
less energy. So this is the mechanical efficiency.
So this way usually the efficiencies of the
fluid machines are defined. Now we come to
a very very important chapter or very very
important section is the principle of similarity
applied to fluid machines. So before I discuss
this principle of similarity applied to fluid
machines, I like to have a recapitulation
or a brief review of the concept of principle
of similarity in physical processes.
Okay. To brush up your knowledge that already
you have gathered in your basic fluid mechanics
course. As you know that most of the engineering
problems or the problems related to practical
applications, their solutions are determined
from experiments. This is because of the complex
nature of the practical problems or engineering
problems in practice, where besides the analytical
method, even the most updated CFD tool cannot
take care into consideration of all the aspects
and features that happen in practice are associated
with the problem.
So therefore we always depend on experiments,
even if you do theory, these are being calibrated
from the experiments. Theories are done, the
CFD analysis are made to guide the experiments
in which direction the experiments will go
and to predict the qualitative trend. But
to have a quantitative final solutions of
those engineering problems, we always do the
experiments and their solutions are obtained
mostly from these experimental studies.
Now because of economic advantage, saving
of time and ease of investigations, it is
not possible in almost all instances to do
the experiments in laboratory under the identical
conditions in relation to the operating parameters
and the geometry of the system that prevail
in practice, it is not possible. So therefore
we have to do the experiments under altered
conditions that happen in practice. For examples,
if you have to do an experiment to find out
the pressure distributions over an aircraft
wheel, we have to do the studies but where
we cannot have the aerofoil of the dimensions
that an actual aircraft does have.
We cannot have the fluid velocities, the condition
of pressure and temperature similar to that
happen in practice. Similarly for example
if we think of simulating or studying the
flow through pipes which will generate data
to predict the performance of the pipe networking
practice, we cannot do the experiment with
such big dimensions of pipe, big diameters
of pipe, length of pipe and sometimes we cannot
have that flow rate, we cannot generate that
flow rate in the laboratory.
For example to find out the drag in a ship
or submarine by our, in our, we do experiments,
if you want to do experiments in our laboratory,
we cannot have a ship or submarine of its
actual dimension and we cannot create a ship
in our laboratory. So therefore laboratory
tests are always performed under altered conditions
that happen in actual practice. But at the
same time the results from these tests have
to be used to predict the performance of the
actual system under actual operating conditions.
So now in relation to these, there are two
very pertinent questions come up.
One is that, which ensures us that the results
from these altered conditions, tests at altered
conditions in the laboratory will be, can
predict the performance parameters of the
actual system which are operating under different
conditions in practice. How we can compare
these, what has to be satisfied to make comparisons
or to use these results for predictions of
parameters in practice, Numberminusone.
And number two question is that if a physical,
if a physical process depends upon a number
of independent controlling parameters, then
to find out the influence of each and every
parameter, we have to do a number of experiments
by varying each and every parameter by keeping
other parameters fixed. So which involves
huge cost, huge time, okay. Can we do this
way to save time and cost that vary some less
number of variables and can predict the performance,
parameters, the influence of the variable,
other variables on the performance parameters
of the system.
Now these two questions are very important
and a positive clue in answering these two
questions lie in the principle of physical
similarity. That means this can be done, that
means the results from the laboratory at altered
conditions can be used to predict the performance
of the same process in practice at different
conditions and at the same time we can reduce
the number of experiments by varying only
few independent controlling parameters so
that the influence of other controlling parameters
can be determined from that, provided the
physical similarity exists between the two
problems.
So therefore I come to this physical similarity.
Now physical similarity, the first line of
understanding is that, physical similarity
has to be sought between two problems defined
by the same physics. First of all we have
to understand that the physical similarity,
always, if we say that the two problems are
in hundred percent or identical similarity,
have physical similarity but the problems
have to be defined by the same physics. So
first requirement of physical similarity,
that physical similarity has to be sought
between the problems governed by the same
physics.
For example, the flow governed by gravity
force and inertia force, for example a surface
wave in a sea and another process that flow
of fluid through a pipe which is governed
by inertia force, viscous force and pressure
force cannot be made similar under any identical
condition. We can make similar situations
of pipe flow problems at different operating
conditions of the surface, problems relating
to surface wave at a different operating conditions.
So the physics of the problem has to be same.
And in that case, the physical similarity
to be maintained or to be obtained, we have
three types of similarities have to be ensured.
One is the geometrical similarity, one is
geometrical, another is kinematic similarity
and another is dynamic similarity, another
is dynamic similarity. Geometrical, kinematic
and dynamic similarities. Now geometrical
similarity is the similarity of dimensions
which tells that the system dimensions in
actual case and in the laboratory case should
be such that the ratio of the corresponding
dimensions should be same and it is maintaining
the same shape. And this we know since our
childhood, an example I am giving.
Let us consider the two cylinders whose length
is one L and the diameter D, if we have got
another cylinder whose length is L by two
and the diameter is D by two, they are geometrically
similar. Similarly if we have a parallelepiped
if we have a parallelepiped of this A, this
is B and this is C and if we have a small
parallelepiped where A, B, C are reduced by
the same proportions, say this is A by three,
this is B by three and this is C by three,
then they are geometrically similar. Now what
is kinematic similarity, kinematic similarity
is the similarity of motion that requires
that the ratio of the velocities, particular
component of velocity at the different systems
should be same at corresponding point.
That means the ratio of the velocities of
a particular velocity, that means particular
velocity component of both the system and
the in the laboratory, both the system, system
in the laboratory and that in practice should
be at the same ratio at the corresponding
points.
Let us define this way, let us have U, V and
W, three components of velocity, X Y Z. If
we have a system one, if we have a system
two. And we sought the kinematic similarity
between these systems, then U for system one
divided by U for system two, this ratio has
to be same at corresponding points of the
system one and system two. Similarly V of
system one divided by V of system two will
be same at corresponding points, similar is
the case of W. And accordingly one can write
this also, U, sorry, this is U by V of system
one divided by U by V of system two at the
corresponding points has to be also same for
both the systems.
So therefore you see the ratios of the corresponding
velocities will be same at corresponding points
between the two systems and accordingly the
ratios of any two components of velocity system
one and system two will be same at the corresponding
points of the two systems. This ensures the
kinematic similarity. Similarly the dynamic
similarity will ensure the similarity of forces.
So dynamic similarity is the similarity of
forces, dynamic similarity is the similarity
of forces, similarity of forces, dynamic similarity
is the similarity of forces, similarity of
forces. Now problems, physical problems are
governed, all physical problems or physical
processes are governed by the different forces.
Let us consider few such forces. Inertia force,
I am writing in short, I am telling inertia
force, viscous force, gravity force, surface
tension force, elastic force and so on.
A particular problem, a particular process
may not be governed by all the forces, some
processes may be governed by inertia force,
viscous force, pressure force, another force
is the pressure force, inertia force, viscous
force, pressure force, some may be governed
by inertia force, gravity force, some may
be governed by inertia force, viscous force,
gravity force. So to ensure the dynamic similarity,
the ratio of the forces at the corresponding
points between the system and, between the
two systems, system one and system two between
which the similarity is sought for has to
be same.
So to make, to make this, we have to find
out the parameters that define the inertia
force and viscous force. Now here we know
the ratios of velocities we get in terms of
velocities but when you say that the ratio
of inertia force by viscous force has to be
same for both the system, then we have to
know that the ratio of inertia to viscous
force is given by which parameters, which
variables of the fluid flow. Usually we deal
with the velocity, pressure, density, viscosity,
the fluid property, or the characteristic
geometrical dimensions of the problem.
So therefore it is customary to express these
ratios of forces in terms of those variables
concerned with the fluid flow problems. So
this is done by scaling the forces. So how
to done, now inertia force, if we scale inertia
force is equal to mass times acceleration,
acceleration. Now what is mass, mass is, can
be scaled as rho into, L is the characteristic
dimension of the system rho L cube. And acceleration
is the rate of change of velocity, so if we
take the characteristic velocity as V, so
acceleration is V divided by the time and
time also we can write in terms of L by V,
in terms of the characteristic length and
the characteristic velocity.
So that this becomes is equal to rho, this
L square V square. So therefore we see we
can scale inertia force in terms of characteristic
length and characteristic velocity as rho
into L square V square where rho is the density
of the fluid, L is the characteristic length
and V is the characteristic velocity. Similarly,
if we see the viscous force, viscous force
is proportional, can be written as shear stress
times the area, surface area on which the
shear stress is acting. Okay. Now this shear
stress from the Newton’s law of viscosity
can be written in terms of the mu into velocity
gradient, for a very simple case it is shear
rate but in a oneminusdimensional flow, where
U is a function of Y only and there is U component
of velocity only, then mu into D U D Y into
the area. So therefore the scaling is mu times
the velocity gradient. This will be only the
velocity gradient. So mu into velocity gradient,
okay so therefore what you write, velocity
gradient is V is the characteristic dimension
and what is Y, Y is the length dimension L
and what is area, area is L square.
So this can be expressed in terms of area
and the velocity characteristic, velocity
and length as mu V L. So therefore inertia
force by viscous force by scaling is proportional
to rho L square V square by mu V L which equals
to rho LV by… This is the L we are writing,
the same L why I am writing the big L, the
same L, rho LV by mu. So therefore we see
if we have to keep the inertia force by viscous
force, this ratio same for both the systems
between which the physical similarity is sought,
we have to keep the combination of these variables
same in both the cases.
And the combination of these variable rho,
L, V, rho is the density of the fluid, L is
the characteristic dimension, geometrical
dimension, V is the characteristic velocity
and mu is the viscosity is known as Reynolds
number. So in the similar way we can derive
various dimensionless numbers.
Another example I tell you that the problems
or process which is governed by both inertia
force and gravity force. Now gravity force
is scaled as the mass into acceleration due
to gravity, mass into G and mass is rho L
cube G. So therefore the gravity force or
inertia force by gravity force will be rho
L square V square divided by rho L cube G.
So that becomes equal to V square by GL and
therefore for the problems which are governed
by the inertia force than gravity force, their
ratio has to be same at all corresponding
points between the two systems to maintain
that V square by G L has to be kept constant.
And this ratio, the square root of this, V
by root over GL or sometimes root over GL
by V is known as Froude number. There is no
such convention, sometime this is defined,
sometimes
reciprocal of this is also defined as the
Froude number. So therefore this is known
as Froude number but Reynolds number as a
classical convention, it has to be this, inertia
force by viscous force, not the viscous force
by inertia force. So inertia force by gravity
force scale like that.
So this way by scaling the different force
we can find out the similarity parameters
which have to be made same for both the systems
to maintain the dynamic similarity that the
ratio of the forces at corresponding points
will be same. Now it is not always possible
to do this that the maintaining the similarity,
we have to always sought, we have to go for
scaling of all forces. So therefore to make
it simple, one can find out the nondimensional
parameters which keep the principle of similarity,
make the two systems under physical similar
condition by one theorem known as Buckingham’s
pie theorem.
So again I tell you we have seen that to maintain
the similarity between the two systems, we
have to make the ratio of the corresponding
geometrical dimensions same, the ratio of
the corresponding velocities at corresponding
points have to be same, ratio of the different
forces at corresponding points between the
systems have to be same. All these are nondimensional
number and they are usually expressed in terms
of the dimensional variables involved in fluid
flow like pressure, velocity, density, the
property pressure, velocity and the property
of the fluid density, viscosity and the geometrical
dimensions.
For kinematic similarity ratio of velocities
and the geometrical similarity ratio of physical
dimension, it is straightforward, apparent.
But for ratio of forces, this has to be done
by scaling the forces and sometimes it becomes
difficult. So a easy method of finding out
those nondimensional number which keeps all
the similarity has been given by Buckingham
and known as Buckingham pie theorem. Buckingham’s
pie theorem.
Now according to this theorem, Buckingham’s
pie theorem one can tell that if a problem
is defined by a number of variables, number
of physical dimensional variables, let X one,
X two, X three, M number of, M number of physical
variables M. Number of variables, dimensional
variables M. Then Buckingham first from his
intuitive thinking found that due to dimensional
homogeneity, the number of independent nondimensional
terms governing this problem or the process
will be less than the number of dimensional
counterparts, that is M.
M is the number of dimensional variables controlling
the process and the number of nondimensional
variables controlling the process and all
these nondimensional variables will combine
some of the dimensional terms will be less
than that of the dimensional counterparts
M. And this was given by Buckingham and he
told if N is the number of nondimensional,
number of nondimensional variables, independent
variables, rather independent variables, you
might name number of here also, independent
variables controlling the problem. Number
of independent variables, number of independent
variables, this is known as pie terms.
Nondimensional independent, if N is the number
of pie terms, number of pie terms, then he
told that N is the number of pie terms, sorry,
here N I will not write, I will write the
number of pie terms, pie terms is the number
of nondimensional independent variables, this
will be equal to M minus N where N is the
number of fundamental units, fundamental dimensions,
number of fundamental dimensions in which
the variables are expressed.
Rather it can be told this way, if a problem
is defined by M number of independent dimensional
variables, then because of dimensional homogeneity,
there will be a less number of nondimensional
variables controlling the process which is
less than the number of the counterpart dimensional
variables M. And if these M variables can
be expressed in terms of N fundamental dimensions
like mass, length, time, temperature, like
this, then the number of these nondimensional
independent variables pie will be M minus
N where N is the number of fundamental dimension
in which these M variables can be expressed.
Now this can be worked out, now how to find
out this number of pie terms. This to find
out, we have to do this. Let us consider a
problem is first defined by its number of
M number of dimensional physical variables.
Now this line of definition of the problem,
a problem is defined by this, means here it
has to be made, Zero, an implicit functional
relationship. Now first line of this Buckingham’s
pie theorem is that we have to write the problem,
this mathematical statement of the problem
by an implicit functional relation where these
are the independent dimensional variables
governing this process.
And what are the dimensional, pertinent dimensional
independent variables governing the process,
that you have to know from the physics of
the problem or physics of the process. So
after that if we write this thing, if this
is the M, M is the number of variables, M
is the number of variables. Now we see if
these variables are expressed by N number
of fundamental dimensions, N is the number
of fundamental dimensions, then by Buckingham’s
pie theorem, the number of pie terms, so pie
terms, number of pie terms will be M minus
N.
Now how to do it, how to find out this M minus
N pie terms. Now you first select N number
of, N quantities, N number of quantities as
a repeating variable. Let X one, X two up
to X N, that means repeating variables with,
these are known as repeating variables. We
choose any arbitrarily, any N number of physical
dimensional variables, any arbitrarily N number.
But there should be a caution. In this repeating
variable, N number of repeating variables
where N is the number of fundamental dimensions,
no variables should be an output parameter
of the process, that is number one caution.
Number two caution is that all the variables
chosen taken together must have all the fundamental
dimensions. We should not choose the variables
in such a way that in none of the variable,
one of the fundamental dimensions is missing.
That means inclusive of all variables, all
the fundamental dimensions should be there.
And another caution is that no two, no two
variables should have the same quantity. That
means no two variables should be either geometrical
dimensions or velocity, so these two same
physical entities should not be taken.
So if you take so, and you can find out from
this, the pie term. How, you make each pie
term as raise all the repeating variables
raised to the power some unknown X N to the
power N number of indices and taking the rest
X N plus one. The rest of the M minus N variables.
So therefore pie two will be formed as X,
these A B C sets will be different for different
pie terms. These sets will be different for
different pie terms and it will include X.
Like that we will be having M minus N pie
terms, pie M minus N will be X one A, X two
B, X three C to X N N into X N plus one, N
plus two X M.
So therefore N minus M, N plus one M, okay.
So we say, this way we can form the pie terms.
Now all A B, A to N sets are different for
different pie terms. What we will do now for
each pie term, we now replace the dimensional
formula of each variable and indices are there.
This is nondimensional quantity. That means
indices of all fundamental dimensions will
be zero. So if we equate each N fundamental
dimension indices to zero, we get N number
of equations and there are N variables, we
solve for A to N from all the pie terms.
So therefore we get the explicit, the explicit
term for this pie term. Okay, so this way
the pie terms can be found out. Let us have
an example, let us consider the principle
of similarity in pipe flow. Let us consider
an example of pipe flow problem. A pipe flow
problem. A pipe flow, the fully developed
flow in a pipe, fully developed flow in a
pipe. Now in a fully developed, this is a
recapitulation, that is why I am going little
fast, fully developed flow we can express
that the problem as that pressure drop, rather
we write this way, you will be under, you
will be able to understand it better.
That in a fully developed pipe flow, pressure
drop per unit length of the pipe is usually
a function of velocity of flow, some characteristic
velocity of flow, some characteristic dimensions
of the pipe which we call as hydraulic diameter
and the density of the fluid as its property
and the viscosity. And this can be expressed
as this, implicit functional relationships
of Delta P by, this is the output parameter.
V, D H, rho, mu, zero. A pipe flow problem
is governed by the pressure force and viscous
force.
And in a fully developed flow, inertia force
is zero. So therefore Delta P by L, V, D H,
rho, mu defines as in dimensional variables
a pipe flow problem. So a pipe flow problem
can be defined implicitly by this functional
relationship. Now how to find out the, apply
the pie theorem, so here the number of variables,
number of variables, number of variables are
one, two, three, four, five. So number of
fundamental dimensions, number of fundamental
dimensions, number of fundamental dimensions,
that is N, that is number of variables M,
N the number of fundamental dimensions is
equal to, these are expressed as mass, length
and time.
All these parameters are expressed in terms
of three fundamental dimensions. So number
of pie terms therefore, number of pie terms
therefore equal to five minus three is equal
to two, so therefore two pie terms. Now we
see that there are number of, N is three,
so we can have three repeating variables.
This is the output variable, this should not
be there and there are how many choices, there
are four dimensional variable choices four
C three, that means we can have many sets
of repeating variables.
One set is V, D H, rho, another set is V,
D H, this is comma, mu, another set is D H,
rho, mu, another set is V, rho, mu. So incidentally
all these sets include all the fundamental
dimensions. We can take these three as repeating
variables, we can take these three, we can
take these three, we can take these three.
Let us take these three are the repeating
variables V, D H and rho. And if we take this
as the repeating variable V, D H and rho.
Then what we have, we have pie one is equal
to, now we can write here, we have pie one
equals to V D H rho we have taken V to the
power A, D H to the power B, rho to the power
C, V, DH, rho, we have taken, so left parameter
is Delta P by L and mu. So first we take Delta
P by L and this second pie term we take as
V A to the power A, D H B to the power, rho
to the power C mu. So therefore if we now
equate this, pie one is what M to the power
zero, L to the power zero, T to the power
zero, V is L, T to the power minus one A,
L to the power B, M L to the power minus three
to the power C.
And what is Delta P by L? M Delta P, Delta
P is the pressure were M L to the, M L square,
pressure is M L to the power minus one T to
the power minus two and that divided by L.
That means M L to the power minus three T
to the power minus two, okay. So therefore
Delta P is M L to the power minus one , divided
by L, M L to the power minus two, sorry, M
, L to the power minus two. What is Delta
P, Delta P is the pressure, that is M L to
the power minus one, T to the power… Okay.
Then divided by L. So therefore if we now
equate the L, M and T indices, we will get
here A is equal to minus two, if you do it,
B is equal to one and C is equal to minus
one.
If we equate L, L A plus B minus three C is
equal to zero, similarly for T, minus A minus
two is equal to zero, A is equal to minus
two, this way you get this. And if you do
the similar procedure for this thing that
pie two, M zero L zero T zero is the similar
procedure that L T to the power minus one
to the power A, L to the power B, M L to the
power minus three C and mu is the M L to the
power minus one T to the power minus one,
then you get this. In the second situation,
A is equal to minus one, B is equal to minus
one and C is equal to minus one.
And if you put this in the pie term, the value
of A, B and C, you get pie one as Delta P
by L, that is the pie one as you get D H by
rho V square. And pie two if you put this,
you get pie two is equal to mu by rho, V,
D H. Okay. So this you get pie, if you put
the value of A, B, C here and this you get
as pie two, if you put A B C value here. So
pie one will be Delta P by L D H rho by V
square and pie two will be mu by rho V D H.
Thank you.
