Fluid Mechanics
Internal and External Flows
Part A
Friction factor, Pipe losses, Boundary Layer, Over external bodies, Flow Separation and control methods, Lift generation, Flow simulation methodology
Part B
Siphon, Transmission of power, Drag and lift, Characteristics of bodies
This is basic course in mechanical engineering both graduate and post graduate level.
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Aditya Deshpande
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2. 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
Fluid Mechanics- Aditya Deshpande 2
3. 1. Continuity equation
2. Bernoulli's theorem
3. Bernoulli's theorem for real fluids
PERQUISITE
Fluid Mechanics- Aditya Deshpande 3
4. 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
5. 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
Fluid Mechanics- Aditya Deshpande 5
6. 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)
Fluid Mechanics- Aditya Deshpande 6
7. 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
Fluid Mechanics- Aditya Deshpande 7
8. (Darcy weisbach equation)
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 ]
Fluid Mechanics- Aditya Deshpande 8
9. 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,
(Darcy weisbach equation)
continued….
P1- P2=
𝑓′ 𝑃𝐿𝑉2
𝐴Fluid Mechanics- Aditya Deshpande 9
10. 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.
(Darcy weisbach equation)
continued….
hf=
4𝑓𝐿𝑉2
2g 𝑑Fluid Mechanics- Aditya Deshpande 10
11. 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
Fluid Mechanics- Aditya Deshpande 11
12. 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
13. 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
Fluid Mechanics- Aditya Deshpande 13
14. 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
Fluid Mechanics- Aditya Deshpande 14
15. 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
16. 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
17. 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)
Fluid Mechanics- Aditya Deshpande 17
18. Loss of head due to Sudden
Expansion of pipe
Flow at Sudden Enlargement
1
1
2
2
Fluid Mechanics- Aditya Deshpande 18
19. Head loss due to sudden enlargement of pipe
he=
(𝑽 𝟏
−𝑽 𝟐
) 𝟐
𝟐𝒈
Where,
V1=velocity at section 1-1
V2=velocity at section 2-2
Loss of head due to Sudden
Expansion of pipe
Fluid Mechanics- Aditya Deshpande 19
20. Loss of head due to Sudden
Contraction of pipe
Flow at sudden contraction
1
1
2
2
Fluid Mechanics- Aditya Deshpande 20
21. Head loss due to sudden contraction of pipe
hc=
𝟎.𝟓 𝑽 𝟐
𝟐
𝟐𝒈
Where,
V2=velocity at section 2-2
Loss of head due to Sudden
Contraction of pipe
Fluid Mechanics- Aditya Deshpande 21
22. 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
23. 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
Fluid Mechanics- Aditya Deshpande 23
24. Loss of head due to obstruction
in pipe
Loss of head due to obstruction
Fluid Mechanics- Aditya Deshpande 24
25. 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
Loss of head due to obstruction
in pipe
Fluid Mechanics- Aditya Deshpande 25
26. Loss of head at Entrance of pipe
Fig. Loss of head at Entrance of pipeFluid Mechanics- Aditya Deshpande 26
27. 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
Fluid Mechanics- Aditya Deshpande 27
28. Loss of head at Exit of pipe
Fig. Loss of head at Exit of pipeFluid Mechanics- Aditya Deshpande 28
29. 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
Fluid Mechanics- Aditya Deshpande 29
30. 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 30
32. 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
Fluid Mechanics- Aditya Deshpande 32
34. 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
Fluid Mechanics- Aditya Deshpande 34
35. 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
Fluid Mechanics- Aditya Deshpande 35
36. 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
Fluid Mechanics- Aditya Deshpande 36
41. The negative sign indicates pressure force is acting in
downward direction.
Expression For Drag & Lift
continued…
Fluid Mechanics- Aditya Deshpande 41
43. 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
Fluid Mechanics- Aditya Deshpande 43
44. 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
Problems and Answers
Fluid Mechanics- Aditya Deshpande 44
46. 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
Problems and Answers
Fluid Mechanics- Aditya Deshpande 46
47. 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
Problems and Answers
Fluid Mechanics- Aditya Deshpande 47
50. 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
Problems and Answers
Fluid Mechanics- Aditya Deshpande 50
51. 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
Problems and Answers
Fluid Mechanics- Aditya Deshpande 51
53. 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.
Problems and Answers
Fluid Mechanics- Aditya Deshpande 53
54. 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
Problems and Answers
Fluid Mechanics- Aditya Deshpande 54
57. 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.
Problems and Answers
Fluid Mechanics- Aditya Deshpande 57
58. 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
Problems and Answers
Fluid Mechanics- Aditya Deshpande 58
62. 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
Fluid Mechanics- Aditya Deshpande 62
63. 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
Fluid Mechanics- Aditya Deshpande 63
66. 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
Fluid Mechanics- Aditya Deshpande 66
67. Depends upon
1. Total head available at end of pipe
2. Weight of liquid flowing through pipe
Transmission of power through Pipes
Fluid Mechanics- Aditya Deshpande 67
68. 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
70. 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
71. 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
Fluid Mechanics- Aditya Deshpande 71
72. 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
Fluid Mechanics- Aditya Deshpande 72
74. 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’.
Fluid Mechanics- Aditya Deshpande 74
76. 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
Fluid Mechanics- Aditya Deshpande 76
78. 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
Fluid Mechanics- Aditya Deshpande 78
79. 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
Fluid Mechanics- Aditya Deshpande 79
86. 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
Fluid Mechanics- Aditya Deshpande 86
88. Unstreamlined Body Streamlined Body
Sphere with large wake sphere with small wake
Methods of avoiding separation
Fluid Mechanics- Aditya Deshpande 88
89. Making Slot, Suction, Blowing
Making slot Making suction
Making blowingFluid Mechanics- Aditya Deshpande 89