Fluid Mechanics - Flows

Prepared by:
Aditya Deshpande
Guided by:
Prof. S. S. Shinde
Fluid Mechanics- Aditya Deshpande 1

Part A (8 hrs.)
Friction factor, Pipe losses, Boundary Layer, Over
external bodies, Flow Separation and control
methods, Lift generation, Flow simulation
methodology.
Part B (Self study)
Siphon, Transmission of power, Drag and lift,
Characteristics of bodies.
SYLLABUS

1. Continuity equation
2. Bernoulli's theorem
3. Bernoulli's theorem for real fluids
PERQUISITE

Energy losses
Major energy
losses
This is due to friction
and is calculated by
1.Darcy-Weisbach
formula
2.Chezy’s formula
Minor energy
losses
This is due to
a. Sudden expansion of pipe
b. Sudden contraction of pipe
c. Bend in pipe
d. An obstruction in pipe
e. At entrance
f. At exit of pipe
g. Due to pipe fittingsFluid Mechanics- Aditya Deshpande 4

1. When liquid is flowing through pipes
Velocity of liquid layer adjacent to wall of
pipe is zero.
2. Velocity of liquid goes on increasing from
wall and thus shear stresses are produced
in whole liquid due to viscosity.
Loss of energy of fluid through
pipes

Consider a uniform horizontal pipe having steady flow
Let,
p1=press. At sect 1-1
V1=velocity Of flow at 1-1
L=pipe length between 1-1 and 2-2
D=diameter of pipe
F’=frictional resistance/ wetted area per unit velocity
Hf=loss of head due to friction
Similarly, P2, V2
Expression for loss of head due to friction
in pipes
(Darcy weisbach equation)

Applying Bernoulli’s equation between 1-1 and 2-2
Total head at 1-1=
(total head at 2-2)+ (friction loss of head between 1-1 and
2-2)
Here z1=z2 as pipe is horizontal V1=v2 as dia. of pipe is same
𝑝1
𝜌𝑔
=
𝑝2
𝜌𝑔
+ ℎ𝑓 so, hf =
𝑝1
𝜌𝑔
−
𝑝2
𝜌𝑔
L
2
2
2
2
2
1
1
1
H
g2
V
z
g
P
g2
V
z
g
P




Fig. No.1 Uniform Horizontal Pipe

continued….
But, hf is head loss due to friction and intensity of pressure
will be reduced in direction of flow by frictional resistance.
Frictional resistance=
frictional resistance per unit wetted area per unit velocity x
wetted area x V2
So,
F1=f’ x πd L x V2 [ wetted area= πdL , V=V1=V2 ]
F1= f’ x PL V2 [ P=perimeter=P ]

Forces acting on fluid between 1-1 & 2-2 are pressure force
1. Pressure force at section 1-1= p1 x A
2. Pressure force at section 2-2=p2 x A
where A= area of pipe, Frictional force F1 as shown in fig.
Resolving all forces in horizontal direction, we have
p1 x A - p2 x A – F1 = 0
(p1-p2) x A=F1=f’PLV2 or,
continued….
P1- P2=
𝑓′ 𝑃𝐿𝑉2
𝐴Fluid Mechanics- Aditya Deshpande 9

But we know p1-p2=𝝆 x g x hf
equating value of (p1-p2) we get
𝝆 x g x hf =
𝑓′ 𝑃𝐿𝑣2
𝐴
𝝆
ℎ𝑓 =
𝑓′
𝜌 𝑋 g
𝑃
𝐴
x LV2
But we know
𝑃
𝐴
=
4
𝑑
So hf =
𝑓′
𝜌 𝑋 g
4
𝑑
x LV2 =
𝑓′
𝜌𝑋 g
x
4𝐿𝑉2
𝑑
Put
𝑓′
𝜌
=f/2 We get f is known as friction factor.
continued….
hf=
4𝑓𝐿𝑉2
2g 𝑑Fluid Mechanics- Aditya Deshpande 10

As we know, ℎ𝑓 =
𝑓′
𝜌 𝑋 g
𝑃
𝐴
x LV2 from above equation
Now ratio of
𝐴
𝑃
is called hydraulic mean depth or hydraulic
radius and given by m.
So, m=
𝐴
𝑃
=
𝑑
4
Substituting
𝐴
𝑃
=m we get,
Hf=
𝑓′
𝜌 𝑋 g
1
𝑚
xLV2
Chezy’s formula for loss of
head due to friction in pipes

Chezy’s formula
continued…
 V2=
𝜌 𝑋 𝑔
𝑓′
x m x
ℎ𝑓
𝐿
 So that, V=
𝜌 𝑋 𝑔
𝑓′
x m x
ℎ𝑓
𝐿
=
𝜌 𝑋 𝑔
𝑓′
x =
𝑚 𝑥 ℎ𝑓
𝐿
 Let, is constant called Chezy's constant
 which is loss of head per unit length of pipe.
 Putting these in above equation,
ℎ𝑓
𝐿
= i
𝜌 𝑋 𝑔
𝑓′
= 𝑐
V=C 𝑚𝑖 ……………………Chezy’s formulaFluid Mechanics- Aditya Deshpande 12

Question on friction head loss
Q. Find the head loss due to friction in pipe of dia. 300 mm
and length 50m, through which water is flowing at velocity
of 3m/s using
1. Chezy’s formula
2. Darcy weisbach formula
Take C=60 and kinematic viscosity for water =0.01 stoke

Given:
 Diameter of pipe=d=300mm=0.3m
 Length of pipe=L=50m
 Velocity of flow=V=3m/s
 Chezy’s constant=C=60
 Kinematic viscosity = v = 0.01stoke
= 0.01 cm2/s
=0.01 x10-4m2/sec
To find:
head loss using
1. Chezy’s formula
2. Darcy Weisbach formula

Solution:-
 Chezy’s formula:
V=C 𝑚𝑖
M=d/4=0.30/4
=0.075m
3=60 0.075 𝑥 𝑖
i=(3/60)^2 x (0.075)-1 = 0.033
But i=
ℎ𝑓
𝐿
=
ℎ𝑓
50
Equating two values of i,
ℎ𝑓
50
=0.0333 so
hf=1.665mFluid Mechanics- Aditya Deshpande 15

 Darcy- Weisbach Formula
hf=
4𝑓𝐿𝑉2
2𝑔𝑑
Where f = coeff. of friction is function of Re
Re=V X d/v=(3 x 0.30)/(0.01x10-4)
Re=9 x105
Value of f =0.079/(Re1/4)
f =0.079/((9 x105)1/4)
f =0.00256
Hf=
4 𝑋 0.00256 𝑋 50 𝑋 9
2 𝑋 9.81 𝑋 0.3
Hf=0.7828mFluid Mechanics- Aditya Deshpande 16

Loss of energy due to change of velocity of flowing
fluid in magnitude or direction
This is due to
1. Sudden expansion of pipe
2. Sudden contraction of pipe
3. Bend in pipe
4. An obstruction in pipe
5. At entrance
6. At exit of pipe
7. Due to pipe fittings
Minor energy losses(head)

Loss of head due to Sudden
Expansion of pipe
Flow at Sudden Enlargement
1
1
2
2

Head loss due to sudden enlargement of pipe
he=
(𝑽 𝟏
−𝑽 𝟐
) 𝟐
𝟐𝒈
Where,
V1=velocity at section 1-1
Expansion of pipe

Contraction of pipe
Flow at sudden contraction
1
1
2
2

Head loss due to sudden contraction of pipe
hc=
𝟎.𝟓 𝑽 𝟐
𝟐
𝟐𝒈
Where,
Contraction of pipe

Loss of head due to bend in pipe
1. Sharp bends result in
separation downstream of
the bend.
2. The turbulence in the
separation zone causes
flow resistance.
3. Greater radius of bend
reduces flow resistance.
Bend in pipesFluid Mechanics- Aditya Deshpande 22

Head loss due to bend of pipe
hb=
𝑲𝑽 𝟐
𝟐𝒈
Where,
V=velocity of flow
K= coeff. Of bend depends on
1. Angle of bend,
2. Radius of curvature,
3. Dia. Of pipe
Loss of head due to bend in pipe

Loss of head due to obstruction
in pipe

Head loss due to obstruction
hobst =
𝑽 𝟐
𝟐𝒈
𝑨
𝑪𝒄 𝑨−𝒂
− 𝟏
𝟐
Where V=Velocity of liquid
a=Max. area of obstruction
A=Area of pipe
Cc= Coeff. of contraction
in pipe

Loss of head at Entrance of pipe
Fig. Loss of head at Entrance of pipeFluid Mechanics- Aditya Deshpande 26

Occurs when liquid enters a pipe connected to large tank or
reservoir.
Similar to loss of head due to sudden contraction and
depends upon form of entrance.
For sharp edge entrance this loss is more than rounded
entrance.
hi=
𝟎.𝟓𝑽 𝟐
𝟐𝒈
Where, V=Velocity of fluid in pipe
Loss of head at Entrance of pipe

Loss of head at Exit of pipe
Fig. Loss of head at Exit of pipeFluid Mechanics- Aditya Deshpande 28

Occurs due to velocity of liquid at outlet of pipe
Which is dissipated in the form of free jet if outlet of pipe
is free or it is lost in the tank or reservoir if outlet of pipe
is connected to outlet of reservoir
ho=
𝑽 𝟐
𝟐𝒈
where V= velo. At outlet of pipe
Loss of head at Exit of pipe

Occurs at various pipe fittings such as valves, couplings,
etc.
hf=
𝑲𝑽 𝟐
𝟐𝒈
Where,
V=Velocity of flow
K=Coefficient of pipe fitting
Loss of head in various Pipe
Fittings

Fluid Mechanics- Aditya Deshpande
Siphon
31

Use for flow of liquids through tubes ,but refers specifically
to a tube in an inverted U shape
Siphon causes a liquid to flow uphill, above the surface of
reservoir, without pumps, powered by fall of liquid as it
flows down tube under pull of gravity, and is discharged at
a level lower than surface of the reservoir it came from.
Siphon

1. To carry water from one reservoir to
another reservoir separated by hill or ridge.
2. To take out the liquid from tank which is
not having outlet.
3. To empty a channel not provided with any
outlet sluice.
Use of Syphon

1. Drag(FD): component of total force(FR) in direction
of motion
2. Lift(FL): component of total force (FR) in direction
perpendicular to direction of motion
Drag And Lift

Example of Drag:
Wind resistance to a moving car,
water resistance to torpedoes etc
Example of Lift:
Weight (in the case of an airplane in cruise)
Drag And Lift

Force On Stationary Body

Expression For Drag & Lift

Lift always acts perpendicular to drag.
Fdrag = 1/2(CdAv2)
Flift = 1/2(ClAv2)
Fair
resistance
Fdrag
Flift
continued…

continued…

The negative sign indicates pressure force is acting in
downward direction.
continued…

Problem 1 :
A circular disc 3 m in diameter is held normal to 26.4 m/s
wind velocity. What force is required to hold it at rest?
Assume density of air=1.2 kg/m3 and coefficient of drag of
disc = 1.1.
Problems and Answers

Given data
Diam. of disc= d =3m
Area=π/4 x d2 = 7.0685 m2
Velocity of wind = U= 26.4 m/s
Density of wind = ρ= 0.0012 gm/cc
= 0.0012 x 106 / 1000 = 1.2 Kg/m3
Coeff. of drag = CD= 1.1
To find FD

Solution:

Problem 2
A square plate of side 2m is moved in a stationary air of
density 1.2Kg/m3 with a velocity of 5Km/hr. If the
coefficient of drag and lift are 0.2 and 0.8 respectively,
determine :
1. The lift force
2. The drag force
3. The resultant force and
4. The power required ton keep the plate in motion

Given data:
Side of square plate = 2m Area = 2x2=4m2
Density of air = 1.25 kg/m3
Velocity of air = 40 Kmph
CD= 0.2
CL= 0.8
To find :
1. The lift force
2. The drag force
3. The resultant force and
4. The power required ton keep the plate in motion

Solution:

Problem 3
Find drag on a solid sphere 500mm in diameter, held
completely immersed in the flow of sea water. Velocity of flow
is 1.15m/s and R.D. of sea water is 1.025, Assume drag
coefficient= 0.60

GIVEN:
Diameter of sphere = d= 500mm = 0.5m
Area = A=π/4 x 0.52 =
Velocity of flow =1.15m/s
R.D. of water =1.025
CD = 0.6
TO FIND : Drag force

Solution:
Drag force=79.85 N

Problem 4
Find the drag force exerted by a parachute, 3 m diameter, at sea
level when the speed is 25 m/s. At what speed will the same
braking force be exerted by this parachute at elevation 2 km.
CD=1.2 which remains constant. Density of air at sea level is
1.225 kg/m3 and changes at rate of 0.109 kg/m3 per km.

Given data :
Diameter=3 m
speed is 25 m/s
CD=1.2
Density of air =1.225 kg/m3
To Find:
1. The drag force exerted by a parachute
2. The speed of braking force to be exerted by parachute at
elevation 2 km

Problem 5
A kite has a plan area of 0.25m2 and is flying in a wind with
velocity 25 kmph. The kite has a net weight of 1.2N. When,
the string is inclined at an angle, of 15o two the vertical, the
tension in string was found to be 3 N. Evaluate coefficient of
lift and drag. Take density of air as 1.15 kg/m3.

Given data:
A=0.25m2
Net weight of 1.2N
Angle=15o
Spring tension=3N
Density of air=1.15 kg/m3
Wind velocity=25 kmph
To find: coefficient of lift and drag

CL=0.591
Solution:

CD= 0.112

Skin Friction Drag, Pressure Drag

Streamline Body
1. Body surface coincides with streamlines when
the body is placed in flow.
2. Separation of flow will take place only at trailing
edge.
3. Boundary will start at leading edge and become
turbulent from laminar, but not separate upto
rear most part of body
4. Wake formation zone- very small and pressure
drag will be small.
Total Drag = Frictional Drag

Bluff Body
1. Body surface does not coincide with streamline
lines
2. Flow is separated much ahead of trailing edge of
surface
3. Very large wake formation zone.
(approx.) as frictional drag will be very small
(negligible)
Total Drag = Pressure Drag

Whenever fluid flows through say pipe or nozzle
channel power is transmitted.
Here we are interested in transmission of power
through pipes and nozzles.
Transmission of power

Depends upon
1. Total head available at end of pipe
2. Weight of liquid flowing through pipe
Transmission of power through Pipes

L= Length of pipe
D= diam. of pipe
H= Total head at inlet of pipe
V= Velocity of flow in pipe
hf=Loss of head due to friction
f= coeff. of frictionFluid Mechanics- Aditya Deshpande 68

Power transmitted at outlet of pipe
P=
𝝆 𝐱 𝒈
𝟏𝟎𝟎𝟎
×
π
𝟒
× d2 × V × (H-
𝟒𝒇𝒍𝑽 𝟐
𝟐𝒈𝒅
)
Efficiency of power transmission
ɳ =
𝐏𝐨𝐰𝐞𝐫 𝐚𝐭 𝐨𝐮𝐥𝐞𝐭 𝐨𝐟 𝐩𝐢𝐩𝐞
𝐩𝐨𝐰𝐞𝐫 𝐚𝐭 𝐢𝐧𝐥𝐞𝐭 𝐨𝐟 𝐩𝐢𝐩𝐞
=
𝐰𝐭.𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 𝐩𝐞𝐫 𝐬𝐞𝐜. 𝐱 𝐇𝐞𝐚𝐝 𝐚𝐭 𝐨𝐮𝐭𝐥𝐞𝐭
𝐰𝐭.𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 𝐩𝐞𝐫 𝐬𝐞𝐜. 𝐱 𝐇𝐞𝐚𝐝 𝐚𝐭 𝐢𝐧𝐥𝐞𝐭
=
𝐖 𝐱 (𝐇−𝐡𝐟)
𝐖 𝐱 𝐇
=
𝐇−𝐡𝐟
𝐇

Condition For Maximum Efficiency for
power transmission through pipes
H=3hf or, hf= H/3
The power transmitted through a pipe is maximum when the
loss of head due to friction is one third of total head at inlet.
Maximum efficiency of transmission of power
ɳ=
𝑯−𝒉𝒇
𝑯
But hf=H/3
ɳ=
𝐻−𝐻/3
𝐻
=1-
1
3
=
2
3
or, ɳ (max.)=66.7%Fluid Mechanics- Aditya Deshpande 70

L= Length of pipe D= Diam. Of pipe
H= Total head at inlet of pipe
V= Velocity of flow in pipe
hf=Loss Of head due to friction
f= Coeff. Of friction in pipe
d= Diameter of nozzle
v= Velo. Of flow at outlet of nozzle

Power transmitted at outlet of
pipe
P=(ρ 𝐱 g)/1000 𝐱
𝝅
𝟒
𝐱 d2 𝐱 V 𝐱 (H-
𝟒𝒇𝒍𝑽 𝟐
𝟐𝒈𝒅
)
Where, P is in kW

ɳ =
𝐏𝐨𝐰𝐞𝐫 𝐚𝐭 𝐨𝐮𝐥𝐞𝐭 𝐨𝐟 𝐧𝐨𝐳𝐳𝐥𝐞
𝐩𝐨𝐰𝐞𝐫 𝐚𝐭 𝐢𝐧𝐥𝐞𝐭 𝐨𝐟 𝐩𝐢𝐩𝐞
=
𝐰𝐭.𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 𝐩𝐞𝐫 𝐬𝐞𝐜. 𝐱 𝐇𝐞𝐚𝐝 𝐚𝐭 𝐨𝐮𝐭𝐥𝐞𝐭
𝐰𝐭.𝐨𝐟 𝐰𝐚𝐭𝐞𝐫 𝐩𝐞𝐫 𝐬𝐞𝐜. 𝐱 𝐇𝐞𝐚𝐝 𝐚𝐭 𝐢𝐧𝐥𝐞𝐭
=
𝟎.𝟓 𝛒 𝐚 𝐯 𝟑
𝛒 𝐱 𝐠 𝐱 𝐐 𝐱 𝐇
=
𝐯 𝟐
𝟐 𝐱 𝐠 𝐱 𝐇
=
𝟏
(𝟏 +
𝟒 𝒇 𝑳
𝑫
𝐱
𝒂 𝟐
𝑨 𝟐)
Efficiency of power transmission

Condition For Maximum Efficiency for
power transmission through nozzle
The power transmitted through a nozzle is maximum when
the loss of head due to friction is one third of total head at
inlet.
hf= H/3
To find diameter of nozzle for max. power transmission
through nozzle
𝑨
𝒂
=
𝟖 𝒇 𝑳
𝑫
𝟏
𝟐
Find ‘a’ and then find ’d’.

Boundary Layer Theory

A thin layer of fluid in the vicinity of boundary
whose velocity is affected due to viscous shear is
called as Boundary Layer
The region normal to the surface, in which velocity
gradient exists is known as Boundary Layer.
Boundary Layer

Boundary Layer Example

1. The distance from the leading edge
2. Viscosity of fluid
3. The free stream velocity
4. Density of fluid
Factors affecting the growth of
boundary layer

1. Calculation of friction drag of bodies in a flow.
2. Calculation of pressure drag formed because of
boundary layer separation.
3. Answers the important question of what shape a
body must have in order to avoid separation.
Importance of boundary layer
theory

Boundary layer separation

Boundary Layer

Boundary Layer and separation
gradientpressure
favorable,0


x
P
gradientno,0


x
P
0, adverse
pressure gradient
P
x



Flow accelerates
Flow decelerates
Constant flow
Flow reversal free shear
layer highly unstable
Separation pointFluid Mechanics- Aditya Deshpande 82

1. Large amount of energy is lost
2. Bodies are subjected to lateral vibrations
3. Pressure drag is increased and hence additional
resistance to movement of the body is developed
Effects of separation

Examples of boundary layer
separation

Unstreamlined Body Streamlined Body
Sphere with large wake sphere with small wake
Methods of avoiding separation

Making Slot, Suction, Blowing
Making slot Making suction
Making blowingFluid Mechanics- Aditya Deshpande 89

Practical example of boundary
separation

Thank You

Fluid Mechanics - Flows

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