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Welcome back to the video lecture on fluid
mechanics. In the first lecture, we have discussed
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the introductory aspects of fluid mechanics
We have discussed about the various fluid
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properties, various theories used in fluid
mechanics and how it will be used in our this
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video course; we discussed the various fundamental
theories which applicable in the case of fluid
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mechanics; we discussed the various flow visualization
techniques, used in fluid mechanics and the
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flow lines like stream lines, path lines and
streak lines. Today, in this lecture, first
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we will discuss about the classification of
fluids.
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As I mentioned earlier, the fluids can be
classified according to various fluid properties,
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fluid behavior or the dimensions on which
they will be dealing with the fluids or the
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fluid problem which we are dealing. So, mainly
the fluids fluid flow can be classified according
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to the rhelogical consideration, spatial dimensions,
dilational tensor, then motion characteristics,
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the temporal variations and fluid types. Here,
we will discuss in detail about the various
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types of fluid flow according to various aspects.
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As I mentioned earlier, the fluids can be
either gases or liquids. The first classification
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of fluid flow is whether it is gas gaseous
flow or liquid flow. The first classification,
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as you can see in this slide say, the fluid
flow can be either gaseous flow or say liquid
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type fluids. Secondly, according to whether
the fluids can be compressed or whether it
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cannot be compressed say, accordingly we can
classify the fluids in to compressible fluids
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and incompressible fluids; mainly, the gases
which will be generally dealing with will
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be compressible in nature and some of the
liquids are may be some compressible depending
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up on the type of the fluids and some other
liquids we can consider as incompressible.
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Even though sometimes we can compress to every
small level but still it can be considered
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as incompressible. For example, water is considered
as incompressible even though, may be to certain
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level it can be compressed by applying the
pressure.
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As I mentioned, say, the fluid flow or fluid
can be again classified. Fluid flow can be
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classified as steady state fluid flow or transient
fluid flow. The fluid, since it is deforming
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with respect to space and time, if it is with
respect to time there is no variation at all,
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then the fluid flow is said to be steady.
That means with respect to time, there is
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no variation of the fluid properties but the
fluid properties like velocity or depth a
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flow all this fluid properties are constants
with respect to time and there is no variation
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with respect to time.
In the case of UN study flow or transient
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flow the fluid property like velocity, pressure
and depth, all these parameters are varying
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with respect to time. So, these types of fluids
are called unsteady fluid flow or transient
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flow. As we discussed earlier the viscosity
is an important fluid property. So, accordingly,
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whether the fluid flow which we are dealing
with has got viscosity or the viscosity is
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negligible or there is no viscosity, then
accordingly we can classify the fluids into
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viscous fluids or viscous flow and non-viscous
flow or invest flow. So according to the variation
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or according to the fluid has viscosity fluid
is called viscous fluid flow and then if viscosity
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is not considered then that kind of fluid
is called non-viscous or invest fluid flow.
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If there is any circulation or any rotation
aspect as far as the fluid flow is concerned
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we can classify the fluid into rotational
fluid or irrotational fluid. In the case of
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rotational fluid, there can be circulation
there can be vorticity and all other parameters
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but as far as irrotational fluid is concerned
circulation is not there and vorticity is
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0. So, those fluids are called irrotational
fluid or irrotational fluid flow.
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According to the space we are considering
3 dimension, x, y and z. We can classify the
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fluid as 3 dimension fluid flow or 2 dimension
fluid flow or 1 dimension fluid flow. As all
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of you know fluid flow is definitely 3 dimensional
in nature as you can see any type of fluid
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flow is 3 dimension nature.
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But many times, we can consider 3 dimension
fluid flow as 2 dimensions, for example: here
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you can see say, a river. We are considering
a river section like this; if you are considering
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here as z axis, this as x axis, here say,
y axis, so this is xyz axis. Here you can
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see that when we consider the fluid flow with
respect to this river section like this x
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verses z, then this other longitude dimension
is not considered. So, in this case, the fluid
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flow can be considered as 2 dimension; we
may be interested what is happening with respect
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to this x direction and z direction even though
the fluid flow is definitely 3 dimensional
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in nature. We are simplifying the 3 dimension
fluid flow into 2 dimension since it will
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be very easy to analyze and when interrupt
the results also it will be very easier. So,
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depending upon the case, depending upon the
problem and depending upon the fluid flow
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problem we can consider a 3 dimensional flow
as 2 dimensional. So that the accuracy of
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the results are not much affected.
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We can consider the river flow in some times
as 1 dimensional flow say for example, if
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you are considering the river is flowing like
this. In this case, we are dealing from one
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section to another section, say, here 1 1,
2 2, 3 3; we are dealing from what is happening
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from section one to section two and section
three; in this case, obviously, the fluid
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flow is 3 dimensional in nature. We can consider
the flow as what is happening with respect
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to this longitudinal direction as far as this
river flow is concerned. We can consider in
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this case, the fluid flow as 1 dimension.
All the fluid flow which we are dealing with
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is 3 dimensional in nature but depending upon
the problem to simplify the problem we can
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consider the fluid flow as 2 dimension or
whether we can consider sometimes as 1 dimension.
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So this simplification makes the problem analyze
much simpler and the interpretation of results
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also much easier without losing much across.
Most of the problems even though 3 dimensional
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in nature, we may be solving as 2 dimensional
fluid flow problem or 1 dimensional fluid
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flow problem.
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As we have seen the previous slide, the fluids
can be either gases or liquids based up on
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the molecular behavior as you can see in this
slide. Then the second case: the fluids when
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we are considering continuum or discrete fluids.
A continuum means individual molecular properties
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are negligible so that we are considering
as a continuum. In most of the fluid flow
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problems you will be considering as continuum
and in other aspects, we can also consider
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discrete fluid; in the case of discrete fluids,
each molecule is treated separately and then
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what is happening with respect to that molecule
is studied. So, two aspects are there in fluid
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flow analysis: the continuum analysis or discrete
fluid flow analysis. Most of the time we will
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be dealing with the continuum fluid flow analysis
and in very rare cases we deal with the discrete
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fluid flow analysis.
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In third case, we can again say the fluids
can be either perfect your ideal fluids or
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the fluids can be real fluid. Real fluids,
as I mentioned earlier does not slip past
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a solid wall and most of the fluids which
we are dealing with is real fluids. So, the
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third case is perfect verses real fluids.
Te fourth case, as I mentioned is Newtonian
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verses non Newtonian fluid. In the Newtonian
fluids we consider the qualification of viscosity
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than on coefficient of viscosity mu constant
for a fixed fluid temperature and pressure.
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For example, water is considered as Newtonian
fluids and then we have already seen earlier
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non Newtonian fluids, when the shear stress
verses shear strain various is not linear
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those fluids are non Newtonian fluids and
then the non Newtonian fluids as I mentioned
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in the mu varies, say, for example, milk is
a non Newtonian fluid
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In the fifth case the fluid can be classified
as compressible or incompressible fluid. As
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we have seen earlier in the compressible fluid
the density changes with applied pressure;
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if you will apply some pressure to the fluid
the density changes., for example we consider
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just some air then we can see in a container
if you put some pressure in this container
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the density inside changes, so this is what
is called compressible fluid. In incompressible
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fluid, density will not change by external
force, so this kind of fluid is called incompressible
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fluid, say, for example, water. When we apply
some pressure the density is not changing
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accordingly, so that type of fluid is called
incompressible fluid. The sixth classification
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as I mentioned in the fluids can be classified
as that the steady state or the unsteady.
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In the steady fluid flow the properties like
velocity, the pressure or independent of time
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or in unsteady fluid flow the property is
depending on the time.
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We have already discussed the 1 dimensional,
2 dimensional and 3 dimensional flows according
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to the spatial variations. In the eighth classification,
we have already seen the rotational verses
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irrotational flow. Irrotational flow- there
is no rate of angular deformation of any fluid
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particle, for example, potential flow can
be considered as irrotational flow. The rotational
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flow-there is a rate of angular deformation
with respect to the fluid flow. These are
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some of the important classifications of fluids
and fluid flow used in fluid mechanics. Based
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up on these classifications the study of the
fundamental principles also slightly varies.
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In the coming lectures, we will be discussing
in detail with respect to various types of
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fluids; also how it will be dealing, what
would be the principle, what are the changes
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as per the principles compared to the incompressible
real fluids or compressible fluids. That we
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will be discussing in later lectures. Now,
finally in this introductory lecture we will
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discuss how we can solve a fluid flow problem.
If we get a fluid flow problem how can we
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solve the problem? The problem may be to find
out the velocities or the problem may be to
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find out the pressure distribution or the
viscosity changes or density changes or any
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type of engineering fluid flow problem. If
we get how we can approach the problem and
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then how we can solve the problem, these questions
we will be asked generally.
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Here this flow chart briefly indicates how
the fluid flow problem can be solved. In general,
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as for as fluid flow analysis is concerned
there can be two types of investigations:
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first one is it can be a theoretical investigation;
second one is the experimental investigation.
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In theoretical investigations, what we are
doing is based up on the available theory
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or based up on the fundamental principles
or based up on the available literature published
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papers we are trying to approach the problem
and we are trying to solve it theoretically;
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it may be theoretical background or some exact
solutions are available or analytical solutions
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are available or it may be some numerical
metrologies which we can directly apply for
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the problem. This is actually a mathematical
way of approach since we are trying to solve
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the problem theoretically.
In case of excremental investigations, what
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we are doing? We are trying to replicate the
problem in the laboratory to certain scale
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so a scaling is to be done; most of the real
field problem will be very large in nature;
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we cannot replicate the same in the laboratories.
What we can do is that we can reduce the dimensions;
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we can change it to a certain scale so that
the problem will be small in nature and we
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can easily deal in the laboratory. So, experimental
investigation is very much used in fluid mechanics.
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In the last two hundred to three hundred years
experimental fluid mechanics is very much
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developed in the solution of various problems.
As far as the experimental investigations
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are concerned there are certain limitations.
The limitations are: it is very expensive
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in nature and we have to do the scaling; through
the scaling certain aspects of the problems
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may not be able to replicate in the laboratory.
So certain level of the reality of the problem
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may be last in the replications of the laboratory
and then running the model in the laboratory,
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the accuracy may be reduced. So, experimental
investigations have certain limitations but
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if theoretical investigations or analytical
solutions are not available then it has got
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some limitations, the accuracy again may be
reduced. Now, as far as theoretical investigations
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are concerned how are we approaching the problem?
If the engineering fluid flow problem is given
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then first we can understand the problem in
a physical nature, what is the problem, what
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are the inputs data available and what are
the outputs expected what the physical principles
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are based up on the problem which we have
to investigate. So this analysis is called
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the physical analysis.
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The physical analysis can be done based up
on the representation of the problem by a
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client or so; we can go to the real problems
for example, if we are constructing a bridge
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across a river then
we have to construct pillars and then we have
to understand the flow behavior across the
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river section so that we have to do fluid
flow analysis as far as the section is concerned
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where the bridge is constructed. We may go
to the field and see how the river is flowing,
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what are the important parameters which we
have to deal here. We have to analyze and
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we have to understand the problem physically.
So we may have to go to the field or we have
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to analyze in our office depending up on the
data available given by the client.
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For example when we are going to construct
a bridge across the river, we have to see
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depending up on the monsoon or depending upon
the rain fall through out the year how the
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water level will be varying, how it is increasing
or how it is decreasing and then what will
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be the flow velocity at the particular location.
We are going to construct the bridge the bridge
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and then the bank of the river has any motion
problems or all these things we have to analyze
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physically. This is called physical analysis.
In the case of fluid mechanics analysis this
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physical analysis can be based up on the force
concept or energy concept. The force concept
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means we will analyze what are the forces
acting on the particular problem which we
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have dealing with and then how it is going
to affect the fluid flow properties or the
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particular variable which we are going to
understand and the second concept is called
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energy concept, how energy variations are
like here as they would be the energy creations;
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we will be analyzing how it will be affecting
the particular problem so the physical analysis
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is very important in the solution of any fluid
flow problem. Physical analysis generally
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can be there upon the force concept or based
up on the energy concept. After this physical
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analysis is over we can understand the problem
in which way we have to approach the problem
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and we can try to make a mathematical formulation
of the problem. After this physical analysis
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next step is the mathematical analysis. In
the mathematical analysis with respect to
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the physical analysis we already know what
will be the domain of the problem which we
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will be dealing.
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As I mentioned, we are going to construct
a bridge across the river. So when the pillars
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will be constructed what will be the flow
conditions? All these things will be analyzed.
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We know the domain of the problem and then
also we can easily identify the data. For
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example, flow of the variations at various
seasons with respect to throughout the days
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of the year how the flow depth will be varying.
All these data are available; then the boundary
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conditions with the problem is concerned at
particular location of the domain say the
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boundary, then how the flow will be the conditions
like depth variations or velocity variation
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can be identified and then based upon the
particular problem we can get the mathematical
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equations; mathematical equations can be derived
based upon the theoretical analysis based
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upon the first concept or the energy concept.
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So, once the mathematical equations are derived
and then the boundary conditions are required
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for unsteady state problem or transient problems.
But in the case of transients problems or
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unsteady state problems you have to also prescribe
the initial conditions, time t is equal to
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0 but the conditions may be the depth, velocity
or pressure all these parameters will be varying.
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From time t equal to 0, transient analysis
or unsteady state analysis of the fluid flow
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problem is starting. This initial condition
is very much important in a fluid flow analysis.
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The given equations are prescribed; we have
already prescribed the boundary conditions
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of the problem; with respect to this the mathematical
analysis is complete. Again with respect to
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a figure here I describe, for example, if
you consider the potential flow problems,
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it can flow in a homogeneous isotropic axis
which is given by del square phi is equal
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to 0, the Laplace equations. If you consider
a rectangle domain like this then we know
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the length and the breath. So now the dimensions
of the domain are known and the domain nature
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00:23:51,559 --> 00:23:58,309
is known. Now, for example we deal with the
potential problem, the genuine equations is
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00:23:58,309 --> 00:24:05,309
del square phi is equal to 0, the Laplace
equation. This is the steady state problem.
201
00:24:05,929 --> 00:24:12,929
So, the boundary conditions can be the potential
phi is prescribed as boundary phi is equal
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00:24:18,100 --> 00:24:25,100
to phi1 here and phi is equal to phi2 here.
So the boundary condition on this way in this
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00:24:25,610 --> 00:24:32,610
domain is ABCD. In the particular case, the
boundary conditions on AD is phi equal to
204
00:24:39,309 --> 00:24:46,309
phi1and similarly, on BC we can prescribe
phi is equal to phi2. Now, the given equations
205
00:24:49,600 --> 00:24:54,269
are known and the boundary conditions are
known. On boundary AB, for example, if we
206
00:24:54,269 --> 00:25:01,269
know the variation or the flux, del phi by
del n which is equal to 0 on the boundary
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00:25:08,529 --> 00:25:15,529
AB and CD, here del phi by del n is equal
to 0. That means no flow boundary condition
208
00:25:17,419 --> 00:25:24,419
on AB and CD. So there cannot be flux and
cannot cross that means this side AB and CD
209
00:25:25,840 --> 00:25:29,630
are imperil in nature.
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00:25:29,630 --> 00:25:36,070
These boundary conditions are called nature
boundary conditions that mean, what exists
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00:25:36,070 --> 00:25:42,139
in nature. So this flow cannot cross this
can imperil the boundary this side AB and
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00:25:42,139 --> 00:25:49,139
on this side CD, so these conditions are called
nature conditions or and the boundary conditions
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00:25:50,759 --> 00:25:57,320
which we prescribed on AD and BC are called
divisional boundary conditions or direct boundary
214
00:25:57,320 --> 00:25:59,970
conditions.
So as far as this potential flow problem is
215
00:25:59,970 --> 00:26:06,970
concerned the mathematical analysis or mathematical
statement is over since we have only described
216
00:26:07,149 --> 00:26:11,649
the domain, we have already described the
boundary condition and we have already described
217
00:26:11,649 --> 00:26:18,149
the given equations. Now, with respect to
this, we can solve this problem either we
218
00:26:18,149 --> 00:26:23,519
can get analytical solution depending upon
the problem or we can solve the problem numerically
219
00:26:23,519 --> 00:26:26,200
so depending upon the case.
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00:26:26,200 --> 00:26:32,880
This is about the mathematical analysis and
as I mentioned in the experimental investigation
221
00:26:32,880 --> 00:26:39,880
we have to replicate the real problem in the
laboratory and then with respect to certain
222
00:26:41,570 --> 00:26:48,570
scale we will be running the experiment what
is happening in the field that will be done
223
00:26:48,950 --> 00:26:55,299
in the laboratory. Then we will be taking
some readings sometimes it can be say either
224
00:26:55,299 --> 00:27:01,690
the velocity measurements are the depth variations
or pressure measurements or this measurements
225
00:27:01,690 --> 00:27:07,649
will be done experimentally in the laboratory.
Then to have some specific relationships we
226
00:27:07,649 --> 00:27:12,960
can give some dimensional analysis as mentioned
in this slide here. Dimensional analysis is
227
00:27:12,960 --> 00:27:17,399
also very important in experimental investigation
as well as mathematical investigations. So,
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00:27:17,399 --> 00:27:24,019
dimensions with respect to the bucking phi
theorem or various theorems, we will discuss
229
00:27:24,019 --> 00:27:26,399
about this dimensional analysis later.
230
00:27:26,399 --> 00:27:33,399
In summary, in this introductory lecture on
fluid mechanics, we have discussed the various
231
00:27:39,039 --> 00:27:46,039
aspects of fluid flow and the various fluid
properties. We have discussed the fundamental
232
00:27:47,090 --> 00:27:52,799
principles which will be used in the fluid
flow analysis like the mass consideration
233
00:27:52,799 --> 00:27:58,840
of momentum, consideration of energy and then
we also discussed the various fluid visualization
234
00:27:58,840 --> 00:28:05,840
techniques like using dies, using smoke or
other metrologies. We have discussed various
235
00:28:13,159 --> 00:28:18,980
metrologies of representing of a fluid flow
like a stream lines, potential line, streak
236
00:28:18,980 --> 00:28:25,980
lines path lines will have discussed and then
we have discussed based up on the various
237
00:28:28,279 --> 00:28:34,450
fluid properties we have classified the flows
with respect to space, one D flow, two D flow
238
00:28:34,450 --> 00:28:41,409
and three D flow or with respect to the viscosity,
that is, viscous flow or non-viscous flow
239
00:28:41,409 --> 00:28:47,440
and then finally, we have discussed about
how we can approach a fluid flow problem or
240
00:28:47,440 --> 00:28:51,740
solution to a fluid flow problem.
As I mentioned, we can approach the fluid
241
00:28:51,740 --> 00:28:57,830
flow problem theoretically or experimentally
and then in theoretical investigation we can
242
00:28:57,830 --> 00:29:03,990
derive a mathematical model by describing
the domain, then given equation in boundary
243
00:29:03,990 --> 00:29:10,990
condition. So this is about the introductory
lecture on this video course on fluid mechanics
244
00:29:11,509 --> 00:29:16,450
In this second chapter on fluid mechanics
video course, we will be discussing mainly
245
00:29:16,450 --> 00:29:18,620
on fluid statics.
246
00:29:18,620 --> 00:29:25,620
The main objectives in this section is to
introduce the concepts of fluid pressure,
247
00:29:26,690 --> 00:29:33,690
forces on solid surfaces, buoyant forces and
related theories. Secondly, we shall see the
248
00:29:33,879 --> 00:29:40,279
determination of fluid pressure and forces
for various cases. Then we will emphasis in
249
00:29:40,279 --> 00:29:47,120
the importance of fluid statics as far as
fluid mechanics is concerned.
250
00:29:47,120 --> 00:29:54,120
Now, as I mentioned earlier, static fluids
means fluid is not moving that means fluid
251
00:29:57,629 --> 00:30:04,629
is in rest. So, you can see there is some
water in a small basin. Here fluid is now
252
00:30:09,960 --> 00:30:14,960
at rest; there is no movement as far as the
fluid is concerned; it is as far as the boundary
253
00:30:14,960 --> 00:30:21,960
is concerned the fluid is at rest so that
we can say the fluid is in statics conditions
254
00:30:23,169 --> 00:30:30,169
and the various theories on statics will be
applicable as far as a static fluid is concerned.
255
00:30:34,350 --> 00:30:41,350
In solid mechanics, most of the problems we
will be analyzing are in that condition. So,
256
00:30:43,700 --> 00:30:50,389
most of the theories in solid mechanics are
applicable for fluids at rest or static fluid.
257
00:30:50,389 --> 00:30:57,389
In static fluids you can see that since fluid
is not moving it is at rest here, so that
258
00:30:57,919 --> 00:31:04,919
there is no shearing force. Whenever the fluid
is trying to move only between the layers
259
00:31:05,200 --> 00:31:11,190
there will be shearing force but as far as
fluid is at rest or static fluid there is
260
00:31:11,190 --> 00:31:18,039
no shearing force. We need not have to worry
about the shearing force and now before going
261
00:31:18,039 --> 00:31:25,039
to more details of fluid statics we will just
review some aspects of forces.
262
00:31:27,750 --> 00:31:34,750
We can classify the forces into body forces
and surface forces. A body force acts through
263
00:31:42,750 --> 00:31:49,750
a distance, for example, when there is a gravitational
force the force of gravity is not directly
264
00:31:52,649 --> 00:31:58,059
acting, but it is acting from a distance so
the action is through the distance and also
265
00:31:58,059 --> 00:32:05,059
if you consider the electro magnetic force
most of the time it will be acting at a distance.
266
00:32:07,700 --> 00:32:12,529
This kind of forces like gravity force and
electromagnetic force acting on fluids and
267
00:32:12,529 --> 00:32:17,909
that kind of study which we are dealing will
be related to the body force. The second kind
268
00:32:17,909 --> 00:32:24,909
of force is called the surface force; it is
due to the virtue of direct or in between
269
00:32:25,460 --> 00:32:31,379
the molecules. What is happening as far as
the fluid is concerned is if you apply a force
270
00:32:31,379 --> 00:32:38,379
here on the fluids, then between the virtue
and direct contact between the fluid molecules
271
00:32:38,999 --> 00:32:45,309
only there is a force. That force is called
the surface force and as the stress is concerned
272
00:32:45,309 --> 00:32:50,950
we can also classify, as we have discussed
earlier, stresses as normal stresses and shear
273
00:32:50,950 --> 00:32:57,490
stresses. Normal stress means when we are
considering direct normal force on the body
274
00:32:57,490 --> 00:33:03,289
that kind of stress is called in a normal
stress and if a shearing force is applied
275
00:33:03,289 --> 00:33:09,249
that kind of stress is called shear stress.
We will be dealing with normal stresses and
276
00:33:09,249 --> 00:33:16,249
shear stresses in fluid mechanics and also
we will be dealing with xyz directions.
277
00:33:19,399 --> 00:33:26,399
Here, this is x this is y this is z. So the
normal stress can be in x direction, y direction
278
00:33:31,620 --> 00:33:36,649
or z direction or with respect to x and y,
with respect to x and z or with respect to
279
00:33:36,649 --> 00:33:43,649
y and z. Like that the stresses will be varying,
for example, if sigma is the stress in normal
280
00:33:44,669 --> 00:33:51,669
stress then if we consider a body like this
in 3 dimension then if you consider x, y and
281
00:34:04,820 --> 00:34:11,820
z the normal stress we would be considering
if it is acting from x direction, the normal
282
00:34:12,980 --> 00:34:19,980
stress can be sigma x and it is acting from
the y direction it can be sigma y and from
283
00:34:20,120 --> 00:34:27,120
the z direction it can be sigma z. Then with
respect to x y plane if it is acting, it can
284
00:34:27,440 --> 00:34:34,440
be sigma xy or sigma yx or if it is with respect
to same x and z axis, then it can be sigma
285
00:34:39,000 --> 00:34:46,000
xz or sigma zx; then if it is with respect
to y and z it can be sigma yz or sigma zy.
286
00:34:54,649 --> 00:35:01,649
Like that we can have stress tensor. So, with
respect to xyz direction we can classify the
287
00:35:02,770 --> 00:35:09,770
stresses in x direction, y direction and z
direction or xz xy or yz direction.
288
00:35:10,440 --> 00:35:17,440
Stress tensor is very much used in many of
the fluid analysis. In a very similar way,
289
00:35:17,720 --> 00:35:24,720
the shear stress is concerned for example
tau is the shear stress; if it is depending
290
00:35:26,060 --> 00:35:33,060
upon the problem if you consider again a cube
of three D then the shear stress also will
291
00:35:40,680 --> 00:35:47,680
be acting if a shear force is applied in the
other direction and the shear force is applied
292
00:35:48,780 --> 00:35:55,780
either so this other face also. According
to this we can have the shear stress in xy
293
00:35:57,109 --> 00:36:04,109
plane or tau yx then tau yz or tau zy or tau
xz or tau zx.
294
00:36:12,770 --> 00:36:19,770
According to the sheer force acting it will
be varying with respect to xyz. The stress
295
00:36:22,790 --> 00:36:29,760
tensor for normal stresses and shear stresses
will be utilizing the shears stress tensor
296
00:36:29,760 --> 00:36:36,760
with respect to xyz direction and then as
far as force is concerned, force will be especially
297
00:36:39,579 --> 00:36:46,579
if static fluid is concerned force will be
acting upon externally. So, the force between
298
00:36:48,490 --> 00:36:54,730
fluid and boundaries for example if you consider
it acting always at right angle boundary,
299
00:36:54,730 --> 00:37:01,730
so here in this slide you can see there is
a boundary here and then say the force is
300
00:37:01,740 --> 00:37:07,770
acting. With respect to the boundary generally,
the force will be acting at right angle so
301
00:37:07,770 --> 00:37:14,770
you can see that in the fluids that is the
fluid is at rest. When we are applying a force
302
00:37:15,540 --> 00:37:21,059
we will be considering this acting at right
angles and all the fluid static analysis will
303
00:37:21,059 --> 00:37:28,059
be accordingly. So as I mentioned, most of
the theories which have been developed in
304
00:37:28,170 --> 00:37:35,170
the cases fluid mechanics are also applicable
in the case of static fluid or fluid at rest.
305
00:37:35,430 --> 00:37:42,430
Fluid at rest, we can consider each element
for example, this basin water is at rest so
306
00:37:44,569 --> 00:37:51,569
when we consider the fluid statics or static
fluid here. If we consider each element then
307
00:37:52,540 --> 00:37:58,309
the fluid element should be in equilibrium;
there is no moment and the fluid is at rest.
308
00:37:58,309 --> 00:38:05,309
So, if there are n forces acting then finally
each fluid element should be in equilibrium.
309
00:38:07,309 --> 00:38:14,309
If you consider any solid with respect to
solid face, if the solid is at rest and in
310
00:38:16,609 --> 00:38:23,609
equilibrium then say we can consider with
respect to, for example, say this a solid
311
00:38:23,910 --> 00:38:30,910
and if it at rest and if some forces are acting,
we know that the body is in equilibrium then
312
00:38:30,920 --> 00:38:37,640
in solid mechanics we use the equilibrium
condition. For example, if the forces in x
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00:38:37,640 --> 00:38:44,130
directions sigma fx should be equal to 0;
this is one of the condition which will be
314
00:38:44,130 --> 00:38:51,130
generally used in solid mechanics and then
sigma fy is the forces in y direction should
315
00:38:52,900 --> 00:38:56,670
be 0 and then sigma fz should be 0.
316
00:38:56,670 --> 00:39:03,430
So, these three conditions as far as forces
are concerned, considering the solid equilibrium
317
00:39:03,430 --> 00:39:10,329
and then we consider each element in equilibrium,
so that we can say all the sum of the forces
318
00:39:10,329 --> 00:39:16,839
in x directions is equal to 0; sigma fx is
equal to 0; sigma fy is equal to 0 and sigma
319
00:39:16,839 --> 00:39:22,180
fz is equal to 0.
So these three equations as far as the all
320
00:39:22,180 --> 00:39:29,160
the brace, some of the forces sigma fx, sigma
fy sigma fz are used in the solid mechanics.
321
00:39:29,160 --> 00:39:36,160
These equations, the solid at rest, are applicable
for static fluid or fluid statics. We can
322
00:39:37,010 --> 00:39:43,790
use these equations sigma fx is equal to 0,
sigma fy is equal to 0 and sigma fz is equal
323
00:39:43,790 --> 00:39:50,790
to 0. Also in solid mechanics, we know that
for a solid at rest but in equilibrium condition,
324
00:39:54,520 --> 00:39:59,809
when there is a force acting then the sum
of the moments of force at any point must
325
00:39:59,809 --> 00:40:04,020
be 0 as far as the particular solid is concerned.
326
00:40:04,020 --> 00:40:10,440
This equation is also very much used in a
fluid mechanics for fluid at rest. If we consider
327
00:40:10,440 --> 00:40:17,210
that the sum of the moments, if you apply
any force and then its moment with respect
328
00:40:17,210 --> 00:40:23,990
to that force acting upon the fluid at rest
and then with respect to about any particular
329
00:40:23,990 --> 00:40:29,869
point sigma mx that means the sum of the moments
of force should be equal to 0.
330
00:40:29,869 --> 00:40:35,410
These fundamental equations or sigma fx is
equal to 0 and sigma fy is equal to 0 sigma
331
00:40:35,410 --> 00:40:42,369
fz is equal to 0, sigma Mx or the sum of the
moment of forces about any point must be 0.
332
00:40:42,369 --> 00:40:48,160
These four equations are very much valid in
the case of static fluid also. These equations
333
00:40:48,160 --> 00:40:53,690
generally, we use in solid mechanics and directly
these equations are applicable in the case
334
00:40:53,690 --> 00:41:00,690
of static fluid. Now, another important aspect
as far as fluid statics is concerned is the
335
00:41:02,300 --> 00:41:02,640
pressure.
336
00:41:02,640 --> 00:41:09,569
Pressure, as I mentioned earlier, pressure
is the average of normal stresses. So if there
337
00:41:09,569 --> 00:41:16,569
is force acting, force per unit area is termed
as pressure and generally we will be considering
338
00:41:19,280 --> 00:41:26,280
the average of normal stresses as bulk stresses
and that term is defined as pressure. It is
339
00:41:26,380 --> 00:41:33,290
the force per unit area and the general used
unit in the system international is one Pascal
340
00:41:33,290 --> 00:41:38,920
which is equal to Newton per meter square
and its dimension is M L to the power minus
341
00:41:38,920 --> 00:41:44,839
1 and T to the power minus 2 as shown in this
slide and say, for example, another unit which
342
00:41:44,839 --> 00:41:51,839
is for bar used is 1 bar is equal to 10 to
the power 5 Newton per meter square. This
343
00:41:53,299 --> 00:41:57,619
pressure is also very important as far as
fluids are concerned. We will be discussing
344
00:41:57,619 --> 00:42:04,440
further about the pressure shortly for fluid
in static state and another important aspect
345
00:42:04,440 --> 00:42:11,440
for fluid statics is equilibrium. Equilibrium
condition in solid mechanics which we generally
346
00:42:12,589 --> 00:42:18,559
use is stable equilibrium for example, this
cylinder here is said to be stable if it is
347
00:42:18,559 --> 00:42:24,020
place like this. So that it is stable; there
is no change; so it is when if we apply more
348
00:42:24,020 --> 00:42:30,540
force it will come back to its normal proportion
with respect to small forces. That is called
349
00:42:30,540 --> 00:42:36,780
stable equilibrium but if you consider this
cylinder if you put it in the other way like
350
00:42:36,780 --> 00:42:43,780
this then you can see with a small force it
is quit unstable, so this solid in this position
351
00:42:45,210 --> 00:42:51,670
it is said to be in unstable equilibrium;
then third equilibrium condition used is called
352
00:42:51,670 --> 00:42:58,670
neutral equilibrium. So when it is lying like
this on this flow it is a neutral equilibrium;
353
00:42:58,760 --> 00:43:05,160
whatever force is applied it comes back to
the same condition. So it is said to be neutral
354
00:43:05,160 --> 00:43:09,380
equilibrium. Equilibrium conditions namely
stable equilibrium, unstable equilibrium and
355
00:43:09,380 --> 00:43:15,380
neutral equilibrium are very much solid mechanics.
These equilibrium conditions are also very
356
00:43:15,380 --> 00:43:20,730
much valid in the case of statics fluid which
will be discussing later.
357
00:43:20,730 --> 00:43:27,380
Now, we will be discussing about the fluid
pressure. One of the important theories is
358
00:43:27,380 --> 00:43:34,380
developed in seventeenth century called the
Pascal law for pressure at any point. This
359
00:43:36,520 --> 00:43:43,520
law has been developed by balky Pascal by
various experiments. Here, if you consider
360
00:43:46,799 --> 00:43:53,150
the fluid at rest and now we are considering
this slide fluid element and then with respect
361
00:43:53,150 --> 00:43:59,240
to the fluid elements, Pascal offset with
respect to various experiment that the pressure
362
00:43:59,240 --> 00:44:02,470
at any point is same in old directions.
363
00:44:02,470 --> 00:44:09,210
We consider a fluid element like this; the
pressure at any point is same in all directions.
364
00:44:09,210 --> 00:44:15,530
For example, if you consider the water in
this basin, if you consider the pressure at
365
00:44:15,530 --> 00:44:22,530
any point in this; since the fluid is statics
there is no moment; if you consider the pressure
366
00:44:23,270 --> 00:44:29,440
the pressure will be the same old directions.
This is the law derived by Pascal and it is
367
00:44:29,440 --> 00:44:36,440
called a Pascal’s law. Here you can see
a fluid element in this slide. If you consider
368
00:44:36,559 --> 00:44:43,559
fluid element with respect to this phase,
the pressure will be py in this direction
369
00:44:44,180 --> 00:44:49,309
py into delta x into delta z; in this direction
this is the force and the other direction
370
00:44:49,309 --> 00:44:56,309
it is delta x into delta y and this direction
it is x direction which we are dealing is
371
00:44:57,900 --> 00:45:00,819
px into delta x into delta z.
372
00:45:00,819 --> 00:45:07,819
So, if you consider fluid element like this,
then according to the Pascal law when a certain
373
00:45:09,770 --> 00:45:14,869
pressure is applied at any point in a fluid
at rest the pressure is equally transmitted
374
00:45:14,869 --> 00:45:21,869
in all direction and to every other points
in the fluid. If you consider this with respect
375
00:45:22,240 --> 00:45:28,250
to the fluid element which we consider sigma
fy can be written as sigma fy is equal to
376
00:45:28,250 --> 00:45:35,250
py into delta x in to delta z minus ps in
to delta x delta x sin theta, that is, equal
377
00:45:36,150 --> 00:45:43,150
to rho into delta x delta y delta z by 2 into
ay, where rho is the density of the fluid
378
00:45:44,680 --> 00:45:51,680
and ay is the acceleration of the fluid element
which we are considering. Here, we are using
379
00:45:51,950 --> 00:45:55,480
the Newton’s second law: the total force
is equal to mass into acceleration.
380
00:45:55,480 --> 00:46:02,480
Similarly, in z direction, sigma fz is equal
to pz into delta x into delta y minus ps in
381
00:46:04,730 --> 00:46:11,730
to delta x delta x cos theta minus gamma delta
x delta y delta z by 2 since the weight of
382
00:46:12,900 --> 00:46:17,780
the element is considering this term which
is equal to the mass into the acceleration
383
00:46:17,780 --> 00:46:24,510
in z direction. So, here mass is rho into
delta x into delta y into delta z by 2 and
384
00:46:24,510 --> 00:46:31,130
az is the acceleration.
Now, if you simplify the force, sigma fy the
385
00:46:31,130 --> 00:46:37,460
total occupied force in y direction which
is equated to the mass into acceleration and
386
00:46:37,460 --> 00:46:43,589
sigma fz which is equated to mass into acceleration
in dz direction. So if you simplify this you
387
00:46:43,589 --> 00:46:50,589
will get py minus ps is equal to rho into
ay delta y by 2. Since, delta y can be written
388
00:46:52,430 --> 00:46:59,190
as delta x cos theta and delta z can be written
as delta x sin theta, the second equation
389
00:46:59,190 --> 00:47:06,190
can be written as pz minus ps is equal to
rho into az plus gamma into delta z by 2.
390
00:47:06,960 --> 00:47:13,960
Now, with respect to the Pascal law if you
consider the fluid at a particular point then
391
00:47:20,900 --> 00:47:26,930
you can see that delta x, delta y and delta
z will be tending to 0. So that we can see
392
00:47:26,930 --> 00:47:33,930
py is equal to pz is equal to ps as shown
in this slide. That means the pressure at
393
00:47:35,790 --> 00:47:40,470
any particular point is same in all directions.
So, here, if you consider the pressure at
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00:47:40,470 --> 00:47:47,470
particular point at all directions it will
be same according to the Pascal’s law
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00:47:49,750 --> 00:47:56,099
After discussing the Pascal’s law, we will
be discussing the pressure variations at various
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00:47:56,099 --> 00:48:03,099
vertical direction are shown in a direction
and then we trying to derive general equations
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00:48:03,450 --> 00:48:06,079
which can be applied in all the cases.
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00:48:06,079 --> 00:48:13,079
So, if we consider the vertical element cylinder
of fluid like this, on this phase the pressure
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00:48:20,349 --> 00:48:27,349
is even and the area is a, here the pressure
is p2 and the area of cross section is A and
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00:48:28,250 --> 00:48:35,099
the fluid density is rho and from the datum
this distance is z1 to this phase and z2 is
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00:48:35,099 --> 00:48:40,599
to the second phase; you can see that the
pressure decreases with height in vertical
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00:48:40,599 --> 00:48:47,599
direction. We can write the equation p1 into
A minus p2 into A minus rho g A into z2 minus
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00:48:50,450 --> 00:48:57,380
z1 that is equal to 0. So this gives the pressure
variation in the vertical direction.
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00:48:57,380 --> 00:49:04,380
Similarly, if you consider the pressure variation
in horizontal direction here a horizontal
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00:49:04,720 --> 00:49:09,910
fluid element, say, cylinder of fluid is considered
here and the area of cross section is same
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00:49:09,910 --> 00:49:16,910
A; the fluid density is rho and on this left-hand
side phase the pressure is p1 and the pleft
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00:49:19,569 --> 00:49:26,569
multiplied by A is equal to pright A. Since
it will be in horizontal direction on this
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00:49:29,109 --> 00:49:34,799
position in the total pressure the total force
is equal to this position, so that pressure
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00:49:34,799 --> 00:49:40,549
in horizontal direction is constant; accordingly,
the pressure variation in horizontal direction
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00:49:40,549 --> 00:49:42,180
is constant
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00:49:42,180 --> 00:49:49,180
So, with respect to this, if we consider the
pressure intensity of fluid, say, for various
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00:49:50,630 --> 00:49:57,230
shapes you can see that with respect to in
this slide the shapes are varying from the
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00:49:57,230 --> 00:50:04,230
rectangular, then the circular, then say conical
differentiates. Now, if you consider an excess
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00:50:05,030 --> 00:50:10,670
line this is at a height x, so here the fluid
pressure will be same as the surface of the
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00:50:10,670 --> 00:50:17,240
water; it is ps and then if you consider this
excess line for a fluid with specific gamma
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00:50:17,240 --> 00:50:23,589
you can see that the fluid pressure density
p is equal to gamma into h plus the fluid
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00:50:23,589 --> 00:50:30,589
surface pressure ps. So, this will be same
at all the points in this line excess. So,
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00:50:30,780 --> 00:50:36,020
this is the pressure intensity of fluid at
this particular section excess.
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00:50:36,020 --> 00:50:43,020
So, these principles we can utilize in many
areas. Now, if you consider a piston type
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00:50:44,450 --> 00:50:51,450
mechanism here, force verses area. If we consider
a small piston here and for small cylinder
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00:50:52,680 --> 00:50:58,119
here and larger cylinder which is interconnected
through in the mechanism like this, the pressure
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00:50:58,119 --> 00:51:04,819
density is same at level p and q. So, the
pressure intensity is at same the level p
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00:51:04,819 --> 00:51:11,079
and q. Pp is equal to Pq. This is the small
cylinder and this is the large cylinder of
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00:51:11,079 --> 00:51:16,380
area A2. Here, area of small cylinder area
force section is A1 and pressure intensity
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00:51:16,380 --> 00:51:23,380
is same p at this point and at this point
on lead of the cylinder and then you can see
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00:51:23,599 --> 00:51:27,920
that since the area of force section A1 is
smaller here and here the area of force section
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00:51:27,920 --> 00:51:34,920
is larger, if you get the total force f1 will
be equal to p into A1 on this phase here and
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00:51:36,780 --> 00:51:43,780
here it will be f2 is equal to p into A2.
Since the pressure intensity is same we can
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00:51:43,930 --> 00:51:50,630
transmit large quantity force. Here f2 is
equal to p into A2 is much larger compare
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00:51:50,630 --> 00:51:56,089
to f1. So, the transmission of pressure through
stationary fluid is possible with respect
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00:51:56,089 --> 00:52:02,780
to this mechanism since the pressure intensity
is same and area of force section A2 is much
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00:52:02,780 --> 00:52:04,450
larger compare to A1.
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00:52:04,450 --> 00:52:10,819
So, for larger area we apply larger force
f2. This principle is very much applicable
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00:52:10,819 --> 00:52:15,700
in many of the mechanisms like hydraulics
utilizing these principles. We can transmit
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00:52:15,700 --> 00:52:22,700
the larger pressure by applying more force
here and we can obtain larger force here.
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00:52:24,700 --> 00:52:30,059
This principle is used in lifts, hydraulic
jacks, pressure extra. What we are doing here
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00:52:30,059 --> 00:52:36,420
is we are applying a small force here f1 that
will give correspondingly large force; input
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00:52:36,420 --> 00:52:43,190
will be small and larger force output is produced
here which is used in lifts hydraulics jacks
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00:52:43,190 --> 00:52:49,910
pressures and many other hydraulics equipments.
So this principle is very much very important
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00:52:49,910 --> 00:52:55,220
in fluid statics which is based upon the theory
that the transmission of pressure through
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00:52:55,220 --> 00:52:55,960
stationary fluid.
442