Fluid Mechanics, Chapter Five

8/8/2019 Fluid Mechanics, Chapter Five

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MAKERERE UNIVERSITY

FACULTY OF TECHNOLOGY

PETROLEUM GEOSCIENCE & PRODUCTION

PGP 2106: FLUID MECHANICS

CHAPTER FIVE

THE ENERGY EQUATION AND ITS APPLICATION

CLASS NOTES

Prepared By:Edmund Tumusiime

scien ce


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5.1 IntroductionIn this chapter, study is made on the energy transfers within a flowing

fluid, and also the prediction of fluid flow phenomena. Bernoulli's

equation will be developed and demonstrated in a more general form that

can accommodate apparent energy losses due to frictional andseparation effects, by application of conservation of energy principle.

The transfer of energy into, or out of a fluid flow system, by introduction

of mechanical devices such as fans, pumps, or turbines, is considered

leading to the introduction of the general steady flow energy equation.

The representation of apparent energy losses due to friction separation

effects will be defined and the application of the energy equation to the

measurement of the flow rate and velocity is demonstrated for a range of

pipe flow and free surface flow conditions.

5.2 Mechanical energy of a flowing fluid

Consider an element of fluid (Fig 5.1) in motion

The element will possess potential energy due to its elevation 'z' above

some chosen horizontal datum, and kinetic energy due to its velocity 'v'

like any other object.

If the weight of element is 'mg', then

Potential energy of element = mgz

Fig 5.1: Energy of a

flowing fluid


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Potential energy per unit weight = z

Kinetic energy of element = 2

2

1mv

Kinetic energy per unit weight =g

v2

2

A steady flowing fluid can also do work due to the force generated when

the fluid pressure acts on a given area in the flow. If the pressure at

section AB of area 'A' is 'p', then

Force exerted on AB = pA

After the weight 'mg' of fluid has moved along the stream tube, section ABwill have moved to A'B'.

Volume passing AB =g

mg

=

m

Therefore, Distance AA =A

m

Work done = (Force) x (Distance AA)

=A

mpA

=

pm

And, Work done per unit weight =g

p

..................(5.1)

Note: The quantityg

p

is known as the flow work or pressure energy


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5.3 Bernoulli's TheoremStates that; 'for steady flow of a frictionless fluid along a streamline, the total energy

per unit weight remains constant from point to point.

i.e.

Hzg

v

g

p

2

2

whereg

p

= pressure head (meters)

g

v

2

2

= velocity head (meters)

z = potential head (meters)

H = total head (meters)

If 1&2 are any two points in a stream, then

2

222

1

211

22z

g

v

g

pz

g

v

g

p

.(5.2)

(T.E/wt)1 = (T.E/wt)2

Note: The above equation (5.2) assumes that no energy has been

supplied to or taken from the fluid between points 1&2. Energy could

have been supplied by introduction of a pump. Equally, energy could

have been lost by doing work against friction or in a machine such as a

turbine.

In such a case, Bernoulli's equation can be modified to include these

conditions.

Pr.E per unitweight +

K.E per unit

weight +

P.E per unitweight =

T.E per unitweight = Constant


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thus,

whz

g

v

g

pqz

g

v

g

p 2

2

221

2

11

22

(5.3)

h = loss per unit weight

w = work done per unit weight

q = energy supplied per unit weight

Question

1. a) A tapering pipe of 2m length is placed in vertical position in such amanner that its short end (10cm diameter) is at top and the big end(20cm diameter) is at the bottom. If the discharge through the pipe is

30litres/sec, find the difference of pressures between the two ends.

b) If a differential manometer with mercury is connected between the

top and bottom ends, and if gasoline of specific gravity 0.8 flows

through the pipe, then calculate the manometer reading.

c) If the pipe in part a) is inclined at 300 to the horizontal, then what

will be the pressure difference?

5.4 Kinetic Energy correction factor The Bernoulli's equation was derived assuming uniform velocity across

the inlet and outlet sections. In a real fluid flowing in a pipe or over a

solid surface, the velocity will vary from the solid boundary, increasing

with increase in distance from the solid boundary. The kinetic energy perunit weight of the fluid will increase in a similar manner.

If the cross-section of flow is assumed to be composed of a series of small

elements of area A , and the velocity normal to each element is 'u', then

Mass passing through element in unit time = uA


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K.E per unit time passing through the element = 3

2

1uA

Total K.E passing per unit time = Au 3

2

1

Total weight passing in a unit time = Agu

Thus, taking into account the variation of velocity across the stream,

True K.E per unit weight =

Au

Au

g

3

2

1;

which is not the same asg

u

2

2

; where AA

Qu ,

is the mean velocity

Therefore, True K.E per unit weight =g

u

2

2

where is the Kinetic energy correction factor, whose value depends on the

shape of the cross-section and velocity distribution.

5.5 Representation of energy changes in a fluid systemThe changes of energy and its transformation from one form to another,

which occur in a fluid system, can be represented graphically. In a real

fluid system, the total energy per unit weight will not remain constant.

Unless energy is supplied to the system at some point by means of a

pump, it will gradually decrease in the direction of motion due to losses

resulting from friction and disturbance of flow at changes of pipe section,

or as a result of changes of direction.

Consider the figure below,


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The flow of water from the reservoir at A to the reservoir at D is

assisted by a pump, which develops a head hp thus providing an

addition to the energy per unit weight of hp.

At the surface of the reservoir A, the flow has no velocity and is at

atmospheric pressure (zero gauge pressure), so that the total energy per

unit weight is represented by the HA of the surface above datum.

As the fluid enters the pipe with velocity u1, there will be loss of energy

due to disturbance of the flow at the pipe entrance and a continuous loss

of energy due to friction as the fluid flows along the pipe, so that the total

energy line will slope downwards.

Fig 5.2:Energy changes in a fluid system


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At B, there is a change of section with an accompanying loss of energy,

resulting in change of velocity to u2. The total energy line will continue

to slope downwards but with a greater slope since12 uu and friction

losses are related to velocity.

At C, the pump will put energy into the system and the total energy line

will rise by an amount hp. The total energy falls again due to friction

losses and losses due to disturbance at entry to the reservoir, where the

total energy per unit weight is represented by the height of the reservoir

surface above datum (the velocity of the fluid being negligible; and hence

zero, and pressure atmospheric).

If a piezometer tube were to be inserted at point 1, the water would not

rise to the level of total energy line, but to a level g

u2

2

1 below it, since

some of the total energy is in form of kinetic energy. Thus, at point 1, the

energies present are:

Potential energy =1z

Pressure energy = gp 1

Kinetic energy =g

u

2

2

1

the three energies adding up to the total energy at that point.

The line joining all points to which the water would rise, if an open stand

pipe (piezometer tube) were inserted is known as Hydraulic gradient line,

and runs parallel to the Total energy line at a distance below it equal to

the velocity head.

Since the loss of energy due to friction and separation of the stream from

its boundaries depend on velocity of the stream, the losses can be

encapsulated in the kinetic energy equation as

2

2

1uK ;


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where K is a constant that depends upon the conduit

parameters such as length, diameter, roughness, or fitting

type, and u is the local flow velocity.

Thus, the energy equation applied to points 1 and 2 gives,

2

2

2

221

2

112

1

2

1

2

1Kugzvpgzvp

When applied to reservoirs 1 and 2 with an open surface, the pressures

1p and 2p at the reservoir open surface may be taken as zero since the

atmospheric pressure is the gauge pressure frame of reference.

Also, if the surface area of the reservoirs are very large as compared to

the cross-section areas of the connecting pipe, the velocities 1v and 2v

may be disregarded compared with the pipe flow velocity u .

Thus, the steady energy equation reduces to

2212

1Kuzzg

...(5.2)

5.6 The SiphonConsider now flow in a pipe which rises above the hydraulic gradient (Fig

5.3),


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The pressure in portion PQ will be below atmospheric, and will form a

Siphon. Under reduced pressure, air or other gasses may be released

from solution, or a vapour pocket may form and interrupt the flow.

In such a case, if the control volume is chosen between points 1 and 2,

and the steady energy equation applied at the extremities, it would give

misleading results. The application of steady energy equation between

points 1 and the siphon (A), allows the practicability of the siphon to be

assessed.

Applying the principle of equation (4.2) to a siphon to assess its

practicability, between points 1 and A, gives,

22

1

2

112

1

2

1

2

1Kugzvpgzvp

AAA

Where the friction and separation loss term 2

2

1uK refer to loss

between 1 and A.

If the pipe between the two points is assumed constant diameter, then

the

local velocity in the loss term = Velocity at A =Au

giving, KuzzgpA 12

1 221 ; u = pipe flow

velocity


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5.7 MEASUREMENT OF FLOW AND FLOW VELOCITY5.7.1The Pitot tubeThe piptot tube is used to measure velocity of the stream and consists of

a simple L-shaped tube facing into the oncoming flow (Fig 5.4)

If the velocity of the stream at A is u, a particle moving from A to the

mouth of the tube at B will be brought to rest so that u0 is zero.

Applying Bernoullis equation,

(Total energy per unit weight at A) = (Total energy per unit weight at B)

Giving,g

p

g

u

g

p

g

u

0

2

02

22

Fig 5.4: The Pitot tube


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Such that

g

p

g

p

g

u

02

2

.(5.3)

But zg

p

and zhg

p

0

Thus, from (5.3), we have zg

uzh

2

2

Such that ghu 2

Note: When the pitot tube is used in a channel, the value of h can be

determined directly (Fig 5.4 (a)), but if it is to be used in a pipe, the

difference between the static pressure and the pressure at the impact

hole must be measured with a differential pressure gauge, using static

pressure tapping in the pipe walls (Fig 5.4 (b))

While, theoretically the measured velocity ghu 2 , pitot tubes may

need calibration. Thus, the true velocity is given by ghCu 2 , where C

is the coefficient of the instrument, and h is the difference of head

measured in terms of fluid flowing.

5.7.2The VenturimeterThe venturimeter (Fig 5.5) is a measuring device used to determine the

volume rate of flow through a pipeline. It uses the concept of pressure

difference to determine the quantity of flow passing per unit time for a

particular configuration.


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As shown above, it consists of a short converging conical tube leading

into a cylindrical portion, called the Throat, of smaller diameter than

that of the pipeline, which is followed by a diverging section in which the

diameter increases again to that of the pipeline.

The pressure difference from which the volume rate of flow can be

determined is measured between the entry section 1 and the throat

section 2, often by means of a U-tube manometer.

Assuming no loss of energy, and applying Bernoullis equation to

sections 1 & 2, give:

g

v

g

pz

g

v

g

pz

22

2

212

2

111

21

2121

22 2 zz

gppgvv

(5.4)

For continuity of flow,2211 vAvA Or 1

2

12 v

A

Av


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Giving from (5.4),

1

2

2

12

1A

Av

21

212 zzg

ppg

Such that

2121

2/12

2

2

1

21 2 zz

gppg

AAAv

And Volume rate of flow,11vAQ =

gH

AA

AA2

2/12

2

2

1

21

Where, 2121 zz

g

ppH

known as departure from the hydraulic gradient is

determined by equating pressures at the datum level (i.e. X-X in the

above case)

If the area ratio mA

AA

2

1

Then,

gH

m

AQ 2

12/12

1

.(5.5)

Determination of HTo determine H, we equate pressures at level X-X in both limbs.

Thus, ghhzzgpzzgp man 2211

Expanding and re-arranging gives,

121

21

manhzzg

ppH

Substituting into equation (5.5) gives the gives the equation for the flow

rate as

12

12/12

1

mangm

AQ (5.6)


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The arrangement is cheap compared with the cost of a venturimeter, but

there are substantial energy losses. The theoretical discharge can be

calculated from equation (5.5) but the actual discharge is less. A

coefficient of discharge must therefore be introduced (Cd = 0.65 for a

sharp edged orifice)

5.8.1Theory of small orifices discharging into the AtmosphereDefinition: An 'Orifice' is an opening, usually circular, in the side or base

of a reservoir, through which fluid is discharged in form of a jet, usually

into the atmosphere. The volume rate of flow discharged through an

orifice will depend upon the head of the fluid above the level of the

orifice, and, it can therefore be used as a means of measurement.

The term 'small orifice' is applied to an orifice which has a diameter or

vertical dimensions, which are small compared to the head producing

flow, so that it can be assumed that this head does not vary appreciably

from point to point across the orifice.

(Fig 5.7) shows a small orifice in the side of a large tank containing liquid

with a free surface open to the atmosphere.

At point A on the free surface, the pressure A

p is atmospheric and, if

the tank is large compared to the orifice, the velocity A

v will be

negligible; and hence zero.

Fig 5.7: Flow through a small orifice


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At some point B in the jet, just outside the orifice, the pressure Bp will

again be atmospheric, and the velocity Bv will be that of the of the jet v.

Taking the datum for potential energy at the centre of the orifice andapplying Bernoullis equation between points A and B, and assuming

there is no loss of energy,

(Total energy per unit weight at A) = (Total energy per unit weight at B)

i.e. BBB

AAA z

g

v

g

pz

g

v

g

p

22

22

substituting for Hzz BA , 0Av , vvB , and BA pp , give;

velocity of the jet gHv 2 (5.6)

This is a statement of Torricellis Theorem, that the velocity of the issuing

jet is proportional to the square root of the head producing flow.

Note: Equation (5.6) applies to any fluid, 'H' being expressed as head ofthe fluid flowing through the orifice.

Theoretically, if 'A' is the cross-sectional area of the orifice, then

Discharge

gHAQ 2 ..(5.7)

In practice, the actual discharge is considerably less than the theoretical

given by equation (5.7), which must therefore be modified by introducing

the coefficient of discharge Cd, so that


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Actual discharge, QActual = CdQtheoretical

Or gHACQdActual 2 .(5.8)

Note: There are two reasons for the difference between the theoreticaland actual discharges; namely,

The velocity of the jet is less than that given by equation (5.6),

because there is loss of energy between points 'A' and 'B'.

therefore, Actual velocity at 'B' =

gHCvCvv

2

where v

C is the coefficient of velocity, which has to be determined

experimentally and is of order 0.97

Considering the contraction of the jet (Fig 5.8). The particles of the

fluid at the orifice converge at the orifice, and the area of the issuing

jet at 'B' is less than the area of the orifice at 'C'.

In the plane of the orifice, the particles have a component of velocity

towards the centre and the pressure at 'C' is greater than atmospheric. It

is only at 'B', a small distance outside the orifice, that the paths of the

Fig 5.8: Contraction of the issuing jet


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particles have become parallel. The section through 'B' is called 'vena

contracta'.

Therefore, Actual area of the jet at B = ACc

where Cc is the coefficient of contraction which can be determinedexperimentally, and depends on the profile of the orifice. For a sharp

edged orifice, of the form shown above, it is order 0.64

hence, Actual discharge = ( Actual area at B) x (Actual

velocity at B)

= gHCACvc

2

=

gHACCvc

2 . (5.9)

Comparing equations (5.8) and (5.9),

vcdCCC

Note: The values of Cc and Cv are determined experimentally, and

values are available for standard configurations in British Standards

Specifications (BSS).To determine Cd, it is only necessary to collect or otherwise measure the

actual volume discharged from the orifice in a given time, and compare

with the theoretical discharge.

such that,

Question

A rectangular orifice in the side of a tank is 1.5m broad and 0.75m deep.

The level of water in the tank is 750mm above the top edge of the orifice.

Calculate the discharge through the orifice in litters per second if the

coefficient of discharge is 0.6

Coefficient of discharge = Actual Measured dischargeTheoretical discharge


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5.8.2Theory of Large orificesIf the vertical height of the orifice is large, so that the head producing

flow is substantially less at the top of the opening than at the bottom, the

discharge calculated for the small orifice, will not be the true value, since

the velocity will vary substantially from top to bottom of the opening.

Such an orifice is termed as 'Large orifice'

The method adopted for this case is to calculate the flow through a thin

horizontal strip across the orifice (Fig 5.9), and then integrate from top to

bottom of the opening to obtain the theoretical discharge, from which the

actual discharge can be determined if the coefficient of discharge is

known.

Area of strip = hB

Velocity of flow through the strip = gh2

Discharge through the strip, VelocityAreaQ

= hhgB 2/12

For the whole orifice, integrating from1Hh to 2Hh , gives

Total Discharge, dhhgBQ

H

H

2

1

2/12

= 2/31

2/3

22

3

2HHgB ..(5.10)

Fi 5.9: Flow throu h a lar e ori ice


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5.9 Elementary theory of Notches and Weirs

A 'Notch' is an opening in the side of a measuring tank or reservoirextending above the free surface. It is, in effect, a large orifice which has

no upper edge, so that it has a variable area depending upon the level of

the free surface.

A 'Weir' is a notch on large scale, used for example to measure the flow of

a river. It may be sharp-edged or have a substantial breadth in the

direction of flow.

The method of determining the theoretical flow through a notch is the

same as that adopted for the large orifice. For a Notch of any shape (Fig

5.10),

Considering a horizontal strip,

Area of strip = hb

Velocity through the strip = gh2

Discharge through the strip, ghhbQ 2 ..(5.11)

Integrating from h = 0 at the free surface to h = H at the bottom of the

Notch, we have;

Theoretical

H

dhbhgQ0

2/12 ...(5.12)

Fi 5.10:Dischar e throu h a notch


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5.10 POWER OF A STREAM

The fluid flowing can do work as a result of its pressure p , velocity v and elevation z and the total energy per unit weight is given by:

zg

v

g

pH

2

2

Power = Energy per unit time =Weight

Energyx

time

Weight

If Q is the volume rate of flow, then

Weight per unit time = gQ

Giving Power

z

g

v

g

pgQgQHP

2 ..(5.15)

Questions

1) Just inside a fire whose, the gauge pressure is 4 bar. Estimate thevelocity in the jet of diameter 50mm just outside the nozzle, taking the

inside hose diameter as 100mm. Determine also how much high the

jet from the hose might rise, if the hose is pointed vertically upwards.

2) Water discharges from a tank via a pipe, which runs out horizontallyfrom the bottom of the tank. If the water in the tank is 20m deep, and

the head loss in the pipe is known to be 10m, calculate the velocity of

the water on exit from the pipe.


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If the end of the pipe is now placed at a distance of 10m below the

bottom of the tank, and the head loss in the pipe is now given as 12m,

calculate the new velocity of the water on exit from the pipe.

Comment on the two values of exit velocity

3) Water is supplied to a building from a tank (in which the depth is1.5m) on the 10th storey. Consider the flow from a tap located 1m

above the floor of the 5th storey, whose floor level is 15m below the

10th storey level. The water emerges from the tap with a velocity of

5m/s, in a jet of diameter 1cm.

a. What is the head loss in the piping system conveying water fromthe tank to the outlet from the tap?

b. A 26mm pipe leads to the tap and starts 0.5m below it. The gaugepressure measured at the beginning of the pipe is equivalent to a

head of 5m of water. What is the head loss between the tank pipe

and the outlet from the tap?

Documents

Fluid Mechanics, Chapter Five