Good morning, I welcome you all to this session
of fluid mechanics. If you recall in the last
class, we mentioned about the class of problem,
that when a fluid body is moving with uniform
velocity or with uniform acceleration, what
will be the force field generated in the fluid
body, which we will be discussing today. So,
first of all you must know that when fluid
body is moving with uniform velocity, that
is velocity is uniform throughout the velocity
field is no longer equate to be find out,
because velocity is same. Since, that case
we can find out a pressure field from the
velocity field by the application of Bernoulli’s
equation, in case of an inviscid fluid.
And in case of a viscous fluid also since
the velocities are uniform, there is no shear
stress developed in the fluid, because there
is no velocity gradient. So, shear stress
cannot come into picture. So, in either cases
the fluid will behave as an inviscid fluid
or you can tell it is an irrotational fluid,
where pressure field is found straight from
the Bernoulli’s equation, that p by rho
plus v square by 2 is constant in the flow
field. Of course, with the elevation head,
then we can find out the constant with some
reference boundary condition. This is a very
simple class of problem with uniform velocity,
simply a rigid body translation or a rigid
body rotation.
Now, if we consider a uniform translation,
what is the case? That means if, a fluid body
is translated uniformly, this happens for
example, a fluid body contained in a tanks,
solid tank in practice, that tank is accelerated
in certain direction with a uniform acceleration,
in some direction, it may be along the coordinate
direction. So, in any direction if the tank
is accelerated uniformly, what will be the
force field generated or the pressure field
generated in the fluid body? So, this is the
precise problem for today. So, let us now
analyze this problem with uniform translation.
Let us consider a axis like this o x, o y,
o z with this coordinate axis, let us consider
a tank like this. Let us now, not tank, let
us consider a fluid body. Let us consider
a fluid body now, appropriate to the coordinate
axis, in a Cartesian coordinate system. Now,
we take a general case that the fluid mass,
that means this fluid mass, which we have
taken here separate fluid mass is accelerated
in both x y and z direction. These are the
component of acceleration, that means it is
accelerated in x direction by a x, accelerated
in the y direction by a y and accelerated
in the z direction by a z. These are the components
or acceleration in x y and z direction.
Well now, let us define as usual d x is this
dimension, d y is the dimension of the parallelepiped
along the y direction and let this is d z,
that means d z is this one, this one is, let
this one is d z. this one is d z. That means
the parallelepiped of dimension d x d y and
d z. Now, how to find out the pressure field?
So, simple thing is that, the way we found
out the force equilibrium condition, hydrostatic
force equilibrium condition, it is the same
one only that is, inertia force is acting
because of the acceleration, this is extremely
simple.
If you define the pressure at the center,
it is better more generalized approach, I
did earlier that p p plus del p, del rather
p is defined at the center uniquely for this
parallelepiped. So, then what is the force
on this plane? If we now again define, to
recognize the plane A, B, C, D and this one
is E, this one is F, this is G, I have not
drawn this with dotted line. These are the
back lines, from if you see from this. However,
so the plane A, F, H, D is the x plane and
C, G, E, B are the x planes. Now, what is
the pressure in this x plane and what is the
pressure in this x plane, in this direction?
Pressure in this x plane, which is separated
from the centre by a, d x by 2 distance in
the positive directions, will be p plus del
p del x d x by 2. Neglecting the higher order
term, that means this is the increment of
p due to a change in d x by 2 into positive
direction. Therefore, pressure here will be
p minus, that is, this is also an increment
of p from that to this plane for a change
in x. That is separated by a distance minus
d x by 2; that means d x by 2 in the negative
direction. That is why the negative sign is
there mathematically. Similarly, for this
y direction, there will be pressure forces
for the y plane. That means for the y plane,
that means the pressure force is in D, C,
B, A similarly, A, G, E, F.
For A, G, E, F this pressure forces, this
force, that means this force perpendicular
to the face A, G, E, F, which is at a distance
d y by 2 from the centre of this parallelepiped,
along the y direction. Therefore, this magnitude
will be p plus, I think you follow it. Del
p del y into d y by 2, that means the pressure
in this plane H, G, E, F that is the back
plane, which is displaced from the point p
along the y direction by d y by 2, because
this is d y. So, this is centre point, so
d y by 2 plus d y by 2, so it is increased.
Similarly, the front plane that D, C, B, A
or A, B, C, D y plane, which is separated
from p or distance from p in the minus d y
by 2, that means in the opposite direction
of the positive y direction.
So, mathematically this will be p minus del
p del y into d y by 2, all right? Similarly,
the pressure at these planes, this pressure
will be, what this will be? p plus del p del
z into d z by 2. Similarly, pressure here,
this pressure will be, p minus del p del z.
That means with respect to a pressure p, defined
at the centre of the parallelepiped, we can
specify the pressure at each faces. That means
2 x faces, 2 y faces. x faces means faces
perpendicular to x direction, 2 y faces faces
perpendicular to y direction and two z faces.
So, 6 faces, the pressures are specified like
this.
Now, simply we find out the force balance
equation. Now, in x direction, what is the
force balance? x direction, the force balance
is the net force acting in the positive x
direction, must equal to the mass times the
acceleration in positive x direction. Obviously
by Newton’s second law of motion, that means
net force acting in the x direction, will
be the force. That means this pressure into
this area minus pressure into this area, so
that will be p minus del p del x, very simple.
d x by 2 into the area, that is area of the
x plane A, F, H, D is d y d z minus the force
exerted in this direction, due to the pressure
in C, G, E, B plane.
That is p plus del p del x into d x by 2 area
is same, because it is parallelepiped. The
area is same for 2 x plane and that that must
be equal to rho into d x d y d z times a x.
So, therefore we see from these equation,
del p del x is equal to minus rho a x. So,
this is one very important equation. Similarly,
if we make the y direction, y direction, if
we make the y direction balance, force balance,
so it will be p similar way minus del p del
y d y by 2 into, what is y direction area?
d x d z, that means this is the force on the
plane A, B, C, D pressure is p minus del p
del y d y by 2 times d x d z minus the force
acting on the G, H, E, F plane.
Due to this pressure, which is p plus del
p del y d y by 2 in the similar fashion p
plus del p del y d y by 2 and the area is
same d x d z, very simple is equal to rho
times d x d y d z. That is the mass of the
elemental volume times the acceleration uniform
acceleration in y direction.
So, this gives rise to the equation del p
del y is equal to minus rho a y. If you solve
this, you will get. So, in x and y direction,
we get del p del x minus rho x and y direction
del p del y minus rho a y. now, what is in
z direction? Now, in z direction if you see,
z direction, here we can write, some space
is there.
Here, we can write for z direction, z direction
what is the net force acting on this volume
in the positive z direction, please? p, so
minus del p del z into d z by 2, d x d y,
well minus force due to this pressure on this
surface, this is the force due to this pressure
on, A, F, E, B surface minus the force due
to the pressure on D, A, G, C surface. That
is p plus del p, any difficulty you please
tell me. d z by 2 into d x d y, all right?
d x d y, you can see that d x d y, then another
force, what is that one?
Wait, that is acting downwards, so minus rho
d x d y d z times the g is equal to again
rho d x d y d z times a z. That means, this
is the acceleration, uniform acceleration
of the body in the positive direction of z
axis it is given upward. So, if you solve
this one, if you solve this one, you get,
if you solve this one, what you get?
You get, well you get, if you solve this one,
if you solve this one you get del p del z
is equal to minus rho d x d y d z, yes. rho
g is equal to minus rho g, sorry, is equal
to minus rho into g plus a z minus del p del
z. This is minus del p del z, if I take there,
then it comes here minus rho into g plus a
z from here, this is clear? Now, therefore
we can write that in this case, what we get
in a fluid body, which is very important,
x, y and z. That means if a fluid mass is
uniformly accelerating in both x, y, z, with
an acceleration component a x a y and a z,
then the pressure distribution in the fluid
mass in x, y, z direction, now follows these
equation.
If you recall or just we have derived del
p del y is minus rho a y and del p del z is
minus rho into g plus a z. Now, if we see
these equation and compare with the hydrostatic
case, now you see in hydrostatic condition
del p del x was 0. That means there was no
pressure distribution in x. Similarly, y direction
in y, if z direction is vertical that means
if a x is 0, hydrostatic condition a y is
0 and a z is 0. If there is no acceleration
or in case it moves with uniform velocity,
then also a x, a y, a z 0. So, whether a fluid
moves with uniform velocity or does not move
at all, there is no variation of pressure
in any of the horizontal direction, but only
it varies in the z direction, which becomes
equal to minus rho g physically.
It means that it supports the weight of the
fluid, the pressure difference created in
the z direction supports the weight of a fluid
element, but when it moves with an acceleration
a x, definitely the pressure will vary. For
example, if we consider a fluid element like
this, so pressure should essentially vary
in the x direction, which should take care
for it is force acting on the body to cause
an acceleration a x. So, this physically implies
this thing. So, pressure should essentially
vary in the y direction, so that the net pressure
force in y direction should take care of the
acceleration in the y direction. Similarly,
for z direction the pressure should essentially
vary to take care of the weight of the fluid
plus the acceleration.
So, simply these are the equations of tank.
Now, in most practical cases what happens
is, that if we have a tank, now you see this
practical case, if there is a free surface
like this, initially if this tank is now accelerated.
For example, let us consider the acceleration
in this a x direction, now what happens? You
will see the constant pressure lines will
be like this, it changes like this. This is
the constant pressure lines. So, to solve
this type of equations, let us derive in general.
In this case, what should be the equations
or planes equation of the planes for constant
pressure?
Let us consider this for a two dimensional.
First for simplicity, let us consider a two
dimensional problem, that means x and z. Let
us consider a two dimensional problem, that
means in x and z. If we see a two dimensional
problem with a x 0, a z 0, so we see del p
del z is minus rho g. So, in this case of
two dimensional problem, we can solve this
with, well we can solve this like this, that
in a two dimensional problem of this nature.
Let us write, d p is equal to del p del x.
Let us for simplicity take a two dimensional
case, del p del z d z that means p is a function
of x z. Then along a constant pressure line,
we are going to search for a constant pressure
line in the x z plane. If a two dimensional
case a x is 0, a z is 0, z is the vertical
direction, so constant pressure lines will
be horizontal, that we know. Now, here let
us find out in general in case, when there
is an acceleration, what is del p del x? It
is minus rho a x d x. What is del p, del z?
Del p del z is minus rho, as we have seen
already a z d z.
So, we can integrate it to get the value of
pressure at any point, in terms of x and z,
but our first idea is to find out the equation
of constant pressure line, along the constant
pressure line d p will be 0. Therefore, we
can write d z d x is equal to minus rho rho
cancels, a x by g plus a z. That means therefore,
we can draw lines, whose slope is like this,
that means these are the constant pressure
line, that means in a practical problem of
tank, the constant pressure lines will be
like that. That means if there is a free surface,
the free surface will be like that. So, this
is the slope with this, so this is theta.
So, that tan theta is equal to minus a x by
g plus a z.
Now, when a x is 0, that means there is no
acceleration in this direction, even if there
is an acceleration in z direction. So, the
constant pressure lines will be horizontal
because there is no pressure variation in
the x direction, because del p del x is rho
a x del p del x is minus rho x. So, a x is
0, there is no pressure variation, that means
it is simply given the horizontal line. Try
to understand, as the constant pressure line
but the del p del z will be given by rho g
plus a z. That means as if g is increased
by an amount g plus a z when a z is the acceleration
in the vertical direction.
But if there is no acceleration in the vertical
direction a z is 0, only in the horizontal
direction, then tan theta is equal to minus
a x by g. In that case a z is 0, that means
rho a x d x minus rho g d z is equal to d
p. Therefore, if a tank is accelerated for
example, in this direction, in one of the
horizontal direction, the initial free surface,
which is horizontal will now take a shape
like this. So, this will be the constant pressure
lines, so the free surface will be p is equal
to p atmospheric. So, you have understood
this? Now, with this understanding some very
important conclusions will come, that when
a x is 0, the surface is…
Now, again if this a x is 0, when a x is 0,
1 square, there is a z in that direction,
then the liquid free surface will remain horizontal,
as it is. But the pressure distribution will
change because d p is minus rho g plus a z
d z, all right? That means, if we integrate
it p is equal to minus rho a x x minus rho,
these are constant g and a z z plus C some
constant arbitrary constant. So, when a x
is 0, this is 0, so p is simply given by minus
rho g plus a z z plus C. That means, if we
consider a free surface form any datum plane,
if we, or exact, if we take this as the axis
o x o z.
So, if this be z 0, and if another point we
take z, by applying this. Here if it is p
atmospheric, a simple application of this
equation, without a x we can write p at any
point minus p atmospheric is simply rho into
g plus a z into z 0 minus z because you will
put z is equal to z 0, where p is equal to
p atmospheric, defining z 0 is this coordinate
taking this as axis. So, C will be found out
p atmospheric plus rho into g plus a z into
z 0. Then we will put that value of C and
find out the value of pressure p at any other
coordinate z. We will get, that means if this
be the depth h, that means from the free surface
we know the pressure is more than the atmospheric
pressure by rho g h.
In case of a static fluid or fluid moving
with uniform velocity, but in case of a fluid
moving with uniform acceleration in z direction.
So, contribution of z direction acceleration
is to change the g, that means it is increasing
g by g plus a z and take care of that, all
right? If there is no acceleration in x direction,
there is no variation of pressure in the x
direction. So, lines will be horizontal, but
if there is an a x, then the lines will not
be horizontal because the pressure at this
two points will be varying. As we have told,
then the constant pressure line is like.
That means, in that case at two same horizontal
point, the pressure will be changing, pressure
will be changing because this is the free
surface then. So, here pressure is equivalent
to this depth and here pressure is equivalent
to this depth, whether it is depth into rho
g or rho g plus a z, depends upon the a z.
But this will be more than this therefore,
two horizontal points are not in the same
pressure because there will be another constant
pressure line, through this point, which is
higher than this.
So, this two difference in pressure represents
a net force in the x direction, that has to
be, to balance the inertial force or the force
required to cause the acceleration a x. Therefore,
this thing should be very clear from physical
sense and mathematical sense this is the constant
pressure line. Now, another interesting results
comes out.
If a z is minus g, that means if the tank
forget about a x a x is 0, if a tank is accelerated
downwards with a acceleration equal to acceleration
due to gravity, that means if the tank is
allowed to fall freely, the free fall of the
tank in that case, we can say a z is equal
to minus g according to a mathematical nomenclature.
In that case del p del z, I will not write
here from the differential equation itself
del p del z is equal to 0 del p del z is equal
to 0 from the basic differential equation
a z is minus g, that means from this equation,
you can put that a z is minus g.
That means p minus p atmospheric is 0 for
any value of z, that means in that case the
pressure at each and every point remains same.
Similarly, if there is no acceleration a x
in these direction, then del p del x is also
0 minus rho a x, that means pressure also
does not change in that direction. That means,
if there is a free surface and the open tank
and if you allow the tank to fall freely,
the pressure throughout the liquid bodies
atmospheric pressure. pressure neither changes
along this pressure, neither changes along
z.
So, pressure neither changes along z comes
simply from the equation, that del p del z
is minus rho g plus a z. You know a z minus
g automatically 0 physical explanation is
very simple. That the mass or weight of the
body to balance it the pressure differential
is not required because the weight of the
body causing, it is free fall. A body falls
freely because of its weight, that means the
force by which the gravitation is attracted
force of gravitation or the force by which
attracts the body.
That means it is because of its weight, this
is falling freely. So, if the fluid body is
allowed to fall freely the pressure differential
in z direction seizes to exist, that means
there is no pressure differential. So, pressure
is uniform along the z direction, that is
del p del z is 0, this is most important fact,
you can experience it in your house or hall.
One of our colleague experience that, if you
have a tank, that is written in my book and
there is a hole made and you carefully fill
up the tank with these hole closed by your
finger, then you let the tank fall freely.
If it fall freely, you just let it down go
down from your from the roof of your house,
you will see the water will not leak. V is
0, why water will not leak because here is
the atmospheric pressure. So, water usual
case leaks, when is a static condition or
any other acceleration, when the pressure
here will be more than, the pressure here
is p, if it is more than p atmosphere, then
only water will leak
So, if you fill up a container and make a
hole at the bottom the water will come out
like anything because the pressure here is
rho g h plus the p atmosphere, if p atmosphere
is pressure. But when it falls freely the
pressure throughout is atmospheric pressure,
because there is no pressure gradient. If
there pressure is p atmosphere, so here also
p atmosphere, so water will not leak from
the hole. So, this is a very practical application
of it, all right, Simple? Now, let us solve
a problem, it is very simple. Let us solve
a problem example.
One determine the equation of free surface,
this problem is of course, an worked out example
in my book, but still I have given some extension
to this problem. So, first part is worked
out, which is a straight forward application
of the equation, that we have derived just
now. Determine the equation of free surface
of water in a tank moving with a constant
acceleration of 0.5 g along the x axis. as
shown in figure. I will show the figure, you
write, that means a tank is moving with a
constant acceleration of 0.5 g along the x
axis and water is contained in the tank. The
first part is the determine the equation of
free surface of water.
Next part is that, to determine the equation
of free surface. You are, you do not require
any other data, next part if the tank is 4
meter long in x direction, that means in the
horizontal direction 2.5 meter, that I will
show the x direction. Means, the direction
in which it is being accelerated 2.5 meter
wide, that means in a perpendicular direction
and contains water up to an initial depth
of 2 meter, contains water up to an initial
depth of 2 meter, find the hydrostatic forces
on each faces of the tank.
Please please if the tank is 4 meter long,
that means I should wait 2.5 meter wide and
contains water up to an initial depth of 2
meter, find the hydrostatic forces on each
faces of the tank? So, this problem, if you
solve the problem tells like this, okay?
Let me draw the problem, this is the problem,
that means initial depth is 2 meter, that
means this is 2 meter, which is not required
for the first part, but still this is x direction.
Let us consider this is x direction, that
means in this direction it is accelerated
a x 4 meter and 2.5 meter wide, which is perpendicular
to this plane. Now, initially first part is
that it is given a x is equal to 0.5 g. Well,
so the, we know the surface will be like this,
constant pressure surface we just draw it
qualitatively without knowing anything, but
better, let us write the equation del p del
x is minus rho a x del p del y is minus rho
a y and what is del p del z? Del p del z is
is equal to minus rho g plus a z, well g plus
a z.
So, here there is no a y 0, there is no one
dimensional, that means acceleration in this
direction. So del p del x is minus, no sorry,
a z is 0, so del p del z is rho g. So, del
p del x is minus rho x. So, here del p del
x is minus rho x and del p del z as usually,
even if there is no acceleration, it is a
minus sign. But we can straight forward use,
that d z d x that means, d p is del p del
x d x into del p del z d z, as I have done
earlier, just you can see that I have done
earlier. Just here you see that d p is del
p del z this one. So, where I have written
is d z d x is minus a x by g plus a z, that
means I can use this formula straight forward
d z d x. That means I can again deduce this
formula from here in general a z. If it make
a z 0 for the specific problem, that means
for the specific problem it is minus a x by
g. So, it is very simple.
So, in this case it is minus a x by g, that
means it is nothing but minus del p del x
by del p del z. That is del p del x by del
p del z, you can derive it. You should not
write a formula directly because this will
give you del z del x or d z d x, that is minus
a x by g. So, therefore, d z d x is equal
to minus 0.5 g. that means equal to minus
0.5. So, if I define this as theta, this then
tan theta is equal to minus 0.5 and this gives
a value of theta. Better I write the value
153.45 degree. I have solved it; that means
this angle is 26.56 degree, which will be
required. 26.56 degree, so this is the value.
So, now equation of a, this will be required
afterwards, now we cannot use this. Now, better
we find out the equation of free surface,
you integrate this. z is equal to 0.5 x. if
you integrate this. The value of tan theta
will come afterwards, 0.5 x plus C. So, this
is the equation, of the free surface. That
means the equation of this line, which makes
a slope tan theta minus 0.5, these are the
parallel equations of the free surfaces with
C as the parameters. Now, tactfully one can
find out C and can make it 0. If we take this
surface of intersection, if we take at the
middle for symmetry, if we take x here and
z here, then we can define that at x is equal
to 0, z is equal to 0, if we take here axis.
So, that C is 0, so z plus 0.5 x is equal
to 0 is the equation of the free surface.
z plus 0.58 is equal to 0. so along the free
surface always z plus 0.5 x is equal to 0
or you can write z plus 0.5 x is equal to
C. So, the value of C depends upon a arbitrary
reference for choice of the coordinate axis,
this part is all right? This part is all right?
Now, next part, well I can draw the figure
again for the next part. For the next part,
let us consider this is, again for next part.
Let us consider this again 2 meter and this
is 4 meter now, the free surface we have already
determined like this. So, water is carrying,
carried like this. This is now the water,
this is the earlier free surface, this was
the free surface earlier. So, this was the
earlier free surface, now the free surface
has taken a shape like this, where this is
the liquid, this angle. As I have already
found out theta is equal to 26.65 degree.
Now, we have to find out the forces on each
plane. Now, tell me first, what should be
the force on the horizontal plane, bottom
plane? Bottom plane is the very good, so weight
of the liquid, so whichever way it is accelerated
in the horizontal direction, a x, it does
not matter the total weight of the liquid
will be the force here. But if it could have
been accelerated in the z direction, then
the force will be weight plus the acceleration.
That means, as if we will consider g is increased
by g plus a z or minus a z downward, if there
is a free fall, what is the force on the bottom?
0. You must know this thing very carefully.
So, now the bottom plane the weight is simply
we will equate the weight, that means rho,
that means rho is 10 to the power 3 into V.
V is 4 into 2.5 into 2, it is the total volume
initial depth was 2. So, V into rho into 9.8,
I have taken care of all these things, V into
rho into 9.84 into 2 into 2.5 9.8 Newton.
So, this becomes, is equal to 196 kilo Newton.
This is as simple as anything, now what are
the forces on this two planes? First of all
consider leading plane, leading face. What
is leading face? This face that means, face
leading means the acceleration is taking place,
the leading face. What will be the component
forces? If you see this face, in this direction,
this is a face like this, where this dimension
is 2.5 meter. I see from this direction this
is the face, so this is the free surface.
So, free surface is this, so this height.
Let this height is, what is this height? This
is not 2 because water touches, so free surface
is parallel here because there is no acceleration
in this direction. So, free surface is horizontal
here from this point, this is not 2. So, this
value is what? This value, let this is H 1,
this H 1 is what H 1 is? I write H 1, what
is H 1 2 minus 4 2 into tan of 26.65. 2 is
the half of this; that means this minus this,
what is this value? This value is 1 because
it is 0.5, correct? So, you see if the problem
is said like that, students will not be having
any difficulty.
So, therefore this is the force on a vertical
surface, which is submerged up to a depth
of 1 meter, H 1 is one meter and whose width
is 2.5 meter. So, therefore, it is rho g into
the pressure at the centroid or the centre
of area into 0.5 into 2.5 into 1 Newton, so
this becomes also very simple calculation,
12.25 kilo Newton. Similarly, for this trailing
face. Trailing face please tell trailing face,
trailing face is the same thing, only difference
is the
3
Very good. Now, this is 3 meter H 2. H 2 is
equal to 3 meter because 2 plus, this quantity
2 plus tan 26.25 degree. which is 0.5, that
is 2 into 0.51, so 2 plus
1
Very good.
So, this remains as it is, the width perpendicular
direction 2.5 meter. So, what is this value?
10 to the power 3 9.8 into 1 into 2.5 into
2
1.5
1.5 because 3 centre is 1.5 rho g 1.5 into
3
3 into 2.5 area. I am sorry, so do not, do
that mistake. I may be escaped as a teacher,
but you in the examination… So, 1.5 is the
centre of area, so 2.5 into 3. So, after that,
if you do a mistake, the deduction of mark
will be less, because your concept is clear,
because some silly mistake your concentration
was not there, you have calculated wrong.
But here, if you write 1, the way I gave written,
then I will think that you have not understood,
centre of area is middle of that. That should
be 1.5 and area is 2.5 into 3.
So, then you can, so the mistake in calculations
you can afford to do, but you should not.
The probably I have got the result 110.25
kilo Newton. So, now remaining other two lateral
faces are 3, where the forces will be identical.
That means a vertical face like this, where
the free surface of water is adhering like
that. This is the water, so how to find this?
Integrating.
By integrating, very good simple by integration.
Very good by simply integrating, I can use
another thing, well good by simple integrating.
That means you can take a strip like that.
For this purpose, I define this as the x axis
and this axis z axis. Therefore, for this
strip, the force d F is… What if this is
z this coordinate of the free surface with
respect to this x axis is z, that means this
is z. So, centre of area is z by 2, so rho
g z by 2 and its area is, if we define this
thickness as d x because this is the x direction
that is very simple, z into d x. So, simply
F for entire surface will be d F is equal
to integration.
Rho g take outside incompressible, that is
a liquid and z square by 2 d x. 2 also I can,
could have taken outside, though this should
be H 2 minus H 1, form this I am going the
positive direction. Another constant is that
d z d x is equal to… What is d z d x minus
tan theta? Here if I define this as the theta,
now again it is as the theta, so tan theta,
so simply d z d x. My no question of pressure
head this is geometry because d x is increasing,
d z is decreasing with increasing z z is increasing,
x z is falling.
So, d x by d z is equal to minus tan theta.
That means if I now substitute it, it should
be, F is equal to rho g by 2. That means we
can take H 1 H 2, change d x is d z by tan
theta, that means you take tan theta out d
z d x is d z by tan theta. That means z square
d z ,that means simple rho g z cube 6 3 into
2 6 tan theta into H 1 cube H 1 H 2 we know,
sorry H 2 cube and H 1 cube H 2 has gone up,
because of this minus this tan theta is tan
of 26.65 degree. From geometry, we have taken
this theta, this thing is this, is the force
acting on each of the lateral faces. That
means front face and there is another back
face, any problem?
This is the equation, please, no problem?
So, with full confidence on you, I can leave
you today for this midterm examination. So,
this is up to this, In next class, I will
discuss few other problems of this section
and I will start the next section of this
course, Applications of conservation equations
in fluid flow problems in practice. So, thank
you.
Good morning, I welcome you all to this session
of fluid mechanics, today we will start a
new chapter, a new section, that is fluid
flow applications. We have already discussed
the conservation equations; that is the conservation
of mass, equations for, conservation of mass,
conservation of momentum and conservation
of energy and we have solved some problems
related to finite control volumes, both inertial
and non-inertial control volumes. Now, in
this section, chapter, we will discuss few
practical problems of fluid flow applications,
which will be solved on the basis of the equations
of for conservation of mass momentum and energy.
The applications of those equations, better
we can say that, it is the applications of
conservation equations to fluid flow problems.
At the outset, we will discuss little bit
about the Bernoulli’s equation or Bernoulli’s
theorem, which we have discussed earlier,
a little elaboration on that. If we recall
we earlier recognized the Bernoulli’s equation,
as the equation of mechanical energy or mechanical
energy equation in case of a fluid flow, which
tells about the conservation of energy in
a fluid flow. We have recognized that, if
the fluid is inviscid, obviously the mechanical
energy remains constant.
If there is no energy added from outside,
no mechanical energy is taken out of the fluid
and if the fluid is inviscid, that means viscous
action is not present, that fluid is inviscid,
no viscosity, that means no friction. So,
mechanical energy cannot be dissipated in
terms of intermolecular energy or heat as
you can tell. So, in that case total mechanical
energy remains constant. We have seen that,
in general these mechanical energy total mechanical
energy remains constant along a streamline.
We have recognized different forms of mechanical
energy, as the pressure energy, kinetic energy
and the potential energy.
If you recall the equation, you can tell that
p by rho, which represents the pressure energy
per unit mass plus V square by 2, which represents
the kinetic energy per unit mass. We have
to always recall in this form plus g z, where
g z represents the potential energy per unit
mass here. Of course, we consider only the
Earth’s gravitational force field as the
only body force field. Therefore, the potential
energy per unit mass will be g z, where g
is the acceleration due to gravity and z is
the elevation of the point concerned, where
we are considering the energy per unit mass
from an arbitrary datum.
So, p by rho plus V square by 2 plus g z thus
constitutes, the total mechanical energy and
which is equal to constant for an incompressible
steady flow because this form of the equation,
comes only for a steady incompressible flow.
That is of course, important otherwise integral
d p by rho, will come for a compressible flow,
but p by rho this term comes for an incompressible
flow. So, for an incompressible steady flow
p by rho plus v square by 2 plus g z, simply
the sum of the 3 3 components of the mechanical
energy, is constant along a streamline.
If you recall the derivation of this equation,
we did it by integrating the Waeler’s equation
along a streamline. Probably you can recall
this, that integrating the Waeler’s equation
along a streamline. Therefore, the constant,
which came out of these integration, was a
constant. That means it was valid only along
a streamline. Therefore, essentially the mechanical
energy varies from streamline to streamline
in general. But now we will extend this, in
case of an irrotational flow; that means if
we add a further constraint in the flow.
That if the flow is irrotational, already
you have started, what is an irrotational
flow in fluid kinematics? That a flow is irrotational,
when the rotation at each and every point
is 0 and you know the rotation of a fluid
element at each point is defined as the Carl
of the velocity vector, Carl V in different
coordinate system. We can expand this Carl
V physically, the rotation means the arithmetic
average of the angular velocity of the two
linear segments meeting at that point, which
was initially perpendicular.
So, rotation is 0, means the flow is irrotational.
So, if we add a separate or additional constant
of irrotationality, that means if flow becomes
irrotational in which its steady incompressible,
then these mechanical energy is constant throughout
the flow field. That means p by rho plus v
square by 2 g plus g z is equal to constant,
not only along a streamline, but at any point
in the entire flow field. This we will derive
first, which is a very simple derivation,
let us do it.
Let us see this, let us first take a two dimensional
case, a two dimensional flow. A two-dimensional
flow, in a Cartesian coordinate system in
a Cartesian coordinate system, in a Cartesian,
rectangular that is Cartesian coordinate.
A Cartesian coordinate system and in this
case, let us consider only x and y, no z.
So, y we consider and this we consider the
y axis, is in the vertical direction. y directed
positive upwards, vertically upwards, so that
g is acting like this. So, if this we consider
as the frame of coordinate and also we consider
a steady and incompressible flow, if we consider
a steady and incompressible flow. steady and
incompressible flow
Then if we recall the equation of motion,
that is the Waeler’s equation, we can write
that for a steady the temporal, that you will
not come u del u del x plus v del u del y,
is equal to minus 1 upon rho del p del x,
if you recall? So, this is D U D t, big D
U D t, big D U D t that means this is equal
to minus 1 upon rho del p del x this is the
x direction. So, this has been expanded del
u del t is 0 for steady state. So, I have
not written del u del t is 0 for steady state
and rho is minus 1 upon rho. Then this is
the form x direction equation, y direction
equation. Similarly, will be D v D t if you
recall, is equal to minus 1 upon rho del p
del y.
So, if you expand this, in case of steady
flow, that means del v del t is 0. Then you
get u del v del x plus v del v del y is equal
to minus 1 upon rho del p del y. So, this
two equations, are the equations for incompressible
steady flow, with respect to a two dimensional
Cartesian coordinate system. Of course, here
I am wrong, minus rho del p del y, here g
is here, acting. So, we have considered the
y axis, such a way that vertically upwards
directing positive. So, g is in the negative
direction. So, body force term will come,
which will be here minus g because these are
the force per unit mass, because here these
left hand term is the acceleration per unit
mass.
So, first term on the right hand side is,
the pressure force per unit mass. So, per
unit mass the body force will be minus g why
minus per unit mass it will be g m g divided
by m g and it is in the negative direction
of the positive y axis. Therefore, minus g
this comes as the body force. If you recall,
the body force here, the body force is 0 in
the direction of x there is no body force.
So, we have chosen the y axis, in such a way
that it is directed upwards, vertically upwards,
so that the gravity force come as the body
force, so per unit mass basis it appears as
minus g. I am sorry, so this term will come
now. What we will do? We simply know that,
the basic approach for finding out the energy
equation is to, multiply the momentum equation
with the displacement.
So, what we do? If we multiply this equation
x direction equation by d x and multiply the
y direction equation by d y, both this sides
and add it, what we get? u del u del x d x,
so we first take the u term common, then from
this equation it will be del u del x d x and
from this equation it will be, u del v del
x d x, u del v del x d y sorry, u del u del
x d x plus v del v u del v del x d y plus
v del u del y d x plus…
