FLUID DYNAMICS
J3008/5/1
BENDALIR DINAMIK (FLUID DYNAMICS)
5.4.4
Meter Venturi Condong (Inclined Venturi Meter)
 = spec. wt of
liquid in pipeline
A2, v2,
p2 and
z2
A1, v1,
p1 and
z1
g
Z1
( z1-y )
Z2
X
P
Q
y
Figure 5.5
From Bernoulli’s Equation,
2
2
v
p
v
p
z1  1  1  z 2  2  2
2g 
2g

 p  p 2 

2
2
v2  v1  2 g  1
  z1  z 2 
  

For continuity of flow,
Q1 = Q2
A1v1  A2 v2
or
v2 
A1
v1  mv1
A2
——————(1)
FLUID DYNAMICS
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Nisbah luas muka keratan, m = area ratio =
A1
A2
Masukkan dalam (1) and dapatkan v1
 p  p 2 

2
2
m 2 v1  v1  2 g  1
  z1  z 2 
  

v1 
1
m
2
  p1  p 2 

   z1  z 2 
2 g 

   

1
Kadaralir sebenar (Actual discharge) Qs = Cd  A1  v1
C d  A1
Qs 
m
2

1
  p1  p 2 

   z1  z 2  ——— (2)
2 g 

   
Where Cd = pekali kadaralir (coefficient of discharge.)
Untuk U-tube,
Bahagian kiri,
p P  p1  wz1  y 
Bahagian kanan,
p Q  p 2   ( z 2  y  x)  wHg x
PP = PQ
p1   z1  y   p 2   ( z 2  y  x)  wHg x
p1   z1   z 2  p 2   z 2   y   x  w Hg x
p1  p 2

QS 
  Hg

 z1  z 2  x
 1
 

C d A1
m
2

1
  Hg

(2 gx)
 1
 

FLUID DYNAMICS
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Example 5.5
Satu meter venturi condong mengukur aliran minyak berketumpatan bandingan 0.82
dan mempunyai diameter 125mm pada salur masuk dan 50mm pada leher. Bahagian
leher berada 300mm lebih tinggi daripada bahagian salur masuk. Jika pekali kadaralir
ialah 0.97 dan perbezaan tekanan 27.5 kN/m2. Tentukan kadaralir sebenar minyak?
2
1
z1
Qactual 
z2
C d  A1
m
2
  p1  p 2 

   z1  z 2 
2 g 

   

1
3.1420.125
 0.01226 m 2
A1 
4
2
Cd = 0.97
p1  p2  27.5  10 3 kN / m 2
  0.82  9.81 10 3 N / m 2
z1  z 2  0.3 m
m=
d1
2
d2
2
2
 125 

  6.25
 50 
Therefore,
QS 
0.97  0.01226
6.25
2

1

 27.5  10 3

 = 0.01535m 3 / s

0
.
3
2  9.81
3
 0.82  9.81  10


FLUID DYNAMICS
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Example 5.6
Satu meter venturi dipasangkan pada sebatang paip berdiameter 7.5 cm di mana
diameter pada bahagian leher 2.5 cm. Tentukan kadaralir yang melalui paip tersebut,
ketika turus tekanan setinggi 41.2 cm air. Anggap Cd = 0.97.
(jwp : Qs = 1.362 x 10-3 m3/s)
Example 5.7
Satu meter venturi mengalirkan air berdiameter 1.2 m dan lehernya 0.6 m.
Perbezaan tekanan di antara bahagian masukan dan leher diukur dengan jangka
tekanan bezaan di mana ianya menunjukkan bacaan 51 mm. Cari kadaralir dan halaju
air melalui leher.
(jwp : Qs = 0.292 m3/s , v = 1.03 m/s)
Example 5.8 (SOALAN FINAL JUL 04)
Satu meter venturi mempunyai diameter utama 65 mm dan diameter bahagian
leher 26 mm. Apabila mengukur aliran sesuatu cecair yang berketumpatan 898 kg/m3,
bacaan pada satu tolok tekanan pembeza raksa ialah 71 mm. Hitungkan aliran yang
melalui meter dalam unit m3/jam. Ambil pekali kadaralir 0.97 dan graviti tentu raksa
13.6.
(jwp : Qs = 2.317 x 10-3 m3/s , = 8.34 m3/j)
(15 m)
Example 5.9 (SOALAN FINAL JAN 04)
Garispusat salur masuk sebuah meter venturi mendatar ialah 0.2 m dan
garispusat pada bahagian leher ialah 0.1 m. Ia digunakan untuk mengukur kadaralir
minyak yang berketumpatan bandingan 0.8. Perbezaan manometer raksa/minyak yang
digunakan adalah menunjukkan bacaan 0.2 m. Anggap pekali discas, Cd = 0.9.
Tentukan :
i.
ii.
iii.
Halaju aliran minyak.
Kadaralir teori minyak.
Kadaralir sebenar minyak.
(jwp : v = 2.046 m/s, QT = 0.064 m3/s, QS = 0.058 m3/s)
(10 m)
FLUID DYNAMICS
5.4.5
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Small Orifice
Section 1 :
A1, v1, p1
Section 2 :
A2, v2, p2
X
Figure 5.7
Section 1, given :
A1
= area of section 1
v1
= velocity of section 1
p1
= pressure of section 1
Section 2, given :
A2
= area of section 2
FLUID DYNAMICS
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v2
= velocity of section 2
p2
= pressure of section 2
From Bernoulli’s Equation,
Total energy at section 1 = Total energy at section 2
2
p1
2
v
p
v
 1  2  2
 2g  2g
——————(1)
v 2  v1
p  p2
 1
2g

——————(2)
2
2
z1 = z2 because the two parts are at the same level
We know that,
Q  A v
For continuity of flow,
Q1 = Q2
or
A1v1 = A2v2
So,
A1v1
A2
v2 =
——————(3)
Putting (3) into (2),
v 2  v1
p  p2
 1
2g

Av
v2 = 1 1
A2
2
Then,
2
——————(2)
——————(3)
FLUID DYNAMICS
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2
v1
2g
 A1 2
 p1  p 2
 2  1 

 A2

So,
v1 
 p  p2 
2g 1

  
 A1 2

 2  1
A

 2

But,
H=
p1  p 2
A1
2
m=
A2
2

And,
So,
v1 
2 gH
m2 1


To determine the actual discharge, Qactual ;
Qactual  Cd  A1  v1
So,
Qactual  C d  A1
2 gH
m 2 1


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.
Solution to Example 5.7
Given,
FLUID DYNAMICS
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d1
d2
Cd
= 100 mm = 0.10 m
= 250 mm = 0.25
= 0.65
= 0.9
= 750 mm = 0.75 m
oil
p1 - p2
So,
A1 
d 2
4
2
3.1240.25

 0.049m 2
4
 Hg

 x
 1
 oil
  oil

13.6 
 0.75
 1
 0.9

 10.58 m
H
m
p1  p 2
d1
2
d2
2
=
0.252
0.102
 6.25
Therefore,
2 gH
m2 1
Qactual = C d  A1

Qactual = 0.65  0.049

2  9.81  10.58
6.25 2  1

 0.074m / s
3
5.4.5.1
Types of orifice
1. Sharp-edged orifice, Cd = 0.62

FLUID DYNAMICS
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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
FLUID DYNAMICS
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x
h
A
x
y
B
Figure 5.8
From Figure 5.8 ,
x
y
h
Cv
t
= horizontal falls = velocity  time = v  t
1
1
= vertical falls = gravity  time 2 = g  t
2
2
= head of liquid above the orifice
v
= Coefficient of Velocity = C v 
2 gH
= time for particle to travel from vena contracta A to point B
Coefficient of Velocity, Cv =
Cv 
Example 5.8
v
2 gH
Actual velocity at vena contacta
Theoretical velocity
FLUID DYNAMICS
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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.
Solution to Example 5.8
x  vt
and
y=
1
g t
2
Eliminating t these equation give,
2v 2 y
x
g
y
h
Cv
t
= 1.8 – h
= head of liquid above the orifice
v

2 gH
= time for particle to travel from vena contracta A to point B
Putting y  1.8  h and v  Cv
So,
x

2 Cv
2gh
2 gh   1.8  h 
2
g
C v 4 gh1.8  h 
x
g
2
 2Cv h1.8  h 
Thus x will be a maximum when h1.8  h is maximum or,
h1.8  h   1.8  2h  0
h
So,
h  0.9 m
Example 5.9
FLUID DYNAMICS
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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.
Pipe Area, A1
P1
P2
V1
V2
Orifice area A2
C
X
C
Figure 5.9
Solution to Example 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,
p1
2
2
v
p
v
 1  2  2
 2g  2g
——————(1)
v 2  v1
p  p2
 1
2g

——————(2)
2
2
z1 = z2 because of the two parts are at the same level
We know that,
Q  A v
For continuity of flow, Q1 = Q2
FLUID DYNAMICS
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or
A1v1 = A2v2
So,
v2 =
A1v1
A2
——————(3)
Putting (3) into (2),
v 2  v1
p  p2
 1
2g

Av
v2 = 1 1
A2
2
2
——————(2)
——————(3)
Then,
2
2
 p1  p 2
v1  A1
 2  1 
2 g  A2


So,
v1 
 p  p2 
2g 1

  
 A1 2

 2  1
A

 2

This equation can therefore be written,
v1 
A
A2
2
1
 a2
2

 p  p2 
2g 1

  
——————(4)
So,
Actual disch arg e  coefficient of disch arg e  theoretical disch arg e
Qactual  Cd  A1  v1
Putting v1 into (5)
——————(5)
FLUID DYNAMICS
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Qactual  Cd  A1 
A2
A
2
1
 a2
2

 p  p2 
2g 1

  
——————(6)
but,
m=
A1
2
A2
2
so putting m into (6),
Qactual  Cd  A1 
C d A1
m
2
 p  p2 
2g 1

  

1
Considering the U-tube gauge, where pressures are equal at level CC
p1   x  p 2   q x
p1  p 2

 p p 
 x 1 2 
  
Putting x  760 mm  0.76 m and,
 g 13.6

 15.1

0.9
p1  p2

 0.76  14.1  10.72 m of oil
C d  0.65
A1 
d 2
4
 0.0497 m 2
0.252  15.1
A1 d1
 2 
A2 d 2
0.102
2
m
m 2  6.17
Qactual 
0.65  0.0497
6.17
2  9.81  10.72
 0.00524  14.5  0.0762 m3 / s
FLUID DYNAMICS
5.4.6
J3008/5/15
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.
FLUID DYNAMICS
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h
H
B
A
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
2
2
p1 v1
p
v
——————(1)

 2  2
 2g  2g

Now d 

Thus, the equation (1)
v 2 p0  p

 h or v  2 gh
2g


Although theoretically
v
p
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
p
that h  d  0 .

2gh
The actual velocity is then given by
coefficient of the instrument.
Example 5.10
, pitot tubes may require calibration.
v  C 2 gh
where C is the
FLUID DYNAMICS
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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.
Solution to Example 5.10
vair  C 2 gH
p water  p air
ghwater  ghair
hwater   water  hair  air
hwater  0.006 
 water
1000
 0.006 
 air
1.25
 4.8 m
So,
v  0.94 2  9.81  4.8
 9.12 m / s
FLUID DYNAMICS
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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.
FLUID DYNAMICS
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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 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. =
Qactual  Cd  A1  v1 and
Qactual = C d  A1
2 gH
m2 1


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 v 
2gh
, pitot tubes may require calibration.
6. The actual velocity is given by v  C
instrument.
2gh
where C is the coefficient of the
FLUID DYNAMICS
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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. 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.
FLUID DYNAMICS
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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