Practice English Speaking&Listening with: Lec-13 Dynamics of Fluid Flow

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Welcome back to the video course on fluid mechanics. We were discussing about the dynamics

of fluid flow; we have seen the derivation of the Eulerís equation and how to derive

the Bernoulliís equation that also we have seen in the last lecture.

Today, we will further see the applications of Bernoulliís equation and then many other

practical cases where this equation can be utilized effectively that we will discuss

today. For Bernoulliís equation as you can see in this slide, the Bernoulliís equation

can also be derived from conservation of energy. We have seen the energy per unit weight plus

the pressure energy plus unit weight plus kinetic energy per unit weight plus potential

energy per unit weight is equal to total energy per unit weight of the system. We have already

seen that the total energy in the system does not change as per the conservation of energy;

we can say that total head does not change so that this way also we can derive the Bernoulliís

equation. Also, we have seen this between two points if you take total head at 1 is

equal to total head at 2 if you consider in a pipe flow.

We can utilize this principle in many forms including, considering the total energy of

the system including the work done per unit weight or the energy supplied per unit weight

as we can see in this slide.

So, this Bernoulliís equation is applicable to many problems, but if the flow is irrotational

we can stay along a streamline, but if the flow is irrotational we can also use this

across the streamline; some of the restrictions we will discuss later. First, we will see

some of the important applications of the Bernoulliís equation. Let us consider a streamline

around a blend body in this slide.

Here, you can see that the flow is coming in this direction and with the blend body

effect; there will be a stagnation point like this. We can find out this in case 1. The

application of Bernoulliís equation: case 1 - stagnation pressure. We can isolate a

point in the field where velocity is 0 that is so called stagnation point. Here, in this

particular case, in this figure, this point 2 will be the stagnation point and by considering

the flow here at this location, at section 1 and section 2. Between these two sections

if you apply the Bernoulliís equation along a streamline from 1 to 2, if you consider

the datum like this horizontal so that flow is horizontal, so z1 is equal to z2. So applying

the Bernoulliís equation along the streamline between section 1 and 2, we can write P1 by

rho g plus u1 square by 2g plus z1 is equal to P2 by rho g plus u2 square by 2g plus z2,

here P1 is the pressure at section 1 , P2 is the pressure at section 2, u1 is the velocity

at section 1 , u2 is the velocity at section 2, rho is the density of the liquid and g

is the acceleration due to gravity. We can see that we are considering the stagnation

point; due to this stagnation point the velocity at that point will be 0 so that this equation

we can write P1 by rho, if you multiply both sides by g then P1 by rho plus u1 square by

2 is equal to P2 by rho. Since z1 is equal to z2 this is canceled and u2 since we are

considering stagnation point we can write u2 is equal to 0. So that P1 by rho plus u1

square by 2 is equal to P2 by rho.

From this we can write P2 is equal to P1 plus 1/2 rho u1 square. By considering the point

1 we can find out the pressure at the stagnation point so that is called stagnation pressure.

Now, we have applied the Bernoulliís equation between two points: one is stagnation point

and the other one is the particular section 1 so that we could find the stagnation pressure.

This is one of the basic application of the Bernoulliës equation in fluid mechanics and

there are number of other applications. Based upon this stagnation pressure let us consider

a pitot tube. Pitot tube is generally used to measure the mean flow velocity especially

in closed conduit such as pipes.

Here you can see in the slide a pipe is there; the flow direction is this and then we want

to find the pressure at the central line. If you want to find the pressure at this particular

central line velocity what we can do is we can immerse the pitot tube like this at this

particular point 2 where we want to find the velocity. We should have another point also

where we can introduce a piezometric tube like this. So this is the piezometric tube

here at section 1 and at section 2 we introduce a the pitot tube so you can see the fluids

level that there will be a difference since it is mainly due to the velocity effect at

the centre line. The head at the piezometric level will be h1 from central line and on

the pitot it will be h2 from the central line. Now, if you apply the Bernoulliís equation

between section this section 1 1 and section 2 2 we can write this P2. You can see that

at section 1 1 the pressure is P1 so P1 plus 1/2 rho u1 square will be the total head at

the section 1; at section 2 2 it will be P2 is equal to P1 plus 1/2 rho u1 square so that

P2 can be written as rho g h2 from which we can get the velocity at the centre line for

the pipe loss, that is, u is equal to square root of 2 g h2 minus h1. This is the way while

introducing a piezometer and then a pitot tube we can find the mean velocity of flow

like in a pipe as shown in this figure.

We have found the central line velocity by considering two points, 1 1 and 2 2 and at

point 1 we introduced a piezometer and point 2 we introduced a pitot tube and then we are

trying to find central line velocity. We can get the Bernoulliís equation between section

1 and 2 as this figure and finally u is equal to square root of 2 g h2 minus h1, where h2

minus h1 is the head level between the piezometer this section 1 and the pitot tube introduced

at section 2.

In earlier case what we discussed should have one piezometer and a pitot tube. But instead

of this arrangement, we have another type of arrangement to find out the mean velocity

of fluid, pitot static tube in this figure. You can see that a pitot static tube is introduced

with respect to the pipe where the velocity is found. In this mechanism, it combines tubes

and connector to a manometer, manometer is shown here and the central line velocity where

we are trying to measure velocity this manometer tube in introduced at this location. There

is an opening with respect to this tube at section 1 1 and you can see that when manometer

is connected, the fluid level difference will be like this at a, b as shown in this slide.

So, the pressure at the central line P2 is static. We can write P2 is equal to P1 plus

1/2 rho u1 square with respect to this section 1 1 as we have seen earlier. If you consider

the manometer here this Pa is equal to the pressure, Pa is equal to P2 plus rho g x.

This is fluid level, rho g is the unit weight of the specific weight of the liquid and X

is the height difference between this level as shown this line and this line is X.

Pa is equal to P2 plus rho g and Pb is equal to the pressure at section B. At particular

location b, Pb is equal to P1 plus P1 with respect to the fluid pressure at this section

1 P1 plus rho g X minus h. This difference rho g into X minus h plus the density of the

manometer liquid into g into, this is h, so rho g into h.

With respect to the conditions Pa is equal to Pb, we can equate this both equations.

So that P2 plus rho g X is equal to P1 plus rho g into X minus h plus rho manometer liquid

density into g h. If you use this relationship as in the previous equation you can write

P1 plus rho h g into manometer minus rho is equal to P1 plus rho u1 square by 2.

We can find P2 with respect to this equation and then we can substitute that here so that

will give u1, the velocity. For the central line velocity square root of 2 g h rho, the

density of manometer liquid minus density of the pipe divided by rho. So u1 is equal

to square root of 2 g h into rho m minus rho divided by rho. Like this a pitot static tube

is used to measure the mean velocity of flow in a pipe instead of as we have seen in the

previous case here we have to use piezometer as well as a pitot tube but here the mechanism

is pitot static tube and we can find the velocity as shown here.

The third case which we will be discussing is application of Bernoulliís equation for

venturimeter. We can see that this venturimeter is used to find the discharge in a pipe and

this is the pipe which we want to find the discharge and for this venturimeter arrangement

there will be a converging section as you can see here; there will be a diverging section

like this. At this particular point you can see a minimum cross sectional area and then

we will be using here a manometer like this and it will be connected between this particular

section, that is, before the divergence starts this particular section 1 and then particular

section 2 as shown in this figure and then with the manometer liquid you can see the

a fluid levels at a height difference of h. With respect to this convergence section and

divergence section we want to find the discharge and then that is the pressure difference which

we measure between the discharges obtained from the pressure difference measurement between

section 1 and 2.

If you apply the Bernoulliís equation along a streamline from point 1 to 2 in the previous

figure from point 1 to point 2 then we 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, where P1 and P2 are the pressure

at section 1 and 2 and u1 and u2 111110 are the velocities at section 1 and 2 and z1 and

z2 are the high difference with respect to the datum here, this is z1 and z2. From the

continuity equation, you can write Q is equal to u1 A1 is equal to A2 u2 with respect to

the velocity and area of cross section we can write Q the discharge is same. So Q is

equal to A1 u1 is equal to A2 u2 from which we can write u2 is equal to u1 A1 by A2. We

substitute for u2 in this equation so that we can write P1 minus P2 by rho g plus z1

minus z2 is equal to u1 square by 2 g into A1 by A2 whole square minus 1, from which

we can find the velocity, central line velocity that means at this location we can find the

velocity by using the venturimeter. So once the velocity is known we can just multiply

by Q is equal to A1 u1 that will be discharged through the pipe. This is another mechanism

which we generally use to either find the velocity or you need to find the discharge

depending upon the case in a pipe flow. This is another application called venturimeter.

As far as these kind of hydraulic equipments are concerned, venturimeter is concerned we

can see that theoretical discharge is Q hydraulic discharge is equal to u1 A1 but actual discharge

you can see that there will be a reduction with respect to this measurement. So the actual

discharge will not be equal to theoretical discharge.

The actual discharge is obtained by actual discharge, Qactual. We have to multiply by

a coefficient of discharge Cd, so that Qactual is equal to Cd by in Cd into Q theoretical.

So that we can write Qactual discharge is equal to Cd into u1 into A1. We can write

with respect to the previous equation which we derived for u1. The Qactual is equal to

the coefficient of discharge multiplied by A1 into A2, so A1 is the cross sectional section

at 1 and A2 is the cross section at the converging point, A2 multiplied by square root of 2 g

into P1 minus P2 by rho g plus z1 minus z2 by A1 square minus A2 square. This is the

actual discharge. With respect to this kind of measurement there is a difference between

the actual discharge and theoretical discharge. We have to multiply by the coefficient of

discharge. So, Qactual is equal to Qtheoretical into coefficient of discharge.

Now, we can say in terms of the manometer readings we can write P1 plus rho g z1 is

equal to P2 plus density of manometer is equal t into g h plus rho g z2 minus h. This we

can write as P1 minus P2 by rho g plus z1 minus z2 is equal to h into rho manometer

by rho minus 1. Actual discharge can be written as Cd into A1 into A2 square root of 2 g h

rho manometer by rho, the density of the fluid in the pipe minus 1 divided by A1 square minus

A2 square.

Like this by using venturimeter we can measure the velocity or the discharge in a pipe flow.

Now the application as far as venturimeter is concerned we will see further applications

of Bernoulliís equation. We have seen for the venturimeter; we have also seen another

mechanism for flow measurement, the discharge of velocity like the various flow measurement

equipments are shown here orifice as shown in this figure.

Then nozzles and venturimeter which we have already discussed are some of the equipments

used for flow measurements which we utilize the Bernoulliís equations and continuity

equation together. So that we can find the flow of velocity over the discharge at particular

sections of a pipe line especially the mechanism is used for pipe line. This is one of the

applications of the Bernoulliís equation

We will go to further applications of the Bernoulliís equation. Fourth one is the flow

through orifice. You can see that this orifice is a mechanism. If there is a tank like this

and if there is a small hole like this and so water will be coming from the tank as in

the case of a jet, so this is called an orifice. There are different forms of orifice. This

is one of the simple most forms of the orifice; we can utilize to find the time to empty a

tank or to particular discharge measurement can be utilized. Here for the flow through

orifice, you can see that whenever the fluid is coming out of the orifice or this is small

hole which is called orifice, then all the streamlines are converging to the opening

at the orifice. Then you can see that there is a location where the streamlines are converging

and the area of cross section of the jet is minimum. This section is so called Vena contracta,

where streamlines contract after orifice to minimum and then become parallel like this.

You can see that now the streamlines become parallel. At this particular location so called

vena contracta there will be a velocity and pressure; the velocity and pressure will be

uniform across the this particular location of the vena contracta, so this is so called

orifice. As we can see that this is also the orifice, as I mentioned, you can utilize for

the velocity or the discharge measurement.

The same principle is utilized there also. Now, we are considering the vena contracta

and between the tank which we are considering here, if we consider the surface of the tank

at this particular location point 1 and then if you consider the orifice location of the

particular vena contracta which we are considering at section 2, we are considering two positions

namely, position 1 and position 2. Between position 1 and 1we can see that jet is flowing

freely to the atmosphere; the pressure at this location will be 0 and at this is an

open surface. At location 1, the pressure will be 0. So, p1 is equal to 0; p2 is equal

to 0 and this is the open surface of the tank. There the velocity u1 is equal to 0 and then

we can see that datum head z1. If you consider this centre of the orifice as the datum then

z1 is equal to h and z2 is equal to 0. If we can apply the Bernoulliís equation between

section 1 and 2, between the section position points 1 and 2, we can see that we will get

h is equal to u2 square by 2 g or we can write u2 is equal to square root of 2 g h. So the

velocity at this location u2 at location 2 will be u2 that is equal to square root of

2 g h.

Now, as we have seen earlier as far as coefficient of discharge, again the actual velocity will

be defined from the theoretical velocity. We have to multiply by a factor called coefficient

of velocity for these kinds of problems. Due to the friction, the actual velocity will

be different. So when the fluid is coming out of the orifice there will be friction

with respect to the atmosphere and the then due to the friction effect the actual velocity

will be slightly different. We have to multiply by a coefficient so called velocity. So, the

actual velocity is equal to coefficient of velocity into the theoretical velocity.

This coefficient of velocity for these kinds of problem varies from 0.7 to 0.9 depending

upon the various conditions like orifice locations or orifice diameter and other conditions.

So, the actual velocities obtained as the coefficient of velocity multiplied by the

theoretical velocity is flow through orifice.

Finally, for the orifice you can see that if you consider the vena contracta here then

with respect to the vena contracta the actual area of the jet is the area of vena contracta.

So, the jet which we are considering out of the orifice the actual area of cross section

of the jet is the area of the vena contracta. So Aactual is equal to we have to multiply

by a coefficient of contraction into A into area of cross section of the orifice. So,

actual area is equal to coefficient of contraction multiplied by the cross sectional area of

the orifice. Finally, now if you want to find the actual discharge Qactual is equal to area

of cross section into u. So, Qactual is equal to Aactual that means Aactual is C into Aorifice

into u theoretical. Finally, Qactual is equal to coefficient of discharge into area of cross

section of the orifice into the theoretical velocity. This gives the actual discharge

from the orifice. This has got many practical applications as I mentioned. Even flow measurement

orifice can be used so many other applications are there in fluid mechanics for this flow

through orifice. We are now using the three coefficients: one is the coefficient of contraction

to find the actual area; second one is the coefficient of velocity to find the actual

velocity and the actual discharge; finally, we are getting Cd is equal to C into Cv that

means coefficient of discharge is equal to coefficient of contraction multiplied by coefficient

of velocity so that is Cd, finally, we get the actual discharge. This is another application

of the Bernoulliís equation.

Now, just find the time for emptying a tank. As I mentioned with this Bernoulliís equation

again can be utilized. If you want to find how much time is taken for a liquid in a tank

to come from level h1 to h2. If you want to find Bernoulliís equation can be utilized

here and then say if h is the level difference between h1 minus h2 we can write the discharge

Q is equal to A into the velocity, since the level is going down we are using minus here.

So this is equal to minus A into del h by del t.

With respect to time this head is changing; so Q is equal to minus A into del h by del

t. This we want to find here, the time for emptying the tank from h1 to h2. From this

equation we can write delta t is equal to minus A into, we will write the discharge

equation which we have seen earlier, delta t is equal to minus A divided by Cd into A

0 into root two g into delta h by root h.

Now, we can integrate this expression with respect to the levels h1 to h2. So, t is equal

to minus A into Cd into A0 into root 2 g into integral h1 to h2 del h by square root of

h. If we integrate this expression we will get finally the time for emptying a tank h1

to h2 will be equal to minus 2 A divided by Cd into A0 root 2 g into square root of h2

minus square root of h1. Here again we have to use the Bernoulliís equation for emptying

a tank from one level to another level. So like this various applications are there.

Earlier, we have seen just an orifice which is directly emptying the liquid to or water

to the atmosphere. If we are considering a submerged orifice as shown in this figure

you can see there is an orifice here at location 2 and then through this orifice the liquid

is passing from tank 1 to tank 2, very adjacent tanks. Finally after some time this become

a submerged orifice. To find the discharge passing through submerged orifice and the

velocity of flow through the submerged orifice we can apply the Bernoulliís equation from

this point 1 to this point 2 at the centre of the orifice. We can write from Bernoulliís

equation 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. We can see that since due to that atmosphere pressure, p1 rho g will

be equal to 0 and here the velocity will be 0. So u1 square by 2 g also is 0 and here

z1 is equal to h1 if you take the datum as the centre line of the orifice.

So h1 is z1 and then p2 by rho g, you can write as rho g h2 by rho g plus u2 square,

the velocity is u2 so u2 square by 2 g since the datum is taken z2 is equal to 0. Finally,

we can get this velocity of flow u2 is equal to square root of 2 g into h1 minus h2. h1

minus h2 is actually the level difference between this level and this level, so this

gives the h1 minus h2. The velocities are found and then we can find the discharge of

Q is equal to, we can use the continuity equation, Q is equal to A- area of cross section area

of the orifice into velocity and finally the actual discharge is equal to we have to multiply

by the coefficient of discharge area of cross section of the orifice into the velocity which

we are calculating here. So this is the case of a submerged orifice.

Now, we will see another case so called confined flows, if you consider which is a confined

with respect to this tank as shown here. The flow is coming in this direction and then

flow is going through the other nozzle at location 2. If you want to find for this confined

flow the discharge or any other parameter or velocity of flow at section 2 again we

can use the Bernoulliís equation as shown between the section 1 here and section 2.

When we apply the Bernoulliís equation, we can write p1 plus 1/2 rho V1 square plus gamma

z1 is equal to p2 plus 1/2 rho V2 square plus gamma z2. You can see this is open to atmosphere.

So, p1 is equal p2 is equal to 0 and z1 is equal to h and z2 we are taking it as the

datum; so z2 is equal to 0. Finally, we get 1/2 V1 square plus g h is equal to 1/2 V2

square and then we can find the velocity. From that we can get the discharge. This is

as far as if you consider a confined flow as shown in this figure. Now, there are many

other applications for Bernoulliís equation. Some of the open air flow types again we will

discuss here.

Next one is sluice gate. If you want to find the discharge passing through sluice gate

so here you can see this figure here. In this figure, you know that the sluice gate is generally

in a reservoir or different chemical plants or water supply. You just use this gate to

pass particular amount of discharge and if you want to find how much discharge pass through

this particular opening again we can use the Bernoulliís equation and continuity equation

for a sluice gate.

So sluice gate is here and the liquid level is at this level and the flow is coming in

this direction. After the gate again as we have seen again a vena contracta will be formed

that means we know cross sectional area streamlines will be parallel and then we will be applying

the Bernoulliís equation between the section 1 and section 2 of this area of cross section

here between section 1 and 2. Applying the Bernoulliís equation and continuity equation,

if you apply the continuity equation between section 1 and 2, we can write Q is equal to

A1V1 is equal to, b is the width of this opening or with respect to this equation. So, b into

V1 z1 that is equal to A2 V2 that is b V2 z2, where b is the thickness or the width

which we are considering.

Now we can apply the Bernoulliís equation between point 1 and point 2. So that we can

write p1 plus 1/2 rho V1 square plus gamma z1 is equal to p2 plus 1/2 rho V2 square plus

gamma z2, here p1 and p2 is atmospheric pressure. So, p1 and p2 have to be considered.

From this equation, we can find out the velocity, for example, here V2. Then, we can get an

expression for the discharge passing through the gate, that is, Q is equal to z2d into

square root of 2 g into z1 minus z2 divided by 1 minus z2 by z1 whole square, this is

z1. The bottom of the sluice gate is considered as the datum and this is z1 here and z2 is

depth of at this particular location section 2 and z1 is the depth at section 1.

This way here also p1 equal to p2 equal to 0 and then by using the Bernoulliís equation

and continuity equation we can find the discharge passing through this sluice gate. This is

one of the applications as far as open channel flow is concerned and then another application

is application for the Bernoulliís equation for weir or notch. So, here we can see that

weir is here so we want to find how much is the discharge passing over the weir um or

a notch.

This is rectangular sharp crested weir. If you want to find, we can say here again this

particular figure you would be considering this location a streamline like this. Then,

between section 1 and 2, we will be considering, this is 2 and this is 1, we will be considering

the Bernoulliís equation. If you consider this particular point p1 is equal to this

height, depth of flow with respect to this point 1 p1 is equal to gamma h and then here

it is atmospheric pressure so we can neglect the pressure.

So p2 is equal to 0. If you consider between section 1 and 2, if you apply the Bernoulliís

equation we can show that Q is equal to C1 into h b root 2 g h or this is equal to C1

into b into square root of 2 g into h to the power 3 by 2, where C1 is a constant as far

as this weir or notch is constant. Like this we can apply this Bernoulliís equation combination

with the continuity equation to large varieties of problem in practical cases.

Again, here if you consider a general equation for the weir or notch the general equation

you can see is the sharp crested weir and then you consider a small strip of elemental

strip through a notch. You can see that this velocity u is equal to square root of 2 g

h and discharge through the strip delta Q is equal to area of cross section into velocity

V into delta h into square root of 2 g h with respect to this figure here.

Then, we can just integrate between 0 to H; H is the depth of flow with respect to crest

of the weir as shown in this figure. So, this is the depth of flow. We can integrate from

0 to H, so that Qtheoretical the theoretical discharge is square root of 2 g integral 0

to h b H to the power 1/2 d h and then we can integrate, the theoretical discharge is

equal to b root 2 g integral 0 to 2 H square root of h d H that is equal to 2 by 3 b root

2 g H to the power 3 by 2 and the actual discharge as we have seen here again. We have to use

a coefficient of discharge. So Qactual is equal to Cd into 3 by 2 b root 2 g, H to the

power 3 by 2. So this is another application of the Bernoulliís equation in combination

with the continuity equation.

Similarly, if you consider a V notch as shown in this figure, again, this V notch is just

a triangular weir, here this b width at this particular location can be found at depth

h from the surface of the water in the total depth is H; so b is equal to two times H minus

H into tan theta by 2 is the angle of the V notch. Then, Qtheoretical is equal to 2

into root 2 g tan theta by 2 integral 0 to H minus h into square root of h d H.

Finally, the equation is this is equal to 8 by 15 root 2 g tan theta by 2 H to the power

5 by 2. In a triangular weir or V notch the discharge will be varying with respect to

H to the power the total depth of floor H to the power 5 by 2. Actual discharge is equal

to the coefficient of discharge multiplied by the theoretical discharge as shown in this

slide.

Like this by using the Bernoulliís equation and the continuity equation we can solve many

practical problems in hydraulics or in the closed quantity flow like the pipe flow and

also open channel as we have seen in the notches, weirs or the sluice gate problem. We can solve

different kinds of problems using the simplified form of the Bernoulliís equation and the

continuity equation you can see all this problem is very simple approach. Bernoulliís equation

is a simple equation so that we can easily without many complexities can easily approximate

what kind of flow is coming and then we can apply the Bernoulliís equation between two

sections between two points on a streamline. We can find either the discharge or the velocity

or the pressure between the two sections.

These are the most important applications of the Bernoulliís equation. So, before going

further with the derivation of other fundamentals momentum equation we will just see some aspect

of the total energy with respect to hydraulic grade line and the energy line.

We have already seen with respect to the Bernoulliís equation total head is equal to the pressure

head plus velocity head plus datum head. This total energy will be there of course, there

will be losses also of the internal force, internal energy will also be there. That also

we have to consider and that is the total energy. While finding this energy at various

section of a pipe line or a open channel flow depending upon the case which we are doing

in the problem, we can represent two lines called energy line and hydraulic grade line.

So energy line and hydraulic grade line are actually the graphical forms of the Bernoulliís equation. The basic principle

as we have seen is some of the various energies of fluid remains constant as the fluid moves

from one section to another. The energy is conserved when we consider a pipe flow like

this. If you consider this as a pipe, between one section and another section, we can see

that the various energies remain constant as the fluid moves from one section to another

section. Now, the equation which we derived Bernoulliís equation is derived for steady,

inviscid, incompressible flow, total energy remains constant along a streamline. This

is the basic equation. The energy line concept and hydraulic grade line concept is also based

upon the conservation of energy that means the total energy remains constant along a

streamline. If you consider the Bernoulliís equation the pressure head p by gamma plus

the velocity head v square by 2 g plus z the datum head is equal to constant on a streamline.

So that we can write this is equal to H. Thus, pressure head plus velocity head plus datum

head is equal to total head H as shown in this slide here.

Now if you consider this figure taken from the Munson et al Fundamentals of Fluid Mechanics

it is slightly modified. You can see if you consider a pipe flow like this and now if

you take the piezometer level at this location and at section 1 1 and at location 2 2, you

can see that piezometer indicates the pressure head and datum head. With respect to piezometer

we can write p by gamma plus z, that is the piezometric head. So this as the static pressure

does not measure the velocity head. This gives the piezometric head at section 1 1 and here

at section 2 2, p2 by gamma plus z2 that is the piezometric head. So the hydraulic gradient

line is actually the locus of series of piezometric taps at different locations. If you consider

this as a pipe at different locations if you just find the piezometric head, that is actually

the velocity head plus the datum head.

If you just draw the locus of a series of piezometric taps, you have a long pipe line

as you can see here which we are considering; with respect to this is the datum. So, with

reference to this pipe line we can have a series of piezometer and then we can plot

the piezometric heads. You can if you just put the locus of this piezometric head that

gives the hydraulic grade line. So, this is the fluid flow, this is the pipe flow, this

is the datum which we are considering, this is z3, if you can measure to the centre line

this is z2, this is z1 to the centre line. So, when we plot the locus of the piezometric

heads that is the hydraulic grade line does not include the velocity head. Here we are

not considering the velocity head at various locations which we have seen. So that gives

the hydraulic grade line. Hydraulic grade line is the locus of a series of piezometric

taps. For example, now we consider a pipe flow like this and we have three sections:

section 1 1, section 2 2 and section 3 3 as shown in this figure.

If you can introduce a piezometer at the center line as we have seen earlier. With respect

to the piezometer we can see that it will be considering the velocity head also. So,

for these three sections if you find the piezometric levels you can see that this is varying like

this. With respect to this, if you just plot the locus where the pitot tubes levels, here

this is the level, this is the level, and this is the level. So this line gives the

energy lines. So now energy lines include the datum head, the pressure head and the

velocity head. If you consider a pipe line at various locations and if you introduce

the pitot tubes and if you just get the levels of the pitot tubes and then if you join those

levels with a line, that line is called energy line. As shown here with reference to hydraulic

grade line the difference is that there is extra velocity head. For hydraulic grade line

we consider only the datum head and the pressure head, we are using locus of the levels with

reference to piezometric tools and various equations but as the energy line is concerned

we are considering the velocity head also. You can see in this figure we are introducing

the pitot tubes at various locations and then we finally get a line so called energy line.

This hydraulic grade line and energy line concepts are very useful especially for pipe

flow analysis and open channel flow analysis. This is coming here and again we are using

the Bernoulliís equation of total energy so that principle of conservation of energy

is used.

This concept of energy line and hydraulic grade line is used in many problems especially

pipe line problems also sometimes in open channel flow. In all these problems what we

have seen so far when we measure the velocity we are getting the mean velocity or say the

velocity measurement is with respect to most of the 1 dimensional problem, V is the average

of the mean velocity.

Here, we have to do a correction with respect to this. Since we are considering the mean

flow if you consider a pipe then you can see that with reference to this we are considering

the mean flow velocity. So, with respect to the mean flow velocity the average velocity

at a cross section is taken; the actual velocity may not be uniform. The concept which we are

using here is that the mean velocity is taken in such a way and then it is multiplied by

all other area of cross section. Q is equal to area of cross section into velocity, V

is the mean velocity. This mean velocity is not actually considered in a pipe flow like

this with reference to the center line. The mean velocity you can see the velocity is

varying like this; it will be maximum at the center line; then, it will be minimum 0 at

both sides of the pipe valve.

The mean velocity concept will not give a correct value as far as you find the various

measurements like discharge. So, it will not give a correct value, we have to use a correction

factor. So as far as kinetic energy is concerned we have the actual velocity is non-uniform

as you can see in this pipe flow or even the open channel flow so what we are considering

will be non-uniform flow. For the non-uniform flow we have to use a correction factor called

kinetic energy correction factor which is so called alpha.

We can obtain this alpha. With respect to mean velocity we can say what is the discharge

or with respect to the continuity equation we can derive this alpha is equal to integral

v cube dA divided by V cube A, where small v is the varying velocity. You can see that

the velocity is varying. So you can consider various sections like this. Then, it is the

integral of V cube, v is the velocity at any point and then capital V is the mean flow

velocity. The energy correction factor alpha is equal to integral small v cube dA divided

by V cube A area of cross section of the pipe or the channel section which you are considering

so alpha is equal to integral v cube dA by V cube A, where V is the mean flow velocity,

A is the area of cross section, small v is the velocity varying from various section

which you are considering.

When we are consider the V square by 2 g term we have to use this. We have to multiply by

the kinetic energy correction factor alpha so that we get a correct velocity head or

the correct kinetic energy. If you consider the flow to be uniform you can say that alpha

is equal to 1. No need of this correction and if it is greater than 1, we can say that

it will be greater than 1 for all other forms. That means for non-uniform flow condition

it will be greater than 1 and then for laminar flowing pipes this we can derive as equal

to 2. Finally, if you use the kinetic energy correction factor the energy equation becomes

p1 by gamma plus alpha V1 square by 2 g plus z1 that is equal to p2 by gamma plus alpha

V2 square by 2 g plus z2.

We have to utilize this correction factor and finally the equation become p1 by gamma

plus alpha V1 square by 2 g plus z1 is equal to p2 by gamma plus alpha V2 square by 2 g

plus z2. To get a correct value, this energy kinetic energy correction factor to be used,

we multiply with respect to the velocity head V1 square by 2 g or V2 square by 2 g when

we use the Bernoulliís equation or the energy equation as shown in this slide.

Finally, to conclude the Bernoulliís equation or the energy equation, as I mentioned earlier,

we have to see the various losses whether any of the various energy levels are added

or taken out or external work done. All these things are to be considered while we solve

real practical field problem.

General equation for conservation of energy for incombustible fluid between two sections

we can write as p1 by gamma plus alpha V1 square by 2 g plus z1 plus qw plus H E is

equal to P2 by gamma plus alpha V2 square by 2 g plus z2 plus e2 minus e1, where this

qw is the heat added per unit weight of fluid so that effects is to be considered. e1 and

e2 are the internal energies, it is there where the flow is considering and H E is the

external work done.,

This is the final form of the energy equation even though we have seen the simple form of

the Bernoulliís equation but when we solve practical problems we have to see what is

the work added or energy is added or any losses of energy due to various aspects, like internal

energy aspects or any heat added. All these things we have to consider and final form

of the equation when we consider the conservation of energy as shown here.

Now, e2 minus e1 minus qw can be written as reversible plus irreversible head. Irreversible

head is can be written as head loss and it is the energy loss per unit weight of fluid

due to friction and other causes. For incompressible fluid we can write total head at 1 plus heat

added due to machine like pumps or turbine and then minus head loss is equal to total

head

Finally, the energy equation can be written as H1 plus HE minus HL is equal to H2 and

if you want to find the work done over a fluid, power input into a fluid is equal to gamma

Q Hm in watts, where gamma is the unit weight of fluid in Newton per meter cube, Q is discharge

in meter cube per second and Hm is the head added to flow in meter. So this is the general

form of the energy equation which is used to solve many of the practical flow problems.

We have now seen various application of the Bernoulliís equation and we have seen the

general energy equation. Later, we will be discussing various limitations of the Bernoullliís

equation and we will solve some of the example problems related to Bernoulliís equation.

The Description of Lec-13 Dynamics of Fluid Flow