Unit 5-Fluid Mechanics

FLUID DYNAMICS J3008/5/1

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 Bernoulli’s 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.

UNIT 5


5.1 Energy of a flowing fluid

A liquid may possess three forms of energy:

5.1.1 Potential energy If a liquid of weight W is at a height of z above datum line

Potential energy = Wz Potential energy per unit weight = z The potential energy per unit weight has dimensions of Nm/N and is measured as a length or head z and can be called the potential head.

5.1.2 Pressure energy

When 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 pressure p on cross-section is pa.

If a weight W of liquid passes the cross-section

Volume passing cross-section =ωW

Distance moved by liquid = a

Wω

Work done = force × distance =a

Wapω

×

= ωpW

pressure energy per unit weight =ωp =

gpρ

Similarly the pressure energy per unit weight p/W is equivalent to a head and is referred to as the pressure head.

INPUT


5.1.3 Kinetic energy

If a weight W of liquid has a velocity v,

Kinetic energy = 2

21 v

gW

Kinetic energy per unit weight =g

v2

2

The kinetic energy per unit weight g

v2

2

is also measured as a length and

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 = g

vpz2

2

++ω

5.2 Definition of Bernoulli’s Equation

Bernoulli’s 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:

tconsg

vpzH tan2

2

=++=ω

Figure 5.1


By Bernoulli’s Theorem, Total energy per unit weight at section 1 = Total energy per unit weight at section 2

g

vpz

gvp

z22

222

211

1 ++=++ωω

z = potential head

ωp = pressure head

g

v2

2 = velocity head

H = Total head 5.3 The limits of Bernoulli’s Equation

Bernoulli’s Eqution is the most important and useful equation in fluid mechanics. It may be written,

ωω2

22

11

21

1 22p

gv

zp

gv

z ++=++

Bernoulli’s 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).

Do you know :

The Bernoulli equation is named in honour of Daniel

Bernoulli (1700-1782). Many phenomena

regarding the flow of liquids and gases can be analysed by simply using the Bernoulli equation.


5.4 Application of Bernoulli’s Equation Bernoulli’s equation can be applied to the following situations. 5.4.1 Horizontal Pipe Example 5.1

Water flows through a pipe 36 m from 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

gvpz2

2

++=ω

( )81.92

8.481.91000

104103623

×+

××

+=

NJ /96.78=

36 m

Figure 5.2


5.4.2 Inclined Pipe Example 5.2

A bent pipe labeled MN measures 5 m and 3 m respectively 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.

Solution to Example 5.2 Using Bernoulli’s Equation:

g

vpz

gvpz NN

NMM

M 22++=++

ωω …(1)

Discharge at section M = Discharge at section N

NM QQ = NNMM avav ×=× …(2) From (2),

N

MMN a

avv ×=

( ) ( )( )2

2

05.02.01=

sm /16=

5 m 5 m

3 m

Figure 5.3

N

M


Given 2/5 cmkgpM =

0001.0

81.95×=

2/5.490 mkN=

From (1),

( ) ωω

×⎥⎥⎦

⎤

⎢⎢⎣

⎡−++

−= NM

MNMN zzp

gvv

p2

22

( ) ( ) 9810359810

49050081.92

2161×⎥⎦

⎤⎢⎣⎡ −++

×−

=

2/382620 mN=


ACTIVITY 5A TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

5.1 Define and write the Bernoulli’s Equation.

5.2 Water is flowing along a pipe with a velocity of 7.2 m/s. Express this as a velocity

head in meters of water. What is the corresponding pressure in kN/m2?


FEEDBACK ON ACTIVITY 5A

5.1 Bernoulli’s 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.

tconsg

vpzH tan2

2

=++=ω

5.2 velocity head of water = ( )( ) m

gvH 64.2

81.922.7

2

22

===

64.2==ωpH

ω64.2=p ( )981064.2= 2/4.25898 mN= 2/9.25 mkN=


5.4.3 Horizontal Venturi Meter

Venturi 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.

Figure 5.4

INPUT

Entry

Converging Cone

Throat Diverging Section

Section 1

ρ1 v1

Section 2

ρ2 v2

Direction of

Leads gauge filled with liquid in pipeline,

x

Spec.wt. of gauge liquid= ωg


The arrangement and mode of action of a Venturi Meter ( Figure 5.4 )

The venture meter consists of a short converging conical tube leading to a cylindrical portion called “throat” which is followed by a diverging section.

The entrance and exit diameter is the same as that of the pipe line into which it is inserted. The angle of the convergent cone is usually 21o, the length of throat is equal to the throat diameter, and the angle of the divergent cone is 5o to 7o to ensure the minimum loss of energy.

Pressure tappings are taken at the entrance and at the throat, either from the single holes or from a number of holes around the circumference connecting to an annular chamber or Piezometer ring, and the pressure difference is measured by a suitable gauge.

For continuity of flow velocity v1 at the entry (section 1) will be less than the velocity v2 at the throat (section 2) since a1v1 = a2v2 and a1 is greater than a2. The kinetic energy in the throat will be greater than at the entrance and since by Bernoulli’s theorem the total energy at the two sections is the same, the pressure energy at the throat will be less than that at the entrance. The pressure difference thus created is dependent on the rate of flow through the meter.

Description: The Venturi tube consists of a plain tube with a smooth constriction in its bore at the middle. It is found in carburetors, fluid flow meters, and aircraft airspeed indicators. This version is made of glass and has three side tubes for attaching a manometer. When high pressure air is applied from the air tank, the water level in the three manometer legs is clearly different. Adapted from : http://demoroom.physics.ncsu.edu:8770/html/demos/353.html


Derivation for the theoretical discharge through a horizontal venture meter and modification to obtain the actual discharge.

From Figure 5.4 Putting ;

Bernoulli’s Equation for section 1 and 2 gives :

ωω2

22

11

21

1 22p

gv

zp

gv

z ++=++

Ignoring losses for horizontal meter z1 = z2

ω

212

12

2

2pp

gvv −

=−

——————(1)

For continuity of flow, 2211 vAvA = , giving

12

12 v

AAv =

where 4

2dA π=

p1 = pressure of section 1

v1 = velocity of section 1

A1 = area of section 1




ω1 = liquid in pipeline (specific weight, spec.wg)

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

g = gravity (9.81 m/s2)

z = height above datum


Substituting in equation (1)

⎟⎠⎞

⎜⎝⎛ −

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

ω21

22

212

1 21 ppgAAv

So, ( ) ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

−=

ω21

22

21

21 2 ppg

AA

Av

Discharge, Qtheory = ( )gHAA

AAvA 22

22

1

2111

−= ——————(2)

Where H = ω

21 pp − = pressure difference expressed as a head of the liquid

flowing in meter venturi.

If area ratio, 2

1

AA

m = equation (2) becomes,

12

21 −=

mgHAQtheory

The theoretical discharge Q can be converted to actual discharge by multiplying by the coefficient of discharge Cd found experimentally.

Actual discharge, 1

221 −

=×=m

gHACQCQ dtheorydactual —————(3)

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

( )ωω −=− gxpp 21

H = ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−121

ωω

ωgx

pp


Example 5.3

A venture tube tapers from 300 mm in diameter at the entrance to 100 mm in 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),

1

221 −

=m

gHAcQ dactual

So,

Actual discharge, Qactual = 181

693.081.920706.098.0−××

×

Qactual = sm /0285.0 3

x = 55 mm

ωω g

= 13.6

H = 0.055 ×12.6 = 0.0706 m2

Cd = 0.98

A1

( ) 22

0706.04

3.0142.3 m==

m 9

412 2

22

21

2

1 =⎟⎠⎞

⎜⎝⎛===

dd

AA


Example 5.4

A horizontal venturi meter measures the flow of oil of specific gravity 0.9 in a 75 mm diameter pipe line. If the difference of pressure between the full bore and the throat tapping is 34.5 kN/m2 and the area ratio, m is 4, calculate the rate of flow, assuming a coefficient of discharge is 0.97. Solution to Example 5.4

From Equation (3),

12

21 −=

mgHAcQ dactual

The difference of pressure head, H must be expressed in terms of the liquid following through the meter,

H = ωp

= 3

3

1081.99.0105.34××

×

= oilofm92.3

A1 ( ) 2

2

00441.04

075.0142.3 m==

m = 4

Cd = 0.97

So,

Actual discharge, Qactual = 116

92.381.9200441.097.0−××

×

Qactual = sm /0106.0 3


5.4.4 Inclined Venturi Meter

Derivation 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 z1 respectively and at the throat, the corresponding values are A2, v2, p2 and z2.

From Bernoulli’s Equation,

ωω

22

21

12

11 22

pg

vz

pg

vz ++=++

( )⎭⎬⎫

⎩⎨⎧

−+⎟⎠⎞

⎜⎝⎛ −

=− 21212

12

2 2 zzpp

gvvω

——————(1)

Figure 5.5

Z2 Z1

X

y

( z1-y )

P Q

A1, v1, p1 and z1

A2, v2, p2 and z2

ω = spec. wt of liquid in pipeline

Spec.wt = ωg


For continuity of flow,

2211 vAvA = or

112

12 mvv

AA

v ==

where

m = area ratio =2

1

AA

Substituting in equation (1) and solving for v1

( )⎭⎬⎫

⎩⎨⎧

−+⎟⎠⎞

⎜⎝⎛ −

=− 21212

12

2 2 zzpp

gvvω

( )( ) ⎥

⎦

⎤⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

−+⎟⎠⎞

⎜⎝⎛ −

−= 21

21

21 21

1 zzpp

gm

vω

Actual discharge, Qactual = 11 vACd ××

( )( ) ⎥

⎦

⎤⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

−+⎟⎠

⎞⎜⎝

⎛ −

−

×= 21

21

2

1 21

zzpp

gm

ACQ d

actual ω ——— (2)

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 level PQ are the same in both limbs,

For left limb, ( )yzwppz −+= 12 For right limb, xwxyzpp gz +−−+= )( 22 ω


Thus, Pressure for left limb = Pressure for right limb ( ) =−+ yzp 12 ω xwxyzp g+−−+ )( 22 ω =−+ 212 zzp ωω xwxyzp g+−−+ ωωω 22

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−+

−121

21

ωω

ωgxzz

pp

Equation (2) can therefore be written

( ) ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

−= 12

12

1

ωω gd

actual gxm

ACQ


Example 5.5

A vertical venturi meter measures the flow of oil of specific gravity 0.82 and has an entrance of 125 mm diameter and throat of 50 mm diameter. There are pressure gauges at the entrance and at the throat, which is 300 mm above 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. Solution to Example 5.5

In equation (2),

( )

( ) ⎥⎦

⎤⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

−+⎟⎠

⎞⎜⎝

⎛ −

−

×= 21

21

2

1 21

zzpp

gm

ACQ d

actual ω

This is independent of z1 and z2, so that the gauge reading x for a given rate of flow, Qactual does not depend on the inclination of the meter.

Then,

( )

( ) ⎥⎦

⎤⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

−+⎟⎠

⎞⎜⎝

⎛ −

−

×= 21

21

2

1 21

zzpp

gm

ACQ d

actual ω

1

2

z2 z1


So,

A1( ) 2

2

01226.04

125.0142.3 m==

23

21 /105.27 mkNpp ×=− 23 /1081.982.0 mN××=ω mzz 3.021 −=−

m = 25.650

125 2

22

21 =⎟

⎠⎞

⎜⎝⎛=

dd

Cd = 0.97 Therefore,

( )

( ) ⎥⎦

⎤⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

−+⎟⎠

⎞⎜⎝

⎛ −

−

×= 21

21

2

1 21

zzpp

gm

ACQ d

actual ω

( )( ) ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

×××

×−

×= 3.0

1081.982.0105.2781.92

125.6

01226.097.03

3

2actualQ = sm /01535.0 3


Example 5.6

The water supply to a gas water heater contracts from 10mm in 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?


In the Figure 5.6 the diameter, pressure and velocity at A are d1, p1 and v1 ; and at B are d2, p2 and v2.

By Bernoulli’s theorem, for horizontal pipe,

ωω2

221

21

22p

gvp

gv

+=+

This equation can therefore be written,

gvvpp

2

21

2221 −

=−ω

d1 d2

A B

p1 ,v1 p2 v2

Figure 5.6


For continuity of flow,

2211 vAvA = or

2

22

1

21

44v

dv

d⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛ ππ

then,

22

212

1 vdvd ×=× So,

2

2

112 ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

ddvv

Putting smv /5.41 = , mmd 101= , mmd 72=

2

2 7105.4 ⎟

⎠⎞

⎜⎝⎛=v

sm /18.9=

and

gvvpp

2

21

2221 −

=−ω

81.925.418.9 22

21

×−

=−ω

pp 226.3 m=

ω

21 pp − 226.3 m=

21 pp − ω×= 226.3 m

Pressure difference, 2321 /1081.926.3 mNpp ××=−

3/9.31 mkN=


Area of piston = 32

/4

mkNdπ

( ) 22

000314.04020.0 m==

π

We all know that,

Force, F = p A× Where, p = pressure and A = area

So,

Force on piston = N1.10000314.0109.31 3 =××


ACTIVITY 5B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT 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 ω and ωg ?

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!


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)

4. What is denoted by ω and ωg ? ω denotes the specific weight of lead gauge filled with liquid in pipeline and ωg denotes the specific weight of gauge liquid.

Just Kidding ! You are already on earth, your answers are correct, Just sit there and continue your studies.

Converging cone

Diverging section

throat

entry


5.4.5 Small Orifice

The 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).

INPUT

X

Section 1 : A1, v1, p1

Section 2 : A2, v2, p2

Figure 5.7


Section 1, given :

Section 2, given :

From Bernoulli’s Equation,

Total energy at section 1 = Total energy at section 2

gvp

gvp

22

222

211 +=+

ωω ——————(1)

ω21

21

22

2pp

gvv −

=−

——————(2)

z1 = z2 because the two parts are at the same level

We know that,

vAQ ×=

For continuity of flow, Q1 = Q2

or A1v1 = A2v2

So,

v2 = 2

11

AvA ——————(3)








Putting (3) into (2),

ω21

21

22

2pp

gvv −

=−

——————(2)

v2 = 2

11

AvA ——————(3)

Then,

ω

212

2

21

21 1

2pp

AA

gv −

=⎥⎦

⎤⎢⎣

⎡−

So,

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎠⎞

⎜⎝⎛ −

=

1

2

22

21

21

1

AA

ppgv ω

But,

H = ω

21 pp −

And,

m = 22

21

AA

So,

( )12

21 −=

mgHv

To determine the actual discharge, Qactual ; 11 vACQ dactual ××=

So,

( )12

21 −×=

mgHACQ dactual

Where Cd = coefficient of discharge.

Example 5.7


A meter orifice has a 100 mm diameter 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 ωoil = 0.9 p1 - p2 = 750 mm = 0.75 m

So,

4

2

1dA π

=

( ) 22

049.04

25.0124.3 m==

⎥⎦

⎤⎢⎣

⎡−=

−= 121

oil

Hg

oil

xpp

Hωω

ω

⎥⎦⎤

⎢⎣⎡ −= 1

9.06.1375.0

m58.10=

m 22

21

dd

= = ( )( )2

2

10.025.0

25.6=

Therefore,

Qactual = ( )12

21 −×

mgHACd

Qactual = ( ) 125.658.1081.92049.065.0 2 −

×××

sm /074.0 3=

5.4.5.1 Types of orifice


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


Figure 5.8

From Figure 5.8 ,

x = horizontal falls = velocity × time = tv× y

= vertical falls = 2

21 timegravity × = tg ×

21

h = head of liquid above the orifice Cv = Coefficient of Velocity =

gHvCv 2

=

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

Coefficient of Velocity, Cv = velocitylTheoretica

contactavenaatvelocityActual

gHvCv 2

=

Example 5.8

hx

y AB


A tank 1.8 m high, standing on the ground, is kept full of water. There is an orifice in its vertical site at depth, h m below the surface. Find the value of h in order the jet may strike the ground at a maximum distance from the tank.


x tv×=

and

y = tg ×21

Eliminating t these equation give,

g

yvx22

=

y = 1.8 – h h = head of liquid above the orifice Cv gH

v2

=

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

Putting hy −= 8.1 and ( )ghCv v 2=

So,

( )[ ] ( )

ghghC

x v −×=

8.1222

( )g

hghCx v −=

8.142

( )[ ]hhCv −= 8.12

Thus x will be a maximum when ( )hh −8.1 is maximum or, ( )[ ] 028.18.1

=−=− h

hhh

So, mh 9.0=

Example 5.9


An orifice meter consists of a 100 mm diameter in a 250 mm diameter 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.

Figure 5.9


Let v1 be the velocity and p1 the pressure immediately upstream of the orifice, and v2 and p2 are the corresponding values in the orifice. Then, ignoring losses, by Bernoulli’s theorem,

gvp

gvp

22

222

211 +=+

ωω ——————(1)

ω21

21

22

2pp

gvv −

=−

——————(2)

z1 = z2 because of the two parts are at the same level

We know that,

vAQ ×= For continuity of flow, Q1 = Q2

Pipe Area, A1

P1 P2 V1

V2

X Orifice area A2 C C


or A1v1 = A2v2

So,

v2 = 2

11

AvA ——————(3)

Putting (3) into (2),

ω21

21

22

2pp

gvv −

=−

——————(2)

v2 = 2

11

AvA ——————(3)

Then,

ω21

22

21

21 1

2pp

AA

gv −

=⎥⎦

⎤⎢⎣

⎡−

So,

⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎟⎠⎞

⎜⎝⎛ −

=

1

2

22

21

21

1

AA

ppg

v ω

This equation can therefore be written,

( ) ⎟⎠⎞

⎜⎝⎛ −

−=

ω21

22

21

2

1 2 ppgaA

Av ——————(4)

So,

edischltheoreticaedischoftcoefficienedischActual argargarg ×=

11 vACdQactual ××= ——————(5)

Putting v1 into (5)


××= 1ACdQactual ( ) ⎟⎠⎞

⎜⎝⎛ −

− ω21

22

21

2

2pp

gaA

A ——————(6)

but,

m = 22

21

AA

so putting m into (6),

( )⎟⎠

⎞⎜⎝

⎛ −

−××=

ω21

2

11 2

1

ppg

m

ACACdQ d

actual

Considering the U-tube gauge, where pressures are equal at level CC

xpxp qωω +=+ 21

⎟⎠⎞

⎜⎝⎛ −

=−

ωω2121 ppxpp

Putting mmmx 76.0760 == and,

1.159.06.13==

ωω g

oilofmpp 72.101.1476.021 =×=−ω

65.0=dC

22

1 0497.04

mdA ==π

( )( )

1.1510.025.0

21

2

2

22

21 ====

dd

AAm

17.62 =m

( )72.1081.9217.6

0497.065.0××

×=actualQ

sm /0762.05.1400524.0 3=×=


5.4.6 Simple Pitot Tube

Figure 5.10 Pitot Tube

a

b

- 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.


Figure 5.11 Simple Pitot Tube

Actual Velocity, V From Figure 5.11, if the velocity of the stream at A is v, a particle moving

from A to the mouth of the tube B will be brought to rest so that v0 at B is zero. By Bernoulli’s Theorem : Total Energy at A = Total Energy at B or

gvp

gvp

22

222

211 +=+

ωω ——————(1)

Now ωpd = and 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 ω

0pdh =+ .

Thus, the equation (1) ghvorhppg

v 202

2==

−=

ω

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

The actual velocity is then given by ( )ghCv 2= where C is the

coefficient of the instrument.

A B

h

H


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 mm water. The weight density of air is 1.25 kg/m3. Calculate the air velocity if coefficient of the pitot tube, C = 0.94.


gHCvair 2= airwater pp = airwater ghgh ρρ = airairwaterwater hh ωω ×=×

25.1

1000006.0006.0 ×=×=air

waterwaterh

ωω

m8.4= So,

8.481.9294.0 ××=v sm /12.9=


ACTIVITY 5C

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT 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 ( )ghv 2= , pitot tubes may require______________.

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

instrument.


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 tap upstream 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. =

11 vACQ dactual ××= and Qactual = ( )12

21 −×

mgHACd

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 tube and the impact tube.

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

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

instrument.


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. The

difference 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 mm when 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 has one orifice pointing upstream and the other perpendicular to it. If the pressure difference 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 mm diameter, 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.


FEEDBACK ON SELF-ASSESSMENT

Answers :

5.1 40.7 mm

5.2 0.977

5.3 (a) 40.3 mm

(b) 9.5 m/s

(c) 12.15 dm3/s

Unit 5-Fluid Mechanics

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