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Welcome to this lecture number 28 on this NPTEL course on Fluid Mechanics for chemical
engineering undergraduate students. So, far in this course we have discussed two major
approaches towards analyzing fluid flow problems that, occur in chemical engineering.
One is the integral balances integral balances or microscopic balances. Here, we write integral
balances of mass momentum and energy over a control volume; and when we do this we can
choose the control volume to be to comprise of various flow equipments such as, pumps
or pipes or compressors or valves and. So, on and we can do this microscopic balances,
but to solve this microscopic balances importantly, we need some inputs. The inputs to solve the
microscopic balances are viscous losses, viscous losses that occur, in various parts of the
system like flow in pipes or flow through a wall or flow through a sudden expansion,
there are always viscous losses; we need that as an input and we also need the forces exerted
on solid surfaces as an input. Unless we provide these as an input to the
microscopic or integral balances, we may not be able to solve them fully accurately. But
in many occasions, because we do not have a knowledge, accurate knowledge of this, we
can neglect this and get a approximate answer to various flow problems; and this we highlighted
in some of the examples that we did before, but these are not entirely accurate, because
we have neglecting some important factors such as, viscous losses or forces viscous
forces exerted by the solid drag forces exerted by the solid on the fluid surfaces since we
neglect them, we may get only approximate answers.
But this is one extreme, the other thing that we did was differential balances or microscopic
balances or microscopic balances. Now, the end result of all the analysis on in microscopic
balances can be summarized in a very simple way for an incompressible Newtonian fluid
and it is deceptively simple, because even though we can write down the equations in
a very simple way for the flow of an incompressible liquid.
The solution of this, so called navier stokes equations are notoriously difficult, because
in a very very general setting, the navier stokes equation are non-linear coupled partial
differential equations and solving them in their most general end form is very very difficult
task. So, we had to end up making simplifications, such as flows only in one direction, the flow
is steady and the flow velocity varies only in one direction and so on.
We had to make several simplifying assumptions and we can get the solution. But the advantage
of differential balances is that, you do not need any input other than saying that what
is the viscosity of the fluid and what is the density of the fluid. And once you know
what is the viscosity as an input from an experiment and density as an input and once
you provide suitable boundary conditions to the flow in principle one can solve them,
but in practice the solution is very difficult; so, we had to make all these simplifying assumptions,
but we also pointed out that the solutions, that one gets by making this simplified assumptions
are not unique.
For example for flow in a pipe, we obtain a solution called the hagen poiseuille velocity
profile using the differential balances, so this is the pipe, but the solution that we
got is not always observed in reality, because the assumptions that we make fail when the
velocities are sufficiently high or when the Reynolds number is greater than 2000. This
simple solution, which is obtained by solution not observed in experiments.
Now, therefore, even though for Reynolds number smaller than 2000, we can get an accurate
description of how much pressure drop is required to make the fluid flow at a given flow rate,
we cannot have the same luxury, when the Reynolds number is greater than 2000, because the simplifying
assumptions that we made to solve the navier stokes equations, they are not correct.
The flow becomes more complicated or specifically turbulent, so it does not enough to solve
the navier stokes equations for such simplifying approximations, one has to in principle one
has to solve full navier stokes equations, which is a very very difficult problem. In
principle, one has to use very sophisticated numerical logarithms and high performance
computers to be able to solve the navier stoke equations in turbulent regime.
So, we have these opposing requirements on the one hand integral balances are relatively
simple to solve, but they need some input such as a viscous losses, while differential
balances, they do not need any inputs, external inputs only you have to give is the viscosity
of the fluid and the density of the fluid through a single experiment; and then everything
is specified, but their solution is very very difficult unless one is prepared to go to
very sophisticated computation. So, but in industrial practice in chemical
engineering, often the equipment the process equipments that one has they are exceptionally
complex. So, one cannot often one cannot always hope to use differential balances to solve
for example, the flow problems in chemical engineering.
Although one can make an attempt it is not always feasible, the main reason is that in
chemical engineering applications, the flow in chemical engineering or chemical process
industries, flow is often turbulent and flow is often multi-phase by which, I mean you
may have fluid flow I mean a flow of suspension of solids and fluid together. So, and we may
have to estimate what is the pressure drop to make such a complex suspension flow in
a pipe and for which, we cannot use the differential balance, because differential balances work
only for a simple Newtonian fluid. So, and the flow is often theologically complex,
even if it is effectively a single phase flow, you may have flow of a polymer molten plastic
which is a very very complex liquid in terms of its rheology you may not described as simple
Newtonian fluid. So, it is not sufficient for us to just use the differential balances
in many industrial setting; although we will hope to use the differential balances as much
as possible because as they are very very accurate than the in the results, but the
solution of the differential balances is very very difficult.
So, the third way of analyzing problems is through experimentation in industrial design.
So, we have to carry out experiments for example, if we say fluid is flowing in the turbulent
regime in a pipe, we cannot know what is the flow rate that is required what is the flow
rate that will come out if you apply a given pressure drop for laminar flow we know the
answer, we already solved the answer from differential balances.
For turbulent flow we do not know the answer from first principles to from the navies stokes
equations. So, if you want to design a pump to make a fluid flow at a given flow rate
or if you want to predict how much fluid will flow, what is flow rate of fluid for a given
pressure drop in the turbulent regime, then clearly we have to resort to experiments.
Now, doing experiments in the laboratory is often simplified and it is much made much
more organized and logical by doing, what is called dimensional analysis and that is
topic of our discussion today. So, in the absence of any solution to a problem through
differential balances, we have to resort to experiments; and experiments are made much
much more logical and organized and simplified, if we use the principles of dimensional analysis;
and that is the goal of our current discussion to illustrate the principles underlying the
underlying dimensional analysis and their applications.
Now, I am going to illustrate first, the nature of dimensional analysis by using a very simple
problem, the problem is drag force on a sphere. Imagine a sphere moving with a constant velocity,
you have a sphere of radius r a nice rigid sphere and it is moving with a constant velocity
u in a fluid and the fluid has viscosity mu and density rho.
In many applications, we want to know, what is the force, resistant force that is experienced
by this sphere? Due to the fact that the sphere is moving through a fluid, it has to…
When the sphere is moving through a fluid, it has to push the fluid around it in order
to move and when the fluid solid surface such as this, this is a solid and the fluid flows
around the solid surface, there is a relative motion between the fluid and solid, that means,
that they are viscous stresses, whenever there is a relative motion between two fluid elements
or between a fluid and solid elements, there are viscous stresses, and when you consider
the sum total of all viscous traces exerted by the fluid on the solid surface, you get
the net force that opposes the motion of the sphere, that is called the drag force, that
is the drag force. The drag force on a solid substance objects
such as, a sphere is the net force that opposes the motion of the solid surface, a solid object,
because when the solid object moves through a fluid, it has to push the fluid around and
when the fluid moves around the solid's surface it exerts viscous shear stress and that is
the concept of the drag force. Now, suppose you want to know, what is a drag
force on a sphere and what type of experiments that we should do? So, let us first do a simple
let us first estimate this problem.
Suppose, you want to know drag force on, you want to have experiments that correlate the
drag force on a sphere, on the various physical parameters, that affect the problem. So, first
we want to ask, what are the variables in which the drag force should be a function
of? So, let us denote the drag force to be F the drag force is denoted by the symbol
f. If F is a drag force what is it a function of? It is a function this is of various variables.
We may imagine that it is the function of the radius r or let us say diameter D, D is
the diameter of the sphere, we may imagine that it is the function velocity of the sphere,
because this sphere moves with a higher velocity we may expect that the force experienced by
the sphere is different compared to when its moving at lower velocity, we may imagine that
it is a function of viscosity mu, density rho; we may have other variables such as the
surface tension or interfacial tension between fluid and solid, but this is a judgment that
one has to make. What are the variables on which, the given
physical observer such as drag force should be function of?
You have many variables that affect the problem, but we expect at this point of time that,
these are the most relevant physical variables upon which, the drag force will be a function
of. And if there are more variables in the problem then, it will show up later while
doing experiments. So, these are the variables that, we think are the most important variables
that will affect the drag force on a sphere. Now, if you want to find out how the drag
force is going to vary with all these parameters, you have four parameters, the diameter of
the sphere, the velocity of the sphere, it is called the velocity V. So, we have four
parameters upon which the drag force can depend on and we want to find the parametric variations
of the force on all the four parameters. Now, let us imagine how we will do the experiments
in a simplistic way. Well first, we will say that in order to find
how F is affected by v, you fix D mu rho and let us say we measure 10 different values
of velocity, we have 10 different values of velocity of V of V and we measure 10 different
values of F, correspondingly 10 values. Then, so all we would say now all we would have
is for a given diameter viscosity intensity how does the force vary with the velocity.
Now, you may say that well, now we want to know how the force varies with the diameter,
so you fix the velocity, viscosity, density and then find F as a function of of D, again
you may have 10 different values. So, to find how F depends on just two parameters
such as velocity and density, you already have 10 different values of velocity and 10
different sorry velocity and diameter, you already have 10 different values of velocity
and 10 different values of diameter you already have 100 data points.
So, you may even it is not even easy to plot such 100 data points as a function of these
two variables diameter and velocity, but you have other parameters, you have F as suppose
you want to find the force, how the force depends drag force depends on the viscosity,
so you have to fix the velocity diameter density and then find F as a function of viscosity,
again you have let us said 10 different values. That is you may choose 10 different fluids
of different viscosities and then find F and finally, you may have to find how to F here
varies as a function of density. So, you fix velocity diameter and viscosity and find F
as a function of rho again you may have 10 values say. So, if you consider all the data
points that, you want to generate to find how F depends on these four parameters namely
the diameter, velocity, diameter, viscosity and density you have 10 times 10 times 10
times 10, 10 to the 4 data points
So, So, you have to do 10 to the 4 experiments to find all these 10000 experiments, which
is extremely tedious and cumbersome and even if you generate those 10 to the 4 data points,
how are going to make sense out of this 10 to the 4 data points? Because you have four
independent variables upon which F is function of upon which F depends on, so how are we
even going to plot, such a large number of data points with the in which F depends on
four independent variables, so it is not at all an easy task.
Secondly, doing 10 to the 4 experiments is also not easy, because if each experiments
takes about 30 minutes, then you can say you can say that you can estimate that 10 to the
4 experiments will take 10 to the 4 times 30 minutes, which will be a very very large
number of time it is about 2 years. So, to find how the drag force depends on
four variables and with just 10 data points for each variable, we already find that we
have to do many many experiments with order of 10 to the 4 and it is going to take a long
long time to complete the data. Now, even after completing the experiment making sense
out of the data in order to see any trend or to analyze the data is also extremely difficult
and cumbersome. So, clearly there has to be better way to do experiments and this is where
dimensional analysis plays an important role.
What dimensional analysis does is that, I am first going to make a claim and then I
will of course, show how this comes about what dimensional analysis tells is that for
this drag force problem, if you do dimensional analysis correctly then, you can write the
functional relationship between, you have the 5 variables F, D, V, rho, mu.
The functional relationship among these five variables can be written as a functional relationship
between just two groups, but these groups are non-dimensional. You can show, we will
show, we will prove that, F this functional relationship among these five variables can
be written as V square is a function of rho V D by mu, these two groups are non-dimensional
these two groups are non-dimensional. By non-dimensional, we mean that, that if
you check the dimensions of this groups they will be mere numbers, they will not have any
physical dimensions such as length, mass or force or acceleration anything associated
with them, they will be pure numbers such pure numbers are called non dimensional groups
or dimensionless groups, they are also called dimensionless groups.
We have some some terms, which are purely non dimensional or dimensionless and we will
later show that the relationship between all these 5 variables the force, velocity, diameter,
viscosity density can be written as a non-dimensional as a relation between just two non-dimensional
groups, which is a great simplification. Now, we will also point out that, the form
of this function the form of the function is still undetermined is not determined by dimensional analysis.
The form of the function is not determined by dimensional analysis. We still have to
do experiments, but remember that, in contrast to doing 10 to the 4 data points or 10 to
the 4 experiments, we simply have to to do, just you have to find the functional dependence
of one variable on another variable, one group on another group we simply have to do ten
experiments. So the functional relationship among 5 variables
is now reduced to functional relationship among only two non-dimensional groups. Now,
first of all it leads to greater, I mean great amount of simplification because, firstly
the number of experiments that you have to do decreases by an order of three out some
angle from 10 to the 4 experiments, we are now to only 10 experiments, which are fairly
easy to do in a laboratory. Secondly, in the previous case if you want
to know how force depends on viscosity, we have to do 10 different fluids, because you
have to find 10 different fluids of 10 different viscosity and only then you will be able to
find how force depends on all these 10 different viscosities or 10 viscous fluids or 10 different
viscous fluids. Now, here since only we have saying that there
is a relationship among these two groups, we can choose the fluid to be anything that
is conveniently available to us, for example we may use water or air which is conveniently
available. So, we can use the same fluid, we can just use the same sphere and we can
make it fall at a different velocity move at different velocity or we can use the same
velocity and so, when you choose the given fluid fixed for a given fluid.
The viscosity and density are already fixed, if you choose let us air or water, so the
only parameters you have in order to generate 10 data points are either the diameter of
the sphere or the velocity at which the sphere is moving. So, depending on experimental convenience
we can choose either one of this and generate 10 data points just by varying let us say
the diameter of the sphere. Let us say we cannot change the velocity for
some reason. We have experimental set of where the sphere is moving only at a given velocity
let us say 10 meters per second or 1 meter per second. Then, the only variable that we
have to vary is the diameter and let us say we choose steel balls of different diameters
10 different diameters. So, we can achieve, we can easily carry out this experiments and
write everything in terms of this non dimensional groups.
Now, once you find this non dimensional functional relation between this non dimensional group
which is a non-dimensional force and this non dimensional group which is essentially
the Reynolds number, which we pointed out in the context of transition from laminar
to turbulent in a pipe. We will come to this shortly the interpretation
of Reynolds number, but once you find this function relationship between two these two
non-dimensional groups, then in principle you can find the force the dependence of drag
force on any sphere in a fluid with any any property, any velocity, any viscosity and
any density and this sphere velocity can also anything and sphere diameter can be anything.
So, but so, let us say now, we have found that F by rho V square D square is a function
of rho V D by mu and let us say we have generated, how F varies with F by rho V square D square
varies with rho V D by mu. Let us say we have this data points, let us say this is the data
experimental data. Now, if somebody comes and tells you that, they have the motion of
a sphere and let us say we have generated this experimental data with water and a steel
bowls of some let us say millimeter diameter. If somebody comes and says that, they want
to find out the drag force on a sphere of a diameter, let us say 10 centimeters and
the more the fluid in which it is flowing is not water, but it is very viscous oil.
Then, how are we going to do that how are we going to compute the drag force.
All we want to know is what is the velocity at which the sphere is moving? We need to
know this, because if somebody wants to compute the drag force on an object, they should first
tell you what is the velocity at which the object like the sphere is moving? So, somebody
is telling you that a steel ball of 10 centimeter of diameter is moving in a very viscous liquid
like castor oil. And diameter is 10 centimeters and the velocity
is given to us, let us say it is 1 centimeter per second. Then, we can compute the Reynolds
this group, because we know the density of the once you say it is a given oil then, viscosity
and density are fixed, we know the diameter of the sphere, we know the velocity of the
sphere. Then, we know this group, so let us say that
value of the group is here. Now, we have to simply walk across to the y coordinate and
find what is this group?
So, what this graph is going to tell you what is F by rho V square D square? But we already
know what is rho which is the density of the castor oil which, already know D which is
the velocity with which the big sphere is moving and we also we already know the diameter
of the sphere and this is the only unknown, which can be computed.
Because, we already know what is this combination F by rho V square D square from this graph,
but this graph was obtained for a completely different system. It was obtain for motion
of let us say 1 mms or 5 mms steel ball in water and that is how you made this graph
by suitably non dimensional zing the variables. But now, once you have done that that graph
that you obtain and which you plotted in terms of these non dimensional groups is valid for
any problem provided the fluid is Newtonian. And you can find out the drag force on any
object of any dimension moving through any fluid with any viscosity. So, that is the
power of dimensional analysis, that first of all it is able to reduce the amount of
experimentation. Secondly, now you are able to do experiments
on things, that are easily available in a laboratory, let us you have a very tiny steel
ball of 5 m diameter and you have water readily available in a laboratory. You can do the
experiment, but or you have to report the experimental data or reanalyze or regroup
the experimental data in terms of these non-dimensional group F by F divided by rho V square D square
and rho V D by mu. Once you do that, then the results of such
a graph are valid for any sphere of any diameter moving with any velocity in any fluid, so
long as the fluid is Newtonian because in your lab you have used water as the only as
the as the test fluid to get these results, so this is once very very important simplification
that one gets.
Now, suppose let us say you do not have the data for F, you have not done many experiments
to get this data, but still you you want to know, what is the drag force on sphere with
diameter D 1, let us say D prototype and the velocity of this sphere is V prototype moving
in a liquid of viscosity mu prototype and density rho prototype.
What is the drag force experienced by this? Let us say, this is a very very huge sphere
moving in a very viscous liquid, we want to know this, but we do not have access to the
data like before, we do not have have this non dimensional relationship between F by
rho V square D square and rho V D by mu. Suppose you will not have this how are we going to
estimate this drag force, we want this, we want to compute this in some application.
Now, in order to this we can scale down the problem in the following way. First compute
the Reynolds number rho p V p D p of the prototype, this is the non-dimensional group call the
Reynolds number which I have we which we have encountered before, first compute the Reynolds
number of the prototype.
Now, construct a laboratory model construct a laboratory model in which, you choose a
liquid that is conveniently available to you a model liquid rho m with density m, you choose
a velocity that is feasible to you, choose a sphere of diameter that is feasible to you
in such a manner that this Reynolds number is identical to the prototype Reynolds number.
R e model should be equal to R e prototype that is you should choose these values rho
V D mu such that, with the Reynolds numbers of the model and prototype are the same. Then,
what dimensional analysis is going to tell you, which we will show shortly that, F by
rho V square D square is a function only of rho V D by mu.
If that is the case, then once you fix the Reynolds number. So, the non-dimensional force
drag force on an object is a function only of Reynolds number. Since we have fixed the
Reynolds number of model to be the same as the Reynolds number of prototype, what this
equation is telling us is that F by rho V square D square for the model should be the
same as F by rho V square D square for the prototype.
Now, for the model we can measure the drag force, so this becomes F model by rho model
V model square D model square, so you measure F for the model, you measure this, you measure
this. Now, this should also be equal to F prototype by rho prototype V prototype square
D prototype square. Now, we know, what is the density of the liquid in the prototype
in the real situation, we know, what the velocity with the sphere is moving, we know, what is
the diameter of the prototype sphere particle. So, since and we also know, the density of
the model liquid D velocity at which you are making the sphere move in the model and the
diameter of the sphere, so the only unknown can be computed.
So, you need not have the entire data as to how F by rho V square D square depends on
the non-dimension group rho V D by mu that is Reynolds number. So, long as we know what
is the Reynolds number at which the prototype is operating and if you match the Reynolds
number of the prototype with the Reynolds number at which, you are carrying out experiment
in the in the lab and in which you can easily measure the force, then it is much easier
to compute the force that will be experience the drag force that we experienced by the
prototype particle just from this relation. And why is this possible all this is possible
is, because of dimensional analysis because dimensional analysis tells us that, the functional
relation between F rho V D mu can be written as F by rho V square D square is function
of rho V D by mu, this is made possible by dimensional analysis.
So, first thing many many simplifications come by dimensional analysis, one is the notion
of reducing the number of experimentation to obtain a functional relationship between
many variables, secondly, once you once dimensional analysis tells us that, the relationship between
the drag force and various parameters is only through this non dimensional form then, in
order to find the force on a real situation, a prototype situation where as a larger sphere
is moving let us say through a very viscous liquid, all we have to do is find the Reynolds
number of the prototype. Now, do a similar experiment in the lab, by
matching the same Reynolds number, but by using fluids and spheres that are readily
available to us and velocities that are easily feasible in the lab. As long as you match
the Reynolds number then the non-dimensional force on the model on the prototype must be
identical, because at a given Reynolds number there is a unique relation between these two
non-dimensional groups. So, once I know what is the non-dimensional
force, this group on the model then, that should be the same on the prototype then by
knowing the velocity prototype diameter of the prototype and density of the prototype
liquid, we can calculate what is the drag force experienced by a prototype. So, this
is the most important lesson that one gets from dimensional analysis. So, far we have
told you the advantages and consequences and applications of dimensional analysis, but
we have yet not yet told you how to achieve this relation, this non dimensional relation.
How to get this? Now, that is what we are going to do next and that is through a principle
called Buckingham’s pi theorem. Suppose, you have a functional relationship among 5
variables, let us say the functional relationship is denoted by this small letter g, so we have
this 5 variables force, diameter, density, velocity, viscosity. And this is the function
we want to determine through experiments. In general this is the functional relationship
In general I can generalize this to a functional relationship among n variables, so now I am going to generalize the discussion
and not restrict ourselves to the drag force, but of course, from time to time I will draw
comparisons between the general formulation as well as the specific problem of drag force
on a sphere. So, in general let us say there are q 1 q
2 q 3. So, on up to q n variables and so, there are n dimensional parameters and you want to find a relationship among
these n dimensional parameters. Now, what the pi theorem tells is is that, you can write
this functional relationship among n dimensional parameters as a functional relationship among
n minus 1 non n minus m non dimensional groups.
Where pi 1, so let me explain the notation pi 1 pi 2 pi 3 and so on up to pi n n minus
m, they are non-dimensional groups, they are non-dimensional groups they are dimensionless
groups in other words; and m is the number of fundamental dimensions in the problem.
What do you mean by number of fundamental dimensions? The fundamental dimensions is
in any physical problem are mass, length, time, temperature and so on, so these are
the fundamental dimensions, present in a problem in a physical problem; so, an n m is the minimum
number of fundamental dimensions fundamental dimensions present in the problem, used to describe all the parameters.
So, let us now go to the sphere problem sphere drag problem, you had how many variables you
had g as a function of F V D mu rho is 0, so you had n equals 5 variables and the fundamental
dimensions are mass, length and time. If you look at dimensions of all the parameters then,
you will find that there are only 3 fundamental dimensions, so m is 3 with these 3 fundamental
dimensions, we can describe the dimensions of all the variables, there is no need to
go. This is the minimum number of fundamental
dimensions, that is that are used to describe all the variables dimensions of all the physical
variables present present in the problem, so that is 3 for this case, so what pi theorem
is telling us is that the number of dimensionless or non-dimensional groups is 5 minus 3 is equal to 2 and those two groups
an illustration or an…
An example of those two groups is what are what we wrote before F by rho V square and
rho V D by mu, examples of such two groups, non dimensional groups are these things, are
these two groups. So, the pi theorem for the problem of drag
force on a sphere tells us that, the functional relationship among 5 dimensional physical
variables namely F, V, rho, mu, D can be represented as a functional relationship among only two
non-dimensional groups and the pi theorem will also tell you, how to derive those two
non-dimensional groups. So, but again the pi theorem or dimensional
analysis does not tell you what is functional, what is the nature of the function between
the two non-dimensional groups, that is not told by dimensional analysis, that still has
to be carried out, that still has to be determined only through experimentation, but it does
the dimensional analysis does result in a great amount of simplification by reducing
the number of variables in the problem by from 5 to 2, initially we had 5 dimensional
variables, now we have only two dimensionless or non-dimensional variables.
Now, these n minus m pi groups the non-dimensional groups are called sometimes called as pi groups are independent,
in the sense you cannot get one pi group by combining the other pi groups, once you have
found n minus m pi groups for example, here you have these two pi groups, you cannot get
one pi group from another they are functionally related, but you cannot just manipulate one
from that, they are independent because ultimately you will have to experiments to find how this
is function of that. All I am saying is that you cannot write in
a more complicated problem, you cannot write pi 1 as pi 2 by pi 3, because you have pi
1, pi 2, pi 3 as three independent groups, you cannot get one in terms of other readily.
So, So, all these three all in general n minus m groups are independent groups in that one
cannot write one in terms of the other. Now, suppose you do find that after doing experiments.
Suppose, you find that you have two only two groups and let us say you have pi 1 as a function
of pi 2 and after doing experiments, you find that pi 1 is equal to 1 over pi 2, this implies
that pi 1 pi 2 is equal to 1 or some constant over pi 2, is this constant. We will show
later that this implies that you have overestimated the number of physically relevant dimensional
variables in the problem; and one of those dimensional variables will automatically drop
out of the functional relationship if it, so happens.
If the physical circumstances are such that one of the variables is irrelevant then, you
will find that these two groups are not they are not simply they are very very simply related
and you can combine these two groups to get a third non dimensional another non dimensional
group and you will find that, one of the dimensional variables becomes an irrelevant variable in
the problem, so I will point point this out little later when we actually when we implement
dimensional analysis.
But now let us go through how to determine the pi groups, so all I have said is a claim.
All I have said is a claim, that given a set of m dimensional variables and you have given
a set of n dimensional variables and you have m fundamental dimensions minimum number of
fundamental dimensions required to describe the dimensions of those n dimensional variables
then Buck Ingham’s pi theorem tells you that, you can reduce these n variables to
n minus m non dimensional groups, now and these groups are often called as pi groups
traditionally, historically. How are we going to determine these pi groups?
So, there is a well set methodology set of rules, that one follows and one can easily
get the pi groups, let me go through it step by step. First list all the dimensional parameters
involved in the problem in the problem, let it be n. Now in the case of drag force past
a sphere we said that force is the of course, objective is to get the force, it could be
function of velocity, diameter, viscosity, density; but we neglected the interfacial
tension and we probably neglected the specific capacity things like this because, we felt
that those variables are irrelevant to the problem.
Now, how are we going to check whether the hypothesis is correct, it will be proved right
by experimentation only and right now it is purely a physically motivated guess that we
are saying that these are the physical variables that are relevant to the problem.
Now, initially the students may find it difficult to find out to to judge, what are the physically
relevant variables to the to a given problem, but it is a matter of physical and engineering
judgment as to what these are, but with practice and with experience then one gains enough
experience to write down, what are the relevant dimensional groups that that affect the given
problem. Now, the next step is select a set of fundamental
dimensions, for example mass, length and time. Mostly in our case, we can choose mass length
and time as a set of fundamental dimensions; occasionally people may use instead of mass
a force as a fundamental dimension, but for all practical purposes you have to use only
mass length and time in fluid mechanics problems. Suppose, you are doing heat transfer then
you may have to add temperature as a fundamental dimension because within the realm of classical
thermodynamics temperature is another fundamental dimension. Of course, if you go to molecular
theories such as statistical thermodynamics or kinetic theory then, temperature is essentially
measure of average kinetic energy of molecules, so that becomes not a fundamental dimension,
but within the realm of continuum classical thermodynamics, it is a fundamental dimension,
so in heat transfer we may have to use temperature also.
Now, let let this be m, the number of fundamental dimensions. Now, we have to the third step
is to list the dimensions of the n physical variables in terms of fundamental dimensions. So, we
have to write, what is the dimension of suppose you choose mass length and time, we have to
first write what is the dimension of force in terms in m l t dimensions and so on.
Now, the forth step is to select m parameters, dimensional parameters, that include all fundamental
dimensions all three dimensions. Not all three in general all m and of course, m is less
than n, because there are n numbers of physical variables; and out of these n numbers of physical
variables you are choosing m physical variables, which have all the three fundamental dimensions.
Now, the next step is to construct the fundamental dimensions, in terms of the m, these are called
the repeated variables, so let us give a name to them these are called the repeating variables,
in terms of the m repeating variables, once you do this you take the remaining n minus
m dimensional variables and non-dimensionalize them non-dimensionalize them with the dimensions constructed out of
from step 5. You have already constructed all the physical
fundamental dimensions in terms of m repeating variables, so all you have to do is take the
remaining variables and none dimensionalize them and finally, check whether if you have
done everything correctly this n minus m dimension groups will be purely non-dimensional, they
will not have any dimensions. We will stop at this point and we will illustrate this
method in the next lecture for drag force fastest sphere. Thank you.