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.