Unit -2b Fluid Dynamics [Compatibility Mode].pdf

Syllabus
Fluid dynamics - equations of motion - Euler's equation along a
streamline - Bernoulli's equation – applications - Venturi meter, Orifice
meter and Pitot tube. Linear momentum equation and its application.
• Second Half of Unit -2
»Bernoulli Equation(Energy equation)
»Applications:
»Applications:
Venturimeter
Orificemeter
Pitot tube
»Momentum Equation
Force on bends
Torque exerted in Lawn Sprinkler

Equations Of Motion
In a fluid flow, the following forces are present:
Gravity force, Fg
Pressure force ,Fp
Force due to viscosity,Fv
Force due to turbulence,Ft
Force due to compressibility,Fc

Equations Of Motion
According to Newton’s second law of motion ,
the net force Fx acting on a fluid element in the
direction of x is equal to mass m of the fluid
element multiplied by the acceleration ax in the
x-direction.
x-direction.
Fx = m. ax
Therefore the net force :
Fx = ( Fg )x + (Fp) x + (Fv) x + (Ft) x + (Fc) x

Equations Of Motion
Fx = ( Fg )x + (Fp) x + (Fv) x + (Ft) x + (Fc) x
a) If the force due to compressibility, (Fc) is negligible,
the resulting net force
Fx = ( Fg )x + (Fp) x + (Fv) x + (Ft) x
this equation of motion are called Reynold’s
Equation Of Motion.
Equation Of Motion.
b) For flow where (Ft) is negligible, the resulting
equation of motion are known as Navier - Stokes
Equation.
c) If the flow is assumed to be ideal, viscous force
(Fv) is zero and equation of motion are known as
Euler’s Equation Of Motion.

Euler’s Equation Of Motion
Based on Law of Conservation of Energy

Euler’s Equation Of Motion
Sum of all Forces = Mass x Acceleration
Pr.Force
Body Force

Bernoulli’s Equation From Euler’s Formula
= constant
If flow is incompressible, ρ is constant
= constant
Dividing through out by g
 
 
 vdv
gdz
dp

2
2
v
gz
p



Dividing through out by g
= constant - Bernoulli’s equation
Where:
= pressure energy per unit weight of fluid or
pressure head
v2/2g = kinetic energy per unit weight or kinetic head
Z = potential energy per unit weight or potential head
g
z
g
p v
2
2



g
p


Bernoulli’s Equation
The following are the assumptions made :
The fluid is ideal, i.e viscosity is zero
The flow is steady
The flow is incompressible
The flow is irrotational

Example 1
Water is flowing through a pipe of 5cm diameter under a
pressure of 29.43 N/cm2 (gauge) and with mean velocity of 2
m/s. Find the total head or total energy per unit weight of the
water at a cross-section, which is 5m above the datum line.
Given:
Diameter of pipe = 5cm = 0.5 m
Diameter of pipe = 5cm = 0.5 m
pressure : p = 29.43 N/cm2
velocity : v = 2 m/s
Datum Head : z = 5m
Required : total head or total energy per unit weight of the
water at a cross-section

Example 1
Solution :
Total head = pressure head + kinetic head + datum head
pressure head = = = 30 m
kinetic head = = = 0.204 m
g
p

81
.
9
1000
104
43
.
29


v
2
22
kinetic head = = = 0.204 m
Total head = =
= 35.204 m
g
v
2 81
.
9
2
2

z
g
g
p v 

2
2

5
204
.
0
30 


Example 2
The water is flowing through a pipe having diameters 20cm
and 10cm at sections 1 and 2 respectively. The rate of flow
through pipe is 35 litres/s. The section 1 is 6m above datum
and section 2 is 4m above datum. If the pressure at section 1 is
39.24 N/cm2, find the intensity of pressure at section 2.
(Neglect the losses).

Example 2
Given:
At section 1 , D1 = 20cm = 0.2m
A1 = = 0.0314 m2
p1 = 39.24 N/cm2
= 39.24 x 10 4 N/m2
2
.
0
2
4

= 39.24 x 10 N/m
z1 = 6m
At section 2 , D1 = 10cm = 0.1m
A1 = = 0.0314 m2
z2 = 4m
1
.
0 2
4


Example 2
Rate of flow : Q = 35 lit/s =
= 0.035 m3/s
Required : Intensity of pressure at section 2
Solution :
Q = A V = A V
1000
35
Q = A1V1 = A2V2
= 1.114 m/s
= 4.456 m/s
0314
.
0
035
.
0
1
1


A
V
Q
00785
.
0
035
.
0
2
2


A
V
Q

Example 2
Applying Bernoulli’s equation at section 1 & 2
4
81
.
9
2
456
.
4 2
81
.
9
1000
2
6
81
.
9
2
114
.
1 2
81
.
9
1000
104
24
.
39









 p
z
g
V
g
p
z
g
V
g
p
2
2
2
2
2
1
2
2
1
1







81
.
9
2
81
.
9
1000
81
.
9
2
81
.
9
1000 



012
.
5
9810
2
063
.
46 

p
051
.
41
9810
2 
p
p2 = 40.27 x104N/m2
Answer :Intensity of pressure at section 2 = 40.27 N/cm2

Bernoulli’s Equation for real fluid:
The Bernoulli’s Equation was derived on
the assumption that fluid is inviscid and therefore frictionless.
But all the real fluid are viscous and hence offer resistance to
flow. Thus there are always some losses in fluid flows and
hence these losses are taken into consideration.
hence these losses are taken into consideration.
Bernoulli’s Equation for the real fluids are :
hL : loss of energy between two points
h
z
V
p
z
V
p
L
g
g
g
g





 2
2
2
2
1
2
1
1
2
2 


Example 3
A pipe of diameter 400 mm carries water at a velocity of 25 m/s.
The pressure at the points A and B are given as 29.43 N/cm2
respectively while the datum head at A and B are 28 m and 30 m.
Find the loss of head between A and B.

Example 3
Given :
Dia. Of pipe, D = 400 mm = 0.4 m
Velocity, V = 25 m/s
Required: loss of head between A and B
Solution:
At point A, pA = 29.43 N/cm2 = 29.43 * 104 N/m2
zA = 28 m
zA = 28 m
vA = v = 25 m/s
Total energy at A,
= 30 + 31.85 + 28 = 89.85 m
Z A
g
v A
g
p A



2
2
E A

28
81
.
9
2
252
81
.
9
1000
104
43
.
29
EA 






Example 3
At point B, pB = 22.563 N/cm2 = 22.563 * 104 N/m2
zB = 30 m
vB = v = vA = 25 m/s
Total energy at B, Z B
g
v B
g
p B



2
2
E B 
252
104
563
.
22 
= 23 + 31.85 + 30 = 84.85 m
Loss of energy:hL = EA – EB = 89.85 – 84.85 = 5.0 m
Answer : loss of head between A and B = 5.0 m
30
81
.
9
2
252
81
.
9
1000
104
563
.
22
EB 






Syllabus
Fluid dynamics - equations of motion - Euler's equation
along a streamline - Bernoulli's equation – applications -
Venturi meter, Orifice meter and Pitot tube. Linear
momentum equation and its application.
»Applications:
Venturimeter
Orificemeter
Pitot tube
»Momentum Equation

Practical applications of bernoulli’s equation
Bernoulli’s equation is applied in all problems of
incompressible fluid flow where energy consideration are
involved. Its application to the following measuring devices
are:
1. Venturimeter.
1. Venturimeter.
2. Orificemeter. To measure discharge(rate of flow)
3. Rotameter
4. Pitot tube. – To measure velocity in a flow of liquid.

Pictures
Venturimeter Rotameter

Venturimeter
It is used for measuring the rate of a flow of a fluid
flowing through the pipe. The basic principle is that
by reducing the cross-sectional area of the flow
passage, a pressure difference is created and the
measurement of pressure difference enables the
determination of the discharge through pipes. It
consists of three parts.,
consists of three parts.,
(i) A short converging part.
(ii) Throat.
(iii) Diverging part.

Venturimeter
Expression for rate of flow through venturimeter
Let d1 = diameter at inlet or at section (1)
p1 = pressure at section (1)
v1 = velocity of fluid at section at section (1)
a1 = area at section (1) =
And d1 , p1 , v1 , a1 = corresponding values at section (2)
d 2
1
4


Venturimeter
Applying Bernoulli’s equation at section (1) & (2)
As pipe is horizontal, hence z1 = z2
z
g
V
g
p
z
g
V
g
p
2
2
2
2
2
1
2
2
1
1







V
p
V
p 2
2
2
2
1
1



- (i)
Now applying continuity equation at section (1) & (2)
g
V
g
p
g
V
g
p
2
2
2
2
1
1





g
V
g
V
g
p
p
2
2
1
2
2
2
2
1




h
g
p
p



2
1
g
a
v
a
g
V
h
2
1
2
2
2
2
2
2


1
2
2
1
2
2
1
1
a
or
Q v
a
v
v
a
v
a 



Venturimeter
Actual discharge will be less than theoretical discharge
Where Cd = co-efficient of venturimeter and its value is less than 1
Value of h is given by differential u – tube manometer
gh
a
a
a
a
Cd
Qact 2
2
2
2
1
2
1



Value of h is given by differential u – tube manometer
Case i: Let the differential manometer contains a liquid which is
heavier than the liquid flowing through the pipe.
Let ℎ= . ℎ ℎ
= . ℎ
= ℎ ℎ −

To find piezometric head h
Pressure
Manometers
Pressure at two
points
Pressure head at
two points
Difference in manometer
(Specific gravity of manometeric liquid
is greater than specific gravity of liquid
flowing in pipe)
Difference in manometer
(Specific gravity of manometeric
liquid is less than specific gravity of
liquid flowing in pipe)
Possible datas given in problems

Venturimeter
Case i: This case related to the inclined venturimeter having differential u-
tube manometer. Let the differential manometer contains heavier
liquid, then ‘h’ is given by
 
1
2
2
1
1






















S o
S h
x
g
g
h z
p
z
p


Case ii: This case related to the inclined venturimeter having differential u-






























S
S
o
l
g
g
h x
z
p
z
p 1
2
2
1
1


Case ii: This case related to the inclined venturimeter having differential u-
tube manometer. Let the differential manometer contains lighter
liquid, then ‘h’ is given by
z
g
V
g
p
z
g
V
g
p
2
2
2
2
2
1
2
2
1
1







Case iii: Incase velocity given in problem, substitute in Bernoulli equation and
find h

Example 1
A horizontal venturimeter with inlet and throat diameters 30cm and
15cm respectively is used to measure the flow of water . The
reading of differential manometer connected to the inlet and the
throat is 20cm of mercury. Determine the rate of flow. Take Cd =
0.98.
Given :
dia at inlet : d1 = 30cm
dia at inlet : d1 = 30cm
area at inlet : a1 = = = 0.0706m2
dia at throat : d2 = 15cm
area at throat : a2 = = 0.017 m2
Cd = 0.98
reading of differential manometer = x = 20 cm of mercury
required : rate of flow
Solution :
d 2
1
4
 302
.
0
4

152
.
0
4


Example 1
Difference of pressure head :
Where:
Sh= sp. gr of mercury = 13.6
S0= sp. gr of water = 1
 
1


S o
S h
x
h


= 2.52 m of water
The discharge through venturimeter :
= 0.12496m3/s = 125 lit/s





 
 1
1
6
.
13
20
.
0
h
gh
a
a
a
a
C d
Q act 2
2
2
2
1
2
1



52
.
2
81
.
9
2
2
0176
.
0
2
0706
.
0
0176
.
0
0706
.
0
98
.
0 






Example 2
A horizontal venturimeter with inlet diameter 20cm and throat diameter 10cm is
used to measure the flow of water. The pressure at inlet is 17.658 N/cm2 and the
vacuum pressure at the throat is 30cm of mercury. Find the discharge of water
through venturimeter. Take Cd = 0.98.
Given : dia at inlet : d1 = 20cm
area at inlet : a1 = = 0.031416cm2
dia at throat : d2 = 10cm
area at throat : a = = 0.007854 cm2
202
.
0
4

102
.
0

area at throat : a2 = = 0.007854 cm2
Cd = 0.98
p1 = 17.658 N/cm2 =17.658 x 104 N/m2
= 18 m of water (ρ for water =1000kg/m3)
= -30 cm of mercury
= -0.3 m of mercury = -0.3 x 13.6 = -4.08 m
Required : discharge of water through venturimeter
Solution :
102
.
0
4

1000
81
.
9
104
658
.
17
1



g
p

g
p

2

Example 2
Differential head : = 18-(-4.08) = 22.08 m of water
Discharge :
g
g
h
p
p


2
1


gh
a
a
a
a
Cd
Q 2
2
2
2
1
2
1



08
.
22
81
.
9
2
007854
.
0
031416
.
0
007854
.
0
031416
.
0
98
.
0 2
2






=0.165 m3/s = 165.55lit/s
Answer : Discharge of water through venturimeter = 165.55lit/s

Problem -3
In a vertical pipe conveying oil of specific gravity of 0.8, two pressure gauges have been
installed at A and B where the diameters are 16 cm and 8 cm respectively. A is 2 meters
above B. The pressure gauges readings have shown that the pressure at B is greater
than A by 0.981N/cm2. Neglecting all losses, calculate the flow rate. If the gauges are
replaced by tubes containing same liquid and connected to a U-tube containing
mercury. Calculate the difference of level of mercury in the two limbs of U-tube.
Given Data:
Specific gravity of oil = 0.8
Specific gravity of oil = 0.8
Area at section A (a1) = Πd1
2/4= 3.14 x 0.162/4=0.02 m2
Area at section B (a2)= Πd2
2/4= 3.14 x 0.082/4=0.005 m2
Datum head at A Z1 = 2m
Datum head at B Z2 = 0m
Difference in pressure p2 –p1 = 0.981 N/cm2 = 0.981 x 104 = N/m2
Difference in pressure head (p2 –p1)/ρg = (0.981 x 104 ) / (800 x 9.81)
= 1.25 m
But we need, (p2 –p1 )/ρg = - 1.25m

Step 1:Find h:
So h = -1.25 + 2 = 0.75 m
Step 2: Find
Qth = 0.0198m3/s
Step 3 : Second Part : TO FIND DIFFERENCE OF MERCURY LEVELS
IN U TUBE MANOMETER ie x



















 z
p
z
p
g
g
h 2
2
1
1


gh
a
a
a
a
Qth 2
2
2
2
1
2
1


IN U TUBE MANOMETER ie x
0.75 = x ((13.6/0.8)-1)
So x = 0.04687 m = 4.687 cm
 
1


S o
S h
x
h

Syllabus
»Applications:
Venturimeter( Cd = 0.96 to 0.98)
Orificemeter (Cd = 0.62 to 0.68)
Pitot tube (C = 0.98)
Pitot tube (Cv = 0.98)
»Momentum Equation
Forces on pipe bends (Momentum eqn)
F= ρQ (V2-V1)
Forces on sprinkler system
(Moment of Momentum)
F= ρQ (V2r2 –V1r1)

Momentum equation
 It is based on the law of conservation of momentum or
on the momentum principle, which states that the net
force acting on a fluid mass is equal to the change in
momentum of flow per unit time in that direction.
 The force acting on a fluid mass ‘m’ is given by the
 The force acting on a fluid mass ‘m’ is given by the
Newton’s second law of motion.
F = m x a
where ‘a’ is the acceleration acting in the same
direction as force F.

Momentum equation
dt
dv
m
F
dt
dv
a


F = m x a
Net Force F = mass x acceleration
F = Density x volume x acceleration
F = Density x volume x velocity
time
F = Density x Discharge x change
 
dt
mv
d
F
dt

This is the momentum principle
F.dt = d(mv) which is known as the impulse-momentum equation and
states that the impulse of a force ‘F’ acting on a fluid mass ‘m’ in a
short interval of time ‘dt’ is equal to the change of momentum d(mv)
in the direction of force.
in velocity
F = ρQv

Momentum equation
Force exerted by a flowing fluid on a pipe-bend
By using Impulse-momentum equation:
Let V1=velocity of flow at section (1)
P1=pressure intensity at section (1)
A1=area of cross-section of pipe at section(1)
V2,P2,A2 = are for section (2)

v2
p A
p2A2
FY
p2A2 cos θ
v2Sinθ
v2Cosθ
p2A2 sin θ
Net Force F = ρ Q V
Along x direction,
p1A1 – p2A2 cos θ – Fx = ρ Q (V2 cos θ- V1)
Along y direction
-p2A2 sin θ – Fy = ρ Q V2 sinθ
Resultant Force
Direction of Force tan θ1 = Fy/Fx
+ve Direction
+ve Direction
v1
θ
p1A1
Fx
 
F
F
F y
x
R
2
2



Problems
• A 45o reducing bend is connected to a pipe line, the
diameters at the inlet and outlet of bend being 600 mm
and 300 mm respectively. Find force exerted by water
on bend if the pressure intensity at inlet is 8.829 N/cm2
and pressure intensity at outlet is 5.45 N/cm2. The Rate
of flow is 600 litres/sec
of flow is 600 litres/sec
Given Data : Angle of bend θ= 45
Diameter at sec.1 d1= 0.6 m Find A1
Diameter at sec. 2 d2 = 0.3 m Find A2
Pressure intensity at sec 1 p1 = 8.829 x 104 N/m2
Pressure intensity at sec 2 p2 = 5.45 x 104 N/m2
Discharge Q = 0.6 m3/s

• Use continuity Equation:
Q = A1V1 = A2V2
Find V1 and V2
• Force along x direction
p1A1 – p2A2 cos θ – Fx = ρ Q (V2 cos θ- V1)
– Find F x
Answers: V1 = 2.122m/s , V2 = 8.488 m/s
Answers:
– Find F x
• Force along x direction
-p2A2 sin θ – Fy = ρ Q V2 sinθ
– Find Fy
• Resultant Force
 
F
F
F y
x
R
2
2


Answers:
Fx= 19.911kN
Fy = - 6.322 kN
Fr = 20.89 kN

Example
250 litres of water is flowing in a pipe having a
diameter of 300mm. If the pipe is bent by 135, find
the magnitude and direction of the resultant force
on the bend. The pressure of water flowing is 39.24
N/cm2 .
N/cm .
figure

Unit -2b Fluid Dynamics [Compatibility Mode].pdf

Example
Given : Since pipe diameter is same, V1 = V2,
Substituting, V1 = V2 and Z1 = Z2 we get,
Pressure P1=P2 = 39.24 N/cm2
= 39.24 x 104 N/m2
Discharge : Q = 250 litres/s = 0.25 m3/s
D1 = D2 =300 mm = 0.3m
A1 = A2 =
V1 = V2 = 3.54 m/s
Required: magnitude and direction of the resultant force on
07068
.
0
25
.
0

A
Q

Example
Solution :
Force along x-axis:
P1A1 + P2A2 cos 45 - Fx =  Q (-V1 +V2cos 45 )
F = P A + P A cos 45+ Q (-V +V cos 45 )
Fx = P1A1 + P2A2 cos 45+ Q (-V1 +V2cos 45 )
= 48855.4 N
 









 45
cos
537
.
3
537
.
3
25
.
0
1000
)
45
cos
1
(
07068
.
0
24
.
39 10
4

Example
Force along y-axis :
- Fy - P2A2 sin 45 =  Q (V2 sin 45 )
Fy = - Q(V2sin 45 ) - P2A2 cos 45
= -20236.3 N
  






 45
sin
07068
.
0
24
.
39
45
sin
537
.
3
25
.
0
1000 10
4
= -20236.3 N
= 52880.6 N
   
3
.
20236
4
.
48855
2
2
2
2



 F
F
F y
x
R
'
30
22
4142
.
0
4
.
48855
3
.
20236
tan 




F
F
x
y


Moment Of Momentum Equation
Moment of Momentum equation is derived from moment of
momentum principle which states that the resulting torque acting
on a rotating fluid is equal to the rate of change of moment of
momentum.
let V1 = velocity of fluid at section 1
r1 = radius of curvature at section 1
1
Q = rate of flow of fluid
ρ = density of fluid
V2 & r2 = velocity and radius of curvature at section 2
Momentum of fluid at section 1 = mass x velocity
= ρQ x V1 /s
moment of momentum per sec at sec -1 = ρQ x V1 x r1

Moment Of Momentum Equation
Similarly moment of momentum per sec at sec -2
= ρQ x V2 x r2
Rate of change of moment of momentum
= ρQV2r2 - ρQV1r1
= ρQ(V2r2 - V1r1)
According to moment of momentum principle :
According to moment of momentum principle :
Resultant torque = rate of change of moment of momentum
T = ρQ(V2r2 - V1r1)
This eq is known as moment of momentum equation. It is
applied :
 Analysis of flow problems in turbines and centrifugal pumps
 For finding torque exerted by water on sprinkler.

Example
A lawn sprinkler as shown in fig. has 0.8cm diameter nozzle at the end
of a rotating arm and discharge water at the rate of 10 m/s velocity.
Determine the torque required to hold the rotating arm stationary. Also
determine the constant speed of rotation of the arm , if free to rotate.
Given :
Dia. Of each nozzle : 0.8 cm = 0.008 m
Area of each nozzle = = 5.026 x 10 5 m2
 
008
.
0
2
4

Velocity of flow at each nozzle = 10 m/s
Required : torque & speed of rotation of the arm
Solution :
Discharge: Q = Area X velocity
= 5.026 x 10 5 x 10 = 5.026 x 10 4 m3/s
 
4

Example
Torque exerted by water moment of momentum
coming through nozzle = of water through A
= rA x ρ x Q x VA
= 0.25 x 1000 x 5.026 x 10 4 x 10
= 1.26 (clockwise)
coming through nozzle = of water through B
= rB x ρ x Q x VB
= 0.20 x 1000 x 5.026 x 10 4 x 10
= 1.0 (clockwise)
Torque exerted by water on sprinkler :
= 1.26+ 1 = 2.26 Nm

Example
Speed of rotation of arm , if free to rotate:
Let ω = speed of rotation of the sprinkler
The absolute velocity of flow of water at the nozzles A & B:
V1 = 10 - 0.25 x ω
V2 = 10 - 0.2 x ω
V2 = 10 - 0.2 x ω
Torque exerted by water coming out at A on sprinkler :
= rA x ρ x Q x V1
= 0.25 x 1000 x 5.026 x 10 4 x (10 - 0.25 x ω)
= 0.125 (10 - 0.25 x ω)
Torque exerted by water coming out at B on sprinkler :
= rB x ρ x Q x V2
= 0.2 x 1000 x 5.026 x 10 4 x (10 - 0.2 x ω)
= 0.1 (10 - 0.2 x ω)

Example
Torque exerted by water = 0.125 (10 - 0.25 x ω) + 0.1 (10 - 0.2 x ω)
Since moment of momentum of the flow entering is zero and no external torque
is applied on sprinkler, so the resultant torque on the sprinkler must be zero.
0.125 (10 - 0.25 x ω) + 0.1 (10 - 0.2 x ω) = 0
ω = 43.9 rad/s
 9
.
43
60
60 


N
= 419.2 r.p.m
Answer : Torque = 2.26 Nm & speed of rotation of the arm = 43.9
rad/s



2
9
.
43
60
2
60 


N

References
1. Bansal, R.K., “Fluid Mechanics and Hydraulics Machines”,
5th edition, Laxmi Publications Pvt. Ltd, New Delhi, 2008
2. Modi P.N and Seth "Hydraulics and Fluid Mechanics
including Hydraulic Machines", Standard Book House New
including Hydraulic Machines", Standard Book House New
Delhi. 2015.

Unit -2b Fluid Dynamics [Compatibility Mode].pdf

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Unit -2b Fluid Dynamics [Compatibility Mode].pdf