regulation 2017 academic year : 2018-2019 ce8394 fluid ... · subject handler: mr.m.k.karthik &...

REGULATION : 2017 ACADEMIC YEAR : 2018-2019

JIT-JEPPIAAR/MECH /Mr.D.Arunkumar & Mr.M.K.Karthik/IIrd Yr/SEM 03 /CE8394/FLUID MECHANICS AND

MACHINERY/UNIT 1-5/QB+Keys/Ver1.0

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CE8394 FLUID MECHANICS AND MACHINERY L T P C

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OBJECTIVES

1. The properties of fluids and concept of control volume are studied

2. The applications of the conservation laws to flow through pipes are studied.

3. To understand the importance of dimensional analysis

4. To understand the importance of various types of flow in pumps.

5. To understand the importance of various types of flow in turbines.

UNIT I FLUID PROPERTIES AND FLOW CHARACTERISTICS 12

Units and dimensions- Properties of fluids- mass density, specific weight, specific volume, specific gravity,

viscosity, compressibility, vapor pressure, surface tension and capillarity. Flow characteristics – concept of control

volume - application of continuity equation, energy equation and momentum equation.

UNIT II FLOW THROUGH CIRCULAR CONDUITS 12

Hydraulic and energy gradient - Laminar flow through circular conduits and circular annuli-Boundary layer

concepts – types of boundary layer thickness – Darcy Weisbach equation –friction factor- Moody diagram-

commercial pipes- minor losses – Flow through pipes in series and parallel.

UNITIII DIMENSIONAL ANALYSIS 12

Need for dimensional analysis – methods of dimensional analysis – Similitude –types of similitude - Dimensionless

parameters- application of dimensionless parameters – Model analysis.

UNIT IV PUMPS 12

Impact of jets - Euler’s equation - Theory of roto-dynamic machines – various efficiencies– velocity components at

entry and exit of the rotor- velocity triangles - Centrifugal pumps– working principle - work done by the impeller -

performance curves - Reciprocating pump- working principle – Rotary pumps –classification.

UNIT V TURBINES 12

Classification of turbines – heads and efficiencies – velocity triangles. Axial, radial and mixed flow turbines. Pelton

wheel, Francis turbine and Kaplan turbines- working principles - work done by water on the runner – draft tube.

Specific speed - unit quantities – performance curves for turbines – governing of turbines.

TOTAL: 60 PERIODS

OUTCOMES:

Upon completion of this course, the students will be able to

CO1 Apply mathematical knowledge to predict the properties and characteristics of a fluid. UNIT 1

CO2 Can analyse and calculate major and minor losses associated with pipe flow in piping networks. UNIT 2

CO3 Can mathematically predict the nature of physical quantities UNIT 3

CO4 Can critically analyse the performance of pumps. UNIT 4

CO5 Can critically analyse the performance of turbines. UNIT 5

TEXT BOOK:

1. Modi P.N. and Seth, S.M. "Hydraulics and Fluid Mechanics", Standard Book House, New Delhi 2013.

REFERENCES:

1. Graebel. W.P, "Engineering Fluid Mechanics", Taylor & Francis, Indian Reprint, 2011

2. Kumar K. L., "Engineering Fluid Mechanics", Eurasia Publishing House(p) Ltd., New Delhi 2016

3. Robert W.Fox, Alan T. McDonald, Philip J.Pritchard, “Fluid Mechanics and Machinery”, 2011.

4. Streeter, V. L. and Wylie E. B., "Fluid Mechanics", McGraw Hill Publishing Co. 2010

OUTCOMES:

Upon completion of this course, the students will be able to

• Apply mathematical knowledge to predict the properties and characteristics of a fluid.

• Can analyse and calculate major and minor losses associated with pipe flow in piping networks.

• Can mathematically predict the nature of physical quantities

• Can critically analyse the performance of pumps

• Can critically analyse the performance of turbines.




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Subject Code: CE8394 Year/Semester: II /03

Subject Name: Fluid Mechanics and Machinery

Subject Handler: Mr.M.K.Karthik & Mr.D.Arunkumar

UNIT I FLUID PROPERTIES AND FLOW CHARACTERISTICS

Units and dimensions- Properties of fluids- mass density, specific weight, specific volume, specific

gravity, viscosity, compressibility, vapor pressure, surface tension and capillarity. Flow characteristics –

concept of control volume - application of continuity equation, energy equation and momentum

equation.

PART * A

Q.No. Questions

1. Define fluids. BTL1

Fluid may be defined as a substance which is capable of flowing. It has no definite shape of its

own, but confirms to the shape of the containing vessel.

2

What are the properties of ideal fluid? (APRIL2012) BTL1

Ideal fluids have following properties:

• It is incompressible

• It has zero viscosity

• Shear force is zero

3

What are the properties of real fluid? (APRIL2012,2014) BTL1

Real fluids have following properties

• It is compressible

• They are viscous in nature

• Shear force exists always in such fluids.

4

Define density and specific weight..BTL1

Density is defined as mass per unit volume (kg/m3)

Specific weight is defined as weight possessed per unit volume (N/m3)

5

Define Specific volume and Specific Gravity. BTL1

Specific volume is defined as volume of fluid occupied by unit mass (m3/kg)

Specific gravity is defined as the ratio of specific weight of fluid to the specific weight of

standard fluid.

6

Define Surface tension and Capillarity. BTL2

Surface tension is due to the force of cohesion between the liquid particles at the free surface.

Capillary is a phenomenon of rise or fall of liquid surface relative to the adjacent general level

of liquid.

7

Define Viscosity.BTL2

It is defined as the property of a liquid due to which it offers resistance to the movement of one

layer of liquid over another adjacent layer.

8 Define Kinematic viscosity. (NOVEMBER2012,2015) BTL2

It is defined as the ratio of dynamic viscosity to mass density. (m²/sec)

9 Define Relative or Specific viscosity. BTL2

It is the ratio of dynamic viscosity of fluid to dynamic viscosity of water at 20°C.




3- 3

10

Define Compressibility. BTL2

It is the property by virtue of which fluids undergoes a change in volume under the action of

external pressure.

11

Define Newton’s law of Viscosity. (NOVEMBER 2012) BTL1

According to Newton’s law of viscosity the shear force F acting between two layers of fluid is

proportional to the difference in their velocities du and area A of the plate and inversely

proportional to the distance between them.

12

What are Newtonian and Non-Newtonian fluids? (APRIL2011) BTL2

Fluids which obey the Newton’s law of Viscosity are known as Newtonian fluids and the fluids

which do not obey the Newton’s law of Viscosity are known as Non-Newtonian fluids.

13

What is cohesion and adhesion in fluids? (APRIL2013) BTL2

Cohesion is due to the force of attraction between the molecules of the same liquid.

Adhesion is due to the force of attraction between the molecules of two different liquids or

between the molecules of the liquid and molecules of the solid boundary surface.

14

State momentum of momentum equation. (APRIL2014, NOVEMBER2016) BTL1

It states that the resulting torque acting on a rotating fluid is equal to the rate of change of

moment of momentum

15

What is momentum equation? (APRIL2012) BTL1

It is based on the law of conservation of momentum or on the momentum principle. It states

that, the net force acting on a fluid mass is equal to the change in momentum of flow per unit

time in that direction.

16

What are the types of fluid flow? (NOVEMBER2015) BTL1

• Steady & unsteady fluid flow

• Uniform & Non-uniform flow

• One dimensional, two-dimensional & three-dimensional flows

• Rotational & Irrotational flow

17

Name the different forces present in fluid flow.BTL1

• Inertia force

• Viscous force

• Surface tension force

• Gravity force

18

When in a fluid considered steady? BTL2

In steady flow, various characteristics of following fluids such as velocity, pressure, density,

temperature, etc at a point do not change with time. So it is called steady flow.

19

State the application of Bernoulli’s equation. BTL1

It has the application on the following measuring devices.

• Orifice meter.

• Venturimeter.

• Pitot tube.

20 What is importance of kinematic viscocity? (NOVEMBER 2014) BTL1

Since the density is a strong function of pressure and temperature, so is the kinematic viscosity it




3- 4

is generally a preferred unit when we deal with motion of fluid under the influence of gravity.

PART * B

1

Water flows at the rate of 200 litres per second upwards through a tapered vertical pipe.

The diameter at the bottom is 240mm and at the top 200mm and the length is 5m. The

pressure at the bottom is 8 bar and the pressure at the top side is 7.3bar. Determine the

head loss through the pipe. (13M) (APRIL2010) BTL2

Answer: Page 126 – R.K.Bansal

d1=0.24m, d2=0.2m, A1=0.04523m2, A2=0.03141m2, z1=0, z2=5m, Q=0.2m3/s, P1=8x105

N/m2, P2 =7.3 x105 N/m2.

(i) Q = A1V1 = A2V2

(ii 𝑃1

𝜌𝑔+

𝑣12

2𝑔+ 𝑧1 =

𝑃2

𝜌𝑔+

𝑣22

2𝑔+ 𝑧2 + ℎ𝑓 (2M)

Q = A1V1 (1M)

V1= Q/A1= 0.2/0.04523

V1= 4.4218 m/s, (2M)

V2= Q/A2= 0.2/0.03141

V2=6.3763 m/s (2M)

Z1=Z2=0

hf=1.067m (6M)

2

The water level in a tank is 20m above the ground. A hose is connected to the bottom of the

tank, and nozzle at the end of the hose is pointed straight up. The tank is at sea level, and

the water surface is open to the atmosphere. In the line learning from the tank to nozzle is a

pump, which increases the pressure of the water. If the water jet rises to a height of 27

from the ground, determine the minimum pressure rise supplied by the pump to the water

line. (13M) (APRIL2012,2016) BTL2


z1=0, z2=7m,

p1=0, p2 =0,

V1= 0, V2=0 (2M) 𝑃1

𝜌𝑔+

𝑣12

2𝑔+ 𝑧1 + ℎ𝑝 =

𝑃2

𝜌𝑔+

𝑣22

2𝑔+ 𝑧2 (1M)

hp=7m (3M).

Pp=ρghp

Pp=68.67 x103 N/m2 (7M)

3

If the velocity discharge of a fluid over a plate is given by u=ay2+by+c with the vertex 0.2

from the plate, where the velocity is 1.2m/s. Calculate the velocity gradient and shear

stress at a distance of 0m, 0.1m and 0.2m from the plate, if the viscosity of the fluid is

0.85Ns/m2. (13M) (APRIL2010) BTL2


U=120cm/sec,

µ=0.85Ns/m2,

u=ay2+by+c (1M)




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i)y=0, u=0 (1M)

ii)y=0.2, u= 120m/s (2M)

iii)y=0.2, du/dy=0.

c=0, b=12, a=-0.3

u=-0.3y2+12y (1M)

du/dy=-0.6y+12

y=0,

du/dy=12 s-1,

τ=10.2 N/m2. (4M)

y=10, du/dy=6 s-1,

τ=5.1 N/m2(4M)

4

A pipe 200m long slopes down at 1in 100 and tapers from 600mm diameter at the higher

end to 300mm diameter at the lower end, and carries 100lps of oil having specific

gravity 0.8, if the pressure gauge at the higher end reads 60kN/m2., determine the

velocities at the two ends and also the pressure at the lower end. (13M) (APRIL2012)

BTL2


d1=0.5m, d2=0.25m

z1=5m

Q=0.07m3/s,

p1=2.5x105 N/m2

L1=L2=150m (1M)

(i) Q = A1V1 = A2V2

(ii) 𝑃1

𝜌𝑔+

𝑣12

2𝑔+ 𝑧1 =

𝑃2

𝜌𝑔+

𝑣22

2𝑔+ 𝑧2 + ℎ𝑐 (1M)

V1=0.3565m/s (2M)

V2=1.4260 m/s. (2M)

hc=0.5V22/2g = 0.0518m (3M)

P2 = 297.5883 x103 N/m2(4M)

5

Write the assumptions and derive the bernoulli’s equation from the euler’s equation.

Assumptions: (13M) (APRIL2015,2016) BTL2


Assumption:

• flow is steady and continuous

• flow is incompressible

• flow is non viscous

• flow is irrotational (4M)

Bernoulli’s principle says that the faster a fluid is moving the less pressure it exerts.

Pressure force = pdA in the direction of flow

Pressure force 𝑃 +𝜕𝑃

𝜕𝑠 𝑑𝑆𝑑𝐴 opposite to the direction of flow

pdA - 𝑃 +𝜕𝑃

𝜕𝑠 𝑑𝑆𝑑𝐴 – ρgdAdS cosθ = 0 (3M)




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𝐴𝑆 = 𝑑𝑣

𝑑𝑡 (2M)

𝑃

𝜌𝑔+

𝑣2

2𝑔+ 𝑧 = constant (4M)

6

Calculate the dynamic viscosity of oil which is used for lubrication between a square plate

of size 0.8m x 0.8m and an inclined plane with the angle of deflection 30 . The weight of the

square plate is 330N and it is slide down the inclined plane with a uniform velocity of

0.3m/s. the thickness of the oil film is 1.5mm. (13M) (NOVEMBER 2011,2015) BTL2

Answer: Page 204 – R.K. Bansal

b=0.2m

m=30kg, θ=30°,

µ=2.3 x10-3 Ns/m2

t=0.02mm(1M)

F=Wsinθ (1M)

τ=µ du/dy. (1M)

U=31.9891 m/s. (4M)

F=147.15N (6M)

7

A 6m long pipe is inclined at an angle of 20º with the horizontal. The smaller section of the

pipe which is at lower level is of 100 mm dia and the larger section is of 300 mm dia. If the

pipe is uniformly tapering and the velocity of the water at the smaller section is 1.8m/s.

Determine the difference of pressures between two sections. BTL2 (13M)


L=6m

Θ = 20º

D1=100mm

D2=300mm

V1=1.8m/s.

Q = A1V1 (1M)

A1= π/4 d2 (1M)

A1= 581.77 m2 (1M)

Q = 1.047m3/s (3M)

A1V1=A2V2 𝑃1

𝜌𝑔+

𝑣12

2𝑔+ 𝑧1 =

𝑃2

𝜌𝑔+

𝑣22

2𝑔+ 𝑧2

P1-P2=18.44 N/m2 (7M)

8

A 30 cm x 15 cm venturimeter is provided in a vertical pipe line carrying oil of specific

gravity 0.9, the flow being upwards. The difference in elevation of the throat section and

entrance section of the venturimeter is 30 cm. The differential U tube mercury manometer

shows a gauge deflection of 25 cm. Calculate: (i) the discharge of oil. (ii) The pressure

difference between the entrance section and the throat section. Take Cd=0.98 and specific

gravity of mercury as 13.6. (13M) (APRIL2012) BTL2


S= 0.9

D1=30 cm

D2=15 cm




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𝑃1

𝜌𝑔+

𝑣12

2𝑔+ 𝑧1 =

𝑃2

𝜌𝑔+

𝑣22

2𝑔+ 𝑧2 + ℎ (2M)

h= 3.53 m of oil (3M)

A1= π/4 d2

A1= 0.0706 m2 (2M)

Q = A1V1

Discharge Q= 0.1489 m3/s (3M)

P1-P2=33.8 KN/m2 (3M)

PART * C

1

A 400 mm diameter shaft is rotating at 200 rpm. in a bearing of length 120 mm. If the

thickness of oil film is 1.5 mm and the dynamic viscosity of the oil is 0.7 N.s/m2 determine:

(i) Torque required to overcome friction in bearing (ii) Power utilized in overcoming

viscous resistance.(15M) (APRIL2007) BTL2

Answer: Page 12 – R.K.Bansal,

D= 400mm

N= 200 rpm

L=120mm

dy= 1.5mm (1M)

dynamic viscosity = 0.7 N.s/m2

𝑢 = πDN

60

u=4.19 m/s (3M)

shear force ,F= 294.85N (3M)

Torque, T= Fx D/2= 58.97 N-m(3M)

Power utilized, P = 2πNT/60 = 1235 w (3M)

2

A horizontal venturimeter with inlet and throat diameter 300 mm and 100 mm respectively

is used to measure the flow of water. The pressure intensity at inlet is 130 Kn/m2 while the

vacuum pressure head at throat is 350 mm of mercury. Assuming that 3% head lost

between the inlet and throat. Find the value of coefficient of discharge for the venturimeter

and also determine the rate of flow. (15M) (APRIL2009,2010,2013) BTL2


D1= 300mm

D2= 100mm

hf = 3%

(6M)




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(9M)

3

A vertical venturimeter 40 cm x 20 cm is provided in a vertical pipe to measure a flow of oil

of relative density 0.8. The difference in elevations of the throat section and the entrance

sections in 1 m, the direction of flow of oil being vertically upwards. The oil-mercury

differential gauge shows deflection of mercury equal to 40 cm. Determine the quantity of

oil flowing the pipe. Neglect losses. (15M) BTL2


D1= 40 cm

D= 20 cm

S=0.8

Z1-Z2= 1m

h= 40 cm 𝑃1

𝜌𝑔+

𝑣12

2𝑔+ 𝑧1 =

𝑃2

𝜌𝑔+

𝑣22

2𝑔+ 𝑧2 + ℎ (1M)

𝑃1/ 𝜌𝑔 = 13.25 (2M)

𝑃2/ 𝜌𝑔 = −4.76 𝑚 (2M)

H= 18.01 m (2M)

Cd=0.985(3M)

Q=0.146 m3/s(5M)

UNIT II FLOW THROUGH CIRCULAR CONDUITS

Hydraulic and energy gradient - Laminar flow through circular conduits and circular annuli-Boundary

layer concepts – types of boundary layer thickness – Darcy Weisbach equation –friction factor- Moody

diagram- commercial pipes- minor losses – Flow through pipes in series and parallel.

PART * A

Q.No. Questions

1.

Mention the general characteristics of laminar flow. BTL1

• There is a shear stress between fluid layers

• No slip at the boundary

• The flow is rotational

• There is a continuous dissipation of energy due to viscous shear

2

What is Hagen poiseuille’s formula? BTL1

𝑃1−𝑃2

pg = hf =

32 µ𝑈𝐿

𝜌𝑔𝐷2

The expression is known as Hagen poiseuille formula.

Where, 𝑃1−𝑃2

pg = Loss of pressure head; U = Average velocity




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µ = Coefficient of viscosity; D = Diameter of pipe, L = Length of pipe.

3

List the factors influencing the frictional loss in pipe flow. BTL1

Frictional resistance for the turbulent flow is

• Proportional to velocity varies from 1.5 to 2.0.

• Proportional to the density of fluid.

• Proportional to the area of surface in contact.

• Independent of pressure.

• Depend on the nature of the surface in contact.

4

Write athe expression for head loss due to friction in Darcy formula. BTL1

hf = 4𝑓𝐿𝑉

2𝑔𝐷

2

Where f = Coefficient of friction in pipe L = Length of the pipe

D = Diameter of pipe V = velocity of the fluid

5

What do you understand by the terms a) Major energy losses b) Minor energy losses. BTL2

Major energy losses: -

This is loss due to friction and is calculated by Darcy-Weisbach formula and

Chezy’s formula.

Minor energy losses:-

• Sudden expansion in pipe.

• Sudden contraction in pipe.

• Bend in pipe.

• Obstruction in pipe.

• Various pipe fittings.

6

Give an expression for loss of head due to sudden enlargement of the pipe. BTL1

he = (𝑉1−𝑉2)

2𝑔

2

Where,

he = Loss of head due to sudden enlargement of pipe .

V1 = Velocity of flow at section 1-1

V2 = Velocity of flow at section 2-2

7

Give an expression for loss of head due to sudden contraction. BTL1

hc = 0.5 𝑉2

2𝑔

Where,

c = Loss of head due to sudden contraction.

V = Velocity at outlet of pipe.

8

Give an expression for loss of head at the entrance of the pipe. BTL1

hi = 0.5 𝑉2

2𝑔

Where,

hi = Loss of head at entrance of pipe.




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V = Velocity of liquid at inlet and outlet of the pipe.

9

Define the term Hydraulic gradient line [HGL]. (NOVEMBER 2015) BTL2

Hydraulic gradient line is defined as the line which gives the sum of pressure head and

datum head of a flowing fluid in a pipe with respect the reference line.

10

What is Syphon? Where it is used? BTL2

Syphon is a long bend pipe which is used to transfer liquid from a reservoir at a higher

elevation to another reservoir at a lower level.

Uses of syphon : -

• To carry water from one reservoir to another reservoir separated by a hill ridge.

• To empty a channel not provided with any outlet sluice.

11

What are the basic educations to solve the problems in flow through branched pipes? BTL1

• Continuity equation.

• Bernoulli’s formula.

• Darcy Weisbach equation.

12

What is Dupuit’s equation? BTL1

𝐿1

𝑑15 +

𝐿2

𝑑25 +

𝐿3

𝑑35=

𝐿

𝑑5 Where

L1, d1 = Length and diameter of the pipe 1



13

Define kinetic energy correction factor. BTL1

Kinetic energy factor is defined as the ratio of kinetic energy of the flow per sec based on

actual velocity across a section to the kinetic energy of the flow per sec based on average

velocity across the same section. It is denoted by (α).

14

What is Hydraulic mean depth? BTL1

It is the ratio of cross-sectional area of the flow and wetted perimeter, where wetted perimeter is

the perimeter of the pipe or the channel which remains in contact with the flowing fluid.

15

What do you understand by the transmission efficiency of a pipe? BTL1

It is the ratio of the power available at the end of the pipe to power available at the entry of the

pipe.

μ = Power available at the end of the pipe/ Power available at the entry of the pipe

16 Mention the condition for maximum efficiency. BTL1

The condition for maximum efficiency is given by (H-(H/3))/H

17

What do you meant by viscous flow? BTL1

A flow is said to be viscous if the Reynold’s number is less than 2000 (or) the flows in layers

i.e., Re<2000.

18 State the Relationship between Shear stress and pressure gradient. BTL1

The relationship between shear stress and pressure gradient indicates that the pressure gradient in

the direction of flow is equal to the shear gradient in the direction normal to the direction of flow.

19 State the assumptions made in finding the equivalent length of a compound pipe. BTL1

A compound pipe consisting of several pipes of varying diameters and length may be replaced by




3- 11

a pipe of uniform diameter is known as equivalent pipe.

• The material of the pipe is same, and hence the co efficient is same.

• The minor losses are neglected.

20

Define the term Total Energy line [TEL].BTL2

Total energy line is defined as the line which gives the sum of pressure head, datum head

and kinetic head of a flowing fluid in a pipe with respect to some reference line.

PART * B

1

A fluid of viscosity 0.7 Pa.s and specific gravity 1.3 is flowing through a pipe diameter

120mm. the maximum shear stresss at the pipe is 205.2N/m2. Determine the pressure

gradient, reynold number and average velocity? 13M (APRIL2015) BTL2


µ=0.7 Pa-s

S=1.2

D=0.12m

R=0.06m (1M)

τmax=200 N/m2.

τmax=(-dp/dx) R/2 (2M)

Uavg=(-dp/dx)R2/8µ

Re=ρUavg D/µ

dp/dx = -6.667 x103 N/m2/m (3M)

Uavg=4.2859 m/s (3M)

Re=881.6708(4M)

2

For a flow of viscous fluid flowing through a circular pipe under laminar flow conditions

show that the velocity distribution is a parabola. And also show that the average velocity is

half of the maximum velocity. (13M) (APRIL2011) BTL2


τ=(-dp/dx) r/2

τmax=(-dp/dx) R/2(4M)

u= (-dp/dx)1/4µ(R2-r2)

u max =(-dp/dx) R2/4µ

u = umax(1-r2/R2) (9M)

3

Three pipes of 400mm,200mm and 300mm diamters have lengths 400m, 200m and 300m

resp. they are connected in series to make a compound pipe. The ends of this compound

pipe are connected with two tanks whose difference of water levels is 16m. if the coefficient

of friction for these pipes is same and equal to 0.005. determine the discharge through the

compound pipe neglecting the minor losses. (13M) (NOVEMBER2016) BTL2


d1=0.2m, d2=0.4m, d3=0.6m

L1=400m, L2=800m, L3=1200m

H=20m

Q = A1V1 = A2V2 = A3V3




3- 12

V2 = 0.25 V1

V3 = 0.1111 V1(4M)

H=hf1+hf2+hf3

Hf=4flv2/2gD

H=51.5926 v12/2x9.81

V1=2.7578 m/s(4M)

Q=86.63 lps(5M)

4

The velocity distribution in the boundary layer is given by u/U = 2(y/δ) – (y/δ)2, δ being the

boundary layer thickness. Calculate the following: i. Displacement thickness, ii.

Momentum thickness, iii. Energy thickness. (NOVEMBER2016) (13M) BTL2


u/U = 2(y/δ) – (y/δ)2

δ - boundary layer thickness

(6M)

(7M)

5

Derive the expression for shear stress and velocity distribution for the flow through

circular pipe and using that derive the Hagen Poiseuille formula. (NOVEMBER2016)


The derivation of the Hagen-Poiseuille equation for laminar flow in straight, circular pipes is

based on the following two assumptions.

a) The viscous property of fluid follows Newton’s law of viscosity, that is, τ=μ(du/dy).

b) There is no relative motion between fluid particles and solid boundaries, that is no slip of

fluid particles at the solid boundary. (3M)




3- 13

(3M)

g

pp 21

= hf

2gD

uL32

(7M)

6

Calculate discharge through a pipe of diameter 200 mm when the difference of pressure

head between the two ends of a pipe 500 mm apart is 4m of water. Take the value of

f=0.009. (13M) BTL1


d=0.2m

L= 500m

f=0.009

Q = AV

A = π/4 d2 = 0.0314 m2(3M)

hf=4flv2/2gD,

V= 0.934 m/s (2M)

hf = 0.872 (4M)

Q = 0.0314x0.934= 0.0293 m3/s (4M)

PART * C

1

Derive an expression to find the loss of head due to friction in pipes. (15M)

(NOVEMBER2016) BTL1


It is also called as Darcy’s weibach’s equation. The Darcy–Weisbach equation is a phenomenological equation, which relates the head loss, or pressure

loss, due to friction along a given length of pipe to the average velocity of the fluid flow for an

incompressible fluid. (5M)

(4M)




3- 14

Where,

hf = head loss due to friction

f- coefficient of friction in pipe material

l- length of pipe v- velocity of liquid

g- acceleration due to gravity d-diameter of pipe.

hf=4flv2/2gd (6M)

2

A pipe of 0.6 m diameter is 1.5 km long. In order to augment the discharge, another line of

the same diameter is introduced parallel to the first in the second half of the length.

Neglecting minor losses. Find the increase in discharge, if friction factor hf = 0.04. The

head at inlet is 300 mm. (15M) (APRIL2012) BTL2


D= 0.6m

L=1.5m

hf=0.04

(2M)

A1= 0.282 m2

V1= 1.3 m/s (2M)

V2= 2.4 m/s(2M)

Q1= 0.02 m3/s (3M)

Q2= 0.25 m3/s (3M)

Q = 0.27 m3/s (3M)

3

A crude oil of kinematic viscosity 0.4 stoke is flowing through a pipe of diameter 300 mm at

the rate of 300 litres per sec. Find the head lost due to friction for a length of 50 m of the

pipe. (15M) BTL2


d=0.3m




3- 15

L= 50m

Q = 300 lit/s (2M)

Q = 0.3 m3/s

V= 4.24 m/s (5M)

hf=4flv2/2gD (1M)

hf = 3.61 m (7M)

UNIT III DIMENSIONAL ANALYSIS

Need for dimensional analysis – methods of dimensional analysis – Similitude –types of

similitude - Dimensionless parameters- application of dimensionless parameters – Model

analysis.

PART * A

Q.No. Questions

1.

State the methods of dimensional analysis. BTL1

The methods of Dimensional analysis are:

1. Rayleigh’s method

2. Buckingham’s π theorem

2

State Buckingham’s π theorem. (NOVEMBER2016) BTL1

It states that, “If there are n variables (independent and dependent variables) in a physical

phenomenon and if these variables contain m fundamental dimensions (M, L,T), then the

variables are arranged into (n-m) dimensionless terms. Each term is called π-term”.

3

Define the term dimensional homogeneity. BTL1

Dimensional homogeneity means the dimensions of each terms in an equation on both sides are

equal.

4

Differentiate between fundamental units and derived units. Give examples. BTL1

Fundamental units are physical quantities from which every other unit can be generated. They

are a set of seven basic units.

Example: Length, Mass, Time.

Derived units are obtained by combining one or more fundamental units.

Example: Speed(or) Velocity, Volume, Density, etc.

5

State the limitations of dimensional analysis. BTL1

1. Dimensional analysis does not give any due regarding the selection of variables.

2. The complete information is not provided by dimensional analysis.

3. The values of coefficient and the nature of function can be obtained only by

experiments or from mathematical analysis.

6 Define Similitude. (NOVEMBER2016) BTL1

Similitude is defined as the complete similarity between the model and prototype.

7

State Froude’s model law. BTL1

The law states ‘The Froude’s number is same for both model and prototype. Only Gravitational

force is more predominant force.




3- 16

8

What is meant by Model and Prototype? BTL1

The model is the small scale replica of the actual structure or machine. The actual structure or

machine is called Prototype.

9

How are hydraulic models classified? BTL1

The hydraulic models are classified as:

1.Undistorted models

2. Distorted models.

10

Define Euler’s Number. (APRIL 2016) BTL1

It is defined as the square root of the ratio of the inertia force of a flowing fluid to the pressure

force.

Mathematically, it is expressed as: Eu = √𝑭𝒊

𝑭𝑷 =

𝑽

√𝒑/𝝆

11

Define Weber’s Number. BTL1

It is defined as the square root of the ratio of the inertia force of a flowing fluid to the surface

tension force.

`Mathematically, it is expressed as: We = √𝑭𝒊

𝑭𝒔 =

𝑽

√𝝈/𝝆𝑳

12

List the types of similarity. BTL1

1. Geometric Similarity

2. Kinematic Similarity

3. Dynamic Similarity

4. Complete Similarity

13

Define Geometric Similarity. BTL2

The model must be the same shape as the prototype. Each dimension must be scaled by the

same factor.

14

Define Kinematic Similarity. BTL1

This similarity requires that the length and time scales be similar between the model and the

prototype implying that velocities at corresponding points be similar

15

Define Mach’s Number. BTL1

It is the ratio of inertia force to the elastic force of flowing fluid.

M = √𝑭𝒊

𝑭𝒆

PART * B

1

A ship model of scale 1/50 is towed through sea water at a speed of 1 m/s. A force of 2N is

required to tow the model. Determine the speed of the ship and the propulsive force on the

ship, if the prototype is subjected to ware resistance only.(13M) (APRIL2015) BTL2





3- 17

(6M)

(7M)

2

Using buckhingam’s π theorem, show that the velocity of flow through a cicrular

orifice is given by V=√2gH (D/H, µ/ρvH). (13M) (APRIL2011) BTL2


V=f(H,D,µ,ρ,g)

The dimensions are V = LT-1

H = L D = L

µ = ML-1T-1 (1M) ρ = ML-3 (1M)

g = LT-2 (1M) Then

n=6

m=3 no.of π terms = n-m = 3

π1=Ha1.ρb1.gc1.v (2M) π2=Ha2.ρb2.gc2.D (2M)

π3=Ha3.ρb3.gc3.µ (2M) solving we get, V=√2gH ᴓ(D/H, µ/ρvH) (4M)

3

Model of an air duct operating with water produces a pressure drop of 10kN/m2 over 10m

length. If the scale ratio is 1/50. Density of water is 1000kg/m3 and air density is 1.2kg/m3.

Viscosity of water is 0.001Ns/m2 and viscosity of air is 0.00002 Ns/m2. Estimate

corresponding drop in a 20m long air duct. (13M) BTL2


Lp/Lm= 1/50, µp/ µm =0.001/0.0002, ρp/ρm = 1.2/1000




3- 18

(6M)

(7M)

4

Using buckhingam’s π theorem, derive an expression for the drag force R on a partially

submerged body moving with a relative velocity V in a field; the other variables are linear

dimension L, height of surface roughness K, fluid density and acceleration due to gravity g.

(13M) BTL2


R=f(µ,ρ,l, v,g)

The dimensions are

R = MLT-2 (1M) µ = ML-1T-1 (1M)

ρ = ML-3 (1M) l = L

V=LT-1 g = LT-2

Then n=6

m=3

no.of π terms = n-m = 3 π1=la1.Vb1. ρ c1.R (2M)

π2=la2.Vb2. ρ c2. µ (2M) π3=la3.Vb3. ρ c3. g(2M)

solving we get, R= ρl2v2 ᴓ(µ/ρVl, lg/V2) (4M)

5

An object of diameter 900mm is to move in air at 60m/s. Its drag is to be estimated from

tests on a half scale model in water. The drag on the model is 1140N. Estimate the speed of

the model and drag on the full scale object. Assume density of air as 1.2kg/m3, viscosity of




3- 19

air 1.86x10-5 Ns/m2, water as 1.01x10-3 Ns/m2, and density of water as 1000 kg /m3. (13M)


Vp=60m/s, µp=µair=1.86x10-5 Ns/m2, ρp= ρair=1.2kg/m3

(6M)

(7M)

6

Find the discharge Q of a centrifugal pump depends upon the mass density of fluid, speed

of the pump and diameter of the impeller, manometric head and viscosity of the fluid.

(13M) BTL1


Q=f(N,D,g,H,µ,ρ) The dimensions are

D = L N = T-1

P = ML-3 Q = L3T-1

G = L T-2

H = L µ = ML-1T-1(3M)

Then n=7

m=3 no.of π terms = n-m = 4

π1=Da1.Nb1. ρ c1.Q(2M) π2= Da2.Nb2. ρ c2.g(2M)

π3= Da3.Nb3. ρ c3.H(2M) π4= Da4.Nb4. ρ c4. µ(2M)

solving we get, Q= ND3 ᴓ(gH/N2D2, µ/ ρ ND2) (2M)

PART * C




3- 20

1

A pipe of diameter 1m is required to transport an oil specific gravity 0.9 and viscosity 3 x

10-2 poise at the rate of 3000 liters/s. Tests were conducted on a 15 cm diameter pipe using

water at 20oC. Find the velocity and the rate of flow in the model. Viscosity of water at

20oC is 0.01 poise. (15M) (APRIL2014) BTL1


(5M)

(10M)

2

A model of submarine is scaled down to 1/20 of the prototype and is to be tested in a wind

tunnel where free stream pressure is 2 MPa and absolute temperature is 50oC. The speed

of the prototype is 7.72 m/s. Determine the free stream velocity of air and the ratio of the

drags between model and prototype. Assume kinematic viscosity of sea water as 1.4 x 10-6

m2/s and viscosity of air as 0.0184 CP. (15M) (APRIL2012) BTL2


(3M)




3- 21

(12M)

UNIT IV PUMPS

Impact of jets - Euler’s equation - Theory of roto-dynamic machines – various efficiencies–

velocity components at entry and exit of the rotor- velocity triangles - Centrifugal pumps–

working principle - work done by the impeller - performance curves - Reciprocating pump-

working principle – Rotary pumps –classification.

PART * A

Q.No. Questions

1. What is the Slip in a reciprocating pump? (APRIL 2015) BTL1

Slip is the difference between the theoretical discharge and actual discharge of the pump.

Slip= Qth-Qact.

2

What is meant by Priming? (NOVEMBER2016) BTL1

The delivery valve is closed and the suction pipe, casing and portion of the delivery pipe up to

delivery valve are completely filled with the liquid so that no air pocket is left. This is called as

priming.

3

What are the main parts of a reciprocating pump? BTL1

1. A cylinder with a piston, Piston rod, connecting rod and a crank.

2. Suction pipe, Delivery pipe.

3. Suction valve and

4. Delivery valve.

4

How will you classify the reciprocating pumps? (APRIL 2010) BTL1

The reciprocating pump may be classified as,

1. According to the water in contact with one side or both sides of the piston.

2. According to the number of cylinders provided.

Classification according to the contact of water is

(1) Single acting (2) Double acting.




3- 22

According to the number of cylinders provided they are classified as,

(1) Single Cylinder pump.

(2) Double cylinder pump.

(3) Triple cylinder pump.

5

Define Mechanical efficiency. BTL1

It is defined as the ratio of the power actually delivered by the impeller to the power supplied

to the shaft.

6 Define Overall efficiency. BTL1

It is the ratio of power output of the pump to the power input to the pump.

7

Define Speed ratio and Flow ratio. BTL1

Speed ratio:

It is the ratio of peripheral speed at outlet to the theoretical velocity of jet

corresponding to manometric head.

Flow ratio:

It is the ratio of the velocity of flow at exit to the theoretical velocity of jet

corresponding to manometric head.

8

What are the various types of casing used in centrifugal pumps?

1. Volute casing

2. Vortex casing

3. Volute casing with guide blades.

9

Why negative slip occurs in reciprocating pump? BTL1

If the actual discharge is more than the theoretical discharge the slip of the pump will be

negative. Negative slip occurs only when delivery pipe is short, Suction pipe is long and pump

is running at high speed.

10

What is indicator diagram? BTL1

Indicator diagram is nothing but a graph plotted between the pressure head in the cylinder and

the distance traveled by the piston from inner dead center for one complete revolution of the

crank.

11

What is meant by Cavitation? (NOVEMBER2013) BTL1

It is defined phenomenon of formation of vapor bubbles of a flowing liquid in a region where

the pressure of the liquid falls below its vapor pressure and the sudden collapsing of theses

vapor bubbles in a region of high pressure.

12

Define rotary pumps. BTL1

Rotary pumps resemble like a centrifugal pumps in appearance. But the working method differs.

Uniform discharge and positive displacement can be obtained by using these rotary pumps; It

has the combined advantages of both centrifugal and reciprocating pumps.

13

What is an air vessel? BTL1

An air vessel is a closed chamber containing compressed air in the top portion and liquid at the

bottom of the chamber. At the base of the chamber there is an opening through which the liquid

may flow into the vessel or out of the vessel.

14 What is the purpose of an air vessel fitted in the pump? (NOVEMBER2016) BTL1




3- 23

1. To obtain a continuous supply of liquid at a uniform rate.

2. To save a considerable amount of work in overcoming the frictional resistance in the suction

and delivery pipes, and

3. To run the pump at a high speed without separation.

15

What is the work saved by fitting an air vessel in a single acting, double acting pump?

BTL1

Work saved by fitting air vessels in a single acting pump is 84.87%, In a double acting pump the

work saved is 39.2%.

16

What is meant by Pump? BTL1

It is defined as the hydraulic machine in which converts the mechanical energy into hydraulic

energy, which is mainly in the form of pressure energy.

17

Mention the main components of Centrifugal pump. BTL1

1. Casing

2. Impeller

3. Suction pipe, strainer & Foot valve

4. Delivery pipe & Delivery valve

PART * B

1

A centrifugal pump having outer diameter equal to two times the inner diameter and

running at 1000rpm works against a head of 40m. The velocity of flow through the impeller

is constant and equal to 2.5 m/s. the vanes are set back at angle of 40 at outlet. If the outer

diameter of the impeller is 500mm and width at outlet is 50mm determine (i) vane angle at

the inlet, (ii) manometric efficiency, (iii) work done by impeller on water per second.

(13M) (APRIL2012) BTL2

Answer: Page: 953 – R.K.Bansal

D2= 2 D1, N=2000rpm, Hm=75m, Vf1=Vf2=3m/s, ˉᴓ =30°, D2=0.6m, B2=0.05m, D1=0.3m

𝑈1 = πD1N

60

𝑡𝑎𝑛𝜃 = Vf1

𝑢𝑙

𝑈2 = πD2N

60


(u2 − Vw2)

Q=πD2B2Vf2

𝑊𝐷 = ρQ

(u2. Vw2)

𝐻 = gHm

(u2. Vw2)

U1= 18.8495m/s (2M)




3- 24

θ= 9.0430°

U2= 37.6991m/s (2M)

Vw2=32.5029 m/s (2M)

Discharge,Q=0.2827m3/s (2M)

Work Done =3464008 W (2M)

Efficiency η=60.06% (3M)

2

A single acting reciprocating pump running at 50rpm delivers 0.01 m3/s of water. The

diameter of the piston is 200mm and stroke length 400mm. determine (i) theoretical

discharge (ii) coefficient of discharge (iii) slip and percentage of the pump.



N=50rpm,

Qact=0.01m3/s,

D=0.2m,

L=0.4m

A=πD2/4,

𝑄𝑡ℎ = ALN/60

𝐶𝑑 =𝑄𝑎𝑐𝑡

𝑄𝑡ℎ

𝑠𝑙𝑖𝑝 = 𝑄𝑡ℎ − 𝑄𝑎𝑐𝑡(1M)

A=0.03141m2 (2M)

Theoretical discharge, 𝑄𝑡ℎ = 0.01047m3/s (3M)

Coefficient of discharge, Cd=0.955, Slip = 4.1x10-4 (3M)

%slip = 4.489% (4M)

3

Explain the construction and working of Reciprocating pumps with a neat sketch.(13M)

(APRIL2015) BTL2


Principle: Reciprocating pump operates on the principle of pushing of liquid by a piston that

executes a reciprocating motion in a closed fitting cylinder. (2M)

Components of reciprocating pumps:-

(i) Piston or plunger: – a piston or plunger that reciprocates in a closely fitted cylinder.

(ii) Crank and Connecting rod: – crank and connecting rod mechanism 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.

(iii)Suction pipe: – one end of suction pipe remains dip in the liquid and other end attached to

the inlet of the cylinder.

(iv) Delivery pipe: – one end of delivery pipe attached with delivery part and other end at

discharge point.

(v) Suction and Delivery value: – suction and delivery values are provided at the suction end

and delivery end respectively. These values are non-return values. (4M)




3- 25

(3M) WORKING OF RECIPROCATING PUMP

(i) Operation of reciprocating motion is done by the power source (i.e. electric motor or i.c

engine, etc).

(ii) 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 connecting

rod and piston).

(iii)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 (4M)

4

A centrifugal pump has an impeller 500mm diameter running at 400rpm. The discharge at

the inlet is entirely radial. The velocity of the flow at inlet is 1m/s. the vanes are curved

backwards at outlet at 30 to the wheel tangent. If the discharge of the pump is 0.14m3/s.

calculate the impeller power and torque on the shaft.



D2=0.5m,

R2=0.25m,

N=400rpm,

α=90°,

Vf2=1m/s,

ᴓ=30°,

Q=0.14m3/s

𝑈2 = πD2N

60




3- 26


(u2 − Vw2)

𝑇 = ρQ

(u2. Vw2)

𝑃 = 2πNT

60

U2=10.4719 m/s (3M)

Vw2=8.7399 m/s (3M)

Torque, T=305.8965 N-m (3M)

Power, P=12.8133 kW (4M)

5

Explain the performance characteristics curves of centrifugal pump (NOVEMBER2015)

(13M) BTL2


The curves which are plotted from the series of a number of tests on the centrifugal pump are known as characteristics or performance curves. It refers to the graphical representation of variation in head, power and efficiency of pump drawn to a common base line of flow rate. (5M)

The following four types are the characteristic curves used for centrifugal pumps.

(i) Main characteristic curves,

(2M) (ii) Operating characteristic curves,

(2M) (iii) Constant efficiency curves




3- 27

(2M) (iv) Constant head and constant discharge curves.

(2M)

6

With neat sketch describe about the various components of centrifugal pump and its

working principle. (13M) (NOVEMBER 2016) BTL2


Working principle of a centrifugal pump remains the same, based on the impeller and suction.

Working of the centrifugal motor is pretty simple. The rotation of the impeller creates a

very low pressure at its inlet, called the eye of the impeller. The fluid gets carried along

the impeller towards the casing. (3M)

Some of the most common components found in centrifugal pumps are:

(i) Pump main housing.

(ii) Impeller.

(iii) Impeller seal.

(iv) Impeller bearings.

(v) Motor.

(vi) Coupling.

(vii) Shaft-drive. (2M)




3- 28

(5M)

Vane pumps use vanes (flat blades) that slide in and out as they rotate, moving the fluid

from the inlet to the outlet and flinging it out at speed. Impeller pumps use a wheel with

curved blades called an impeller, which is a bit like a multi-bladed propeller fitted snugly

in the middle of a closed pipe. (3M)

PART * C

1

The internal and external diameters of the impellers of a centrifugal pump are 300mm and 600mm resp. The pump is running at 1000rpm. The vane angles of the impeller at inlet and outlet are 20 and 30 resp. the water enters the impeller radially and velocity of flow is constant. Determine the work done by the impeller per unit weight of water. (15M) (NOVEMBER2012) BTL2


D1=0.2m,

D2=0.4m,

N=1200rpm,

θ=20°,

ᴓ=30°,

α=90°,

Vw1=0, Vf1=Vf2.

𝑈1 = πD1N

60


𝑢𝑙

𝑈2 = πD2N

60




3- 29


(u2 − Vw2)

𝑊𝐷 = ρg

(u2.Vw2) (2M)

U1=12.5663 m/s (2M)

U2=25.1327m/s (2M)

Vf1=4.5737m/s (2M)

Vf2=4.5737m/s (2M)

Vw2 =17.6444m/s (2M)

Work Done = 45.9839 W (3M)

2

Discuss the working of lobe pumps with a neat sketch. (15M) (NOVEMBER 2012) BTL2


Lobe Pump:

(i) Lobe pumps are similar to external gear pumps in operation in that fluid flows around

the interior of the casing.

(ii) As the lobes come out of mesh, they create expanding volume on the inlet side of the

pump. Liquid flows into the cavity and is trapped by the lobes as they rotate.

(iii) Lobe pumps are used in a variety of industries including pulp and paper, chemical,

food, beverage, pharmaceutical, and biotechnology. Rotary pumps can handle solids

(e.g., cherries and olives), slurries, pastes, and a variety of liquids. (8M)

(7M)

3 The impeller of a centrifugal pump having external and internal diameter 450mm and

225mm respectively. Width at outlet is 45mm and running at 1250rpm, works against a




3- 30

head of 50m. The velocity of flow through the impeller is constant and equal to 2.8 m/s. The

vanes are set back at an angle of 40° at the outlet. Determine (i) inlet vane angle, (ii) work

done by the impeller on water per second, (iii) manometric efficiency. (15M) (NOVEMBER

2014) BTL2


D1=0.45m,

D2=.225m,

N=1250rpm,

θ=40°,

Vf1= Vf2= 2.8m/s (2M)

𝑈1 = πD1N

60


𝑢𝑙

𝑈2 = πD2N

60

𝑊𝐷 = ρQ

(u2.Vw2) (2M)

U1= 8.5663 m/s

U2=10.7m/s (2M)

Vw2 =17.6444m/s (2M)

Work Done = 136.9839 W (3M)

Manometric efficiency = 63.7% (4M)

UNIT V TURBINES

Classification of turbines – heads and efficiencies – velocity triangles. Axial, radial and mixed flow

turbines. Pelton wheel, Francis turbine and Kaplan turbines- working principles - work done by

water on the runner – draft tube. Specific speed - unit quantities – performance curves for

turbines – governing of turbines.

PART * A

Q.No. Questions

1.

Give an example for low head, medium head and high head turbine. BTL1

Low head turbine – Kaplan turbine

Medium head turbine – Modern Francis turbine

High head turbine – Pelton wheel.

2

What is impulse turbine? Give example. BTL2

In impulse turbine all the energy converted into kinetic energy. From these the turbine will

develop high kinetic energy power. This turbine is called impulse turbine.

Example: Pelton turbine

3 What is reaction turbine? Give example. BTL2




3- 31

In a reaction turbine, the runner utilizes both potential and kinetic energies. Here portion of

potential energy is converted into kinetic energy before entering into the turbine.

Example: Francis and Kaplan turbine.

4

What is axial flow turbine? BTL2

The water flows parallel to the axis of the turbine shaft is called axial flow turbine

Example: Kaplan turbine.

5

What is mixed flow turbine? BTL2

In mixed flow water enters the blades radially and comes out axially, parallel to the turbine shaft.

Example: Modern Francis turbine.

6

What is the function of spear and nozzle? BTL2

The nozzle is used to convert whole hydraulic energy into kinetic energy. Thus the nozzle

delivers high speed jet. To regulate the water flow through the nozzle and to obtain a good jet of

water spear or nozzle is arranged.

7

Define gross head and net or effective head. BTL1

Gross Head:

The gross head is the difference between the water level at the reservoir and the level at the

tailstock.

Effective Head:

The head available at the inlet of the turbine.

8 What is hydraulic efficiency? BTL1

It is defined as the ratio of power developed by the runner to the power supplied by the water jet.

9

Define mechanical efficiency. BTL1

It is defined as the ratio of power available at the turbine shaft to the power developed by the

turbine runner.

10

What is volumetric efficiency? BTL1

It is defined as the volume of water actually striking the buckets to the total water supplied by the

jet.

11

Define overall efficiency. BTL1

It is defined as the ratio of power available at the turbine shaft to the power available from the

water jet.

12

Differentiate between Kaplan turbine and propeller turbine.BTL1

The difference between the Propeller and Kaplan turbines is that the Propeller turbine has fixed

runner blades while the Kaplan turbine has adjustable runner blades.

It is a pure axial flow turbine uses basic aerofoil theory.

13

List down the main components of pelton wheel. BTL1

➢ Nozzle and flow regulating arrangements

➢ Runner and buckets




3- 32

➢ Casing, and

➢ Breaking jet.

14

What do you mean by Net Positive Suction Head (NPSH)? (NOVEMBER2015) BTL1

NPSH can be defined as two parts:

NPSH Available (NPSHA): The absolute pressure at the suction port of the pump.

NPSH Require (NPSHR): The minimum pressure required at the suction port of the pump to

keep the pump from cavitating.

15

Define negative slip in reciprocating pump. BTL1

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

Negative slip occurs when delivery pipe is short suction pipe is long and pump is running at

high speed.

PART * B

1

A Pelton wheel, working under a head of 500 m develops 13 MW when running at a speed

of 430 rpm. If the efficiency of the wheel is 85%, determine the rate of flow through the

turbine, the diameter of the wheel and the diameter of the nozzle. Take speed ratio as 0.46

and coefficient of velocity for the nozzle as 0.98. (13M) (APRIL2014) BTL2


H=500m,

P=13MW,

N=430rpm

η= 0.85

Kv=0.98

N=0.46

𝑢 = πDN

60

η = p

ρgQH

𝑢 = 𝐾𝑣. √2gH (1M)

U = 45.56 m/s (4M)

Rate of flow =3.11m3/s (4M)

Diameter = 2.02 m (4M)

2

A Pelton wheel works under a gross head of 510 m. One third of gross head is lost in

friction in the penstock. The rate of flow through the nozzle is 2.2 m3/sec. The angel of

deflection of jet is 165°. Find the (i) power given by water to the runner (ii) hydraulic

efficiency of Pelton wheel. Take CV = 1.0 and speed ratio = 0.45. (13M)(APRIL2017) BTL1





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Vw1 = 𝑉1 = 𝐶𝑣. √2gH

𝑢 = 𝐾𝑢. √2gH

Vr1 = 𝑉1 − 𝑢

𝑊𝐷 = ρQ

(Vw1 + Vw2)𝑢

η = 2(Vw1+Vw2)𝑢

v2 (1M)

Vw1= 81.67 m/s (2M)

U = 36.75 m/s (2M)

Vr1= 44.92 m/s (2M)

Work done/s = 713986.3 Nm/s (3M)

η= 97.3% (3M)

3

A 137 mm diameter jet of water issuing from a nozzle impinges on the buckets of a Pelton

wheel and the jet is deflected through an angle of 165 by the buckets. The head available

at the nozzle is 400m. Find: (a) Force exerted on the buckets and (b) Power developed.

Assume Cv as 0.97, speed ratio as 0.46 and reduction in velocity while passing through the

buckets as 15%. (13M)(APRIL2010) BTL1


Vw1 = 𝑉1 = 𝐶𝑣. √2gH

𝑢1 = 𝐾𝑢. √2gH

Vr1 = 𝑉1 − 𝑢1

𝑃 = Fx . u

1000

Fx = ρav1(Vw1 − Vw2)𝑢 (1M)

Vw1= 85.93 m/s (2M)

U1 = 40.75 m/s (2M)

Vr1= 45.18 m/s (2M)

Vr2= 38.40 m/s (2M)

Fx = 104206 N (2M)

P = 4246.4 kW (2M)

4

A Pelton turbine is required to develop 9000 KW when working under a head of 300 m the

impeller may rotate at 500 rpm. Assuming a jet ratio of 10 and an overall efficiency of 85%

calculate (i) Quantity of water required, (ii) Diameter of the wheel, (iii) No of jets, (iv) No

and size of the bucket vanes on the runner. (13M)(APRIL2011) BTL1


𝑉1 = 𝐶𝑣. √2gH





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𝑢 = πDN

60

Vr1 = 𝑉1 − 𝑢

η = 𝑃

ρgQH

m = 𝐷

d (1M)

V1= 75.56 m/s (2M)

U = 34.52 m/s (2M)


Discharge=3.59 m3/s (2M)

No. of jets = 3.48 (2M)

No. of bucket = 15+0.5 m = 20

Width = 0.659 mm

Depth = 0.158 mm (2M)

5

A pelton wheel turbine develops 3000kW power under a head of 300m. The overall

efficiency of the turbine is 83%. If the speed ratio = 0.46, Cv = 0.98 and specific speed is

16.5, then find diameter of the turbine and diameter of the jet. (13M)(APRIL2016) BTL1


𝑉1 = 𝐶𝑣. √2gH


𝑢 = πDN

60

Po = 𝑃

ρgQH1000

Ns = 𝑁

𝐻24

V1= 75.12 m/s (2M)

U = 34.95 m/s (2M)


Discharge=1.23 m3/s (3M)

N = 375 rpm (4M)

PART * C

1

A hub diameter of a Kaplan turbine, working under a head of 12m, is 0.35 times the

diameter of the runner. The turbine is running at 100rpm. If the vane angle of the runner

at outlet is 15deg. And flow ratio 0.6, find (i) diameter of the runner, (ii) diameter of the

boss, and (iii) Discharge through the runner. Take the velocity of whirl at outlet as zero.

(15M)(APRIL2012) BTL1





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H =12 m

N = 100 rpm

𝐹𝑙𝑜𝑤 𝑟𝑎𝑡𝑖𝑜 = 𝑉𝑓1/√2gH = 0.6

tan 𝜙 = 𝑉𝑓1

𝑢2

U2 = 34.33 m/s (3M)

Diameter, Db = 2.3 m (4M)

Flow velocity,Vf1= 9.2 m/s (4M)

Discharge Q = 271.7 m3/s (4M)

2

A reaction turbine works at 450 r.p.m. under a head of 120 m. Its diameter at inlet is 1.2 m

and the flow area is 0.4 m2. The angles made by absolute and relative velocities at inlet are

20° and 60° respectively with the tangential velocity. Determine: (i)the volume rate of flow,

(ii) the power developed, and (iii) the hydraulic efficiency. (15M)(APRIL2016) BTL2


tan 𝜃 = 𝑉𝑓1

𝑉𝑤1 − 𝑢1

tan 𝜃 = 𝑉𝑓1

𝑉𝑤1

𝑢1 = πDN

60

𝑄 = 𝜋𝐷𝐵. 𝑉𝑓1

W = ρQ(Vw1. u1)

ηH = Vw1.u1

gH (1M)

u1 = = 28.27 m/s (2M)

Vf1 = 0.364 Vw1 (3M)

Vw1 = 35.79 m/s (3M)

Discharge = 5.211 m3/s (3M)

Hydraulic efficiency ,ɳ = 85.95% (3M)




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