Welcome back to the video course on fluid mechanics. In the last lecture we were discussing
about the dynamics of fluid flow we were discussing about the energy equation and we were discussing
about the corresponding Bernoulliís equations and its applications.
Today we will before further proceeding to other topics we will be discussing more about
the limitations of the Bernoulliís equations which we have already seen earlier, as we
have seen here large number of applications of Bernoulliís equations the general energy
equation with respect to the Bernoulliís equations already we have discussed here it
is p1 by gamma plus alpha v1 square by 2 g plus z1 plus q w plus h e is equal to p2 by
gamma alpha v2 square by 2 g plus z2 plus e2 minus e1.
We have already seen considering with respect to two sections of flow between for example
if you consider a pipe flow like this, we have already seen here between sections one
and two here the total energy equation or general energy equation is written as p1 by
gamma plus alpha v1 square by 2 g plus z1 plus q w plus h e is equal to p2 by gamma
plus alpha v2 square by 2 g plus z2 plus e2 minus e1.
Various terms like q w is the heat added for with respect to the system which we discuss
q w is the heat added per unit weight of fluid e1 e2 are the internal energy per unit weight
of the fluid at respective states and h e is the external work done.
We have already seen this e2 minus e1 minus q w is equal to reversible and irreversible,
irreversible head is the head loss h l is the energy loss per unit weight of fluid due
to the friction and other causes, for incompressible fluid total head at 1 plus heat added due
to machine or head loss is equal to total machine minus head loss is equal to total
head.
Finally, the general energy equation we can write as h1 plus h e minus hl is equal to
h2 and the work done over a fluid is equal to power input into power input flow that
is equal to p is equal to gamma q h m in watts and where gamma is the unit weight of fluid
q is the discharge in meter cube per second and h m is the head added to the flow in meter.
We have already seen a large number of applications of the Bernoulliís equation and now we have
already discussed the general energy equations, the general energy equation is the Bernoulliís
equation which we have derived is, for in viscid flow the Bernoulliís equation can
be generalized with respect to the equations which we have seen as general energy equation.
Now the Bernoulliís equation which we have discussed, have number of limitations since
we have put certain assumptions with respect to the derivations of the Bernoulliís equations.
The limitations important limitations are first one is the compressibility effects of
the fluid., the Bernoulliís equation which we have derived or we have discussed far,
the compressibility effect is not taken into account. We have assumed that the fluid is
incompressible if the compressible fluid fluids are considered then, we have to slightly modify
the Bernoulliís equation. The Bernoulliís equation can be modified for the compressibility
effects and al the Bernoulliís equation which we have discussed for steady state whenever
the problem, we have to deal with the unsteady conditions then the equation is to be modified
for the unsteady flow.
In the next slides here with the limitations of the Bernoulliís equation which we have
already seen earlier. Here you can see the Bernoulliís equation for the unsteady case.
Here we can see the ro into del v by del t. The steady state equation is modified like
this rho into del v by del t d s plus d p plus half ro d v square plus gamma d z is
equal to 0 along a streamline and similarly we can rewrite as p1 plus half rho v square
plus gamma z1 is equal to rho integral s1 to s2 del v by del t d s plus p2 plus half
rho v2 square plus gamma z2 along a streamline.
Bernoulliís equation which we have seen earlier discussed is steady state as shown here we
can modify whenever we consider the transient or the unsteady conditions. These have some
of the limitations of the Bernoulliís equation and further limitations are like the rotational
effect. Whenever the fluids which we consider, has got the rotational effects the rotational
effects care should be taken while applying for flow across the streamline.
We have derived the Bernoulliís equation by considering a streamline. Whenever the
rotational effects are there for the problem which we consider we have to take care the
while applying for flow across a streamline. Also other limitations include the equation
the Bernoulliís equation is not valid for flows with mechanical devices for example
whenever a pipe line system is when we consider a problem with pumps turbines etc specific
care should be taken that the Bernoulliís equation can be applied. As we have already
seen the one of the important assumptions in the derivation of the Bernoulliís equation
is fluid is inside whenever viscous fluids are considered still Bernoulliís equation
can be applied, but there will be it is one of the limitations it is not 100 percent correct
that for viscous fluid these equations the assumption the Bernoulliís equation is derived
based upon the assumption is that fluid is inside these are me of the important limitations
on the use of Bernoulliís equations.
Now we will see few of the example problems with respect to the Bernoulliís equations
we have discussed far before going to other topics. The Bernoulliís equations the example
number two earlier, we have already seen one of the example the next example is here we
consider a problem, water is flowing through a smooth pipe of uniform diameter here there
is a pipe the diameter is 20 centimeter at section one at elevation 5 meter. Here we
consider two section one at 5 meter from datum this is the datum the pressure is 30 kilo
Pascal here the pressure is 30 kilo Pascal. Section two at elevation 7 meter here at this
section pressure is 15 kilo Pascal and velocity is 1 meter per second. We want to find the
head loss between the sections between section 1 and two and the flow direction.
This is one of the simple one simple applications of the Bernoulliís equation. Here a pipe
flow is there water is going through the pipe and the velocity is given at section two and
the pressures are given at section one and two. We have to find out the head loss between
section one and two and the flow direction. This is one of the simplest applications of
the Bernoulliís equation.
To solve this problem first we will use the continuity equation if you use the continuity
equations we are considering two sections continuity equation can be applied with respect
to the conservation of mass. We apply the continuity equation q is equal to a1 v1 is
equal to a2 v2 here only the diameter is known the diameter is 20 centimeter q is equal to
pi by 4 into 0.2 square velocity v1 is given as 1 meter per second pi by 4 into 0.2 square
into 1 that gives 0.0314 meter cube per second, here since the pipe is assumed the uniform
diameter v1 is equal to v2 and a1 is equal to a2 and al this specific weight of the liquid
is 998. This can be converted to 998 into 9.81 by the d rho is given as 998 kilo gram
per meter cube.
That can be converted as gamma is equal to gamma is rho g 998 into 9.81 by 1000, 9.79
kilo Newton per meter cube. We are considering atmospheric pressure and hence we are taking
it as 0 at section one if you apply the Bernoulliís equation if you consider section one here.
If you consider the Bernoulliís equation here we can see as per the Bernoulliís equation
p1 by gamma plus v1 square by 2 g plus z1 p1 by gamma already it is given that the pressure
is thirty Pascal p1 by gamma is 30 by 9.79 then v1 is given as 1 per second p1 by gamma
plus v1 square by 2 g plus z1 becomes 30 by 9.79 plus 1 square by 2 into 9.81 plus 5 that
will be obtained as 8.15 meter.
If you consider similar way the section two which we consider here for section total head
h2 is equal to p2 by gamma plus v2 square by 2 g plus z2 p2 is given as the pressure
is section two is given as 15 kilo Pascal p2 by gamma is 15 by 9.79 plus v2 is 1, 1
square by 2 into g 9. 81 plus z2 is 7. This we get as 8.583 meter here you can see this
h2 is greater than h1 since 8.58 there is greater than 8.115 the flow occurs from section
two to one.
We can see that the pressure head here at section two is much higher than section one.
The flow takes place from section two to one and head loss is h2 minus h1 is 0.468 meter.
This is one of the simple application of the Bernoulliís equation we can apply the Bernoulliís
equation for large number of problems. We will also discuss few more examples here example
number three which we discuss is here a tank is filled with oil of relative density 0.9
and siphon of diameter 10 centimeter is used to empty the oil as shown in figure here this
is the figure here.
Here we have got a tank and a siphon is used to empty the oil from this tank like this
using a siphon and question is the oil surface in the tank is at an elevation of 5 meter
and siphon discharges to atmosphere at an elevation of 2 meter. Center line of siphon
at its highest d is at an elevation of 6 meter the losses in pipe to higher is 0.7 meter
and 0.9 meter from higher point to outlet.
we have to find out the discharge in the pipe and the pressure at point d. problem is here
we have water tank we are using a siphon to take the oil out from the tank with respect
to this here you can see we have to find out the discharge in the pipe throughout the siphon
we have to find out the pressure at point d. This is the problem; here it is given that
the density of oil rho is 0.9 d is the diameter of the pipe is 10 centimeter.
If you consider now consider using the Bernoulliís equation here you consider three points, one
is the liquid level or the oil level in the tank this is point one, we consider point
two as the outlet of the siphon here this is point two, we will also consider the highest
point of the siphon that is point three, we will consider these three points. We will
use the Bernoulliís equation to solve this problem we want to find out the discharge
passing through this pipe also the pressure at this point this is the point in the problem.
If you consider point one, we can see that here this tank is open to atmosphere that
we can say that the pressure p1 is equal to 0 here the there is no velocity of fluid v1
is equal to 1 the elevation head is given as z1 as 5 meter for this problem. At section
two where the at the outlet of the siphon, we can see that the pressure is given as p2
is equal to 0 that it is the flow into the atmosphere p2 is equal to 0 we can see here
the elevation head is given as 2 meter.
We will apply the Bernoulliís equations between section one and two if you consider the Bernoulliís
equation between this 0.12, here or section two and section one using the Bernoulliís
equation we can write p1 by gamma plus v1 square by 2 g plus z1 is equal to p2 by gamma
plus v2 square by 2 g plus z2 plus head loss between point one to point two.
This is the Bernoulliís equation here you can see that we have already seen p1 is equal
to the pressure at point 1 is atmosphere we consider it 0 p 1 by gamma is 0 velocity at
this location is al 0 v1 square by 2 g is 0 and z1 is already given as 5 meter 0 plus
0 plus 5 is equal to at section 2 p2 by gamma p2 is already 0 this al this term al 0 and
plus v2 square by 2 g here the velocity at this section to be v2 v2 square. By 2 g and
the elevation head is given as 2.
Plus 2 plus we have got the head loss. the head loss it is already in the problem it
is given here the head losses the losses in pipe to higher point is 0.7 and 0.9 from higher
point to the outlet. These are the head loss from section one to two 0.7 plus 0. 9 this
al added we use the Bernoulliís equation between section one and section two.
If you use this here now we get write v 2 square by 2 g is equal to, we can get an expression
for the velocity v 2 square by 2 g is equal to 5 minus 3.6 that is equal to 1.4 meter.
Finally we get the velocity at the outlet v 2 here we get v 2 we get the value of v
2 as v 2 is equal to 5.241 meter per second and finally the first part of the question
is we want to find out the discharge. Discharge is equal to area of cross section into velocity
area of cross section is the diameter of the pipe is 10 centimeter pi by 4 into 0.1 square
into the velocity v2 5.241 that is gives the discharge.
Finally, we get the discharge as 0.0412 meter cube per second. Then, the second part of
the question here is we are asked to find the pressure at highest point d to find out
the pressure at point d. We will again consider here the figure, we will consider the Bernoulliís
equation between this point one section one and section three between this point and this
point, we will consider between one and three. If you apply the Bernoulliís equation again
p1 by gamma v1 square by 2 g plus z1 is equal to p3 by gamma plus v3 by 2 g plus z3 plus
h l that means the head loss between section one and two one and three, here you can see
that since the pipe siphon is of same size area of cross section is same the velocity
also you can see that v3 is equal to v2 we have already found out v 2 as 5.241 v3 is
equal to v2 that is equal to 5.241 meter per second.
Finally, if you write the Bernoulliís equation between sections one and three, we can write
0 plus p1 by gamma is 0 v1 is 0, 0 plus 0 plus 5 that is equal to p3 by gamma plus v3
square by 2 g v3 is already found 5.241 5. 241 square by 2 into 9.81 plus 6 is the z3
here we can see that this section is 6 meter elevation is given as 6 meter in the datum
plus 6 plus. The head loss between points one to three is given as 0.7 meter, plus 0.7
meter this is the Bernoulliís equation between section one and three. Finally we can get
v3 by gamma as v 3 by gamma is equal to 5 minus 6 minus 0.7 minus v square 5.241 square
by 2 into 9.81 this is equal to minus 3.1 meter pressure at d finally we get this will
be 3 by gamma gives the d that is equal to minus 27.315 kilo Pascal.
This is the pressure at d like this we can use the Bernoulliís equations for a given
problem to find the discharge or the pressure or whatever the variable depending upon the
problem. Before closing this we will see one more example related to the Bernoulliís equation.
the 4th example related to the Bernoulliís equation is here we have got a water tank,
we use a pump here to pump water from this tank here water is pumped from a tank at the
rate of 0.5 cubic meter per second, here the discharge the pumping rate is 0.5 cubic meter
per second and the pump supplied an energy of 20 kilo watt here the energy supplied by
the pump is 20 kilo watt we are asked to find the pressure at intensity at a.
The location of the pump we want to find out the pressure intensity at a. here even though
there is a pump attached with the problem pump is there but still we can use the Bernoulliís
equation here first the pump energy is given as 20 kilo watt pump energy is equal to, we
have already seen the equation for pump energy as gamma q into delta h here pump energy is
20 kilo watt gamma q delta h is 20 kilo watt. From which we can write the with respect to
pumping what will be the head delta h is equal to 20,000 by gamma for water. If you take
9810 and here the discharge is given as 0.5 20 thousand divided by 9810 into 0.5 from
which we can get delta h that means with respect to pumping we get is 4.07 meter of water.
From that we get the head which can take care by the pump that is equal to 4.07 meter of
water from that we get the head which can take care by the pump that is equal to 4.07
meter, we have to find the velocity at the location a here already the we have already
calculated the head here from that we can try to find out the velocity already the discharge
is given mew here velocity at a is equal to discharge by cross sectional area.
Here this pipe diameter is 15 centimeter here from which velocity at a, is equal to v a,
is equal to q by a, we get 28.29 meter per second. If you consider here at the exit of
this pipe with respect after the pumping is located here water is going through this exit
at d.
Here the diameter is given as 10 centimeter the velocity at d we can find vb is equal
to 15 by ten square into v a since to conserve the continuity equation we can see that v
b into a is equal to vb into a is equal to a 1 is equal to or a 2 is equal to v a into
a1 if you use this we can see that.
We can write v b is equal to 15 by ten whole square into v is equal to 63.65 meter per
second. Finally, we will apply the Bernoulliís equation between point a and c here the tank
top of the water level here you can see that the pressure will be pressure is atmosphere
we can take it as 0.
You can see here there is no flow takes place the velocity can be considered as 0 if you
consider the Bernoulliís equation between points c and a that, we can find the write
the equation as here the z1 the Bernoulliís equation is applied that will give here the
velocity all the values are known this is 2 plus you can see that between this level,
the location a and location c the level difference is 2 meter that, we can that 2 plus it is
z the location with respect to datum 2 plus 0 plus 0 this term the pressure is 0.
This is 0 the velocity is 0 again 0 2 plus 0 plus 0 is equal to p a by gamma plus the
velocity here v 2 square by 2 g that is 28.29 which we already got or v a square 28.29 square
by 2 into 9.81 that gives p a by gamma p a by gamma. Finally we get as minus 38.79.
We have seen few applications how we can solve various problems by using the Bernoulliís
equation. Finally, we have already seen the Bernoulliís equation we have derived by certain
assumptions like fluid is in viscid and flow is incompressible, we have already seen the
limitations and al we have seen the various applications of the Bernoulliís equations.
We have tried to solve few of the numerical examples various typical examples of this
kinds of where the Bernoulliís equation can be applied, we have seen with respect to the
energy equations and the Bernoulliís equations and now we will go to the next topic which
is the linear momentum equations
Here we have already seen the energy equation the Bernoulliís equation now we will discuss
the momentum equation linear momentum equations. In all these we have already seen that the
Newtonís laws we have directly applying this momentum most of the time we will be dealing
with now we are discussing the dynamics of fluid flow. We are dealing with moving fluids
all the time the moving fluids exert the force either one way or another way. The moving
fluids the force which is exerted by the moving fluids have got lot of practical importance
where we can use it for various purposes the momentum or the force created by the moving
fluids;.
For example, if you consider the lift force on a air craft by air moving over the wing
al we can easily see many times jet actions many times we use jet action. For example
in turbines and me of the other kinds of problems where in engineering, we use jet action jet
what is jet action fluid is coming through a pipe, we are making it just like a nozzle
we are reducing is diameter the liquid will be coming at a force to a plate here you can
see that here there is a plate the liquid is coming at a force you can see this is the
jet action.
This jet action al the force is exerted by the moving fluid the moving fluids exert forces
here we can see here there is a nozzle is there and jet is put on a plate like this,
you can see that there is a force and this force we can utilize for various engineering
purposes this moving fluids exert forces with respect to this we can use the momentum equation
to find out how much is the force coming.
Here, we use the Newtonís second law to analyze this moving fluid we derived the linear momentum
equation with respect to the Newtonís second law linear momentum equation is it is this
momentum equation statement of the Newtonís second law, as I mentioned moving fluids exert
forces force is equal to mass into acceleration from this the Newtonís second law, we can
derive the momentum equation it is actually a statement of the Newtonís second law
if you consider special properties of fluids like sum of forces we can with respect to
movement of fluids we can relates sum of forces acting on an element of fluid to its acceleration
or rate of change of momentum. As per linear momentum equation by using the Newtonís second
law we can write the sum of forces acting on a element of fluid to its acceleration
or rate of to its acceleration or rate of change of momentum force is equal to mass
into acceleration, we can equate it to the rate of change of momentum taking place between
if you consider two section or a control volume.
That control volumes what changes takes place this we can equate to the rate of change of
momentum that is that way we can derive the linear momentum equations from the Newtonís
second law we can write the rate of change of momentum of a body is equal to the resultant
force acting on the body and takes place in the direction of force. This gives the Newtonís
second law the Newtonís second law states that the rate of change of momentum of a body
is equal to the resultant force acting on the body using this Newtonís second law,
we can we can see that here it deals with system momentum and forces.
If you consider a open system or closed system the Newtonís second law deals with the momentum
and the momentum occurring with respect to control volume or the between the sections
which we consider and the forces acting on with respect to control volume. As we have
already seen the control volume concept.
We have to consider both surface forces as well as body force acting on the control volume
control volume we have already seen most of the problems which we will be solving here.
We are using the control volume for the control volume we have to consider what happens within
the body the volume itself the surfaces what happen, we have to consider the linear momentum
equation here we have to consider we are using the Newtonís second law.
Force is mass into acceleration then we are equating that to the momentum changes we have
to see the system the control volume of the system which we are considering its momentum
and the forces as the forces or the moment are concerned what happens on the surfaces
as well as the inside the body forces. Here in this slide if you consider typical control
volume for flow in a tunnel shaped pipe this is pipe coming it is just opens like this
fluid is flowing in this direction. If you to consider the linear momentum equation for
a finite control volume like this
If you consider section between this section and this section we can find out the vector
sum of all external forces acting on a fluid mass that will be equal to the rate of change
of linear momentum. As per Newtonís second law we can find out
the vector sum of all the external forces and that will be equal to the rate of change
of linear momentum finally linear momentum equation is obtained, as we can find out the
rate of change of linear momentum f is equal to the force is equal to the vector sum of
all external forces equal to dm by dt or the rate of change of linear momentum, if you
consider this control volume what are the forces acting on this control volume, we can
equate the with respect to the fluid incoming fluid and outgoing fluid within this control
volume. We can find out the rate of change of linear momentum equate to the vector sum
of all external forces within this control volume.
If you since we are considering here the forces when we deal with the forces any body or any
control volume which we deal we have to deal with the internal forces as well as external
forces. Here the forces which we are considering the external forces which we are considering
first one is the boundary forces what the if you consider the control volume what are
the forces on the boundary that includes the boundary forces the normal forces with respect
to the flow coming inside and flow going outside there will be normal forces. We have to consider
this normal forces of course the internal forces of the body or field forces we have
to consider.
The forces which we have to consider include the boundary forces the normal forces and
the body or field forces. These are the forces when we consider the linear momentum equation
these are for a control volume these are the forces which we have to consider finally with
respect to this we can write the equation as.
With respect to the Newtonís second law, finally, the equation already shown here it
is written here d by dt, if you consider the system integral of the system v rho d v is
equal to sigma f system you can see the total the adverse me of the forces is equal to the
rate of change of momentum as we have seen the rate of change of momentum is equal to
resultant force acting on the body we can write d by d t integral system for the system
v ro d v is equal to sigma f system. Finally, this we can be written as I mentioned we have
to consider the control volume inside the volume what happens on the surfaces what happen.
We can this term we can write as del by del t of the on the control volume v rho d v plus
the control surface integral control surface v rho v dot n d a that is on the control surface
that is equal to sigma the forces of the system that means sigma f contents of the control
volume. This gives finally the equation which we are looking for with respect to the linear
moment of equation for finite control volume derived based upon the Newtonís second law.
Now with respect to the equation which we derived if you consider here the linear momentum
equation for finite control volume, shown here, considers a pipe flow like this. On
this section we consider two sections here at this section if the area of cross section
is a one and the velocity is u1 and the density is rho 1 at section two if the area of cross
section is a two the velocity is u2 and the density is rho 2.
Force is equal to as per the equation which we force is equal to rate of change of momentum
force f is equal to rate of change of moment is given as rho 2 a2 u2 delta t u2 minus at
section 2 minus rho 1 at section 1 rho 1 a1 u1 delta t u1 divided by delta t this gives
the rate of change of momentum. Here you can see this is the momentum between section one,
two and the time difference delta t.
Finally, we can write f is equal to q rho u2 u2 minus u1 if you consider this approximation
after simplification, we can write the force is equal to q is the discharge passing through
this pipe. q is constant since due to the conservation of mass between section one and
2 q is constant and if you assume the density ro is constant we can write rho 2 is equal
to rho 1. If you consider that it will be mainly the f is equal to ro q capital q is
the discharge ro q into u2 minus u1 that gives the velocity difference between section two
and one f is equal to the force is equal to q rho u2 minus u1. Similarly, if you consider
as a1 dimension case and if you consider 2 dimension system we have to consider in x
and y direction, we have to equate the force in x direction with respect to the change
in momentum in x direction force in y direction, we have to equate to the rate of change of
momentum in y direction.
Similar way we can write for two dimension system for a two dimension system if you consider
the fx the force is in x direction is equal to rate of change of momentum is equal in
x direction that is equal to rate of change of mass into change in velocity in x direction
rate of change of mass if any mass change takes place. If you put a dash is the rate
of change of mass change in velocity in x direction.
With respect to this figure we can write as u2 cos theta2 minus u1 cos theta1 if theta1
is the angle here and theta2 is the angle here with respect to the x direction we can
write as fx is equal to rate of change of momentum in x direction that is equal to rate
of change of mass into change in velocity in x direction.
Change in velocity in x direction can be written as u2 cos theta2 minus u1 cos theta1 into
rate of change of mass. This is fx, finally, we can write as u2 cos theta2 is written as
u2 x in this direction in x direction and u1 cos theta1 is written as u1 x finally fx
is equal to m dash u2 x minus u1 x.
That is equal to m dash the rate of change of mass if you write the density into the
discharge flows through the control volume rho q into u2 cos theta2 minus u1 cos theta1
finally this can be written as fx is equal to rho q u2 x minus u1 x.
For two dimensional case can write in x direction the force in x direction is equal to the rate
of change of momentum in x direction that is fx is equal to rho q u2 x minus u1 x. similar
way we can write for the rate of change of the rate of change of momentum in y direction
f y is equal to ro q sin u2 sin theta2 minus u1 sin theta1 that is equal to rho q u2 y
minus u1 y.
With respect to this figure u2 y is the velocity in y direction and u1 y is the velocity for
section one and u2 y is the velocity in section two in the direction of y. Finally for f y
we can write as that means the rate of change of momentum in y direction is equal to rho
q in the into velocity change in y direction u2 y minus u1 y this is the case for two-dimension
system. Earlier we have seen the case of one-dimension system now we have seen two-dimension system
similar way we can write for three-dimensional system also now the we have already seen the
when we discuss the energy equation or when we discuss the Bernoulliís equation or the
total energy equation we have seen that most of the time you will be considering the average
velocity.
With respect to the average velocity, when we consider the control volume or the sections
which we are considering, v is the average velocity at a cross section but you can see
that velocity is varying from one section to another. Here we can see that here if you
consider a pipe flow here a pipe flow between section one and two the pipe flow is here
you can see that velocity variation if you plot it will be like this you can see the
velocity will be maximum at this center line this is velocity variation.
Velocity will be maximum at the center line of the pipe and it will be 0 on the on the
side wall of the pipe but the equations which we have derived is the average velocity at
a cross section. But in actual sense actually reality we can see that actual velocity will
be non-uniform you can see that it will be varying from one section to another.
As we have already seen in the case of energy equation and the Bernoulliís equation we
have applied a energy correction factor. We are discussing the momentum equation linear
momentum equation for the momentum also since we are considering the average velocity at
the section we have to apply the apply a correction factor called momentum correction factor momentum
calculated using the average velocity must be corrected by a correction factor beta.
We have to correct the momentum which we have calculated using a correction factor called
beta.
This beta as we have already seen in the case of energy correction factor we can in a very
similar way we can derive this momentum correction factor beta as beta is equal to one by a integral
a v by capital v whole square where capital v is the average velocity and small v is the
velocity at any section which we consider as shown here this is small v and capital
v is the average velocity which we consider for the case.
We have to apply for the momentum equation which we consider we have to apply a correction
factor called momentum correction factor and the momentum correction factor beta is defined
as beta is equal to if a is the area of cross section beta is equal to 1 by a integral a
about a v by v whole square where small v is the varying velocity and v capital v is
the average velocity which we consider. If you consider the flow as uniform flow then
means the velocity is same throughout the section in that case beta will be equal to
one for non-uniform flow you can see that this will be beta, will always be greater
than one depending upon the case whether it is depending upon the cross section, depending
upon either it is pipe flow or open channel flow this beta will be varying and whether
it is lamina flow or turbinal flow also the beta value will be varying which we will be
discussing later.
Finally if you apply this momentum correction factor to the apply momentum principle to
flow through the generalized system like here you can see the generalized system here we
consider, here this is the system control volume which we are considering and here section
one and section two and here if the velocity is v1 and the discharge passing through the
system is q and here at section two the velocity is v2.
Finally we have formed beta the momentum correction factor we have to apply with respect to this
we have to finally we can write the general equation the generalized equation can be written
as sigma f is equal to momentum out minus momentum in sigma that is equal to the algebraic
sum of the force that in the x direction can be written as sigma fx is equal to rho 2 a2
v2 into the velocity in x direction vx2 minus rho 1 a1 v1 into vx1. In the direction with
respect to this figure sigma fx is equal to rho 2 a2 v2 into v x 2 minus rho 1 a1 v1 into
vx1.
We have to apply the correction factor beta we have to use the momentum coefficient beta
if you use the momentum coefficient; finally the equation should be written as like this
sigma fx is equal to rho q beta2 vx2 minus beta1 vx1.
beta2 is the momentum correction factor at section two and beta1 is momentum correction
factor at section one we can write sigma fx is equal to rho q beta2 vx2 minus beta1 vx1
this in the x direction similar way we can write the equation in y direction. We can
solve the generalized momentum equation is shown here we can use the momentum principle
by using the generalized equation. As we have seen the Bernoulliís equation the energy
equation there are large number of applications are there for momentum equation also, here
now we will discuss some of the important applications of the momentum equations.
Some of the important applications can be classified like first one is force due to
flow of fluid around a pipe bend enlargement and contraction these are some of the applications.
We can apply the momentum equation for pipe bend enlargement or pipe contractions al in
the open channel case also bends or enlargements or expansions we can contraction we can use
and second case we can use the momentum equation force on nozzle at the outlet of a pipe whenever
jet action is there. We can use the momentum equations and this is the impact of jet on
a plane surface or inclined surface we can use the momentum equations.
Force due to flow around a curved vane like in a turbine we can use the momentum equation
momentum equation which we have derived or which we have seen far has got large number
of practical engineering applications. We will be discussing some of the applications
here the momentum equations. First one is the application of momentum equation force
due to flow around a pipe bend. Here we will discuss a pipe bend here you can see a bended
pipe like this here.
We want to know how much force is required to force should be there to support or the
thrust block must withstand is there is a thrust block, here the force the flow is taking
place in this bend pipe, we want to find how much is the thrust block how much thrust should
be put that.
We can support the pipe the first application the force due to flow around a pipe bend then
we can solve in the following steps first we will consider control volume, we have to
consider a control volume with respect to flow or with respect to the case which you
are considering, we will calculate the total force the including the pressure force and
body force we can find out the resultant force.
As we have already seen as per the momentum equations we can equate the rate of change
of momentum to the force once the force is given we can find out the momentum also we
can directly apply the case. To find out the force due to flow around a
pipe bend as shown here first step is you consider the control volume here we consider
the control volume that means the section between one and two we consider the control
volume like this here, there is an angle theta with respect to here and at section one the
velocity is u1 and section two the velocity is u2 and x axis is in this direction y axis
is in this direction the total force, which if you consider the total force with respect
to section one and two. we can see here in the x direction, as we have already seen earlier
before total force in x direction will be rho q is the discharge passing through the
pipe and ro is the density rho q into u2 x minus u1 x the velocity in x direction at
section two and velocity at in x direction at section one is u1 x rho q into u2 x minus
u1 x. If you consider u1 x is equal to u1 itself here and u2 x can be with if u2 is
this direction u to x can be written as u2 x is equal to u2 cos theta.
Finally, we can write f 2 x equal to ftx is equal to rho q u2 cos theta minus u1 the force
in x direction is equal to rho q u2 cos theta minus u1. Similarly the total force we can
write f t y is equal to rho q into u2 y minus u1 y in y direction if you consider u2 y u1
y rho q into u2 y minus u1. Here with respect to this figure if you since u1 is in this
horizontal x direction u1 y will be u1 sine 0 that is equal to 0 and u2 y is u2 sine theta.
Finally, f t y the total force in y direction will be f t f t y is equal to rho q u2 sine
theta the we have already considered a control volume and calculated the total force with
respect to the fluid movement between section one and two the pressure force, we can calculate
the pressure force here between section one and two the pressure force is the change in
pressure force pressure force at 1 minus pressure force at 2 the pressure force p1 is the pressure
intensity at section one and p2 is the pressure intensity at section two. We can write the
pressure force in the x direction can be written as f p x is equal to p1 a1 cos 0 minus p1
p2 a2 cos theta here this angle is 0 here angle is theta p1 a1 cos 0 minus p2 a 2 cos
theta that is equal to p1 a1 minus p2 a2 cos theta.
Similarly in y direction we can write the pressure force in y direction p1 a1 sine 0
minus p2 a2 sin theta sine 0 is 0 this is equal to minus p2 a2 sin theta if you consider
the body force here mainly it is due to gravity. Here the pipe is the bend pipe is put way
such that the body force we donít have to consider here the gravity force not considered
body force is 0. Finally find out the resultant force that the resultant force will be the
summation of the total force. If you consider in the x direction the summation of total
force in x direction that means rho q u2 cos theta minus u1 plus the pressure force in
x direction that is p1 a1 minus p2 a2 cos theta
This gives the fx force in x direction y direction will be f t y the total force in y direction
is equal to rho q u2 sin theta pressure force in y direction is minus p2 a2 sine theta . That
gives the resultant force in y direction and finally the fx is plotted here fy in this
direction finally we can find out the resultant force will be square root of fx square plus
fy square and the angle will be fy divided by fx tan phi is equal to fy by fx
We can find out the resultant force and its direction you have already seen one of the
example how we can apply the momentum equation linear momentum equation which we have derived.
We can apply for a particular case the force due to flow around a pipe bend Here we have
found the force the resultant force acting on the particular problem that will be equal
to the rate of change of momentum. That gives the linear momentum equation using this we
can find out the resultant force. For this particular problem how much thrust or how
much force should be used if you want to know how much force to support or it will put as
a thrust block to withstand the pipe that is obtained by the resultant force.
We can apply the linear momentum equations for various cases we will be discussing few
more cases. Finally force due to flow around a pipe then it is f t x in y direction frx
plus fpx plus fbx that is already obtained as ftx is already found the pressure force
is found the body force is 0. Finally this expression gives the force in x direction
and y direction is fty minus fpy minus 0. This is rho q u2 sine theta plus p2 a2 sine
theta that gives the force in y direction.
We found the force in x direction y direction we can find out the resultant force that is
the thrust required to keep the pipe the that pipe the thrust block should withstand this
force. This is one of the simple application of the momentum linear momentum equation.
Further we will be discussing some more applications of this linear momentum equations in the next
lecture.
