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 =
ω
W
Distance moved by liquid = a
W
ω
Work done = force × distance = a
W
ap
ω
×
=
ω
p
W
pressure energy per unit weight =
ω
p
= g
p
ρ
Similarly the pressure energy per unit weight p/W is equivalent to a head and is
referred to as the pressure head.
5.1.3 Kinetic energy
INPUTINPUT

If a weight W of liquid has a velocity v,
Kinetic energy =
2
2
1
v
g
W
Kinetic energy per unit weight =
g
v
2
2
The kinetic energy per unit weight
g
v
2
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
vp
z
2
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:
tcons
g
vp
zH tan
2
2
=++=
ω
By Bernoulli’s Theorem,
Total energy per unit weight at section 1 = Total energy per unit weight at section 2
Figure 5.1

g
vp
z
g
vp
z
22
22
2
2
11
1 ++=++
ωω
z = potential head
ω
p
= pressure head
g
v
2
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,
ωω
2
2
2
1
1
2
1
1
22
p
g
v
z
p
g
v
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).
5.4 Application of Bernoulli’s Equation
Bernoulli’s equation can be applied to the following situations.
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.

5.4.1 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
g
vp
z
2
2
++=
ω
( )
81.92
8.4
81.91000
10410
36
23
×
+
×
×
+=
NJ /96.78=
5.4.2 Inclined Pipe
Example 5.2
36 m
Figure 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
.
Using Bernoulli’s Equation:
g
vp
z
g
vp
z NN
N
MM
M
22
++=++
ωω
…(1)
Discharge at section M = Discharge at section N
NM QQ =
NNMM avav ×=× …(2)
From (2),
N
MM
N
a
av
v
×
=
( ) ( )
( )2
2
05.0
2.0
1=
sm /16=
Given 2
/5 cmkgpM =
0001.0
81.95×
=
2
/5.490 mkN=
5 m 5 m
3 m
Figure 5.3
N
M

From (1),
( ) ω
ω
×








−++
−
= NM
MNM
N zz
p
g
vv
p
2
22
( ) ( ) 981035
9810
490500
81.92
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.
tcons
g
vp
zH tan
2
2
=++=
ω
5.2 velocity head of water =
( )
( )
m
g
v
H 64.2
81.92
2.7
2
22
===
64.2==
ω
p
H
ω64.2=p
( )981064.2=
2
/4.25898 mN=
2
/9.25 mkN=
5.4.3 Horizontal Venturi Meter
INPUTINPUT

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.
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
Entry
Converging
Cone
Throat
Diverging Section
Section 1
ρ1
v1
a1
Section 2
ρ2
v2
a2
Direction of flow
Leads gauge
filled with
liquid in
pipeline,
spec.wt = ω
x
Spec.wt. of
gauge liquid=
ωg

Derivation for the theoretical discharge through a horizontal venture meter and
modification to obtain the actual discharge.
From Figure 5.4
Putting ;
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

Bernoulli’s Equation for section 1 and 2 gives :
ωω
2
2
2
1
1
2
1
1
22
p
g
v
z
p
g
v
z ++=++
Ignoring losses for horizontal meter z1 = z2
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

ω
21
2
1
2
2
2
pp
g
vv −
=
−
——————(1)
For continuity of flow, 2211 vAvA = , giving
1
2
1
2 v
A
A
v =
where
4
2
d
A
π
=
Substituting in equation (1)





 −
=







−
ω
21
2
2
2
12
1 21
pp
g
A
A
v
So,
( ) 










 −
−
=
ω
21
2
2
2
1
2
1 2
pp
g
AA
A
v
Discharge, Qtheory = ( )gH
AA
AA
vA 2
2
2
2
1
21
11
−
= ——————(2)
Where H =
ω
21 pp −
= pressure difference expressed as a head of the liquid
flowing in meter venture.
If area ratio,
2
1
A
A
m = equation (2) becomes,
1
2
21
−
=
m
gH
AQtheory
The theoretical discharge Q can be converted to actual discharge by
multiplying by the coefficient of discharge Cd found experimentally.
Actual discharge,
1
2
21
−
=×=
m
gH
ACQCQ dtheorydactual —————(3)
If the leads of U-tube are filled with water,
( )ωω −=− gxpp 21

H = 





−=
−
121
ω
ω
ω
g
x
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
2
21
−
=
m
gH
AcQ dactual
So,
x = 55 mm
ω
ωg
= 13.6
H = 0.055 ×12.6 = 0.0706 m2
Cd = 0.98
A1
( ) 2
2
0706.0
4
3.0142.3
m==
m 9
4
12
2
2
2
2
1
2
1
=





===
d
d
A
A

Actual discharge, Qactual =
181
693.081.92
0706.098.0
−
××
×
Qactual = sm /0285.0 3
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.
From Equation (3),
1
2
21
−
=
m
gH
AcQ dactual
The difference of pressure head, H must be expressed in terms of the liquid
following through the meter,
H =
ω
p
= 3
3
1081.99.0
105.34
××
×
= oilofm92.3

A1
( ) 2
2
00441.0
4
075.0142.3
m==
m = 4
Cd = 0.97
So,
Actual discharge, Qactual =
116
92.381.92
00441.097.0
−
××
×
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.
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

From Bernoulli’s Equation,
ωω
2
2
2
1
1
2
1
1
22
p
g
v
z
p
g
v
z ++=++
( )






−+




 −
=− 21
212
1
2
2 2 zz
pp
gvv
ω
——————(1)
For continuity of flow,
2211 vAvA =
or
11
2
1
2 mvv
A
A
v ==
where
m = area ratio =
2
1
A
A
Substituting in equation (1) and solving for v1
( )






−+




 −
=− 21
212
1
2
2 2 zz
pp
gvv
ω
( )
( ) 











−+




 −
−
= 21
21
2
1 2
1
1
zz
pp
g
m
v
ω
Actual discharge, Qactual = 11 vACd ××

( )
( ) 











−+




 −
−
×
= 21
21
2
1
2
1
zz
pp
g
m
AC
Q 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
ω
ω
ω
g
xzz
pp
Equation (2) can therefore be written
( ) 













−
−
= 12
12
1
ω
ωgd
actual gx
m
AC
Q

Example 5.5
A vertical venture 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.
1
2
z2z1

In equation (2),
( )
( ) 











−+




 −
−
×
= 21
21
2
1
2
1
zz
pp
g
m
AC
Q 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
2
1
zz
pp
g
m
AC
Q d
actual
ω
So,
A1
( ) 2
2
01226.0
4
125.0142.3
m==
23
21 /105.27 mkNpp ×=−
23
/1081.982.0 mN××=ω
mzz 3.021 −=−
m = 25.6
50
125
2
2
2
2
1
=





=
d
d
Cd = 0.97
Therefore,
( )
( ) 











−+




 −
−
×
= 21
21
2
1
2
1
zz
pp
g
m
AC
Q d
actual
ω
( )( )












−
××
×
×
−
×
= 3.0
1081.982.0
105.27
81.92
125.6
01226.097.0
3
3
2
actualQ =
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?
d1 d2
A B b
p1 ,v1 p2 v2
Figure 5.6

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,
ωω
2
2
21
2
1
22
p
g
vp
g
v
+=+
This equation can therefore be written,
g
vvpp
2
2
1
2
221 −
=
−
ω
For continuity of flow,
2211 vAvA =
or
2
2
2
1
2
1
44
v
d
v
d








=






 ππ
then,
2
2
21
2
1 vdvd ×=×
So,
2
2
1
12 





=
d
d
vv
Putting smv /5.41 = , mmd 101= , mmd 72 =
2
2
7
10
5.4 





=v
sm /18.9=
and

g
vvpp
2
2
1
2
221 −
=
−
ω
81.92
5.418.9 22
21
×
−
=
−
ω
pp 2
26.3 m=
ω
21 pp − 2
26.3 m=
21 pp − ω×= 2
26.3 m
Pressure difference, 23
21 /1081.926.3 mNpp ××=−
3
/9.31 mkN=
Area of piston = 3
2
/
4
mkN
dπ
( ) 2
2
000314.0
4
020.0
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
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)
Converging cone
Diverging section

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.
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.
throat
entry
INPUTINPUT

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, given :
Section 2, given :
X
Section 1 :
A1
, v1
, p1
Section 2 :
A2
, v2
, p2
Figure 5.7

From Bernoulli’s Equation,
Total energy at section 1 = Total energy at section 2
g
vp
g
vp
22
2
22
2
11
+=+
ωω
——————(1)
ω
21
2
1
2
2
2
pp
g
vv −
=
−
——————(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
A
vA
——————(3)
Putting (3) into (2),
ω
21
2
1
2
2
2
pp
g
vv −
=
−
——————(2)
v2 =
2
11
A
vA
——————(3)
Then,
ω
21
2
2
2
1
2
1
1
2
pp
A
A
g
v −
=





−

So,








−





 −
=
1
2
2
2
2
1
21
1
A
A
pp
g
v
ω
But,
H =
ω
21 pp −
And,
m = 2
2
2
1
A
A
So,
( )1
2
21
−
=
m
gH
v
To determine the actual discharge, Qactual ;
11 vACQ dactual ××=
So,
( )1
2
21
−
×=
m
gH
ACQ 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
oil = 0.9
p1 - p2 = 750 mm = 0.75 m
So,
4
2
1
d
A
π
=
( ) 2
2
049.0
4
25.0124.3
m==






−=
−
= 121
oil
Hg
oil
x
pp
H
ω
ω
ω






−= 1
9.0
6.13
75.0
m58.10=
m 2
2
2
1
d
d
= =
( )
( )2
2
10.0
25.0
25.6=
Therefore,
Qactual =
( )1
2
21
−
×
m
gH
ACd
Qactual =
( ) 125.6
58.1081.92
049.065.0 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
h
x
yAx
B

Figure 5.8
From Figure 5.8 ,
x = horizontal falls = velocity × time = tv ×
y
= vertical falls =
2
2
1
timegravity × = tg ×
2
1
h = head of liquid above the orifice
Cv
= Coefficient of Velocity = gH
v
Cv
2
=
t = time for particle to travel from vena contracta A to point B
Coefficient of Velocity, Cv = velocitylTheoretica
contactavenaatvelocityActual
gH
v
Cv
2
=
Example 5.8
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 ×
2
1
Eliminating t these equation give,
g
yv
x
2
2
=

y = 1.8 – h
h = head of liquid above the orifice
Cv gH
v
2
=
t = time for particle to travel from vena contracta A to point B
Putting hy −= 8.1 and ( )ghCv v 2=
So,
( )[ ] ( )
g
hghC
x v −×
=
8.122
2
( )
g
hghC
x v −
=
8.14
2
( )[ ]hhCv −= 8.12
Thus x will be a maximum when ( )hh −8.1 is maximum or,
( )[ ] 028.1
8.1
=−=
−
h
h
hh
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,
g
vp
g
vp
22
2
22
2
11
+=+
ωω
——————(1)
ω
21
2
1
2
2
2
pp
g
vv −
=
−
——————(2)
z1 = z2 because of 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
A
vA
——————(3)
Putting (3) into (2),
Pipe Area, A1
P1
P2
V1
V2
X
Orifice area A2
C C

ω
21
2
1
2
2
2
pp
g
vv −
=
−
——————(2)
v2 =
2
11
A
vA
——————(3)
Then,
ω
21
2
2
2
1
2
1
1
2
pp
A
A
g
v −
=





−
So,








−





 −
=
1
2
2
2
2
1
21
1
A
A
pp
g
v
ω
This equation can therefore be written,
( )





 −
−
=
ω
21
2
2
2
1
2
1 2
pp
g
aA
A
v ——————(4)
So,
edischltheoreticaedischoftcoefficienedischActual argargarg ×=
11 vACdQactual ××= ——————(5)
Putting v1 into (5)
××= 1ACdQactual
( )





 −
− ω
21
2
2
2
1
2
2
pp
g
aA
A
——————(6)
but,
m = 2
2
2
1
A
A
so putting m into (6),

( )





 −
−
××=
ω
21
2
1
1 2
1
pp
g
m
AC
ACdQ d
actual
Considering the U-tube gauge, where pressures are equal at level CC
xpxp qωω +=+ 21





 −
=
−
ωω
2121 pp
x
pp
Putting mmmx 76.0760 == and,
1.15
9.0
6.13
==
ω
ωg
oilofm
pp
72.101.1476.021
=×=
−
ω
65.0=dC
2
2
1 0497.0
4
m
d
A ==
π
( )
( )
1.15
10.0
25.0
2
1
2
2
2
2
2
1
====
d
d
A
A
m
17.62
=m
( )72.1081.92
17.6
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
g
vp
g
vp
22
2
22
2
11
+=+
ωω
——————(1)
 Now
ω
p
d = 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
0p
dh =+ .
 Thus, the equation (1) ghvorh
pp
g
v
2
0
2
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.
Example 5.10
A B
h
H

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
1000
006.0006.0 ×=×=
air
water
waterh
ω
ω
m8.4=
So,
8.481.9294.0 ××=v
sm /12.9=
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 ( )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 =
( )1
2
21
−
×
m
gH
ACd
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
A Pitot-static probe connected to a water manometer is used to measure the
velocity of air. If the deflection (the vertical distance between the fluid levels in
the two arms) is 7.3 cm, determine the air velocity. Take the density of air to be
1.25 kg/m3
.
Pressure in a manometer = ρfluidgh
The height change in the water of the manometer is due only to the dynamic
pressure entering the device.
V = 33.8 m/s

Unit 5 Fluid Mechanics

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (20)

Ähnlich wie Unit 5 Fluid Mechanics

Ähnlich wie Unit 5 Fluid Mechanics (20)

Mehr von Malaysia

Mehr von Malaysia (20)

Kürzlich hochgeladen

Kürzlich hochgeladen (20)

Unit 5 Fluid Mechanics