Semester-III

Fluid Mechanics and

Machinery

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 The aim of this publication is to supply information taken from sources believed to be valid and
reliable. This is not an attempt to render any type of professional advice or analysis, nor is it to
be treated as such. While much care has been taken to ensure the veracity and currency of the
information presented within, neither the publisher nor its authors bear any responsibility for
any damage arising from inadvertent omissions, negligence or inaccuracies (typographical or
factual) that may have found their way into this book.

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 B.E./B.Tech. DEGREE EXAMINATION,
NOV/DEC 2013
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY
(Regulation 2008/2010)
(Common to Aeronautical Engineering, Automobile
Engineering, Production Engineering, Mechatronics
Engineering, Mechanical and Automation Engineering and
Fourth Semester Manufacturing Engineering, Industrial
Engineering and Industrial Engineering and Management)
(Common to PTCE 3213 – Fluid Mechanics and Machinery
for B.E. (Part-Time) Third Semester – Manufacturing
Engineering Regulation 2009)
Time: Three hours Maximum: 100 marks
Any missing data can be suitably assumed with justification.
Answer ALL questions
PART A (10 × 2 = 20 marks)

  1.  What is cavitation? What causes it?

  2.  Differentiate between kinematic viscosity of liquids and gases with

respect to pressure.

  3.  How does surface roughness affect the pressure drop in a pipe if the flow
is turbulent?

  4.  A piping system involves two pipes of different diameters (but of identical
length, material, and roughness) connected in parallel. How would you
compare the flow rates and pressure drops in these two pipes?

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 4 B.E./B.Tech. Question Papers

  5.  The excess pressure Dp inside a bubble is known to be a function of the

surface tension and the radius. By dimensional reasoning determine how
the excess, pressure will vary if we double the surface tension and the
radius.

  6.  Determine the dynamic pressure exerted by a flowing incompressible

fluid on an immersed object, assuming the pressure is a function of the
density and the velocity.

  7. What is meant by hydraulic efficiency of turbine?

  8. Define – manometric head and write its mathematical equation.

  9.  Draw the relationship between discharge and crank angle for a single
acting pump.

10.  Mention the significance of ‘back leakage’.

PART B (5 × 16 = 80 marks)

11.  (a)  (i)  Water enters the bottom of the cone in Fig. 1. at a uniformly
increasing average velocity V = Lt. If d is very small, derive an
analytic formula for the water surface rise h(t), assuming h = 0 at
t = 0.  (8)

Cone

θ θ
Diameter d
h(t)

V = Lt

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 Fluid Mechanics and Machinery (Nov/Dec 2013) 5

(ii)  According to Torricelli’s theorem, the velocity of a fluid draining

from a hole in a tank is V = (2gh)1/2, where h is the depth of water
above the hole. Let the hole have area A0 and the cylindrical tank
have bottom area Ab. Derive a formula for the time to drain the
tank from an initial depth h0. (8)
Or
(b)  (i)  A simple turbomachine is constructed from a disk with two
internal ducts which exit tangentially through square holes, as in
the Fig. 2. Water at 20°C enters the disk at the center, as shown.
The disk must drive, at 250 rev/min, a small device whose
retarding torque is 1.5 Nm. What is the proper mass flow of
water, in kg/s? (8)

2 cm

2 cm

32 cm Q

(ii)  A two-dimensional sheet of water, 10 cm thick and moving at 

7 m/s, strikes a fixed wall inclined at 20° with respect to the jet
direction. Assuming frictionless flow, find the normal force on
the wall per meter of depth, and the widths of the sheet deflected
upstream and downstream along the wall. (8)

12.  (a)  (i)  Consider turbulent flow (f = 0.184 Re–0.2) of a fluid through a
square channel with smooth surfaces. Now the mean velocity of
the fluid is doubled. Determine the change in the head loss of the
fluid. Assume the flow regime remains unchanged. What will be
the head loss for fully turbulent flow in a rough pipe? (10)

(ii)  Discuss on various minor losses in pipe flow. (6)

Or

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 6 B.E./B.Tech. Question Papers

(b)  (i)    Two tanks of fluid (r =998 kg/m3 and m= 0.001 kg/ms.) at 20°C
are connected by a capillary tube 4 mm in diameter and 3.5 m
long. The surface of tank 1 is 30 cm higher than the surface of
tank 2. Estimate the flow rate in m3/h. Is the flow laminar? For
what tube diameter will Reynolds number be 500? (10)

(ii)  Explain the concept of boundary layer in pipes for both laminar
and turbulent flows with neat sketches. (6)

13.  (a)  (i)  Derive an expression for the shear stress at the pipe wall when an
incompressible fluid flows through a pipe under pressure. Use
dimensional analysis with the following significant parameters:
pipe diameter D, flow velocity V, and viscosity m. and density r
of the fluid. (10)

(ii)  Use dimensional analysis and the MLT system to arrange the
following into a dimensionless number: L, r, m and s. (6)
Or
(b)  (i)  Consider viscous flow over a very small object. Analysis of the
equations of motion shows that the inertial terms are much
smaller than viscous and pressure terms. Fluid density drops out,
and these are called creeping flows. The only important
parameters are velocity U, viscosity m, and body length scale d.
For three-dimensional bodies, like spheres, creeping flow
analysis yields very good results. It is uncertain, however, if
creeping flow applies to two-dimensional bodies, such as
cylinders, since even though the diameter may be very small, the
length of the cylinder is infinite. Let us see if dimensional
analysis can help. (1) Apply the Pi theorem to two-dimensional
drag force F2-D as a function of the other parameters. Be careful:
two-dimensional drag has dimensions of force per unit length,
not simply force. (2) Is your analysis in part (1) physically
plausible? If not, explain why not. (3) It turns out that fluid
density r cannot be neglected in analysis of creeping flow over
two dimensional bodies. Repeat the dimensional analysis, this
time including r as a variable, and find the resulting non-
dimensional relation between the parameters in this problem. 
 (10)

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 Fluid Mechanics and Machinery (Nov/Dec 2013) 7

(ii)  When fluid in a pipe is accelerated linearly from rest, it begins as

laminar flow and then undergoes transition to turbulence at a
time ttr which depends upon the pipe diameter D, fluid acceleration
a, density r and viscosity m. Arrange this into a dimensionless
relation between ttr and D. (6)

14.  (a)  (i)  A Kaplan turbine delivers 10 MW under a head of 25 m. The hub
and tip diameters are 1.2 m and 3 m. Hydraulic and overall
efficiencies are 0.90 and 0.85. If both velocity triangles are right
angled triangles, determine the speed, guide blade outlet angle
and blade outlet angle. (10)

(ii)  Discuss – characteristics of centrifugal pump at constant speed.

 (6)
Or
(b)  (i)  A centrifugal pump with an impeller diameter of 0.4 m runs at
1450 rpm. The angle at outlet of the backward curved vane is 25°
with tangent. The flow velocity remains constant at 3 m/s. If the
manometric efficiency is 84% determine the fraction of the
kinetic energy at outlet recovered as static head. (10)

(ii)  Explain the load efficiency characteristics of hydraulic turbines

with a diagram. (6)

15.  (a)  Discuss on the following: Working of double acting pump, indicator
diagram, acceleration head, friction head. (16)
Or

(b)  With neat sketches, discuss about the rotary positive displacement
pump.

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 B.E./B.Tech. DEGREE EXAMINATION,
MAY/JUNE 2013
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY

Time: Three hours Maximum: 100 marks

Answer ALL questions
PART A (10 × 2 = 20 marks)
  1. Define relative or specific viscosity.

  2. What do you understand by impulse momentum equation?

  3. Mention general characteristic of laminar flow.

  4. What do you mean by flow through parallel pipes?

  5. Give the dimensions foe the following; surface tension and dynamic
viscosity.

  6. State Froude’s Model law.

  7. Define hydraulic efficiency and axial thrust of a roto-dynamic hydraulic

machine.

  8. Distinguish between Impulse & reaction turbine.

  9. What is negative slip in reciprocating pump? What are the causes for it?

10. What are the advantages of air vessel?

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 Fluid Mechanics and Machinery (May/June 2013) 9

PART B (5 × 16 = 80 marks)

11. (a)  (i)  A liquid is compressed in a cylinder having a volume of 0.012 m3

at a pressure of 690 N/Cm2. What should be the new pressure in
order to make its volume 0.0119 m3? Assume bulk modulus of
elasticity (K) for the liquid = 6.9 × 104 N/Cm2

(ii)  A 15 cm diameter vertical cylinder rotates concentrically inside

another cylinder of diameter 15.10 cm. Both cylinders are 25 cm
high. The space between the cylinders is filled with a liquid
whose viscosity is unknown. If a torque of 12 NM is required to
rotate the inner cylinder at 1000 rpm. Determine the viscosity of
the fluid.
Or

11. (b)  (i)  State Bernoulli’s theorem and assumption for steady flow of an
incompressible fluid.

(ii)  The water is flowing through a taper pipe of length 100 m having
diameter 600 mm at the upper end and 300 mm at the lower end,
the rate of 50 lps. The pipe has a slope of 1 in 30. Find the pressure
at the lower end, if the pressure at higher level is 19.62 N/cm2.

12. (a)  For a flow of viscous fluid flowing through a circular pipe under
laminar flow condition show that the velocity distribution is a
parabola and also show that average velocity is half of maximum
velocity.
Or

12. (b)  A horizontal pipe line 40 m long is connected to a water tank to one
end and discharge freely into atmosphere at other end for the first
25 m of its length from the tank, the pipe is 150 mm diameter and its
diameter is suddenly enlarged to 300 mm. The height of water level
in the tank is 8m above the centre of pipe considering all losses of
head which occur, determine the rate of flow, take f = 0.01 for both
section of pipe.

13. (a)  State Buckingham’s p theorem. The discharge of a centrifugal pump

(Q) is dependent on N (speed of pump), d (diameter of impeller),

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 10 B.E./B.Tech. Question Papers

g  (acceleration due to gravity), H (manometric head developed by

pump) and r and m (density and dynamic viscosity of the fluid).
Using the dimensional analysis and Buckingham’s p theorem, prove
 gH µ 
that Q = Nd 3 f  2 2 , 
 N d Nd 2 ρ 

Or

13. (b)  (i)  What are the similarities between model and prototype? Mention
the application of model testing.

(ii)  A spillway model is to be built to a geometrically similar scale of

1
across a flume of 600 mm width. The prototype is 15 m high
50
and maximum head on it is expected to be 1.5 m.
(i)  What height of model and what head on the model should be
used?
(ii)  If the flow over the model at a particular head is 12 lps, what
flow per metre length of the prototype is expected?
(iii)  If the negative pressure in model is 200 mm, what is negative
pressure in prototype? Is it practicable?

14. (a)  A francis turbine with an overall efficiency of 75% is required to

produce 148.25 kw power. It is working under a head of 7.62 m. The
peripheral velocity = 0.26 2 gh and the radial velocity of flow at
inlet is 0.96 2 gh . The wheel runs at 150 rpm and the hydraulic
losses in the turbine are 22% of available energy. Assume radial
discharge. Determine (i) guide blade angle (ii) wheel vane angle at
inlet (iii) Diameter of wheel at inlet (iv) Width of wheel at inlet.
Or

14.  (b)  The internal and external diameter of an impeller of a centrifugal

pump which is running at 100 rpm are 200 mm and 400 mm reply.
The discharge the pump is 0.04 m3/sec and velocity of flow is cost
and equal to 2 m/s. The diameter of section and delivery pipes 150
and 100 mm and section & delivery heads are 6m (abs) and 30 m
cabs of water. If outlet vane angle is 45° and power required to drive

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 Fluid Mechanics and Machinery (May/June 2013) 11

the pump is 16.186 kw. Determine (i) vane angle at inlet (ii) overall
efficiency of pump and (iii) manometric efficiency of pump.

15. (a)  The cylinder of a single acting reciprocating pump is 15 in diameter

and 30 m in stroke. The pump is running at 30 rpm and discharge
water to a height of 12 m. The diameter and length of delivery pipe
are 10 cm and 30 m. If a larger vessel is fitted in delivery pipe at a
distance of 2 m from centre of pump, find the pressure head in
cylinder.

(i)  at beginning of delivery and

(ii)  in middle of delivery. Take f = 0.01

Or

15. (b)  (i)  Explain in detail the working principle and construction of rotary
pumps with neat sketch.
Rotary pumps resemble like centrifugal pumps in appearance but
the working method differs. Uniform discharge and positive
displacement can be obtained by using rotary pumps.

(ii)  Calculate the work saved by filtering air vessel for a double
acting single cylinder reciprocating pump.

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 Solutions
PART A

  1.  It is the ratio of dynamic viscosity of fluid to dynamic viscosity of water
at 20°C
Dynamic viscosity of fluid
Relative or Sp. viscosity =
Dynamic viscosity of water at 20°C

  2.  Impulse momentum equation states that the impulse of fore acting
on a fluid mass is a short interval of time is required to the change of
momentum in direction of force.
  3. 
i.  There is a shear stress between fluid layers
ii.  ‘No slip’ at the boundary
iii.  The flow is rotational
iv.  Loss of energy is proportional to first power of velocity and viscosity.
  4.  When a main pipes, which again join together to form a single pipe and
continues as a main line. These pies are said to be pipes in parallel.
  5.  Surface tension = MT-2
Dynamic viscosity = ML-1T-1
  6.  Only gravitational force is more predetermining force. The law states
that “The froude number is same for both model and prototype”.
  7.  Hydraulic efficiency is defined as ratio of power developed by the runner
to power supplied by water jet.
Axial thrust is defined as the thrust caused by the pressure acting against
the cross-sectional area of an impeller. This thrust must be accomodated
by the pump’s thrust bearing.

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 Fluid Mechanics and Machinery (May/June 2013) 13

8.

Impulse turbine Reaction turbine

1.  All the potential energy is 1.  Only a portion of the fluid energy
converted to kinetic energy by is transformed to kinetic energy
nozzle
2.  Blades are only in action 2.  Blades are in action at all the time
when they are in front of
nozzle
3.  Water may be allowed to 3.  Water is admitted over the
enter a part or whole of wheel circumference of the wheel
circumference

  9.  Slip is difference between theoretical discharge and actual discharge.

When Qact may be higher Qthe and the slip is negative. It is possible when
the delivery pipe is short and pump runs at high speed.
10.  The air vessels, amount of work to be supplied to the reciprocating pump
is reduced.

PART B
11.  (a)  (i)  Given
Initial volume V1 = 0.012 m3
Initial pressure P1 = 690 N/Cm2
Final volume V2 = 0.0119 m3
Bulk modulus K= 6.9 × 104 N/Cm2
Solution
dp
Bulk modulus K =
( /v1 )
dv
P1 − P2
=
V1 −V2
V

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 14 B.E./B.Tech. Question Papers

690 − P2
6.9×10 4 =
0.012 − 0.0119
0.012
575 = 690 − P2

P2 = 690 − 575

P2 = 115 N/Cm 2

11.  (a)  (ii)  Given data:

Diameter of inside cylinder d = 15 cm = 0.15 m
Diameter of outer cylinder D = 15.10 cm = 0.151 m
Height of both cylinders l = 25 cm = 0.25 cm
Torque of inner cylinder = T = 12 N-m
Speed of cylinder N = 100 rpm
Solution:
p dN
Tangential velocity u =
60
p × 0.15×100
=
60

u = 0.785 m / sec = du

Area of contact of fluid with inner cylinder

A = p dl
= p × 0.15× 0.25
A = 0.117 m 2
Clearance between cylinders
D − d 0.151− 0.15
Dy = =
2 2
= 5×10−4 m
Shear stress
du 0.785
τ=µ = µ× = 1570 µ
dy 5×10−4

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 Fluid Mechanics and Machinery (May/June 2013) 15

Shear force
F = τ A = 1570 µ × 0.1178
= 184.946 m
Torque
d 0.15
T = F × = 184.946 m × = 12
2 2
m = 0.865 N-s/m 2

11.  (b)  (i)  Bernoulli’s theorem states that ‘In an ideal, incompressible fluid
when the flow is steady and continuous, the sum of pressure
energy, kinetic energy and potential energy are constant”.
P V2
+ + Z = costant
w 2g

Assumptions done:
  The liquid is ideal and incompressible
  The flow is steady and continuous
  The velocity is uniform over cross section
  The only forces acting on fluids are gravity and pressure forces
  All frictional losses are neglected

11.  (b)  (ii)  Given data:

D1 = 600 mm = 0.6 m
D2 = 300 mm = 0.3 m
Q = 50 litres/sec
P1 = 19.62 N/cm2 = 19.62 × 104 N/m2
Slope = 1:30
L = 100 m

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 16 B.E./B.Tech. Question Papers

Solution:

D1 = 0.6m

0
1:3
pe m
s l o
= 100 p1 = 19.62 × 104 N/m2
D2 = 0.3m gth
len

p 2 p
A1 = D1 = × 0.6 2 = 0.283 m 2
4 4
p 2 p
A2 = D2 = × 0.32 = 0.0706 m 2
4 4
50
Q = 50 litres/sec = = 0.05 m3 /sec
1000
Datum passes through centre of lower end.
Z2 = 0
Slope is 1:n 30
1 10
\  Z1 = ×100 = m = 3.333 m
30 3

Q 0.05
V1 = = = 0.176 m/sec
A1 0.283

Q 0.05
V2 = = = 0.707 m/sec
A2 0.0700
Using Bernoulli’s equation
P1 V12 P V2
+ + Z1 = 2 + 2 + Z 2
w 2g w 2g

19.62×10 4 0.1772 P (0.177) 2

× + 3.333 = 2 × +0
1000 2× 9.81 1000 2× 9.81
P2
199.53 = + 0.0254
1000

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 Fluid Mechanics and Machinery (May/June 2013) 17

P2
= 199.50
1000
P2 = 199.50 ×103 N/m 2

P2 = 19.95 N/m 2

12.  (a)  Solution:

τ*2πr∆x
R
r

∆x
 
pπr2  p + ∂p ∆x  πr 2
 ∂x 

Velocity Distribution:
Shear stress is given by
du
τ=µ
dy

y measured from the pipe wall.

\  y = R − r
dy = −dr

du du
τ=µ = −µ .
−dr dr
from element value.
∂p r
τ =−
∂x 2
du ∂p r
−m =−
dr ∂x 2
du 1 ∂p
= r
dr 2m ∂x

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 18 B.E./B.Tech. Question Papers

Integrating the above

1 ∂p 2
u= r +C
4m ∂x
where C is a constant and can be found by the boundary condition
r = R, u = 0
1 ∂p 2
\  0 = R +C
4m ∂x
1 ∂p 2
C =− R
4m ∂x
1 ∂p 2 1 ∂p 2
\  u = r − R
4m ∂x 4m ∂x

−1 ∂p 2
u= [R − r2 ]
4m ∂x

∂p
In the above equation m, and R are constant. \ u varies with r2
∂x
which is a equation of parabola. The velocity distribution across the
section of pipe is parabolic velocity distribution.

velocity distribution

(ii)  Average velocity is half of maximum velocity

When r = 0 maximum velocity is obtained
1 ∂p 2
\  U max = − R
4m ∂x
Average velocity u is obtained by dividing the discharge of fluid
across section of by area of pipe. Fluid flowing per second through
elementary ring
dq = Velocity at a radius × area of ring element

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 Fluid Mechanics and Machinery (May/June 2013) 19

= u × 2π rdr
1 ∂p 2
=− ( R − r 2 ) × 2π rdr
4µ ∂x

R R 1 ∂p 2
Q = ∫ dQ = ∫ − ( R − r 2 ) × 2π rdr
0 0 4µ ∂x

1  −∂p  R
× 2π ∫ ( R − r )rdr
2 2
= 
4 µ ∂x  0

1  −∂p  R
=  × 2π ∫ (R R 2 − r 3 )dr
4 µ ∂x  0

R
1  −∂p   R2 r 2 r 4 
=  × 2π  − 
4µ ∂x   2 4  0
1  −∂p   R4 R4 
=  × 2π  − 
4µ ∂x   2 4
1  −∂p   R 4 
=   2π
4µ  ∂x   4 

π  −∂p  4
Q =  R
8µ  ∂x 

Area velocity
π  −∂p  4
 R
Q 8µ  ∂x 
U = =
Area π R2

1  −∂p  2
U =  R .
8m  ∂x 

1 ∂p 2
− R
U 4 m ∂x
\  max = = 2.0 .
U 1  −∂p  2
 R
8m  ∂x 

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 20 B.E./B.Tech. Question Papers

U max
\   U=
2
\  Average velocity is half of maximum velocity.

12.  (b)  Given data:

Total length of pipe L = 40 m
Length of 1st pipe L1 = 25 m
Diameter of 1st pipe d1 = 150 mm = 0.15 m
Length of 2nd pipe L2 = 40 - 25 = 15 m
Diameter of 2nd pipe d2 = 300 mm = 0.3 m
Height of water H = 8 m
Coefficient of friction f = 0.01

v2
v1
L1= 25 m
d1= 0.15 m L2 = 15 m
d2 = 0.3 m

Using Bernoulli’s theorem

P1 V12 P V2
+ + Z1 = 2 + 2 + Z 2 + all losses
w 2g w 2g

P2 V2 2
0 +0 +8 = + + 0 + all losses Z1 = 8 Z 2 = 0
w 2g

V2 2
8= 0+ + hi + h f1 + he + h f2
2g

V12
hi = loss at entrance = 0.5
2g
4 fL1 V12
h f1 = head loss due to friction pipe 1 =
2 gd1

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 Fluid Mechanics and Machinery (May/June 2013) 21

(V1 −V2 ) 2
he = loss due to enlargement =
2 gd2

4 fL2 V2 2
h f2 = loss due to friction pipe 2 =
2 gd2
1 1 = A2V2
AV

p 2
d2 ×V2  d 2
A2V2
V1 = = 4 =  2  V2
A1 p 2  d1 
d1
4
2
 0.3 
=   ×V2 = 4V2
 0.15 

0.5V12 0.5×( 4V2 ) 2 8V2 2

\  hi = = =
2g 2g 2g

4 × 0.01× 25× 4V2 2

h f1 =
2× g × 0.15

V2 2
= 106.67
2g

(V1 −V2 ) ( 4V2 −V2 ) 2 9V2 2

he = = =
2g 2g 2g

4 fL2 V2 2 4 × 0.01×15×V2 2
h f1 = =
2 gd2 2× 0.3× g

V2 2
=2
2g
Sub all values in (1)
V2 2 8V2 2 V 2 9V 2 2V 2
8= + + 106.67 2 + 2 + 2 .
2g 2g 2g 2g 2g

V2 2
= 126.67
2g

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 22 B.E./B.Tech. Question Papers

8× 2× 9.81
\  V2 2 =
126.67

V2 = 1.113 m/sec
Rate of flow
p
Q = A2V2 = ×(0.3) 2 ×1.113
4
= 0.07867 m3 /sec

Q = 0.07867 m3 /sec

13.  (a)  Solution:

Buckingham’s p theorem states that if there are n variables in a
dimensionally homogeneous equation and if these variables contain
m-dimensions, then they are grouped into (n - m) dimensionless
independent p term.
Discharge Q is a function of N, D, g, H, m, r.
P = f ( ω , V , D , µ, ρ , C )  (1)

f1 ( P , ω, V , D , µ, ρ, C ) = 0  (2)
No of variables
Q = L3T-1  N = T-1  D = L  g = LT-2 
H = L  m = ML-1T-1  r = ML-3
No. of fundamental dimensions
m=3
\  No. of p terms = n - m = 7 - 3 = 4.
\  equation (2) can be written as
f1(p, p2, p3, p4) = 0 (3)
a1 b1 c1
\  π1 = D N ρ ⋅Q
π2 = D a2 N b2 ρ c2 ⋅ g
π3 = D a3 N b3 ρ c3 ⋅ H
π4 = D a4 N b4 ρ c4 ⋅ µ

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 Fluid Mechanics and Machinery (May/June 2013) 23

p1 term
π1 = D a1 N b1 ρ c1 ⋅ Q
M 0 L0T 0 = La1 (T −1 )b1 ( ML−3 )c1 ( L3T −1 )
= M c1 + La1 −3c1 +3T −b1 −1
equating powers
0 = c1 , a1 − 3c1 + 3 = 0, − b1 −1 = 0

\  c1 = 0 , b1 = −1 , a1 = 3c1 − 3

a1 = −3

π1 = D −3 N −1ρ 0 ⋅ Q

Q
p1 =
ND 3
p2 term
π2 = D a2 N b2 ρ c2 ⋅ g

M 0 L0T 0 = La2 (T −1 )b2 ( ML−3 )c2 ( LT −2 )
For M : 0 = c2  For L : 0 = a2 - 3c2 + 1
For T: 0 = -b2 - 2
\  c2 = 0   b2 = −2   a2 = 3c2 −1

a2 = −1

\  π2 = D −1 N −2 ρ 0 ⋅ g

g
p2 =
DN 2
p3 term
π3 = D a3 N b3 ρ c3 ⋅ H

M 0 L0T 0 = La3 (T −1 )b3 ( ML−3 )c3 L

For M : 0 = c3  For L : 0 = a3 - 3c3 + 1

a3 = +3c3 - 1

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 24 B.E./B.Tech. Question Papers

For T: 0 = -b3
\  c3 = 0   b3 = 0   a3 = −1

\  π3 = D −1 N 0 ρ 0 ⋅ H

H
p3 =
D

p4 term
π4 = D a4 N b4 ρ c4 ⋅ µ

M 0 L0T 0 = La4 (T −1 )b4 ( ML−3 )c4 ML−1T −1
equating power on both sides
For M : 0 = c4 + 1 For L : 0 = a4 - 3c4 - 1
a4 = 3c4 + 1
= -3 + 1
For T: 0 = -b4 - 1
\  c4 = −1   b4 = −1   a4 = −2

\  π4 = D −2 N −1 ρ−1µ

µ
π4 =
D2 N ρ

 Q g H µ 
\  f1 =  , , , =0
 ND DN D ρ ND 2 
3 2

p1 × p2 dimensionless.
 Q gH µ 
\  f1 =  , 2 2, =0
 ND N D ρ ND 2 
3

 gH µ 
Q = ND 3φ  2 2 , 
 N D ρ ND 2 

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 Fluid Mechanics and Machinery (May/June 2013) 25

13.  (b)  (i)  The similarities between model & prototype are
(i) Geometric similarity
A model and its prototype are geometrically similar, if the
ratio of corresponding length dimensions are equal.
Lp bp Dp
= = = Lr = length scale ratio
Lm bm Dm
Ap L p × bp
Area scale ratio = = = Lr 2 and
Am Lm × bm
3 3 3
Vp  L p   bp   D p 
Volume scale ratio =   =   =   = Lr 3 .
Vm  Lm   bm   Dm 

(ii) Kinematic similarity

It is the similarity of motion. If the acceleration ratios and
acceleration vectors points are same in the same direction,
the two flows are kinematically similar.
Tp
Time scale ratio ⇒ Tr =
Tm
Lp / Tp Lr
Velocity scale ratio, Cr = =
Lm / Tm Tr

Lp / Tp 2 Lr
Acceleration scale ratio ar = 2
=
Lrm / Tm Tr 2

L13 / T p Lr 3
Discharge scale ratio Qr = =
Lm 3 / Tm Tr

(iii) Dynamic similarity

It is similarity of forces. The flows in the model and prototype
are of dynamic similar.
( Fi ) p ( Fv ) p ( Fg ) p
= = = Fr = force ratio
( Fi ) m ( Fv ) m ( Fg ) m

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 26 B.E./B.Tech. Question Papers

Applications of the model testing

i. Civil engineering structures such as dams, weirs, canals
ii. Design of harbour, ships and submarines.
iii. Aeroplanes rockets and machines etc.

13.  (b)  (ii)  Solution:

Given data:
Scale ratio for length Lr = 50
Width of model Bm = 60 mm = 0.6 m
Flow over model Qm = 12 litres/sec
Pressure in model hm = -20 cm of water
Height of prototype Hp = 15 m
Head on prototype Hp* = 1.5 m

Height and Head of model

Let the height of model = Hm
head on model Hm*
Hp Hp *
Linear scale ratio Lr = = = 50
Hm Hm *

Hp 15
Height of model ⇒ H m = = = 0.3 m
50 50
Hp * 1.5
Head of model H m * = = = 0.03 m
50 50
Width of prototype B p = Lr × Bm = 50 × 0.6 = 30 m

Discharge ratio
It is given by the equation
Qp
= Lr 2.5 = (50) 2.5 = 17677.67
Qm
Q p = Qm ×17677.67 = 12×17677.67

Q p = 212132.04 lit/sec

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 Fluid Mechanics and Machinery (May/June 2013) 27

Discharge per meter length of prototype.

Qp 212132.04
= =
Length of protype Width of protype

212132.04
=
30

Qp = 7071.078 litres/sec
L
Negative pressure head in prototype
hp = Lr × hm
= 50 ×(−20)
= −1000 cm

hp = −10.0 m

14.  (a)  Given data:

Overall efficiency h0 = 80% = 0.8
Power developed P = 148.25 kw
Head H = 7.62 m
Speed = N = 150 rpm.
Hydraulic losses = 22% of available energy.
VW2 = 0   V f1 = V f2 [Radial discharge]

Solution:
Peripheral velocity of wheel = 0.26 2 gh
u1 = 0.26 2 gh
= 0.26 2× 9.81× 7.62
u1 = 3.18 m/s
Radial velocity at inlet = 0.96 2 gh
V f1 = 0.96 2× 9.81× 7.62
V f1 = 11.74 m/s

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 28 B.E./B.Tech. Question Papers

Hydraulic efficiency
Head at inlet − Hydraulic losses
hn =
Head at inlet
H − 0.22H 0.78H
= = = 78%
H H
Vw1 µ1
ηn =
gH
Vw1 ×3.18
0.78 =
9.81× 7.62
Vw1 = 18.34 m/s

u2

φ
v2 = vf = vf
1 2
vr2

v1
vr1 vf1
α θ

u1
vw1

Guide blade angle (µ)

V f1 11.74
tan α = =
Vω1 − µ1 18.34

∝= 32.62°

Vane angle at inlet (q)

V f1 11.74
tan θ = = = 0.774
Vω1 − µ1 18.34 − 3.18

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 Fluid Mechanics and Machinery (May/June 2013) 29

q = 37.75°
π D1 N
Velocity µ1 =
60
p × D1 ×150
3.18 =
60

D1 = 0.404 m = 405 mm

Overall efficiency
P
h0 =
wQH

148.25
0.8 =
9.81× Q × 7.62
Q = 2.48 m3 /sec
Q = p D1 B1V f1
2.48 = p × 0.405× B1 ×11.74
B1 = 0.16602 cm
B1 = 166.02 mm

14.  (b)  Given data:

Speed N = 1000 rpm
Internal diameter D1 = 200 mm = 0.2 m
External diameter D2 = 400 mm = 0.4 m
Discharge Q = 0.04 m3/sec
Velocity of flow V f1 =V f2 = 2m/sec
Suction pipe diameter Ds = 1500 mm = 0.15 m
Delivery pipe diameter Dd = 100 mm = 0.10 m
Suction head hs = 6 m cabs
Delivery head hd = 30 m cabs
Outlet vane angle f = 45°
Power required P = 16.186 kw

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 30 B.E./B.Tech. Question Papers

u2
vw
2

β φ
v2 vf2
vr2

vr
1
vf1 = v1
θ
u1

(i) Vane angle of impeller at inlet (q)

V f1 2
tan θ = =
µ1 µ1

π D1 N π × 0.2×100 r
µ1 = = = 10.47 m/sec
60 60
2
tan q = = 0.191
10.47

q = 10.81°
(ii) Over all efficiency of pump (h0)
WH m / 75
h0 =
P
WQH m 1000 × 0.04 × H m
= =
75 P 75×16.186

P V 2  P V 2 
H m =  0 + 0 + Z 0  −  i + i + Zi 
 w 2 g   w 2 g 

Z 0 = Zi

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 Fluid Mechanics and Machinery (May/June 2013) 31

 V 2  V2
H m = 30 + d  − 60 + s 
 2 g   2 g 

Discharge 0.4 0.04

Vd = = = = 5.09 m/sec
Area of delivery pipe p 2 p
( Dd ) (0.1) 2
4 4
0.04 0.4 0.04
Vs = = = = 2.26 m/sec
Area of suction p 2 p 2
( Ds ) × 0.15
4 4

 (5.09) 2   ( 2.26) 2 
H m = 30 +  − 6 + 
 2× 9.81  2× 9.81
= 31.32 − 6.26 = 25.06 m.

1000 × 0.04 × 25.06

\  h0 =
75×16.186

h0 = 82.57%

Manometric efficiency of pump (hman)

π D2 N π × 0.04 ×1000
µ2 = =
60 60
m2 = 20.94 m/sec
From outlet triangle
V f2 2
tan φ = =
µ2 −Vω2 20.94 ×Vω2

2
20.94 −Vw2 =
tan 45

Vw2 = 20.94 − 2

Vw2 = 18.94 m/sec

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 32 B.E./B.Tech. Question Papers

gH m 9.81× 25.06
η man = =
Vω2 µ2 18.94 × 20.94

h man = 61.98%

15.  (a)  Given data:

Diameter D = 15 cm = 0.15 m
Stroke length = L = 30 cm = 0.3 m
L 0.3
Tank radius r = = = 0.15 m
2 2
Speed N = 30 rpm
Delivery head hd = 12 m
Diameter of delivery pipe dd = 10 cm = 0.1 m
Length of delivery pipe l = 30 m
Length of the air vessel from centre of cylinder
ld′ = 2 m
Length of delivery pipe above air vessel
ld = l − ld′ = 30 − 2 = 28 m

p 2 p
Area A = D = × 0.152 = 0.01767 m 2
4 4
Angular speed
2π N 2π ×30
ω= = = π rad/sec
60 60
Area of delivery pipe
p 2 p
ad = dd = × 0.12 = 0.007854.
4 4
Coefficient of friction f = 0.01

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 Fluid Mechanics and Machinery (May/June 2013) 33

Solution:
Pressure head in cylinder at beginning of delivery stroke is
2 2
ld′ A 4 fld  A ωr  1  A ωr 
= hd + + ω2r + × ×  +  × 
g ad 
dd 2 g  ad π  2 g  ad π 
2
2 0.01767 4 × 0.01× 28  0.01767 π × 0.15 
= 12 + + × π 2 × 0.15×  × 
9.81 0.007854 0.1× 2× 9.81  0.007854 π 
2
1  0.01767 π × 0.15 
+  × 
2× 9.81  0.007854 π 
= 12 + 0.6709 + 0.065 + 0.0058
= 12.75 m

(ii)  Pressure head in cylinder in middle of the delivery stroke.

2 2 2
4 f ld′  A  4 fld  A ωr  1  A ωr 
= hd + × ωr  + × ×  +  × 
dd × 2 g  ad  dd 2 g  ad π  2 g  ad π 
2
4 × 0.01× 2  0.01767 
= 12 +  × π × 0.15 + 0.065 + 0.0058
0.1× 2× 9.81 0.007854 
= 12 + 0.0458 + 0.065 + 0.0058
= 12.116 m

15.  (b)  (i)  The various types are

  External Gear Pump
  Internal Gear Pump
  Lobe Pump
  Vane pump

External Gear Pump

It has two identical intermeshing spur wheels A and B. One wheel
is connected to driving shaft and other one is driven. The liquid
is entrapped between teeth and allowed to discharge into delivery
pipe when the gear rotates.
Thee rotating gears itself set up sufficient pressure to force the
liquid.

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 34 B.E./B.Tech. Question Papers

Internal Gear Pump

Two intermeshing wheels are internally meshed. A crescent
shaped partition is made between these two wheels. This space
acts as a seal between suction and discharge. During rotation the
space between inner and outer wheel increases and liquid fills the
space. Due to continuous rotation the fluid is pushed to delivery
end.

Lobe Pump
This pump resembles like a gear pump in action but not in
appearance. The wheels inside the casing have two or more lobes.
The liquid is entrapped & flows to the delivery end when lobe
rotates. The lobe itself build up sufficient pressure to force the
liquid.

Vane Pump
A disc is eccentrically rotating inside the pump casing. Slots
containing vanes are provided in the disc. The vanes are moved
against the casing due to centrifugal force. Due to continuous
rotations the liquid is entrapped and forced to the delivery end
with sufficient pressure.

Internal gear pump

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 Fluid Mechanics and Machinery (May/June 2013) 35

External gear pump

Vane pump

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 36 B.E./B.Tech. Question Papers

Lobe pumps

15.  (b)  (ii)  Work done against friction without air vessels
The work done is calculated by assuming no loss of head due to
friction at beginning & end of stroke friction head is maximum
in middle of the stroke.
Loss of head due to friction
4 flV 2
Hf =  (1)
2 gd
A
V= wr ; sub in (1)
a
 A 2
4 fl  wr 
a 
Hf = .
2 gd
Area under profile due to friction is parabola
2
= × base × height
3
2
Work done = ×Stroke × Head due to friction
3
work done per stroke

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 Fluid Mechanics and Machinery (May/June 2013) 37

2
2 4 fl  A 
W1 = ×  wr 
3 2 gd  a 
Power required
2
w AN w AN 2 4 fl  A 
P = W1 × = × ×  wr 
60 60 3 2 gd  a 
Work done against friction with air vessel
Velocity of piston with air vessel due to S.H.M
2
4 flV 2 4 fl  A ωr 
Hf = =  
2 gd 2 gd  a π 

2
4 fl  A ωr 
Work done = Stroke × h f = h f ×  
2 gd  a π 
Work saved against friction by filling air vessel
W1 −W2  W 
Work saved = ×100 = 1− 2 ×100
W1  W 
1

 4 fl  Aωr 
2 
 × 
 
 aπ   L

2 gd
= 1− ×100
 2  
2 
 × 2×
4 fl  A ω r  
 3 2 gd  aπ  
 
 3 
= 1− 2 ×100
 2π 

Work saved = 84.8%

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 B.E./B.Tech. DEGREE EXAMINATION,
NOV/DEC 2012
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY

Time: Three hours Maximum: 100 marks

Answer ALL questions
PART A (10 × 2 = 20 marks)

  1. Define Newton’s law of viscosity

  2. Write the continuity equation

  3. Write the Hagen Poiseville equation for laminar flow

  4. What is the use of Moody’s Diagram?

  5. Define Reynold’s Number

  6. State the Buckingham-p Theorem

  7. What is meant by hydraulic efficiency of turbine?

  8. Define flow ratio of reaction radial flow turbine

  9. Mention the main parts of centrifugal pump

10. Define slip of Reciprocating pump

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 39

PART B (5 × 16 = 80 marks)

11. (a)  The space between two square slat parallel plate is filled with oil.
Each side of the plate is 600mm. The thickness of the oil films is
12.5mm. The upper plate, which moves at 2.5m/s requires a force of
98.1N to maintain the speed. Determine.
(i) The dynamic viscosity of the oil in poise
(ii) The kinematics viscosity of the oil in strokes if the specific
gravity of the oil is 0.95.
Or

11. (b)  Derive the Euler’s equation of the motion and deduce the expression
to Bernoulli’s equations

12. (a)  (i)  A plate of 600mm length and 400mm wide is immersed in a fluid
of specific gravity 0.9 and kinematic viscosity of oil is 
1 × 10-4 m2/s, the fluid is moving with the velocity of 6m/n.
Determine.
(1) Boundary layer thickness; (2) Shear stress at the end of the
plate and (3) Drag force on the sides of the plate.

12. (a)  (ii)  Derive Chezy’s formula for loss of head due to friction in pipes.
Or

12. (b)  A 150 mm diameter pipe reduces in diameter abruptly to 100 mm

diameter. If the pipe carries water at 30 liters per second. Calculate
the pressure loss across the construction. Take coefficient of
contraction as 0.6.

13. (a)  Derive on the basic of dimensional analysis suitable parameters to

represent the thrust developed by a propeller. Assume that the thrust
P depends upon the angular velocity w, speed of advance V, diameter
D, dynamics viscosity m, mass density r, elasticity of the fluid
medium which can be denoted by the speed of sound in the 
medium C.
Or

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 40 B.E./B.Tech. Question Papers

13. (b) A pipe of diameter 1.5m is required to transport an oil of specific

gravity 0.9 and viscosity 3 × 10-2 poise at the rate of 3000 lps. Tests
were conducted on 15 cm diameter pipe using water at 20°C. Find the
velocity and rate of flow in the model. Viscosity of water at 
20°C = 0.01 poise.

14. (a)  The following data is given for a francis turbine Net head H = 60 m,
Speed N = 700 rpm Shaft Power = 294.3 Kw, overall
efficiency = 84%, hydraulic efficiency = 93%, flow ratio = 0.20,
breadth ratio n = 0.1, outer diameter of runner = 2 × inner diameter
of runner, velocity of flow is constant at inlet & outlet and discharge
is radial at outlet. Determine.

(i) Guide blade angle (ii) Runner vane angle at inlet and outlet 
(iii) Diameter of runner at inlet & outlet (iv) width of wheel at outlet.
Or

14. (b)  A pelton wheel has a mean bucket speed of 10 m/s with a jet of water
flowing at the rate of 0.7m3/sec under a head of 30m. If the buckets
deflect the jet through an angle of 160°. Calculate the power given by
water to the runner and hydraulic efficiency of turbine. Assume
coefficient of velocity of 0.98.

15. (a)  The cylinder bore diameter and stroke of a single acting reciprocating
pump are 150mm and 300mm respectively. The pump runs at 50 rpm
and lifts water to a height of 25 m. The delivery pipe is 22 m long,
100 mm diameter. Find the theoretical discharge and theoretical
power required to run the pump. If the actual discharge is 
4.2 litres/sec. Find the percentage of slip.
Or

15. (b)  The internal and external diameters of the impeller of a centrifugal
pump are 200 mm and 400 mm respectively. The pump is running at
1200 rpm the vane angle of impeller at inlet & outlet are 20° & 30°.
The water enters the impeller radially and velocity of flow is constant.
Determine the work done by the impeller per unit weight of water.

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 Solutions
PART A
1.  It states that shear force F acting between two layers of fluid is proportional
to the difference in their velocities and Area of the plate and inversely
proportional to the distance between them.

2.  C1A1V1 = C2A2V2

where C-density
A-Area
V-Velocity.
32µUL 128µQL
3.  P1 − P2 = =
D2 λD 4
4.  Moody’s Diagram is used to determine friction factor for any turbulent
flow problem.

5.  It is defined as ratio of the Inertia force to the viscous force of a flowing
fluid, denoted by Re
Inertia force ev 2 L2 eVL Lv
Re = = = or
viscous force µLV µ γ

6.  It states that if there are p variables in a dimensionally homogeneous

equation and if these variables contain ‘m’ fundamental dimension, then
they are grouped into (n - m) dimensionless independent p-terms.
7.  It is defined as the ratio of power developed by the runner to the power
supplied by the water jet.
Power developed by runner
ηn =
Power supplied by the water jet

8.  The ratio of the velocity of flow at inlet (vf1) to the theoretical velocity
vf1
( )
2gh is known as flow ratio k f =
2 gh
. The value varies from 0.15
to 0.3.
9.  The centrifugal pump consists of following parts Impeller, Casing, Suction
pipe, Strainer Foot valve, Delivery pipe and delivery value.

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 42 B.E./B.Tech. Question Papers

10.  The difference between the theoretical discharge and actual discharge is
called slip of the pump.
Slip = Qth - Qact

PART B
11.  (a)  Given data:
Thickness of oil film dy = 12.5 mm
Side of plate = 6.00 mm
Velocity du = 2.5 m/s
Force on the plate = 98.1N
Specific gravity = 0.95.

dy

Solution:
Area of the plate = 600 × 600
= 360000
=0.36 m2.
Shear stress

F
τ=
A

98.1
=
0.36

τ = 272.5 N/m 2

du
τ=µ
dy

2.5
272.5 = µ
12.5×10−3

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 43

µ = 1.3625 N-s/m 2

µ = 1.3625 Poise
Density of oil = Sp. gravity × Density of water
= 0.95 × 1000 = 950 kg/m3.
Kinematic viscosity
µ
γ=
C

1.3625
=
950

γ = 1.43×10−3 m 2 / sec

11.  (b)  A steady flow of an ideal fluid along a streamline with a small element
of the flowing fluid LM of cross section dA and length dS as shown.
Direction of flow
stream line
ds
S
M
θ ds θ dz
ptdp
P dW
dALL dw

Let,
P = Pressure on the element at A.
P + dp = Pressure on element at M and
V = velocity of fluid element.
We know that the net force acting on the fluid element in the direction
of flow
= P.dA - (p + dp)dA.
= -dPdA.
we also know that the weight of the fluid element.
dw = rg×dA×ds
From the geometry, we find that component of weight of the fluid
element in the direction of flow.

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 44 B.E./B.Tech. Question Papers

= −ρg ⋅ dA ⋅ ds ⋅ cos θ
 dz 
= −ρg ⋅ dA ⋅ ds ⋅  
 ds 
= −ρg ⋅ dA ⋅ dz
The resultant force on fluid element in direction of S
= - dPdA - r×g×dA×dz
\  Mass of the fluid element
= r×dA×ds
The acceleration of fluid element
dv dv ds dv
a= = × = v⋅
dt ds dt ds
According to Newton’s second law of motion
Force = Mass × Acceleration
dv
(−dpdA) − (ρgdAdz ) = ρdAds × v ×
ds
Dividing both sides by r×dA.
dp
− − gdz = v × dv
ρ

dp
+ vdv + gdz = 0
ρ

The above equation is known as Euler’s equation of motion.

Bernoulli’s Equation from Euler’s Equation
dp
+ vdv + gdz = 0
ρ
Integrating the above equation
1
ρ∫
dp + ∫ gdz + ∫ v ⋅ dv = constant .

p v2
+ gz + = constant
ρ 2

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 45

dividing by g.
p v2
+z+ = constant
ρg 2g

p v2
+ + z = constant
w 2g

p1 v 21 p v2
+ + z1 = 2 + 2 + z2
w 2g w 2g

The above equation is called as Bernoulli’s equation.

12.  (a)  (i)  Given data:
L = Length of plate = 600 mm
W = Width of plate = 400 mm
S = specific gravity = 0.9
g = Kinematic viscosity = 1 × 10-4 m2/s
V = Velocity = 6 m/s
Solution:
Regnold’s number
UL 6 ×600 ×10−3
Re = =
γ 1×10−4
= 36 ×103 < 5×105

Leminer Flow
(i)  Boundary layer Thickness
5L
δ lem =
Re
5×600 ×10−3
=
36 ×103
= 0.0158m

δ lem = 15.8 mm

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 46 B.E./B.Tech. Question Papers

  (ii) Shear stress at the end of the plate

du
τ = µ y=L
dy
Dynamic viscosity
µ = γe
Density
r = sp.gravity × Density of water.
= 0.9 × 1000

ρ = 900 kg/m3

µ = 1×10−4 × 900

µ = 0.009 N-s/m 2

du
τ = µ y=L
dy
6
= 0.09× [du = 0 dy = δ lem ]
0.0158

τ = 34.17 N/m 2

(iii) Drag force on one plate.

FD = τ × B × L

= 34 ⋅17×600 ×10−3 × 400 ×10−3

FD = 8.2008 N.

12.  (a)  (ii)  Major loss of energy can be found by Chezy’s formula
Chezy’s formula is given as
V = C mi
V = Velocity of flow
C = Chezy’s coefficient
M = Hydraulic mean depth.

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 47

Area of the Pipe

=
Wetted Perimeter
π 2
D
A 4 D
= = =
P πD 4
I = loss of head per unit length of pipe.
hf D hf
= \ V =C
L 4 L
4 LV 2
hf =
C⋅D
12. (b) Given data:
Q = 30 LPS
D1 = 150 mm
30 3
D2 = 100 mm   = m /sec
1000
Cc = 0.6
Q = 0.03 m3 /sec
Solution:
Q = AV
1 1 = A2V2

π
A2 = ×(100 ×10−3 ) 2
4
A2 = 7.853mm 2
Q = A2V2
0.03 = 7.852V2
V2 = 3.81m / sec

Q = AV
1 1

π
A1 = ×(150 ×10−3 ) 2
4
A2 = 0.0176 m 2
Q = AV
1 1

0.03 = 0.0176 ×V1

V1 = 1.704 m / sec

Head loss due to sudden contraction

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 48 B.E./B.Tech. Question Papers

2
V 2 2  1 
hc =  −1
2 g  Cc 

2
(3.81) 2  1 
=  −1
2× 9.81  0.6 

hc = 0.3288
Applying Bernoulli’s theorem
P1 V 21 P2 V 2 2
+ = + + hc
w 2g w 2g

P1 (1.704) 2 2 P (3.81) 2 2
+ = 2 + + 0.3288
9810 2× 9.81 9810 2× 9.81
P1 P
− 2 = 1.0686 − 0.148
9810 9810
= 0.9206

( P1 − P2 ) = 9.031 KN/m 2

13.  (a)  Thrust P is a function of: ω, V , D, µ, ρ, C

P = f (ω, V , D, µ, ρ, C )

f1 ( P , ω, V , D, µ, ρ, C ) = 0
Total number of variables, n = 7
The dimensions are
Thrust P = MLT −2
ω = T −1

V = T −1
D=L
m = ML-1T-1
r = ML-3
C = LT-1
number of fundamental dimensions
m=3

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 49

\ Number of p terms = n - m = 7 - 3 = 4.
f1(p, p2, p3, p4) = 0
D, V, r can be taken as repeating variables.
\ π1 = D a1 V b1 ρ c1 ⋅ P

π2 = D a2 V b2 ρ c2 ⋅ ω
π3 = D a3 V b3 ρ c3 ⋅ µ
π4 = D a4 V b4 ρ c4 ⋅ C
p1 term
π1 = D a1 V b1 ρ c1 ⋅ P
M 0 L0T 0 = La1 ( LT −1 )b1 ( ML−3 )c1 ⋅ MLT −2
= M c1 +1 + La1 +b1 −3c1 +1T −b1 −3c1 −2
equating M, L, T power on both sides
0 = c1 + 1
0 = a1 + b1 − 3c1 + 1
0 = −b1 − 3c1 − 2

\ c1 = −1, b1 = −2, a1 + b1 − 3c1 + 1 = 0

a1 − 2 + 3 + 1 = 0
a1 = −2

π1 = D −2V −2 ρ−1 ⋅ P

P
π1 =
D 2V 2 ρ

p2 term
π2 = D a2 V b2 ρ c2 ⋅ ω
M 0 L0T 0 = La2 ( LT −1 )b2 ( ML−3 )c2 ⋅ T −1
= M c2 + La2 +b2 −3c2 T −b2 −1
equating power on both sides
c2 = 0
a2 + b2 − 3c1 = 0
−b2 −1 = 0

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 50 B.E./B.Tech. Question Papers

c2 = 0 b2 = −1 a2 = −b2 + 3c2
a2 = 0

\ π2 = D1 V −1 ρ 0 ω

Dω
π2 =
V
p3 term

π3 = D a3 V b3 ρ c3 ⋅ µ
= La3 ( LT −1 )b3 ( ML−3 )c3 ⋅ ML−1T −1
= M c3 +1 + La3 +b3 −3c3 T −b3 −3c3 −1
equating power on both sides
0 = c3 + 1
0 = a3 + b3 − 3c3 −1
0 = −b3 −1
c3 = −1 b3 = −1 a3 = −b3 + 3c3 + 1
= −1 − 3 + 1

= −1

\ π3 = D1 V −1 ρ 0 µ

µ
π3 =
DV ρ

p4 term

π4 = D a4 V b4 ρ c4 ⋅ C
M 0 L0T 0 = La4 ( LT −1 )b4 ( ML−3 )c4 ( LT −1 )
= M c4 La4 +b4 −3c4 +1T −b4 −1
equating power on both sides
c4 = 0
a4 + b4 − 3c4 + 1 = 0
−b4 −1 = 0

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 51

c4 = 0 b4 = −1 a4 = −b4 + 3c4 −1
= 1 −1

a4 = 0

\  π4 = D 0 V −1 ρ 0C

C
π4 =
V
f1(p, p2, p3, p4) = 0

 P Dω µ C 
f1  2 2 , , ,  = 0
 D V ρ V DV ρ V 

P  Dω µ C 
= φ  , , 
22
DV ρ  V DV ρ V 

 Dω µ C 
P = D 2V 2 ρφ  , , 
 V DV ρ V 
13.  (b)  Given data:

Specific gravity of oil = 0.9 = Sp

Viscosity of oil = 3 × 10-2 poise = mp

Rate of oil flow Q = 3000 liters/sec = 3 m3/s

Diameter of Protype Dp = 1.5 m

Diameter of model Dm = 15 cm = 0.15m

Viscosity of water at 20°C mm = 0.01 poise

1
= 0.01×
10
= 0.001 Ns/m 2

Solution

(i) velocity of flow in model (Vm):

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 52 B.E./B.Tech. Question Papers

ρ mVm Dm ρ PVP DP
=
µm µP
Vm ρ P DP µ m
=
VP em Dm µ P
 9.81× 0.9 
  1.5 0.001

=  9.81 × ×
 9.81 / 9.81 0.15 0.003
0.9 1.5 0.001
= × ×
1 0.15 0.003
Vm
=3.
VP

QP 3
VP = = = 1.697 m/s
AP π 2
×(1.5)
4
Vm = 3×VP
= 3×1.697

Vm = 5.092 m/s

  (ii) Rule of flow in the model Qm

Qm = Am ×Vm
π
= × D 2 m ×Vm
4

π
= ×(0.15) 2 ×5.092
4
= 0.09 m3 /sec.

Qm = 90 liters/sec

14.  (a)  Given data:

Net head H = 60 m
Speed N = 700 rpm
Shaft Power Hp = 294.3 Kw
Overall efficiency h0 = 84% = 0.84

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 53

Hydraulic efficiency hh = 93% = 0.93

flow ration kf = 0.2
breadth ratio n = 0.1
D1 = 2 × D2
Vf1 = Vf3
Solution:
Overall efficiency
P
η0 =
WQH

294.3
0.84 =
9.81× 2×60

Q = 0.595 m3 /s
flow ratio
vf1
kf =
2 gh

vf1
0.2 =
2× 9.81×60

vf1 = 6.86 m/s

Discharge Q = Area × velocity

= πD1 B1 × vf1
B 
= πD1 × 0.1D1 × vf1  1 = n = 0.1
 D1 
 
2
0.595 = πD1 × 0.1×6.86
D1 = 0.525 m

D1 = 525 mm

B1 = 0.1 D1

B1 = 52.5 mm

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 54 B.E./B.Tech. Question Papers

Diameter of wheel at outlet

D1
D2 =
2
0.525
= = 0.2625
2
D2 = 262.5 mm
Tangential of wheel at intlet
πD1 N
u1 =
60

π× 0.525× 700
=
60
u1 = 19 ⋅ 242 m/s
we know that
V ω1u1
ηh =
gh

V ω1 ×19.242
0.93 =
9.81×60
V ω1 = 28.448 m/s
Tangential velocity at outlet
πD2 N
u2 =
60

π× 0.2625× 700
=
60
u2 = 9 ⋅ 621 m/s
Guide blade angle (µ)
Vf1 6.86 6.86
tan ∝= = =
V ω1 28.448 28.448

∝= 13.55°

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 55

u2

φ
v2 = vf1=vf2
vr2

v1
vf1
vr1

α θ

u1

vw1

Runner vane angle (q)

Vf1
tan θ =
V ω1 − u1

6.86
=
28.448 −19.242
θ = 36.69°
Runner blade angle at outlet (f)
Vf 2 Vf1
tan φ = =
u2 u2

6.86
=
9.6211

φ = 35.48°

14.  (b)  Given data:

u = u1 = 10 m/s u2 = Angle of deflection = 160°
Cv = 0.98

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 56 B.E./B.Tech. Question Papers

H = 30 m
Q = 0.7m3/sec
P = ? = rgQH
(V ω1 ± V ω1 )u
ηh =
gH
Angle f = 180 - 160
= 20°
Velocity of jet
V1 = Cv 2 gH = 0.98 2× 9.81×30

V1 = 23.77 m / sec

Vr1 = V1 − u1 = 23.77 −10 = 13.77 m/ sec

Vω1 = V1 = 23.77 m/ sec

u2 vw2

φ β
v2 vf2
vr2

165°
u1 vr1

v1 = vw1

from outlet velocity triangle

Vr2 = Vr1 = 13.77 m/ sec
Vω2 = Vr2 cos φ − u2
= 3.77 cos 20 −10

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 57

Vω2 = 2.94 m/ sec

Work done by set per second is given by

= eaV1 Vω1 + Vω2  × u

 
1000
= × 0.7[23.77 + 2.94]×10[Q = 9v1 ]
9.81
= 19059.12 kgf-m / sec
Horse power is given to turbine
Work done
=
75

19059.12
=
75
P = 254.12 h. p

Hydraulic efficiency of turbine is given by

2 Vω + Vω  + u
ηh =  1 2 2 
V1

2[23.77 + 2.94]×10
=
( 23.77) 2

η h = 94.54%

15.  (a)  Given data:

D = 150 mm = 0.15m
L 0.3
L = 300 m = 0.3 m r = = = 0.15 m
2 2
N = 50 rpm
hd = 25m
Q = 4.2 litres/sec = 4.3 × 10-3 m3/sec
ld = 22 m dd = 100mm = 0.1 m

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 58 B.E./B.Tech. Question Papers

Solution:
(i) Theoretical discharge
LAN
Qth =
60
π
0.3× ×(0.15) 2 ×50
= 4
60

Qth = 4.4178×10−3 m3 /sec

Power required to drive pump

P = WQth ( hs + hd )
= 9810 × 4.4178×10−3 (0 + 25)
= 1083 ⋅ 48 W

Pthe = 1.08 kW
(ii) Percentage slip
Qth − Qact
% slip = ×100
Qth

4.4178×10−3 − 4.2×10−3
= ×100
4.4178×10−3

% slip = 4.93%

15.  (b)  Given data:

Speed N = 1200 rpm
Inlet vane angle q = 20°
Outlet vane angle f = 30°
Internal diameter D1 = 200 mm = 0.2m.
External diameter D2 = 400 mm = 0.4m.
Solution:
Tangential velocity at inlet
πD1 N π× 0.2×1200
u1 = =
60 60

u1 = 12.56 m / sec

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 Fluid Mechanics and Machinery (Nov/Dec 2012) 59

Tangential velocity at outlet

πD2 N π× 0.4 ×1200
u2 = =
60 60

u2 = 25.13 m/sec

u2

vw2

β φ
vf2
v2 vr2

v1 = Vf1
vr2
α
θ
u1

from inlet triangle

vf1 vf1
tan θ = ⇒ tan 20° =
u1 12.56

vf1 = 4.57 m/sec

vf1 = vf2 = 4.57 m/s

from outlet triangle.
vf 2
tan φ =
u2 −Vω2

4.57
tan 30° =
25.13 −Vω2

Vω2 = 17.21 m/s

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 60 B.E./B.Tech. Question Papers

work done per N of water

Vω2 u2
=
g

17.21× 25.13
=
9.81
W = 44.08 N-m

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 B.E./B.Tech. DEGREE EXAMINATION,
MAY/JUNE 2012
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY
(Regulation 2008)
(Common to Aeronautical Engineering, Automobile
Engineering and Production Engineering)
Time: Three hours Maximum: 100 marks
Answer ALL questions
PART A (10 × 2 = 20 marks)

  1. Two horizontal plates are placed 12.5 mm apart, the space between them
being filled with oil of viscosity 14 poise. Calculate the shear stress in
the oil if the upper plate moved with a velocity of 2.5 m/s. Define specific
weight.

  2. Calculate the height of capillary rise for water in a glass tube of diameter
1 mm.

  3. What are the minor losses? Under what circumstances will they be
negligible?

  4. What factors account for the energy loss in laminar flow?

  5. Under what circumstances, will the Buckingham-Pi theorem yield

incorrect number of dimensionless group?

  6. If two systems (model and prototype) are dynamically similar, is it

implied that they are also kinematically and geometrically similar?

  7. What precautions are to be taken while starting and closing the centrifugal
pump?

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 62 B.E./B.Tech. Question Papers

  8. Differentiate reaction turbine and impulse turbine.

  9. What is the function of non-return valve in a reciprocating pump?

10. What factors govern the speed of reciprocating pump?

PART B (5 × 16 = 80 marks)

11.  (a)  (i)  The space between two large flat and parallel walls 25 mm apart
is filled with a liquid of absolute viscosity 0.7 Pa.sec. Within this
space a thin flat plate, 250 mm × 250 mm is towed at a velocity
of 150 mm/s at a distance of 6 mm from one wall, the plate and
its movement being parallel to the walls. Assuming linear
variations of velocity between the plate and the walls, determine
the force exerted by the liquid on the plate. (8)

(ii)  Eight kilometers below the surface of the ocean the pressure is
81.75 MPa. Determine the density of sea water at this depth if the
density at the surface is 1025 kg/m3 and the average bulk modulus
of elasticity is 2.34 GPa. (8)
Or
(b)  (i)  A 0.3 m diameter pipe carrying oil at 1.5 m/s velocity suddenly
expands to 0.60 m diameter pipe. Determine the discharge and
velocity in 0.6 m diameter pipe. (4)

(ii)  Derive the momentum equation for steady flow. (12)

12.  (a)  (i)  A pipe line 20 cm in diameter, 70 m long, conveys oil of specific
gravity 0.95 and viscosity 0.23 N.sec/m2. If the velocity of oil is
1.38 m/s, find the difference in pressure between the two ends of
the pipe. (8)

(ii)  Oil of mass density 800 kg/m3 and dynamic viscosity 0.02 poise
flows through 50 mm diameter pipe of length 500 m at the rate of
0.19 litres/sec. Determine

(1)  Reynolds number of flow

(2)  Centre line velocity

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 Fluid Mechanics and Machinery (May/June 2012) 63

(3)  Pressure gradient

(4)  Loss of pressure in 500 m length

(5)  Wall shear stress and

(6)  Power required to maintain the flow. (8)

Or
(b)  (i)  Obtain expression for Darcy-Weishbach friction factor f for flow
in a pipe. (6)

(ii)  A smooth pipe carries 0.30 m3/s of water discharge with a head
loss of 3.0 m per 100 m length of pipe. If the water temperature
is 20°C, determine the diameter of the pipe. (10)

13.  (a)  The power required by the pump is a function of discharge Q, head
H, acceleration due to gravity g, viscosity m, mass density of the fluid
r, speed of rotation N and impeller diameter D. Obtain the relevant
dimension less parameters.
Or
(b)  Model tests have been conducted to study the energy loss in a
pipeline of 1 m diameter required to transport kerosene of specific
gravity 0.80 and dynamic viscosity 0.02 poise at the rate of 
2000 litre/sec. Tests were conducted on a 10 cm diameter pipe using
water at 20°C. What is the flow rate in the model? If the energy head
loss in 30 m length of the model is measured as 44.0 cm of water,
what will be the corresponding head loss in the prototype? What will
be the friction factor for the prototype pipe?

14. (a)  A centrifugal pump discharges 2000 1/s of water per second
developing a head of 20 m when running at 300 rpm. The impeller
diameter at the outlet and outflow velocity is 1.5 m and 3.0 m/s
respectively. It vanes are set back at an angle of 30° at the outlet,
determine
(i)  Manometric efficiency
(ii)  Power required by the pump.
If inner diameter is 750 mm, find the minimum speed to start the
pump.

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 64 B.E./B.Tech. Question Papers

Or
(b)  An inward flow reaction turbine discharges radially and the velocity
of flow is constant, show that the hydraulic efficiency can be
expressed by

1
η =
0.5 tan 2 α
1+
tan α
1−
tan θ

Where a and q are the guide and vane angles at inlet.

15.  (a)  (i)  Determine the percentage of work saved in one cycle when an air
vessel is provided on the delivery side of a single cylinder single
acting reciprocating pump. (8)

(ii) Explain the working principle of reciprocating pump with neat

diagram in detail and state its advantages and disadvantages over
centrifugal pump. (8)
Or

(b)  (i)  A single cylinder double acting reciprocating pump has a piston
diameter of 300 mm and stroke length of 400 mm. When the
pump runs at 45 rpm, it discharges 0.039 m3/s under a total head
of 15 m. What will be the volumetric efficiency, work done per
second and power required if the mechanical efficiency of the
pump is 75%? (10)

(ii)  With an example, explain in detail the working principle and

construction of rotary pumps with neat diagram. (6)

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 Solutions
PART A

  1.  dy = 12.5 mm  du = 2.5 m/s  m = 14 poise

= 1.4 N-s/m2
du
  \ τ = µ
dy
525
= 1.4 × = 0.28×10−3 N/m 2
12.5×10−3
τ = 0.28 KN / m 2

  Specific weight is defined as weight

W
w=
V
4s
  2. h=
wd
w = 9810   s = 0.075 N/m
4 × 0.075
h= = 0.03 m
9810 ×1×10−3
h = 0.03 m
  3.  The loss of energy caused on account of the change in velocity of flowing
fluid is called minor losses. It may be neglected when the pipes are long.
  4.  The inertia force and viscosity account for energy loss in laminar flow.
  5.  If the variables are wrongly taken then resulting functional relationship
is erroneous.
Lr
  6.  The time scale ratio is Fr = , so when the systems are dynamically
Vr
similar, then they are kinematically and geometrically similar.
  7.  The delivery value should be closed during starting and stopping. No air
pocket should be left while starting. Delivery value should be opened
after impeller reaches normal speed.

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 66 B.E./B.Tech. Question Papers

8.

Impulse turbine Reaction turbine

1.  All the potential engines are 1.  Only a portion of fluid energy is
converted into kinetic energy by transferred into kinetic energy
nozzle before enters the turbine
2.  Blades are only in action when 2.  Blades are in action at all times
they are in front of nozzle

  9.  The non-return values are provided to avoided reverse flow of water in
the suction and delivery side.
10.  The velocity of the piston and stroke length of the piston governs the
speed of reciprocating pump.

PART B
11.  (a)  (i)  Given data:
Distance between walls h = 25 mm = 0.025 m
Plate area = 250 × 250 mm2 = 0.0625 m2
Dynamic viscosity m = 0.7 poise = 0.07 N-S/m2
Velocity du = 150 mm/s = 0.15 m/s

6 mm
25 mm 0.15 m/s
19 mm

Shear stress

du u du u
τ=µ =µ , τ=µ =µ
dy t1 dy t2
u u 
τ = τ1 + τ 2 = µ  + 
 t1 t 2 

Force
F = Shear stress × Area

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 Fluid Mechanics and Machinery (May/June 2012) 67

1 1
= mu  +  × A
 t1 t 2 

1 1
= muA  + 
 t1 t 2 
t1 = 6 mm = 0.006 m t 2 = 19 mm = 0.019 m.
 1 1 
\  Force F = 0.07× 0.15× 0.0625  + 
 6 ×10 −3
19×10 
−3

F = 0.1439 N

11.  (a)  (ii)  Given data:

P2 = 81.75 Mpa = 81. 75 × 106 N/m2
r1 = 1025 Kg/m3.
K = 2.34 GPa = 2.34 × 109 N/m2
at surface pressure is atmospheric pressure.
P1 = 1.01 bar = 1.01 × 105 N/m2
dp dp
Bulk modulus K = − =
 dv   d r 
   
 v   r 

dr = r2 − r1

dp
K=
 r2 − r1 
 
 r1 

81.75×106 −1.01×105
2.34 ×10 9 =
 r2 −1025 

 1025 

8.369×1010
2.34 ×10 9 =
r2 −1025

r2 −1025 = 1060.76 kg/m3

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 68 B.E./B.Tech. Question Papers

11.  (b)  (i)  Given data:

d1 = 0.3 m
V1 = 1.5 m/s
d2 = 0.6 m
To find Q = ? V2 = ?
Solution:
Continuity equation
A1V1 = A2V2
p p
× 0.32 ×1.5 = × 0.6 2 ×V2
4 4
V2 = 0.375 m/s
p 2
Q = AV
1 1 = d1 ×V1
4
p
= × 0.32 ×1.5
4

Q = 0.106 m3 /sec
11.  (b)  (ii)  The force acting on a fluid mass ‘m’ is given by Newton’s second
law of motion
F=m×a
where a is acceleration acting in the same direction of force ‘F’
dv
a=
dt
dv
\  F = m ⋅ [m is constant and can be taken inside the
dt differential]
d ( mv )
=
dt
The above equation is known as momentum principle.
F ⋅ dt = d ( mr )

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 Fluid Mechanics and Machinery (May/June 2012) 69

The above equation is impulse-momentum equation and states

that impulse of force F acting on momentum d(mv) is direction
of force.
12.  (a)  (i)  Given data
D = 20 cm = 20 × 10-2 m.
L = 70 m
Sp.gravity s = 0.95
m = 0.23 N-s/m2
V = 1.38 m/s
Discharge Q = A × V
p
= × D 2 ×V
4
p
= ×( 20 ×10−2 ) ×1.38
4
Q = 0.0434 m3 / sec
From Hagen poiuseulle’s equation
128µQL
P1 − P2 =
πD4

128× 0.23× 0.0434 × 70
=
π ×( 20 ×10−2 ) 4
P1 − P2 = 17.8×103 Pa

12.  (a)  (ii)  Given data:

r = 800 kg/m3
m = 0.02 poise = 0.002 N
D = 50 mm = 50 × 10-3 m
L = 500 m
Q = 0.19 litres/sec = 1.9 × 10-4 m3/sec.
Solution
ρuD
(i)  Reynold’s number Re =
µ

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 70 B.E./B.Tech. Question Papers

Q 1.9×10−4
U= = = 4.838×10−3 m/s
A p −3
×50 ×10
4
800 × 4.838×10−3 ×50 ×10−3
\   Re =
0.002
Re = 96.76
(ii)  Pressure gradient
∂ρ 128µQ
=
∂x πD4

128× 0.002×1.9×10−4
=
p ×(50 ×10−3 )

∂r
= −2.477
∂x

(iii)  Maximum velocity

1 ∂ρ 2
U max = − R
4 µ ∂x

1
=− ×(−2.477) ×( 25×10−3 ) 2
4 × 0.002
U max = 0.1935 m/s
(iv)  Loss of pressure in 500 m length

32muL
P1 − P2 =
D2

32× 0.002× 4.838×10−3 ×500
=
(50 ×10−3 ) 2
P1 − P2 = 61.93
(v)  Wall shear stress
∂ρ R P1 − P2 R
τ0 = − ⋅ = ⋅
∂x 2 L 2
−3
61.93 25×10
= ×
500 2

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τ 0 = 1.548×10−3 N/m 2
(vi) Power required for flow
P = Q( P1 − P2 )

= 1.9×10−4 (61.93)
P = 0.011767 watts

4 FLV 2
12.  (b)  (i)  Head loss h f =
2 gD
f - Darcy weishbach coeff of friction from the above equation
16
f = for Re < 2000
Re
0.079
f = for Re varying from 4000 × 106
(Re)0.25

12.  (b)  (ii)  Given data:

Q = 0.3 m3/s
L = 100 m
T = 20°C
g = 0.0098 strokes = 0.0098 × 10-4 m2/sec for smooth pipe.
0.0791
f =
1
(Re)
4
UD
Re =
r
Q 0.3 0.38
U= = = 2 .
A p 2 D
D
4
0.38 / D 2 × D
Re =
0.0098×10−4
389767.2
Re =
D

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0.0791
f = 1
 389767.2 4
 
 D 
= (3.16 ×10−3 )( D1/ 4 )
4 fLV 2
hf =
2 gD
 0.38 
4 ×(3.16 ×10−3 )( D1/ 4 ) ×100 × 2 
D 
3=
2× 9.81× D
0.25
0.0245 D
3=
D3
122.5 = D −2.75
1
= −2.75
D
1
D −2.75 =
122.5
D = 0.174 m

13.  (a)  Power developed.

P = f (Q, H , g , µ, ρ, N , D ).

\  f1 ( P , Q, H , g , µ, ρ, N , D ) = 0
No of variables
P = ML2T-3, Q = L3T-1, N = T-1, D = L, g = LT-2, H = L,
m = ML-1T-1, r = ML-3
Number of fundamental dimensions m = 3
Number of p terms = n - m = 8 - 3 = 5.
\  f1(p1, p2, p3, p4, p5) = 0
each p terms contains m + 1 variables.
D, N and r are repeating variables.
π1 = D a1 N b1 ρ c1 ⋅ P .

π2 = D a2 N b2 ρ c2 ⋅ Q.

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 Fluid Mechanics and Machinery (May/June 2012) 73

π3 = D a3 N b3 ρ c3 ⋅ g .
π4 = D a4 N b4 ρ c4 ⋅ µ.
π5 = D a5 N b5 ρ c5 ⋅ H .

p1 term
π1 = D a1 N b1 ρ c1 ⋅ P
M 0 L0T 0 = La1 (T −1 )b1 ( ML−3 )c1 ( ML2T −3 )
= M 1+c1 La1 −3c1 + 2T −b1 −3
equating M ⇒ 0 = 1 + c1

L ⇒ 0 = a1 − 3c1 + 2

T ⇒ 0 = −b1 − 3

c1 = 0   a1 − 3c1 + 2 = 0

b1 = −3 a1 + 3 + 2 = 0

a1 = −5

π1 = D −5 N −3 ρ−1 ⋅ P.

P
π1 = .
DN ρ

p2 term
π2 = D a2 N b2 ρ c2 ⋅ Q
= La2 (T −1 )b2 ( ML−3 )c2 L−3T −1
M 0 L0T 0 = M c2 La2 −3c2 −3T −b2 −1
0 = c2  a2 - 3c2 - 3 = 0
-b2 - 1 = 0
c2 = 0   b2 = −1   a2 = 3

\  π2 = D 3 N −1 ρ 0 ⋅ Q

QD 3
p2 =
N

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 74 B.E./B.Tech. Question Papers

p3 term
π3 = D a3 N b3 ρ c3 ⋅ g
= La3 (T −1 )b3 ( ML−3 )c3 LT −2
M 0 L0T 0 = M c3 La3 −3c3 +1T −b3 −2
c3 = 0  a3 - 3c3 + 1 = 0  -b3 - 2 = 0

c3 = 0   a3 = −1   b3 = −2

\   π3 = D −1 N −2 ρ 0 ⋅ g

g
p3 =
ND
p4 term
π4 = D a4 N b4 ρ c4 ⋅ µ
= La4 (T −1 )b4 ( ML−3 )c4 ML−1T −1
M 0 L0T 0 = M c4 +1 La4 −3c4 −1T −b4 −1
0 = c4 + 1   a4 − 3c4 −1 = 0   −b4 −1 = 0

c4 = −1   a4 = −2   b4 = −1

\  π4 = D −2 N −1 ρ−1µ

µ
π4 = 2
D Nρ

p5 term

π5 = D a5 N b5 ρ c5 ⋅ H
M 0 L0T 0 = La5 (T −1 )b5 ( ML−3 )c5 L
= M c5 La5 −3c5 +1T −b5
0 = c5   0 = a5 − 3c5 + 1   0 = −b5

c5 = 0   a5 = −1   b5 = 0

\  π5 = D −1 N 0 ρ 0 ⋅ H

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H
p5 =
D

 Q QD 3 g µ H
f1 =  , , , 2 , =0
 DN ρ N ND D N ρ D 
 
P  QD 3 g µ H
= φ  , , 2 , 
DN ρ  N ND D N ρ D 
 QD 3 g µ H
P = DN ρφ  , , 2 , 
 N ND D N ρ D 

13.  (b)  Given data

Sp = 0.8
Dp = 1 m
mp = 0.02 poise
Qp = 2000 lit/sec = 2 m3/sec
Dm = 10 cm = 0.1 m
Model fluid = Water
Prototype fluid = Kerosene
mm = 0.01 poise at 20°C
Wp = Density of kerosene = Sp × 1000 = 0.8 × 1000 = 800 kgf/m3

Wp800
rp = = = 81.54 N/m3
g 9.81

W 1000
rm = m =
g 9.81

ρm Vm Dm ρ p V p D p
=
µm µp

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Vm ρ p D p µm
= × ×
V p ρm Dm µ p
 800 9.81   1   0.01 
=  × × 
1000 9.81  0.1  0.02 
Vm
=4
Vp

Qp 2
Vp = = = 2.546 m/s
Ap p 2
(1x )
4

Vm = 4 ×V p
= 4 × 2.546
Vm = 10.184 m/s

Qm = AmVm
p
= ×(0.1) 2 ×10.184
4
= 0.0799 m3 /sec

Qm = 79.9 litres/sec

Lm = 30 m H m = 44cm of water

Qp
Linear scale ratio Lr =
Qm
3000
=
79.9
= 37.5

Hp
= Lr
Hm
H p = Lr H m
= 37.5× 44 ×10−2
H p = 16.5 m of water

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 Fluid Mechanics and Machinery (May/June 2012) 77

Lp
Lr = 37.5 =
Lm

\  L p = 37.5× Lm = 1125 m

4 fL pV p 2
Hp =
2 gD p
4 f ×1125×( 2.546) 2
\  16.5 =
2× 9.81×1
f = 0.01109

14.  (a)  Given data:

Q = 2000 lit/sec = 2 m3/sec

H = 20 m

N = 300 rpm

D2 = 1.5 m

V f2 = 3 m/s

f = 30°
π D2 N
µ2 = 60

π ×1.5×300
=
60
µ2 = 23.562 m/s

V f2 3
tan f = ⇒ tan 30 =
U 2 −Vw2 23.56 −Vw2

23.56 −Vw2 = 5.196

Vw2 = 18.36 m/s

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Manometric efficiency

gH
hm =
Vw2 u2
9.81× 20
=
18.36 × 23.562
hm = 45.35%

Power required by the pump

Assuming overall efficiency of 85%

P
h0 =
wQH
P
0.85 =
9.81× 2× 20
P = 333.54 kw

D1 = 750 mm

Minimum speed for starting the pump

120 × η max ×Vw2 × D2

N min =
π[ D2 2 − D12 ]
120 × 0.4535×18.36 ×1.5
=
π[1.52 − (750 ×10−3 ) 2 ]
1498 ⋅ 726
=
5.3014
N min = 282.72 rpm

14.  (b)  Solution

Velocity of radial and constant
V f = V f2 Vw2 = 0 V f = V2
1 2

from inlet triangle

V f1
tan ∝ = ⇒ V f1 = Vw1 tan ∝
Vw1

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u2
φ
vf2 = v2
vr2

v1 vf1
vr1

α θ

Vf
tan ∝ = 1

Vw1 − u1
Vf
(Vw1 − u1 ) = 1

tan q
Vw1 tan ∝
=
tan q
Vw tan ∝
u1 = Vw1 − 1
tan q
 tan ∝ 
u1 = Vw1 1− 
 tan q 
W.k.t.

V2 2 1
H− = (Vw1 u1 )
2g g
1 V2
H= Vw1 u1 + 2
g 2g
2
1 Vf
= Vw1 u1 + 1 [V2 = V f2 = V f1 ]
g 2g

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Sub V f1 and u1
2
1 1− tan ∝  (Vw1 tan ∝)
H = Vw1 ×Vw1  +
g  tan q  2g

Vw1 2  tan ∝ tan 2 ∝ 
H= 1− + 
g  tan q 2 

now hydraulic efficiency

Vw1 u1 Vw1 u1
ηh = =
gH V  tan ∝ tan 2 ∝ 
2

g⋅
w1
1− + 
g  tan θ 2 
 tan ∝ 
Vw1 ×Vw1 1− 
 tan θ 
=
 tan ∝ tan 2 ∝ 
Vw21 1− + 
 tan θ 2 
tan ∝
1−
= tan θ
 tan ∝ tan 2 ∝ 
1−− + 
 tan θ 2 
1
ηh =
1
1+
0.5 tan 2 ∝
 tan ∝ 
1− 
 tan θ 

15.  (a)  (i)  Solution

The velocity of water in delivery pipe with air vessel.

2 A ωr
V=
a π
Friction head loss
4 flV 2
Hf =
2 gd
2
4 fl  2 A ωr 
=  
2 gd  a π 

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Work done per stroke

2
4 fl  2 A ωr 
W2 =  
2 gd  a π 

We also know that the velocity of water in delivery pipe without
air vessels is
2
2 4 fl  A ωr 
W1 =  
3 2 gd  a π 

Saving in work done per stroke

W1 −W2
= ×100
W1
W2
= 1− ×100
W1
 4 fl  2 A ωr  
2
  
 2 gd  a π  
= 1− 2
×100
 2 4 fl  A ωrr  
   
 3 2 gd  a π  
4 / π2
= 1− ×100
2/3
6
= 1− 2 ×100
π

Work saving = 39.2%

15.  (a)  (ii)  Reciprocating pump consists of

•  A Piston
•  Suction and delivery pipe
•  Crank and Connecting rod mechanism
In single acting cylinder reciprocating pump liquid acts on
one side of piston only. It is usually placed above liquid level
in pump. When crank rotates clockwise from inner dead centre
(IDC) to outer dead centre (ODC), the piston moves outward to

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 82 B.E./B.Tech. Question Papers

right and vacuum is created on left side of piston. This vacuum

causes suction values to open and liquid is forced from jump
into left of piston.

Delivery pipe

Qi

hd
Delivery value Connecting road
IDC Crank
ODC
Piston cylinder
C
A

hs

Suction value
Suction pipe

Sump level

When crank rotates from ODC to IDC, the piston moves inward
to the left and a high pressure is built up in the cylinder. The
increase in pressure closes the suction value and opens the
discharge value, and delivers the liquid.

Advantages and disadvantage of reciprocating pump against

centrifugal pump.

Wear and tear more due to more moving parts

Procedure low heat

Used for height lift

More efficient

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 Fluid Mechanics and Machinery (May/June 2012) 83

15.  (b)  (i)  Given data:

D = 300 mm = 0.3 m

L = 400 mm = 0.4 m

N = 45 rpm

Qact = 0.039 m3/sec

H = 15 m

hm = 75%

p
×(0.3) 2 × 0.4 × 45× 2
2 ALN 4
Qthe = = = 0.042 m 2 /sec
60 60
Volumetric efficiency
Qact
hv = ×100
Qthe
0.039
= ×100
0.042
hv = 91.95%
Work done per second

WALN
=2 (H )
60
p
9810 × × 0.32 × 0.4 × 4.5× 2
= 4 ×15
60
= 6241 N-m

Input Power done

P = w Q + n( H )

= 9810 × 0.042×15 = 6.18 Kw

Mechanical efficiency = 75%

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 84 B.E./B.Tech. Question Papers

P0 P
h mech = ⇒ 0.75 = 0
Pi 6.18

Output power = 47.635 Kw

15.  (b)  (ii)  Rotary pumps resemble like centrifugal pumps in appearance
but the working method differs. Uniform discharge and positive
displacement can be obtained by using rotary pumps.
The various types are
•  External Gear Pump
•  Internal Gear Pump
•  Lobe Pump
•  Vane pump

External Gear Pump

It has two identical intermeshing spur wheels A and B. One
wheel is connected to driving shaft and other one is driven. The
liquid is entrapped between teeth and allowed to discharge into
delivery pipe when the gear rotates.
These rotating gears itself set up sufficient pressure to force the
liquid.

Internal Gear Pump

Two intermeshing wheels are internally meshed. A crescent
shaped partition is made between these two wheels. This space
acts as a seal between suction and discharge. During rotation
the space between inner and outer wheel increases and liquid
fills the space. Due to continuous rotation the fluid is pushed to
delivery end.

Lobe Pump
This pump resembles like a gear pump in action but not in
appearance. The wheels inside the casing have two or more
lobes. The liquid is entrapped & flows to the delivery end when
lobe rotates. The lobes itself build up sufficient pressure to force
the liquid.

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Vane Pump
A disc is eccentrically rotating inside the pump casing. Slots containing
vanes are provided in the disc. The vanes are moved against the casing due
to centrifugal force. Due to continuous rotations the liquid is entrapped and
forced to the delivery end with sufficient pressure.

Internal gear pump

External gear pump

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Vane pump

Lobe pumps

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 B.E./B.Tech. DEGREE EXAMINATION,
APRIL/MAY 2010
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY
Time: Three hours Maximum: 100 marks
Answer ALL questions
PART A (10 × 2 = 20 marks)
1. A soap bubble is formed when the inside pressure is 5 N/m2 above the
atmospheric pressure. If surface tension in the soap bubble is 0.0125 N/m,
find the diameter of the bubble formed.

2. The converging pipe with inlet and outlet diameters of 200 mm and
150 mm carries the oil whose specific gravity is 0.8. The velocity of oil
at the entry is 2.5 m/s, find the velocity at the exit of the pipe and oil flow
rate in kg/sec.

3. Define boundary layer and give its significance.

4. Find the loss of head when a pipe of diameter 200 mm is suddenly

enlarged to a diameter of 400 mm. Rate of flow of water through the pipe
is 250 litres/s.

5. A centrifugal pump delivers 20 litres/s of water against a head of 850 mm

at 900 rpm. Find the specific speed of pump.

6. What do you understand by fundamental units and derived units?

7. The mean velocity of the buckets of the Pelton wheel is 10 m/s. The jet
supplies water at 0.7 m2/s at a head of 30 m. The jet is deflected through an
angle of 160° by the bucket. Find the hydraulic efficiency. Table Cv = 0.98.

8. The following data refer to a centrifugal pump which is designed to run

at 1500 rpm. D1 = 100 mm, D2 = 300 mm, B1 = 50 mm, B2 = 20 mm,
V0 = 3 m/s. β2 = 60°. Find the velocity of flow at outlet.

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 4.4
5.4 B.E./B.Tech. Question
B.E./B.Tech. Question Papers
Papers

9. Define slip of reciprocating pump.

10. Mention the working principle of an air-vessel.

PART B (5 × 16 = 80 marks)
11. (a) A drainage pipe is tapered in a section running with full of water.
The pipe diameters at the inlet and exit are 1000 mm and 500 mm
respectively. The water surface is 2 m above the centre of the inlet
and exit is 3 m above the free surface of the water. The pressure at
the exist is 250 mm of Hg vacuum. The friction loss between the
inlet and exit of the pipe is 1/10 of the velocity head at the exit.
Determine the discharge through the pipe.
Or
(b) A pipe of 300 mm diameter inclined at 30° to the horizontal is
carrying gasoline (specific gravity = 0.82). A venturimeter is fitted
in the pipe to find out the flow rate whose throat diameter is 150 mm.
The throat is 1.2 m from the entrance along its length. The pres-
sure gauges fitted to the venturimeter read 140 kN/m2 and 80 kN/m2
respectively. Find out the coefficient of discharge of venturimeter if
the flow is 0.20 m2/s.

12. (a) For a turbulent flow in a pipe of diameter 300 mm, find the discharge
when the centre-line velocity is 2.0 m/s and the velocity at a point
100 mm from the centre as measured by pitot-tube is 1.6 m/s.
Or
(b) For a town water supply, a main pipe line of diameter 0.4 m is
required. As pipes more than 0.35 m diameter are not readily avail-
able, two parallel pipes of same diameter are used for water supply.
If the total discharge in the parallel pipes is same as in the single
main pipe, find the diameter of parallel pipe. Assume coefficient of
discharge to be the same for all the pipes.

13. (a) Using Buckingham’s theorem, show that the velocity through a
circular orifice in a pipe is given by v = 2 gH f {d / H , μ / ρcH }
where v is the velocity through orifice of diameter d and H is the
head causing the flow and ρ and µ are the density and dynamic vis-
cosity of the fluid passing through the orifice and g is acceleration
due to gravity.
Or

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Fluid Mechanics and Machinery (April/May 2010) 4.5
5.5

(b) The efficiency (η) of a fan depends on ρ (density), µ (viscosity)

of the fluid, ω (angular velocity), d (diameter of rotor) and Q (dis-
charge). Express η in terms of non-dimensional parameters. Use
Buckingham’s Π theorem.

14. (a) In an inward radial flow turbine, water enters at an angle of 22° to
the wheel tangent to the outer rim and leaves at 3 m/s. The flow
velocity is constant through the runner. The inner and outer diam-
eters are 300 mm and 600 mm respectively. The speed of the runner
is 300 rpm. The discharge through the runner is radial. Find the
(i) Inlet and outlet blade angles
(ii) Taking inlet width as 150 mm and neglecting the thickness of
the blades, find the power developed by the turbine.
Or
(b) A Kaplan turbine working under a head of 20 m develops 15 MW
brake power. The hub diameter and runner diameter of the turbine
are 1.5 m and 4 m respectively. The guide blade angle at the inlet is
30°. η0 = 0.9 and η0 = 0.8. The discharge is radial. Find the runner
vane angles and turbine speed.

15. (a) The diameter and stroke of a single acting reciprocating pump are
120 mm and 300 mm respectively. The water is lifted by a pump
through a total head of 25 m. The diameter and length of delivery
pipe are 100 mm and 20 m respectively. Find out
(i) Theoretical discharge and theoretical power required to run the
pump if its speed is 60 rpm
(ii) Percentage slip, if the actual discharge is 2.95 I/s and
(iii) The acceleration head at the beginning and middle of the deliv-
ery stroke.
Or
(b) Explain the working of the following pumps with the help of neat
sketches and mention two applications of each.
(i) External gear pump
(ii) Lobe pump
(iii) Vane pump
(vi) Screw pump

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 Solutions
PART A
1. Pressure = 5 N/m2
Surface tension, = 0.0125 N/m
8θ
P=
d
8θ 8 × 0.125
d= = = 2 cm
P 5

2. D1 = 0.2 m, V1 = 2.5 m/s

D2 = 0.15 m,V2 = ?
A1 V1 = A2 V2
(0.2)2
V2 = × 2.5
(0.15)2
V2 = 4.44 m/s
M = P1 A1 V1 = P2 A2 V2
π
M = 800 × (0.2)2 × 2.5
4
M = 62.83/kg/8.

3. A boundary layer is the layer of fluid in the immediate vicinity of a

bounding surface where the effects of viscosity are significant.

4. D1 = 200 mm = 0.2 m
π π
A1 = × D12 = (0.2)2
4 4
= 0.03141 m2
D2 = 400 mm = 0.4 m
π
A2 = × (0.4)2 = 0.12564 m 2
4
q = 250 liters/s = 0.25 m3/s

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 2010)
Fluid Mechanics and Machinery (April/May 2010) 4.7
5.7

θ 0.25
V1 = = = 7.96 m/s
A1 0.03141
θ 0.25
V2 = = = 1.99 m/s
A2 0.12564

(V1 − V2 ) 2 (7.96 − 1.99)2

he = =
2g 2 × 9.81
he = 1.816 m of water.

5. q = 20 litls = 0.2 m3/s

Hm = 850 mm = 0.85 m
N = 900 rpm
N θ 900 0.02
Ns = =
H m3/ 4 (0.85)3/ 4

Ns = 143.78

6. Fundamental units cannot be broken down. Some SI fundamental units

are cm, kg.
Derived units are combinations of base units. For example, a meter per
second is a derived unit for velocity.

7. U = U1 = U2 = 10 m/s
q = 0.7 m3/s
Angle of deflection = 160°
Angle, f = 180° − 160° = 20°
Coefficient of velocity, Cv = 0.98
H = 30 m
V1 = Cv 2 gH
= 0.98 2 × 9.81 × 30

V1 = 23.77 m/s
Vω1 = V1 = 23.77
Vrl = V1 − U1 = 23.77 − 10
Vrl = 13.77 m/s
Vr 2 = Vr1 = 13.77 m/sec.

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 4.8
5.8 B.E./B.Tech. Question
B.E./B.Tech. Question Papers
Papers

Vω 2 = Vr 2 cos φ − U 2

= 13.77 cos 20° − 10

Vω 2 = 2.94 m/s.

Hydraulic efficiency
2[Vw1 + Vw 2 ]
Nh = × U = 94.54 %
V12

8. D1 = 100 mm, B1 = 50 mm
D2 = 300 mm, B2 = 20 mm
Vf 1 = 3 m/s, B2 = 60°
N = 1500 rpm
θ θ
Vf 1 = =
A1 π D1 B1

3 = 0.0471 m3/s
θ
Vf 2 =
A2
0.04 − 71
= = 2.998 m/s
π × 0.1 × 0.05

9. The difference between the theoretical discharge and the actual discharge
of the pump is called slip of the reciprocating pump.

10. The Air-vessel stores excess quantity of water during the first half of the
delivery stroke and supplies excess quantity of water to the delivery pipe
during the second half of the stroke.

PART B
11. (a) The resisting force K depends upon
• Length, = l,
• Velocity, = V,
• Viscosity, = µ,
• Density, = p,
• Bulk modulus, = K.

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Fluid Mechanics and Machinery (April/May 2010) 4.9
5.9

K Al a ⋅ V b ⋅ µ c ⋅ p d ⋅ K e ... (1)

where A is the non-dimensional constant.

Substituting the dimensions on both sides of the equation (1)
MLT −2 = ALa ( LT −1 ) b ⋅ ( ML−1T −1 )c ⋅ ( ML−3 )d ⋅ ( ML−1T −2 )e .

Equating the powers of M, L, T on both sides,

Power of M, 1=c+d+e
Power of L, 1 = a + b − c − 3d − e
Power of T, − 2 = − b – c – e
There are five unknowns but equations are only three. Expressing
the three unknowns in terms of two unknowns (µ and K).
∴ Express the value of a, b, and d in terms of c and e.
Solving
d =1-c–e
b = 2 - c – 2e
a = 1 – b + c + 3d + e
= 1 – (2 – c - 2e) + c + 3 (1 – c - e) + e
= 1 – 2 + c +2e + c + 3 – 3c – 3e + e
=2–c
Substituting these values in (1) we get

R = Al 2 − c ⋅V 2 − c − 2 e ⋅ μ c ⋅ ρ1− c − e ⋅ K e
= Al 2 ⋅V 2 ⋅ ρ(l − c ⋅V − c ⋅ μ c ⋅ ρ − c ) ⋅ (v −2 e ⋅ ρ − e ⋅ K e )
ρ p
⎛ μ ⎞ ⎛ K ⎞
= Al 2V 2 ρ ⎜ ⋅
⎝ ρVL ⎟⎠ ⎜⎝ ρ v 2 ⎟⎠
⎡⎛ M ⎞ ρ ⎛ K ⎞ ρ ⎤
Rp = AρI 2V 2φ ⎢⎜ ⎟ ⋅⎜ 2⎟
⎥⋅
⎢⎣⎝ ρVL ⎠ ⎝ ρV ⎠ ⎥⎦

(b) d1 = 300 mm = 0.3 m

d2 = 150 mm = 0.15 m
S = 0.82
Z1 = 01 Z2 = 1.2 m

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 4.10
5.10 B.E./B.Tech. Question
B.E./B.Tech. Question Papers
Papers

P1 = 140 KN/m2
P2 = 80 KN/m2
q = 0.20 m2/s
⎛ p ⎞ ⎛p ⎞
h = ⎜ 1 = Z1 ⎟ − ⎜ 2 + Z 2 ⎟
⎝ pg ⎠ ⎝ pg ⎠

P = S × Pwater = 6.82 × 100 = 820 kg/m3

⎛ 140 × 103 ⎞ ⎛ 80 × 103 ⎞
h=⎜ − + 1.2⎟
⎝ 820 × 9.81⎟⎠ ⎜⎝ 820 × 9.81 ⎠

6.25 m of gasoline
Theoretical discharge
a a
θth = 1 2 2 gh
a12 - a22
π π
a1 = × d12 = × 0.32 = 0.0707 m 2
4 4
π π
a2 = × d22 = × 0.152 = 0.177 m 2
4 4
θ act
cd =
θ th
0.0707 × 0.0177
θ th = × 2 × 9.81 × 6.25
(0.0707)2 − (0.0177)2
= 0.2 m3 /s
0.2
cd = =1
0.2
Coefficient of discharge of venturimeter cd = 1

12 (a) D = 0.3 m
0.3
R= = 0.15 m
2
Umax = 2 m/s
Velocity U = 1.6 m/s
Y = R − r = 0.15 − 0.10 = 0.05 m

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 2010)
Fluid Mechanics and Machinery (April/May 2010) 4.11
5.11

U max − U
= 5.75log10 ( Rly )
U*
2 − 1.6 ⎛ 0.165 ⎞
= 575log10 ⎜
U* ⎝ 0.05 ⎟⎠

= 5.75 log 103

= 2.7434
0.4
U* = = 0.1458 m/s
2.7434
u −U
= 5.75log10 ( y /R ) 43.75
u*

Y = R, U becomes Umax
U max − U
∴ = 5.75log10 = U ( R /R) + 3.75
U*

= 3.75
2 −U
= 3.75
0.1458
U = 2 − (0.1458 × 3.75)

U = 0.4533 m/s
Discharge θ = Area × AV. Velocity
π
= D 2 × U − 1 Substituting the value
4
q = 0.1027 m3/s

(b) d = 0.4 m
4 f LV 2 4 f LV 2
h= =
2 gd 2 g (0.4)
V* = Velocity of each parallel pipe
D* = Diameter of each parallel pipe
4 f LV*2
h* =
2 g × d*

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 4.12
5.12 B.E./B.Tech. Question
B.E./B.Tech. Question Papers
Papers

h = h*
4 f LV 2 4 f × L × V*2
=
2 g (0.4) 2 g × d*
V 2 V*2 V 2 0.4
⇒ = ⇒ 2 = → (1)
0.4 d* V* d*

q single pipe = q parallel (1) + q parallel (2)

V × A = 2 × V * × A*
π π
V× (0.4)2 = 2 × V* × ( d* )2
4 4
π
2 × d *2
V 4 2d *2
= =
V* π 0.16
(0.4)2
4
Square on both sides
V2 4 d*4
= → (2)
V*2 0.0256

Comparing equations 1 and 2,

0.4 4 d*4
=
d* 0.0256

0.4 × 0.0256
d*5 = = 0.00256
4
d* = (0.00256)1/5 = 0.303 m.

13. (a) Given:

V = f (H, D, m, r, g)
(or)
f1 (V, H, D, m, r, g) = 0 (1)
From the above equation
Total number of variables, n = 6
Dimensions of each of the 6 variables:
V = LT −1 H = L
D=L m = M L −1T −1
r = ML−3 g = LT −2

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Fluid Mechanics and Machinery (April/May 2010) 4.13
5.13

Thus, number of fundamental dimensions, m = 3

∴ Number of p − terms = n − m = −3 = 3
Now, equation (1) becomes.
f1(p1, p2, p3) = 0 (2)
• Each p term contains m + 1 variables, where m = 3.
• So it contains 3 repeating variables.
• Since V is a dependent variable it should not be selected as
repeating variable.
Choosing H, g, r as repeating variables, the 3p terms are.
π 1 = H a ⋅ g b ⋅ ρ c ⋅V
1 1 1

π 2 = H a2 ⋅ g b2 ⋅ ρc ⋅ D 2

π3 = H ⋅ g ⋅ ρ ⋅ μ
a3 b3 c3

First p term: π1 = H a1 ⋅ g b1 ⋅ ρ c1 ⋅V
Substituting dimensions on both sides,
b1
M ° L°T ° = La1 ( LT −2 ( ML−3 )c1 ⋅ ( LT −1 )

Equating the powers on both sides.

Power of M, 0 = C, ∴C1 = 0
Power of L, 0 = a1 + b1 −3c1 + 1 a1 + b1 = −1
1
Power of T, 0 = −2b1 − 1, b1 = −
2
∴ a1 = −1 + /2 = −1/2
a1 = −1/2
∴ p1 becomes,
V
π1 = H −1/ 2 ⋅ g −1/ 2 ⋅ ρ° ⋅V =
gH

Second p term: π 2 = H a2 ⋅ g b2 ⋅ ρ c2 ⋅ D
Substituting the dimensions on both sides.
M ° L°T ° = La2 ⋅ ( LT −2 ) 2 ⋅ ( ML−3 )c2 ⋅ L
b

Equating the powers of M, L, T

M 0 = C2 ∴C2= 0
L a2 + b2 – 3C2 + 1 = 0

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 4.14
5.14 B.E./B.Tech. Question
B.E./B.Tech. Question Papers
Papers

a2 + b2 = −1
T − 2b2 = 0
b2 = 0 ∴ a2 = −1

D
Thus, π 2 = H −1 ⋅ g ° ⋅ ρ° ⋅ D =
H
Third p terms: p 3 = H a3 ⋅ g b3 ⋅ ρ c3 ⋅ m
Substituting the dimensions on both sides.
M ° L°T ° = La3 ⋅ ( LT −2 )b3 ⋅ ( ML−3 )C3 ⋅ ML−1T −1

Equating the powers

M 0 = c3 + 1
∴C3 = −1
L a3 + b3 – 3C3 − 1 = 0
a3 + b3 = −2
T – 2b3 – 1 = 0
b3 = −1 / 2 ∴ a3 = −2 + 1 / 2 = − 3/2

a3 = − 3/2

Now, π 3 = H −3/ 2 ⋅ g −1/ 2 ⋅ ρ −1 ⋅ μ

Multiplying and dividing by V
μ
=
Hf gH
μV μ
π3 = = ⋅ π1
HfV gH HfV

Substituting the values of p1, p2 and p3 in equation (2),

⎛ V D V ⎞
ρ1 ⎜ , π1
, =0
⎝ gH H HfV ⎟⎠
Or
V ⎡D μ ⎤
= φ ⎢ , π1 ⎥
gH ⎣H HfV ⎦

⎡D μ ⎤
⇒ V = 2 gH φ ⎢ , ⎥
⎣ H ρ VH ⎦
Multiplying by a constant does not change the character of p terms.

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 2010)
Fluid Mechanics and Machinery (April/May 2010) 4.15
5.15

(b) Given
h is a function of r, µ,w, D and Q
h = r (r, µ,w, D, Q)
(or)
r1 = (h, r, µ,w, D, Q) = 0 (1)
Hence, total number of variables n = 6
Number of fundamental dimensions, m = 3
Dimensions of each variable are
h = Dimensionless w = T −1
r = ML−3 D=L
m = ML T−1 −1
Q = L3 T −1
∴m=3
Number of p terms = n −m = 6 − 3 = 3
∴ (1) becomes, r1 (p1, p2, p3) = 0
Each p term contains m + 1 variables and m number of repeating
variables.
Choosing D, W and r as repeating variables, we have
π1 = D a ⋅ ω b ⋅ ρ c ⋅ η
1 1 1

π 2 = D a ⋅ ω b ⋅ ρc ⋅ η
2 2 2

π 3 = D a ⋅ ω b ⋅ ρc ⋅ Q
3
3 3

First p term (p1)

π1 = D a ⋅ ω b ⋅ ρ c ⋅ η
1 1 1

M ° L°T ° = La1 ⋅ (T −1 ) 1 ⋅ ( ML−3 ) ⋅c1 ⋅M ° L°T °

b

Equating powers on both sides,

Power of M, 0 = c1 + 0 ⇒ c1 = 0
Power of L, 0 = a1 + 0, ⇒ a1 = 0
Power of T, 0 = −b1 + 0, ⇒ b1 =0
Substituting the values of a1 b1 and c1 in p we get,
π 1 = D ° ⋅ ω ° ⋅ ρ° ⋅ η = η
Second p term
π 2 = D a ω b ⋅ ρc ⋅ μ
2 2 2

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 4.16
5.16 B.E./B.Tech. Question
B.E./B.Tech. Question Papers
Papers

Substituting dimensions on both sides,

b2
M ° L°T ° = La2 ⋅ (T −1 ⋅ ( ML−3 )c 2 ⋅ ( ML−1T −1 )

Equating powers on both sides,

Power of M, 0 = c2 + 1 ⇒ c2 = −1
Power of L, 0 = a2 – 3 c2 − 1
∴a2 = 3C2 + 1
= −3 + 1 = −2
Power T, 0 = −b2 – 1 ⇒ b2 = −1
Substituting the values of a2 b2 and c2 in p2,
μ
π 2 = D −2 ⋅ ω −1 ⋅ ρ −1 ⋅ μ =
D ωρ
2

Third p term
π3 = D a ⋅ ω b ⋅ ρc ⋅ Q
3 3 3

Substituting the dimensions on both sides,

M ° L°T ° = La3 ⋅ (T −1 )b3 ⋅ ( ML−3 )c3 ⋅ L3T −1

Equating the powers of M, L and T on both sides,

Power of M, 0 = c3 ⇒ c3 = 0
Power of L, 0 = a3 = 3c3 – 3 = −3
⇒ a3 = 3c3 – 3 = −3
Power of T, 0 = −b3 −1,
b3 = −1
Substituting the values of a3, b3, and c3 in p3,
Q
π 3 = D −3 ⋅ ω −1 ⋅ ρ° ⋅ Q =
D 2ω
Substituting the values of p, p2 and p3 in equation (2),
⎛ μ Q ⎞ ⎡ μ Q ⎤
f1 ⎜ η, 2 , 2 ⎟ = 0 (or) η = φ ⎢ 2 , 2 ⎥
⎝ D ωρ D ω ⎠ ⎣ D ωρ D ω ⎦

14. (a) Given

Guide blade angle, α = 22°
Velocity of the flow, V f = V f = 3 m/s
1 2

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Fluid Mechanics and Machinery (April/May 2010) 4.17
5.17

D1 = 300 mm = 0.3 m
D2 = 600 mm = 0.6 m
Speed, N = 300 rpm
Discharge at outlet is radial,
∴ β = 90° and Vω = 0
2

Width of the wheel at inlet, B1 = 150 mm

FM − 47 = 0.15 mm
Solution
Tangential velocity of wheel at inlet,
π D1 N π × 0.3 × 300
u1 = = = 4.71m/s
60 60
Tangential velocity of wheel at outlet,
π D2 N π × 0.6 × 300
u1 = = = 9.43 m/s
60 60
Absolute velocity of water at inlet,
Vf
3
V1 = 1 = = 8.0084 m/s
sin α sin 22

Velocity of whirl at inlet,

Vω = V1 cos α = 8.0084 × cos 22
1

Vω = 7.4253 m/s
1

Discharge through runner, Q = π D1 B1 Vf

1

= p × 0.3 × 0.15 × 3 = 0.4241 m3/s

For runner blade angles:
From inlet velocity triangles,
Vf 3
tan θ = 1
= = 1.1048
Vω −u
7.4253 − 4.71
1 1

q = tan−1(1.1048) = 47.85°
q = 47.85°

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 4.18
5.18 B.E./B.Tech. Question
B.E./B.Tech. Question Papers
Papers

From outlet velocity triangles,

Vf 3
tan φ = 2
= = 0.3181
u 9.43
φ = tan −1 (0.3181)
φ = 17.65°
u2

f V2 = Vf 1 = Vf 2

Vr 2

V1

Vr1 Vf 1
a q
u1
uw1

Power developed,
P = ρ QVω × u1 = 1000 × 0.4241 × 7.4253 × 4.71
1

P = 14.83 kW

Result
(i) θ = 47.85°
φ = 17.65°
(ii) P = 14.83 KW
(b) Given data
Head, H = 20 m
Power, P = 15 MW
= 15000 kW
Outer diameter of runner, D0 = 4 m

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 2010)
Fluid Mechanics and Machinery (April/May 2010) 4.19
5.19

Hub diameter Db = 1.5 m

Guide blade angle, a = 30°
Hydraulic efficiency, hh = 0.9 = 90%
Overall efficiency, h0 = 0.8 = 80%
Discharge is radial
∴ β = 90° & Vω = 0
2

θ=? φ=? N =?
Solution
P
η0 =
H
15000 × 103
0.80 =
1000 × 9.81 × Q × 20
θ = 95.56 m3 /s
We know that,
π
θ= ( D02 − Db2 ) × V f
4 1

π
95.56 = (4 2 − 1.52 ) × Vf
4 1

Vf = 8.8487 m/s
1

u2

Vw1 = 0 b f
V1 = V2
Vf 2

V1
Vr1
Vf1
q a
Vw1

u1

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Papers

From inlet velocity triangle,

Vf
tan α = 1

Vω
1

8.8487
tan 30° =
Vω
1

Vω = 15.33 m/s
1

Vω u1
Hydraulic efficiency, η = 1

h
gH
15.33 × u1
0.9 =
9.81 × 20
u1 = 11.5186 m/s

Vf
From velocity triangle, tan θ = 1

Vw − u1
1

8.8487
=
15.33 − 11.5186
= 2.3216
θ = tan −1 (2.3216)
θ = 66.69°
For Kaplan turbine, u1 = u2 = 11.5186 m/s
Vf = V f = 8.8487 m/s
1 2

From outlet velocity triangle,

Vf 8.8487
tan φ = 2
= = 0.7682
u2 11.5186
φ = tan −1 (0.7682)
φ = 37.53°

π D0 N
Peripheral velocity, u1 =
60
π ×4×N
11.5186 =
60
N = 54.9973 rpm

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Fluid Mechanics and Machinery (April/May 2010) 4.21
5.21

Result
(i) Runner vane angles, Inlet θ = 66.69°
Outlet, φ = 37.53°
(ii) Speed of the turbine, N = 54.9973 rpm

15. (a) Given data

D = 120 mm = 0.12 m
L = 300 mm = 0.3 m
L 0.3
∴ r= = = 0.15 m
2 2
N = 60 rpm
Qactual = 2.95 litres/s = 0.00295 m3/s
hd = 25 m ld = 20 m
dd = 100 mm = 0.10 m
Solution
(i) Theoretical discharge,
ALN
Qth =
60
π
× 0.122 × 0.3 × 60
= 4 = 0.00339 m3 /s
60
Power required to run the pump,
P = wQth (hs + hd)
= 9810 × 0.00339 (0 + 25)
P = 0.8313 kW
(ii) Percentage slip
Qact
% slip Qth − × 100
Qth
0.00339 − 0.00295
= × 100
0.00339
% slip = 12.98%
(iii) Acceleration head at the beginning of the stroke

ld A 2 ⎡∵ θ = 0° ⎤
had = ω r⎢ ⎥
g ad ⎣cos 0 = 1⎦

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Papers

π
(0.12)2 2
20 4 ⎛ 2π × 60 ⎞
= × ×⎜ ⎟ × 0.15
9.81 π × (0.1)2 ⎝ 60 ⎠
4
= 17.38 m of water
Acceleration head at the middle of the stroke:
ld A 2
had = ω r cos 90°
g ar
had = 0

(b) (i) External gear pump

It has two identical intermeshing spur wheels A and B. These
two wheels are meshed with a fine clearance inside the casing.
When the fluid flows, it will make joint with the point of teeth
contact. Among these, one of the wheels is connected to the
driving shaft and the other one is driven. Before starting the
pump it is filled with liquid.

A B

The liquid is entrapped between teeth and allowed to discharge

into the delivery pipe when the gear rotates. These rotating
gears itself set up sufficient pressure to force the liquid. So,
each and every tooth on the wheels acts as a piston of a recip-
rocating pump.

(ii) Lobe pump

This pump resembles like a gear pump in action but not in
appearance. The specially designed wheels are arranged inside
the casing. The wheels have two or more lobes. Here also, the
liquid is filled in the pump before it starts.

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5.23

A B

Lobe Pumbs

Lobe pumps
The liquid makes a tight joint at the point of contact. The liquid
is entrapped and flown to the delivery end when the lobes rotate.
The lobes itself build up sufficient pressure to force the liquid.
The main drawback is that the discharge is not uniform.
(iii) Vane pumps

Receiver

Rotating Drum

Vanes

Caring

Inlet
Lobe pumps

A disc is eccentrically rotating inside the pump casing. Slots

containing vanes are provided on the disc. The vanes are moved
against the casing due to centrifugal force when the disc rotates.
So, the liquid tight seal is formed. Due to continuous rotation of
disc, the liquid is entrapped and forced to the delivery end with
sufficient pressure. The vanes are generally hinged. In some
cases, springs are used to press the vanes against casing.

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 B.E./B.Tech. DEGREE EXAMINATION,
NOV/DEC 2009
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY
Time: Three hours Maximum: 100 marks
Answer ALL questions
PART A (10 × 2 = 20 marks)
1. What is the variation of viscosity with temperature for fluids?

2. Find the height of a mountain where the atmospheric pressure is 730 mm

of Hg at normal conditions.

3. List the causes of minor energy losses in flow through pipes.

4. What is T.E.L?

5. What is dynamic similarity?

6. Define Froude’s number.

7. Define specific speed.

8. What are the different types of draft tubes?

9. Can actual discharge be greater than theoretical discharge in a recipro-

cating pump?

10. Which factor determines the maximum speed of a reciprocating pump?

PART B (5 × 16 = 80 marks)
11. (a) Explain the properties of a hydraulic fluid.
Or
(b) A 0.5 shaft rotates in a sleeve under lubrication with viscosity 5
poise at 200 rpm. Calculate the power lost for a length of 100 mm,
if the thickness of the oil is 1 mm.

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5.25

12. (a) Explain the losses of energy in flow through pipes.

Or
(b) Determine the equivalent pipe corresponding to 3 pipes in series
with lengths and diameters l1, l2, l3, d2, d3 respectively.

13. (a) Express efficiency in terms of dimensionless parameters using den-

sity, viscosity, angular velocity, diameter of the rotor and discharge
using Buckinghain Π theorem.
Or
(b) Classify models with scale ratios.

14. (a) Determine the hp of the Pelton wheel with tangential velocity
20 m/s. Head 50 m, discharge 0.03 m3/s, side clearance angle 15°.
Take CV as 0.975.
Or
(b) A reaction turbine at 450 rpm, head 120 m, diameter at inlet 120 cm,
flow area 0.4 m2 has angles made by absolute and relative velocities
at inlet 20° and 60° respectively. Find volume flow rate, H.P. and
efficiency.

15. (a) Determine the minimum speed for starting a centrifugal pump.
Or
(b) Explain the characteristic curves of a centrifugal pump.

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 Solutions
PART A
1. Viscosity of liquids decrease when the temperature increases. Viscosity
of gases increases with increase in temperature.

2. Refer answer 5 from April/May 2010 Question paper.

3. Minor energy losses inflow through pipes can happen with

• Sudden contraction
• Sudden enlargement
• Sudden obstruction in the pipe
• Loss of energy at the exit of pipe.

4. The total energy losses of the pipe plotted at different sections are joined
together to obtain the total energy line.

5. A relationship existing between two fluids flow when they have identi-
cal types of forces that one parallel at all the corresponding points, with
magnitudes related by a constant scale factor. Dynamic similarity makes
it possible to scale results from model tests to predict corresponding
results for the full scale prototype.

6. The Froude number is a dimensionless number defined as the ratio of

a characteristic velocity to a gravitational wave velocity. It also can be
defined as the ratio of a body’s inertia to gravitational forces.

V
Fr =
C

7. Specific speed NS is a non-dimensional number used to classify pump

impellers as to their type and proportions. In imperial units it is defined
as the speed in revolutions per minute at which a geometrically similar
impeller would operate if it were of such a size as to deliver one gallon
per minute against one foot of hydraulic head.

8. Different types of draft tubes are:

• Straight divergent tube,
• Simple elbow type.

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Fluid Mechanics and Machinery (Nov/Dec 2009) 4.27
5.27

9. In some cases when the slip of the pump is a negative slip, the q act may
be higher than q and the Negative slip is possible when the delivery pipe
is short, suction pipe is long and pump is running at high speed.

10. Factors are:

• Atmospheric head
• Delivery head
• Acceleration head
• Suction head

PART B
11. (a) Properties of Hydraulic Fluid

Viscosity
Viscosity is a measure of a hydraulic fluid’s resistance to flow. It is a
hydraulic fluid’s most important characteristic and has a significant
impact on the operation of the system.
When a hydraulic oil is too thin (low viscosity), it does not seal suf-
ficiently. This leads to leakage and wear of parts. When a hydraulic
oil is too thick (high viscosity), the fluid will be more difficult to
pump through the system and may reduce operating efficiency.
All hydraulic fluids must be able to retain optimum viscosity during
operation in cold or hot temperatures, in order to consistently and
effectively transmit power.

Compressibility
Compressibility is a measure of the amount of volume reduction due
to pressure. Although hydraulic oils are basically incompressible,
slight volume reductions can occur under certain pressure ranges.
Compressibility increases with pressure and temperature and has
significant effects on high-pressure fluid systems. It causes servo
failure, efficiency loss, and cavitation; therefore, it is important for a
hydraulic oil to have low compressibility.

Wear Resistance
Wear resistance is a hydraulic fluid’s ability to reduce the wear rate
in frictional boundary contacts. Antiwear hydraulic fluids contain
antiwear components that can form a protective film on metal surfaces

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to prevent abrasion, scuffing, and contact fatigue. Antiwear additives

enhance lubricant performance and extent equipment life.
Oxidation Stability
Oxidation stability is a hydraulic oil’s resistance to heat-induced
degradation caused by a chemical reaction with oxygen. Hydraulic
oils must contain additives that counteract the process of oxidation,
improve the stability and extend the life of the fluid. Without these
additives, the quality of the hydraulic oil will deteriorate quickly.
Thermal Stability
Thermal stability is the ability to resist breakdown at elevated tem-
peratures. Antiwear additives naturally degrade over time and this
process can be accelerated at higher temperatures. The result of poor
thermal stability is the formation of sludge and varnish which can
clog filters, minimize flow and increase downtime. In addition, as
these antiwear agents decompose at high temperatures, acids are
formed which attack bronze and yellow metals in piston pumps and
other hydraulic system components. Hydraulic oils can be formu-
lated with very high levels of thermal stability to minimize these
issues and help extend the life of the hydraulic fluid and the compo-
nents of the hydraulic system.
Filterability
Water can react with additives in hydraulic fluids forming oil insol-
uble material. These contaminants can precipitate from the lubricant
and block filters, valves and other components resulting in decreased
oil flow or the system going on bypass. Blockage can eventually
result in unplanned downtime. Hydraulic fluids are designed to be
filtered with modern filtration systems without fear of the additive
being depleted or removed from the system. This enables systems
to stay clean without sacrificing critical performance requirements
such as antiwear, rust protection or foam inhibition.
Rust and Corrosion Protection
In many systems, water can enter as condensation or contamination,
and mix with the hydraulic oil. Water can cause rusting of hydrau-
lic components. In addition, water can react with some additives
to form chemical species which can be aggressive to yellow met-
als. Hydraulic oil formulations contain rust and corrosion inhibi-
tors which prevent the interaction of water or other chemical species
from attacking metal surfaces.

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Foam Resistance
Foam results from air or other gases becoming entrained in the
hydraulic fluid. Air enters a hydraulic system through the reservoir
or through air leaks within the system.
A hydraulic fluid under high pressure can contain a large volume of
dissolved or dispersed air bubbles. When this fluid is depressurized,
the air bubbles expand and produce foam. Because of its compress-
ibility and poor lubricating properties, foam can seriously affect the
operation and lubrication of machinery.
Proper foam inhibitors modify the surface tension on air bubbles so
they more easily break up.
Demulsibility
Water that enters a hydraulic system can mix or emulsify with the
hydraulic oil. If this ‘wet’ fluid is circulated through the system, it can
promote rust and corrosion. Highly refined mineral oils permit water
to separate or demulsify quickly. However, some of the additives used
in hydraulic oils promote emulsion formation, preventing the water
from separating and settling out of the fluid. Demulsifier additives
are incorporated to promote water separation from hydraulic fluids.
Hydrolytic Stability
When hydraulic fluids come into contact with water, the water can
interact with the additive system of the hydraulic oil resulting in
the formation of acids. Hydraulic fluids that lack hydrolytic stability
hydrolyze in the presence of water to form oil insoluble inorganic
salts that can block filters and valves inhibiting oil flow. This can
result in hydraulic system failure. Properly formulated hydraulic flu-
ids are designed to contain additives that are resistant to interactions
with water, helping to extend the life of the equipment.
Seal Compatibility
Leaking hydraulic fluids can cause many issues from simple house-
keeping problems to more serious safety concerns and lubrication
failures. Most hydraulic systems utilize rubber seals and other elas-
tomers to minimize or prevent hydraulic oil leakage. Exposure of
the elastomer to the lubricant under high temperature conditions can
cause the rubber seals to harden, crack and eventually leak. On the
other hand, hydraulic oil exposure can seal to swell excessively pre-
venting hydraulic valves and pistons from moving freely. Hydraulic
oils are tested against a variety of seal materials to ensure that the
hydraulic fluid will be compatible with seals under various conditions.

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(b) µ = 5 poise
Ns
= 0.5
m2
Diameter D = 0.5 m.
Speed N = 200 rpm.
Sleve length L = 100 mm = 0.1 m
Thickness of oil film f = 1mm = 10−3 m
π DN π × 0.5 × 200
u= = ≤ 5.23 m/s
60 60
du
τ=μ
dy
5.23
τ = 0.5 × = 2615 N/m 2
10 −3

Shear force

F = τ × Area
= 2615 × πD × L
= 2615 × π × 0.5 × 0.1
= 410.76 N
Torque
D 0.5
T =F× = 410.76 × = 102.69 Nm
2 2
2π NT 2π × 200 × 102.69
Power lost = = = 2150.75 w.
60 60

12. (a) Pipe Flow

Losses in Pipe
It is often necessary to determine the head loss, hL, that occur in a
pipe flow so that the energy equation, can be used in the analysis of
pipe flow problems.
The overall head loss for the pipe system consists of the head loss
due to viscous effects in the straight pipes, termed the major loss and
denoted hL-manjor.
The head loss in various pipe components, termed the minor loss
and enoted hL-minor.

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5.31

That is,
hL = hL-major + hL-minor

The head loss designations of “major” and “minor” do not necessar-

ily reflect the relative importance of each type of loss.
For a pipe system that contains many components and a relatively
short length of pipe, the minor loss may actually be larger than the
major loss.

Major Losses
The head loss, hL-major is given as;
lV 2
hL -major = f
D2 g
where f is friction factor.
Above mention equation is called the Darcy-Weisbach equation. It
is valid for any fully developed, steady, incompressible pipe flow,
whether the pipe is horizontal or on hill.
Friction factor for laminar flow is;
64
f =
Re
Friction factor for turbulent flow is based on Moody chart.
It is because, in turbulent flow, Reynolds number and relative rough-
ness influence the friction.
ρVD
Reynolds number Re =
μ
ε
Relative roughness =
D
(Relative roughness is not present in the laminar flow.)
The Moody chart is universally valid for all steady, fully developed,
incompressible pipe flows.
The following equation form Colebrook is valid for the entire non-
laminar range of the Moody chart. It is called Colebrook formula.

1 ⎛ε/D 2.51 ⎞
= −2.0 log ⎜ + ⎟
f ⎝ 3.7 Re f ⎠

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 0.1 5.32
4.32
0.09 Wholly Turbulent Flow

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0.07 0.05

32 32
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03 ∋
f 0.004
D
B.E./B.Tech.

0.025
0.002
0.02 0.001
B.E./B.Tech. Question

Laminar 0.0008
0.0006
Flow 0.0004
0.015
Question Papers
Papers

Smooth 0.0002
Transition Range 0.0001
0.00005
0.01
0.009
0.008 0.00001
2(103) 4 6 8 2(104) 4 6 8 2(105) 4 6 8 2(106) 4 6 8 2(107) 4 6 8
103 104 105 106 107
rVD
Re = m

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 2009)
Fluid Mechanics and Machinery (Nov/Dec 2009) 4.33
5.33

Minor Losses
The additional components such as valves and bend add to the over-
all head loss of the system, which in turn alters the losses associated
with the flow through the valves.
Minor losses termed as,

V2
hL - min or = KL
2g

Where KL is the loss coefficient.

Each geometry of pipe entrance has an associated loss coefficient.
Entrance flow conditions and loss coefficient.

Reentrant KL = 0.8 Sharp-edged KL = 0.5

(a) (b)

Slightly-rounded KL = 0.2 Well-rounded KL = 0.04

(c) (d)

A1 A1
Condition: = 0 or =∞
A2 A2

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Exit flow conditions and loss coefficient.

(2)

(1)

Reentrant KL = 1.0 Sharp-edged KL = 1.0

(a) (b)

Slightly-rounded KL = 1.0 Well-rounded KL = 1.0

(c) (d)

A1 A1
Condition: = 0 or =∞
A2 A2
Losses also occur because of a change in pipe diameter.
For sudden contraction:
0.6
V22
A1 A2 hL = KL
2g
0.4

KL

0.2

0
0 0.2 0.4 0.6 0.8 1.0
A2/A1

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2 2 2
⎛ A⎞ ⎛ A⎞ ⎛ 1⎞
K L = ⎜1 − 2 ⎟ = ⎜1 − 2 ⎟ = ⎜1 − ⎟
⎝ A1 ⎠ ⎝ Ac ⎠ ⎝ Cc ⎠

For sudden expansion:

1.0

V12
0.8 A1 A2 hL = KL
2g

0.6
KL
0.4

0.2

0
0 0.2 0.4 0.6 0.8 1.0
A1/A2

2
⎛ A⎞
K L = ⎜1 − 1 ⎟
⎝ A2 ⎠

(b) Equivalent Pipe

A compound pipe consisting of several pipes of varying diameters
and length may be replaced by a pipe of uniform diameter which is
known as equivalent pipe.
For equivalent pipe, the loss of head and discharge are equal to the
loss of head and discharge of compound pipe.
Let L1, L2, L3 ⇒ Length of compound pipe
D1, D2, D3 ⇒ Diameters of compound
4 f ⎡ LV 2
LV2 LV2 ⎤
H= ⎢
1 1
+ 2 2 + 3 3 ⎥ (1)
2 g ⎣ D1 D2 D3 ⎦
Q = A1V1 = A2 V2 = A3 V3
Q Q 4Q
V1 = = =
A1 π D 2 π D12
4 1
Similary
4Q
V2 =
π D22

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Papers

4Q
V3 =
π D32
Substituing V1, V2, V3 in (1),

4f ⎡ L ⎛ 4Q ⎞ 2 L ⎛ 4Q ⎞ 2 L ⎛ 4Q ⎞ ⎤
H= ⎢ 1⎜ ⎟ + D ⎜ π D2 ⎟ + D ⎜ π D2 ⎟ ⎥
2 3

2g ⎢⎣ D1 ⎝ π D12 ⎠ 2 ⎝ 2 ⎠ 3 ⎝ 3 ⎠⎥⎦
2 ⎡ L L3 ⎤
4 × 16 + Q L2
H= ⎢ 5 + 5 + 5⎥
1
(2)
π × 2 g ⎣ D1 D2 D3 ⎦
2

Head loss in the equivalent pipe,

4 fLV 2
H=
2 gn D
where,
4Q
V=
π D2
2
4 f L ⎛ 4Q ⎞
H=
2 g D ⎜⎝ π D 2 ⎟⎠
4 × 16 f Q 2 ⎡ L ⎤
H=
π 2 × 2 g ⎢⎣ D 5 ⎥⎦
Substituting value of H in equation (2),

4 × 16 f Q 2 ⎡ 2 ⎤ 4 × 16 f Q 2 ⎡ L1 L2 L3 ⎤
= ⎢ 5 + 5 + 5⎥
π 2 × 2 g ⎢⎣ D 5 ⎥⎦ π2 × 2 g ⎣ D1 D2 D3 ⎦
L L L L3
= 15 + 25 + 5
D 5
D1 D2 D3

where L = L1 + L2 + L3
The above equation is known as Dupit’s equation.

13. (a) Given,

h is a function of ρ, µ, ω, D and Q
h = f (ρ, µ, ω, D, Q)
f1 (η, ρ, µ, ω, D, Q) (1)
Hence total number of variables, n + 6

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5.37

Number of fundamental dimensions m = 3

Dimensions of each variable are
η = Dimension less
ρ = ML−3
µ = ML−1 T −1
ω = T −1
D=L
Q = L3 T −1
∴m = 3
Number of Π terms = n − m = 6 − 3 = 3
∴ (1) becomes.
f1 (π1, π2, π3) = 0
Each π term contains m + 1 variables and m number of repeating
variables.
Choosing D, ω and ρ and repeating variables, we have
π1 = D a ⋅ ω b ⋅ ρ c ⋅ η
1 1 1

π2 = D ⋅ω ⋅ ρ ⋅ μ
a2 b2 c2

π 3 = D a ⋅ ω b ⋅ ρc ⋅ Q
3 3 3

First p term (p1)

π 1 = D a ⋅ ω b ⋅ ρ c ⋅ η.
1 1 1

M 0 L0T 0 = La1 ⋅ (T −1 )b1 ⋅ ( ML−3 )c1 ⋅ M 0 L0T 0

Equating powers on both sides,
Power of M , 0 = c1 + 0 ⇒ c1 = 0
Power of L1 0 = a1 + 01 ⇒ a1 = 0
Power of T1 0 = −b1 + 0 ⇒ b1 = 0

Substituting the values of a1 b1 and c1 in π we get,

π1 = D 0 ⋅ ω 0 ⋅ ρ 0 ⋅ η = η
Second p term
π 2 = D a ⋅ ω b ⋅ ρ c ⋅ μ.
2 2 2

Substituting dimensions on both sides.

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M 0 L0T 0 = La2 ⋅ (T −1 ) ⋅ ( ML−3 )c2 ( ML−1T −1 )

Equating powers on both sides,

Power of M , 0 = c2 + 1 ⇒ c2 = −1
Power of L, 0 = a2 − 3 c2 − 1
∴ a2 = 3 c2 + 1
= −3 + 1 = −2
Power of T , 0 = − b2 − 1 ⇒ b2 = −1

Substituting the values of a2b2 and c2 in T1

μ
π 2 = D −2 ⋅ ω −1 ⋅ ρ −1 μ =
D ωρ
2

Third p term
π 3 = D a ⋅ ω b ⋅ ρc ⋅ Q
3 3 3

Substituting the dimensions on both sides,

M 0 L0T 0 = La3 ⋅ (T −1 )b3 ⋅ ( ML−3 )c3 ⋅ L3T −1

Equating the powers of M, L and T on both sides,

Power of M , 0 = c3 , ⇒ c3 = 0
Power of L, 0 = a3 − 3 c3 + 3
⇒ a3 = 3 c3 − 3 = −3

Power of T, 0 = − b3 − 1,
b3 = − 1
Substituting the values of a3, b3 and c3 π3.
Q
π 3 = D −3 ⋅ D −3 ⋅ ρ a − Q =
D 3ω
Substituting the values π, π2 and π3 is equation (2),
⎛ μ Q ⎞
f1 ⎜ η, 2 , 3 ⎟ =0
⎝ D ω ρ D ω⎠
⎡ μ Q ⎤
η=φ⎢ , 3 ⎥
⎣ D 2
ω ρ D ω⎦

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5.39

(b) Classification of models with scale ratio.

1. Reynolds Model Law
(Re) model = (Re) prototype

ρmVm Lm ρ pV p L p
So, = (1)
μm μp
ρp Vp Lp 1
⇒ × × × =1
ρm Vm Lm (μ p /μ m )
ρrVr Lr
=1
μr

Lr
Time scale ratio, Fr = [∵ F = m × a]
Vr
Vr
= mr ×
Tr

Discharge scale ratio, Qr = (ρAV)r

Qr = ρr ArVr = ρr L2r Vr

2. Froude Model Law

(Fr)m = (Fr)p
Vm Vp
= (2)
g m Lm g p Lp
gm = gp
Vm Vp
=
Lm Lp

Vp Lp
= = Lr
Vm Lm
Vp
= Vr
Vm

∴Vr = Lr

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 4.40
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Tp Lp Vm
Time scale ratio, Tr = = ×
Tm Vp Lm
Lp Vm 1 L
= × = Lr × = r
Lm Vp Vr Lr
Tr = Lr

Acceleration scale ratio, ar = 1

Discharge scale ratio, ar = Lr5/2

3. Euler Model Law

( Eu ) m = ( Eu ) p
Vm Vp
=
Pm /ρm Pp /ρ p
Vm Vp
= (If ρm = ρ p )
Pm Pp

4. Weber Model Law

(We ) m = We
Vm Vp
=
σ m /Pm Lm σ p /ρ p L p
Vm Vp
= (If ρm = ρ p )
σ m /Lm σ p /L p

5. Mach Model Law

( M )m = ( M ) p
Vm Vm
=
K m /ρ m K /ρ p
Vm Vp
= (If ρm = ρ p )
Km Kp

Vm, Km, rm are velocity, elastic force and density of fluid in model
similarly Vp, Kp, rp are corresponding parameters of prototype.

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 2009)
Fluid Mechanics and Machinery (Nov/Dec 2009) 4.41
5.41

14. (a) Given,

H = 50 m
Q = 0.03 m3/s
u = 20 m/s
f = 15˚
Cv = 0.975
Solution
V1 = Cv 2 gH
= 0.975 × 2 × 9.81 × 50 = 29.755 m/s
Power, P = u (V1 − u) (1 + cos f)
= 20 (29.755 − 20) (1 + cos 15)
= 383.55 W
(b) Given,
N = 450 rpm, H = 120 m, D1 = 1.2 m,
A = 0.4 m2, a = 20°, q = 60°.
Solution
π D1 N
u1 =
60
π × 1.2 × 450
= = u = 28.27 m/s
60
From inlet velocity triangle,
Vf
tan α = 1
Vω
1

Vf
tan 20° = 1

Vω
1

V f = Vω
1 1

V f = Vω tan × 20° = 0.364Vω V f

1 1 1 1

We know that,
Vf
tan θ = 1

Vω − u1
1

0.364Vω
tan 60° = 1
− 28.27
Vω
1

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 4.42
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0.364Vω
1.732 = 1

Vω − 28.27
1

1.368 Vω = 48.96
1

u2

f V2 = Vf 1 = Vf 2

Vr2

V1

Vr1 Vf 1
a q
u1
uw1

Vω = 35.789 m/s
1

V f = 0.364 × 35.789 = 13 m/s

1

(i) Volume flow rate Q = A × V f

1

= 0.4 × 13
= 5.2 m3/s
(ii) Power developed P = ωQH
= 9.81 × 5.2 × 120
P = 6135.57 kW
Vω u1
(iii) Hydraulic efficiency ηh = 1
gH
35.789 × 28.27
=
9.81 × 120
ηn = 0.864 (or) 86.4%

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 2009)
Fluid Mechanics and Machinery (Nov/Dec 2009) 4.43
5.43

15. (a) Minimum Speed for Starting a Centrifugal Pump

According to Euler’s equation, the theoretical Eulerian head of a
centrifugal pump,
1
He = (V u − V u )
g ω 2 2 ω1 1
1 ⎡ 2
He = (u − u12 ) + (V22 − V12 ) + (Vr2 − Vr2 ) ⎤⎥
2 g ⎣⎢ 2 1 2 ⎦

During starting of the pump, the fluid flow will not be there. So, the
fluid flow velocities become zero. It means the velocity due to kinetic
energy and flow velocity are zero. Only the pressure head caused by
(u 2 − u12 )
rotation of rotor will be real which is 2 . Pumping of fluid
2g
will start, when the speed of the pump attains the manometric head.
So, the condition will be,
(u22 − u12 )
≥ Hm
2g
Vd2
Where manometric head H m = hs + hd + h f + h f +
s d
2g
If Vd = 0 and no friction losses in both suction and delivery pipes,
Hm = hs + hd
If manometric efficiency ηm is considered, then the minimum speed
to commence the flow will be,
(u22 − u12 )
= ηm H e
2g
2 2
⎛ π D2 N ⎞ ⎛ π D1 N ⎞
⎜⎝ 60 ⎟⎠ − ⎜⎝ 60 ⎟⎠ = 2 gηm H e
2
⎛πN ⎞
⎟ ( D2 − D1 ) = 2 gηm H e
2 2
⎜⎝
60 ⎠
If D1 = 0.5 D2 and ηm = 0.75 are assumed,
84.64 H e
N=
D2

(b) Characteristic Curves of a Centrifugal Pump

The curves which are plotted from the series of a number of tests on the
centrifugal pumps are known as characteristic or performance curves.

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The following four types of characteristic curves are usually pre-

pared for the centrifugal pumps.
• Main characteristic curves
• Operating charactering curves
• Constant efficiency or mescal curves
• Constant head and constant discharge curves
1. Main Characteristic Curves
To obtain main characteristic curves of a pump, it is run at a constant
speed and the discharge is varied over the desired range measure-
ments which are taken for head (Hm) and shaft power (P) for each dis-
charge. Calculations are mode for the pump efficiency (η0). Curves
are then plotted for head, power and efficiency against discharge.
N = 1000 rpm
00
20

Power, P
00
15
Head, Power, Efficiency

2000
m
rp
0 0
10
=
N
20
N
=

00
15
10

00
00
rp
m

Head, H

Discharge, Q

2. Operating Characteristic Curves

Design Conditions
Head, Power, Efficiency

Efficiency h

Shaft Power, P

Head, H

Discharge, Q

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5.45

Pumps are generally designed for maximum efficiency and that

occurs when the pump operates at the design speed. A particular
set of main characteristics corresponding to the design speed is
called operating characteristics. The operating characteristics help
to obtain the design discharge and heads, this corresponding to the
point of maximum efficiency.
3. Constant Efficiency or Mitchell Curves
These curves are also called as Iso efficiency curves. It depicts the
performance of a pump over its entire range of operation. The data
used for these curves is obtained from the main characteristic curves
namely h Vs Q and Hm Vs Q. For a given efficiency, the values
of discharge obtained for various speeds from the below figure.
A curve for the best performance is obtained when the peak points
to various Iso efficiency curves are joined.
Constant
Efficiency Lines

N5
Head, A

Speed Increasing

N4
N3
N2
N1
Best Speed Line
Discharge, Q

4. Constant Head and Constant Discharge Curves

Hmano (Head)
Q (Discharge)

Hmano = Constant Q = Constant

N (Speed) N (Speed)

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 4.46
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The performance of a variable speed pump is for which the constant

speed variation can be obtained by these curves. When the pump has
a variable speed, the plotted graph between Q and N and Hm and N
may be obtained. In the first case, Hm is kept constant and in second
case Q is kept constant.

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 B.E./B.Tech. DEGREE EXAMINATION,
NOV/DEC 2007
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY
Time: Three hours Maximum: 100 marks
Answer ALL questions
PART A (10 × 2 = 20 marks)
1. How does the dynamic viscosity of (a) liquids and (b) gases vary with
temperature?

2. What is the difference between gauge pressure and absolute pressure?

3. Define stream line. What do stream lines indicate?

4. Write the Bernoulli’s equation in terms of head. Explain each terms.

5. What is the physical significance of Raynold’s number?

6. Define the terms drag and lift.

7. Differentiate between the turbines and pumps.

8. Define specific speed.

9. What do you mean by manometric efficiency and mechanical efficiency

of a centrifugal pump?

10. Define slip of a reciprocating pump.

PART B (5 × 16 = 80 marks)
11. (a) (i) When a pressure of 20.7 MN/m2 is applied to 100 litres of
a liquid and its volume decreases by one litre, find the bulk
modulus of the liquid and identify this liquid.

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(ii) State and prove Pascal’s law.

Or
(b) (i) The barometric pressure at sea level is 760 mm of mercury while
that on a mountain top is 735 mm. If the density of air is assumed
constant at 1.2 kg/m3, what is the elevation of the mountain top?
(ii) Calculate the capillar effect in millimetres in a glass tube of
4 diameter, when immersed in (1) water and (2) mercury. The
temperature of the liquid is 20ºC and the values of surface tension
of water and mercury at 20ºC in contact with air are 0.0735 N/m
and 0.51 N/m respectively. The contact angle for water
q = 0 and for mercurcy q = 130º. Take specific weight of water
at 20ºC as equal to 9790 N/m3.

12. (a) (i) Obtain an expression for continuity equation in Cartesian

coordinates.
(ii) The two dimensional stream function for a flow is Y = 9 + 6x –
4y + 7xy. Find the velocity potential.
Or
(b) (i) State if the flow represented by
u = 3x + 4y and v = 2x – 3y
is rotational or irrotational. Find the potential function, if the
flow is irrotational and vorticity, if it is rotational.
(ii) A 300 mm × 150 m venturimeter is provided in a vertical pipe
line carrying oil of relative density 0.9, the flow being upwards.
The differential U tube mercury manometer shows a gauge
deflection of 250 mm, calculate the charge of oil if the coef-
ficient of meter is 0.98.

13. (a) Obtain an expression for Hagen Polseuille Sow. Deduce the condi-
tion of maximum velocity.
Or
(b) A flat plate 1.5 m × 1.5 m moves at 50 km/h in a stationary air of
density 1.15 kg/m3. If the coefficient of drag and lift are 0.15 and
0.75 respectively, determine
(i) the lift force
(ii) the drag force
(iii) the resultant force and
(iv) the power required to set the plate in motion.

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 2007)
Fluid Mechanics and Machinery (Nov/Dec 2007) 4.49
5.49

14. (a) Obtain an expression for the workdone per second by water on the
runner of a Polton wheel. Hence derive an expression for maximum
efficiency of the Pelton wheel giving the relationship between the jet
speed and bucket speed.
Or
(b) (i) A Pelton wheel in having a mean bucket diameter of 1 m and
is running at 1000 rpm. The non-head on the Pelton wheel is
700 m. If the side clearance angle is 15° and discharge through
nozzle is 0.1 m3/s, find
• Power available at the nozzle and
• Hydraulic efficiency of the turbine. Take C = 1.
(ii) A turbine is to operate under a head of 25 m at 200 rpm. The
discharge is 9 m3/s. If the efficiency is 96% determine specific
speed of the machine, power generated and type of turbine.

15. (a) (i) A centrifugal pump delivers water against a net head of
14.5 meters and a design speed of 1000 rpm. The vanes are
curved back to an angle of 30° with the periphery. The impeller
diameter is 300 mm and outlet width 50 mm. Determine the
discharge of the pump if manometric efficiency is 95%.
(ii) A single acting reciprocating pump running at 50 rpm, delivers
0.01 m3/s of water. The diameter of the piston is 200 mm and
stroke length 400 mm. Determine the theoretical discharge of
the pump, coefficient of discharge and slip and the percentage
slip of the pump.
Or
(b) (i) Explain the working principles of same pump and gear pump
with neat sketches.
(ii) A reaction turbine works at 450 rpm under a heat of 120 m.
Its diameter at inlet is 120 cm and the flow area is 0.4 m. The
angles made by absolute and relative velocity at inlet are 20°
and 60° respectively, with the tangential velocity. Determine
the volume flow rate, the power developed and the hydraulic
efficiency.

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 Solutions
PART A
1. (a) The viscosity of liquids decrease with the increase in temperature.
(b) The viscosity of gases increase with the increase in temperature.

2. Absolute Pressure Gauge Pressure

• It is defined as the pres- • It is defined as the pressure
sure which is measured with which is measured with the help
reference to absolute vaccum of measuring instrument, in
pressure. which atmospheric pressure is
taken as datum.
• Absolute pressure = • Gauge pressure = Absolute pres-
atmospheric pressure + Gauge sure − atmospheric pressure.
pressure.
3. A stream line may be defined as an imaginary line within the flow so that
the tangent at any point on it indicates the velocity at that point. Equation
of stream line in 3-D flow

dx dy dz
= =
u v w

Streamline

P2 V 2
4. + + z = Constant
ρg 2g
P
= Pressure Head
ρg
V2
= Kinetic Head
2g
Z = Potential Head.

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Fluid Mechanics and Machinery (Nov/Dec 2007) 4.51
5.51

5. Raynold’s number is defined as the ratio of inertial force of a flowing

fluid and the viscous force of the fluid.
Raynold’s number for pipe flow
ρVd
Re =
μ
r = Density of fluid
V = Velocity
d = dia. of the pipe
µ = Viscosity of fluid.
For laminar flow Re < 2000
For turbulent flow Re > 4000.

6. Drag
The component of the total force (FR) in the direction of motion is called
drag. This component is denoted by (FD). Thus, drag is the force exerted
by the fluid in the direction of motion.
Lift
The component of the total Force (FC) in the direction perpendicular
to the direction of motion is known as lift. This is denoted by FL. Thus,
lift is the force exerted by the fluid in the direction perpendicular to the
direction of motion.

7. Pumps Turbines
• In pumps flow takes place • In turbines flow takes place
from the low pressure to high from high pressure to low
pressure. pressure.
• Pump flow is accelerated • In turbine, there is a decelarated
flow. flow.
• It is energy absorbing • It is energy producing machine.
machine.

8. It is defined as the speed of a geometrically similar pump which would

deliver one cubic meter of liquid per second against a head of the metre.
N Q
NS = .
H max
3/ 4

9. Manometric efficiency (hmano)

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The ratio of the manometric head to the head imparted by the impeller to
the water is known as manometric efficiency.
Output of the Pump
ηmano =
Power Imparted by the Impeller

Mechanical efficiency
The ratio of power available at the impeller to the power at the shaft is
called mechanical efficiency
Power at Impeller
ηm = .
Power at theShaft

10. Slip of a pump is defined as the difference between theoretical and actual
discharge.
Slip = Qth – Qact.

PART B
11. (a) (i) Given
Increase in pressure (dp) = 20.7 MN/m2
= 20.7 × 103 KN/m2
Initial volume (v) = 100 litres
Decrease in volume (dv) = 1 litres.
⎛ dv ⎞ 1
Volumetric strain ⎜ − ⎟ =
⎝ v ⎠ 100
“(−) sign means the volume decreases with increase of pressure”
dp
Bulre modulus (K ) =
⎛ dv ⎞
⎜⎝ − ⎟⎠
V
20.7 × 103
=
⎛ 1 ⎞
⎜⎝ ⎟
100 ⎠
K = 20.7 × 105 KN/m 2 .

(ii) It states that the pressure or intensity of pressure at a point in a

static fluid is equal in all directions. This is proved as:

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 2007)
Fluid Mechanics and Machinery (Nov/Dec 2007) 4.53
5.53

The fluid element is of very small dimension, i.e., dx, dy and ds.
Consider an arbitrary fluid element of wedge shape in a fluid
may at rest as shown in the figure below. Let the width of the
element perpendicular to the plane of paper is unity and Px,
Py and Pz are the pressures or intensity of pressure acting on
the face AB, AC and BC respectively. Let ∠ABC = 0. Then the
forces acting on the element are:
• Pressure forces normal to the surfaces.
• Weight of element in the vertical direction.
The forces on the faces are:
Force on the face AB = px × Area of face AB
= px × dy × 1
Similarly force on the face AC = py × dx × 1
Face on the face BC = pz × ds × 1
Weight of element = (Mass of element) × g
⎛ AB × AC ⎞
= (Volume × ρ ) × g = ⎜ × 1⎟ × PX
⎝ 2 ⎠

where r = density of fluid.

Resolving the forces in x-direction, we have
px × dy × 1 – r (ds × 1) sin (90º − q) = 0
or px × dy × 1 – pz ds × 1 cos q = 0
But from the figure ds cos q = AB = dy
∴ px × dy × 1 – pz × dy × 1 = 0
or px = pz (1)
Y
B

q Pz.ds.1
q
Px.dy.1 dy X

dx
C
A
Z
Py.ΔX.1

Similarly, resolving the forces in y-direction, we get

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dx × dy
p y × d x × 1 − pz × ds × 1cos (90° − θ ) − ×1× p × g = 0
2
dxdy
or p y × d x − pz ds sin θ − × p× g = 0
2
But ds sin q = dx and also the element is very small and hence
weight is negligible.
∴ py dx – pz × dx = 0 or py = pz
From equations (1) and (2), we have
px = py = pz.
The above equation shows that the pressure at any point in x, y
and z directions is equal.
Since the choice of fluid element was completely arbitrary,
which meas the pressure at any point is the same in all directions.

(b) (i) Pressure at sea P0 = 760 mm of Hg

760
= × 13.6 × 1000 × 9.81 N/m 2 = 101396 Np
1000
Pressure at mountain P = 735 mm of Hg.
735
= × 13.6 × 1000 × 9.81 = 98060 N/m 2
1000
Density of air P = 1.2 kg/m3
Let h = Height of the mountain from sea level.
We know that as the elevation above the sea-level increases, the
atmospheric pressure decreases. Here the density of air is given
as constant, hence the pressure at any height ‘h’ above the sea-
level is given by the equation
p = P0 − P × g × h
P0 − P 101396 − 98060
or h = = = 283.33 m.
P×g 1.2 × 9.81
Here pressure head (Z) is given as 760 mm of Hg.
Hence (P/Pg) = 760 mm of Hg.
The density (r) for mercury = 13.6 × 1000 kg/m3.
Hence pressure (P) will be equal to r × g × Z

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5.55

760
i.e., 13.6 × 1000 × 9.81 × N/m2 . mm
1000

(ii) Given:
d = 4 mm = 4 × 10−3 m.
4σ cos θ
Capillary effect = h = .
p× g×d
where, s → Surface tension in kgf/m.,
q → Angle of contact r = density.
Capillary effect for water:
σ = 0.0735 N/m
θ = 0° C
ρ = 9790 N/m3 ⇒ 9790 × 9.81 kg/m3
= 96039.9 kg/m3
0.735 × cos 0°
h = 4× = 7.8 × 10 −4 m.
96039.9 × 9.81 × 4 × 10 −3
∴ h = 0.78 mm
Capillary effect for mercury:
σ = 0.51 N/m θ = 130°
ρ = 13.6 × 1000 = 13600 kg/m 2
0.51 × cos130°
∴h = 4 ×
13600 × 9.81 × 4 × 10 −3
= −2.457 × 10 −3 m
= −2.46 mm (capillary depression)

12. (a) (i) Consider a fluid element of lengths dx, dy and dz in the direc-
tion of x, y and z. Let u, v, and w are the inlet velocity compo-
nents in x, y and z directions respectively. Mass of fluid entering
the face ABCD per second.
= r × Velocity in x-direction × Area of ABCD
= r × u × (dy × dz)
The mass of fluid leaving the face EFGH per second
∂
= ρudydz + ( ρudydz ) dx
∂x

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∴ Gain of mass in x-directions

= Mass through ABCD – Mass through EFGH per second
∂
= ρudydz − ρudydz − ( ρudydz ) dx
∂x
∂
= − ( ρudydz ) dx
∂x
∂
= − ( ρu ) dxdydz
∂x
Similarly the net gain of mass in y-directions
∂
=− ( ρV ) dxdydz
∂y
∂
and in z-direction = − ( ρw ) dxdydz
∂z
∴ Net gain of masses
⎡∂ ∂ ∂ ⎤
= − ⎢ ( ρu ) + ( ρv ) + ( ρw ) ⎥ dxdydz
⎣ ∂ x ∂ y ∂ z ⎦

D H

A E

x dz
C G
B
F
dy
y dx

Since the mass is neither created not destroyed in the fluid ele-
ment, the net increase of mass per unit time in the fluid element
must be equal to the rate of increase of mass of fluid in the ele-
ment. But mass of fluid in the element is r. dx.dy.dz and its rate
of increase with time is
∂ ∂ρ
( ρ dx ⋅ dy ⋅ dz ) or ⋅ dxdydz
∂t ∂t

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Equating the two expressions

⎡∂ ∂ ∂ ⎤ ∂ρ
or − ⎢ ( ρu ) + ( ρv ) + ( ρw ) ⎥ dxdydz = ⋅ dxdydz
⎣ ∂ x ∂ y ∂ z ⎦ ∂t
∂ρ ∂ ∂ ∂
or + ( ρu ) + ( ρ v ) + ( ρ w ) = 0
∂t ∂x ∂y ∂z

(cancelling dx, dy, dz to both sides).

Equation is the continuity equation in Cartesian co-ordinates in
its most general form. This equation is applicable to
• Steady and unsteady flow,
• Uniform and non-uniform flow, and
• Compressible and incompressible fluids.
∂ρ
For steady flow, = 0 and hence equation becomes as
∂t
∂ ∂ ∂
( ρu ) + ( ρv ) + ( ρw ) = 0.
∂x ∂y ∂z
If the fluid is in compressible, then P is constant and the above
equation becomes as
∂u ∂v ∂w
+ + = 0.
∂x ∂y ∂z
This equation is the continuity equation in the three-dimensions.
For a two-dimensional flow, the component w = 0 and hence
continuity equation becomes as
∂u ∂v
+ = 0.
∂x ∂y

(ii) Y = 9 + 6x − 4y + 1xy.
The velocity potential function = f.
− ∂ψ ∂ψ
u= ;v= .
∂y ∂x
∂
∴u = − (9 + 6 x − 4 y + 7 xy )
∂y
= − ( −4 + 7 x )
= (4 − 7 x ) (1)

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v = 0 + 6 + 7y
= (6 + 7 y ) (2)
We know that,
∂φ ∂φ
= −u & = −v
∂x ∂y
∂φ
⇒ = (7 x − 4) (3)
∂x
∂φ
= ( − 6 − 7 y) (4)
∂y
On integrating,

⇒ ∫ dφ = ∫ (7 x − a) dx
7x2
⇒φ = − 4x + c (5)
2
Different (5) with respect to y,
∂φ ∂c
⇒ = .
∂y ∂y
∂φ
But from (4) = −6 − 7y
∂y
∂c
∴ = −6 − 7y
∂y
∴ c = ∫ ( −6 − 7 y ) dy
7 y2
⇒ −6y − .
2
On substituting c value in (5), we get,
7x2 7 y2
φ= − 4x − 6 y − .
2 2
∴ The velocity potential function,
7x2 7 y2
φ= − 4x − 6 y − .
2 2

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(b) (i) If the flow is irrotational means,

1 ⎛ ∂v ∂u ⎞
− = 0.
2 ⎜⎝ ∂x ∂y ⎟⎠
∂v ∂u
⇒ =2: = 4.
∂x ∂y
1 1
⇒ (2 − 4) = ( −2) = −1⋅ ≠ 0
2 2
∴ The flow is rotational.
∴ Potential function exists.
∂φ ∂φ
⇒ = −u & = −v
∂x ∂y
⇒ φ = ∫ ( −3x − 4 y ) dx
3x 2
=− − 4 xy + c
2
Difference with respect to y,
∂φ ∂c
⇒ = − 4x + ⇒ 3y − 2x
∂y ∂y
∂c 3 y2
⇒ = 3y + 2x ⇒ c = + 2 xy
∂y 2
3x 2 3 y2
∴φ = − − 4 xy + + 2 xy.
2 2
3x 2 3 y2
φ=− − 2 xy + .
2 2

(ii) Diameter at inlet d1 = 30 cm

π
∴ Area a1 = (30)2 = 706.85 cm 2
4
Diameter at throat d2 = 15 cm
π
∴ Area a2 = (15)2 = 176.7 cm 2
4
Let section (1) represents inlet and section (2) represents throat.
Then Z2 − Z1 = 30 cm
Specific gravity of oil S0 = 0.9

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Specific gravity of mercury Sm = 13.6

Reading of different manometer, x = 25 cm
The differential head, h is given by
⎡P ⎤ ⎡P ⎤
h = ⎢ 1 + Z1 ⎥ − ⎢ 2 + Z 2 ⎥
⎢⎣ ρ g ⎥⎦ ⎢⎣ ρ g ⎥⎦
⎡S ⎤ ⎡13.6 ⎤
= x ⎢ m − 1⎥ = 25 ⎢ − 1⎥ = 352.77 cm of oil.
⎣ 0S ⎦ ⎣ 0.9 ⎦

a1 a2
• The discharge, Q of oil = Cd × 2 gh
a12 − a22

0.98 × 706.85 × 176.7

= × 2 × 981 × 352.77
(706.85)2 − (176.7)2
101832219.9
= = 148790.5 cm3 /s
684.4
= 148.79 litres/s.
• Pressure difference between entrance and throat section
⎡P ⎤ ⎡P ⎤
h = ⎢ 1 + Z1 ⎥ − ⎢ 2 + Z 2 ⎥ = 352.77
⎣ ρ g ⎦ ⎣ ρ g ⎦
⎡ P1 P2 ⎤
or ⎢ − ⎥ + Z1 − Z 2 = 352.77
⎣ ρg ρg ⎦
But Z 2 − Z1 = 30 cm

⎡P P ⎤
∴ ⎢ 1 − 2 ⎥ − 30 = 352.77
⎣ ρg ρg ⎦
P P
∴ 1 − 2 = 352.77 + 30 = 382.77 cm of oil.
ρg ρg
= 3.8277 m of oil.
(or) (P1 – P2) = 3.8277 × rg
But density of oil = Specific gravity of oil × 1000 kg/m3
= 0.9× 1000 = 900 kg/cm3

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N
∴ ( P1 − P2 ) = 3.8277 × 900 × 9.81
m2
( P1 − P2 ) = 3.3795 N/cm 2 .

13. (a)
+ dm
é ú
t êP × dx ù
ë dx û
r

Let us consider a horizontal circular pipe of radius ‘R’ and a concen-

tric cylindrical fluid element of radius ‘r’ and length dx.
Forces action on fluid
Pressure force (P × pr2) in left face.
∂P
(P + ⋅ dx ) π r 2 on right face.
∂x
Shear force (τ ⋅ 2π r dx ) acting on the periphery of the cylinder
opposing the motion.
For equilibrium condition,
⎛ ∂P ⎞
pπ r 2 − ⎜ p + ⋅ dx ⎟ π r 2 − τ × 2π r × dx = 0
⎝ ∂x ⎠
∂P
− ⋅ dx π r 2 − τ × 2π r × dx = 0
∂x
∂P
− ⋅ r − 2τ = 0
∂x
∂P
− ⋅ r = 2τ
∂x
∂P r
τ=− ⋅ (1)
∂x 2
If r = 0 ; τ = 0
∂P R
If r = R ; τ = − ⋅
∂x 2

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(Max)

Γ(Max)

As per Newton’s law,

du du
τ = μ⋅ = −μ ⋅ (2)
dy dr
Equation (1) and (2)
∂P r du
− ⋅ = −μ ⋅
∂x 2 dr
du 1 ∂P r
= ⋅ ⋅ .
dr μ ∂x 2
Integrating the above equations with respect to r

1 ∂P
∫ du = ∫ dr
μ ∂x
1 ∂P ⎛ r 2 ⎞
u= ⎜ ⎟ +C
μ ∂x ⎝ 2 × 2 ⎠
1 ∂P 2
u= r +C (3)
4 μ ∂x
u = 0; r = 0
u = 0; r = R
1 ∂P 2
0= ⋅R +C
4 μ ∂x
1 ∂P 2
C=− ⋅ ⋅R (4)
4 μ ∂x

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Substitute this in (3)

1 ∂P 2 1 ∂P 2
u⇒ ⋅ ⋅r − ⋅R
4 μ ∂x 4 μ ∂x
1 ∂P 2
u=− ⋅ [R − r2 ] U(Max)
4 μ ∂x
if r = 0 then
1 ∂P 2
μ=− ⋅ ⋅R
4 μ ∂x
The quantity of fluid flowing per second through the pipe, dQ is
obtained by considering the flow through a circular ring element r.

dQ = Area × Velocity
dQ = 2pr dr × u
⎡ 1 ∂P 2 ⎤
= 2π r ⋅ dr ⎢ − ( [ R − r 2 ]⎥
⎣ 2 μ ∂x ⎦
dr

r R

R
Total discharge a = ∫ dQ
0

R πr ⎡ ∂P ⎤
=∫ − dr ⎢ ( R2 − r 2 ) ⎥
0 2μ ⎣ ∂x ⎦
π ∂P ⎡ R 2 2
( R − r )r ⋅ dr ⎤
2 μ ∂x ⎣⎢ ∫0
=− ⋅
⎦⎥
R
π ∂P ⎡ R 2 r 2 r 4 ⎤
=− ⋅ − ⎥
2 μ ∂x ⎢⎣ 2 4 ⎦0
π ∂P ⎡ R 4 R 4 ⎤
=− − ⎥
2 μ ∂x ⎢⎣ 2 4 ⎦
π ∂P 4
Q=− ⋅ ⋅R
8 μ ∂x

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Replacing radius ‘R’ by diameter D

D
R=
2
4
π ⎡ ∂P ⎤ ⎡ D ⎤
Q= −
8μ ⎢⎣ ∂x ⎥⎦ ⎢⎣ 2 ⎥⎦
π ⎡ ∂P ⎤ 4
Q=− D
128μ ⎢⎣ ∂x ⎥⎦
∂P 128μ Q
− =
∂x π D4
Discharge Q
Mean velocity u = =
Area A
⎡ ∂P ⎤ 4
π / 128 ⎢ ⎥ D
⇒− ⎣ ∂x ⎦
π ⋅ D2
4
1 ⎡ ∂P ⎤
u = − μ ⎢ ⎥ D2
32 ⎣ ∂x ⎦
32 μ ⋅ u ⋅ dx
∂p = −
D2
−32 μ ⋅ u
∫ ∂p = ∫ D 2 dx
−32 μ ⋅ u
p1 − p2 = ⎡x − x ⎤
D2 ⎣ 1 2 ⎦
32 μ ⋅ u
p1 − p2 = ⎡x − x ⎤
D2 ⎣ 2 1 ⎦
32 μ ⋅ u
p1 − p2 = [ L]
D2
32 μ ⋅ u
p1 − p2 = L
D2
Pressure head
P1 − P2
hL =
ω
32 ⋅ μ ⋅ a ⋅ L
hL =
ω D2
Hagen Poiseuilley equation.

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(b) Area of the plate A = 1.5 × 1.5 = 2.25 m2

Velocity of the plate
50 × 1000
U = 50 km/hr = m/s = 13.89 m/s
60 × 60
Density of air P = 1.15 kg/m3
Co-efficient of drag CD = 0.15
Co-efficient of lift CL = 0.75
• Lift Force (FL), using equation
PU 2 1.15 × 13.892
FL = C L A × = 0.75 × 2.25 × N
2 2
= 1.87.20 N.
• Drag Force (FD), using equation
PU 2 1.15 × 13.892
FD = C D × A × = 0.15 × 2.25 × N
2 2
= 37.44 N.
• Resultant Force (FR), using equation
FR = FD + FL = 37.44 2 + 187 20 2
2 2

= 1400 + 35025 = 190.85 N.

• Power required to keep the plate in motion

Force in the direction of motion × Velocity
P= KW
1000
F × U 37.425 × 13.89
= D = KW = 0.519 KW.
1000 1000

14. (a) Let H = Net head acting on the pelton wheel

= Hg − hf
4 fLV 2
Where H g = Gross head and h f = .
D * X 2g

where D* = Diameter of Penstock, N = Speed of the wheel in r.p.m

D = Diameter of the wheel, d = Diameter of the jet.
Then V1 = Velocity of jet at inlet = 2 g H

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π DN
u = u1 = u2 =
60
The velocity triangle at inlet will be a straight line where
Vr1 = V1 – u1 = V1 – u
Vw1 = V1; α = 0 and q = 0
From the velocity triangle at outlet, we have
Vr = Vr and Vw = Vr cos φ − u2
1 1 2 2

The force exerted by the jet of water in the direction of motion is

given by equation as
Fx = ρaV1 [Vw + Vw ].
1 2

As the angle b is an acute angle, + ve sign should be taken. Also this

is the case of series of varies, the mass of water striking is ra V1 and
not ra Vr1. In equation, ‘a’ is the area of the jet which is given as
π
a = Area of jet = d2.
4
Now work done by the jet on the winner per second
= Fx × u = ρaV1 [Vw + Vw ] × u Nm/s
1 2

Power gives to the runner by the jet

ρaV1 ⎡⎣Vw + Vw ⎤⎦ × u
= 1 2
KW
1000
u2 Vw2

f b

V2
Vf1
Vr2

Angle of
V1 Defletion
A B
Inlet u1 V1

Vw1

Work done/s per unit weight of water striking/s

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ρaV1 ⎡⎣Vw + Vw ⎤⎦ × u
= 1 2

Weight of Water Striking/s

ρaV1 ⎡⎣Vw + Vw ⎤⎦ × u 1⎡
= 1 2
= V + Vw ⎤ × u
ρaV1 × 8 8 ⎣ w1 2 ⎦

The energy supplied to the jet at inlet is in the form of kinetic energy
1
and is equal to = mV 2
2
1
∴ K.E of jet per second = ( ρaV1 ) × V12
2
Work done per second
∴ Hydraulic efficiency, ηh =
K.E of jet per second
ρaV1 ⎡⎣Vw + V w ⎤⎦ × u 2 ⎡Vw1 + Vw ⎤
= 1 2
= ⎣ 2 ⎦
×u
1 V12
( PaV1 ) × V12
2
Now Vw = V1 , Vr = V1 − u1 = (V1 − u )
1 1

∴ Vr 2 = (V1 − u )
and Vw = Vr cos φ − u2 = Vr cos φ − u = (V1 − u) cos φ − 1
2 2 2

Substituting the values of Vw1 and Vw2 in equation

2 ⎡V + (V1 − u ) cos φ − u ⎤⎦ × u
ηh = ⎣ 1 2
V1
2 ⎡V − u + (V1 − u ) cos φ ⎦⎤ × u 2(V1 − u ) [1 + cos φ ] × u
= ⎣ 1 =
V12 V12

The efficiency will be maximum for a given value of V1 when

d d ⎡ 2u (V1 − u ) (1 + cos φ ) ⎤
(ηh ) = 0 or ⎢ ⎥=0
du du ⎣ V12 ⎦
(1 + cos φ ) d d
or (2uV1 − 2u 2 ) = 0 or ⎡2uV1 2u 2 ⎤⎦ = 0
V12 du du ⎣
⎛ 1 + cos φ ⎞
⎜∵ V 2 ≠ 0⎟
⎝ 1 ⎠
V1
or 2V1 − 4u = 0 or u =
2

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Equation states that hydraulic effieicny of a pelton wheel will be

maximum when the velocity of the wheel is half the velocity of the
jet of water at inlet. The expression for maximum efficiency will be
V
obtained by substituting the value of a = 1 in equation
2
⎛ V 1⎞ V
2 ⎜V1 − ⎟ (1 + cos φ ) × 1
⎝ 2⎠ 2
∴ Maximum ηh =
V12

V1 V
2× (1 + cos φ ) 1
= 2 2 = (1 + cos φ ) .
V12 2

(b) (i) Diameter of wheel D = 1.0 m

Speed of wheel N = 1000 r.p.m
∴Transential velocity of the wheel,
π DN π × 1.0 × 1000
u= = = 52.36 m
60 60
Net head (or) turbine H = 700 m
Side clearance angle φ = 15°
Discharge, Q = 0.1 m3/s
Velocity of jet at inlet V1 = Cv 2gH = 1 × 2 × 9.81 × 700

(∴ Value of Cv is not given. Take it = 1.0)

Or V1 = 117.17 m/s.
• Power available at the nozzle is given by equation as
W ×H ρ×g×Q×H
W .P = =
1000 1000
1000 × 9.81 × 0.1 × 700
= = 686.7 KW .
1000
• Hydraulic efficiency is given by equation as
2(V1 − u ) (1 + cos φ )u
ηh =
V12
2(117.18 − 52.36) (1 + cos15) × 52.36
=
117.19 × 117.19
2 × 64.83 × 1.966 × 52.36
= = 0.9718 = 97.18%.
117.19 × 117.19

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(ii) Head H = 25 m
Speed N = 200 r.p.m
Discharge Q = 9 cumec = 9 m3/s
Efficiency η0 = 90% = 0.90
(Take the efficiency as overall η)
Now using relation
Power Developed P
η0 = =
Water Power ρ × g × Q×H
1000
ρ× g×Q×H
P = η0 ×
1000
0.90 × 9.81 × 1000 × 9 × 25
= = 1986.5 kW
1000
• Specific speed of the machine (NS)
Using equation
N P 200 × 1986.5
NS = = = 159.46 rpm
H 5/ 4 255/ 4
• Power generated P = 1986.5 KW.
• As the specific speed lies between 51 and 255, the turbine is
a Francis turbine.

15. (a) (i) Net head, Hm = 14.5 m

Speed N = 1000 rpm
Vane angle at outlet φ = 30°
Impeller diameter means the diameter of the inmpeller at outlet.
∴ Diameter D2 = 300 mm = 0.30 m
Outlet width B2 = 50 mm = 0.05 m
Manometric efficiency, hman = 95% = 0.95
Tangential velocity of impeller at outlet
π D2 N π = 0.30 × 1000
u2 = = = 15.70 m/s
60 60

gH m
Now using equation hman =
VW2 × u2

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9.81 × 14.5
∴ 0.95 =
VW × 15.70
2

0.95 × 14.5
∴VW = = 9.54 m/s
2 0.95 × 15.70
From outlet velocity triangle, we have
Vf Vf Vf
tan φ = 2
or tan 30° = 2
= 2

(u2 − Vw ) (15.70 − 9.54) 6.16

2

∴Vf = 6.16 × tan 30° = 3.556 m /s

2

∴ Discharge φ = π D B2 × Vf
2 2

= π × 0.30 × 0.55 × 3.55b

=0.1675 m3 /s.

(ii) Speed of the pump N = 50 rpm

Actual discharge Qact = 0.01 m3/s
Diameter of piston D = 200 mm = 0.20 m
π
∴ Aera A = ⋅ (0.2)2 = 0.031416 m2
4
Storage, L = 400 mm = 0.40 m.
• Theoretical dishcarge for single-acting reciprocating pump
is given by equation as
A × L × N 0.031416 × 0.40 × 50
Qth = = = 0.01047 m3 /s.
60 60
• Coefficient of discharge is given by
φact 0.01
Cd = = = 0.955.
φih 0.01047

• Using equation, we get

Slip = φth − φact = 0.01047 − 0.01 = 0.00047 m3 /s

And percentage slip

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(φth − φact ) (0.01047 − 0.01)

= × 100 = × 100
φth 0.01047
0.00047
= × 100 = 4.487%.
0.01047

(b) (i) Gear Pump

It is a rotary pump in which two gears mesh to provide the
pumping action. This type of pump mostly used for cooling
water and pressure oil to be supplied for lubrication to motors,
turbines, machine tools, etc.
Gear pump consists of two identical intermeshing gears work-
ing in a fine clearance inside a casing. One of the gear is keyed
to a driving shaft. The other gear revolves due to driving gear.
The space between teeth and the casing is filled with oil. The oil
is carried round between the gears form the suction pipe to the
delivery pipe.

Casino

Delivery
Suction

The mechanical contact between the gears does not allow the
flow from inlet to outlet directly. The outer radius tips of the
gears and sides of the gears form apart off moving oil.
The oil pushed into the delivery pipe, cannot back into the suc-
tion pipe due to the meshing of the gears.

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Receiver

Rotating Drum

Vanes

Caring

Inlet

Vane Pump
A disc is eccentrically rotating inside the pump casing. Shoes
containing vanes are provided on the disc. The vanes are moved
against the casing due to centrifugal force when the disc rotates.
So the liquid tight seal is formed.
Due to continus rotation of disc, the liquid is entrapped and
forced to the delivery and with sufficient pressure. In some
cases springs are used to press the waves against the casing. But
vanes are generally hinged.

(ii) Speed of turbine N = 450 r.p.m

Head H = 120 m
Diameter at inlet D1 = 120 cm = 1.2 m
Flow area π D × B1 = 0.4 m 2
1

Angle made by absolute velocity at inlet, α = 20°

Angle made by the relative velocity at inlet, q = 60°
Whirl at outlet, VW = 0
2

Tangential velocity of the turbine at inlet,

π D1 N π × 1.2 × 4.50
u1 = = = 28.27 m/s.
60 60
From inlet velocity triangle

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Vf Vf Vf
tan α = 1
or tan 20° = 1
or 1
tan 20° = 0.364
Vw Vw Vw
1 1 1

∴ Vf = 0.364 Vw ... (1)

1 1

u2
Vw2

b
Vr2 f

V2 Vr1

Vane

V1 + L1 V1 = LV1

q L
u1

Vf 0.364 Vw
Also tan θ = 1
= 1
(∵V f = 0.364Vw )
Vw − u1 Vw − 28.27 1 1
1 1

0.364 Vw
or 1
= tan θ = tan 60° = 1.732
Vw − 28.27
1

∴ 0.364 Vw = 1.732(Vw − 28.27) = 1.732Vw − 48.96

1 1 1

or (1.732 − 0364)Vw = 48.96

1

48.96
∴ Vw = = 35.789 = 35.79 m/s.
1
(1.732 − 0.364)

From equation (i),

Vf = 0.364 × Vw = 0.364 × 55.79 = 13.027 m/s
1 1

• Value flow rate is given as

Q = π D B1 × Vf
1 1

But π D × B1 = 0.4 m2 ; Q = 0.4 × 13.027 = 5.211 m3 /s.

1

• Work done per second on the turbine is given by equation

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= ρ ⋅ Q [Vw u1 ] (∵ Vw = 2)
1 2

= 1000 = 5.211 [35.79 × 28.27] = 5272402 Nm /s

∴ Power developed in KW
Work done per second 5272402
= = = 5272.402 KW.
1000 1000
• The hydraulic efficiency is given by equation as
Vw u1 35.79 × 28.27
ηh = 1
= = 0.8595 = 85.95%.
gH 9.81 × 120

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 B.E./B.Tech. DEGREE EXAMINATION,
NOV/DEC 2005
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY
Time: Three hours Maximum: 100 marks
Answer ALL questions
PART A (10 × 2 = 20 marks)
1. What is viscosity? What is the cause of it in liquids and in gases?

2. State Pascal’s law.

3. State the equation of continuity in three dimensional in compressible flow.

4. Define circulation.

5. Differentiate between laminar and turbulent flow.

6. Define lift.

7. Explain specific speed.

8. Classify turbines according to flow.

9. When will you select a reciprocating pump?

10. What is the role of a volute chamber of centrifugal pump?

PART B (5 × 16 = 80 marks)
11. (a) Describe the working and principles of a reciprocating pump.
Or
(b) What is priming in a centrifugal pump? Why is it necessary?

12. (a) (i) Define the term pressure. What are its units?

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(ii) Calculate the capillary effect in millimeters in a glass tube of

4 mm diameter, when immersed in (1) water and (2) mercury
when the temperature of the liquid is 20°C and the value of sur-
face tension of water and mercury at 20°C in contact with air
are 0.0735 N/m and 0.51 N/m respectively. The contact angle
for water q = 0 and for mercury q = 130°. Take specific weight
of water at 20°C as equal to 9790 N/m3 and specific gravity of
mercury is 13.6.
Or
(b) (i) What do you mean by surface tension? If the pressure differ-
ence between the inside and outside of the air bubble of diam-
eter, 0.01 mn is 29.2 kPa, what will be the surface tension at air
water interface? Derive an expression for the surface tension
in the air bubble and form it, deduce the result for the given
conditions.
(ii) Define capillarity.

13. (a) (i) The ‘x’ component of velocity is u = x2 + z2 + 5 and the ‘y’ com-
ponent is v = y2 + z2. Find the simplest component of velocity
that satisfy continuity.
(ii) The two dimensional stream function for a few is Ψ = 9 + 6x –
4y + 7xy. Find the velocity potential.
Or
(b) (i) The resisting force (R) of a supersonic flight can be considered
as dependent upon the length of the air craft ‘l’, velocity ’v’
air viscosity ‘µ’, air density ‘p’ and bulk modulus of air is ‘k’.
Express the functional relationship between these variables and
the resisting force.
(ii) An orificemeter with orifice diameter 15 cm is inserted in a pipe
of 30 cm diameter. The pressure of the upstream and down-
stream of orificemeter is 14.7 N/cm2 and 9.81 N/cm2. Find the
discharge, if Cd = 0.6.

14. (a) Derive Hagen–Poiseuille equation and state the assumptions made.
Or
(b) (i) A pipe line 10 km long delivers a power of 50 kW at its outlet
ends. The pressure at inlet is 5000 kN/m2 and pressure drop per
km of pipeline is 50 kN/m2. Find the size of the pipe and effi-
ciency of transmission. Take 4f = 0.02.

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(ii) The velocity of water in a pipe 200 mm diameter is 5 m/s. The

length of the pipe is 500 m. Find the loss of head due to friction,
if f = 0.008.

15. (a) (i) Give the comparison between impulse and reaction turbine.
(ii) In a hydroelectric station, water is available at the rate of 175
m/s under head of 18 m. The turbine run at a speed of 150 rpm,
with overall efficiency of 82%. Find the number of turbines
required, if they have the maximum specific speed of 460.
Or
(b) (i) With the help of a neat diagram explain the construction and
working of a Pelton wheel turbine.
(ii) What is the condition for hydraulic efficiency of a Pelton wheel
to be maximum?

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 Solutions
PART A
1. Viscosity is a measure of the resistance of a fluid which is being deformed
by either shear stress or tensile stress.

2. Pascal’s law or the principle of transmission of fluid-pressure states that

“pressure exerted anywhere in a confined incompressible fluid is trans-
mitted equally in all directions throughout the fluid such that the pressure
ratio (initial difference) remains the same.

∂u ∂v ∂w
3. + + =0
∂x ∂y ∂z
u, v and w = inlet velocity components
x, y and z = Directions.

4. Circulation is the line intergral around a closed curve of the fluid velocity
circulation is denoted by Γ = ∫

5. Laminar flow is also referred to as streamline or viscous flow. These

terms are descriptive of the flow because, in laminar flow
• layers of water flowing over one another at different speeds with virtu-
ally no mixing between layers,
• fluid particles move in definite and observable paths or streamlines,
and
• the flow is characteristic of viscous (thick) fluid or is one in which
viscosity of the fluid plays a significant part.
Turbulent flow is characterized by the irregular movement of particles
of the fluid. There is no definite frequency as there is in wave motion.
The particles travel in irregular paths with no observable pattern and no
definite layers.

6. A fluid flowing past the surface of a body exerts surface force on it. Lift
is any component of this force that is perpendicular to the oncoming flow
direction.

7. Specific speed Ns is a quasi non-dimensional number used to classify

pump impellers as to their type and proportions. In imperial units it is
defined as the speed in revolutions per minute at which a geometrically
similar impeller would operate if it were of such a size as to deliver one
gallon per minute against one foot of hydraulic head.

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8. They can be classified as follows:

• Tangential flow turbine
• Radial flow turbine
• Axial flow turbine
• Mixed flow turbine

9. It is often used where relatively small quantity of liquid is to be handled

and where delivery pressure is quite large.

10. One of the main purposes of a volute casing is to help balance the
hydraulic pressure on the shaft of the pump. However, this occurs best at
the manufacturer’s recommended capacity. Running volute-style pumps
at a lower capacity than the manufacturer recommends can put lateral
stress on the shaft of the pump, increasing wear-and-tear on the seals and
bearings, and on the shaft itself.

PART B (5 × 16 = 80 marks)
11. (a) Reciprocating Pump
Principle
Reciprocating pump operates on the principle of pushing of liquid
by a piston that executes a reciprocating motion in a closed fitting
cylinder.

Delivery
hd Pipe
Connecting Crank
Delivery Cylinder w
Rod
Valve L Piston Rod

q r
I.D.C. O.D.C.
Suction Stroke

hS Piston
x Delivery Stroke
Suction Valve x

Suction L = 2r
Pipe
I.D.C. = Inner Dead Centre
O.D.C. = Outer Dead Centre
Sump Well

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Construction Details of a Reciprocatin Pump

Components of reciprocating pumps
• Piston or plunger: A piston or plunger that reciprocates in a
closely fitted cylinder.
• Crank and Connecting rod: Crank and connecting rod mecha-
nism operated by a power source. Power source gives rotary
motion to crank. With the help of connecting rod we translate
reciprocating motion to piston in the cylinder.
• Suction pipe: One end of suction pipe remains dip in the liquid
and other end attached to the inlet of the cylinder.
• Delivery pipe: One end of delivery pipe attached with delivery
part and other end at discharge point.
• Suction and Delivery value: Suction and delivery values are
provided at the suction end and delivery end respectively. These
values are non-return values.
Working of Reciprocating Pump
Operation of reciprocating motion is done by the power source
(i.e. electric motor or engine, etc). Power source gives rotary motion
to crank; with the help of connecting rod we translate reciprocating
motion to piston in the cylinder (i.e, intermediate link between con-
necting rod and piston). When crank moves from inner dead centre
to outer dead centre, vacuum will create in the cylinder. When piston
moves outer dead centre to inner dead centre and piston force the
water at outlet or delivery value.
Expression for Discharge of the Pump
ALN
Q=
60
Where
Q: discharge in m3/sec
A: cross-section of piston or cylinder in m2
L: length of stroke in meter
N: speed of crank in r.p.m

(b) All centrifugal pumps must be primed by filling them with water
before they can operate. The objective of priming is to remove a suf-
ficient amount of air from the pump and suction line to permit atmo-
spheric pressure and submergence pressure to cause water to flow

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into the pump when pressure at the eye of the impeller is reduced
below atmospheric as the impeller rotates.
When axial-flow and mixed-flow pumps are mounted with the pro-
pellers submerged, there is normally no problem with re priming of
these pumps because the submergence pressure causes water to refill
the pumps as long as air can readily be displaced. On the other hand,
radial-flow pumps are often located above the water source, and they
can lose prime. Often, loss of prime occurs due to an air leak on the
suction side of the pump. Volute or diffuser pumps may lose prime
when water contains even small amounts of air or vapor. Prime will
not be lost in a radial-flow pump if the water source is above the eye
of the impeller and flow of water into the pump is unrestricted.
In some cases pumps are primed by manually displacing the air in
them with water every time the pump is restarted. Often, by using a
foot value or a check valve at the entrance to the suction pipe, pumps
can be kept full of water and primed when not operating. If prime is
lost, the water must be replaced manually, or a vacuum pump can be
used to remove air and draw water into the pump.
A self-priming pump is one that will clear its passages of air and
resume delivery of liquid without outside attention. Centrifugal
pumps are not truly self-priming. So called self-priming centrifu-
gal pumps are provided with an air separator in the form of a large
chamber or reservoir on the discharge side of the pump. This sepa-
rator allows the air to escape from the pump discharge and entraps
the residual liquid necessary during re priming. Automatic priming
of a pump is achieved by the use of a recirculation chamber which
recycles water through the impeller until the pump is primed, or by
the use of a small positive displacement pump which supplies water
to the impeller.
• Impeller top must be submerged with the process fluid or water
without any air.
• If you want to pump the air in case of a centrifugal pump, pump-
ing water, operating at the speed of 2900 RPM, then you have to
run the pump at a speed of [2900 × 800] as the ratio of specific
gravity of air to water is 1/800. And it is practically impossible.
• If you don’t prime the pump and start the pump, then you will
not get the flow, pump will run dry and ultimately damage the
mechanical seal or gland packing.

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12 (a) (i) Pressure is the force per unit area applied in a direction
perpendicular to the surface of an object. The SI unit pressure is
the pascal (Pa), equal to one newton per square meter (N/m2 or
kg.m–1.s–2).

(ii) d = 4 mm = 0.004 m
Surface tension at 20°C
qwater = 0.0735 N/m
qmercury = 0.05 N/M
Specific weight of water = 9790 N/m3
4 σ cos θ
h=
wd
• Water:

4 × 0.07351 cos 0°
h= θ water = 0
9790 × 0.004
=0.751 × 10-3m
h = 7.51 mm/(rise)
• Capillary effect of mercury
4 × 0.051 × cos130˚
h = θ 130˚
(13.6 × 9790) × 0.004 mercury
= – 2.46 × 10–3m
= – 2.46 mm
h = 2.46 mm.

(b) (i) Surface tension is a property of the surface of a liquid that allows
it to resist an external force. It is revealed, for example, in the
floating of some objects on the surface of water, even though
they are denser than water, and in the ability of some insects
(e.g. water striders) to run on the water surface. This property
is caused by cohesion of similar molecules, and is responsible
for many of the behaviors of liquids. Surface tension has the
dimension of force per unit length, or of energy per unit area.

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π
π× ⋅ d 2 = σ (π d )
4
p×d
σ=
4
29.2 × 103 × 0.01 × 10 −3
σ=
4
σ = 0.073 N/m
(ii) Capillary action, or capillarity, is the ability of a liquid to flow
in narrow spaces without the assistance of, and in opposition to
external forces like gravity. The effect can be seen in the draw-
ing up of liquids between the hairs of a paint-brush, in a thin
tube, in porous materials such as paper, in some non-porous
materials such as liquified carbon fiber, or in a cell. It occurs
because of inter-molecular attractive forces between the liq-
uid and solid surrounding surfaces. If the diameter of the tube
is sufficiently small, then the combination of surface tension
(which is caused by cohesion within the liquid) and adhesive
forces between the liquid and container act to lift the liquid.

13. (a) (i) The continuity equation for in compressible fluid is given by
∂u ∂v ∂w
+ + =0
∂x ∂y ∂z
u = x2 + z2 + 5
v = y2 + z2
∂u
= 2x
∂x
∂v
= 2y
∂y

∂u ∂v
Substituting the values of and in continuity equation
∂x ∂y
∂w
2x + 2 y + =0
∂z
∂w
= −2x − 2y
∂z
Integration of both sides given
∫dw = ∫( – 2x – 2y)dz

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w = – 2xz – 2yz + constant of integration

where constant of integration cannot be a function of z. But it
can be a function of x and y that is f (x,y)
w = ( − 2xz – 2yz) + f (x, y).

(ii) The velocity components v in terms of Ψ are

−∂ψ −∂
u= = (9 + 6 x − 4 y + 7 xy )
∂y ∂y
= −( −4 + 7x)
= 4 − 7x
∂ψ ∂
v= = (9 + 6 x − 4 y + 7 xy )
∂x ∂x
= 6 + 7y

Velocity potential function φ

We know
∂φ
= −u
∂x
= − (4 − 7x)
= − 4 + 7x ...(i)

∂φ
= −v
∂y
= − (6 + 7y)
= − 6 − 7y ...(ii)

Integrating equation (1) we get

∫ dθ = ∫ ( −4 + 7 x)dx
7x 2
φ = −4x+ +C ...(iii)
2
where C is a constant which is independent of ‘X’ but can be
function of Y.
Differentiating equation (3) with respect to ‘Y’, we get
∂φ ∂C
=
∂y ∂Y
But from (2)

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∂φ
= −6 − 7 y
∂y
∂C
= −6 − 7 y
∂y
Integrating this equation, we get
C = ∫ ( −6 − 7 y ) dy
7 y2
C = −6 y − .
2
Substituting the value of C in equation (3) we get
7x 2 7y 2
φ = −4 x + − 6y −
2 2
7x 2
7y 2
φ= − 4x − − 6 y.
2 2

(b) (i) Refer answer 11(a) from April/May 2010 Question paper.
(ii) Pressure difference = P1 – P2
= 14.7 – 9.8
= 4.9 N/cm2 = 49 kN/m2
Pressure
Lead of water =
Specific weight of water
49
= = 5m of water
9.81
Theoretical Discharge = a 2 gh
π
= × 0.152 2 × 9.81 × 5
4
Qth = 0.175 m3 / sec.

Actual discharge = Cd theoretical discharge

= 0.6 × 0.175
Qact = 0.105 m3/sec.

14. (a) Refer answer 13(a) from Nov/Dec 2007 Question paper.
(b) (i) Given data
Length of pipe L = 10 km

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Power transmitted = 50 kw
Pressure at inlet P = 5000 kN/m2
Pressure drop = 50 kN/m2 = 50 × 103 N/mm2
50 × 103
∴Loss of head hf =
1000 × 9.81
= 5.096 m
4f = 0.02

P 5000 × 103
Pressure head at inlet, H = = = 509.6 m
ρg 1000 × 9.81
Head available at the end of the Pipe = H – hf
= 509.6 – 5.096
= 504.5 m
Let the diameter of the Pipe = d
Now Power transmitted is given by,
pg × Q( H − h f )
P= kw
1000
50kw=1000 × 9.81 × Q × 504.5
1000

50 × 1000
Q=
1000 × 9.81 × 504.5
Q = 0.0101
Discharge Q = Area × Velocity
= π/4d2 × V
π/4d2 × V = 0.0101
4 × 0.0101
V=
π d2
0.01286
V=
d2
The head lost due to friction,
4 f × L ×V 2
hf =
d × 2g

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0.02 × 10 × 103 × V 2
5.096 =
d × 2 × 9.81
200V 2
5.096 =
19.62 d
10.19 × V 2
=
d
2
10.19 ⎛ 0.01286 ⎞
= ×⎜
d ⎝ d 2 ⎟⎠
0.001685
5.096 =
d5
1/3
⎛ 0.001685 ⎞
d=⎜
⎝ 5.096 ⎟⎠
d = 0.2013 m
d = 201.3 mm
Efficiency of power transmission is given by
H − hf
η=
H
509.6 − 5.096
=
509.6
= 0.99
η = 99%.

(ii) Dia of Pipe d = 200 mm = 0.20 m

Length of Pipe l = 500 m
Velocity of flow V = 5 m/s
Coefficient of friction f = 0.008
Darcy formula

4 f LV 2
hf =
d × 2g
4 × 0.008 × 500 × 52
hf =
2.2 × 2 × 9.81
hf = 101.94 m.

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15. (a) (i) Impulse turbine Reaction turbine

1. All the potential energy Only a portion of the
is converted into kinetic fluid energy is trans-
energy by nozzle before ferred into kinetic energy
entering to turbine before the fluid enters
runner. the turbine.
2. Flow regulation is pos- Flow regulation is pos-
sible wWithout loss. sible with Loss
3. Flow is regulated by Flow is regulated by
means Of a needle valve means of a guide-vane
fitted into tThe nozzle. assembly.
4. Water may be allowed to Water is admitted over
enter a part or whole of the circumference of the
the wheel circumference. wheel.
5. Wheel does not run full Water completely fills
and air has free access to the vane passages
the buckets. throughout the operation
of the turbine.
6. Unit is installed above Unit is kept entirely sub-
the tail rRace. merged in water below
tail race.
7. Blades are only in action when Blades are in action at all
tThey are in front of nozzle. the time.

(ii) Q = 175 vm3/s

H = 18 m
N = 150 rpm
η0 = 82%
Ns = 460
Number of turbines required
Specific speed of the turbine
N P
Ns =
H 5/ 4
150 P
460 =
(18)5 / 4
Power available at turbine shaft,
2
⎡ 460 × (18)5/ 4 ⎤
p=⎢ ⎥⎦ = 12927.5 kW
⎣ 150

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Power available form turbines = WQH × η0

= 9.81 × 175 × 18 × 0.82
= 25339.23 kW
25339.23
No. of turbines required =
12927.5
= 1.96 ≈ 2 Nos.

(b) (i) INTRODUCTION

Pelton Wheel or Pelton Turbine is a Tangential Flow Impulse
Turbine. This Turbine is used for High Heads and is named after
L.A Pelton. The water strikes the bucket along the tangent of
the tangent of the runner. The energy available at the inlet of
the turbine is only Kinetic Energy. The pressure at the inlet and
outlet is atmospheric pressure. The nozzle increases the kinetic
energy of the water flowing through the penstock. At the out-
let of the nozzle, the water comes out in the form of a jet and
strikes the buckets (vanes) of the runner. Pelton Wheel Turbine
is used for High Heads. Pelton Wheel Turbine has a Specific
Speed less than 30 (S.I) for single jet and between 30 and 60
(S.I) for multi-jet.
Construction
The main parts of Pelton Turbine are:
Nozzle and Flow Regulating Arrangement
The amount of water striking the buckets is controlled by pro-
viding a spear in the nozzle. The spear is a conical needle oper-
ated in the axial direction depending upon the size of the unit.
When the spear is pushed forward, the amount of water striking
the runner is reduced and when the spear is pushed back, the
amount of water striking the runner increases.
Runner and Buckets
Runner consists of a circular disc on the periphery of which a
number of buckets evenly spaced are fixed. The space of the
buckets is of a double hemispherical cup or bowl. Each bucket
is divided into two symmetrical parts by dividing wall which
is known as Splitter. The Splitter divides the jet into two equal
parts. The buckets are shaped in such a way that the jet gets
deflected through 160° or 170°. The buckets are made of cast

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iron, cast steel bronze or stainless steel depending upon the

head at the inlet of the turbine.
Casing
The function of water is to prevent the splashing of water and
to discharge water to tail race. The casing of Pelton Wheel does
not perform any Hydraulic function.
Breaking Jet
When the nozzle is completely closed by moving the spear
in the forward direction, the amount of the water striking the
runner reduces to zero. But the runner due to inertia goes on
revolving for a long time. To stop the runner in a short time, a
small nozzle is provided which directs the jet of water on the
back of vanes. This jet of water is called Breaking Jet.
Break Nozzle

Runner

Buckets
Pen Stock
Nozzle

Water

Casing

Spear Jet of Water

Penstock
Nozzle

Spear Jet of Water

Wheel

This is
Connected to
Runner

Splitter

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Points to remember for Pelton Turbine

• The velocity of the jet at inlet is given by
V1 = Cv 2 gH

Where Cv = co-efficient of velocity = 0.98 or 0.99

H = Net head on turbine
• The velocity of (u) is given by
u = φ 2 gH

Where φ = speed ratio. The value of speed ratio varies from

0.43 to 0.48
• The angle of deflection of the jet through the buckets is taken
at 165° if no angle of deflection is given.
• The mean diameter or the pitch diameter D of the pelton
turbine is given by
π DN 60u
u= ⋅ or ⋅ D =
60 πN
Jet Ratio
It is defined as the ratio of the pitch diameter (D) of the pelton
turbine to the diameter of the jet (d). It is denoted by m and is
given as
m = D/d ( = 12 for most cases)
Number of bucket on a runner is given by
D
Z = 15 + = 15 + 0.5m
2d
Where m = jet ratio
Number of Jets
It is obtained by dividing total rate of flow through the turbine
by the rate of flow of water through a single jet.

(ii) The hydralic efficiency of a pelton wheel will be maximum

when the velocity of the wheel is half the velocity of the jet of
water at inlet.

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 B.E./B.Tech. DEGREE EXAMINATION,
APRIL/MAY 2011
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY
Time: Three hours Maximum: 100 marks
Answer ALL questions
PART A (10 × 2 = 20 marks)
1. Define surface tension and express its unit.

2. What is Darcy’s equation? Identity various terms in the equation.

3. What is viscosity and give its units?

4. Distinguish between a control and differential control volume.

5. Check whether the following equation is dimensionally homogeneous.

Q = Cd ⋅ a 2 gh .

6. Distinguish between Rayleigh’s method and Buckingham’s I theorem.

7. Write the Euler’s equation.

8. A shaft transmits 150 Kw at 600 rpm. What is the torque in Newton

metres?

9. Why is the reciprocating pump called a positive displacement pump?

10. What is a rotary pump? Give its classification.

PART B (5 × 16 = 80 marks)
11. (a) Explain in detail the Newton’s law of viscosity. Briefly classify the
fluids based on the density and viscosity.
Give the limitations of applicability of Newton’s law of viscosity.
Or

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(b) State the law of conservation of mass and derive the equation of con-
tinuity in Cartesian co-ordinates for an incompressible fluid. Would
it alter if the flow were unsteady, highly viscous and compressible?

12. (a) A lubricating oil flows in a 10 cm diameter pipe at 1 m/s. Determine

whether the flows is laminar or turbulent. For the lubricating oil
µ = 0.1 Ns/m2 and r = 930 kg/m3. Calculate also transition and tur-
bulent velocities.
Or
(b) A smooth flat plate with a sharp leading edge is placed along a free
stream of water flowing at 3 m/s. Calculate the distance from the
leading edge and the boundary thickness where the transition form
laminar to turbulent flow may commence. Assume the density of
water as 1000 kg/m3 and viscosity as 1 centipoise.

13. (a) What is the significance and the role of the following parameters?
(i) Reynolds number
(ii) Froade number
(iii) Mach number
(iv) Weber number.
Or
(b) An agitator of diameter D rotates at a speed N in a liquid of density
p and viscosity µ. Show that the power required to mix the liquid is
expressed by a functional form

P ⎛ pND 2 N 2 D ⎞
= f ⎜⎝ μ , g ⎟⎠
pN 2 D 5

14. (a) A gas turbine operates between 1000 k and 650 K temperature lim-
its taking in air 20 kg/s at 125 m/s and discharging at 3000 m3/s.
Estimate the power developed by the turbine. Given Cp = 995 J/kg k.
Or
(b) Explain in detail about the performance curves for pumps and
turbines.

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15. (a) A single acting reciprocatory pump has a plunger of diameter 30 cm

and stroke of 20 cm. If the speed of the pumps is 30 rpm and it
delivers to 6.5 lit/s of water, find the coefficient of discharge and the
percentage slip of the pump.
Or
(b) Write briefly on the following.
(i) Rotary pumps and their classifications.
(ii) Indicator diagram for reciprocating pump.

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 B.E./B.Tech. DEGREE EXAMINATION,
NOV/DEC 2010
Third Semester
Mechanical Engineering
FLUID MECHANICS AND MACHINERY
Time: Three hours Maximum: 100 marks
Answer ALL questions
PART A (10 × 2 = 20 marks)
1. Suppose the small air bubbles in a glass of tap water may be on the order
of 50 µ m in diameter, what is the pressure inside these bubbles?

2. Why is it necessary in winter to use lighter oil for automobiles than in

summer? To what property does the term lighter refer?

3. Cite examples for dimensionally homogeneous and non-homogeneous

equations.

4. Mention the circumstances which necessitate the use of distorted models.

5. What is a laminar sub layer?

6. Define eddy viscosity. How it differs from molecular viscosity?

7. State the principles on which turbo-machines are based.

8. Under what conditions would you suggest use of double-suction pump

and a multistage pump?

9. What is an air vessel? List the objectives that would be fulfilled by the
use of air vessels.

10. What is priming?

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PART B (5 × 16 = 80 marks)
11. (a) (i) A U-tube is made of two capillaries of diameter 1.0 mm and
1.5 mm respectively. The tube is kept vertically and partially
filled with water of surface tension 0.0736 N/m and zero contact
angle. Calculate the difference in the levels of the menisci
caused by the capillary.
(ii) Lateral stability of a long shaft 150 mm in diameter is obtained
by means of a 250 mm stationary bearing having an internal
diameter of 150.25 mm. If the space between bearing and shaft
is filled with a lubricant having a viscosity 0.245 Ns/m2, what
power will be required to overcome the viscous resistance when
the shaft is rotated at a constant rate of 180 rpm?
Or
(b) (i) A pipeline 60 cm in diameter bifurcates at a Y-junction into two
branches 40 cm and 30 cm in diameter. If the rate of flow in the
main pipe is 1.5 m3/s, and the mean velocity of flow in the 30 cm
pipe is 7.5 m/s, determine the rate of flow in the 40 cm pipe.
(ii) Derive the energy equation and state the assumptions made
while deriving the equation.

12. (a) A pipe of 10 cm in diameter and 1000 m long is used to pump

oil of viscosity 8.5 poise and specific gravity 0.92 at the rate of
1200 lit./min. The first 30 m of the pipe is laid along the ground
sloping upwards at 10° to the horizontal and remaining pipe is laid
on the ground sloping upwards 15° to the horizontal. State whether
the flow is laminar or turbulent? Determine the pressure required to
be developed by the pump and the power required for the driving
motor if the pump efficiency is 60%. Assume suitable data for fric-
tion factor, if required.
Or
(b) Two pipes of diameter 40 cm and 20 cm are each 300 m long. When
the pipes are connected in series and discharge through the pipe line
is 0.10 m3/sec, find the loss of head incurred. What would be the loss
of head in the system to pass the same total discharge when the pipes
are connected in parallel? Take f = 0.0075 for each pipe.

13. (a) (i) The resisting force F of a plane during flight can be consid-
ered as dependent upon the length of aircraft (1), velocity (v),
air viscosity (m), air density (r) and bulk modulus of air (K).

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Express the functional relationship between these variables

using dimensional analysis. Explain the physical significance
of the dimensionless groups arrived.
Or
(b) The drag force (F) on a partially submerged body depends on the
relative velogity (V) between the body and the fluid, characteris-
tic linear dimension (1), height of surface roughness (k), fluid
density (r), the viscosity (m) and the acceleration due to gravity (g).
Obtain an expression for the drag force, using the method of dimen-
sional analysis.

14. (a) The head discharge characteristics of a centrifugal pump is given

below.
Discharge (lit/sec) 0 10 20 30 40 50
Head (meters) 25.3 25.5 24.5 22.2 18.7 12.0
The pump delivers fresh water through a 500 m long, 15 cm diam-
eter pipeline having friction coefficient of f = 0.025. The static life is
15 m. Neglecting minor losses in the pipe flow, find (i) the discharge
of the pump under the above conditions (ii) driving power of the
pump motor. Assume a pump efficiency of 72%.
Or
(b) An inward flow reaction turbine having an overall efficiency of 80%
is required to deliver 136 kW. The head H is 16 m and the peripheral
velocity is 3.3 H . The radial velocity of flow at inlet is 1.1 H .
The runner rotates at 120 rpm. The hydraulic losses in the turbine
are 15% of the flow available energy. Determine
• Diameter of the runner,
• Guide vane angle,
• The runner blade angle at inlet and
• The discharge through the turbine.

15. (a) Explain the working principle of single and double acting recipro-
cating pumps with neat diagram in detail. Also explain the effects of
inertia pressure and friction on the performance of the pump using
indicator diagrams with and without air vessel.
Or
(b) Explain the working principle of screw pump and gear pump with
neat diagram in detail.

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