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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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FLUID DYNAMICS J3008/5/2
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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FLUID DYNAMICS J3008/5/3
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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FLUID DYNAMICS J3008/5/4
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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FLUID DYNAMICS J3008/5/5
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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FLUID DYNAMICS J3008/5/6
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.
Solution to Example 5.2
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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FLUID DYNAMICS J3008/5/7
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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FLUID DYNAMICS J3008/5/8
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
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FLUID DYNAMICS J3008/5/10
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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FLUID DYNAMICS J3008/5/12
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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FLUID DYNAMICS J3008/5/13
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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FLUID DYNAMICS J3008/5/14
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.
Solution to Example 5.3
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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FLUID DYNAMICS J3008/5/15
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.
Solution to Example 5.4
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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FLUID DYNAMICS J3008/5/16
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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FLUID DYNAMICS J3008/5/17
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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FLUID DYNAMICS J3008/5/18
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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FLUID DYNAMICS J3008/5/19
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.
Solution to Example 5.5
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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FLUID DYNAMICS J3008/5/20
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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FLUID DYNAMICS J3008/5/21
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
Solution to Example 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
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FLUID DYNAMICS J3008/5/22
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
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FLUID DYNAMICS J3008/5/23
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
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FLUID DYNAMICS J3008/5/24
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 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!
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FLUID DYNAMICS J3008/5/25
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.
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FLUID DYNAMICS J3008/5/26
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
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FLUID DYNAMICS J3008/5/27
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
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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.
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Example 5.7
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FLUID DYNAMICS J3008/5/29
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.
Solution to Example 5.7
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.
Solution to Example 5.8
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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FLUID DYNAMICS J3008/5/33
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
Solution to Example 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
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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
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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.
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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
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 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