Welcome back to the video course on fluid
mechanics. In the last lecture in the topic
of dynamics fluid dynamics we were discussing
about dynamics of fluid flow we were discussing
about the moment principle and its applications.
We have seen that in the case of steady state
system with respect to the linear moment principle
from the Newton’s second law, we have derived
the basic equation as sigma Fx is equal to
rho Q beta2 Vx2 minus beta1 Vx.
Where this sigma F x is the total force acting
upon the control volume external force on
the control volume that is equal to the linear
moment change with respect to system this
we have seen where beta is the moment coefficients
correct or moment correction factor as we
have seen.
We were discussing about the applications
of moment equations we have seen that the
linear moment principle. We can utilize for
various problems like wherever to find out
the force due to flow of fluid around a pipe
bend enlargement and contraction or we can
use the principle to find out the force on
the nozzle on the outlet of the pipe or the
impact of a jet and force due to flow around
a curved vane etcetera this different types
of problem you can use this linear.
Moment principle or linear moment equations,
that we were discussing in the last lecture
and we have seen me of the applications first
case what we have discussed earlier was the
force due to flow around a pipe bend we have
seen.
We will be considering as basic steps the
basic steps involved or we will be considering
control volume, we will be calculating the
total force on the system and including the
pressure force body force we will be finding
the resultant force. We will be equating with
respect to the moment change that is what
we have seen and we have derived the equation
for one of the example force due to flow around
a pipe bend in the last lecture.
The second case we will see here the force
on a pipe nozzle generally we use a nozzle
is used to create a jet we want to find out
with respect to the nozzle what is the force
coming upon the pipe where the nozzle is connected,
as we have seen earlier here al we will be
considering a control volume for example here
we are considering this particular problem
here you can see that a pipe is coming and
nozzle is connected and the control volume
considered is here.
Here the velocity of entry is u1 and velocity
for exit is u2 and the area of cross section
entry at the entrance is A1 and area of cross
section at the entrance is A2, now the control
volume considered is shown in this slide and
the, with respect to the nozzle the fluid
is contracted at the nozzle and the forces
are induced upon the pipe where the nozzle
is connected. First we will consider the total
force, the total force FTx in the x direction
we are considering FT is equal to FTx is equal
to the mass of fluid or rho into Q where rho
is the mass density Q is the discharge coming
from the pipe into the velocity difference
u2 minus u1 where u2 is the exit velocity
and u1 is the entrance velocity.
The total force with respect to this the nozzle
which is acting upon the pipe is FT is equal
to in the x direction FT is equal to FTx is
equal to rho Q into u2 minus u1, here if you
use the continuity equation Q is equal to
A1 u1 is equal to A2 u2, from the continuity
equation. We can get the discharge Q is equal
to A1 u1 is equal to A2 u2 hence we can write
the total force in the x direction FTx is
equal to rho Q square into1 minus A2 minus1
by A1 as shown in this slide, like this we
can find out the total force as per as the
pressure force is concerned here we have we
have already found the total force with respect
to the x direction then next step is you will
be finding the pressure force.
The pressure force F p is equal to in the
x direction Fp is equal to FPx that is the
pressure at1 minus pressure at force at section
two this is section one here section two.
The pressure force the effective pressure
force is equal to pressure at section1 minus
pressure at pressure force at section two
using the energy equation or the Bernoulli’s
equation, we can write with respect to the
section one and section two we can write p1
by rho g plus u1 square by2 g plus z1 is equal
to p2 by rho g plus u2 square by2 g plus z2
plus hf.
Where hf is the losses due to friction here
if we neglect the friction losses here then
hf is equal to 0 that we can write the pressure
force from the Bernoulli’s equation if we
consider this particular case z the data head
is z1 is equal to z2.
The pressure outside is al here when after
the jet is here the from the it is coming
out of the jet here the pressure is atmosphere
as atmospheric that p2 is equal to 0, that
from the continuity equation we can write
here Q is equal to A1 u1 is equal to A2 u2.
Finally, we can write p1 is equal to rho Q
square by2 into1 minus A2 square minus1 minus
A1 square.
This gives the pressure force that means the
pressure here p1 at the section p1 we can
get as p1 is equal to rho Q square by21 minus
A1 square minus A2 square. This gives the
pressure force now this particular case is
concerned there is no body force since the
body force here we neglect finally, resultant
force we can write as total resultant force
is equal to FRx plus FPx plus FBx body force
is 0, P is the pressure force and F is the
resultant force.
We have already found here., finally, we get
the resultant force for the system FTx is
equal to FRx plus FPx plus F B x and finally,
FRx is already can be written as FTx minus
FPx minus 0 this is with respect to the FRx
is obtained like this now finally, FRx is
equal to rho Q square into1 by A2 minus1 by
A1 here which is obtained from the total force
which we have already found FTx.
The pressure force is considered we have already
calculated Q square 2 into1 minus A2 square
minus1 by A1 square into A1 this is gives
the resultant force for the system resultant
force is equal to rho Q square1 divided by
A2 minus1 by A1 minus rho Q square by2 into1
minus1 divided by A2 square minus1 by A1 square.
Where A2 is the cross section at this location
where the nozzle in this location and A1 is
the cross section at this location at section
one, this gives the resultant force, this
resultant force is the force which is acting
upon with respect to the effect of this nozzle.
The force coming on this pipe is the force
on the pipe is given by this resultant force
that force should be that resist the with
respect to the nozzle this is the force which
resist the nozzle is the liquid is passing
through the nozzle.
This is the case for a second case which we
here we discussed is the force on a pipe nozzle
as shown in this figure this is the second
case, now we will be discussing another case
which is the impact of a jet on a plane. As
we have seen there are a number of applications
as far as the linear moment equation is concerned
third case is the impact of a jet on a plane.
Here you can see that a jet is coming impinging
on a on a plate let us assume this case three
third case jet is acting at 90 degrees here
you can see the jet is coming at 90 degree
impinging on this plate on the vertical plate,
we want to find out the force on the plate
and the reaction of the plate this is what
we want to find out with respect to this case
which is the third case the impact of a jet
on a plane.
Here you can see that the pressure force is
0 since you can see here everything is open
to the atmosphere, the jet is coming through
the atmosphere it is impinging on a it is
hitting on a on a flat pipe vertically placed,
here for example if this is the plate the
jet is coming like this vertically with respect
to the this plate now the pressure force here
the pressure force is 0 or atmospheric here
this is the control volume which we consider
here. The control volume is with respect to
this plate and where the jet is impinging
on the plate.
This is the control volume and the body force
al in this particular place it is ignored
the if you assume that velocity here of the
jet is u1, this is the velocity in this upward
direction is u2 and the velocity in the downward
direction is also u2, due to the symmetrical
case. But you can see here with respect to
here the body force is ignored total force
here you can get the velocity of approach
of the jet is u1 and after hitting the flat
plate the velocity is u2 upward direction
downward direction.
The total force FTx is equal to the rho Q
multiplied by u2 x minus u1 x, the we are
finding the force in the x direction of this
direction this is equal to total force is
equal to FTx equal to rho Q into u2 x minus
u1 x the velocity in x direction after the
jet is hit on the on the on the plate and
u1 x is the velocity of approach.
This is equal to rho Q here total force is
equal to minus rho Q into u1 x here you can
see that x direction is concerned x direction
is considered here this plate is placed vertically.
You can see here u2 is the x direction u2
x is here it is 0. since x direction only
the y direction velocity is there u2 is in
the y direction about upward and downward
you can see that x direction velocity u2 x
is equal to 0 since the plate is placed vertically.
Total force FTx is equal to rho Q into u2
x minus u1 x u2 x is 0., this is equal to
FTx is equal to minus rho Q u1 x since due
to the symmetry you can see that here the
force on the on the on the y direction is
equal to FTy is equal to 0 since due to the
symmetry.
We can write that is equal to minus rho Q
into u1 x, this gives the resultant force
finally, for this particular case where jet
is hitting a at 90 degree on a flat plane
here we do not consider the friction on the
plates we have passed that the plate is smooth.
Finally, what we get is the as far as the
x direction the resultant force is concerned
it is given as minus rho Q u1 x where rho
is the density of the fluid considered, Q
is the discharge of the of the jet discharge
of the discharge coming and hitting the plate
and u1 x is the velocity of approach in the
x direction.
This resultant force is obtained by equal
to minus rho Q u1 x this gives if a jet is
hitting on a flat plate the x direction that
is the resultant force here that is equal
to minus rho Q u1 x. This is the third case.
We have analyzed the impact of a jet on a
flat plate or on a plane like this.
This is the third case now we will be discussing
the fourth case fourth case is here the application
of the linear moment equation is considered.
Fourth case is the force on a curved vane
you can see that many problems like for example
we have number of applications like many cases
jets will be coming and hitting on curved
vanes like this., jet will be coming jet action
will be coming like this it will be hitting
on a curved vane like this. There are many
practical cases where these kinds of problems
occur. Now for this particular case we want
to find out the force now the jet is coming
in this direction and we want to.
This is the vane here and this is the jet
we want to find out how much is the force
acted upon acted by this jet on this vane
or the curved vane we consider, here for this
particular case similar to the pipe pressures
here similar to the pipe here the pressures
are equal.
You can see that here everything is atmospheric,
we do not consider the pressure force here
pressure force is atmospheric we can consider
0, which is very similar to the pipe cases
which we have seen pressures are equal and
it is atmospheric the body force also. In
particular case we do not consider the body
force here FBx is equal to 0, now let us consider
a control volume.
In this slide you can see a jet is coming
it is deflected by a curved vane like this,
this is the vane here and jet is coming like
this here the velocity of the incoming jet
is u1 is the velocity. After deflected by
the curved vane it is u2 and x direction is
here y direction is here and the angle with
respect to the where the velocity u2 this
angle is theta with respect to this slide
here.
If you consider for this force on a curved
vane FBy is equal to that means the force
here you can see that body force FBx is equal
to 0. In the x direction the body force is
equal to 0 since x direction is here the total
force. We can find out the total force here
with respect to this particular figure here
is total force is obtained in the x direction
is equal to rho Q into minus u1 cos theta.
Here u2 is in this direction here which is
deflected if you consider this as the normal
direction. The total force with respect to
x direction will be rho Q into u2 minus u1
cos theta here this is u2 and here this is
the angle theta, u2 minus u1 cos theta gives
the total force total force in the x direction
is equal to minus rho Q into u1 minus u2 cos
theta.
Finally, this can be written as that is equal
to minus rho Q square into1 minus cos theta
divided by A which is the area of cross section
since u2 area of cross section of the jet
which is coming here. Since u2 is equal to
u1 is equal to Q by A where Q is the total
discharge this is Q by A that we can write
for this u2 and u1 we can substitute with
respect to this Q by A finally, total force
is obtained that is equal to minus rho Q square
into1 minus cos theta by A.
This gives the force on a curved vane total
force on a curved vane finally, total force
in Y direction will be equal to FTy is equal
to rho Q into u2 sine theta minus 0 here you
can see that this is u2 sine theta and here
it is coming in this direction effectively.
This other direction this is not concerned
there is no y direction there is no velocity
that is FTy is equal to rho Q into u2 sine
theta this can be written as that is equal
to rho Q square by A into sine theta.
Finally, we get FTx in the x direction finally,
we can get FRx that means the resultant force
plus the pressure force any way not considered
here body force is al in the x direction there
is no body force. FTx the resultant force
FT FR x is equal to FTx here this with respect
to the y direction FRy is equal to FTy minus
F B y and the resultant force we can we find
out the resultant is equal to square root
of Fx square plus Fy square.
This gives the resultant force on the curved
vane., now we have seen various cases; first
one is the impact on a jet on a plane, second
case third case is impact first case is the
force due to flow around a pipe bend, second
one is force on a pipe nozzle and third one
we have seen is impact of a jet on a plane.
Fourth one is force on a curved vane as here
we have analyzed and now the fifth case which
we will be discussing here is jets on turbine
blade.
You can see that many cases we will be analyzing
like peloton wheel or turbine problems. You
can see that jet will be coming like this
the flow with respect to jet action you can
see that there will be a wheel with various
vanes like this.
You can see that due to the jet action it
is it will start rotating. It is we can use
this principle of the linear moment equation
which we have seen here we can use the same
principle to analyze the impacts of jets on
turbine blades.
It is depicted in this slide here the jet
of turbine jet of water is coming from through
nozzle like this. This is the jet of water
here the peloton wheel is mounted on axle
and due to the jet action; we can see that
a force is supplied due to that the turbine
wheel is rotating.
Here we will be having either this jet can
be coming directly 90 degrees perpendicular
to the vane or it can also be inclined., here
we will be analyzing how we can find out the
force acted upon the jet on the on the peloton
wheel.
If you consider this u1 most of the time we
can see that this can be either this plates
which is attached with respect to the axle
for the peloton wheel most of the time it
will be curved one like this you can see here
the curved vane type mechanism.
The jet will be coming perpendicular to this
like this. After that it is hitting on the
plate on the vane it is deflected like this
of velocity u2 on both direction into this
way and other way the jet of velocity u1 is
hitting on the vane and it is deflected with
a velocity u2 both direction by like this.
This is the mechanism which is generally used
in a peloton wheel we want to find out the
with respect to this jet action how much is
the force coming on the on the on the peloton
wheel. That is what we want to find out as
we can see here the jet is coming perpendicular
on the horizontal direction it is deflecting
also in the on the horizontal direction. But
it can be al inclined like this; in this slide
you can see that your jet is coming horizontal
direction. But due to these locations or the
nature of the vanes you can see that it can
be al deflected an angle theta like this.
Here u2 is deflected in this direction and
other way it is with respect to angle theta.
For this kinds of problem now let us consider
a control volume as in the previous cases.
First we will be considering a control volume
the control volume is here shown here the
control volume is considered x axis is here
y axis is here.
Now again this case al here this particular
al we do not consider the pressure force the
pressure force is here it is atmospheric pressure
or this 0. We consider the pressure force
as 0 also the body force is not to be considered
here this particular case due to the specific
nature of the problem.
We do not consider the body force you can
see that finally, the total force in the y
direction will be 0 for this due to the symmetrical
due to nature of the particular problem here
the total force in the y direction will be
0 the total force in the x direction.
If you consider this the jet is deflected
at an angle theta like this in the x direction
total force in the x direction that is the
resultant force FTx that is equal to here
with respect to this figure rho into Q by
2. Here if the discharge due to symmetry at
such discharge is Q by 2 is passing Q is coming
and hitting on this and Q by 2 is passing
in this direction and other direction al Q
by 2 is passing.
FTx is equal to rho into Q by2 into u2 x plus
Q by2 into u2 x minus Q into u1 x Q is the
discharge coming to the hitting on the peloton
wheel or the turbine blade the velocity back
is due to at an angle theta let us assume
that in this particular case half of the discharge
is going up this direction and other half
is going down this direction.
That the x direction the resultant force in
the x direction can be written as FTx is equal
to rho into Q by 2 as shown here FTx is equal
to rho into Q by 2 into u2 x plus Q by 2 into
u2 x minus Q into u1 x, where u1 x is the
velocity in x direction and u2 x is the velocity
in the with respect to the x direction and
rho is the density of the liquid or here water
is considered then density of water.
Now the resultant forces obtained as FTx here
you can see that the u1 x the velocity we
are assign for this particular problem the
velocity is coming with respect to the x direction
like this in this direction.
u1 x is equal to minus u1 u2 x here you can
see that u2 x is equal to u2 cos theta., u2
x is equal to u2 cos theta finally, total
the resultant force FTx is equal to rho Q
into u2 cos theta plus u1.
For this particular case FT is the resultant
force is equal to rho Q into u2 cos theta
plus u1 like this we can analyze the force.
Finally, we have found the total force acting
upon this on this on this peloton wheel or
on the turbine blade.
We consider like in a peloton wheel we have
considered the turbine blade, how much is
the force acted upon the turbine blade is
calculated from this expression FTx is equal
to rho Q into u2 cos theta plus u1 where rho
is the density Q is the discharge and jet
u1 is the velocity of the jet u2 is the velocity
after deflecting from the from the turbine
blade.
We can find out the force acted upon the jets
on turbine blade the force jets on turbine
blades can be calculated, this is the fifth
case we have analyzed the application of the
linear moment equation which we have seen
and now the last case we will be discussing
here is force due to a jet hitting an inclined
plane.
As we have seen the jet if the plate here
we have seen that whenever the plate is vertical
placed then the jet is acting like this the
analysis is very simple, we have already seen
we have analyzed here in the previous case
the impact of a jet on a plane.
Plane is it is placed perpendicular the analysis
is much simpler but here in the last case
here we discussed the force due to a jet hitting
at an angle or at an inclined plane. If you
consider a plane like this here a jet is coming
and hitting on the plane like this the jet
position is horizontal jet is acting coming
horizontally but the plane is the plate or
the plane is inclined. You can see in this
slide how the mechanism is this al many problems
you will be of this kind.
Here jet is coming with a velocity u1 it is
hitting on an inclined plane if you consider
the normal to the plane which we consider
this is at angle theta likes this with respect
to the horizontal. After here the velocity
of the jet is u1 in the previous case we have
seen earlier that if it is normally placed
the velocity will be same to the both sides
as u2 and u2 but here you can see that due
to the inclination of the of the plane.
The velocity to this direction u2 is the velocity
in this direction, the other direction which
is u3. It is not symmetrical or the plane
is inclined or the velocity in this direction
is u2 and the downward direction it is u3
here we consider the angle theta.
To analyze this problem, we will be just considering
this problem like this here the same plane
what we consider the jet is in the horizontal
direction and plane is or the plate on which
the jet is hitting is inclined but then the
same case. We can in a very similar way we
can consider the plane is vertical and jet
is inclined here you can see the same problem
it is very similar problem very same problem.
We are considering the plate is or the plane
is vertical like this, the jet is acting inclined.,
we have seen the case is here the actual case
is the plane is or the plate is inclined and
jet is coming horizontal. We are converting
this problem into we place the plate as vertical.
We consider the jet is acting inclined that
is the way which we analyzing this problem.
Here the diagram is rotated for analysis now
we can see this is angle theta and u1 is the
jet velocity is u1 and now x is horizontal
and y is vertical direction and u3 is the
velocity downward u2 is the velocity upward.
If we use the Bernoulli’s equation here
with respect to this figure you can see that
p1 by rho g plus u1 square. As previous case
we will be considering control volume p1 by
rho g plus u1 square by 2 g plus z1 is equal
to p2 by rho g plus u2 square by 2 g plus
z2 is equal to p3 by rho g plus u3 square
by 2 g plus z3.
If you consider three position here and here
three sections from the Bernoulli’s equation
you can write p1 by rho g plus u1 square by
2 g plus z1 is equal to p2 by rho g plus u2
square by 2 g plus z2 is equal to p3 by rho
g plus u3 square by2 g plus z3.
If you assume that height difference is negligible
with respect to the jet action if the height
difference is not much we can assume z1 to
is equal to 2 equal to z3 for analysis purpose
the pressure the jet is acting in atmospheric
pressure. That we can assume that here all
due to the atmospheric pressure here we can
see that u1 is equal to u2 is equal to u3
here velocity with respect to this particular
problem u1 is equal to u2 is equal to u3.
With the Bernoulli’s equation the possibility
is that u1 is equal to u2 since pressure is
atmospheric that is already all this terms
as one z1 is equal to z2 is equal to z3 finally,
we get the velocity this is an approximation
u1 is equal to u2 is equal to u3.
Now by from the continuity equation you can
see that if we apply the continuity equation
the discharge this should be what is hitting
on the plate should be equal to what is going
up and what is going down Q1 is equal to Q2
plus Q3.
For that we can write u1 plus A1 is equal
to u2 plus A2 plus u3 plus A3 or we can write
A1 is equal to A2 plus A3 or finally, we can
write Q1 is equal to A1 u or Q2 is equal to
A2 u and Q3 is equal to what is going down
that discharge will be Q3 is equal to A1 minus
A2 into u Q3 is equal to A1 minus A2 into
U. The pressure force since it is atmospheric
that is considered as 0 and finally, we got
Q1 is equal to A1 u Q2 is equal to A2 u and
Q3 is equal to A1 minus A2 into u and in this
case al we neglect the body force body force
is ignored.
Finally, the total force FTx in the x direction
the total force is equal to rho into Q2 into
u2 x plus Q3 into u3 x minus Q1 into u1 x.
With respect to this figure here, finally,
we get the total force in the x direction
FTx is equal to rho into Q2 into u2 x plus
Q3 into u3 x minus Q1 into u1 x. With respect
to this figure is rotated here you can see
that in x direction u2 x and u3 x are 0 u2
x is 0 u3 x is0 we get u2 x is equal to u3
x is equal to 0 jets are parallel to the plate.
Finally, this expression become since u1 x
is equal to u1 cos theta this angle is theta
u1 x is equal to u1 cos theta finally, we
simplify the equation to FTx the total force
in the x direction is equal to minus rho Q1
u1 cos theta. This will be the force acted
upon by this inclined jet on this plane or
this particular case which we consider the
plane is inclined and jet is horizontal. Finally,
the expression is the total force is FTx is
equal to minus rho Q1 u1 cos theta, like this
we can analyze what will be the force due
to a jet hitting at jet the plane is vertical
and a jet is inclined that both case we can
analyze.
Finally, the resultant force exerted on fluid
is equal to body force is neglected pressure
force is atmospheric, that we don’t consider
in the x direction FRx is equal to minus rho
Q1 u1 cos theta this is the resultant force
exerted on the fluid in y direction FTy you
can see that FTy is equal to 0 in this particular
case Finally, the resultant force is R is
equal to minus FRx which is the reaction or
which is the with respect to the jet what
is the force acted by the plate the reaction
force will be R is equal to in the opposite
direction minus FRx that is equal to rho Q1
u1 cos theta.
This gives the reaction force in the opposite
direction that is minus FRx., like this we
can analyze the resultant force acting on
a acted by a jet hit on an inclined plane
also from what we have seen we can find out
the discharge in each direction Q1 Q2 Q3 that
we have seen here.
That also can be calculated discharge goes
on each direction can be al found in a very
similar way this is the application we have
discussed, what we now we have seen the from
the Newton’s second law we have derived
the linear moment equation. We have seen various
applications this far we have discussed various
applications. Now we from the applications
we can see that some of the interpretations
what we can with respect to the applications
are here the linear moment we have to always
consider the direction, whether it is x direction
or y direction that we can that the linear
moment is direction that is obvious always
we have to which direction it is acting the
linear moment is direction.
That is the first interpretation, most of
the cases we have discussed are steady state
condition for steady state flow the time rate
of change of moment is 0 that is not considered
most of the examples we have considered is
at steady state condition also we have seen
we are using the Newton’s second law.
We are considering the external force we have
to consider the sign that means whether the
force is acting which direction whether plus
or minus the sign is important we have to
consider the external forces. We have to consider
its sign algebraic sign of the external force
to be considered and also here as far as the
control volume which we consider the analysis
is we are starting the analysis by considering
a control volume.
For the control volume considered only external
forces acting on the control volume to be
considered for the problems which we have
seen here also in a very similar way. The
cases which we have analyzed is most of the
time the plate or the vane. All the cases
are fixed it is not the vanes or the plates
are not moving but also the linear moment
equation we can write in a very similar way
for moving control volume.
We can see that the plate which we consider
with respect to the jet is now the jet is
coming and the plate is fixed. We have found
the force but if the plate is moving like
this when the jet is acting simultaneously
the plate is also moving that is the case
where the moving control volume.
The cases which we have considered far is
fixed control volume but some cases like you
have seen the when the wheel is rotating it
is moving control volume or when the plate
is moving with respect to the jet action.
Then we can see that the case is of moving
control volume we have to consider now jet
is coming in with respect to the jet movement
control volume is al moving. Linear moment
equation can be transformed with respect to
the moving control volume for the moving control
volume like moving plates and vanes when a
jet with velocity V strikes a plate moving
with velocity small v fluid mass is no longer
the fluid mass strikes on a stationer plate.
As I mentioned if the velocity here plate
is there and the velocity of the jet is here
V capital V is the velocity the plate is now
moving with a velocity of small v.
As earlier case for the fixed plate which
is not the same it is not the same mass that
is hitting the plate as in the case of a stationer
plate since the plate is al moving, here this
is the plate the initial position and now
jet is coming and hitting like this. But after
some time we can see that the plate is moving
the mass is changing the fluid mass is no
longer the same the fluid mass strikes on
the with respect to stationery plate now plate
is shifted with a.
The plate is moving with velocity small v
meter per second and the jet is of capital
V meter per second then we will be finding
out the mass which really overtakes the plate
per unit time. We have to see that the relative
velocity we have to consider and what mass
will be overtaking the plate per unit time
to be considered when the plate or the vane
is moving. We have to consider the relative
velocity here you can see that capital V minus
small v will be the relative velocity that
will be used in the calculation.
In the next slide here we can see that for
the mass striking the plate is area of cross
section of the jet has an area of cross section
small capital A density is rho then relative
velocity we have to multiply absolute velocity
V capital V minus small v. We have to find
out the relative velocity, mass striking the
plate is Q is equal to area of cross section
into rho into V minus small v the velocity
of the jet minus the velocity of the plate
or the vane we consider.
The calculation relative velocity should be
used with reference to this we will be discussing
few of the examples, here the first case is
we consider a problem like this here there
is a curved vane or a plate or cup like this
a jet is coming and hitting on the on the
vane or the or the cup the vane itself or
the cup itself is moving with a velocity.
We want to find the work done per second by
the jet on the on the cup.
The problem statement is as shown in figure
here figure 6 centimeter diameter jet of water
impinges, here the diameter of the jet is
6 centimeter 6 centimeter diameter jet of
water impinges on a series of hemispherical
cups and is deflected through 180 degrees.
The jet is deflected by 180 degree the velocity
of the jet is10 meter per second and the cup
is moving at a velocity of 4 meter per second
in the x direction calculate the work done
per second by the jet on the cups.
You can see that the jet is coming this direction
and the cup is or the vane is placed curved
vane is placed like this and it moving the
cup or the vane is moving with a velocity
of 4 meter per second. The jet velocity is
10 meter per second and if u1 is the jet velocity
then you can see that it will be after the
jet reaches the cup it will be deflecting
at180 degree.
u3 is the velocity in this direction and u2
is the other downward direction you can see
that since we consider the atmospheric pressure
here, u1 is equal to u2 is equal to u3 for
this particular problem. We want to find out
the work done per second by the jet on the
cup for this particular problem.
Since the jet is moving with velocity of10
meter per second and the cup is moving with
a velocity of4 meter per second the relative
velocity of jet is equal to10 minus4 that
is equal to 6 meter per second.
This particular problem we do not consider
the friction., the mass of the fluid striking
the cup per second is the diameter of the
jet is given that is 6 centimeter area of
cross section of the jet is pi by 4 into 0.06
square into pi d square by 4 into the velocity10
meter per second density is 998., finally,
we get 28.217 kilogram per seconds.
This is the mass of fluid striking the cup
per second now we want to find out the force
exerted by fluid on the cup that is obtained
with respect to the equation which we have
derived. We can write that is equal to the
force exerted on by fluid on the cup is equal
to minus Q rho into the velocity change or
the relative velocity you can see here with
respect to the figure, it is both direction
we have to consider with respect to both directions.
We can see here you can see the force exerted
by fluid will be minus Q into rho where Q
is the discharge rho is the density into minus
V minus v minus V minus v that two times.
We can see it is coming both ways that is
equal to we have already found the mass of
fluid striking the cup is 28.217 into 2 times
into the relative velocity we have calculated
as 6 meter per second into6 that gives as3
hundred and 38.61 Newton.
This is the force exerted by fluid on the
cup is 338.61 Newton and now finally, work
can be found by here you can see that work
is the force in work is equal to force into
displacement. We can see that the with respect
to the cup is moving at 4 meter per second
that we can write the work done per second
is the force exerted by fluid on cup multiplied
by that is 338.61 multiplied by this 4 meter
per second which is the movement of the cup
into 4 divided by 1000 that will give 1.3
by1.354 kilo watt.
This gives the work done per second for this
problem here we have analyzed a problem with
which is relative movement by where the vane
or the cup is moving with respect to the jet
action, here what we have to see what we have
to note here is that we have to find. We have
to calculate while finding out the discharge
the while finding out the force which we discussing
the forced exerted on the on the cup, that
we have to with respect to the relative velocity,
we have to consider the relative velocity
is considered here in this particular case
the cup is al moving in the direction of the
jet.
This is the first example first problem and
second is 
with respect to an inclined plane a jet is
hitting an inclined plane here the problem
statement is as shown in figure here you can
see the figure here.
As shown in this figure a jet is hitting on
an inclined plate like this we have to find
water jet of 5 centimeter having a velocity
of 8 meter per second impinges on a smooth
plate at an angle of 40 degree to the normal
to the plate. We have to find out the impact
of the jet when the plate is stationery case
a and case b when the plate is moving in direction
in the direction of the jet at 3 meter per
second also calculate the work done per unit
time in both cases.
The problem is we have we have a plate put
at an inclination of 40 degree with respect
to the horizontal that a jet is acting jet
is hitting or impinging on the plane at 40
degree. We have to find out the impact of
the jet we have to find out the impact of
the jet when the plate is stationery plate
is not moving that is case a and case b is
when the plate is al moving in a in a direction
of the jet at 3 meter per second, Here it
is already given that the diameter of the
of the jet is 5 centimeter and also since
we as we have seen in the previous discussion
the all the pressure force the pressure is
atmospheric we can that u1 is equal to u2
is equal to u3.
u1 is the jet velocity u2 is the upward direction
u3 is the downward direction u1 is equal to
u2 is equal to 8 meter per second which is
given value and this angle is given as theta
is equal to forty degree.
As we discussed we will be considering a control
volume as shown here this is the control volume
which we consider, from the previous expression
which we have derived the impact of the of
the jet is obtained as resultant force exerted
that is equal to rho Q into u1 cos theta where
rho is the density of the fluid Q is the discharge
of the jet and u1 is the velocity at angle
theta the impact of the impact of the jet
is equal to rho Q into u1 cos theta.
The impact can be calculated for case a when
the plate is stationery impact is equal to
rho Q into u1 cos theta., rho is 998 into
Q is pi by 4 into diameter of the pipe the
jet is 05 centimeter into 0.05 square into
u1 is 8 8 into eight cos 40. This gives the
impact, finally, we can see that the impact
on the plate will be 96.07 Newton in the direction
here in this direction this is the F m this
is in this direction.
This gives the impact of the jet on the inclined
plane that is 96.07 Newton. Work done here
you can see that the plate is stationery since
there is no displacement of the plate work
done is force into displacement is 0. Work
done here in this particular case is 0 and
case two when the plate moves in the x direction,
v is equal to 3 meter per second if you consider
the normal direction and relative velocity
you can see that the relative velocity will
be the plate is moving with respect to 3 meter
per second and jet velocity is8 meter per
second that is equal to 8 minus 3 is equal
to 5 meter per second.
The impact of the jet will be impact is equal
to rho into Q r into u1 r cos theta rho is
998 into pi by 4 into 0.05 square that is
Q r into 5 into 5 cos 40, that gives 37.53,
this will be impact for whenever the plate
is also moving with respect to movement of
the with respect to the jet force. The relative
velocity will be 8 minus 3 into8 minus 3 that
is equal to 5 meter per second with respect
to when the plate is al moving the work done
w is equal to this the force into the displacement
per unit time here 37.53 into cos 40 into
5 that gives 143.75 Newton meter per second.
This is the work done, like this we can solve
most of the problem by considering the linear
moment equation for either a fixed control
volume or for a moving control volume.
