DEPARTMENT OF MECHANICAL ENGINEERING - Vel Tech HighTech Fluid... · Cycle Test-IV (Unit 4) 36 V Classification of turbines TB1, RB4 Class room lecture - Black board C203.5 37 V heads

VELTECH HIGHTECH Dr. RANGARAJAN Dr. SAKUNTHALA

ENGINEERING COLLEGE

DEPARTMENT OF MECHANICAL ENGINEERING

Approved by AICTE, New Delhi & Govt. of Tamil Nadu &

Affiliated to Anna University

REGULATION 2017

CE8394 - FLUID MECHANICS AND MACHINERY

HANDOUTS

Department of Mechanical Engineering,

VelTech HighTech Dr.Rangarajan Dr.Sakunthala Engineering College,

Avadi, Chennai – 600062.

VEL TECH HIGH TECH Dr. RANGARAJAN Dr. SAKUNTHALA ENGINEERING COLLEGE


COURSE DETAILS

Subject Name: Fluid Mechanics and Machinery Course Code: CE8394

Class: Mech / II Year / III Sem A-section Date of Compilation: 01.06.18

COURSE OBJECTIVES:

• The properties of fluids and concept of control volume are studied

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

• To understand the importance of dimensional analysis.

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

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

COURSE OUTCOMES:

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

CO No. Course Outcomes Knowledge

Level

CO203.1 Apply mathematical knowledge to predict the properties and characteristics of a

fluid. k1, k2

CO203.2 Can analyse and calculate major and minor losses associated with pipe flow in

piping networks. k1, k2

CO203.3 Can mathematically predict the nature of physical quantities. k1, k2

CO203.4 Can critically analyse the performance of different types of pumps. k1, k2, k3

CO203.5 Can critically analyse the performance of different types of turbines. k1, k2, k3

Mapping of Course Outcomes with Program Outcomes

CO PO1 PO 2 PO 3 PO 4 PO 5 PO 6 PO 7 PO 8 PO 9 PO 10 PO 11 PO 12

CO203.1 3 3 3 - - 3 - - - - - 3

CO203.2 3 3 3 - - 3 - - - - - 3

CO203.3 3 3 3 - - - 3 - - - - 3

CO203.4 3 3 3 - 3 - 3 - - - 3 3

CO203.5 3 3 3 - 3 - 3 - - - 3 3

C.No PO1 PO 2 PO 3 PO 4 PO 5 PO 6 PO 7 PO 8 PO 9 PO 10 PO 11 PO 12

CO 3 3 3 - 2 2 3 - - - 2 3

Mapping Relevancy

3 – Substantial (Highly relevant)

2 – Moderate (Medium)

1 – Slight (Low)

COURSE DELIVERY METHODS

• Class room lecture - Black board

• PPTs, Videos

• Lab Demonstrations

• Activities like In Plant Training, Industrial Visit and Guest Lecture

Assessment methods

• Continuous Internal Assessment

• Assignments

• Seminars

UNIT I FLUID PROPERTIES AND FLOW CHARACTERISTICS 8

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 8

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.

UNIT III DIMENSIONAL ANALYSIS 9

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

similitude - Dimensionless parameters- application of dimensionless parameters – Model

analysis.

UNIT IV PUMPS 10

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 10

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: 45 PERIODS

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

COURSE DELIVERY PLAN

S.No Date Unit Topic

Text/

Reference

Books

Teaching

Methodology

Course

Outcome

1 I Units and dimensions

TB1, RB4

Class room

lecture - Black

board

CO.203.1

2 I

Properties of fluids- mass density, specific

weight, specific volume, specific gravity,

viscosity, compressibility, vapor pressure,

surface tension and capillarity

CO.203.1

3 I Flow characteristics CO.203.1

Slip Test 1

4 I concept of control volume Class room

lecture - Black

board

CO.203.1

5 I application of continuity equation CO.203.1

6 I application of energy equation PPT CO.203.1

Slip Test 2

7 I application of momentum equation.

Class room

lecture - Black

board

CO.203.1

8 I Problems in unit 1 CO.203.1

Cycle Test-I (Unit 1)

9 II Hydraulic and energy gradient

TB1, RB4

Class room

lecture - Black

board

CO.203.2

10 II Laminar flow through circular conduits and

circular annuli Boundary layer concepts

CO.203.2

11 II types of boundary layer thickness CO.203.2

Slip Test 3

12 II Darcy Weisbach equation –friction factor Class room

lecture - Black

board PPT

CO.203.2

13 II Moody diagram CO.203.2

14 II commercial pipes- minor losses CO.203.2

Slip Test 4

15 II Flow through pipes in series. Class room

lecture - Black

board

CO.203.2

16 II Flow through pipes in parallel. CO.203.2

Cycle Test-II (Unit 2)

17 III Need for dimensional analysis

TB1,RB3

Class room

lecture - Black

board

C203.3

18 III Methods of dimensional analysis C203.3

19 III Similitude C303.3

Slip Test 5

20 III types of similitude Class room

lecture - Black

board

C203.3

21 III Problems

C203.3

22 III C203.3

Slip Test 6

23 III Dimensionless parameters

PPT & Videos

C203.3

24 III Application of dimensionless parameters C203.3

25 III Model analysis C203.3

Cycle Test-III (Unit 1,2&3)

26 IV Impact of jets TB1,RB3 Class room C203.4

27 IV Euler‘s equation lecture - Black

board

C203.4

28 IV Theory of roto-dynamic machines C203.4

Slip Test 7

29 IV various efficiencies– velocity components at

entry and exit of the rotor Class room

lecture - Black

board

C203.4

30 IV velocity triangles

C203.4

31 IV Centrifugal pumps– working principle C203.4

Slip Test 8

32 IV Work done by the impeller - performance

curves PPT & Videos

C203.4

33 IV Reciprocating pump- working principle – C203.4

34 IV Rotary pumps

Class room

lecture - Black

board

C203.4

35 IV classification PPT & Videos C203.4

Cycle Test-IV (Unit 4)

36 V Classification of turbines

TB1, RB4

Class room

lecture - Black

board

C203.5

37 V heads and efficiencies – velocity triangles. C203.5

38 V Axial, radial and mixed flow turbines. C203.5

Slip Test 9

39 V Pelton wheel, Francis turbine and Kaplan

turbines Class room

lecture - Black

board

C203.5

40 V

Working principles - work done by water on

the runner

C203.5

41 V Draft tube. Specific speed - C203.5

Slip Test 10

42 V Unit quantities

PPT & Lab

Demo

C203.5

43 V Performance curves for turbines governing

of turbines

Class room

lecture - Black

board

C203.5

44 V Problems

Class room

lecture - Black

board

C203.5

45 V C203.5

Cycle Test-V (Unit 1,2,3,4&5) CO203.

1,2,3,4,5

VELTECH HIGHTECH Dr.RANGARAJAN Dr.SAKUNTHALA ENGINEERING COLLEGE


CURRICULUM MAPPING

NAME OF THE PROGRAMME: B.E., MECHANICAL ENGINEERING

NAME OF THE COURSE: FLUID MECHANICS AND MACHINERY

COURSE OUTCOMES:

CO No. Course Outcomes

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

CO2 Can analyse and calculate major and minor losses associated with pipe flow in piping

networks.

CO3 Can mathematically predict the nature of physical quantities.

CO4 Can critically analyse the performance of different types of pumps.

CO5 Can critically analyse the performance of different types of turbines.

Program Educational Objectives (PEOs): Our Graduates will,

1. Exhibit technical prowess to solve problems and able to work in multidisciplinary teams lead to

deliver significant societal benefits.

2. Demonstrate professional engineering competence via promotions / positions of increasing

responsibility.

3. Pursue research activities and publish papers, apply for patents, deliver effective conference

presentations.

4. Demonstrate a commitment to the community and the profession through involvement with

community / professional organizations / make contributions towards society’s greater

prosperity.

5. Demonstrate an understanding of the need for life-long learning via progress.

Program Outcomes (POs): Engineering Graduates will be able to:

PO1: Engineering knowledge: Apply the knowledge of mathematics, science, engineering

fundamentals, and an engineering specialization to the solution of complex engineering problems.

PO2: Problem analysis: Identify, formulate, review research literature, and analyze complex

engineering problems reaching substantiated conclusions using first principles of mathematics, natural

sciences, and engineering sciences.

PO3: Design/development of solutions: Design solutions for complex engineering problems and design

system components or processes that meet the specified needs with appropriate consideration

for the public health and safety, and the cultural, societal, and environmental considerations.

PO4: Conduct investigations of complex problems: Use research-based knowledge and research

methods including design of experiments, analysis and interpretation of data, and synthesis of the

information to provide valid conclusions.

PO5: Modern tool usage: Create, select, and apply appropriate techniques, resources, and modern

engineering and IT tools including prediction and modelling to complex engineering activities with an

understanding of the limitations.

PO6: The engineer and society: Apply reasoning informed by the contextual knowledge to assess

societal, health, safety, legal and cultural issues and the consequent responsibilities relevant to the

professional engineering practice.

PO7: Environment and sustainability: Understand the impact of the professional engineering solutions

in societal and environmental contexts, and demonstrate the knowledge of, and need for sustainable

development.

PO8: Ethics: Apply ethical principles and commit to professional ethics and responsibilities and norms

of the engineering practice.

PO9: Individual and team work: Function effectively as an individual, and as a member or leader in

diverse teams, and in multidisciplinary settings.

PO10: Communication: Communicate effectively on complex engineering activities with the

engineering community and with society at large, such as, being able to comprehend and write effective

reports and design documentation, make effective presentations, and give and receive clear instructions.

PO11: Project management and finance: Demonstrate knowledge and understanding of the

engineering and management principles and apply these to one’s own work, as a member and leader in a

team, to manage projects and in multidisciplinary environments.

PO12: Life-long learning: Recognize the need for, and have the preparation and ability to engage in

independent and life-long learning in the broadest context of technological change.

The outcome of this course facilitates to attain the following POs:

CO PO1 PO2 PO3 PO4 PO5 PO6 PO7 PO8 PO9 PO10 PO11 PO12

CO 3 3 3 - 2 2 3 - - - 2 3

Note: 1: Slight (Low); 2: Moderate (Medium); 3: Substantial (High); If there is no correlation, put “-”

1.Explain Euler’s equation and deduce Bernoulli’s theorem

Bernoulli’s equation

In a steady, ideal flow of an incompressible fluid, the total energy at any point is constant. The

total energy consist of pressure energy, kinetic energy and potential energy

Consider a cylindrical element of length ds and cross sectional area dA, in which steady fluid is

flowing

Pressure force at A =P

Pressure force at B =P+dP

The forces acting on the fluid are

1. pressure force

2. Gravity force:

Pressure force = Pressure Area

Pressure force at A,

FA =P dA

Pressure force at B

Gravity force:

Constant z 2g

V

ρg

P 2

=++

VolumeDensity fluid theof Mass =

LengthAreaDensity =

dsdA ρm =

gravity todueon acceleratimass fluid theofWieght =

g mW =

gdsdA ρ =

dsdA g ρW =

PdA FA =

( )dA dPP FB +=

dA dPdA PFB +=

Resultant force in the direction of fluid flow

According to Newton’s second law of motion

Equating equation (1)& (2)

Euler’s equation

WcosθF g =

ds

dZdsdA g ρFg =

dZdA g ρF forceGravity g =

BgA F-F-F =F

( )dAdz g ρ dPdA -PdA-PdA +=

dAdz g ρ dPdA -PdA-PdA +=

..(1).................... dAdz g ρ --dPdA F =

a m =F

ρdAdS m =

dt

dV a =

dt

ds

ds

dV a =

dt

ds V =

Vds

dV a =

dS

dVV a =

ds

dVV ρdAds =F

.(2)....................VdV.......ρdA =F

..(1).................... ρdAgdz --dPdA F =

ρdAgdz --dPdA VdVρdA =

ρdA

gdz - ρ

dP- VdV =

0 VdV gdz ρ

dP=++

0 VdV gdz ρ

dP=++

ds

dZcosθ =

Assumption

➢ The flow is ideal ie viscosity is zero

➢ The flow is steady

➢ The flow is incompressible

➢ The flow is irrotational

3. Derive Hagen poiseuilles.

or

Explain flow of viscous fluid through circular pipe

Consider a horizontal pipe of radius R.

The viscous fluid is flowing through the pipe.

Consider a fluid element of radius r.

Let length of the fluid element be dx.

Let



0 VdV dzg dPρ

1=++

Constant 2

V gz

ρ

P 2

=++

g

Constant 2g

V z

ρg

P 2

=++

Constantz 2g

V

ρg

P 2

=++


1. pressure force

2. Gravity force:



Pressure force at B

Shear force Fs = Shear stress Area

Fs

Fs

(i) Velocity Distribution.

According to Newton’s law of viscosity

y is measured from the pipe wall

Differentiating y with respect to r

πrP F 2

A =

P r πF 2

A =

( ) 2

B r π dPP F +=

dPr πP r πF 22

B +=

dxr π2 τ=

dx τr π2 =

0F-FF SBA =−

( ) 0dx τr π2-r πdPdxPP r π 22 =+−

0dx τr 2π-dPπr-PπrPπr 222 =−

0dx τr 2π-dPdxπr- 2 =

dPr π-dx τr π2 2=

dx r π2

dPr π τ

2

−=

)1.........(..............................dx

dP

2

r -τ =

dy

duμ τ =

rRy −=

10dr

dy−=

drdy −=

From eqn (1)

Boundary condition

At r = R, u = 0

sub C in eqn (2)

Maximum velocity occurs when r =0

Average Velocity

dy

duμ τ =

dr

duμ τ −=

dx

dP

2

r -τ =

dx

dP

2

r

dr

duμ - −=

dx

dP

2

r

dr

duμ - −=

dx

dP

μ 2

drr du =

= drr dx

dP

μ 2

1 du

C2

r

dx

dP

μ 2

1 u

2

+=

2)C........(rdx

dP

μ 4

1 u 2 +=

CRdx

dP

μ 4

1 0 2 +=

2Rdx

dP

μ 4

1- C =

22 R dx

dP

μ 4

1-r

dx

dP

μ 4

1 u =

( ) r-Rdx

dP

μ 4

1 u 22−=

2max R

dx

dP

μ 4

1-U =

Consider a circular ring element of radius r and thickness dr in which is flow take place.

Discharge = area of ring element velocity

Average velocity

Ratio of maximum velocity to Average velocity

udrr 2π dQ =

( )

−= r-R

dx

dP

μ 4

1 u 22

( )

−= r-R

dx

dP

μ 4

1dr r 2π dQ 22

( ) drr r-R dx

dP

μ 2

π dQ 22−=

( ) dr r-Rr dx

dP

μ 2 dQ 32−=

( ) −=

R

0

32 dr r-Rr dx

dP

μ 2 dQ

R

0

422

4

r-

2

Rr

dx

dP

μ 2 Q

−=

−=

4

R-

2

RR

dx

dP

μ 2

π

422

−=

4

R-

2

R

dx

dP

μ 2

π

44

−=

8

2R -4R

dx

dP

μ 2

π

44

−=

8

2R

dx

dP

μ 2

π

4

−=

8

2R

dx

dP

μ 2

π Q

4

2

2

4

Rdx

dP

μ 8

1 u

R π

Rdx

dP

μ 8

π

A

Q u

−=

−

=

=

Ratio of maximum velocity to average velocity = 2.0.

Friction loss

sub P1-P2 in equation (2)

………………………………………………….(3)

Equation (3) is called Hagen Poiseuille Formula.

4.Derive Darcy-Weisbach equation.

or

Expression for loss of head due to Friction in pipes

Let




1. Pressure force

2. Friction force:

Consider a steady fluid flowing through a horizontal pipe

2max R

dx

dP

μ 4

1-U =

2

2

R dx

dP

μ 8

1-

R dx

dP

μ 4

1-

u

Umax=

2u

Umax=

ρg

P-Ph 21

f =

ρg

PPh 21

f

−=

2

_

fD g ρ

L uμ 32h =

2

_

21

D g ρ

L uμ 32

ρg

PP=

−

f2

21 hρgD

L uμ 32

ρg

P-P==



Pressure force at B

d4

πPF 2

1A =

1

2

A Pd4

πF =

d4

πPF 2

2B =

2

2

B Pd4

πF =

2 velocityareacontact f = forceFriction

2VL d πf F = 2LV d fπF =

0FFF BA =−−

0LV d f πPd4

πPd

4

π 2

2

2

1

2 =−−

( ) 0LV d f πP-Pd4

π 2

21

2 =−

( ) 2

21

2 LV d f πP-Pd4

π=

2

2

21πd

LV d f π4P-P

=

d

LV f4P-P

2

21

=

lossFriction Z2g

V

ρg

P Z

2g

V

ρg

P2

2

221

2

11 +++=++

f2

2

221

2

11 hZ2g

V

ρg

P Z

2g

V

ρg

P+++=++

21 Z Z =

21 V V =

f21 h

ρg

P

ρg

P+=

Sub P1-P2 in equation (1)

4. Explain Boundary Layer thickness. Displacement thickness momentum thickness &

Energy thickness.

Boundary Layer Thickness (): It is defined as the distance from the boundary of the solid

body measured in the y – direction to the point, where the velocity of the fluid is approximately

equal to 0.99 times the free steam (U) velocity of the fluid.

Expression for *:

Consider a fluid moving with a free stream velocity(U) approaching a flate plate which is at rest flow of

a fluid having free stream velocity equal to U

Consider section 1-1 at a distance at a distance x from the leading edge. let the thickness of

boundary layer be

The velocity of fluid on the flate plate is zero .The velocity of fluid on the boundary is

U.. Thus velocity varies from zero to U ,

Displacement Thickness (*)

It is defined as the distance, measured perpendicular to the boundary of the solid

body, by which the boundary should be displaced to compensate for the reduction in flow rate on account

of boundary layer formation. It is denoted by *.

Expression for *:

Consider an elemental strip of thickness dy. The elemental strip is distance of y from the

plate. At this strip, the velocity of fluid is u

f21 h

ρg

PP=

−

)1.......(....................ρg

PPh 21

f

−=

d ρg

LV f4 h

2

f

=

d g 2

LV 4f h

2

f

ρ

f

2

f =

=

thicknessbreadth strip elemental of Area =

In the absence of plate, then the fluid is moving free stream velocity (U).

Let the plate is displaced by a distance * .At this distance *,velocity of fluid is U.

Equating (1)& ( 2)

Momentum Thickness ():

Momentum thickness is defined as the distance, measured perpendicular to the boundary of the solid

body, by which the boundary should be displaced to compensate for the reduction in momentum of the

flowing fluid on account of boundary layer formation. It is denoted by .

Expression for :

Consider an elemental strip of thickness dy. The elemental strip is distance of y from the plate. At this

strip, the velocity of fluid is u

bdy =

velocityarea density strip elemental through flowing fluid of rate flow Mass =

ρbdyu=

velociyareadensity plate of absencein strip elemental through fluid of rate flow Mass =

ρbdyU=

ρbdyu-ρbdyU rate flow massin Reduction =

( )u-Uρbdy=

( ) (1)........................................dyu-Uρb rate flow massin Reduction Total 0

=

δb Area =

UVelocity =

velocity areadensity δ through rate flow massin Reduction =

..(2)..............................U.........ρbδ =

( ) dyu-UρbρUbδ0

=

ρUb

dyU

u-Uδ

0

=

dyU

u1δ

0

−=


bdy =

Let the plate is displaced by a distance θ .At this distance θ, velocity of fluid is U.

Equating (3)& ( 4)


ρbdyu=

Velocity rate flow mass fluid thisof secondper Momentum =

uρbdyu =

2ρbdyu =

Velocity rate flow masslayer boundary of absencein secondper Momentum =

ρbdyuU =

2ρbdyu-ρbdyuU secondper momentumin Reduction =

( )2u-uUρbdy=

( ) (3)........................................δ

0dy2u-uUρb secondper momentumin Reduction Total =

θb Area =

UVelocity =

velocity areadensity rate flow mass =

ρbθU=

velocity flowrate mass θ distance a through secondper momentumin Reduction =

UρbθU =

.(4)........................................ρbθU2=

( ) dyu-uUρbρbθU

δ

0

22

=

ρbU 2

dyU

u-uUθ

δ

0

2

2

=

dyU

u1

U

uθ

0

−=

Energy Thickness (**):

Energy thickness is defined as the distance, measured perpendicular to the boundary of the solid

body, by which the boundary should be displaced to compensate for the reduction in kinetic of the

flowing fluid on account of boundary layer formation. It is denoted by **.

Expression for **:

Consider an elemental strip of thickness dy. The elemental strip is distance of y from the plate. At

this strip, the velocity of fluid is u

Let the plate is displaced by a distance ** .At this distance **, velocity of fluid is U.


bdy =


ρbdyu=

2Velocity rate flow mass2

1 fluid thisofenergy Kinetic =

2uρbdyu2

1 =

bdy3ρu2

1 =


1layer boundary of absencein fluid thisofenergy Kinetic =

2Uρbdyu2

1 =

2ρbdyuU2

1=

3ρbdyu2

1-2ρbdyuU

2

1energy Kineticin Reduction =

= 2u-2Uρbudy

2

1

=

δ

0dy2u-2Uρbu

2

1energy Kineticin Reduction Total

(5)........................................δ

0dy2u-2Uuρb

2

1

=

= δb Area

Equating (5)& ( 6)

4.a.1) .The internal and external diameter of the impeller of a centrifugal pump are 300

mm and 600 mm respectively. The pump is running at 1000rpm. The vane angles of the

impeller at inlet and outlet are 20 and 30 respectively. The water enters the impeller

radially and velocity of flow is constant. Determine the work done by impeller per unit

weight of water . Sketch the velocity triangle

UVelocity =

velocity areadensity rate flow mass =

Uρbδ =


1 δ distance aough energy thr kineticin Reduction =

2UρUbδ2

1=

.(6).................... 3Uρbδ2

1 =

( )=

δ

0

22 dyu-Uρb2

13Uρbδ2

1

ρb2

1

( )=

δ

0

22 dyu-Uu 3Uδ

−=

δ

0

2

22 dy

U

u1uU3Uδ

U3

dyU

u1

U

uU δ

δ

0

2

2

3

2

−=

dyU

u1

U

u

0

2

2

−=

Given Data

Internal diameter of the impeller, D1=300 mm= 0.3m;

External diameter of the impeller, D2=600 mm = 0.6 m;

Speed N = 1000 rpm,

Vane angles of the impeller at inlet , = 20

Vane angles of the impeller outlet, = 30

To find

Work done by impeller per unit weight of water

Solution:

60

1000 3.0 πu1

=

m/s 7.15u1 =

60

N D πu 2

2 =

60

1000 6.0 πu2

=

m/s 4.31u2 =

Runner vane angle at inlet,

1

f

u

V θtan 1=

15.7

V 20tan 1

f=

2015.7tan V1

f =

m/s 5.17V1

f =

12ff VV =

m/s 5.17=

Runner vane angle at oulet,

2

2

w2

f

Vu

V tan

−=

2wV4.13

5.7103tan

−=

03tan

5.71V4.13

2w =−

81.9V4.132

w =−

2wV81.94.13 =−

m/s 51.21V2

w =

Work done by impeller per second /N weight of water, W:

g

uV WWorkdone

2w2=

9.81

14.321.51

=

68.5 WWorkdone =

4.a.2) The internal and external diameter of an impeller of a centrifugal pump which is

running at 1000 rpm are 200mm and 400mm respectively. The discharge through pump is

0.04m3/s and velocity of flow is constant and equal to 2.0 m\s .The diameter of the suction

and deliver pipes are 150mm and 100mm respectively ad suction and delivery heads are

6m (abs) and 30m (abs)of water respectively. If the outlet vane angle is 45 and power

required to drive the pump is 16.186kW,

Determine;

i. Vane angle of the impeller at inlet,

ii. The overall efficiency of the pump, and

iii. Manometric efficiency of the pump.

Given Data

Internal diameter of the impeller, D1=200 mm= 0.2m;


Velocity of flow, m/s 3VV21

ff ==

Speed N = 1000 rpm,

Discharge ,Q =0.04 m3/s,


Diameter of suction pipe ds =100 mm= 0.1m;

Diameter of delivery pipe dd =150 mm= 0.15m;

Suction head hs= 6m

Delivery head hd= 30m


To find

i. Vane angle of the impeller at inlet,

ii. The overall efficiency of the pump, and


Solution:

60

1000 2.0 πu1

=

m/s 47.10u1 =

60

N D πu 2

2 =

60

1000 4.0 πu 2

=

m/s 94.20u2 =


2

2

w2

f

Vu

V tan

−=

2wV94.20

254tan

−=

54tan

2V94.20

2w =−

2V94.202

w =−

2wV294.20 =−

m/s 94.18V2

w =


1

f

u

V θtan 1=

191.010.47

2 θtan ==

( )191.0tan θ 1−=

=10.8

Overall efficiency,

powerShaft

powerWater η pump of efficiency Overall o =

Q H g ρpowerWater =

+

+=

2g

Vh-

2g

Vh H

2

ss

2

dd

2g

Vh-

2g

VhH

2

ss

2

dd −+=

Q=AsVs

d4

πA 2

ss =

1.04

π 2=

2

s m 00785.0A =

SV00785.004.0 =

m/s 09.5VS =

Q=AdVd

d4

πA 2

dd =

15.04

πA 2

d =

Ad=0.01766 m2

dV 0.0176604.0 =

m/s 26.2Vd =

9.812

2.266-

9.812

5.0903H

22

−

+=

m 06.25H =


0.4 25.069.811000 =

Water power =9833.544 W

powerShaft

powerWater η o =

31016.186

9833.544

=

6074.0η o =

Manometric efficiency

22w

manu V

H gη ,efficiency Manometric =

20.948.941

25.069.81η man

=

6198.0ηman =

Manometric efficiency 6198.0ηman =

4.a.3) A centrifugal pump running at 920 rpm and delivering 0.32m3/s of water against a head of

28m, the flow velocity being 3m/s. if the manometric efficiency is 80% determine the diameter

and width of the impeller. The blade angle at outlet is 25.

Given Data

Speed N = 920 rpm,

Discharge ,Q =0.32m3/s,

Head , H = 28m

Velocity of flow, m/s 3VV21

ff ==

Manometric efficiency manη =80%=0.8


To find

i. Diameter of the impeller

ii. Width of the impeller

Solution:

22w

hu V


22w u V

289.810.8

=

0.8

289.81u V 22w

=

85.343u V 22w =

Vane angle at oulet,

2

2

w2

f

Vu

V tan

−=

2w2 Vu

352tan

−=

25tan

3Vu

2w2 =−

43.6Vu2

w2 =−

2w2 V43.6u =−

43.6uV 2w2

−=

85.343u V 22w =

( ) 85.343u 6.43-u 22 =

085.3436.43u-u 2

2

2 =−

On comparing

0cbxax 2 =++

a= 1 b= -6.43 c= -343.85

u2= 22.02 m/s

60

N D πu 2

2 =

60

920 D π22.02 2

=

2D209

6022.02=

m 46.0D2 =

Diameter of impeller m 46.0D2 =

Discharge

2f22 V B D πQ =

3 B 0.46 π32.0 2 =

2B3 0.46 π

32.0=

m 0.074B2 =

Width of impeller m 074.0B2 =

4.a.4) A centrifugal pump is running at 1000 r.p.m. The outlet vane angle of the impeller

is 45 and velocity of flow at outlet is 2.5 m/s. the discharge through the pump is 200 litres/s

when the pump is working against a total head of 20m .if the manometric efficiency of the

pump is 80% determine

i. the diameter of the impeller, and

ii. the width of the impeller at outlet.

Given Data

Speed N = 1000 rpm,

Discharge ,Q =200litre/s

/sm10002 33−=

=0.2m3/s,

Head , H = 20 m

Velocity of flow, m/s 5.2VV21

ff ==



To find

i. Diameter of the impeller

ii. Width of the impeller at outlet

Solution:

22w

hu V


22w u V

029.810.8

=

0.8

029.81u V 22w

=

245.25u V 22w =


2

2

w2

f

Vu

V tan

−=

2w2 Vu

2.554tan

−=

45tan

2.5Vu

2w2 =−

5.2Vu2

w2 =−

2w2 V5.2u =−

5.2uV 2w2

−=

245.25u V 22w =

( ) 245.25u 2.5-u 22 =

0-245.252.5u-u 2

2

2 =−

On comparing

0cbxax 2 =++

a= 1 b= -2.5 c= -245.25

u2= 16.96 m/s

60

N D πu 2

2 =

60

1000 D π16.96 2

=

2D1000

6016.96=

m 324.0D2 =

Diameter of impeller m 324.0D2 =

Discharge

2f22 V B D πQ =

2.5B 0.324 π2.0 2 =

2B2.5 0.324 π

2.0=

m 0786.0B2 =

Width of impeller m 0786.0B2 =

4a.5) Find the power required to drive a centrifugal pump which delivers 0.04m3/s of water

to a height of 20m through a 15 cm diameter pipe and 100m long. The overall efficiency

pump is 70% and co-efficient of friction f=0.15 in the formula 2gd

4flVh

2

f

=

Given Data

Speed N = 1000 rpm,

Discharge ,Q =0.04 m3/s,

Hieght, Hs = 20 m

Co-efficient of friction f=0.15

Diameter of pipe,D =15 cm

Length of pipe,L =100 m

Overall efficiency pump oη = 70%

To find

Power required to drive pump


0.4 25.069.811000 =


powerShaft

powerWater η o =


2g

Vh HH

2

fs ++=

Hs= 20 m

2gd

4fLVh

2

f =

Q=AV

d4

πA 2=

15.04

πA 2=

A=0.01766 m2

V 0.0176604.0 =

m/s 26.2V =

0.159.812

2.261000.0154h

2

f

=

m 41.10hf =

9.812

26.241.01 20H

2

++=

H = 30.67 m

Water power 0.0430.97 9.811000 =


powerShaft

12152.6287.0 =

7.0

12152.628powerShaft =

Shaft power =17360.89 W

Power required to drive pump =17360.89 W

4.a.6) A centrifugal pump as an impeller 500mm in diameter running at 400 rpm. The

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

are curved backwards at outlet at 30 to the wheel tangent. If the discharge of the pump is

0.14 m3/s, calculate the impeller power and the torque on the shaft.

60

N D πu 2

2 =

60

400 5.0 πu 2

=

m/s 47.10u2 =


2

2

w2

f

Vu

V tan

−=

2wV47.01

130tan

−=

30tan

1V47.01

2w =−

795.1V47.012

w =−

2wV795.147.01 =−

2wV675.8 =

2wV675.8 =

Impeller power

Q H g ρP =

g

2w uV H 2=

81.9

10.478.675

=

m .259 H =

0.149.25 9.811000P =

0.149.25 9.811000P =

W 12703.95P =

Torque on the shaft

60

TNπ2P

=

60

T004π212703.95

=

T400π2

6012703.95 =

Nm 43.303T =

Torque Nm 43.303T =

4.a.7)A centrifugal pump is to discharge 0.118 m3/s at a speed of 1450 r.p.m. against a head

of 25 m. the impeller diameter is 250 mm, its width at outlet is 50 mm and manometric

efficiency is 75%. determine the vane angle at the outer periphery of the impeller.

Given Data

Speed N = 1450 rpm,


Head , H = 25m


External diameter of the impeller, D2=250 mm = 0.25 m

Width of the impeller at outlet, B2=50 mm = 0.05 m

To find

i. Vane angle of the impeller at outlet,

Solution:

60

N D πu 2

2 =

60

1450 25.0 πu 2

=

m/s 98.18u2 =

Discharge

2f22 V B D πQ =

V.05 0.25 π118.02f=

2fV.05 0.25 π

118.0=

m/s 3V2f =

m/s 3VV21 ff ==

Manometric effciency

22w

hu V


18.98 V

259.810.75

2w

=

18.980.75

529.81V

2w

=

m/s 23.17 V 2w =


2

2

w2

f

Vu

V tan

−=

23.7198.81

3

−=

7143.1tan =

( ) 59.74 7143.1tan -1 ==

4.a.8) A centrifugal pump delivers water against a net head of 14.5 m and a design speed

of 1000 r.p.m. against a head of 25 m. The Vanes are curved back to an angle of 30 with

the periphery. the impeller diameter is 300 mm and outlet width is 50 mm .Determine the

discharge if manometric efficiency is 95%.

Given Data

Speed N = 1000 rpm,


Head , H = 14.5m




Vane angle of the impeller at outlet, =30

To find

i. Discharge

Solution:

60

N D πu 2

2 =

60

1000 3.0 πu2

=

m/s 7.15u2 =


22w

hu V


7.51 V

5.419.810.95

2w

=

7.510.95

5.149.81V

2w

=

m/s 54.9 V 2w =


2

2

w2

f

Vu

V tan

−=

54.97.51

V tan30 2

f

−=

16.6

V tan30 2

f=

m/s 556.3V 2

f =

m/s 3.556VV21 ff ==

Discharge

2f22 V B D πQ =

556.3.05 0.3 π =

/sm 1675.0Q 3=

4.a.9 )A centrifugal pump having outer diameter equal to two times the inner diameter

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

the impeller is constant and equal to 2.5m/s . The vanes are set back at an 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 inlet

ii. work done by impeller on water per second, and

iii. manometric efficiency.

Given Data

Speed N = 1000 rpm,

Head , H = 40m



External diameter of the impeller, D2= 2 Internal diameter of the impeller,

12 D2D =

1D25.0 =

m 0.25D1 =

External diameter of the impeller, D1 = 0.25 m


Velocity of flow, m/s 5.2VV21

ff ==

To find

i. Runner vane angle at inlet,

ii. Work done by impeller on water per second


Solution:

60

N D πu 1

1 =

60

1000 25.0 πu1

=

m/s 09.13u1 =

60

N D πu 2

2 =

60

1000 5.0 πu2

=

m/s 18.26u2 =

Discharge

2f22 V B D πQ =

5.2.05 0.5 π =

/sm 1963.0Q 3=


2

2

w2

f

Vu

V tan

−=

2wV18.26

2.504tan

−=

04tan

2.5V18.26

2w =−

979.2V18.262

w =−

2wV979.218.26 =−

m/s 201.23V2

w =


1

f

u

V θtan 1=

191.013.09

2.5 θtan ==

( )191.0tan θ 1−=

=10.8

Work done by impeller on water per second

22w u V Q ρsecondper on water impeller by doneWork =

26.1823.2010.19631000 =

=119233.03 W


22w

manu V


26.18201.23

049.81η man

=

646.0ηman =

Manometric efficiency 646.0ηman =

To find

i. Discharge

Solution:

60

N D πu 2

2 =

60

1000 3.0 πu2

=

m/s 7.15u2 =


22w

manu V


7.51 V

5.419.810.95

2w

=

7.510.95

5.149.81V

2w

=

m/s 54.9 V 2w =


22w

hu V


18.98 V

2259.810.75

2w

=

18.980.75

529.81V

2w

=

m/s 23.17 V 2w =


2

2

w2

f

Vu

V tan

−=

23.7198.81

V tan30 2

f

−=

16.6

V tan30 2

f=

m/s 556.3V 2

f =

m/s 3.556VV21 ff ==

Discharge

2f22 V B D πQ =

556.3.05 0.3 π =

/sm 1675.0Q 3=

4.a.10 The outer diameter of an impaller of a centrifugal pump is 400mm and outlet

widthis 50mm. the pump is running at 800 r.p.m. and is working against a total head

of15m.the vanes angle at outlet is 40 and manometric efficiency is 75%.

determine

i. velocity of flow at outlet

ii. velocity of water leaving the vane

iii. angle made by the absolute velocity at outlet with the direction of motion at outlet,

and

iv. discharge

Given Data

Speed N = 800 rpm,

Head , H = 15 m

Outer diameter of the impeller, D2=400 mm = 0.4 m

Width at outlet, B2=50 mm = 0.05 m


Manometric efficiency 75.0%75ηman ==

To find

i. Velocity of flow at outlet,

ii. Velocity of water leaving the vane

iii. Angle made by the absolute velocity at outlet

iv. Discharge

Solution:

60

N D πu 2

2 =

60

800 4.0 πu 2

=

m/s 75.16u2 =


22w

manu V


75.61 V

519.810.75

2w

=

75.610.75

159.81V

2w

=

m/s 71.11 V 2w =


2

2

w2

f

Vu

V tan

−=

7111.75.61

V tan40 2

f

−=

04.5

V tan40 2

f=

m/s 23.4V 2

f =

Velocity of water leaving the vane, V2

2

w

2

f2 22VVV +=

22 11.714.23 +=

V2 =12.45 m/s

Angle made by the absolute velocity at outlet, β

tan β=

2

2

w

f

V

V

36.0 12.137

79.17==

β = ( )36.0tan 1−

β =19.8

Discharge

2f22 V B D πQ =

23.4.05 0.4 π =

/sm 265.0Q 3=

RECIPROCATING PUMP

Piston or Plunger Diameter → D

Stroke length →L

Actual discharge, →Qact

Head , →H

Length of suction pipe, → ls

Diameter suction pipe, → ds

Length of delivery pipe, → ld

Diameter delivery pipe, → dd

Suction head →hs

Delivery head →hd

Coefficient of friction →f

Theoretical discharge single acting reciprocating pump

Theoretical discharge single acting pump,60

ALNQ the =

Theoretical discharge double acting reciprocating pump

Theoretical discharge for double acting reciprocating pump,60

2ALNQ the =

Area of piston or plunger

2D4

πA =

Coefficient of discharge

the

actd

Q

QCdischarge, oft Coefficien =

Slip

actthe QQSlip −=

Percentage of slip

100Q

QQslip of Percentage

the

actthe

−=

Power required to drive the double acting pump

Q H g ρ PPower act=

hh H ds +=

2

length troke=r , radiusCrank

S

60

N π2ω =

Area of piston or plunger, 2D4

πA =

Area of suction pipe ,2

ss d4

πa =

Area of delivery pipe ,2

d d4

πa d=

Pressure head due to acceleration head in the Suction stroke, has

rcosθωa

A

g

lh 2

s

sas =

Friction loss in suction pipe 2

d

dfd ωrcosθ

a

A

g

l 4fh

=

At the beginning of suction stroke 0θ =

At the middle of suction stroke 09θ =

At the end of suction stroke 018θ =

At Maximum pressure head due to acceleration 1cosθ =

Pressure head in the cylinder in the suction stroke

Pressure head in the cylinder =hs+has+hfs

Absolute Pressure head in the cylinder in suction stroke

Absolute Pressure head = Atmospheric pressure head - pressure head

Pressure head due to acceleration head in the delivery stroke , had

rcosθωa

A

g

lh 2

d

dad =

Friction loss in delivery pipe 2

d

dfd ωrcosθ

a

A

g

l 4fh

=

At the beginning of delivery stroke 0θ =

At the middle of delivery stroke 09θ =

At the end of delivery stroke 018θ =


Pressure head in the cylinder in the delivery stroke

Pressure head in the cylinder =hd+had+hfd

Absolute Pressure head in the cylinder in delivery stroke

Absolute Pressure head = Atmospheric pressure head + pressure head

4.b.1) 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.

Given:

Speed, N = 60rpm,

Actual discharge , Qact =0.01m3/s

Diameter of piston, D = 200 mm =0.2 m

Stroke length L = 400 mm =0.4 m

To find

i) Theoretical discharge of the pump,

ii) Coefficient of discharge

iii) slip

iv) percentage slip

Solution:

i) Theoretical discharge for single acting reciprocating pump

Theoretical discharge single acting pump,60

ALNQ the =

2D4

πA =

22.04

πA =

2m 0314.0A =

60

500.40.0314Q the

=

Qthe=0.01047 m3/s

Theoretical discharge Qthe=0.015 m3/s

ii) Coefficient of discharge

the

actd

Q

QCdischarge, oft Coefficien =

0.9550.01047

0.01Cd ==

Coefficient of discharge, Cd =0.955

iii) Slip

actthe QQSlip −=

01.001047.0 −=

=0.00047 m3/s

Slip=0.00047 m3/s

iv) Percentage of slip

100Q


the

actthe

−=

10001047.0

01.001047.0

−=

=4.489 %

Percentage of slip =4.489 %

4.b.2) A double –acting reciprocating pump running at 40 r.p.m is discharging 1.0m of

water per minute. The pump has a stroke of 400mm .the diameter of the piston is 200mm.

the delivery and suction head are 20m and 5m respectively. Find the slip of the pump and

power required to drive the pump.

Given:

Speed, N = 40rpm,

Actual discharge , /minm 1Q 3

act =

s 60

m 1 3

=

s/m 0.0166Q 3

act =

Diameter of piston, D = 200 mm =0.2 m

Stroke length L = 400 mm =0.4 m

Suction head hs= 5m

Delivery head hd=20m

To find

i) Slip

ii) Power required to drive the pump

Solution

i. Slip

actthe QQSlip −=

Theoretical discharge for double acting reciprocating pump,60

2ALNQ the =

2D4

πA =

22.04

πA =

2m 0314.0A =

60

600.40.03142Q the

=

Qthe=0.01675 m3/s

Theoretical discharge Qthe=0.01675 m3/s

actthe QQSlip −=

01666.001675.0 −=

=0.00009 m3/s

Slip =0.00009 m3/s

ii. Power required to drive the double acting pump

Q H g ρ PPower act=

hh H ds +=

205 +=

m 25 H =

W4107.9375

250.016759.8110002P

=

=

Power required to drive the double acting pump =4107.9375 W

4.b.3 The cylinder bore diameter of single acting reciprocating pump is 150mm and its

stroke is 300mm. The pump runs at 50 r.p.m. and lifts water through a height of 25m .The

delivery pipe is 22m long and 100mm in diameter .Find the theoretical discharge and the

theoretical power required to run the pump . if the actual discharge is 4.2 litres/s. find the

percentage slip.Also determine the acceleration head at the beginning and middle of the

delivery stroke.

Given:

Diameter of reciprocating pump, D = 150 mm =0.15 m

Stroke length of reciprocating pump, L = 300 mm =0.3 m

Actual discharge, Qact =4.2 litre/s

/sm102.4 33−=

=0.0042m3/s,

Speed, N = 50rpm,

Head , H = 25 m

Length of delivery pipe, ld =22 m

Diameter delivery pipe, dd =100mm =0.1m

To find :

i) Power required to drive the pump

ii) Percentage of slip

iii) Acceleration head at the beginning of delivery stroke,

iv) Acceleration head at the middle of delivery stroke,

Solution:

i) Power required to drive the pump

theQ H g ρP =

Theoretical discharge for single acting pump

60

ALNQ the =

2D4

πA =

215.04

πA =

2m 17670.0A =

60

500.30.001767Q the

=

Qthe=0.0044175 m3/s

0044175.02581.91000P =

P= 1083.39 W

Power required to drive the pump P= 1083.39 W

ii) Percentage of slip

100Q


the

actthe

−=

1000044175.0

0042.00044175.0

−=

=4.92%

Percentage of slip =4.92 %

Pressure head due to acceleration head in the delivery stroke , had

rcosθωa

A

g

lh 2

d

dad =

2


S

m 15.02

0.3=r =

Area of delivery pipe ,2

d d4

πa d=

21.04

π =

2

d m 00784.0a =

ld=22m

2m 001767.0A =

60

N π2ω =

60

50 π2ω

=

rad/s 236.5ω=

cosθ 0.155.236007854.0

001767.0

9.81

22h 2

ad =

cosθ 0.75 2had =

v) Acceleration head at the beginning of stroke,

cosθ 0.75 2had =


cos0 0.75 2had =

m 0.75 2had =

Acceleration head at the beginning of stroke, had =20.75 m

vi) Acceleration head at the middle of stroke,

cosθ 0.75 2had =

At the middle of delivery stroke 09θ =

cos90 0.75 2had =

0had =

Acceleration head at the end of stroke, had =0 m

4.b.4) The length and diameter of a suction pipe of a single acting reciprocating pump

are 5m and 10cm respectively. The pump has a plunger diameter of 15 cm and stroke

length of 35 cm. The center of the pump is 3 m above the water surface in the sump.

The atmospheric pressure head is 10.3m of water and the pump is running at 35 rpm.

Determine

(i) Pressure head due to acceleration at the beginning of suction stroke.

(ii) Maximum pressure head due to acceleration .

(iii) Pressure head in the cylinder at the beginning and at the end of the stroke

Given:

Length of Suction pipe, ls =5m

Diameter Suction pipe, ds =10 cm =0.1m

Diameter of plunger, D = 15 cm =0.15 m

Stroke length , L = 35 cm =0.35 m

The centre of the pump is 4 m above the water surface in the sump

Suction head hs= 3 m

Atmospheric pressure head , hatm =10.3m of water

Speed, N = 40 rpm,

To find :

(i) Pressure head due to acceleration at the beginning of suction stroke.

(ii) Maximum pressure head due to acceleration .

(iii) Pressure head in the cylinder at the beginning and at the end of the stroke

Solution:


rcosθωa

A

g

lh 2

s

sas =

2


S

m 0.1752

0.35=r =

Area of Plunger ,2

D4

πA =

215.04

π=

2m 01767.0A =


ss d4

πa =

21.04

π =

2

s m 00785.0a =

ls =5m

60

N π2ω =

60

35 π2ω

=

rad/s 665.3ω=

cosθ0.175665.300785.0

01767.0

9.81

5h 2

as =

cosθ 2.695has =

i) Pressure head due to acceleration at the beginning of suction stroke.

cosθ 2.695has =


cos0 695.2has =

m 695.2has =

Pressure head due to acceleration at the beginning of suction stroke has=2.695 m

ii) Maximum pressure head due to acceleration .

cosθ 2.695has =


1 2.695has =

m 2.695has =

Maximum Pressure head due to acceleration has=2.695 m

iii) Pressure head in the cylinder at the beginning of the stroke

Pressure head in the cylinder =hs+has

Suction head hs= 3 m


cos0 695.2has =

m 695.2has =

Pressure head in the cylinder =3+2.695 =5.695 m

During suction stroke, pressure head is less than atmospheric pressure head


=10.3-5.695

= 4.605 m

Pressure head in the cylinder at the beginning of the stroke = 4.605 m

iv) Pressure head in the cylinder at the end of the stroke

Pressure head in the cylinder =hs+has

Suction head hs=4 m

At the end of delivery stroke 801θ =

cos180 695.2has =

m 695.2has −=

Pressure head in the cylinder =3- 2.695 =0.305 m

During suction stroke, pressure head is less than atmospheric pressure head


=10.3-0.305

= 9.995 m

Pressure head in the cylinder at the beginning of the stroke = 9.995 m.

4.b.5A single acting reciprocating pump has a stroke length of 15cm .the suction pipe is

7 metre long and the ratio of the suction diameter to the plunger diameter is 3/4.The

water level in the sump is 2.5 metres below the axis of the pump cylinder and the pipe

connecting the sump and pump cylinder is 7.5 cm diameter. If the crank is running at

75 r.p.m, determine the pressure head on the piston

i. in the beginning of the suction stroke

ii. in the end of the suction stroke

iii. in the middle of the suction stroke

Take co efficient of friction as 0.01

Given:

Stroke length , L = 15 cm =0.15 m

Length of Suction pipe, ls =7 m

The pipe connecting the sump and pump cylinder is 7.5 cm diameter

Note: The pipe which connects sump and pump cylinder is suction pipe

Diameter Suction pipe, ds =7.5 cm =0.75m

Diameter of plunger, D = 15 cm =0.15 m

4

3

diameter pipePlunger

diameter pipeSuction =

4

3

D

ds =

4

3

D

0.75=

D3

40.75=

m 0.1D =

Diameter of plunger, D = 0.1 m

The water level in the sump is 2.5 metres below the axis of the pump cylinder

Suction head hs= 2.5 m

Atmospheric pressure head , hatm =10.3m of water

Speed, N = 75 rpm,

To find :

(iv) Pressure head due to acceleration at the beginning of suction stroke.

(v) Maximum pressure head due to acceleration .

(vi) Pressure head in the cylinder at the beginning and at the end of the stroke

Solution:


rcosθωa

A

g

lh 2

s

sas =

2


S

m 0.0752

0.15=r =

Area of Plunger ,2

D4

πA =

21.04

π=

2m 00785.0A =


ss d4

πa =

2075.04

π =

2

s m 004415.0a =

ls =5m

60

N π2ω =

60

75 π2ω

=

rad/s 85.7ω=

cosθ0.07585.7004415.0

00785.0

9.81

7h 2

as =

cosθ 5.87has =

2

ss

sfs sinθr ω

a

A

d 2g

l f 4h

=

2

fs sinθ 0.07585.7004415.0

00785.0

0.075 9.812

7 0.01 4h

=

θsin 0.208h 2

fs =

Pressure head in the piston at the beginning of the suction stroke

Pressure head in the piston =hs+has+ hfs


cosθ 5.87has =


cos0 5.87has =

m 5.87has =

θsin 0.208h 2

fs =


0sin 0.208h 2

fs =

0h fs =

Pressure head in the piston =2.5+5.87+0=8.37 m

Pressure head in the piston at the middle of the suction stroke



cosθ 5.87has =

At the beginning of middle stroke 09θ =

cos90 5.87has =

0has =

θsin 0.208h 2

fs =

At the middle of suction stroke 09θ =

90sin 0.208h 2

fs =

m 0.208h fs =

Pressure head in the piston =2.5+0+0.208=2.708 m

Pressure head in the piston at the end of the suction stroke



cosθ 5.87has =


cos180 5.87has =

m 5.87has −=

θsin 0.208h 2

fs =


180sin 0.208h 2

fs =

0h fs =

Pressure head in the piston =2.5-5.87+0= -3.37 m

PELTON TURBINE

Notation

Shaft power, →S.P.

Speed →N

Head→H

Gross Head→Hg

Friction Head→Hf

Bucket diameter → D

Jet diameter → d

Discharge → Q

Discharge of one jet→ q

Velocity of jet→ 1V

Bucket Speed or Runner Speed →u1

Hydraulic efficiency→h

Overall efficiency→ o

Co efficient of velocity→ vC

Side clearance angle →

Velocity of jet

Velocity of jet, 2gHCV v1 =

1w VV1=

Runner Speed

Runner speed, 60

NDπu1

=

or

Runner speed, 2gH ratio Speedu1 =

Relative Speed

Relative Speed

11r u-VV1=

12 rr V V =

2

2

r

2w

V

uVcos

+=

Power given by water to runner

( ) 1ww u VV Q ρrunner oby water tgiven Power 21

+=

Power available at the nozzle

gHQ ρ nozzle at the availablePower =

The hydraulic efficiency of the turbine

( ) V

u VV 2η ,efficiency Hydraulic

2

1ww

h

2

21+

=

Discharge of one jet

1Vaq =

Overall efficiency

powerWater

powerShaft η turbineof efficiency Overall o =

Water Power

Q H g ρpower Water =

Number of jet

q

Q

jet one of Discharge

Discharge Totaljet ofNumber ==

Net Head

Net head=Gross Head-Friction

H=Hg-hf

Side clearance angle,

Side clearance angle,= 180-Angle of Deflection

=80-Angle of Deflection

Size of buckets

Width of buckets =5d.

Depth of buckets =1.2d

Number of buckets on the wheel

2d

D15+=Z

1. A pelton wheel has a mean bucket speed of 10 metres per second with a jet of water

flowing at the rate of 700 litres/s under a head of 30 metres. The buckets deflect the je

through an angle of 160. Calculate the power given by water to the runner and the

hydraulic efficiency of the turbine. Assume co-efficient of velocity as 0.98.

Solution, Given :

Speed of bucket, u = u1 = u2 = 10 m/s

Discharge Q = 700 litres/s

= 310700 − m3/s

=0.7 m3/s,

Head of water, H = 30 m

Angle of deflection = 160

Side clearance angle, = 180 – 160 = 20

Co-efficient of velocity Cv = 0.98

Velocity of jet


309.81298.0V1 =

V1=23.77 m/s

m/s 23.77VV 1w1

==

Relative Speed

11r u-VV1=

10-77.32V1r=

m/s 77.31V1r=

m/s 77.31V V12 rr ==

From outlet velocity triangle,

2

2

r

2w

V

uVcos

+=

13.77

01V20cos 2w +

=

01V94.122w +=

2wV1094.12 =−

m/s 94.2V2w =



+=

( )10 94.7723. 7.00001 +=

=186970 W


( ) V


2

1ww

h

2

21+

=

( )

23.77

10 94.223.77 22

+=

hη =0.954

2 A Pelton wheel is having a mean bucket diameter of 1m and is running at 100 .r.p.m the

net head on the Pelton wheel is 700m. if the side clearance angle is 15 and discharge

through nozzle is 0.1 m3/s, find

i. Power available at the nozzle,

ii. Hydraulic efficiency of the turbine

Solution,

Given :

Diameter D = 1.0 m

Discharge Q =0.1 m3/s,

Head , H = 700 m

Side clearance angle, = 15

Speed, N = 100 r.p.m.

Velocity of jet


7009.8121V1 =

V1=117.19 m/s

m/s 117.19VV 1w1

==

Runner Speed

Runner speed, 60

NDπu1

=

60

00011πu1

=

m/s 36.52u1 =

Relative Speed

11r u-VV1=

52.36-19.117V1r=

m/s 83.45V1r=

m/s 83.45V1r=

m/s 83.45VV12 rr ==


2

2

r

2w

V

uVcos

+=

54.83

01V15cos 2w +

=

01V15cos54.832w +=

01V52.9612w +=

2wV1052.961 =−

m/s 961.42V2w =

Power available at the nozzle

gHQ ρ nozzle at the availablePower =

1.070081.90001 =

=686700 W


( ) V


2

1ww

h

2

21+

=

( )

117.19

52.36 961.42 117.19 22

+=

hη =0.9718

3 A pelton wheel is to be designed for the following specification ; shaft power =11,772kw;

head = 380 meters; speed= 750r.p.m ; overall efficiency =89%; jet diameter is not exeed

one –sixth of the wheel diameter.

Determine;

i. The wheel diameter

ii. The number of jets required,

iii. Diameter of the jet

Solution,

Given :

Shaft power, S.P. = 11772 kW

= W1011772 3

Head, H = 380 m

Speed N= 750r.p.m ;

89.0%89η efficiency Overall o ==

6

1

Diameter Wheel

DiameterJet =

6

1

D

d

1

=

Co-efficient of velocity Cv = 0.985

Speed ratio =0.45

To find:

i. The wheel diameter,D1

ii. The number of jets required, N

iii. Diameter of the jet, d

Velocity of jet,


3809.812985.0V1 =

V1= .0558 m/s

.05m/s58VV 1w1

==

Runner Speed


3809.812 45.0u1 =

m/s 85.38u1 =

60

NDπu 1

1

=

60

750Dπ85.83 1

=

1D750

6085.83=

m 989.0D1 =

Wheel Diameter m 989.0D1 =

6

1

D

d

1

=

6

1

0.989

d=

165.0909.06

1d == m

Jet Diameter d =0.165 m

Discharge of one jet

1Vaq =

2d4

πa =

20.1654

π=

2m 0.021371=

05.850.021371q =

/sm 818.1q 3=

Discharge of one jet, /sm 818.1q 3=

powerWater

10 117720.84

powerWater

powerShaft η turbineof efficiency Overall

3

o

=

=

0.84

10 11772power Water

3=

Water Power =14014285.7 W


Q 083 9.811000 14014285.7 =

Q083 9.811000

14014285.7 =

/s3.672m Q 3=

Total Discharge /s3.672m Q 3=

powerWater

10 11772

q

Q

jet one of Discharge

Discharge Totaljet of N

3=

==umber

q

Q=

1.818

3.672= =2

Number of jet =2

4. The penstock supplies water from a reservoir to the Pelton wheel with a gross head 500

m. one-third of the gross head is lost in friction in the penstock. The rate of flow of water

through the nozzle fitted at the end of the penstock is 2.0 m3/s. the angle of deflection of the

jet is 165. Determine the power given by the water to the runner and also hydraulic

efficiency of the pelton wheel. Take speed ratio = 0.45 and Cv = 1.0.

Gross head, Hg = 500 m

Head lost in friction 3

Head Gross =

m 7.1663

500

3

hh

g

f ===

Net head H=Hg-hf

= 500 – 166.7

H= 333.30 m

Discharge, Q = 2.0 m3/s

Angle of deflection = 165

Side clearance Angle = 180 – 165 = 15

Speed ratio = 0.45

Co-efficient of velocity, Cv = 1.0

Velocity of jet

velocity of jet, 2gHCV v1 =

333.39.8121V1 =

V1=80.86 m/s

m/s 80.86VV 1w1

==

Runner Speed


333.39.812 45.0u1 =

m/s 387.36u1 =

Relative Speed

11r u-VV1=

36.387-86.80V1r=

m/s 473.44V1r=

m/s 473.44V V12 rr ==


2

2

r

2w

V

uVcos

+=

44.473

387.36V15cos 2w +

=

387.36Vcos15 44.4732w +=

387.36Vcos15 44.4732w +=

2wV387.3642.957 =−

m/s 57.6V2w =



+=

( )36.387 57.68680. 20001 +=

=6362630 W


( ) V


2

1ww

h

2

21+

=

( )

80.86

36.387 57.680.86 22

+=

hη =0.9731

5. A Pelton wheel is to be designed for a head of 60 m when running at 200 r.p.m. the

Pelton wheel develops 95.6475 kW shaft power. The velocity of the buckets = 0.45 times the

velocity of this overall efficiency = 0.85 and co-efficient of velocity is equal to 0.98.

Solution, Given :

Head, H=60m

Speed N = 200 r.p.m.

Shaft power, S.P = 95.6475 kW

Velocity of bucket u = 0.45 x Velocity of jet

Overall efficiency, 0=0.85

Co-efficient of velocity Cv=0.98

Design of Pelton wheel means to find diameter of jet (d), diameter of wheel (D), Width and

depth buckets and number of buckets on the wheel.

i. Velocity of jet, V1 = Cv 2 0.98 2 9.81 60 33.62 m/sgH = =

Bucket velocity, u = u1 = u2 = 0.45 x V1 = 0.45 x 33.62 = 15.13 m/s

But ; where D=Diameter of wheel60

DNu

=

D 200 60 15.13 15.13= D= =1.44 m, Ans

60 200or

ii. Diameter of the jet (d)

overall efficiency 0 = 0.85

0

0

. 95.6475 95.6475 1000But

..

1000

95.6475 1000 =

1000 9.81 60

95.6475 1000 95.64 Q=

1000 9.81 60

S P

W PW P g Q H

Q

= = =

=

275 10000.1912 m / 2

0.85 1000 9.81 60

=

But the discharge, Q = area of jet x Velocity of jet

2 2

10.1912 33.24 4

4 0.19120.085 m=85mm Ans.

33.62

d V d

d

= =

= =

iii. Size of buckets

Width of buckets =5 x d = 5 x 85 = 425 mm.

Depth of buckets =1.2 x d = 1.2 x 85 = 102 mm. Ans.

iv. Number of buckets on the wheel is given by equation, as

1.44

15 15 15 8.5 23.5 say 24. Ans.2 2 .085

DZ

d= + = + = + =

FRANCIS TURBINE

Notation


Speed →N

Head→H

Outer diameter of runner → D1

Inner diameter of runner → D2

Discharge → Q

Velocity of flow→1

fV



Guide blade angle,→

Vane angles at inlet → θ

Vane angle at outlet →

Formula

Overall efficiency

powerWater


Water Power


Discharge

1f11 V B D πQ Discharge =

Hydraulic efficiency

H g

u Vη ,efficiency Hydraulic

1w

h1=

Flow ratio

2gH

Vratio Flow 1

f=

Speed ratio

2gH

uratio Speed 1=

Breadth ratio

1

1

D

B ratioBreadth =

60

N D πu 2

2 =

Guide blade angle,

1

1

w

f

V

V tanα =

Runner vane angles at inlet, θ

1w

f

uV

V θtan

1

1

−=

Runner vane angles at oulet,

2

f

u

V tan 2=

Speed of turbine ,N

60

N D πu 1

1 =

Speed of turbine ,Ns

60

N D πu 1

1 =

4

5s

H

S.PN N =

where Shaft Power ,S.P in kW

2.The following data is given for a francis turbine. Net head h=60m; speed N=700 r.p.m

shaft power=294.3 kW; oη =84% ; hη =93%; flow ratio =0.20; breadth ratio n =0.1

outer diameter of the runner=2 inner diameter of runner. The thickness of vanes occupy

5% of circumferential area of the runner, velocity of flow is constant at inlet and outlet

and discharge is radial at outlet

Determine

i. Guide blade angle

ii. Runner vane angles at inlet and outlet

iii. Guide blade angle

iv. Runner vane angles at inlet and outlet

Given Data

Shaft power, S.P. = 148.5 kW

= W10 148.5 3

Head, H = 7.62 m

Overall efficiency, o = 84% =0.84

Hydraulic efficiency, h =93 % =0.93

Flow ratio =0.6

breadth ratio n = 0.1

Outer diameter of runner →D1


Outer diameter of runner = 2 Inner diameter of runner

D1 = 2 D2

thickness of vanes = 5% of circumferential area of the runner

velocity of flow is constant 21

ff VV =

To find

a. Guide blade angle

b. Runner vane angles at inlet and outlet

c. Guide blade angle

d. Runner vane angles at inlet and outlet

Solution

Velocity of jet

velocity of jet, gH2ratio flowV1 =

609.8122.0V1 =

V1=6.826 m/s

1

1

D

B ratioBreadth =

1

1

D

B1.0 =

B1=0.1D1

powerWater

10 294.30.84

powerWater


3

o

=

=

0.84

10 294.3power Water

3=



Q 60 9.811000 10350.357 3 =

Q60 9.811000

10350.357

3

=

/sm 0.5952 Q 3=

Discharge /sm 0.5952 Q 3=

flow ofVelocity flow of area Actual Discharge =

Thickness of vanes = 5% of circumferential area of the runner

B πD95.0flow of area Actual 11=

1f11 V B D π95.0Q =

6.862 D0.1 D π95.05952.0 11 =

2

1D 6.862 0.1 π95.0

5952.0=

2

1D 29507.0 =

m 54.0D 1 =

11 D 1.0B =

m 0.054 54.0 1.0B 1 ==

21 2D D =

2D2 0.54 =

m 0.27 D2 =

60

N D πu 1

1 =

60

700 54.0 π

=

m/s 79.19u1 =

60

N D πu 2

2 =

60

700 27.0 π =

m/s 896.9u2 =

H g


1w

h1=

60 9.81

19.79 V0.93 1

w

=

1

wV19.79

60 9.81 0.93=

m/s 66.27V1

w =

Guide blade angle,

1

1

w

f

V

V tanα =

27.66

6.862=

248.0 tanα=

( )248.0tan α -1= =13.928


1w

f

uV

V θtan

1

1

−=

872.079.1966.27

6.862 θtan =

−=

( ) 09.41872.0tan θ -1 ==


2

f

u

V tan 2=

896.9

682.6 tan =

( ) 34.74 6934.0tan -1 ==

2.The following data is given on a Francis turbine; Net head =50m, speed = 600rpm; Shaft

Power = 400hp =84% = 90% flow ratio= 0.2; breadth ratio= 0.1; oη =84% ; hη =90%;

flow ratio =0.20; breadth ratio n =0.1 outer diameter of the runner=2 inner diameter of

runner. The thickness of vanes occupy 5% of circumferential area of the runner, velocity

of flow is constant at inlet and outlet and discharge is radial at outlet

Determine

i. Guide blade angle

ii. Runner vane angles at inlet and outlet

iii. Guide blade angle

iv. Runner vane angles at inlet and outlet

Given Data

Shaft power, S.P. = 400 HP

1HP= 735.35 W

= W75.735 148.5

S.P. = 294300 W

Head, H = 50 m


Hydraulic efficiency, h =90 % =0.9

Flow ratio =0.2

breadth ratio n = 0.1

Outer diameter of runner →D1


Outer diameter of runner = 2 Inner diameter of runner

D1 = 2 D2

thickness of vanes = 5% of circumferential area of the runner

velocity of flow is constant 21

ff VV =

To find

a. Guide blade angle

b. Runner vane angles at inlet and outlet

c. Guide blade angle

d. Runner vane angles at inlet and outlet

Solution

Velocity of jet

velocity of jet, gH2ratio flowV1 =

509.8122.0V1 =

V1=6.264 m/s

1

1

D

B ratioBreadth =

1

1

D

B1.0 =

B1=0.1D1

H g


1w

h1=

60 9.81

19.79 V0.9 1

w

=

1

wV19.79

60 9.81 0.93=

m/s 66.27V1

w =

powerWater

2943000.84

powerWater

powerShaft η turbineof efficiency Overall o

=

=

0.84

294300power Water =



Q 50 9.811000 10350.357 3 =

Q50 9.811000

10350.357

3

=

/sm 0.714 Q 3=

Discharge /sm 0.714 Q 3=

flow ofVelocity flow of area Actual Discharge =

Thickness of vanes = 5% of circumferential area of the runner

B πD95.0flow of area Actual 11=

1f11 V B D π95.0Q =

6.264 D0.1 D π95.0714.0 11 =

2

1D 6.264 0.1 π95.0

714.0=

2

1D 382.0 =

m 618.0D 1 =

11 D 1.0B =

m 0.0618 618.0 1.0B 1 ==

21 2D D =

2D2 0.618 =

m 0.309 D2 =

60

N D πu 1

1 =

60

600 618.0 π

=

m/s 415.19u1 =

60

N D πu 2

2 =

60

600 309.0 π =

m/s 7075.9u2 =

H g


1w

h1=

50 9.81

19.415 V0.9 1

w

=

1

wV19.415

50 9.81 0.9=

m/s 737.22V1

w =

Guide blade angle,

1

1

w

f

V

V tanα =

22.737

6.264=

2755.0 tanα=

( )2755.0tan α -1= =15.4


1w

f

uV

V θtan

1

1

−=

88.1415.19737.22

264.6 θtan =

−=

( ) 6288.1tan θ -1 ==


2

f

u

V tan 2=

7075.9

264.6 tan =

( ) 32.83 645.0tan -1 ==

KAPLAN TURBINE

Notation


Speed →N

Head→H

Hub or Boss Diameter → D0

Diameter of runner → D0

Discharge → Q

Velocity of flow→1

fV



Guide blade angle,→

Vane angles at inlet → θ

Vane angle at outlet →

Formula

Overall efficiency

powerWater


Water Power


Hydraulic efficiency

H g


1w

h1=

Flow ratio

2gH

Vratio Flow 1

f=

Speed ratio

2gH

uratio Speed 1=

Remember

In Kaplan turbine

21 uu =

Guide blade angle,

1

1

w

f

V

V tanα =


1w

f

uV

V θtan

1

1

−=


2

f

u

V tan 2=

Speed of turbine ,N

60

N D πu o

1 =


4

5s

H

S.PN N =


1. A Kaplan turbine working under a head of 20 m develops 11772 kW shaft power. The

outer diameter of the runner is 3.5 m and hub diameter 1.75 m. The guide blade angle at

the extreme edge of the runner is 35. The hydraulic and overall efficiencies of the

turbines are 88% and 84% respectively. If the velocity of what is zero at outlet, determine:

(i) Runner vane angles at inlet and outlet at the extreme edge of the runner, and

(ii) Speed of the turbine.

Solution:

Given:

Head, H = 20 m


= W1011772 3

Outer dia. of runner Do = 3.5 m

Hub diameter, Db = 1.75 m

Guide blade angle, = 35

Hydraulic efficiency, h = 88%

Overall efficiency, o = 84%

Velocity of whirl at outlet 0V2w =

To find:

(i) Runner vane angles at inlet, θ

(ii) Runner vane angles at outlet,

(iii) Speed of the turbine, N.

Solution

powerWater

10 117720.84

powerWater


3

o

=

=

0.84

10 11772power Water

3=



Q 20 9.811000 14014285.7 =

Q20 9.811000

14014285.7 =

/sm 71.428 Q 3=

Discharge, /sm 71.428 Q 3=

( )1f

2

h

2

o V DD4

Q −=

( )1f

22 V 75.15.34

71.428 −=

1fV 251.7 71.428 =

1fV 251.7

71.428 =

m/s 9.9V 1f =

m/s 9.9

VV 12 ff

=

=

Guide blade angle,

1

1

w

f

V

V tanα =

1wV

9.935tan =

35tan

9.9V

1w =

m/s 14.14V1

w =

H g


1w

h1=

20 9.81

u 14.140.84 1

=

1u14.14

20 9.81 0.84=

m/s 21.12u1 =

12 uu =

m/s 21.12 u2 =


1w

f

uV

V θtan

1

1

−=

13.521.1214.41

9.9 θtan =

−=

( ) 97.7813.5tan θ -1 ==


2

f

u

V tan 2=

811.021.12

9.9 tan ==

( ) 39.035 811.0tan -1 ==

Speed of turbine ,N

60

N D πu o

1 =

60

N 5.3 π21.21

=

N3.5 π

60 21.21=

N =66.63 rpm

Speed of turbine, N =66.63 rpm

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

diameter of the runner. The turbine is running at 100 r.p.m. If the vane angle of the

extreme edge of the runner at outlet is 15 and flow ratio 0.6, find:

(i) Diameter of the runner,

(ii) Diameter of the boss, and

(iii) Discharge through the runner.

The velocity of whirl at outlet is given as zero.

Solution:

Given

Head, H = 12 m

Hub diameter →Db


Hub diameter = 0.35 Diameter of runner

Db = 0.35 D0

Speed, N = 100 r.p.m

Flow ratio =0.6

Vane angle at outlet, = 15

To find

(i) Diameter of the runner, Do

(ii) Diameter of the boss, Db

(iii) Discharge through the runner, Q

Solution

2gH

Vratio Flow 1

f=

5.59.812

V6.0 1

f

=

1fV5,59.8126.0 =

m/s 2.9V1

f =

m/s 2.9

VV 12 ff

=

=


2

f

u

V tan 2=

2u

9.2 51tan =

51tan

9.2 u 2 =

m/s 34.33 u 2 =

m/s 4.333

uu 21

=

=

Diameter of runner, Do

60

N D πu o

1 =

60

100 D π33.34 o

=

oD100 π

60 33.34=

Do=6.55 m

Diameter of runner, Do=6.55 m

Hub diameter, Db

Db = 0.35 D0

6.550.35Db =

Db=2.3 m

Hub diameter, Db=2.3 m

Discharge through the runner, Q

Discharge , ( )1f

2

h

2

o V DD4

Q −=

( ) 9.2 3.26.554

Q 22 −=

s/m 77.271 Q 3=

3.A Kaplan turbine runner is to be designed to develop 9100 kW. The net available head is

5.6 m .If the speed ratio=2.09, flow ratio=0.68, overall efficiency=86% and the diameter of

the boss is 1/3 the diameter of the runner . find the diameter of the runner, its speed and

the specific speed of the turbine.

Given Data


= W109100 3

Head, H = 5.6 m

Speed ratio = 2.09

Flow ratio =0.68


Hub diameter or Boss diameter→Db


Boss diameter = 3

1 Diameter of runner

Db = 3

1D0

To find:

i. Diameter of the runner,

ii. Speed of turbine

iii. Specific Speed of turbine

2gH

uratio Speed 1=

6.59.812

u09.2 1

=

1u6.59.81209.2 =

m/s 95.21u1 =

2gH

Vratio Flow 1

f=

6.59.812

V68.0 1

f

=

1fV6.59.81268.0 =

m/s 12.7V1

f =

m/s 12.7

VV 12 ff

=

=

powerWater

powerShaft η , turbineof efficiency Overall =

powerWater

10 91000.6

3=

0.6

10 9100power Water

3=



Q 6.5 9.811000 15166.666 =

Q6.5 9.811000

15166.666 =

/sm 192.5 Q 3=


( )1f

2

b

2

o V DD4

Q −=

3

DD o

b =

9

DD

2

o2

b =

−=−

9

DDDD

2

o2

o

2

b

2

o

2

oD9

11

−=

2

oD9

1-9

=

2

o

2

b

2

o D9

8DD =−

7.12 D9

8

4 192.5 2

o =

2

oD7.128π

94192.5 =

2

oD74.38 =

oD74.38 =

Do=6.21 m

Diameter of the runner Do=6.21 m

3

DD o

b =

m 07.23

6.21D b ==

Speed of turbine ,N

60

N D πu o

1 =

60

N 21.6 π95.21

=

N6.21 π

60 95.21=

N =66.63 rpm



4

5s

H

S.PN N =


S.P. = 9100 kW

4

5s

5.6

910067.5 N =

1.255.6

910067.5 =

rpm 746Ns =

4.A Kaplan turbine develops 24647.6 kW power at an average head of 39 meter .assuming a

speed ratio of 2, flow ratio of 0.6, diameter of the boss equal to 0.35 times the diameter of the

runner and an overall efficiency of 90% , calculate the diameter, speed and specific speed of

specific speed of the turbine.

Given Data

Shaft power, S.P. = 24647.6 kW

= W10 24647.6 3

Head, H = 39 m

Speed ratio = 2.

Flow ratio =0.6


Hub diameter or Boss diameter→Db


Boss diameter = 0.35 Diameter of runner

Db = 0.35D0

To find:

i. Diameter of the runner,

ii. Speed of turbine

iii. Specific Speed of turbine

2gH

uratio Speed 1=

399.812

u2 1

=

1u399.8122 =

m/s 32.55u1 =

2gH

Vratio Flow 1

f=

1fV399.8126.0 =

m/s 59.16V1

f =

m/s 6.591

VV 12 ff

=

=

powerWater

10 24647.60.9

powerWater


3

o

=

=

0.9

10 24647.6power Water

3=


399.812

V6.0 1

f

=


Q 39 9.811000 27386.22 =

Q39 9.811000

27386.22 =

/sm 71.58 Q 3=


( )1f

2

b

2

o V DD4

Q −=

ob D 35.0D =

2

o

2

b D 1225.0D =

( )2

o

2

o

2

b

2

o D 1225.0DDD −=−

( ) 2

oD1225.01 −=

2

o

2

b

2

o 0.8775DDD =−

16.59 D8775.04

71.58 2

o =

2

oD16.598775.0π

471.58 =

2

oD163.6 =

Do=2.5 m

Diameter of the runner Do=2.5 m

ob D35.0D =

m 875.05.235.0D b ==

Speed of turbine ,N

60

N D πu o

1 =

60

N 48.2 π32.55

=

N2.5 π

60 32.55=

N =422.61 rpm



4

5s

H

S.PN N =


S.P. = 24647.6 kW

4

5s

93

24647.6422.61 N =

1.2593

24647.6422.61 =

rpm 76.690Ns =

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DEPARTMENT OF MECHANICAL ENGINEERING - Vel Tech HighTech Fluid... · Cycle Test-IV (Unit 4) 36 V Classification of turbines TB1, RB4 Class room lecture - Black board C203.5 37 V heads