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Welcome to this 7 th lecture in this NPTEL
course on fluid mechanics for chemical engineering
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undergraduate students. In the last lecture,
we discuss the fundamental equation of fluid
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statics.
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We were discussing fluids under static conditions,
and we derive a fundamental equation for fluids
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that is, that are present and under the influence
of a gravitational field. So we started by
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taking a volume element, volume element of
a fluid under the influence of gravity, and
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in the limit when, so we took a co-ordinate
system x, y, and z. In the limit when, this
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volume element dimensions of the volume element
shrinks, we derive a fundamental equation
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for fluid statics is equal to 0 minus the
gradient of pressure plus density of the fluid
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times gravity is equal to 0. This is a fundamental
equation that is of use in describing several
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features of fluid statics, fluids under static
conditions.
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It is customer in to point the gravity vector
along the negative z direction as I shown
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here. So there are three unit vectors i, j,
and k, in the along the x, y, and z direction.
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So the acceleration due to gravity vector
is given by minus g times k, where k is a
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unit vector in the positive z direction, but
g is pointing the negative direction. So,
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minus g happens, because of that. So, when
we substitute this, when we referred this
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equation to this co-ordinate system, for this
particular co-ordinate system, this equation
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is very general, because it has no reference
to any co-ordinate system.
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This is general, when applied to the co-ordinate
system shown here, we get minus partial p
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partial x. Now, the vector g can be written
as g x times i plus g y times j plus g z times
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k. It can be dissolved into the three cartesian
directions, and in this co-ordinate system
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g x is 0, and g y is 0, and g z is minus g.
So, we can proceed further by saying that
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d p by minus d p dx is 0, because g x is 0
and minus d p dy is 0, because g y is 0, and
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minus d p dz minus rho g is 0, in the z direction.
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This implies that p is independent of x and
y and it varies only in the z direction. And
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this implies, since p is independent of x
and y, the partial derivative becomes a normal
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derivative minus d p dz is rho g or d p dz
is minus rho g. We can integrate this, if
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rho is a constant and g is normally a constant
under terrestrial conditions, the acceleration
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due to gravity on the surface of this a constant.
Then d p dz is minus rho g can be integrated
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as follows, between any two points, minus
rho g dz. Since rho and g are constants, we
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can pull them out. So, integral d p between
any two points p naught and p is minus rho
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g integral z is z naught to any z dz.
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So, p minus p naught is nothing but minus
rho g times z minus z naught or p minus p
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naught is rho g times z naught minus z, after
sawing the minus sign.
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Now this can also be written as p at any z
minus p naught is rho g times z naught minus
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z. It is customary in fluid mechanics, so
you have the z co-ordinate going up like this.
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Suppose you have a fluid, suppose you have
a water body which is exposed to atmosphere
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air. Where the pressure is p atmosphere, the
pressure of air in the atmosphere is due to
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the weight of the air that is present above
a given elevation. So, at c level the pressure
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of air is conventionally called the atmospheric
pressure. That is precisely because of the
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weight of the air that is present above the
level, so this is known.
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So, if you call this location as z equals
z naught which is the free surface, where
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p is p naught is p atmosphere. Then p at any
location z is p naught which is p atmosphere
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plus rho g times z naught minus z. Z is any
location and z naught is this location, so
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z naught minus z is this depth from the free
surface. So this is conventionally denoted
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by the letter h, so p at any location in the
liquid is p atmosphere plus rho g h. This
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is something that you may be familiar with
from your earlier classes in physics. Where
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the pressure in a column of liquid is an increase
with vertical distance in a linear manner
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and that is precisely because of the fact
that the pressure. Suppose, you take a column
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of liquid and this is the atmospheric pressure,
so pressure always acts normally to a surface.
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So if you look at the pressure on this side,
this pressure will have to be greater than
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atmospheric pressure. Because of the fact
that under static conditions, the fluid will
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have to balance the weight of this liquid
column, which is precisely given by rho g
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h times the area of the element. So that is
the precise physical meaning of this equation.
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And this is valid only for incompressible
fluids, where rho is constant. Now, let me
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just spend a couple of minutes commenting
on the nature of atmospheric pressure. P atmosphere,
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the atmospheric pressure is precisely the
pressure of the air that is present in the
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atmosphere. And so, if you consider c level
that is ground level, then if you take a cylindrical
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column of air, the weight of this air is precisely
the pressure that you will feel at the ground
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level. And at pure vacuum, that is when there
is no air, when you go far away from the ground
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level, far into the atmosphere there is the
density of air will decrease significantly
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the pressure will also decrease. So at pure
vacuum, the pressure of air is 0.
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So there is no air molecules, there is no
pressure and the pressure at the ground level
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is called the atmospheric pressure. And this
atmospheric pressure is roughly 10 to the
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5 pascals, is 10 to 5 newton per meter square
in S I units. The other thing we discussed
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was the role of compressibility. So, if air
is treated compressible, to be an ideal gas,
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so we said that p is rho the specific gas
constants time T, then you had minus d p dz
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is d p dz is minus rho g. Instead of rho,
so this implies rho is p by R g T, so I can
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eliminate d p by dz is minus p by R g T times
g, acceleration due to gravity.
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So I can integrate this in the following way
d p by p integral, this is minus g by R g
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T integral dz. So, logarithm of p is nothing
but minus g by R g T z plus some constant.
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Which can be simplified to write as p is p
naught times exponential of minus g by R g
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T and T is assumed to be constant in this
analysis. The air is a constant temperature,
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so let us call that constant T naught times
z. This p naught is the value, pressure at
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z equals 0. That is the condition, boundary
condition we use to fix this constant. This
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constant is fixed by saying that the pressure
at a z equal to 0 is p naught.
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So p is p naught times exponential of minus
g by R g T naught z. This is an equation that
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is valid, if air is treated compressible.
But we also said that, if the value of this
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exponent g z by R g T naught is small compared
to 1. This is a dimensional less a group,
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so if this number is small compare to 1, you
can Taylor expand and write p is p naught
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times 1 minus. So Taylor expand, e to the
minus x is approximately 1 minus x, if x is
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small. So 1 minus g minus z by R g T naught,
so p is p naught minus p naught by R g T naught
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g z. From ideal gas law, this is nothing but
rho naught, so p is p naught minus rho naught
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g z. So, this is similar to the incompressible
equation that we derived by treating density
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to be constant.
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So, this linear variation is a simplification
of this exponential variation of pressure
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and is valid if g by R g T naught is g z by
R g T naught is less than 0.1 or if z is less
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than 800 meter. We can treat the pressure
variation even in air which is in general
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a compressible system to be a linear variation.
So, the other thing the next thing we will
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do is to apply the fundamental equation of
hydrostatics to what is called manometry.
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Manometry is that branch of fluid mechanics,
which deals with measurement of pressures.
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And the specific device we are going to use
is called a U-tube manometer. So what is the
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construction? Well, it consists of U shaped
tube and one end is exposed to atmosphere.
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The other end is joined to a region, whose
pressure we want to know. So here, they let
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us call this point as A the pressure here
is not known, so pressure unknown. The manometer
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is filled with working liquid is called the
manometric liquid. Now this is open to atmosphere.
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Now the idea is to relate this pressure, unknown
pressure to the atmospheric pressure which
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is known. So, how do we go about doing this?
Let us now draw some label some heights here,
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let us call this height as h 3, let us call
this height as h 4, let us call this height
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as h 2 and between point A and this interfaces
h 1. So we want to now apply the principle
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of the result from fundamental equation of
hydrostatics, that p is p atmosphere plus
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rho g h. Now, if you go from point here to
here, the pressure will increase, this is
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atmospheric pressure, the pressure will increase
because of the weight of air, but that is
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negligible, so we will not worry about this.
From this point, now we are going to between
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these two points, between these two levels
the pressure at this point and the pressure
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at this point is the same. Because the pressure
in this manometric liquid is a function only
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of the elevation and the elevations are the
same to pressure at this point, which is called,
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let us call it B and the pressure at this
point D must be the same. So p B must be equal
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to p D, because the pressure in the manometric
liquid is the function only of the elevation.
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And since these two points are the same elevation,
p B must be equal to p D. If p B is, so how
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do we get p B in terms of p A. Well, p B is
nothing but p A plus rho 1.
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Let us call this liquid rho 1, equate the
density rho 1, acceleration due to gravity
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times, this column height h 1, rho 1, g h
1. This is the pressure at point B. Now p
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D, the pressure at this point D is given by
the pressure at this point which is approximately
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atmospheric pressure plus rho manometric liquid.
Let us call it rho m, so density of this liquid
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the manometric liquid is rho m g. This total
height is h 4, this is h 4 and so this total
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height is h 4, this is h 2, so that this weight
of this, the height of this column is h 4
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minus h 2.
So, we can write p A minus p atmosphere. The
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difference in the value of the pressure at
the point A minus atmospheric pressure is
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nothing but rho m g h 4 minus h 2 minus rho
1 g h 1. So, by measuring these two heights,
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by measuring the height difference h 4 minus
h 2, h 4 minus h 2 is basically this height.
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I am measuring this height that is this and
typically the density of the manometric liquid
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is very large compared to density of the working
fluid. So this is
usually negligible, so we can get the difference
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in the value of pressure at point A from the
atmospheric pressure to be the density of
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the manometric liquid times, the acceleration
due to gravity times, and the height difference
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between the two lengths of the manometer.
So this is a fundamental equation of manometry.
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And by just simply measuring the height, we
measure this to obtain the unknown pressure.
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Now, this difference is called the gage pressure,
as I mentioned in the last lecture. The difference
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between the values of a pressure at a point
from the atmospheric pressure is called the
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gage pressure, and because that is what is
measured by a pressure measuring devices,
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such as the manometers.
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Now how do, how does one measure atmospheric
pressure itself? In order to that, we have,
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what is called a barometer. Usually the manometric
liquid is mercury rho m is the typically mercury,
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which is 13.6 times 10 to the 3 kg per meter
cube, so very large density liquid. Now, a
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barometer is used to measure atmospheric pressure.
So, this is the device that is used to obtain
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what to estimate the atmospheric pressure.
How is it done? Well, the construction of
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the barometer is very simple. You take a trough
of mercury and then in this trough, we invert
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a tube which already has mercury in it. Now
the mercury in this tube will raise.
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So this is mercury, will this raise to in
this tube will rise to a particular height,
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this is air atmospheric pressure, p atmosphere.
Now in this part, it is largely vacuum, it
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has some molecules of mercury vapor. But it
is since a vapor pressure of mercury is very
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very small, this is essentially a vacuum,
there is no pressure here. So the pressure
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at this point is atmospheric pressure, from
the fact that this liquid is exposed to air.
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Now, the pressure at this point must be the
same as the pressure at this point, because
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this is connected by the same liquid is connecting
these two points.
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And since they are at the same elevation,
there cannot be any pressure difference. So
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the pressure at this point is also atmospheric.
So the pressure at this point, which is atmospheric
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pressure is the pressure at point A, which
is let us call it p A which is 0, because
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it is a vacuum plus rho mercury g times h.
Where h is the column height of the column
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of mercury that is present in the tube. So,
p atmosphere which is what we want to calculate
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want this. Can be obtained by this measuring
what is the height of mercury. So, typically
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this height is 760 mm at normal conditions
of temperature and altitudes or 76 centimeters.
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So, the atmospheric pressure is nothing but
density of mercury, this is 13.6 times 10
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to the 3 times 9.8 meter per second square
times 0.76 meters, this is height. When we
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do all these, we get atmospheric pressure
to be 1.03 times 10 to the 5 newton per meter
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square or 1.03 times 10 to the 5 pascals.
So the barometer is the simple construct that
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is used to calculate the atmospheric pressure.
So, the atmospheric pressure in several text
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books or even hand books is denoted in terms
of S I units as 1.013 times 10 to the 5 pascals
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or it is written as 76 mm Hg. Because that
is the height of the mercury column that raises
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to counter balance atmospheric pressure. So,
it is sometimes refer in terms of mm h g and
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this is also trivially called as one atmosphere.
Because it is a normal pressure that is encountered
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in atmosphere, it is of the order of 10 to
the 5 pascals. So, one atmosphere is essentially
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1.013 times 10 to the 5 pascals is also 760
mm of Hg, so mercury column. So, all these
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are used interchangeably while reporting the
values of pressure. Now, that we have done
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all this, now we are going to worry about
the next topic, which is forces under static
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as forces on solid surfaces or forces on sub
merged, surfaces under static conditions.
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So the issue that we are going to understand
is the following. Suppose, you have a liquid
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surface, where it is exposed to atmosphere,
p atmosphere, and within the liquid, there
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is a surface, a planer surface, so we will
look at planer surface for simplicity, so
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we look at a planer surface. So you have a
plane and this plane extends in the third
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direction. So let me put co-ordinate system,
this is the z co-ordinate, this is the y co-ordinate,
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and the x co-ordinate runs in the direction
perpendicular to the board.
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So, if you look at the x y, this surface may
look like this. Now, since and in the y z
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plane, this will look like a line, because
this is a planer surface, this is like a plate
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with some arbitrary shape. Now, we want to
know and this is let us say liquid like water.
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We want to know and gravity is acting in this
direction and this is the liquid surface,
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where the pressure is atmospheric. We want
to know, what is the force that is exerted
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by the fluid on one side of this solid surface,
on this planer surface.
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The question that we are asking, the question
that we want to answer is what is the force?
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F R that is exerted by the fluid. Now, we
also want to know the point at which the point
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x prime, y prime, at which the resultant force
acts. So we want to calculate these two things.
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Now, why is this thing important? Well, this
thing is important in several applications,
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where suppose you are interested in construction
of a dam. So, a dam is something that stores
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or obstructs water, this is water. And this
surface has to be constructed in such a manner,
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that it withstands the force due to the water.
And the reason why hydrostatics is different
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is, because the pressure varies with depth.
So, the pressure at this point is merely atmospheric,
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but as you go down the pressure will increase
linearly.
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00:26:47,100 --> 00:26:51,460
So the force will not be a, the force cannot
be obtained by simply multiplying the pressure
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00:26:51,460 --> 00:26:56,659
by the area. It as to be obtained by integrating
the pressure with respect to the vertical
190
00:26:56,659 --> 00:27:02,990
co-ordinate. So this what we want to do. So
in order to do this, the way we are going
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00:27:02,990 --> 00:27:09,990
to proceed is by taking a tiny strip in the,
you take a tiny area element of length dx
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00:27:17,889 --> 00:27:19,259
dy.
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00:27:19,259 --> 00:27:25,720
So it will appear like a strip here, this
basically a tiny area element. And this tiny
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00:27:25,720 --> 00:27:32,429
area element is at a distance vertical distance
h. And on this area element; the pressure
195
00:27:32,429 --> 00:27:38,429
force will be acting purely normally, because
the fluid is static. So the differential force
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00:27:38,429 --> 00:27:42,840
acting on this area element is purely normal.
And what is this differential force? This
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00:27:42,840 --> 00:27:49,840
is the pressure at this vertical location
h from the free surface times the area dx
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00:27:50,840 --> 00:27:57,840
dy. This is the essential idea of doing the
whole thing. So, dx dy is dA this all we will
199
00:28:03,370 --> 00:28:08,100
do, in order to calculate the effective force.
200
00:28:08,100 --> 00:28:14,590
So the resultant force is nothing but you
take the differential force p dA acting on
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00:28:14,590 --> 00:28:20,710
a tiny slice and integrate over the entire
area that will give you the resultant force.
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00:28:20,710 --> 00:28:27,710
Now, p is nothing but p naught plus rho g
h. So, F R is nothing but integral A p naught
203
00:28:31,679 --> 00:28:38,679
plus rho g h times dA. Now, h so this is y,
this is h, this is theta, so h is basically
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00:28:51,419 --> 00:28:58,419
like this angle, let me this rules slightly
differently. So, this angle is theta so this
205
00:29:04,730 --> 00:29:11,730
is the surface, this angle theta, this is
h, this is y, h is nothing but y sin theta.
206
00:29:17,419 --> 00:29:24,419
So, we want the value at point h. So instead
of h, we will do y, because the co-ordinate
207
00:29:26,460 --> 00:29:33,460
is along the surface that is y. So, we will
say F R is nothing but integral a p 0 plus
208
00:29:36,019 --> 00:29:43,019
rho g y sin theta times dA. This is how one
calculates, the force the resultant force
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00:29:47,580 --> 00:29:53,919
on a surface that is submerged in a fluid.
Well traditionally, the way this force is
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00:29:53,919 --> 00:29:58,499
done, calculated is you could calculate either
by just carrying out this integral or you
211
00:29:58,499 --> 00:30:05,499
define the centroid of a plane. A centroid
of a plane, the plane is in the x y this entered
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00:30:12,480 --> 00:30:19,480
of a surface which is on x y plane is one
over area integral area dA x y.
213
00:30:21,809 --> 00:30:28,210
This is the centroid of the co-ordinates of
the centroid of a surface. So, integral y
214
00:30:28,210 --> 00:30:35,210
dA is nothing but y c times A. So we have
from here, F R is p 0 A upon integration,
215
00:30:38,399 --> 00:30:45,340
p 0 is a constant; it’s usually an atmospheric
pressure plus rho g sin theta. Since, y dA
216
00:30:45,340 --> 00:30:52,340
integral is y c, so we will write this as
y c A or F R is nothing but p 0 plus rho g.
217
00:30:56,740 --> 00:31:03,740
This is sometime refer to us h c rho g h c
A, this is the pressure at the centroid of
218
00:31:09,100 --> 00:31:15,139
the area of the surface.
219
00:31:15,139 --> 00:31:22,139
So F R, the resultant force is the pressure
at the centroid times the area. So this is
220
00:31:23,950 --> 00:31:30,950
the simple result for flow R for the forces
that have been exerted on a planar surface
221
00:31:31,039 --> 00:31:38,039
that is submerged into inside a liquid like
water under static conditions. Now two comments,
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00:31:39,820 --> 00:31:46,820
firstly, this force acts only on one side.
So we are looking at, recall that the geometry
223
00:31:47,570 --> 00:31:53,659
is like this, this is theta, this is the free
surface, this is liquid. This is the force
224
00:31:53,659 --> 00:31:59,259
only acting on one side, the other side is
also comprise of the same liquid, a same amount
225
00:31:59,259 --> 00:32:05,279
of force will act on this side also. But if
the other side is open to some other, it may
226
00:32:05,279 --> 00:32:09,879
be open to atmosphere, then this is the force
that is because of the liquid that is present
227
00:32:09,879 --> 00:32:15,649
on one side of the surface. So it depends
on the problem and context has to what the
228
00:32:15,649 --> 00:32:21,110
other side is. If it is open to atmosphere,
then this will be the force due to the liquid
229
00:32:21,110 --> 00:32:25,240
that is present, and the pressure variation
in the liquid under static conditions. What
230
00:32:25,240 --> 00:32:32,240
is the point of action of the force?
231
00:32:32,860 --> 00:32:37,119
What is the point of action? Well, in order
to find the point of action, we simply take
232
00:32:37,119 --> 00:32:44,119
the moment. Let us call that point of action
as y prime. So y prime, the moment of the
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00:32:44,119 --> 00:32:49,240
force about the point of action must be equal
to the distributed moment y times p times
234
00:32:49,240 --> 00:32:56,240
dA over the entire area. So, y times F R is
nothing but y is rho g sin theta. Now, if
235
00:33:02,909 --> 00:33:09,059
on one side you have liquid and the other
side you have atmospheric air, then you need
236
00:33:09,059 --> 00:33:15,749
not be worry about the atmospheric pressure.
So, p is simply written as the gage pressure,
237
00:33:15,749 --> 00:33:19,860
because the atmospheric, the contribution
due to atmospheric pressure on this side and
238
00:33:19,860 --> 00:33:24,419
this side will cancel. So we can neglect by
atmospheric pressure and write only the gage
239
00:33:24,419 --> 00:33:31,419
pressure. So this is rho g h c times A, so
p is p g is rho g times h which is rho g y
240
00:33:40,700 --> 00:33:47,700
sin theta. So, integral rho g sin theta A
y square dA.
241
00:33:50,720 --> 00:33:57,720
But F R is nothing but p c times the pressure
at this centroid times A. So this is rho g
242
00:33:59,549 --> 00:34:06,549
sin theta times integral y squared. So F R
is nothing but p c times A which is rho g
243
00:34:16,970 --> 00:34:23,970
y c sin theta integral A y square dA. So p
c is nothing but rho g y c sin theta, if the
244
00:34:28,460 --> 00:34:35,460
other side is surrounded by air. So this implies
y prime is nothing but 1 by A y c integral,
245
00:34:41,630 --> 00:34:48,630
so there is this A here is coming, their denominator
integral y square times dA. This is the line
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00:34:48,700 --> 00:34:55,700
of action of the force that is going to act
on us submerged planer surface. So this is
247
00:34:59,070 --> 00:35:06,070
of use in several applications where you are
interested in at the forces that have been
248
00:35:06,360 --> 00:35:13,070
exerted by the fluid under static conditions
on solid surfaces. And this is primarily of
249
00:35:13,070 --> 00:35:20,070
interest in storage of water in dams and so
on. We can also generate, generalize these
250
00:35:20,200 --> 00:35:25,310
two curved surfaces, but I will not go through
this for want of time.
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00:35:25,310 --> 00:35:32,310
So I will go to the next topic, which is buoyancy,
which is also related to forces exerted on
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00:35:33,830 --> 00:35:40,830
curved surfaces. Suppose, you have an object
that is a solid object, that is completely
253
00:35:42,770 --> 00:35:49,770
immersed in a liquid. So you have a free surface
that is atmosphere, you have a liquid like
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00:35:50,540 --> 00:35:57,540
water, so you have a solid object is completely
immersed. Now, let us say you are coating
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00:36:00,990 --> 00:36:07,990
a co-ordinate z like this and gravity is acting
like this. Now, this density has this liquid
256
00:36:08,210 --> 00:36:13,640
has density rho. Now, because of the fact
that is this liquid has a density and acceleration
257
00:36:13,640 --> 00:36:18,730
due to gravity is acting downwards the pressure
here is p atmosphere, the pressure here is
258
00:36:18,730 --> 00:36:25,730
more. So, the pressure exerted on this side
on this submerged solid surface is more than
259
00:36:28,240 --> 00:36:32,930
the pressure that will be exerted on this
side, because the pressure is less. So, this
260
00:36:32,930 --> 00:36:37,450
net force will act up wards is called the
buoyancy force.
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00:36:37,450 --> 00:36:44,450
So, how do we estimate or derive an expression
for a buoyancy force? It is not very difficult.
262
00:36:44,930 --> 00:36:51,930
Simply we have to take a thin cylindrical
volume element, let us call this height as
263
00:36:54,860 --> 00:37:01,860
h. Now, p at h at the bottom is basically
p naught, the pressure at the top plus this
264
00:37:05,380 --> 00:37:12,000
is p naught let us say plus rho g h. This
is the fundamental equation of hydrostatics.
265
00:37:12,000 --> 00:37:19,000
So, the net vertical force
on this volumes cylindrical volume is nothing
but p naught plus rho g, let’s call this
266
00:37:27,700 --> 00:37:34,700
h equals h 2 and lets call the top surface
h equals h 1 so p naught plus rho g h 2 times
267
00:37:36,430 --> 00:37:43,430
dA minus p naught plus rho g h 1 times dA,
so h 2 minus h 1 let us call it h so its rho
268
00:37:49,750 --> 00:37:56,750
g were h dA. This is h dA is the differential
volume of the cylindrical volume element.
269
00:38:03,800 --> 00:38:10,800
So, the net vertical force on this infinitesimal
cylindrical volume element is rho g times
270
00:38:12,740 --> 00:38:15,530
the differential volume.
271
00:38:15,530 --> 00:38:22,220
So the net vertical force on a cylindrical
volume element that is present in a solid
272
00:38:22,220 --> 00:38:29,220
that is sub merged in a fluid the differential
force is called d F z is rho g times dv. This
273
00:38:31,660 --> 00:38:38,660
what we just derive, where dv is the differential
volume of this volume element. And this force
274
00:38:39,580 --> 00:38:45,900
is precisely, because of the fact that the
pressure here and the pressure here are different.
275
00:38:45,900 --> 00:38:50,990
And because of the fact that fluid is under
a gravitational feel and the pressure varies
276
00:38:50,990 --> 00:38:56,720
due to hydrostatic equation, a hydrostatic
force balance. Now to get the force on the
277
00:38:56,720 --> 00:39:01,300
entire object, we simply have to integrate
this differential force over the entire object.
278
00:39:01,300 --> 00:39:08,300
Which is nothing but integrating over the
entire volume rho g dv. Since, rho g is constants,
279
00:39:09,430 --> 00:39:16,430
so you get rho g integral of dv which is nothing
but rho g times the volume of the object.
280
00:39:16,990 --> 00:39:23,990
But let us try to understand this slightly
differently, this is rho times v times g,
281
00:39:24,060 --> 00:39:31,060
here rho is density of the liquid in which
this object is present. So, this is the mass
282
00:39:32,980 --> 00:39:39,980
of the liquid that is displaced by the solid.
283
00:39:48,190 --> 00:39:55,190
So the net vertical force on a solid substrate
solid object that is completely submerged
284
00:39:56,070 --> 00:40:02,700
under in a liquid which is present under gravitational
field, this is called the buoyancy force,
285
00:40:02,700 --> 00:40:09,700
and this is nothing but the weight of the
fluid that is displaced by the solid object.
286
00:40:24,550 --> 00:40:31,550
This of course, the famous Archimedes principle,
so this is called the Archimedes principle.
287
00:40:36,130 --> 00:40:42,950
This is again a consequence of a basic equation
force balance in hydrostatics. And we merely
288
00:40:42,950 --> 00:40:49,950
have to apply this to the context of object
that is immersed. So this is for a fully,
289
00:40:53,090 --> 00:40:56,730
what we derived is for a fully submerged object.
290
00:40:56,730 --> 00:41:03,730
Suppose you have a floating object, suppose
you have an object that is partially submerged,
291
00:41:05,900 --> 00:41:12,900
this is floating. So you have this is atmosphere,
air at atmospheric pressure, this is liquid.
292
00:41:14,170 --> 00:41:21,170
Suppose, you have partially submerged objects,
so only this part is submerged. So this portion
293
00:41:26,770 --> 00:41:33,770
has displaced, this submerged object has displaced
a volume of fluid. And the buoyancy force
294
00:41:35,890 --> 00:41:41,690
will be because of the fact that of due to
that weight of the displaced fluid, so this
295
00:41:41,690 --> 00:41:45,620
is gravity.
Now this solid object is under stable equilibrium
296
00:41:45,620 --> 00:41:51,130
that is its not sinking, it is not moving
down. That means that, the net downward force
297
00:41:51,130 --> 00:41:58,130
on the solid object, this is mass of the solid
object, times acceleration due to gravity
298
00:41:59,880 --> 00:42:05,910
must be equal to the buoyancy force, which
is acting upward. This is the downward force,
299
00:42:05,910 --> 00:42:12,420
this is the upward force, which is buoyancy,
this is the density of the liquid, times acceleration
300
00:42:12,420 --> 00:42:19,420
due to gravity, times the displaced volume.
Because only the displaced volume will contribute
301
00:42:19,620 --> 00:42:26,330
to the net upward force. So, this is the condition
for floating. That the net downward force
302
00:42:26,330 --> 00:42:32,300
must be equal to the net upward force, which
is the buoyancy force which is nothing but
303
00:42:32,300 --> 00:42:36,060
rho times the density of the liquid times
acceleration due to gravity times at displaced
304
00:42:36,060 --> 00:42:37,040
volume.
305
00:42:37,040 --> 00:42:44,040
Now, let us consider a simple example of an
ice berg. Ice icebergs are found in oceans,
306
00:42:45,690 --> 00:42:52,690
these are huge chunks of ice that are present
in sea, in ocean. So this is the water level,
307
00:42:57,110 --> 00:43:04,110
and this whole thing is submerged under water,
this whole thing is submerged under water.
308
00:43:07,220 --> 00:43:13,910
So, let us call this submerged volume V under
water V u w. Let the mass of the ice berg
309
00:43:13,910 --> 00:43:20,910
be M, this is the ice berg mass and the total
volume of ice berg is V total, this is total
310
00:43:22,060 --> 00:43:29,060
volume of ice berg. Now density of ice is
smaller than rho ice is smaller than density
311
00:43:31,820 --> 00:43:38,820
of water. Density of water is 1 gram per cc
and density of ice is 0.92 grams per cc, centimeter
312
00:43:41,360 --> 00:43:47,810
cube. So, if this ice berg is under stable
equilibrium, then the mass of the ice berg
313
00:43:47,810 --> 00:43:51,940
times is the acceleration due to gravity.
The weight of gravity by is nothing but the
314
00:43:51,940 --> 00:43:57,040
buoyancy force; this is the weight of the
displaced fluid which is nothing but the volume
315
00:43:57,040 --> 00:44:03,250
that is submerged under water of the ice berg
times density times acceleration due to gravity.
316
00:44:03,250 --> 00:44:09,480
What is the total mass of the ice berg? it
is nothing but rho ice times V total volume
317
00:44:09,480 --> 00:44:16,480
times g is V under water times rho liquid,
which is let us say water here of course,
318
00:44:17,200 --> 00:44:24,200
times g, g cancels. So V under water by V
total, this the fraction of volume that is
319
00:44:25,710 --> 00:44:32,710
under water is nothing but rho ice divided
b y rho water. This is nothing but 0.92 divided
320
00:44:34,500 --> 00:44:41,400
by 1 is 0.92. So, what this simple example
is telling you is that. When our ice berg
321
00:44:41,400 --> 00:44:48,400
is floating, in a floating ice berg, 92 percent
of the solid mass is under water. So this
322
00:44:59,060 --> 00:45:06,060
is straight forward consequences the buoyancy
principle, the 92 percent of the ice in an
323
00:45:06,530 --> 00:45:10,730
ice berg is completely under water.
324
00:45:10,730 --> 00:45:17,730
We can also derive simple criteria for floating,
when does an object float? When can object
325
00:45:19,170 --> 00:45:26,170
float? Let us consider a simple geometry here,
let us consider a very simple geometry to
326
00:45:35,040 --> 00:45:37,860
get this result.
327
00:45:37,860 --> 00:45:44,860
Which is, take a cylindrical object and this
is water, let us say this cylindrical object
328
00:45:47,570 --> 00:45:54,570
is floating, this is water, and this is air.
So floating happens, when the weight downward
329
00:45:57,110 --> 00:46:01,730
force that is weight acting due to acceleration
due to gravity on the solid object is equal
330
00:46:01,730 --> 00:46:07,950
to the buoyancy force. which is nothing but
so let us call this height that is submerged
331
00:46:07,950 --> 00:46:13,600
as h, so the volume of cylinder that is submerged
is 8 times h, times the density of the fluid
332
00:46:13,600 --> 00:46:20,600
is rho f times g. Now m is nothing but suppose
let us call this whole height as l A l times
333
00:46:24,940 --> 00:46:31,940
rho solid times g is nothing but A h times
rho fluid times g. So, if you cancel A and
334
00:46:32,330 --> 00:46:39,330
g so floatingly can happen. By definition,
floating means h is less than l otherwise
335
00:46:43,680 --> 00:46:46,000
the object will completely immerse under water.
336
00:46:46,000 --> 00:46:51,920
So floating means h is less than l, so if
h is less than l then this equation tells
337
00:46:51,920 --> 00:46:58,920
you that rho s l is rho f h, this implies
that rho s is less than rho l. This is a necessary
338
00:47:02,990 --> 00:47:09,990
condition
for floating. So, this completes the basic
concepts that are that can be obtained by
339
00:47:24,150 --> 00:47:31,150
simple considerations of a fluid under static
conditions. So just to recapitulate, we first
340
00:47:32,040 --> 00:47:38,000
derived the governing equation for fluid under
static conditions, which was simply minus
341
00:47:38,000 --> 00:47:45,000
del p plus rho g is 0. And using this, in
this lecture we derive the fundamental equation
342
00:47:45,740 --> 00:47:50,790
for manometers, the principle of manometry.
And then we introduce a notion of atmospheric
343
00:47:50,790 --> 00:47:55,910
pressure and barometers.
Then we proceeded to derive the forces that
344
00:47:55,910 --> 00:48:01,940
are experienced by a planar surface that is
submerged inside a liquid. And we found that
345
00:48:01,940 --> 00:48:07,360
it can be very easily obtained by integrating
the pressure on a small area element. And
346
00:48:07,360 --> 00:48:11,940
by integrating this over the entire area we
can get the force and we can also find the
347
00:48:11,940 --> 00:48:17,850
line of action of this resultant force on
a solid surface. Next, we proceeded to discuss
348
00:48:17,850 --> 00:48:24,710
the notion of buoyancy on a completely submerged
solid surface. And we can derive the Archimedes
349
00:48:24,710 --> 00:48:31,710
principles from the basic equation of hydrostatics.
By realizing that, the net force down downward
350
00:48:33,280 --> 00:48:37,820
on a on the solid surface is greater than
the net force on the upper surface of the
351
00:48:37,820 --> 00:48:43,100
solid. So, this results to the buoyancy and
we derived the Archimedes principle from the
352
00:48:43,100 --> 00:48:48,640
basic equation of fluid statics. And we also
saw, when can an object float and what are
353
00:48:48,640 --> 00:48:50,980
the necessary conditions under which an object
floats.
354
00:48:50,980 --> 00:48:57,630
So, this completes the discussion on fluid
statics, which is one of the simplest topics
355
00:48:57,630 --> 00:49:02,990
in fluid mechanics, because the subject of
fluid mechanics deals with fluid flow, but
356
00:49:02,990 --> 00:49:07,540
fluid statics is an integral part, because
even under static conditions, the forces that
357
00:49:07,540 --> 00:49:13,540
are being exerted are not simple, because
of the fact that the pressure in a fluid varies
358
00:49:13,540 --> 00:49:19,660
with vertical distance. So, we will stop here,
and we will continue with a new topic in the
359
00:49:19,660 --> 00:49:26,660
next lecture. So, we will see in the next
lecture.