Fluid Mechanics Dynamics

8/13/2019 Fluid Mechanics Dynamics

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FLUID DYNAMICS J3008/5/1

UNIT 5

FLUID DYNAMICS

OBJECTIVES

General Objective : To understand the measurements of fluids in motion

Specific Objectives : At the end of the unit you should be able to :

state and write Bernoulli Equation

state the limits of Bernoullis Equation

apply the Bernoulli Equation to calculate:

- Potential energy- Kinetic energy- Pressure energy

in

- horizontal pipe- inclined pipe- horizontal venturi meter- inclined venturi meter- small orifice

- simple pitot tube

sketch, label and describe fluid motion mechanism in the

horizontal venturi meter.


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INPUT

5.1 Energy of a flowing fluidA liquid may possess three forms of energy:

5.1.1 Potential energyIf a liquid of weight Wis at a height of z above datum line

Potential energy = WzPotential energy per unit weight =z

The potential energy per unit weight has dimensions ofNm/Nand is measured

as a length or headzand can be called the potential head.

5.1.2 Pressure energyWhen a fluid flows in a continuous stream under pressure it can do work. If the

area of cross-section of the stream of fluid is a, then force due to pressurepon

cross-section ispa.

If a weight Wof liquid passes the cross-section

Volume passing cross-section =W

Distance moved by liquid =W

a

Work done = force distance = p a

Wa

=Wp

pppressure energy per unit weight = = g

Similarly the pressure energy per unit weightp/Wis equivalent to a head and is

referred to as the pressure head.


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5.1.3 Kinetic energyIf a weight Wof liquid has a velocity v,

Kinetic energy =1W

v

2

2g

Kinetic energy per unit weight =v

2

2g

The kinetic energy per unit weightv

2

is also measured as a length and2g

referred to as the velocity head.

The total energy of the liquid is the sum of these three forms of energy

Total head = potential head + pressure head + velocity head

Total energy per unit weight = z +p

+v

2

2g

5.2 Definition of Bernoullis EquationBernoullis Theorem states that the total energy of each particle of a body of fluid is

the same provided that no energy enters or leaves the system at any point. The division

of this energy between potential, pressure and kinetic energy may vary, but the total

remains constant. In symbols:

H =z +p

+v

2= constant

2g

Figure 5.1


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By Bernoullis Theorem,

Total energy per unit weight at section 1 = Total energy per unit weight at section 2

p v2 p

2v

2z + 1 + 1=z2

+ +Do you know :1

2g 2g

z = potential head The Bernoulli equation is

p= pressure head

named in honour of Daniel Bernoulli (1700-1782).

2= velocity head

Many phenomena

regarding the flow of2g liquids and gases can be

H = Total head analysed by simply using

the Bernoulli equation.

5.3 The limits of Bernoullis EquationBernoullis Eqution is the most important and useful equation in fluid mechanics.

It may be written,

z +v 2

+p

=z +v

22

+p

21 1

2g 2g 1 1

Bernoullis Equation has some restrictions in its applicability, they are :

the flow is steady

the density is constant (which also means the fluid is compressible)

friction losses are negligible

the equation relates the state at two points along a single streamline (not

conditions on two different streamlines).


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5.4 Application of Bernoullis EquationBernoullis equation can be applied to the following situations.

5.4.1 Horizontal Pipe

Example 5.1

36 m

Figure 5.2

Water flows through a pipe 36 mfrom the sea level as shown in figure 5.2. Pressure in

the pipe is 410 kN/m2 and the velocity is 4.8 m/s. Calculate total energy of every

weight of unit water above the sea level.

Solution to Example 5.1

Total energy per unit weight

=z+p +v2 2g

= 36+410 10

3

+(4.8)2

10009.81 29.81= 78.96J/N


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5.4.2 Inclined Pipe

Example 5.2

M

N

5 m 5 m

3 m

Figure 5.3

A bent pipe labeled MN measures 5 mand 3 mrespectively above the datum line. The

diameter M and N are both 20 cm and 5 cm. The water pressure is 5 kg/cm2. If the

velocity at M is 1 m/s, determine the pressure at N in kg/cm2.


Using Bernoullis Equation:

z +pM +

vM =z +

pN

+

vN

(1)

2g

2g

Discharge at section M = Discharge at section N

QM= Q

vM aM= vN aN (2)

From (2),

vN =vM aM

aN

=(1)(0.2)2(0.05)2

= 16 m/s


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GivenpM =5 kg/ cm2

=5

0.00019.81

= 490.5 kN/ m

2

From (1),

= vM2vN

2 +pM + (z z )

2g

= 1 (16 )2 +490500+ (53) 98102 9.81 9810

= 382620N/m2


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ACTIVITY 5A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT

INPUT!

5.1 Define and write the Bernoullis Equation.

5.2 Water is flowing along a pipe with a velocity of 7.2 m/s. Express this as a velocityhead in meters of water. What is the corresponding pressure in kN/m2?


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FEEDBACK ON ACTIVITY 5A

5.1 Bernoullis Theorem states that the total energy of each particle of a body offluid is the same provided that no energy enters or leaves the system at any

point.

H =z +p

+v2

= constant 2g

5.2 velocity head of water =H=v

2 (7.2)2

= 2.64m=2g 2(9.81)

H =

p=2.64

p =2.64

= 2.64(9810)

= 25898.4N/m2

= 25.9kN/m2


10/43


INPUT

5.4.3 Horizontal Venturi MeterVenturi meter : It is a device used for measuring the rate of flow of a

non-viscous, incompressible fluid in non-rotational and steady-stream lined

flow. Although venturi meters can be applied to the measurement of gas, they

are most commonly used for liquids. The following treatment is limited to

incompressible fluids.

Converging Throat

EntryCone

Diverging Section

Direction of

LeadsSection2

gauge 2

Section filled with v2x

liquid in1

pipeline,1v1 Spec.wt. of

gauge

liquid= g

Figure 5.4


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Derivation for the theoretical discharge through a horizontal venture meter and

modification to obtain the actual discharge.

From Figure 5.4

Putting ;

p1

v1

A1

p2

v1

A1

1

g

g

z

Bernoullis

Equation forsection 1and 2

gives :

z +v 2

+p

=z +v

22 +p

21 1

2g 2g 1 1

Ignoring losses for horizontal meterz1=z2

v2v 2

=pp

2 (1)2 1 1

2g

For continuity of flow,A1v1=A2v2, giving

v2 =AA1

v12

whereA=

4d

2

= liquid in the gauge (specific weight,spec.wg)

= gravity (9.81 m/s2) =

height above datum

= pressure ofsection 1

= velocity ofsection 1

= area ofsection 1

= pressure ofsection 2

= velocity ofsection 1

= area ofsection 1

= liquid in pipeline (specific weight,spec.wg)


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Substituting in equation (1)

2 A2 pp

2v 1 = g 1

A 12

So, v =A p p

2 g 1 21

( A12A22)

Discharge, Qtheory=A v= A1A2 (2gH) (2)1 1 A A

1 2

Where H =p1p2

= pressure difference expressed as a head of the liquid

flowing in meter venturi.

If area ratio, m=A1 equation (2) becomes,A

2

Qtheory =A1

2gH

m21

The theoretical discharge Q can be converted to actual discharge bymultiplying by the coefficient of discharge Cdfound experimentally.

Actual discharge,

Qactual

=

Cd

Qtheory

=

Cd

A1

2gH(3)m21

If the leads of U-tube are filled with water,

p1p2 =x(g )pp

2

H =1

=x

1


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Example 5.3

A venture tube tapers from 300 mmin diameter at the entrance to 100 mmin diameter

at the throat; the discharge coefficient is 0.98. A differential mercury U-tube gauge is

connected between pressure tapping at the entrance at throat. If the meter is used to

measure the flow of water and the water fills the leads to the U-tube and is in contact

with the mercury, calculate the discharge when the difference of level in the U-tube is

55 mm.


Using Equation (3),

Qactual =cdA1 2gHm 21So,

x = 55 mmg= 13.6

H = 0.055 12.6 = 0.0706 m2

Cd = 0.98

1 =3.142(0.3)2

= 0.0706m2

4

A d2 12 2

m =1=

1= = 92

A2 d2 4

Actual discharge, Qactual= 0.980.0706 2 9.810.69381 1

Qactual=0.0285m3/s


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Example 5.4

A horizontal venturi meter measures the flow of oil of specific gravity 0.9 in a 75 mmdiameter pipe line. If the difference of pressure between the full bore and the throat

tapping is 34.5 kN/m2and the area ratio, mis 4, calculate the rate of flow, assuming a

coefficient of discharge is 0.97.


From Equation (3),

Qactual

=

cd

A1

2gH

m 21

The difference of pressure head,Hmust be expressed in terms of the liquid

following through the meter,

H =

p

= 34.51030.99.8110

3

= 3.92 m of oil

A1=3.142

(0.075

)2 =0.00441m24

m = 4

Cd= 0.97

So,

Actual discharge, Qactual= 0.970.00441 2 9.813.9216 1

Qactual=0.0106m3/s


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5.4.4 Inclined Venturi MeterDerivation is an expression for the rate of flow through an inclined

venturi meter. This will show that the U-type of gauge is used to measure the

pressure difference. The gauge reading will be the same for a given discharge

irrespective of the inclination of the meter.

In Figure 5.5, at the entrance to the meter; the area, velocity, pressure

and elevation are A1, v1, p1 and z 1 respectively and at the throat, the

corresponding values areA2, v2,p2andz2.

A1 ,

v1, p1

andz1

A2 ,

v2, p2

andz2

= spec. wt of

liquid in pipeline

Spec.wt = g

Z1( z1-y ) Z2

X

P Qy

Figure 5.5

From Bernoullis Equation,

z +v 2

+p

=z+v2

+p

21 1 2

2g 2g 1 1

2 2 p p 2v

2 v = 2g 1 + (z z ) (1)1

1 2


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For continuity of flow,

A1v1=A2v2

or

v2=A1v = mv

A21 1

where A1

m = area ratio =A

2

Substituting in equation (1) and solving for v1

2 2 p p2v2 v = 2g 1 + z 1z

1

2

v =1

2gp

+( z z1 21 21

(m21)

Actual discharge, Qactual= CdA1v1

Q =

C A

2g

p p

+( z

zactual d 1 1 2 1 2 (2)(m2 1) Where Cd= coefficient of discharge.

Considering the U-tube gauge and assuming that the connections are

filled with the liquid in the pipe line, pressures at levelPQare the same in both

limbs,

For left limb,

pz =p2+w(z1y)

For right limb,

pz =p2+(z2y x)+wgx


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Thus,

Pressure for left limb = Pressure for right limb

p2+(z1y)=p2+(z2y x)+wgx

p2+z1z2 =p2+z2y x +wgx

pp2 g1+z1z2

=x

1

Equation (2) can therefore be written

CdA g

actual1

(m2gx

12

1


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Example 5.5

A vertical venturi meter measures the flow of oil of specific gravity 0.82 and has anentrance of 125 mmdiameter and throat of 50 mmdiameter. There are pressure gaugesat the entrance and at the throat, which is 300 mmabove the entrance. If the coefficient

for the meter is 0.97 and pressure difference is 27.5 kN/m2, calculate the actual

discharge in m3/s.


21

z1 z2

In equation (2),

Q =C A

2gp

+ (z zactual d 1 1 2 1 2

(m2 1)

This is independent ofz1andz2, so that the gauge readingxfor a given rate of

flow, Qactualdoes not depend on the inclination of themeter. Then,

Q =C A

2gp

+ (z zactual d 1 1 2 1 2

(m2 1)


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So,

A1=3.142(0.125)2

=0.01226 m2

4

p p2= 27.510

3 kN/m

2

=0.829.81103N /m2z1z2 = 0.3m

m =d2

=125

= 6.251

22 50

Cd= 0.97

Therefore,

Q =C A p p

+ (z zactual d 1 g 1 2 1 2

(m21)

0.97 0.01226 27.5 10 3 3actual =

((6.25)21)9.81

0.82 9.8110

30.3 =0.01535m /s


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Example 5.6

The water supply to a gas water heater contracts from 10mmin diameter at A (Figure

5.6) to 7 mm in diameter at B. If the pipe is horizontal, calculate the difference in

pressure between A and B when the velocity of water at A is 4.5 m/s.

The pressure difference operates the gas control through connections which is taken to

a horizontal cylinder in which a piston of 20 mm diameter moves. Ignoring friction

and the area of the piston connecting rod, what is the force on the piston?

d1 d2

p1 ,v

1 p

2 v

2

A B

Figure 5.6


In the Figure 5.6 the diameter, pressure and velocity at Aare d1,p1and v1; and at

B are d2,p2and v2.

By Bernoullis theorem, for horizontal pipe,

v12 +p1 =v2

2 +p2

2g 2g

This equation can therefore be written,

p1p2 =v22v1

2

2g


22/43


For continuity of flow,

or

A1v1=A2v2

d2 d 21 v = 2 v2

4 4

then,d

2

2vv= d

2So,

1 1 2

2v2

d1

=

v1

d2

Putting v1=4.5m/s, d1=10mm, d2=7mm

v2=4.510

2

7

= 9.18m/s

and

pp2=

v 2v2

1 2 1

2g

pp2

9.1824.5

2

1= =3.26m

2 2 9.81

p1p2 =3.26m2

p p2

=3.26m2

1

Pressure difference,p p2=3.269.8110

3N/m

2

1= 31.9kN/m

3


23/43


Area of piston =

4d

2kN/ m

3

=(0.020)2=0.000314m24

We all know that,

Force,F = pA

Where,

p = pressure andA = area

So,

Force on piston = 31.91030.000314=10.1N


24/43


ACTIVITY 5B


INPUT!

5.3 To get through the Green Alien you should be able to answer his puzzles !

1. What does a Venturi Meter measure ?

2. Name me two types of Venturi Meter that you

have learnt in this unit.

3. Sketch a Horizontal Venturi Meter for me. (Label the

thoat, entry, diverging section and converging cone)

4. What is denoted by and g?

If you get all the answers right, you will be sent to earth

immediately on the next space shuttle. Only smart people

can go and stay on the earth!


25/43


FEEDBACK ON ACTIVITY 5B

I think I got it all right !

5.3

1. What does a Venturi Meter measure ?

It is a device used for measuring the rate of flow of a non-viscous, incompressible

fluid in non-rotational and steady-stream lined flow.

2. Name me two types of Venturi Meter that you have learnt in this unit.Horizontal Venturi Meter and Inclined Venturi Meter.

3. Sketch a Horizontal Venturi Meter for me. (Label the throat, entry, diverging section

and converging cone)

DivergingConverging

sectioncone

throat

entry

4. What is denoted by and g?

denotes the specific weight of lead gauge filled with liquid in pipeline andgdenotes the specific weight of gauge liquid.

Just Kidding !

You are already on earth, your answers are correct,

Just sit there and continue your studies.


26/43


INPUT

5.4.5 Small OrificeThe Venturi Meter described earlier is a reliable flow measuring

device. Furthermore, it causes little pressure loss. For these reasons it is widely

used, particularly for large-volume liquid and gas flows. However this meter is

relatively complex to construct and hence expensive especially for small

pipelines. The cost of the Venturi Meter seems prohibitive, so simpler device

such as Orifice Meter is used.

The Orifice Meter consists of a flat orifice plate with a circular hole

drilled in it. There is a pressure tap upstream from the orifice plate and another

just downstream. There are three recognized methods of placing the taps and

the coefficient of the meter will depend upon the position of the taps.

The principle of the orifice meter is identical with that of the venturi

meter. The reduction at the cross section of the flowing stream in passing

through the orifice increases the velocity head at the expense of the pressure

head, and the reduction in pressure between the taps is measured by a

manometer. Bernoulli's equation provides a basis for correlating the increase in

velocity head with the decrease in pressure head.

From Figure 5.7 the orifice meter is attached to the manometer. There

are Section 1 (entrance of the orifice) and Section 2 (exit of the orifice also

known as vena contracta).

Section 1 :

A1, v1, p1Section 2 :

A2, v2, p2

X

Figure 5.7


27/43


Section 1, given :

1 = area ofsection 1

v1 = velocity ofsection 1

1 = pressure ofsection 1

Section 2, given :

2 = area ofsection 2

v2 = velocity ofsection 2

2 = pressure ofsection 2

From Bernoullis Equation,

Total energy at section 1 = Total energy at section 2

p+

v2

=2+

v22

(1)1 1

2g 2g

v2

v 2 =pp2 (2)2 1 1

2g

z1=z2because the two parts are at the same

levelWe know that,

Q =A v

For continuity of flow, Q1=Q2

or

A1v1=A2v2

So,

A1v1v2 = (3)A

2


28/43


Putting (3) into (2),

v2v 2

=p

22 1 1

2g

v2 =A1v1A

Then,2

v2 A 2 p p

21 1 1= 12g A 2

2

So,

p p2

2g 1

v1 =2

1 1A 2

2

But,

p1p2

H =

And,

m =

A12

A22

So,

v1=

2gH

(m21)

(2)

(3)

To determine the actual discharge, Qactual;

Qactual

=

Cd

A1

v1

So,

= 2gHQ

actual

Cd A1(m21)

Where Cd= coefficient of discharge.


29/43

Example 5.7


30/43


A meter orifice has a 100 mmdiameter rectangular hole in the pipe. Diameter of the

pipe is 250 mm. Coefficient of discharge, Cd= 0.65 and specific gravity of oil in the

pipe is 0.9. The pressure difference that is measured by the manometer is 750 mm.Calculate the flow rate of the oil through the pipe.


Given,

d1 = 100 mm = 0.10 m

d2 = 250 mm = 0.25

Cd = 0.65

oil = 0.9

p1 -p2 = 750 mm = 0.75 m

So,

A =d21 4

=3.124(0.25)2

= 0.049m2

4

H =p p

2=xHg

1 1

oil oil13.6

= 0.75 10.9

=10.58m

m=d12=(0 .25)

2

d22(0.10)2

=

6.25Therefore,

2gH

Qactual

=

Cd

A1 m 2 1)

Qactual= 0.650.049

2 9.8110.58

(6.252)1= 0.074m

3/s

5.4.5.1 Types of orifice


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1. Sharp-edged orifice, Cd= 0.62

2. Rounded orifice, Cd= 0.97

3. Borda Orifice (running free), Cd= 0.50

4. Borda Orifice (running full), Cd= 0.75

5.4.5.2 Coefficient of Velocity, Cv


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hx

A y

B

Figure 5.8

From Figure 5.8 ,

= horizontal falls= velocitytime= vt

= vertical falls= 1gravity time2 = 1g t

2 2h = head of liquid above the orifice

Cv = Coefficient of Velocity = Cv = v

2gH

t = time for particle to travel from vena contracta A to point B

Coefficient of Velocity, Cv=Actual velocity at vena contacta

Theoretical velocity

Cv

=2

vgH

Example 5.8


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A tank 1.8 mhigh, standing on the ground, is kept full of water. There is an orifice in

its vertical site at depth, hm below the surface. Find the value of hin order the jet may

strike the ground at a maximum distance from the tank.


x =v t

and

y =12gt

Eliminating tthese equation give,

x=2v2

yg

y = 1.8hh = head of liquid above the orifice

Cv=

2vgH

t = time for particle to travel from vena contracta A to point B

Puttingy=1.8 hand v=Cv (2gh)

So,

= 2[Cv (2gh)]2

(1 .8 h)g

=C

24gh(1.8 h)

vg

= 2Cv[h(1.8 h)]

Thusxwill be a maximum when h(1.8h)is maximum or,[h(1.8h)]

=1.82h

=0hSo,

h =0.9m

Example 5.9


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An orifice meter consists of a 100 mm diameter in a 250 mmdiameter pipe (Figure

5.9), and has a coefficient discharge of 0.65. The pipe conveys oil of specific gravity

0.9. The pressure difference between the two sides of the orifice plate is measured by a

mercury manometer, that leads to the gauge being filled with oil. If the difference in

mercury levels in the gauge is 760 mm, calculate the flowrate of oil in the pipeline.

Pipe Area, A1

P1 P2

V1 V2X

Orifice area A2 C C

Figure 5.9


Let v1be the velocity and p1the pressure immediately upstream of the

orifice, and v2 and p2 are the corresponding values in the orifice. Then,

ignoring losses, by Bernoullis theorem,

p+

v2

=2+

v22

(1)1 1

2g 2g

v2 v 2=

p p2

(2)2 1 1

2g

z1=z2because of the two parts are at the same level

We know that,

Q =A v

For continuity of flow, Q1= Q2


35/43


or

A1v1=A2v2

So,

v2 = A1v1 (3)A2

Putting (3) into (2),

v2 v

2=

p p2 (2)2 1 1

2g

v2 =A1v1

(3)

Then,2

v 2A p p 21 1 1= 12gA2

2

So,

p 2

2g 1

v1 = A2

1 1A

2

This equation can therefore be written,

v =2 p p

2 (4)

( A12a22)g 1

1

So,

Actual disch arge =coefficient of disch arge theoretical disch arge

Qactual =Cd A1v1 (5)

Putting v1into (5)


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Qactual =Cd A1A

2 pp

(6)

( A12a22)2g 1

2

but,

m =

A1A2

so putting m into (6),

Q = CdA C A p p

d 1 g 1 2

actual 1 (m21) Considering the U-tube gauge, where pressures are equal at level CC

p1+x =p2+qxpp

2 pp 21 =x 1

Puttingx=760 mm=0.76 mand,

g

=13

0..96

=15.1

p1

p

2 = 0.7614.1=10.72m of oil

Cd = 0.65

A = d2 = 0.0497m 21 4

m =A1

=d

2

=(0.25)2

=15.11

A2d

22 (0.10)

2

m2 = 6.17

Qactual

=0.65 0.0497 (29.8110.72)

6.17

= 0.0052414.5= 0.0762m3/s


37/43


5.4.6 Simple Pitot Tube

b

a

Figure 5.10 Pitot Tube

- The Pitot Tube is a device used to measure the local velocity

along a streamline (Figure 5.10). The pitot tube has two

tubes: one is a static tube (b), and another is an impact

tube(a).

- The opening of the impact tube is perpendicular to the

flow direction. The opening of the static tube is parallel to

the direction of flow.

- The two legs are connected to the legs of a manometer or an

equivalent device for measuring small pressure differences.

The static tube measures the static pressure, since there is no

velocity component perpendicular to its opening.

- The impact tube measures both the static pressure

and impact pressure (due to kinetic energy).

- In terms of heads, the impact tube measures the static

pressure head plus the velocity head.


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h

HA B

Figure 5.11 Simple Pitot Tube

Actual Velocity, V

From Figure 5.11, if the velocity of the stream atAis v, a particle moving

fromAto the mouth of the tubeBwill be brought to rest so that v0atB

is zero.

By Bernoullis Theorem : Total Energy atA= Total Energy atBor

p+

v 2=p

2 +v2

(1)1 1 2

2g 2g

Now d=

pand the increased pressure at B will cause the liquid in the

vertical limb of the pitot tube to rise to a height, h above the free surface so

that h+d=p

0.

Thus, the equation (1) v2 =p0p = h or v= 2gh

2g

Although theoretically v= (2gh) , pitot tubes may require calibration.

The actual velocity is then given by v=C (2gh) whereCisthecoefficient of the instrument.


39/43


Example 5.10

A Pitot Tube is used to measure air velocity in a pipe attached to a mercury

manometer. Head difference of that manometer is 6 mmwater. The weight density of

air is 1.25 kg/m

3

. Calculate the air velocity if coefficient of the pitot tube, C= 0.94.Solution to Example 5.10

vair =C 2gH

pwater

=

pair

ghwater

=

ghair

hwater

water

=

hair

air

hwater=0.006

water=0.006

1000

air1.25= 4.8m

So,

v= 0.94 2 9.814.8= 9.12 m/s


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ACTIVITY 5C


INPUT!

5.4 Fill in the blanks in the following statements.1. The Orifice Meter consists of a flat orifice plate with a circular hole drilled in it.

There is a _____________upstream from the orifice plate and another just

downstream.

2. The reduction of pressure in the cross section of the flowing stream when passing

through the orifice increases the __________________at the expense of the

pressure head. The reduction in pressure between the taps is measured by a

manometer.

3. The formula for Meter Orifice actual discharge, Qactual.=_______________

4. The Pitot Tube is a device used to measure the local velocity along a streamline.

The pitot tube has two tubes which are the_______________and the____________.

5. Although theoretically v=(2gh), pitot tubes may require______________.

6. The actual velocity is given by __________ where C is the coefficient of the

instrument.


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FEEDBACK ON ACTIVITY 5C

5.4

1. The Orifice Meter consists of a flat orifice plate with a circular hole drilled in it. There

is a pressure tapupstream from the orifice plate and another just downstream.

2. The reduction pressure in the cross section of the flowing stream when passing

through the orifice increases the velocity head at the expense of the pressure head.

The reduction in pressure between the taps is measured by a manometer.

3. The formula for Meter Orifice actual discharge, Qactual.=

2gH

Qactual =CdA1v1 and

Qactual

=

Cd

A1 m21

4. The Pitot Tube is a device used to measure the local velocity along a streamline. The

pitot tube has two tubes which are the static tubeand the impact tube.

5. Although theoretically v=(2gh), pitot tubes may require calibration.

6. The actual velocity is given by v=C (2gh)where Cis the coefficient of the

instrument.


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SELF-ASSESSMENT

You are approaching success. Try all the questions in this self-assessment section

and check your answers with those given in the Feedback on Self-Assessment. If you

face any problems, discuss it with your lecturer. Good luck.

5.1 A venturi meter measures the flow of water in a 75 mm diameter pipe. Thedifference between the throat and the entrance of the meter is measured by the

U-tube containing mercury which is being in contact with the water. What

should be the diameter of the throat of the meter in order that the difference in

the level of mercury is 250 mmwhen the quantity of water flowing in the pipe

is 620 dm3/min? Assume coefficient of discharge is 0.97.

5.2 A pitot-static tube placed in the centre of a 200 pipe line conveying water hasone orifice pointing upstream and the other perpendicular to it. If the pressuredifference between the two orifices is 38 mm of water when the discharge

through the pipe is 22 dm3/s, calculate the meter coefficient. Take the mean

velocity in the pipe to be 0.83 of the central velocity.

5.3 A sharp-edged orifice, of 50 mmdiameter, in the vertical side of a large tank,discharges under a head of 4.8 m. If Cc= 0.62 and Cv= 0.98, determine;

(a)the diameter of the jet,(b)the velocity of the jet at the vena contracta,

(c)the discharge in dm3/s.


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FEEDBACK ON SELF-ASSESSMENT

Answers :

5.1 40.7 mm5.2 0.9775.3 (a) 40.3 mm

(b)9.5 m/s

(c)12.15 dm3/s

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Fluid Mechanics Dynamics