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Welcome back to the video lecture on fluid
mechanics. In the last lecture we were discussing
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about the fluid statics; we were discussing
about the courses on solids and then the applications
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of theories from solid mechanics and we are
started with the static fluid theories like
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on a fluid element rest.
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We have seen that the equation sigma fx is
equal to 0 sigma fy is equal to 0 sigma fz
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is equal to 0 at sum of the forces in xyz
directions equal to 0 as we use in lid mechanics
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So similarly the sum of moments of forces
sigma Mx is equal to 0 that also we have seen
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and we are defined the pressure.
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Also we have seen the applications of Pascal
law for pressure at any point as we have seen
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the as for the pascal law pressure at any
point is same in all directions that derivations
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also we have seen in the last lecture.
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And also we have seen the pressure variation
vertical direction and also the horizontal
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direction.
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As we have seen in the last lecture, the pressuring
density of fluid is independent of shape as
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we can see here it is on this excess line
and then.
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We have seen the application of the concept
of force versus area, here as we have discussing
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the last lecture say here force F1 on this
side is p into A1 and other side F2 is equal
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to p into A2. Since p is say on both side
is the level p and q. pp is equal to pq.
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So, transmission of pressure through the stationary
fluid is a large number of applications in
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many practical engineering areas like in use
of hydraulic jacks, pressures etc.,
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So here we can see that the force applied
F is on this side is equal to p in to A1 which
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with small force we can raise or we can get
a large force F2 is equal to p into A2. So
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this application like as I mentioned hydraulic
jacks or lifts pressures etc., are very much
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used in engineering areas. So now just one
application here you can see a man can lift
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the car.
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So here car is placed on this side and here
is also the man is standing and you can see
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just with his weight due to application of
this principle the car is lifting. So that
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is very much obvious from this animation.
So, what is happening is the pressure intensity
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same on both sides but here the area is smaller.
So with respect to that p into A1 is much
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more force using that we can raise this p
into A2 which is very larger force. So, the
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car is raised as shown in this animation.
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Now, we will discuss the general equation
for pressure variation in a static fluid.
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So, for this purpose, we just consider as
shown in this slide you can see here we consider
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a fluid element a cylindrical element of fluid
at an arbitrary orientation like this we consider.
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So on this face of the cylinder the fluid
cylinder say the pressure of the force is
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p into if the pressure intensity is p, force
is equal to p into A and on the other side
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if the pressure is increasing delta p by delta
p then on this side the force is equal to
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p plus delta p into A.
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So, here for the fluid density is row and
now here this instance is z and here z plus
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delta z the force acting on the element are
p into A acting at right angles to the end
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of the face and z this as shown here and on
the other side p plus delta A acting at right
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angle to the end of the cylinder element at
z plus delta z and then there will be the
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weight of the element that will be mg.
Now, we consider all this using the Newton
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second law, we can write p into A minus p
plus delta p into A minus row g A into delta
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z cos theta is equal to 0 if we consider this
direction. So that using the Newton's second
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law that the force is equal to say the total
mass into acceleration here with respect to
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this we can get at equilibrium of the element
the result force in any direction is 0 as
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you have seen in the case of static fluid.
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So using that principle we will get delta
p is equal to minus row g into delta s cos
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theta or we get delta p by delta s is equal
to minus row g cos theta as this theta is
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this angle with respect to this inclination
with respect to vertical theta is represent
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vertical.
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Finally, we get a general expression dp by
dx is equal to minus row g cos theta as shown
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here, theta is this angle g is the acceleration
due to gravity and row is the density of the
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fluid. So this equation gives the general
equation for pressure variation in static
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fluid. So now we have considered an inclined
element or cylindrical fluid element like
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this.
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With respect to this equation we can write
the general equation for any type of say whether
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horizontal vertical or inclined type pressure
variation in static fluid. So now with respect
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to the general equation which we have derived
now if theta is equal to 90, then on horizontal
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plane we can see that it dp by dx as theta
equipped 90 is equal dp by dx that is equal
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to dp by dy is equal to 0 and with respect
to the earlier figure theta is equal to 0
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that means the plane is vertical So for a
vertical plain like this.
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So then the dp by ds is equal to minus row
into g, here we can see that this is a small
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basin where we have stored me water. In the
horizontal plane you can see that the pressure
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is 0 and in the vertical plane you can see
that the pressure when theta is equal to 0
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and dp by ds is equal to minus row g and here
this p, the pressure intensity is a function
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of z only or we can write dp is equal to minus
row g dz or where row g is equal to the specific
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weight So that we can write dp is equal to
minus gamma dz, gives the general equation
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or pressure variation means static fluid.
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Now, we will briefly discuss the relation
between the pressure and head. So for that
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purpose let us consider say a small basin
of fluid or say here water is shown here.
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Now this fluid this is in static condition
the pressure on the surface is atmospheric
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pressure as here shown in this figure.
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We can write p is equal to row in to gz and
plus constant that means if any other than
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atmospheric pressure, generally atmospheric
pressure can be consider as 0.
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So that p is equal to row g into h but if
we consider that atmospheric pressure then
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total pressure is equal to pressure intensity
is equal to row gh plus p atmospheric.
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So this gives a relationship between the pressure
and the head as from in this slide.
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If you consider the atmospheric pressure as
datum then we can say that the pressure above
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or below atmosphere that pressure is called
the gauge pressure. So if we consider the
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atmospheric pressure as 0 or that is the datum,
we can define the gauge pressure as the pressure
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above or below atmospheric pressure that the
p gauge or the gauge pressure is equal to
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row in to gh as we have seen in the previous
slide.
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So as we know the lower limit of any pressure
is 0 for the case of perfect vacuum or if
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you consider as datum, then pressure measured
above this datum is called the absolute pressure.
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So when we consider we can classify the pressure
in to the absolute pressure or gauge pressure,
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the gauge pressure is the pressure above or
below the atmospheric pressure.
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So let the equation is p gauge is equal to
row into gh and at absolute pressure is defined
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as absolute is equal to gauge pressure plus
atmospheric pressure or the datum which the
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absolute pressure. So the absolute pressure
is equal to gauge pressure plus the atmospheric
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pressure.
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So with respect to this, when we consider
the various pressure like gauge pressure or
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absolute pressure or atmospheric pressure
let us consider in this slide as you can see
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if this is the absolute 0 reference as far
as the pressure measurement is concerned say
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if you consider say here the particular points
say 1 then we are measuring as absolute pressure
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and if this line indicates the local atmospheric
pressure reference then with respect to that
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when we measure the pressure at this particular
point it is called the gauge pressure but
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when we are measuring with respect to the
atmospheric pressure and the gauge pressure
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that pressure is used as the absolute pressure.
So we can differentiate the pressure measurement
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with respect to either we are using the gauge
pressure or the absolute pressure as shown
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in this slide.
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Now, we will further discuss about the pressure
variation the pressure variation for incompressible
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fluid, you can see that as we have seen in
the previous slide it is say row g your gamma
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into vz if you consider say from one point
to another in the vertical direction. So we
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can just integrate this to get the total pressure
variation you can just integrate gamma z dz.
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Generally, we have to use a sign convention
whether say the pressure above the atmospheric
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pressure whether it is positive or negative
we can consider say we can put it whether
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it is increasing or decreasing the direction
as positive or negative as shown here. So
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finally if you indicate as we have seen the
previous slide we get p is equal to gamma
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into z. So this we can measure using a manometer
like this the pressure variation can be measured
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say at a particular point p at A can measure
with pressures. So we will be discussing about
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the pressure measurement later parts of the
lecture, now we have seen the pressure can
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be say the gauge pressure or absolute pressure
and then the atmospheric pressure.
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So one atmosphere in the pressure is defined
as say 760 millimeter of mercury or it is
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equivalent to 101.3 kilopascal or again that
is equivalent to 10.34 meter of water quantum.
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So the atmospheric pressure is generally expressed
as either 76 centimeter of mercury or 101.3
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kilopascal or 10.34 meter of water quantum.
So this equation gives the general pressure
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variations and then with respect to that atmosphere
pressure the values are these values given
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here.
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Now, with respect to the pressure variation
we will demonstrate or we will discuss a small
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problem here. So here the problem statement
is first example, 2 kilometer away from the
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sea shore at a spot the sea can be considered
to be in three layers of 100 meter, 200 meter
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and 300 meter depth of salt water with specific
gravity values of salt water in each of the
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constant density layers as 1.01, 1.02 and
1.03 respectively. So, assuming atmospheric
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pressure at the sea surface as 0 determines
the pressure at the interfaces.
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So here I would demonstrate this problem with
respect to this figure here, you can see if
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the sea shore is on this side we are considering
two kilometer away from the sea shore at a
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particular spot here you can see three layers.
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So first layer is about 100 meter and second
layer is 200 meter and third layer is 300
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meter and this is the sea bed or sea bottom.
So now the atmospheric pressure can be consider
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as 0 and at this particular spot the specific
gravity values of salt water each of the constant
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density layer are event for the first layer
the density is 1.01, the second layer the
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density is 1.02 and the third layer the density
is 1.03.
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So we have to determine the pressure at the
various interface you can see here there is
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one interface and here the second interface
and here below the at the position also we
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have to determine the pressure. So we can
just utilize the equation which we have derived
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earlier. So the solution here you can see
say even we want to determine this p1 the
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pressure at this interface. So that will be
equal to the surface pressure the atmospheric
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pressure here p1 is equal to p0 which is the
atmospheric pressure plus row1 g, row1 is
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the density and g is the acceleration gravity
that this is the specific weight of the liquid
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at this layer.
So row1 into g this that is 100 meter if you
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are measuring the depth from the surface this
is at this particular position where we are
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measuring the p1 it will be p1 will be at
this location. So that will be obtained as
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p1 is equal to p0 plus row1 into g into 100
minus 0 since the depth is 100 meter this
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location p1 is equal to p0 plus row1 into
g, row1 g is the specific weight into 100
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minus 0 since we are measuring the datum as
on the surface.
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So from there only the measuring, that is
equal to since atmospheric pressure as per
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the problem it is assumed as p 0 is equal
to 0, 0 plus s1 into row g into 100 here it
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is given as 1.01 as the specific gravity of
the first layer or the constant density layer
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as 1.01g is 9.81 into 1.01.
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So finally multiplied by 100 we get 990.8
kilopascal, at this location the pressure
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will be 990.8 kilopascal. So similarly now
we will determine the pressure at position
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say here this interface between layer 2 and
3, at this location p2 is equal to here this
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location the pressure will be the pressure
with respect to the top layer p1 that we have
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to add plus the specific gravity or the specific
weight of this liquids in this layer.
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So that is given as row2 into g that is the
specific weight with respect to this layer
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and then multiplied with respect to this layer
thickness that is equal to the total distance
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from the surface to the bottom of this layer
is 300 meter.
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So 300 minus 100 meter is the depth of the
first layer row2 g into 300 minus 100 that
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is equal to p1 plus s2 row into g where the
s2 is the specific gravity of the constant
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density layer of the second layer is 1.02.
So, row is 1.02 multiplied by g 9.81 into
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200 this gives the total pressure at this
is the interface between layer 2 and layer
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3. So that total p2 is equal to p1 we have
already determine as 990.8 kilopascal that
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plus this equal to s2 in the row g is about
10, so 10 into 200 total pressure at this
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location will be p2 locate here will
be 2990.8 kilopascal.
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Similarly, we will determine the pressure
at this bottom of the sea p3 is equal to the
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total pressure at this layer, two layers,
first layer and second layer that is all determine
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as p2. p2 plus this layer thickness layer
thickness total see that is say 600 meter
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up to here is 300 meter so 600 minus 300 depth
will be 300 meter it determine this pressure
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here p2 plus row3 into g or into s3. s3 is
given as 1.03 that multiplied by this depth
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300.
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So that is equal to the previous p2 we have
already determined as 2990.8 that plus this
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value is this value is 10.104 into 300 that
gives 6022 kilopascal. The pressure at the
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bottom of the sea at this location will be
6022 kilopascal and the interface between
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layer 2 and layer 3 will be 2990.8 kilopascal
and between layer 192 even will be 990.8 kilopascal.
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So using the general equation which we have
derived earlier you can determine the pressure
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variation like we have illustrated in this
particular problem.
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Now, we will go to the next topic is pressure
measurement, especially fluid mechanics is
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very important since say various pressure
is one of the most important fluid property.
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So we have to determine the pressure in various
cases especially when we consider any kind
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of fluid flow problem whether is static or
dynamics but now we will be discussing which
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static here we will be discussing how to determine
the pressure or how we can measure the pressure
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for various cases. Various types of equipments
or various types of gauges are available for
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measurement first one is called a mercury
barometer.
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So here we can see in this slide the mercury
barometer the pressure is measured as especially
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this mercury barometer used for atmospheric
pressure measurement we can see that the pressure
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will be varying from say at sea level it will
be one say it will be maximum then it will
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be reducing say or the top of the mountain.
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So we have to measure the atmospheric pressure
at various levels or various position we can
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say mercury barometer like this. So we are
say small base in the mercury based and then
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a small tool we can see here using this mercury
barometer here the mercury is used. So the
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atmospheric pressure is measured as this gamma
this depth gamma multiplied by this specific
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weight gamma multiplied by this height h gamma
h plus if any the pressure that us the atmospheric
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pressure using a mercury barometer. So this
is the way which we generally measuring the
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atmospheric pressure. Now we will discuss
the other pressure measuring equipment.
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One of the simple moisture pressure measuring
equipment is called a piezometer tube. So
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here you can see if we have want to measure
the pressure in a pipe like this are in a
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00:22:45,600 --> 00:22:51,299
tank or in the position where we want can
introduce a small piezometer tube. So a piezometer
200
00:22:51,299 --> 00:22:58,299
is just like a tube where we can introduce
it on particular location this small tube
201
00:22:59,740 --> 00:23:06,369
this is called a piezometer and then this
can be connected to various location say like
202
00:23:06,369 --> 00:23:12,610
if you want to connect if you want to measure
the pressure in a pipe then you can just put
203
00:23:12,610 --> 00:23:19,610
like this. Finally, we will be determining
the water column the height here and then
204
00:23:21,499 --> 00:23:28,499
with respect to the equation which we have
derived earlier we can get or we can measure
205
00:23:30,749 --> 00:23:32,499
the pressure.
206
00:23:32,499 --> 00:23:39,499
As you illustrated in this slide using a symbol
piezometer, piezometer is connected here and
207
00:23:41,740 --> 00:23:48,659
then it is open to the atmosphere, the depth
is the raise to the pressure in this pipe
208
00:23:48,659 --> 00:23:54,480
or the container there will be raise in pressure
or raise in water column or liquid column
209
00:23:54,480 --> 00:23:59,730
water or what ever the kind of liquid which
is consider raise in liquid column here than
210
00:23:59,730 --> 00:24:05,749
it is h1 here and then the piezometer from
the piezometer you can return in the pressure
211
00:24:05,749 --> 00:24:12,749
p as gamma into h plus p0 where p0 is the
reference pressure say for example if it is
212
00:24:13,940 --> 00:24:19,580
open then we need it can be atmospheric pressure
which can consider as 0 and pressure measured
213
00:24:19,580 --> 00:24:25,990
is the gauge pressures here the we are measuring
the gauge pressure.
214
00:24:25,990 --> 00:24:32,990
So, the piezometer can be used in many cases
like especially for liquids. The disadvantage
215
00:24:33,929 --> 00:24:39,789
of piezometer is that we cannot measure too
small pressure, it is difficult to measure
216
00:24:39,789 --> 00:24:45,119
or when the pressure is very large we cannot
measure, that is the disadvantage of this
217
00:24:45,119 --> 00:24:51,139
piezometer. The advantages is very simple,
we can directly connect to the pipe or container
218
00:24:51,139 --> 00:24:57,369
or in a system and then see the pressure at
the particular location as demonstrated in
219
00:24:57,369 --> 00:25:02,570
this slide. So the pressure measurement is
using piezometer using the simple equation
220
00:25:02,570 --> 00:25:08,220
and what we are getting is the gauge pressure.
221
00:25:08,220 --> 00:25:15,070
So as we have seen this use of same piezometer
has some disadvantage which we cannot measure
222
00:25:15,070 --> 00:25:21,019
the pressure very accurately, it cannot give
a small variation in pressure or we cannot
223
00:25:21,019 --> 00:25:28,019
measure large pressure. So another type of
equipment we generally used in laboratories
224
00:25:28,509 --> 00:25:35,509
and another place is called U tube manometer.
So the pressure measurement using U tube manometer
225
00:25:35,559 --> 00:25:42,559
here the fluid whose pressure is to measure
we can see that here a U tube like shaped
226
00:25:43,470 --> 00:25:49,929
in U shape and then there will be a manometric
fluid inside in this U tube.
227
00:25:49,929 --> 00:25:56,419
So you can see here this is the manometric
fluid of density row manometer, here you can
228
00:25:56,419 --> 00:26:03,419
see and then it will be a dense fluid and
then what will be doing is there we have to
229
00:26:03,489 --> 00:26:10,470
measure the pressure accurately we can connect
this U tube like this here, we are measuring
230
00:26:10,470 --> 00:26:13,759
the pressure in this pipe or a container like
this.
231
00:26:13,759 --> 00:26:20,340
So we are connecting this U tube like this
with respect to the since due to the pressure
232
00:26:20,340 --> 00:26:27,149
in this pipe or the container the liquid in
the manometer liquid in this manometer U tube
233
00:26:27,149 --> 00:26:34,149
manometer will be rising like this. Now with
respect to this we will be measuring generally
234
00:26:35,519 --> 00:26:42,519
this raise h2 and also this variation h1 we
will be measuring using scales with connected
235
00:26:42,999 --> 00:26:49,029
scales. So from the principles let we say
the pressure at this location b and c should
236
00:26:49,029 --> 00:26:56,029
be same if you use that principle pB is equal
to pC as shown in this slide, here with respect
237
00:26:57,409 --> 00:27:04,379
to this we can write pB is equal to pa that
means the pressure here pa plus row in to
238
00:27:04,379 --> 00:27:08,619
g that means row is the density of fluid in
this pipe.
239
00:27:08,619 --> 00:27:15,619
So row into g is the acceleration o gravity
row into g into h1 that gives the pressure
240
00:27:15,700 --> 00:27:22,700
at this particular location pB. So that is
with respect to pA the pB is equal to pa plus
241
00:27:24,289 --> 00:27:31,289
row g h1 similarly we can find the pressure
at the location c that is obtained as pC is
242
00:27:32,309 --> 00:27:38,190
equal to there will be a since here the manometer
is open to atmosphere.
243
00:27:38,190 --> 00:27:43,470
So this we have to consider the atmosphere
pressure if you have neglecting all with considering
244
00:27:43,470 --> 00:27:49,649
a 0 density, then we know to consider but
here if you consider p atmosphere then pC
245
00:27:49,649 --> 00:27:56,289
is equal to p atmospheric plus the row manometer
that means the density of the the manometer
246
00:27:56,289 --> 00:28:01,139
liquid multiplied by g in to h2 this height
of liquid which we are measuring.
247
00:28:01,139 --> 00:28:08,139
So pC is equal to p atmospheric plus row manometer
into g into h2, now our aim is to measure
248
00:28:10,409 --> 00:28:14,649
the pressure at this particular location here
this point A.
249
00:28:14,649 --> 00:28:20,429
So to measure the pressure at this particular
location A, p we can use this principle pB
250
00:28:20,429 --> 00:28:27,259
is equal to pC look at this then we will get
and if you assume p atmospheric is equal to
251
00:28:27,259 --> 00:28:34,259
0 then we get the pressure is equal to the
density of the manometer liquid multiplied
252
00:28:34,350 --> 00:28:41,129
by g the acceleration between gravity and
this depth h2 which we measured and then minus
253
00:28:41,129 --> 00:28:47,799
row, row is the density of fluid in the pipe
row in to acceleration due to gravity multiplied
254
00:28:47,799 --> 00:28:54,799
by this depth h1. So this gives the pressure
pa that means the pressure at point is equal
255
00:28:57,489 --> 00:29:03,840
to row in manometer g in to h2 minus row in
to g into h1
256
00:29:03,840 --> 00:29:10,840
So the basic principle of manometer here is
you can see that here we have just a U type
257
00:29:14,450 --> 00:29:21,450
pipe like this here we put some liquid either
it can be mercury or any kind of liquid and
258
00:29:25,950 --> 00:29:32,950
then we can connect this through a another
pipe where we have to measure the pressure
259
00:29:34,220 --> 00:29:39,059
So this is the point there will be measuring
this is the simple principle used the case
260
00:29:39,059 --> 00:29:40,950
of manometer.
261
00:29:40,950 --> 00:29:47,950
Now, the pressure measurement using the manometer
you can use it in different ways. So first
262
00:29:47,980 --> 00:29:54,739
one is the last slide just only one limb of
the manometer is connected to a particular
263
00:29:54,739 --> 00:29:58,609
location where we want to measure the pressure
and now if you want to measure the pressure
264
00:29:58,609 --> 00:30:03,200
differentially that means between two position
say here A and B
265
00:30:03,200 --> 00:30:09,879
So for that when we use this manometer, U
tube manometer, it is called the differential
266
00:30:09,879 --> 00:30:14,820
U tube manometer
So here we are determine the pressure at the
267
00:30:14,820 --> 00:30:21,820
pressure difference between A here this location
A and this location B. The one limb of the
268
00:30:23,220 --> 00:30:28,659
U tube manometer connected to the pipe flow
or the container at here through a pipe line
269
00:30:28,659 --> 00:30:34,389
this and other limb of the manometer is connected
to the second pipe or the container and say
270
00:30:34,389 --> 00:30:36,119
with respect to this location B.
271
00:30:36,119 --> 00:30:43,119
If there is a pressure difference between
A and B let means between the positions this
272
00:30:43,989 --> 00:30:50,859
pipe and second pipe the we can write the
equation as pA the pressure at this location
273
00:30:50,859 --> 00:30:57,359
pA we can using scales we can measure the
depth as 1 here as shown in this figure and
274
00:30:57,359 --> 00:31:04,359
also we can measure this depth h2 here say
where the manometer liquid is raised So this
275
00:31:04,519 --> 00:31:11,519
is h2 and then we can measure this depth h3
with respect to this point B this h3 can be
276
00:31:13,169 --> 00:31:20,169
also measure finally we can write the equation
as the A plus gamma one h1 minus gamma2 h2
277
00:31:24,080 --> 00:31:31,080
minus gamma3 h3 is equal to pB
So this is the equation which we use for the
278
00:31:31,989 --> 00:31:37,529
differential U tube manometer, since we know
the this specific weight for the density of
279
00:31:37,529 --> 00:31:44,529
the liquid in this first pipe and also know
the specific weight for the density of the
280
00:31:44,529 --> 00:31:51,529
manometer liquid and we also know the density
of specific weight of pipe two for this container.
281
00:31:54,249 --> 00:32:00,379
So with respect to this h1 h2 h3 are already
measured we can get the pressure difference
282
00:32:00,379 --> 00:32:07,159
between A and B using this equation. If there
is the same pipe line if you want to measure
283
00:32:07,159 --> 00:32:13,129
the pressure between this position and the
earlier position that means between A and
284
00:32:13,129 --> 00:32:19,419
B here you can see the say the fluid density
is same we can connect the manometer both
285
00:32:19,419 --> 00:32:23,679
the links like this say first link can be
connected to this location and second link
286
00:32:23,679 --> 00:32:29,309
connected to the second location and then
this difference can be a h1 h2 h3 can be measured
287
00:32:29,309 --> 00:32:34,720
as in the previous space and then finally
this equation can used to get the pressure
288
00:32:34,720 --> 00:32:40,249
difference between point A and B as demonstrated
in this figure.
289
00:32:40,249 --> 00:32:47,249
So now using the U tube manometer we have
seen how to measure the pressure at a particular
290
00:32:47,440 --> 00:32:54,440
location as say here in the previous slide
and then again we say here seen here at a
291
00:32:56,909 --> 00:33:02,739
particular point if you measure of the pressure
this shown be this shows the slide shows how
292
00:33:02,739 --> 00:33:07,559
we are doing and then the second case you
have seen if you want to measure the pressure
293
00:33:07,559 --> 00:33:13,629
difference between two boils or between say
two pipe lines are between to container or
294
00:33:13,629 --> 00:33:19,059
between two locations any type of locations
we can use the manometer like this or in a
295
00:33:19,059 --> 00:33:24,279
single pipeline with the pressure difference
between two points can be measured like this
296
00:33:24,279 --> 00:33:30,989
by connecting the two links of the manometer
at two locations we want to measure the pressure
297
00:33:30,989 --> 00:33:37,779
difference and then we can use this pa the
points of A and B then pA plus gamma1 h1 minus
298
00:33:37,779 --> 00:33:44,239
gamma2 h2 gamma3 h3 will be the gamma1 is
the specific weight of the first pipe, gamma2
299
00:33:44,239 --> 00:33:50,179
is the specific weight of the manometer liquid
and gamma3 is the specific weight of the fluid
300
00:33:50,179 --> 00:33:52,389
in the second pipe.
301
00:33:52,389 --> 00:33:59,389
So, we can see that using the manometer the
major difficult here is that we have to do
302
00:33:59,940 --> 00:34:05,190
many measurements here; we have to measure
to see the pressure difference three measurements
303
00:34:05,190 --> 00:34:12,190
we have to do p h1 h2 and h3. So if the measurements
are not very accurate then the results will
304
00:34:13,980 --> 00:34:19,520
not be accurate, so more over many measurements
are not good why determining the pressure
305
00:34:19,520 --> 00:34:21,800
very accurately.
306
00:34:21,800 --> 00:34:28,800
So, this kind of problems if we consider a
manometer or this mechanism like this where
307
00:34:30,600 --> 00:34:37,600
the diameter d is much larger compared to
the diameter D on this line. So this line
308
00:34:39,530 --> 00:34:46,530
which is d by D is very smaller, then we can
see that with respect to this say if you measure
309
00:34:48,129 --> 00:34:54,250
this Z2 is measured then between this p1 and
if you wan to measure the pressure difference
310
00:34:54,250 --> 00:35:01,000
between p1 and p2 then we can see with respect
to this figure we can write the equation p1
311
00:35:01,000 --> 00:35:04,220
minus since d by D is very small.
312
00:35:04,220 --> 00:35:11,220
So if d by D is smaller, then we can write
the pressure difference as p1 minus p2 is
313
00:35:11,780 --> 00:35:18,780
equal to row g in to z2. So in this case we
are measuring only at this location only z2
314
00:35:19,750 --> 00:35:25,450
is only measured and equation is p1 minus
p2 equal to row g into z2 so that advantages
315
00:35:25,450 --> 00:35:32,450
only one measurement, but it is less accurate
than what we have seen in the earlier cases.
316
00:35:33,440 --> 00:35:40,440
And also say some times you can use the inclined
or tilted manometer that may be more accurate
317
00:35:41,710 --> 00:35:47,299
compare to the earlier one which we say they
are till inclined or tilted manometer is demonstrated
318
00:35:47,299 --> 00:35:52,849
in this figure here. So you can see here say
now if you want to measure the pressure say
319
00:35:52,849 --> 00:35:59,849
p1 here and then if the pressure here is p2
then the pipe with a manometer liquid like
320
00:36:03,210 --> 00:36:08,130
this. So with respect to this figure we can
write the equation as the pressure difference
321
00:36:08,130 --> 00:36:14,910
is p1 minus p2 is row g z2 that is equal to
we can measure using a scale like this So
322
00:36:14,910 --> 00:36:21,190
that is equal to row g x sine theta where
theta is this angle and x is the measured
323
00:36:21,190 --> 00:36:24,039
distance with using this scale here.
324
00:36:24,039 --> 00:36:30,770
So that is called a inclined manometer or
we can also have a tilted type of manometer
325
00:36:30,770 --> 00:36:37,770
like this where we are connected we are connected
this manometer to pipes location A and B and
326
00:36:38,480 --> 00:36:45,230
then pA minus pB will be this l2 is this distance
measured from measured between this level
327
00:36:45,230 --> 00:36:51,799
and this level and gamma2 is the specific
weight of the manometer liquid inside and
328
00:36:51,799 --> 00:36:58,130
then pA minus pB is equal to gamma2 l2 sin
theta with respect to this figure.
329
00:36:58,130 --> 00:37:04,910
So this kind of inclined or tilted manometer
is useful to measure very small pressure on
330
00:37:04,910 --> 00:37:09,920
the pressure difference. So it is better than
the previous one which we are discussed since
331
00:37:09,920 --> 00:37:16,660
it can measure through more accurate accuracy
from there to be earlier one.
332
00:37:16,660 --> 00:37:23,660
So now this manometer can be used for many
applications say one of the generally used
333
00:37:24,079 --> 00:37:29,730
application is the pressure measurement by
medical doctors, they use this manometer for
334
00:37:29,730 --> 00:37:35,309
blood pressure measurement.
So all of you do experience how they are measuring
335
00:37:35,309 --> 00:37:40,309
the pressure the doctor is measuring the blood
pressure what they are doing is a cuff is
336
00:37:40,309 --> 00:37:47,309
placed around arm and then inflated to a pressure
above pressure in arm artery and the slowly
337
00:37:48,099 --> 00:37:53,559
deflated you can see that me using some mechanism
they are inflating or the inflating in the
338
00:37:53,559 --> 00:38:00,559
arm artery and slowly deflated reading attached
mercury manometer to give peak and peak ions
339
00:38:00,589 --> 00:38:06,309
called systolic and the lowest is called diastolic
pressure in millimeter of mercury
340
00:38:06,309 --> 00:38:13,309
So here also a manometer use which gives the
pressure and this also you can see that this
341
00:38:14,030 --> 00:38:21,030
small u tube type manometer. Also this manometer
we can use in many kinds of say especially
342
00:38:21,520 --> 00:38:27,549
in laboratories we generally used manometer
for pressure measurement But now a days this
343
00:38:27,549 --> 00:38:32,940
manometer since it will be have to measure
the scales and then the accuracy may not be
344
00:38:32,940 --> 00:38:38,980
to the level which we require in many of the
sophisticated problem then we will be using
345
00:38:38,980 --> 00:38:44,829
for say we will be going for mechanical type
or other kinds of manometer.
346
00:38:44,829 --> 00:38:51,829
So like gauges that will be discussing next
slide before that let us see the major advantage
347
00:38:52,910 --> 00:38:58,079
of this manometer. So manometer measurement,
the advantages it is very simple there is
348
00:38:58,079 --> 00:39:03,079
no need of calibration it can just say if
there is a U tube with a manometer liquid
349
00:39:03,079 --> 00:39:09,140
we can just connect to the location where
we want and then that will just using a scale
350
00:39:09,140 --> 00:39:16,140
you can measure the height and that you give
directly the pressure or the pressure difference.
351
00:39:16,660 --> 00:39:23,579
So it is very simple and knows your calibration
but some of the disadvantage here are say
352
00:39:23,579 --> 00:39:30,559
this slow responsity if we want to measure
the response using a manometer is very low.
353
00:39:30,559 --> 00:39:36,079
then it is very difficult to measure small
variations if there is with respect to a small
354
00:39:36,079 --> 00:39:41,460
temperature raise if there is a small raise
in pressure this kinds of differences it will
355
00:39:41,460 --> 00:39:48,460
difficult to observe using a manometer also
and it is not accurate since we have measuring
356
00:39:48,530 --> 00:39:50,030
it using a scale.
357
00:39:50,030 --> 00:39:57,030
So the accuracy is also less and we have to
do as we have seen to three times use a scale
358
00:39:59,380 --> 00:40:04,490
and then measure the heights and all then
we have to use this equations. So these are
359
00:40:04,490 --> 00:40:10,990
some of the disadvantages of this manometer
type of pressure measurement. Now we will
360
00:40:10,990 --> 00:40:17,990
discuss a problem um for differential manometer
how we can use a differential manometer we
361
00:40:18,270 --> 00:40:25,270
are getting the pressure. So in this problem
here in this figure we can see here the problem
362
00:40:28,329 --> 00:40:35,329
So the problem is say in figure the liquid
at p and q or water in the liquid in the manometer
363
00:40:37,849 --> 00:40:44,849
is oil here there is a manometer you can see
it is U tube manometer in the other direction
364
00:40:44,960 --> 00:40:51,839
in the inverse position. So we want to measure
the pressure at p and q this is two pipelines
365
00:40:51,839 --> 00:40:58,839
going in parallel at p we want to measure
the pressure which is the pipe p is containing
366
00:41:00,190 --> 00:41:02,760
water and q also containing water.
367
00:41:02,760 --> 00:41:09,760
So you want to measure the pressure difference
between p and q, the water and the liquid
368
00:41:11,839 --> 00:41:18,690
in the manometer water is there in the pipes
and the liquid in the manometer oil with specific
369
00:41:18,690 --> 00:41:25,690
gravity of 0.9 and the reading in the measurement
h1 is equal to here h1 is equal to 300 millimeter
370
00:41:27,289 --> 00:41:34,289
and h2 is equal to 200 millimeter and this
h3 is measured as 700 millimeter, we have
371
00:41:35,410 --> 00:41:42,410
to find the pressure difference in Pascal's.
So to solve this problem now already we know
372
00:41:44,170 --> 00:41:51,170
this h1 h2 and h3 and we want to find the
pressure difference between p and q in Pascal
373
00:41:52,900 --> 00:41:59,589
and then the density or the specific gravity
of oil in the manometer is also even.
374
00:41:59,589 --> 00:42:06,589
So with respect to this figure we can write
the pressure say at location p can be written
375
00:42:09,109 --> 00:42:16,109
as in terms of hp in terms of meters of water
minus h1 this height multiplied by specific
376
00:42:18,420 --> 00:42:25,420
gravity of the specific water and this is
manometer liquid minus s2 into the specific
377
00:42:29,470 --> 00:42:36,470
weight of oil s into s oil, specific weight
of oil plus h3 specific weight of water.
378
00:42:38,859 --> 00:42:45,859
So this is equal to hp the pressure at location
q in terms of meters of water, now using this
379
00:42:49,799 --> 00:42:56,770
equation which can write hp minus this 300
millimeter 0.3 into one the specific weight
380
00:42:56,770 --> 00:43:03,770
of water 0.31 minus 0.2 into 0.9 the specific
weight of the oil is given as 0.9 with respect
381
00:43:06,500 --> 00:43:09,819
to the water 0.9 times of water.
382
00:43:09,819 --> 00:43:16,819
So 0.2 into 9 plus this is again this h3 is
water 0.7 into 1 that gives hq the pressure
383
00:43:19,190 --> 00:43:26,190
difference between p and q is hp minus hq
is equal to minus 0.22 meter of water this
384
00:43:26,799 --> 00:43:33,069
can be we have, we want to get in terms of
Pascal. So p the pressure difference in between
385
00:43:33,069 --> 00:43:40,069
p and q can be written pp minus pq is equal
to this specific weight of water can be specific
386
00:43:40,470 --> 00:43:47,470
weight multiplied gamma in to hp minus hq
So this gamma equal to row into g this is
387
00:43:48,980 --> 00:43:55,980
equal to 9806 Newton meter cube into minus
0.22. So this is equal to 2157.32 Pascal or
388
00:44:03,839 --> 00:44:10,839
2.157 kilopascal. This way using a differential
manometer we can determine the pressure difference
389
00:44:11,450 --> 00:44:18,450
between p and q. So now we have seen using
a manometer.
390
00:44:20,559 --> 00:44:24,609
How we are measuring the pressure whether
it can be a simple manometer connecting to
391
00:44:24,609 --> 00:44:29,730
only one location or it can be differential
manometer to find the pressure difference
392
00:44:29,730 --> 00:44:35,240
connecting on both links of the U tube manometer
that you get the pressure difference and then
393
00:44:35,240 --> 00:44:41,200
we have seen inclined or tilted type of manometer
say and then we have seen the advantages and
394
00:44:41,200 --> 00:44:47,210
disadvantages of manometer. So as I mentioned
earlier this manometer using manometer the
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00:44:47,210 --> 00:44:53,410
pressure measurement is since we have to measure
the height and then using me equations to
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00:44:53,410 --> 00:44:59,910
convert now a days we have this mechanical
and electronic equipments for measurement
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00:44:59,910 --> 00:45:01,130
of pressure.
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00:45:01,130 --> 00:45:07,930
So the advantages of the mechanical and electronic
pressure measurement equipments are it is
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00:45:07,930 --> 00:45:14,930
very fast, low pressure can be measured and
rapidly varying or small fluctuation can be
400
00:45:16,970 --> 00:45:22,940
also obtained accurately.
So here in this slide we can see different
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00:45:22,940 --> 00:45:29,270
kinds of mechanical type on the left hand
side you can see different kinds of mechanical
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00:45:29,270 --> 00:45:36,270
type of this kinds of say the pressure measuring
gages and it is enlarge portion say whether
403
00:45:38,970 --> 00:45:42,430
it can the electronic type is also shown here.
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00:45:42,430 --> 00:45:47,670
So what we are doing is we can just put this
particular location and that will through
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00:45:47,670 --> 00:45:54,670
put sometimes it will directly give the value
of the pressure as a Pascal or low Pascal
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00:45:55,039 --> 00:46:02,039
or which unit we are looking for and one of
the most commonly used gage for pressure measurement
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00:46:02,220 --> 00:46:09,220
is called bourdon gage here this figure shows
the bourdon gage in this the bourdon gage
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00:46:10,819 --> 00:46:17,819
the pressure is measured say by using a say
you can see here a hollow curved tube of gage
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00:46:20,130 --> 00:46:24,940
is used this tube tense to straighten when
the there is a pressure.
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00:46:24,940 --> 00:46:31,079
So set of linkages as you can see in this
see bourdon gage through a set of linkage
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00:46:31,079 --> 00:46:37,700
that is connected to here that a slight motion
at end of tube that is translated into rotation
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00:46:37,700 --> 00:46:44,700
of this dial here and that indicate the gage
pressure directly So directly this rotation
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00:46:45,680 --> 00:46:51,779
is translated into the gage pressure reading
and then we can directly get the reading,
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00:46:51,779 --> 00:46:58,779
so instead of using the manometer or piezometer
type pressure from measurement. Now a days
415
00:46:59,930 --> 00:47:06,059
we are generally using the pressure measurement
using the mechanical or electronic type from
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00:47:06,059 --> 00:47:13,059
pressure gagging equipment like a bourdon
gage as we are seen. So now we will in detail
417
00:47:14,029 --> 00:47:19,059
discuss about the forces on submerged surface
is in static fluid.
418
00:47:19,059 --> 00:47:23,799
So we have already seen some of the theory
related to this important feature of static
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00:47:23,799 --> 00:47:30,559
fluid which we have discussed so far is say
the first one is hydrostatic or vertical pressure
420
00:47:30,559 --> 00:47:36,420
distribution, we have already seen what is
the hydrostatic pressure distribution. So
421
00:47:36,420 --> 00:47:43,420
if you consider a the small basin or tank
of water like this the hydrostatic pressure
422
00:47:44,990 --> 00:47:47,339
distribution linear pressure distribution.
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00:47:47,339 --> 00:47:52,910
We have already seen that is one of the important
feature which will be using and then the second
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00:47:52,910 --> 00:47:59,140
one is pressure at any equal depths in a continuous
fluid are equal. So say if we consider the
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00:47:59,140 --> 00:48:05,059
fluid in this basin or container the pressure
at any equal depths say here if you consider
426
00:48:05,059 --> 00:48:10,950
here or this location or this location pressure
at equal depths the continuous fluid are equal
427
00:48:10,950 --> 00:48:14,950
and then third important feature which we
are discussed is Pascal law.
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00:48:14,950 --> 00:48:20,089
So according to the Pascal law we have see
the pressure at any point at equally in all
429
00:48:20,089 --> 00:48:27,089
directions and then the last point is from
forces from fluid on a boundary access straight
430
00:48:27,599 --> 00:48:33,470
angle to the that boundary. So forces are
acting a right angle to the boundary this
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00:48:33,470 --> 00:48:37,730
all important features we have already discussed
based up on this features. Now, we will be
432
00:48:37,730 --> 00:48:44,730
discussing the forces on submerge surfaces
in static fluid. So first let us consider
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00:48:46,490 --> 00:48:48,420
an inclined plane like this.
434
00:48:48,420 --> 00:48:55,420
So forces as we have see forces at right angle
to the surface, the full pressure on a surface
435
00:48:55,660 --> 00:49:02,660
is obtained as F is equal to p in to delta
A. So if we consider here a small type of
436
00:49:04,029 --> 00:49:10,510
water then we are considering a small inline
plane like this then the pressure say at various
437
00:49:10,510 --> 00:49:17,319
loca6tion F1 F2 like that we can with respect
to the slide we can see the say f1 is equal
438
00:49:17,319 --> 00:49:24,319
to p1 in to here F1 is equal to p1 in to delta
A1 and F2 is equal to p2 into delta A2 so
439
00:49:27,700 --> 00:49:32,470
like that at various location Fn is equal
to pn into delta An.
440
00:49:32,470 --> 00:49:36,890
So that resultant force we can write the total
force is equal to resultant force this R is
441
00:49:36,890 --> 00:49:43,890
equal to p1 into delta A1 plus p2 into delta
A2 plus p3 into delta A3 like that plus pn
442
00:49:46,660 --> 00:49:53,660
delta An. So this gives the resultant force
and generally as we discuss the resultant
443
00:49:54,380 --> 00:49:59,599
force acts through a center of pressure at
right angle to the plane with respect to this
444
00:49:59,599 --> 00:50:04,980
plane we can just find the center of pressure
where the total resultant force is acting.
445
00:50:04,980 --> 00:50:11,170
So the force on submerged surfaces we can
determine like say the resultant force and
446
00:50:11,170 --> 00:50:13,750
we are determine the center of pressure.
447
00:50:13,750 --> 00:50:19,890
And then now we will see the forces on submerged
surface fluid again say in a horizontal submerge
448
00:50:19,890 --> 00:50:24,579
plane. So if we consider a small tank like
this which we have already seen here at the
449
00:50:24,579 --> 00:50:31,579
bottom say is horizontal the resulting force
is equal to the force on horizontal submerge
450
00:50:33,279 --> 00:50:39,069
plane pressure intensity p it is equal to
all points that we can write resultant force
451
00:50:39,069 --> 00:50:45,760
R is equal to pressure into area plane or
that is equal to p into A. So if we concerned
452
00:50:45,760 --> 00:50:50,480
atmospheric pressure here as 0 then the resultant
force in the case of a horizontal submerged
453
00:50:50,480 --> 00:50:57,480
plane is equal to small p into A. So similarly
we will consider now.
454
00:50:58,329 --> 00:51:05,230
Resultant force and center of pressure on
submerged inclined plane surface here this
455
00:51:05,230 --> 00:51:11,619
inclined plane surface say here you can see
this is a tang of water and then an inclined
456
00:51:11,619 --> 00:51:18,059
plane is considered like this and now we want
to determine the resultant force and center
457
00:51:18,059 --> 00:51:25,059
of pressure on submerged inclined plane surface.
So this we will discuss in the next lecture.
458