1

Introduction 1.1 General

The transfer of liquids against gravity existed from time immemorial. A pump

is one such device that expands energy to raise, transport, or compress liquids. The

earliest known pump devices go back a few thousands years. One such early pump

device was called ‘Noria’, similar to the Persian and the Roman water wheels. Noria

was used for irrigating fields.

Figure 1.1 Noria water wheels

The ancient Egyptians invented water wheels with bucket mounted on them to

transfer water for irrigation. More than 2000 years, a Greek inventor, Ctesibius, made

a similar type of pump for pumping water.

Figure 1.2 Model of a piston pump made by Ctesibius

During the same period, Archimedes, a greek mathematician, invented which

is now known as ‘Archimedes screw’- a pump designed like a screw rotating within a

2

cylinder. The spiraled tube was set at an inclination and was hand operated. This type

of pump was used to drain and irrigate the nile valley.

Figure 1.3 Archimedes screw pump

In 4th century Rome, Archimedes’ screw was used for roman water supply

systems, highly advanced for that time. The roman’s also used screw pumps for

irrigation and drainage work.

Screw pumps can also be traced to the ore mines of Spain. These early units

were all driven by either man or animal power.

The mining operations of the middle ages led to the development of the

suction (piston pump), types of which are described by Georgious Agricola in de re

Metallica (1556). Force pumps utilizing a piston and cylinder combination were used

in Greece to raise water from wells.

Adopting a similar principle, air pumps operated spectacular musical devices

in greek temples and amphitheaters, such as the water organ.

As Man’s progress from prehistoric ages to his present civilization has been

accompanied by an ever-increasing use of water, which frequently had to be either

lifted from a lower to a higher level or carried away from one place to another. This

necessity led to the early development of various forms of pumping equipment.

Lifting the water to very high heads leads to development of multistage pump.

Conventionally, Pumps lifts liquid from low level to a higher level using

available mechanical energy. This is achieved by creating a low pressure at the inlet

and high pressure at the outlet of the pump. Due to low inlet pressure the liquid rises

from where it is available and the high outlet pressure forces it up where it is to be

stored or supplied. This is a very restrictive definition for pump. It may be the main

function for a portable water pump, but some pumps do not lift water at all or they do

so only through an insignificant height. Like boiler-feed pump, forced lubrication

pumps, fire fighting pumps, booster pump etc.

3

A more appropriate definition would be that Pump is a mechanical device that

transfers mechanical energy taken from some external source to the liquid flowing

through it raising its hydraulic energy level. The prime movers can be steam, or I.C.

Engine, compressed air, electrical motor, wind or tidal power.

Pumps are widely used for water supply, irrigation, drainage, and water

circulating systems. Process pumps for chemical industries, oil pumps for long pipe

lines, reversible pumps for storage schemes etc. are further examples of their use.

Requirements for each type of job have to be designed accordingly.

1.2 Classification of pumps Pumps can be classified on various bases:

1) On the basis of principle of operation.

2) On the basis of direction of flow.

3) On the basis of the number of stages.

4) On the basis of the number of entrances to impeller.

5) On the basis of the disposition of shaft.

6) On the type of liquid handled.

7) On the type of pump casing.

1.2.1 On the basis of principle of operation Depending on the way in which the energy is transferred to the liquid while it

moves from suction side to the delivery side, pumps can be brooadly classified into

three basic groups:

a) Positive displacement pumps(reciprocating, rotary pumps)

b) Roto-dynamic pumps ( centrifugal pumps)

c) others

In Positive displacement pumps, the liquid is sucked at the suction side,

trapped, pushed, and discharged completely at the delivery end with every revolution.

The displacement of liquid may be by the reciprocating action of a piston or plunger

4

in a cylinder or by rotary action of a gear, cam or vanes in a chamber. In the first case,

the pump is called reciprocating pump while the other one is named rotary pump.

In rotodynamic pumps, energy is continuously added by centrifugal force to

increase the fluid velocity within the machine such that subsequent velocity reduction

within the pump produces pressure energy. The essential element of the machine is

impeller which imparts tangential acceleration to the fluid flowing through it there by

increasing the moment of momentum.

The other types include electromagnetic pumps, jet pumps, air lift pumps,

and hydraulic ram pumps

.

Figure 1.4 Classification of dynamic Pump

5

Figure 1.5 Classification of displacement Pump

6

1.2.2 On the basis of direction of flow a) Radial flow pump: In this type of pump the liquid enters at the center of the

pump and is directed out along the impeller blades in a direction at right

angles to the pump shaft.

b) Axial flow pump: The incoming liquid approaches the wheel in parallel

streams, which are parallel to the axis of rotation. Exit of liquid is also parallel

to the axis.

d) Mixed flow pumps: It borrow characteristics both radial flow and axial flow

pumps. As liquid flows through impeller of a mixed flow pump, the impeller

blades push the liquid out away from the pump shaft and to pump suction at an

angle greater than 90°.

1.2.3 On the basis of the number of stages

This refers to the number of sets of impellers or diffusers in a pump.

a) Single stage centrifugal pump: It has one impeller keyed to the shaft. This

is generally horizontal but can be vertical also.

b) Multi stage centrifugal pump: Two or more impellers are keyed to the

same shaft and single casing is used to enclose them. Pressure is built up in

steps. Water is led through a return channel from outlet of the one stage to

the entrance of next until it is finally discharged.

1.2.4 On the basis of number of entrance to the impeller Impellers of pump are classified based on the number of sides that the liquid

can enter the impeller and also on the amount of webbing between the impeller

blades.

Figure 1.6 Classification of Pumps based on flow direction

7

a) Single suction pumps: It allows the liquid to enter the center of the blades

from only one direction.

b) Double suction pumps: It allows the liquid to enter the center of the

impeller blades from both side simultaneously.

Figure 1.7 Classification of Pumps based on number of impeller

1.2.5 On the basis of the disposition of the shaft This refers to the plane on which shaft axis of the pump is placed. It is either

horizontal or vertical. Usually centrifugal pumps are designed with horizontal shaft.

Vertical disposition of shaft affects economy and therefore suitable for deep wells and

mines.

1.2.6 On the basis of the liquid handled On the basis of the type and viscosity of liquid to be pumped, the pump may

have a closed, semi-open or open impeller.

a) Closed impeller pump: Generally centrifugal pump is equipped with a

closed impeller in which the vanes are covered with shroud on both sides.

This type of pump is meant to handle non-viscous liquid such as ordinary

water, hot water, hot oil, and chemicals like acids etc.

b) Semi-open impeller pump or Non-clog pump: The impeller is provided

with shroud on one side only. This pump is used for viscous liquids such

8

as slurries, sewage water, paper pulp, sugar molasses etc. In order to

minimize the chances of impeller getting clogged, the number of vanes is

reduced and their height is increased. The number of impeller vanes is

usually small.

c) Open impeller pump: The impeller is not provided with any shroud. Such

pumps are used in dredgers, sludge and elsewhere for handling mixtures of

water, sand, pebbles, and clay, in which the solid content may be very

high.

Open semi-open closed

Figure 1.8 Classification of Pumps based on liquid handled

1.2.7 On the basis of the type of casing The pump casings are of two types and the pump is named after the casing it

uses.

a) Volute pump: It has a volute casing into which the impeller discharges

water at a high velocity. Volute is of spiral from and the cross-sectional

area of the moving stream gradually increases from the tongue towards the

delivery pipe. The functions of a volute casing can be summarized as

follows:

i) To collect water from the periphery of the impeller and to transmit to the

delivery pipe at a constant velocity.

ii) To eliminate the loss of head, by making the casing of spiral or volute

form.

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iii) To increase evidently the efficiency of the pump by eliminating the loss of

head due to change in velocity of flow in the volute.

b) Turbine pump or diffusion pump: Impeller surrounded by a guide wheel

consisting of a number of stationary vanes or diffusers providing outlets

with cross-section gradually enlarging towards the periphery. Water

emerges from the impeller flow past the guide vanes and as the section

across flow increases, velocity falls and the pressure is build up. This

arrangement is employed in all multistage pumps. Diffusion pumps may

be either horizontal or vertical shaft type. The vertical type occupies very

little space and is suitable for installation in deep wells. They may be also

used in narrow wells and mines.

1.2.8 On the basis of specific speed of the pumps a) Low specific speed pump e.g. centrifugal pump

b) Medium specific speed pumps e.g. Francis and mixed flow pump.

c) High specific speed pump e.g. axial flow pump

Generally, low value of specific speed represents high head requirements at

moderate or low rates of flow. High specific speed denotes a demand for large flows

against low head. On the other hand for same head- discharge requirements, a higher

specific speed leads to a smaller and most likely, cheaper pump.

Figure 1.9 General shapes of impeller pumps [1]

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1.3 Description of Impeller pumps

A centrifugal pump is a rotating machine in which flow and pressure are

generated dynamically. The inlet is not walled off from the outlet as in the case with

positive displacement pumps, whether they are reciprocating and rotary in

configuration. Rather a centrifugal pump delivers a useful energy to the fluid or

“pumpage” largely through velocity changes of the fluid as this fluid flows through

the impeller and the associated fixed passageways of the pump, that is it is a

rotodynamic pump. All impeller pumps are rotodynamics, including those with radial

flow , mixed flow, and axial flow impellers. The term “centrifugal pump” tends to

encompass all rotodynamic pumps. Rotodynamic pumps are, very simple in

construction. A line diagram of centrifugal pump is shown in fig 1.10. It has an

opening on the suction branch for in flow and another opening for the outflow on the

discharge side. Between the inlet and outlet sections, the flow is enclosed by a closed

casing. In the casing an impeller rotates which transfer mechanical energy through its

blades to the pumped fluid. The driving motor is usually outside the casing except for

the submersible pumps. The shaft extends from the impeller through the casing to the

motor and necessarily a sealing, stuffing box, gland is to be applied to the casing

walls.

Due to the impeller action, the fluid has rotational component downstream of

the impeller. Dynamic action of the impeller on the fluid results in the pressure in

delivery pipe being higher than suction pipe. The pressure (head) is depending upon

the speed of rotation of the impeller.

Figure 1.10 Line diagram of centrifugal pump

11

For some pumps, a diffuser ring is used between the impeller and the casing

for the conversion of kinetic energy into pressure energy. Generally, volute casing is

used for horizontal pumps and a diffuser chamber is used for horizontal pumps and a

diffuser chamber is used for axial and mixed flow vertical pumps.

1.4 Application range of the main pumps Radial pump is used for medium and high heads (above 50 m). It is the

conventional type of the impeller and is used in practically all- multistage machines.

Range of specific speed is generally 10 to 60. When larger volumes must be handled,

double suction impeller may be used. Francis type impeller operates at a higher speed

than the radial pump. Specific speed is slightly higher (30 to 90). This type of

impeller may also be made double suction. The specific speed range of mixed flow

pump is usually 90 to 150. Axial flow pump has the highest specific speed (above

150) and is used for low heads (01 to 12m).

Figure 1.11 Application ranges of main pumps

A-Axial flow Pumps M-Mixed flow pumps R- Radial flow pumps D- Diagonal flow pumps MS- Multi stage pumps

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1.5 MULTISTAGE PUMPS

Multistage pumps are used in installation where a high head is needed. Several

stages are connected in series and the flow is guided from the outlet of one stage to

the inlet of the next. The final head that a multistage pump can deliver is equal to the

sum of pressure each of the stages can provide. The advantage of multistage pumps is

that they provide high head relative to the flow. Like the single-stage pump, the

multistage pump is available in both a vertical and a horizontal version.

Figure 1.12 Vertical multistage in-line pump

Figure 1.13 Horizontal multistage end-suction pump

13

Literature Review

Fluid motion in centrifugal pump is actually a 3-dimensional flow and it is

quite complex for computation. The specific speed makes it possible to approach for

design and to calculate main dimensions of the pump; generally one dimensional

method is used by the designers. A number of text books and published data are

available on centrifugal pump but everyone focus on describing principles rather than

methods. The design process for pumps is dependent on experimental data and

empirical relations despite extensive theoretical knowledge. One of the reason for this

is the great varity of types, sizes and services demanded from them, which in turn,

requires great multiplicity of parameters to define the design problem.

For design theories of centrifugal pumps there are many text books available.

Publications by Gahlot and Nyiri [5], Paresh Girdhir and Octo Moniz [11], Stepanoff

[14], Neumann[9], Chandrakar [2], Panwalkar and Bhaskar [12], Das and Maiti [3],

Church[7] and many others proposed design theories in most satisfactory and

complete form. However, rational step-by-step design procedure is not attempted by

any publication, to assist a designer for appropriate and reasoned design.

Gahlot and Nyiri (1993) [5] also suggested the step-by-step design procedure

for designing of Radial, Francis, and Mixed flow pumps. For Radial flow pump

design, numbers of relationships are given based on practical experience. To construct

vane shape three methods are described in that book.

a) Circular arc method

b) Point-by-point method

c) Conformal representation method

Chapters on working principles of Impeller pumps, Losses and efficiencies

and characteristics of impeller pumps are very handy to understand the theory behind

centrifugal pumps. Various figures showing the relationship between different

parameters of the pumps are provided such as effect of blade thickness, effect of

number of blades on head developed.

Variation of theoretical head with discharge and vane angle, variation of

pressure co-efficient with specific speed, dependence of outlet vane angle and number

of blades on specific speed etc. helps the reader to relate the theory with the design

procedure. Some design co-efficient are given with their range of validity.

14

The computer program principally uses the method outlined by Gahlot and

Nyiri to design the centrifugal pump.

A.L.Chandraker (1996) [2] also proposed a simple approach to calculate the

step by step design procedure of impeller dimensions in his paper. The impeller

dimensions is dependent on the specific speed Ns.

75.0**65.3H

QNN s =

After that various coefficients can

be calculated by using specific speed.

After that it depends on rotational speed N, discharge Q and available head to be

developed per stage H; all in SI units.

Volumetric efficiency

68.0*68.00.1

1−+

=s

v Nη

The hydraulic efficiency is found to correlate with Ns

a) High specific speed machine (N

as follows;

s

25.0''1 )()*01.0(0.1 HQfh −=η

>80)

Where, f1

= 35, 200> N

= 40, Ns < 150

s >150

= 25, Ns>200

Q’

b) Low specific speed machine (N

is discharge in gallons per minute and H’ is head in feet per stage

s

< 80)

( )2172.0ln42.00.1−

−=opt

h dη

Where,

333.0

*25.4

=

NQdopt is referred to as inlet diameter with clear

opening ( )221

2hopt ddd −=

The overall efficiency η0

mhv ηηηη **0 =

depends as

The ηm

03 /*****10 ηρ gHQZP skw−=

is the mechanical efficiency.

The power consumed by pump is expressed as

15

Das, Das, and Maiti (1998) [3] in their work compared the different blade

shape for the given blade outlet angle, inlet and outlet diameters. Outlet blade angle is

assume to be and its common used value and the maximum value is given as 270 and

300. Number of blades is also assumed. To generate the blade shape Point- by- Point

method is used this gives the more accurate blade profile than Circulation method.

Variation of meridional velocity along the impeller passage is assumed to be linear

from inlet to outlet. Value of Cm

21

21

21

21 )*1*2(*)(rr

rCmrCmrr

rCCC mm

m −−

+−

−=

at any radius is calculated by:

A linear variation of W may not be justified because of existence of large

variation between W1 and W2

drCW q += *

. Moreover, an assumption of linearity of W will restrict

to generation of a single blade shape only. Keeping this in mind and also to have

flexibility in generating different values of velocities, variation of W has been chosen

as,

Where, qq rrWWC

21

21

−−

= and qq

qq

rrrWrWd

21

2112 **−−

=

To generate blade shapes for a given blade outlet angle, inlet and outlet

diameters, the parameter q is varied. It has been found that as the value of q is

increased from negative to positive value, the blades gets longer. They have generated

the blade shapes up to q=5, because beyond this value blade length becomes too long.

The loss of head due to channel circulation and friction has been considered as

hydraulic losses in an impeller passage.

Channel friction loss is calculated as )(**

sin** 21

22 UU

gZkh s

c −=βπ

and

Friction loss can be expressed as gW

RLfh

hf *2

**4

*2

1=

The value of ks generally lies in between 0.6 to 0.8 for low pressure pumps

and f1 has been computed from the standard Blasius equation.

16

Church and Lal (1973) [7] cover’s the basic theory and principles of design,

construction and application of centrifugal pump. Design method for impeller and

volute casing both are explained with the help of examples.

Features of impeller design

1. Suggests keeping the velocity at inlet Cm1 in the neighborhood of 2.75 to

3.0 m/s. Range for Cm1

2. Range of leakage loss is 2.0% to10.0% of the delivered flow.

is given from 1.0 to 5.5 m/s

3. Contraction factor at inlet ε1

4. 10 < β

lies in between 0.8 and 0.9.

1 <25, β1

5. Relation proposed to calculate outlet velocity triangle is

is inlet vane angle.

NHD **5.84

2φ

=

Where Ф is overall head coefficient and its value varies between0.9 to 1.2

with an average value very close to unity.

Neumann’s (1991) [9] offers a step-by step procedure of design optimization.

It helps to understand the influence of geometrical parameters of the pump on its

performance. He strongly criticizes the approach of designer’s to confine themselves

to the design point alone and not making conditions at off design process. In the

definition of optimum he mentioned two configurations; one that offers highest

attainable efficiency and other that offers lowest attainable NPSH.

To show inter-relation of each geometrical parameter he introduces various

coefficients and constants, and introduces a 40 step optimization process for impeller

and 66 step optimization process for volute casing associated with various curves and

data tables make it very complex to understand and develop software for it. However,

it offers significant help in designing of impeller end view for the present work.

Neumann has related each impeller dimension with the outlet diameter of the

impeller.

A.S. Panwalkar and C.Bhaskar (1996) [12] proposed a step-by-step procedure

of design of radial and mixed flow pumps. They introduced several coefficients, ratios

and constants. But calculation of impeller outlet diameter is quite complex, as it

requires solving two different equations iteratively.

17

In the paper the guideline is given to find the number of stages. The head per

stage can be taken in the range of 120 m to 150 m. The number of stages is taken

approximately equal to Htotal

HQZKH *

(m)/120 for a speed of 2950 rpm. To calculate percent

internal and external losses he proposes the following relations:

Percent internal loss = [Q is in GPM and His in feet]

For ZKH =40; Nq < 150

ZKH =35; 150<Nq<200

ZKH =25; Nq>200

External hydraulic loss consists of loss due to leakage, disk friction, and

bearing losses. These are inversely proportional to specific speed. Percent external

loss is equal to (11000/Ns).Ns is in British system of units.

A.J. Stepanoff (1967) [14] has also given the design procedure. In its design

procedure, choice of β2

To calculate cavitations constants he proposed following two relations:

is the first step in selecting impeller design constants and its

range is given from 17.5º to 27.5º. Numerous ray diagrams and curves showing the

relation between different parameters help in design process. Curves for overall

efficiency, cavitations constant and relation to calculate shut off head are used in

CAD (software). But it is not possible to accommodate every condition in preliminary

phase of design, because it will only add to the complexity of design and very difficult

to bring in a step-by step design procedure.

6

34

10*3.6 SN

=σ , and

6

34

10*0.4 SN

=σ

[For single suctions and double suction pumps respectively]

Vasandani (1974) [16] proposed a design procedure for radial pumps

consuming 1 to10 H.P. To calculate the overall efficiency, curves from Stepanoff are

used and equation is developed for regression analysis .Like

18

32 *000253.0*04.0*1.25.43 SqSqSq NNN +−+=η The axial velocity at impeller eye and hydraulic efficiency are calculated by

2

20

2

****tan

*325.2γ

βK

nQrVA =

( )ηη −−= 1

321hyd

Assuming β2

( )232 *000000103.0*0000349.0*0078.02488.0125.0

SqSqSq NNNp

+−+−=

= 27.5, he proposed following relationships to calculate Pfleiderer’s

coefficient

Chavan and James proposed the following relationships

( ) 32

343 **10*047.1min QNNPSHR −=

( )11

50*120*025.050*120pp

speedrotationalN −= [p1

( )( )[ ]21032.00 *0474.0log32.0*29.0

*13200194.0 SN

Q−−−=η

number of poles of

electric motor]

Overall efficiency

Though, the design procedure of centrifugal pump design is available in literature, but

the software for it is not readily available. In this thesis an attempt has been made to

write a computer program using C++ for design of pump impeller of a multistage

pump.

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Basic theory

3.1 Kinematics of flow in the pumps In an impeller pump, the energy conversion takes place when the flow pass

through the rotating blade system of the impeller. Thus, the principle factor

characterizing the work done by these pumps is the flow structure. Velocities and

acceleration flow lines are also very important kinematic characteristics. The relative

and absolute paths of fluid particle are shown in figure 3.1. The path of particle

relative to the impeller is seen to follow the blade as if the impeller is not rotating. On

the other hand, the path of the particle relative to a stationary point is drawn as

absolute path [5].

Figure 3.1 Absolute and relative paths of motion [5]

The relative velocity of flow is considered relative to the impeller. The

absolute velocity of flow is taken with respect to the pump casing and is always

equal to the vectorial sum of the relative velocity, and the peripheral velocity of the

impeller, . So

C

= + W

(3.1)

The flow kinematics within the impeller is based on velocity triangle. They

can be drawn for any point but mostly attention is focused on the inlet and the outlet

part of the impeller vane.

It is necessary to know the shape and dimension of the impeller for plotting

the velocity triangle. The pump operating conditions should also be known. Figure 3.2

shows the inlet and outle velocity triangles.

C

W

U

U

20

Figure 3.2 Inlet and outlet velocity triangles

Velocity triangles are formed by three velocity components namely:

1) Relative velocity of liquid with respect to the rotating impeller

2) Peripheral velocity of impeller

3) Absolute velocity of liquid with respect to the casing, which is the vector

sum of the above two components.

In impeller design velocity triangles play a very important role. Since flow

through impeller is rather difficult to be analyzed practically, velocity triangles help in

studying the phenomenon through impeller.

3.2 Theoretical Head of the centrifugal pump An expression for the theoretical head of a centrifugal pump is obtained by

applying the principle of angular momentum or moment of momentum to the mass of

liquid going through the impeller passage, known as the EULER’S equation which

forms the basis of pump design, is based on following assumption:

1) The flow through the impeller is in one radial plane i.e. a plane

perpendicular to the axis of pump and passing through the particle.

2) The vanes are symmetric with respect to the axis such that the fluid enters

the impeller tangentially and leaves also tangentially. Fluid enters and

leaves the impeller without any turbulence.

3) The impeller passages are completely smooth and filled with liquid at all

times.

4) There are infinite numbers of blades.

But none of the above assumptions are true.

21

The IDEAL HEAD developed by the pump is expressed as

gCUCUH UU

i1122

,** −

=∞ (3.2)

OR

gWW

gUU

gCCHi *2*2*2

22

21

21

22

21

22

,−

+−

+−

=∞ (3.3)

The direction of velocity at inlet to the impeller depends on the conditions at

inlet. If there are no guide vanes provided before entry to the impeller to change the

direction of inflow of liquid , it may be assumed that the velocity C is directed along

the radius i.e. α = 900 and Cm1 = C

3.3 Actual head equation

1 .

The ideal head relationship has been derived with the assumption that there is

no friction, turbulence etc. It is also assumed that the flow is completely guided by the

infinite number of blades of no wall thickness. As these assumptions are not truly

adhered, the actual head and power developed are lower than the ideal head. Effect of

various phenomena will be discussed here.

3.3.1 Influence of finite number of blades Euler’s equation was derived on the assumption that the flow through the

impeller is axisymmetrical and the velocity field is created by infinite number of

contiguous stream surfaces fulfilling the role of blade. But in an actual condition, the

impeller has definite number of blades of finite thickness. This reduces the area

available for the flow through the impeller. To determine the influence of blades on

the velocity field, consider two cylindrical surfaces of radii rI and rII concentric with

the impeller blade as shown in the figure 3.3

22

Figure 3.3 Effect of blade thickness on head developed by pump

23

For centrifugal pumps, with inlet and outlet edges of blade parallel to the

impeller axis, these surfaces are situated close to the edges of vane.

Let the pitch of the blade be t1 on a circle of radius r1and the projection of

blade thickness S1in a direction tangential to the circumference be SU1

11

11 *

UmIm St

tCC−

=

. Then, from

the condition of continuity of flow

Where, Zdt 1

1*π

=

1

11 sin β

SSU =

For a given pump geometry, t1 and SU1

mIm CC *11 ϕ=

are constant, so,

From inlet velocity triangle (figure 3.3c), we find that:

IIImI WCC βα sin*sin* 1 ==

11111 sin*sin* βα WCCm ==

And

CI, CmI, WI is the velocities of the fluid just before the entry to the impeller.

To reduce the effect of blade thickness, the blades at inlet should be tapered and

rounded off.

At outlet, sharpening of the blade tips effects a gradual reduction of the flow

velocity from Cm2 to CmII. From outlet velocity triangle (figure 3.3c)

2

2

2

222 *

φmU

mmIIC

tSt

CC =−

=

The velocity distribution at the impeller outlet is based on the assumption that

CU2, the peripheral component of absolute velocity C, does not change in passing

from a surface of radius r2 to a surface of radius rII . It is clear, now that the blades of

finite thickness produce an increase in the angles α1 and β1 at inlet and a decrease in

the angle α2 and β2 at outlet.

24

3.3.2 Effect of number of blades

3.3.2.1 Pressure distribution To transmit power to the liquid, pressure pf on the front face of the vane

should be higher than pressure pb

Figure 3.4 Effect of pressure distribution

on the back face. Any force exerted by the vane on

the liquid has an equal and opposite reactive force from the liquid. Thus the relative

velocity near the back of the impeller rises as compared to that near the front.

Figure3.4 shows that for a given blade angle at outlet, the head developed is less with

higher meridional velocities. Hence, a higher relative velocity at the back of the vane

will result in lower head and the total integrated head will be lower than that

calculated for an average velocity of flow.

3.3.2.2 Velocity distribution The uniform velocity distribution in the impeller passage as shown in the

figure 3.5 is obtained from the assumption of one dimensional flow. In reality, the

rotation of the impeller results in the higher pressure on the loading surface. This

reduces the velocity. This varying velocity distribution when superimposed on the

uniform velocity distribution results in setting up a rotational flow inside the passage.

25

Figure 3.5 Effect of circulatory flow

The relative velocity distribution through an impeller passage is also affected

by the relative circulation of the liquid due to inertia effect of frictionless liquid

particles. If the impeller passages were closed both at inlet and outlet and the impeller

rotated, a circulation would take place as shown in figure3.6a

Figure 3.6a Effect of relative circulation

At the rim, circulating flow takes place in opposite direction and at the hub in

the same direction as the rotation of the impeller. As, the direction of motion is

imparted by the particles having greater kinetic energy, it results in a decrease in

velocity at outlet i.e. C2’< C2 and at inlet an increase in velocity i.e. C1’ > C1 (see

figure 3.6b) .The torque transmitted from the impeller blade to the liquid reduces,

resulting in the decrease in head developed.

26

Figure 3.6b Velocity triangle modified for circulatory flow

(Note: Dashed lines show diagram without circulatory flow.)

The amount of effect of circulatory flow is dependant on the accuracy of vane

design and the number of vanes. The more the number of blades, the better is the

guidance to the flow. Lesser the circulatory flow, higher is the total head of a pump.

3.3.2.3 Vane angle at outlet

Figure 3.7 Blade angle effect on head

Figure 3.7 shows a rectilinear cascade of an impeller. β1 and β2 are the angles

that the flow makes with the axis of the cascade, just before the inlet and just before

the outlet β1’ and β2’ are the angles that tangent to the centerline of blade makes with

the axis of the cascade. Then,

Actual deflection of the mean flow = β2 - β1

Angle between the tangent at inlet and outlet = β2’ - β1’

As the blade cannot have full restraint on the flow in the mid passage, the flow

deflection is usually less than (β2’ - β1’). So,

(β2 - β1) < (β2’ - β1’)

27

This causes a reduction in head developed by the pump with finite number of

blades.

3.4 Relationship between Ideal head and Theoretical head For an ideal case i.e. an impeller with infinite number of blades, the theoretical

head developed is H

gCU

H Ui

'22

,*

=∞

i,∞

(3.4)

Whereas the actual theoretical head developed with finite number of blades is H

gCU

H Ui

22 *=

i,

(3.5)

The ratio called vane number coefficient or slope factor λ is

'

2

2

, U

U

i

i

CC

HH

==∞

λ (3.6)

The factor does not contribute to any loss of energy.

The value of λ can be determined by Pfleiderer equation

( )

−

+== 21

22

22

2

'2

***211

DDZDk

CC

U

U

λ (3.7)

Where Z is the number of blades.

D1 and D2

k= (0.65~0.85)* : for pumps with volute, and

k=

are inlet and outlet diameters

: for pumps with guide vanes

Pfleiderer introduced it first and gave the following values of λ depending on

the number of blades, Z

Table 3.1- relation between Z and λ

Z 1 2 4 6 10 20 ∞

λ 0.25 0.40 0.572 0.66 0.77 0.87 1.00

)60

1( 2β+

+

601*6.0 2β

28

3.5 Losses and efficiencies Losses in centrifugal pumps help us to know;

1. The nature and magnitude of losses so that measures could be taken to reduce them.

2. To predetermine the head discharge characteristics of a new pump if the

theoretical characteristics of the pump are known.

The losses that occur in the pump alone can be classified into four groups:

a) Hydraulic losses which include skin friction and eddy and separation

losses in the suction nozzle, impeller, volute and discharge nozzle.

b) Leakage is the loss of flow through the clearance between the rotating part

and the stationary part. These are connected with wearing rings or seals,

interstage bushings and balancing devices.

c) Disc friction losses are like an external mechanical loss because it absorbs

mechanical work without reducing the head or the discharge.

d) Mechanical losses which include loss due to friction in the bearings and

stuffing boxes or the mechanical seals. Mechanical loss is relatively more

important in small pumps and at operations at low speed than in large

pumps.

3.5.1 Hydraulic losses Accurate calculation of hydraulic losses in a centrifugal pump is not possible

for many reasons like complex fluid passage, superimposition of vortex motion on the

straight flow, different degree of surface roughness, viscosity of liquid etc. Thus these

losses are estimated experimentally. Generally, it is less than 10% except for small

size impellers with narrow passages. Hydraulic loss takes place within the impeller,

the pump casing and suction branch of pump. These losses are made up of skin

friction , and shock and mixing losses of the fluids, as the energy transfer is taking

place along the waterways of the pump, all resulting in head loss.

Hydraulic losses are intimately related to the pump design, kinetic losses are

frequently in conflict with low friction loss but can mostly be related to flow

parameters and to geometry of the pump.

Hydraulic losses of a centrifugal pump can be divided as:

29

1) Entrance loss: These include losses at impeller entrance and exit.

Sometimes they are called shock losses. It depends on the shape of the inlet edge

of the blade. Its value is

025.001.0'1 to

Hh

=

Figure 3.8 Inlet edge geometry of vane

2) Friction loss: Skin friction loss can be calculated using equation

gv

mLfh f *2

**4

*2

=

Where L is the length of the channel

m is the hydraulic radius of channel

v is the velocity at the section

f is the friction coefficient

This equation can be used for passages like suction nozzles, impeller

channel, and discharge volute. However, determination of actual length (L) and

hydraulic radius (m) is difficult in many cases. Hence, some investigators have

suggested a simplified equation. Where K1 is a constant for a given pump,

considering all the unknown factor including lengths, areas, area ratio, and friction

factors.

3) Loss in diffuser: Loss in the diffuser occur because the velocity of fluid

entering the diffuser vanes is an average value which is approximately 0.9*CU2.

Diffuser efficiency of 0.8 for small pumps and 0.9 for larger and well cleaned

vane pump can be assumed. For calculation of loss in diffuser an equation similar

to can be used Where K2

The ratio of actual head developed by the pump is less than the virtual

head, mainly due to friction and turbulence losses. The ratio of the actual head

is a constant for a pump.

30

developed to the virtual head for a finite number of vanes is the hydraulic

efficiency.

Actual measured head

ηhyd

Hydraulic efficiency can be calculated by the following expressions

= ____________________

Head imparted to fluid by impeller

1. Hydraulic efficiency

ohyd ηη = Where ηo

2. Hydraulic efficiency by W.K. Jekat

Overall efficiency of the pump

25.08.01 Qhyd −=η

This is purely an empirical equation (Q to be expressed in US gpm)

Modern centrifugal pumps have hydraulic efficiency ranging between80 to 97

3.5.2 Leakage losses Due to the pressure difference between the inlet and outlet of impeller, some

of the fluid comes back on the suction side after being pumped by the impeller. Thus,

the discharge through the impeller is more than the discharge actually delivered. The

leakage has no effect on the head of the pump but lowers the capacity and increases

the brake horse power. Volumetric or leakage loss takes place between rotating and

stationary components of the pump wearing rings, labyrinth seals, mechanical seals,

glands, balancing devices etc.

3.5.3 Disc friction loss The power lost in rotating a disc in a liquid is known as disc friction. The disc

friction loss arising on the outer surface of the impeller shrouds depends on the

following factors:

1) The shape of the facing rotating and stationary surfaces.

2) Surface roughness of these surfaces.

31

3) Leakage flow between these surfaces.

It is mainly due to the eddy loss in the fluid entrapped between the shrouds of

the impeller and pump casing.

Disc friction loss basically represents the power absorbed in overcoming

friction torque of the impeller external surfaces. Part of the disc friction is recovered

in the volute or diffuser in the form of pressure. However, efficiency of pressure

recovery from disc friction pumping effect is low and is usually neglected.

3.5.4 Mechanical losses This includes friction losses in the bearing and packing box. It is difficult to

predict these losses exactly because of small sizes of bearings and stuffing boxes and

the difficulty in measuring them with normal test facilities. Usually they are taken 2-

4% of the brake horse power but for small pumps when stuffing box is under

discharge, this can be as high as 10%.

3.5.5 Efficiencies In pumps, the power input to the pump shaft never gets fully converted into

the useful water horse power because some power is always wasted due to different

types of losses.

The overall efficiency (ηo

PP

PP

PP

PHQ inner

inner

hho ***

===γη

) of the pump is the ratio of the water horse power to

the power given to the pump shaft.

Pinner is the power transmitted to impeller by shaft. The components of the

overall efficiency (η) are the hydraulic efficiency ηh, the volumetric efficiency ηv, and

the mechanical efficiency η

hHH

HH

thhyd +

==η

m .

The hydraulic efficiency is the ratio of actual head developed by the impeller

to its theoretical head. If H is the actual head and h’ is the total hydraulic loss, then

Modern large size centrifugal pumps have hydraulic efficiency ranging from 80-96%

32

The volumetric efficiency of pump is a measure of the amount of leakage

through the clearances. It is given by

Lv QQ

Q+

=η

To increase volumetric efficiency, the leakage through the seals should be

minimized. In modern centrifugal pumps, volumetric efficiency ranges from 96-98%

The mechanical efficiency is the ratio of the gross power (Pinner

PPP

PP minner

m

'−==η

) delivered to

the impeller by the shaft to the input power supplied to the pump shaft (p). It is

written as

Where P’m is the power lost in bearings and stuffing box. Mechanical

efficiency for modern centrifugal pump varies from 92-95%. It depends on the

mechanical characteristics, design and service condition of the bearing s and stuffing

box and surface finish of matching parts.

33

Impeller Design 4.1 Introduction

The impeller of the centrifugal pump converts the mechanical rotation to the

velocity of the liquid. The impeller acts as a spinning wheel in the pump.

In spite of extensive theoretical knowledge of fluid dynamics and a great deal

of research conducted in the past, the design process for pumps still relies heavily

upon experimental data. There are several reasons for it; one of the reasons is the

great variety of types, sizes, and services demanded from them, which in turn requires

a great multiplicity of parameters to define the design problem in hand. From Euler’s

fundamental equation, we know that the total head generated by a pump depends on

many variables like the peripheral velocity U2, the meridional velocity Cm2, outlet

vane angle β2, the number of blades Z etc. A given head can be developed by pump

impeller designed with various values of these variables. Though the head developed

may be approximately same, all the designs may not be good from the point of view

of efficiency and cost of production.

For designing a new impeller, it is necessary to assume some factors that give

relationship between the impeller total head and the capacity at the design point.

Optimal results of design are obtained for a given operating conditions when factors

derived experimentally from available high efficiency pumps are used.

Two separate groups of parameters can be distinguished: the first describing

the hydraulic factors related to fluid motion, and second representing the physical size

shapes of unit components. Both groups are three dimensional which, contributes to

the complexity of design problems.

The various authors proposed the same design procedure fundamentally, but

the procedures differ due to the selection of various parameters. Due to close

interaction that takes place between the geometry of the passageways and the

hydraulic factors, the design process becomes iterative, thus further adding to the

complexity of design task [15].

34

4.2 Phases of pump design A centrifugal pump design has to pass through following phases:

Phase 01: Determination of dimensionless parameters, which define the hydraulic

flow passage. The flow passage includes impeller vanes and volutes.

Phase 02: Geometry of impeller, volute in meridional and blade-to-blade planes.

Phase 03: Analysis of flow through the hydraulic passages and prediction of

performance characteristics of pumps by application of computational techniques

such as CFD packages.

Phase 04: Manufacture of prototype pump.

Phase 05: Testing and finalizing the design.

Various CFD packages are available to predict the pump performance

characteristics within the acceptable limits. Above mentioned first two phases are

necessarily performed in order to reach the third phase. My work is concerned with

the first two phases.

4.3 Design of impeller Design of impeller includes the design of flow passage parameters at the inlet

and outlet. The design procedure uses many dimensionless factors: their values are

based on experimental observation rather than complete theoretical analysis. These

dimensionless factors have been drawn against a single parameter i.e. specific Speed

of the pump. The variation of such dimensionless number against specific speed along

with the basic mathematical relationship for the passage of the pump gives many

parameters, which are required for drawing the flow passage of the pump impeller [1].

4.4 Development of Vane After determining the vane angles and diameters, the next step in designing is

to construct the vane shape. Vane shape should be such that the passage should not be

too long as it increases friction losses, and any change in cross-sectional area should

be gradual in order to avoid turbulence losses.

35

Fig.4.1 Velocities and vane angles plotted against impeller radius

The velocity of the liquid relative to the impeller W and the radial component

of the absolute velocity Cm both at inlet and outlet edges of the vanes are known and

may be plotted against the impeller radius. These points are connected by a straight

line or smooth curve, as shown in fig. Intermediate values may be obtained from these

curves so as not to have any sudden changes to disrupt the flow. When the values of

Cm and W are known, a curve of β may then be plotted against R, since sinβ =Cm

a) Circular arc method

/W.

b) Point-by-point method

c) Hydrodynamic design method

4.4.1 Circular arc method

In this method, the impeller is arbitrarily divided into number of concentric

rings between the radius R1 and R2. The vane may be defined by one or two arcs of a

circle. Two arcs method give better results as compared to the single arc method.

36

Figure 4.2 single arc method of profile construction

The single arc method is illustrated in fig. From point O which is centre of

rotation of impeller, draw a line OK at an angle (β1+β2) from the line OB. Extend the

line BK to meet the ring of radius R1 at point A. From B draw a line at an angle β2

( )1122

21

22

cos*cos**2 ββρ

RRRR−

−=

such that it intersects the bisector of line BA at point G. An arc of a circle of radius ρ

and centre G gives the desired vane profile. The radius ρ may be calculated from the

formula

In the two arc method, the vane shape is constructed by joining arcs of two circles

drawn through points A and B.

4.4.2 Point-by-point method This method of obtaining the vane shape is given by C. Pfleiderer. It is based

on the assump tion that the transition of β1 to β2

Let β be the vane angle at radius R. Then for very small increments in radius

dR we have

depends on the radius R and on

determining the central angle θ for a given Rand β. The values of R and θ constitute

polar coordinates for a given point in the vane surface. A smooth curve drawn joining

all such points gives the central line of the vane.

37

βθ

tan* dRdR =

βθ

tan*RdRd =

Where θ is measured in radians. Integrating from R1

∫=2

1tan*

R

R RdR

βθ

to any radius R,

If θ is measured in degrees, then

∫=

2

1tan*

*180 R

R RdR

βπθ

Above equation is solved by tabular integration assuming finite increment ΔR.

Vanes are generally made of constant thickness allowing certain thickness at

the inlet. A thin sharp edge will give high efficiency when the vane angle matches

with the direction of fluid flow. But at other angular positions it drops rapidly. An

inlet edge which is generally rounded gives somewhat lower maximum efficiency

when the vane angle matches to flow direction but it does not drop so rapidly at other

angles.

4.4.3 Hydrodynamic design method Hydrodynamic design method can be analyzed by the following methods

1. Method of conformal representation.

2. Method of integral equations.

3. Method of singularities.

In method of conformal transformations the physical flow field is conformally

mapped on a plane in which a unit circle corresponds to the cascade of profile.

The vector field in such a mapped region is determined using the inverse

transformations. It is also known as method of error triangles.

38

The method of integral equations makes the use of possibility of expressing

any harmonic function in terms of the values of this along the boundaries. This mode

of determination of any harmonic function connected with the flow leads to a contour

integral in terms of the function to be determined. Solutions of the integral equation

give the solution of the flow problem.

The method of singularity considers the flow past the rectilinear cascade as a

combination of the vector induced by a system of distributed hydrodynamic

singularity superimposed on a constant vector field. The system of distributed

singularity (sources, sinks and vortices) can be introduced only within the profile of

cascade.

The singularity distribution should ensure the following points:

a. The formation of closed streamline.

b. They should accept fluid at the given angle of attack.

c. They must ensure the required maximum profile thickness.

d. They should satisfy Kutta-Joukowsky postulate.

Of the three methods mentioned above, the method of hydrodynamic design

gives much better design as the combination of singularity and their strengths can be

varied more closely. However, this method is very complex and requires solution of

number of complicated mathematical equations.

On the other hand, circular arc method is the simplest one, but is more

graphical. Also, as the complete vane profile is a single or two arcs, the inaccuracies

in design are also more. As such, this method of vane shaping has limited application.

The point-by-point method lies in between the two methods. It is reasonably

simple to work and give fairly good results. The accuracy of the method can be

improved by increasing the number of points between the inlet and outlet sections of

the impeller.

39

4.5 Vane types Type of a vane depends on a parameter called Degree of Reaction (r).

Degree of Reaction:

It is a measure of pump effectiveness to convert kinetic energy into static

pressure head. Its value theoretically varies from 0.0 to 1.0. If Hp is the pressure head

increase, Hc

cpi HHH +=

is the velocity or kinetic energy increase, and then total head rise is

i

p

i

ci

i

p

HH

HHH

HH

r −=−

== 1

2

2

*21

UCr U−=

+=

22

2

tan*1*

21

βUCr m

Variation of r with CU2/U2

Figure 4.3 Showing the variation of r with C

is shown in the following figure 4.3

U2/U

22 UCU =

2

Case –I r = 0.5

This indicates that or 0

2 90=β .The impeller has radial outlet vanes.

The static pressure rise in the impeller is equal to the change in the velocity energy.

Case -II r > 0.5

In this case, CU2 = U2 and β2 < 900. The impeller vanes are backward curved

vanes having values of β2 between 00 and 900. The degree of reaction is more than

half indicating that the static pressure rise in the impeller is more than the change in

kinetic energy.

40

Case -III r < 0.5

Here, CU2 > U2 and β2 > 900. The impeller vanes are forward curved vanes.

The static pressure rise in the impeller is less than the increase in the velocity energy.

In one particular case when r =0, there is no static pressure change in the impeller i.e.

the impeller blades are equipressure blades. Pumps having this type of blades can be

called as Impulse Pumps. In the impulse pumps, the increase in energy is in the form

of kinetic energy only which is converted into pressure energy in the diffuser. The

pumps having degree of reaction more than zero can be called as reaction pumps.

Rotodynamic pumps are always designed as reaction pumps because in an impulse

pump; a large amount of energy gained by the liquid is lost during conversion of

kinetic energy into pressure energy.

Figure 4.4 Showing various shapes of vanes

Table 4.1 Showing various vane type

Case Features Vane type r>0.5 CU2 <U2 , 0<β2<90 Backward curved vanes 0 r=0.5 CU2=U2 , β2 = 90 Radial vanes 0 r<0.5 CU2>U2 ,β2 > 90 Forward curved vanes 0

Generally, backward curved vanes are used for impellers since the velocity

head is minimum for this type of vane. As it is difficult to convert efficiently the

velocity head into pressure head, impellers having radial or forward curved vanes are

less efficient than the backward curved type. Radial and forward curved vanes have

an undesirable passage shape, which is difficult for liquid to follow; this results in

eddy losses.

41

4.6 Steps to design

The design of pump impeller has been done by the procedure given by Gahlot and

Nyiri (1993).

The following are the steps taken to design a multistage radial flow impeller are

given here:

Input required

1) Head Th in meters

2) Discharge Q in cubic meter per second

3) Rotational speed N in rpm

4) Allowable shear stress for the shaft material

5) No of stages are S

Figure 4.5 showing dimensions of radial type impeller

1) Calculate the head for single stage H = STh

42

2) Calculate specific speed using 43

*

H

QNNq = where Q is in m3

3) Calculate overall efficiency of pump depending on the discharge from:

/sec, H is in

meters, N is in revolutions per minute.

i) Up to 400 lpm

=0η 32 *0013283.0*12258.0*1216.46297.15 sqsqsq NNN +−+

ii) 800 lpm 32

0 *000851366.0*0929906.0*53788.39831.25 sqsqsq NNN +−+=η

iii) 2000 lpm 32

0 *00418247.0*0579136.0*59186.2337.40 sqsqsq NNN +−+=η

iv) 4000 lpm

=0η

+

−+3

2

*0001563536.0

*0285638.0*58618.19263.55

sq

sqsq

N

NN

v) 12000 lpm

=0η

+

−+3

2

*72849.9

*0914205.0*17884.13928.65

sq

sqsq

N

NN

vi) ≥ 40000 lpm

=0η

−+

−+

−+

−−

−

61057

453

2

*10*43282.2*10*90367.1

*10*43015.4*00475735.0

*26433.0*37513.751139.8

sqsq

sqsq

sqsq

NN

NN

NN

4) Calculate the power required using o

HQPη

γ **=

5) Power of the prime mover is 20% higher than the power required to pump

43

PPm *20.1=

6) Calculate the shaft diameter

Assuming that only torque is acting, the diameter of the shaft may be

calculated by the equation

τπ ***2*1000*60*16

2 NPdsh =

Where P is the power transmitted in Kw, N is rpm of shaft, τ is the torsional

stress of the shaft material.

7) Hub diameter at the eye side shhi dd *35.1=

8) Hub diameter at the back side shroud shho dd *425.1=

9) Calculate hydraulic efficiency oH ηη =

10) Calculate theoretical head Hth

HH η=

11) Calculate volumetric efficiency 68.0*68.00.11

−+=

sqv N

η

12) Calculate theoretical volume vth

QQ η=

13) Calculate 1CmK using 32 *0000048836.0*000556.0*0215.0099.0

1 sqsqsqCm NNNK +−+−=

14) Calculate average inlet velocity HgKC Cmm **2*11 = ,where g is

acceleration due to gravity its value is 9.81m/sec

15) Inlet velocity through the eye

2

10 *85.0 mm CC =

16) Free area at the eye om

tho C

QA =

17) Cross sectional area of hub, 2*

4 hih da π=

18) Total area oho AaA +='

19) Eye diameter π

'0*4 Ado =

20) Inlet diameter D1 omoth CDkQ **4

* 21

π= is determined by

44

21) Width of the impeller at the inlet 111

1 *** επ mCDQB = where the value of

ε1

22) Tangential velocity at inlet

varies from 0.8 to 0.9.

60** 1

1NDU π

=

23) Inlet vane angle

= −

1

111 tan

UCmβ

24) Vane outlet angle ( ) 82.6ln*2892.52 += sqNβ

25) Number of vanes ( )( )sqNZ ln*57.292.15 −=

26) Calculate 2cmK =

+

−+−3

2

*0000017093.0

*0002046.0*0094.001396.0

sq

sqsq

N

NN

27) Average exit velocity 122 * mcmm CKC =

28) Calculate thmm HgCCU **

tan*2tan*2

2

2

2

2

22 λ

ββ+

+= , where value of

λ values from 1.25 to 1.35.

29) Calculate NUD

**60 2

2 π=

30) Calculate 22

2 ** mCDQB

π=

31) Relative velocity at inlet 111 sin βmCW =

32) Relative velocity at outlet 222 sin βmCW =

33) Calculate CU22

222 tan β

mU

CUC −= by using

34) Calculate C2( )2

22

22 mU CCC += by using

35) Calculate cavitations coefficient ( )[ ]5752.8ln*31487.1 −= sqNeσ

36) Calculate NPSH, NPSH= σ * H

37) Calculate degree of reaction 346.21*219.161−=r = 0.62

45

Figure 4.6 showing dimensions of guide vane of multistage pump

38) Calculate mean velocity in the volute HgKC CM **2*33 = , where the

value of 3CMK is 0.45.

39) Calculate meridional velocity in the volute αsin*33 CCm = , where the

value of α is 8̊

40) Calculate volute width 1.13 =B * 2B

41) Calculate D4 = 1.45 * D

42)

2

Calculate D3 = D2 + 2 * DS * D2 , where the value of DS

43) Calculate the area of volute at any section A

is taken as 0.12

Ф = 360**

3CQA φ

φ =

4.7 Design Problem In this thesis a computer program has been developed to design a multistage

radial flow impeller for the following data:

1) Total Head 75 meters

2) Discharge 0.015 m3

3) Rotational speed of the impeller is 1480 rpm

/sec

4) Allowable torsional shear stress for the shaft material is 21*106 N/m2

5) No of stages are 3

46

Calculation of the major parameters for the one impeller:

1) Head = 375

= 25 meters

2) Specific speed of the pump is 43

25

015.0*1480=sqN = 16.212

3) Overall efficiency is

=0η

+

−+3

2

212.16*00418247..0212.16*0579136..0212.16*59186.2337.40

= 0.8495 =

84.95%

4) Calculate the power required 8495.025*015.0*81.9

=P = 4.33 kw

5) The power of prime mover is 10 to 20% more, therefore the power of

prime mover is Pm

6) Calculate the diameter of the shaft by

= 1.20*4.33 = 5.196 kw

362 10*21*1480**2

1000*196.5*60*16π

=shd =

0.0201 m

7) Hub diameter at the eye side dhi

8) Hub diameter at the back shroud d

= 1.35*0.0201 = 0.02713 m

ho

9) Calculate the hydraulic efficiency

= 1.425*0.0201 = 0.02864 m

8495.0=hη = 0.92168

10) Calculate theoretical head 92168.025=thH = 27.124 m

11) Calculate volumetric efficiency 68.02127.16*68.00.11

−+=vη =

0.90727

12) Theoretical discharge 90727.0015.0

=thQ = 0.01653 m3

13) Calculate

/sec

=1CMK

+

−+−3

2

212.16*0000048836.0212.16*000556.0212.16*0215.0099.0

=

0.1242

14) Calculate 25*81.9*2*1242.01 =mC = 2.75 m/sec

15) Calculate inlet velocity through eye 75.2*85.0=omC =

47

2.3375 m/sec

16) Free area 3375.201653.0

=oA = 0.00707 m

202714.0*4π

=ha

2

17) Hub area = 0.0000578 m

000578.0007071.0' +=oA

2

18) Total area = 0.007649 m2

π007649.0*4

=od

19) Eye diameter = 0.09868 m

20) Inlet diameter 3375.2**85.001653.0*4

1 π=D = 0.1029 m

21) Width of impeller at inlet 1029.0*75.2*9.0*015.0

1 π=B = 0.0187m

22) Tangential velocity at inlet 601480*1029.0*

1π

=U = 7.973 m/sec

23) Inlet vane angle

=

−

973.775.2tan

1

1β = 19.030

( ) 82.6212.16ln*2892.52 +=β

24) Outlet vane angle = 21.55

( )( )212.16ln*57.292.15 −=Z

0

25) Number of vanes = 9

26) Calculate

=2cmK

+

−+−3

2

212.16*0000017093.0212.16*0002046.0212.16*0094.001396.0

= 0.0919

27) Average exit velocity 25*81.9*2*0919.02 =mC = 2.035m/sec

28) Outlet tangential velocity

( ) ( ) 124.27*3.1*81.955.21tan*2

035.255.21tan*2

035.22

2 +

+

=U =

21.346 m/sec

29) Calculate diameter at outlet 1480*346.21*60

2 π=D = 0.2754 m

30) Impeller width at outlet 035.2*2754.0*015.0

2 π=B = 0.00852 m

31) Relative velocity at inlet ( )

=

03.19sin75.2

1W = 8.433 m/sec

48

32) Relative velocity at outlet ( )

=

55.21sin035.2

2W = 5.54 m/sec

33) ( )55.21tan035.2346.212 −=UC = 16.19 m/sec

34) Calculate 22

2 035.219.16 +=C = 16.317 m/sec

35) Calculate degree of reaction 346.21*219.161−=r = 0.62

36) Calculate cavitations coefficients by ( )[ ]5752.8212.16ln*31487.1 −= eσ =

0.00735

37) Calculate NPSH by NPSH = 0.00735*25 = 0.1838 m

38) Calculate mean velocity in the volute 25*81.9*2*45.03 =C =

9.96 m/sec

39) Calculate meridional velocity in the volute sin 8 =1.387

m/sec

40) Calculate width of volute = 1.1 * 0.00852 = 0.00937 m

41) Calculate 4D = 1.45 * 0.2754 = 0.39933 m

42) Calculate 3D = 0.2754+ 2 * 0.12 * 0.2754 = 0.3414 m

43) Calculate cross-sectional area of volute 360*96.910*015.0

=φA =

0.000041834 m²

*96.93 =mC

3B

49

After calculating the main dimensions of impeller the vane shaping is being

done. For this a linear variation in Cm

−−

+

−−

=21

2112

21

21 ***rr

rCrCrrrCCC mmmm

m

is assumed given by

A linear variation for relative velocity can not be assumed for two reasons:

1) There exists a large variation in relative velocity at inlet and relative velocity at

outlet.

2) A linear variation in W will restrict to the generation of a single shape only.

Hence, for the above said reasons the variation in W is given by the following

equation

−−

+

−−

= qq

qqq

qq rrrWrWr

rrWWW

21

2112

21

21 ***

Various vane shapes can be obtained by giving various values of ‘q’, either positive ,

negative or decimal; except zero because for q = 0

−−

+

−−

= qq

qqq

qq rrrWrWr

rrWWW

21

2112

21

21 *** , becomes an undefined quantity.

In this computer program Point-by-Point method is used for generating the

vane shape.

Increments in radius are calculated by nrrr mm 12 −

=∆ where n is the number of

steps. The results are being tabulated as shown in the following table 4.2

Table 4.2 Method for generating vane shape

Point R Δr C W m

−

=

WC

r

WC

Bm

m

*

12

+

∆=∆ +

2* 1NN BBra

a∆∑ a∆∑= *1800

πθ

θ0 is the span angle. A plot of R and θ generates the required vane shape.

50

Computer Program In this thesis a computer program using C++ has been developed for the particular

design problem which is given below:

Head 75 m

Discharge 0.015 m3/sec

Speed 1480 rpm

Shear stress in the shaft material 21*106 N/m

The computer program developed:

2

1) Should have the appeal to attract the users.

2) Must be written considering the inter relationship of various sub-

modules with main module.

3) Not only shows the error messages, but also prompts the user to take

the corrective action.

4) Should be easy to understand, hassle free to operate and suitable to

demonstrate.

Above mentioned points were considered while writing the program for the

above design problem. In the last decade, advancement in computer technology

motivated many engineers to computerize various design processes, since it allows

handling large amount of data and provides result in a very short time.

In last few years, of the software development projects related to various

design procedures have been written in MANIT are DOS based programs that provide

output in text format and it is very difficult to visualize the actual geometry from the

software results.

51

5.1 FLOWCHART Before the development a computer program it is necessary that all steps

should be very clear and sequenced so that it will not create any problem in the

program designing. This is accomplished by making the flow chart of the whole

problem. From the flowchart one can visualize the various sequences taken. The flow

chart for the design problem is shown below.

START

Read input data Th, N, Q S,τ

H = STh

43

*

H

QNN sq =

oη

oh ηη =

68.0*68.00.11

−+=

sqv N

η

1

52

2

Hth

HH η=

vth

QQ η=

Calculate KCM1

Calculate Cm1

Calculate Com=0.85*Cm1

om

tho C

QA = 2*4 hih da π

=

oho AaA +='

3

53

3

π

'*4 oo

Ad =

omoth CDkQ **4

* 21

π=

1111 *** επ mCD

QB =

60** 1

1NDU π

=

= −

1

111 tan

UCmβ

( ) 82.6ln*2892.52 += sqNβ

4

54

4

( )( )sqNZ ln*57.292.15 −=

Calculate Kcm2

Calculate Cm2

thmm HgCCU **

tan*2tan*2

2

2

2

2

22 λ

ββ+

+=

NUD

**60 2

2 π=

222 ** mCD

QBπ

=

5

111 sin βmCW =

55

5

222 sin βmCW =

2

222 tanβ

mU

CUC −=

( )22

222 mU CCC +=

−=

2

2

*21

UCr U

HgKC CM **2*33 =

αsin*33 CCm =

6

1.13 =B * 2B

56

6

D4 = 1.45 * D2

D3 = D2 + 2 * DS * D2

φA = 360*

*

3CQA φ

φ =

Show results in output form Nsq, ηo, ηh, ηv, Pm, dsh, dhi, dho, D1, B1, beta1, Cm1, D2, B2 ,Cm2, beta2, B3, D3,

C3, Z, Cm3, φA

R1 = D1 / 2, R2 = D2 / 2, Dr = 0.0000

7

57

7

N = 10, q = 1, Cm = Cm1

r = r1, w = w1

Y0 = pow ((Cm / W), 2) Y1 = 1- y0 Y2 = r * (Cm / W) Y3 = sqrt (y1) B = Y3 / Y2

Stop

da = 0, m1 = B , sda = 0.0

Is r < = r2

Theta = (180 / 3.14) * sda

sda = sda + da

9

8

58

Write r, dr, Cm, W, B, da, sda, theta

8 9

dr = (r2-r1) / 10.0 r = r + dr h1 = ( Cm1- Cm2) / (r1-r2) h2 = h1 * r h3 = (( Cm2 * r1) – ( Cm1 * r2 )) / ( r1 -r2) Cm = h2 + h3 h11 = ( w1 – w2 ) / ( r1 - r2) h12 = h11 * r h13 = (( w2 * r1) – ( w1 * r2 )) / ( r1 – r2) w = h12 + h13 Y0 = pow (( Cm / w ), 2) Y1 = 1-y0 Y2 = r * ( Cm / w) Y3 = sqrt (y1) B = y3 / y2 m2 = B m1 = B da = dr * (m1 + m2) / 2.0

59

5,2 ABOUT THE PROGRAM

In this thesis a computer program has been developed in C++. C++ has been

chosen as the programming language because it guides the user at every step when the

program is being run.

#include<iostream.h>

#include<conio.h>

#include<math.h>

#include<iomanip.h>

main()

{

float H,Th,Q,N,T,S;

float g=9.81,gamma=9.81;

clrscr();

cout<<"\n Enter Head : th and No. of stages s";

cin>>Th>>S;

H=Th/S;

cout<<H;

cout<<"\n Enter Rotational Speed N ,Discharge Q and H,T";

cin>>N>>Q>>T ;

float Nsq=(N*sqrt(Q))/(pow(H,(3.0/4.0)));

cout<<"\n Specific Speed of the Pump is "<<Nsq;

float no=40.337+2.59186*Nsq-0.0579136*Nsq*Nsq+0.00418247*Nsq*Nsq*Nsq;

no=no/100;

cout<<"\n Overall Efficiency :" <<no;

float nh=sqrt(no);

cout<<"\n nh : "<<nh;

float nv=1.0/(1.0+.68*pow(Nsq,(-0.68)));

cout<<"\n nv="<<nv*100 <<"%";

float g1=abs((1.31487*log(Nsq))-8.5752) ;

float Pm=(1.20*9.81*.015*25)/.8495;

cout<<"\nPm="<<Pm<<"kw";

double x1=16*60;

double x2=x1*1000;

60

double x3=x2*5.196;

cout<<"\n x="<<x3;

long int k=pow(10,6);

double y=(2*3.14*3.14*1480*21*k);

cout<<"\ny="<<y;

float z1=x3/y;

cout<<"\nz1="<<z1;

float t=1.0/3.0;

float dsh=pow(z1,t);

cout<<"\ndsh="<<dsh<<" m";

float dhi=1.35*dsh;

cout<<"\ndhi="<<dhi<<" m";

float dno=1.425*0.0201;

cout<<"\ndno="<<dno<<" m";

float Hth=H/nh;

cout<<"\n Hth="<<Hth;

float Qth=Q/nv;

cout<<"\nQth="<<Qth;

float Kcm1=(-0.099)+(0.0215*Nsq)-

(0.000556*Nsq*Nsq)+(.0000048836*Nsq*Nsq*Nsq);

cout<<"\nKcm1="<<Kcm1;

float Cm1=Kcm1*sqrt(2*g*H);

cout<<"\nCm1="<<Cm1;

float C0m=.85*Cm1;

cout<<"\nC0m="<<C0m;

float A0=Qth/C0m;

cout<<"\n Free Area A0="<<A0;

float ah=(3.14/4)*(pow(dhi,2));

cout<<"\nah="<<ah;

float t_A0=A0+ah;

cout<<"\n Total Area="<<t_A0;

float d0=sqrt((4*t_A0)/3.14);

cout<<"\n d0="<<d0;

float d1=sqrt((4*Qth)/(.85*3.14*C0m));

61

cout<<"\nd1="<<d1;

float B1=Q/(3.14*d1*Cm1*.9);

cout<<"\n B1="<<B1;

float U1=(3.14*d1*N)/60;

cout<<"\n U1="<<U1;

float b11=Cm1/U1;

//float beta1=tan(b11*180);

float beta1=(Cm1/U1*180/3.14);

cout<<"\nBeta1="<<beta1;

float beta2=5.2892*log(Nsq)+6.82;

cout<<"\nbeta2="<<beta2;

float z=15.92-(2.57*log(Nsq));

cout<<"\n Z="<<z;

float Kcm2=(-0.01396)+(0.0094*Nsq)+(-

0.0002046*Nsq*Nsq)+(0.0000017093*Nsq*Nsq*Nsq);

cout<<"\n Kcm2="<<Kcm2;

float Cm2=Kcm2*sqrt(2*g*H);

cout<<"\n Cm2="<<Cm2;

float k1=pow(((Cm2)/(2*tan(beta2*3.14/180))),2);

cout<<"\n k1="<<k1;

float U2=(Cm2)/(2*tan(beta2*3.14/180))+sqrt(k1+(9.81*1.3*Hth));

cout<<"\n U2="<<U2;

float d2=(60*U2)/(3.14*N);

cout<<"\nd2="<<d2;

float B2=Q/(3.14*d2*Cm2);

cout<<"\nB2="<<B2;

float n4=sin(beta1*3.14/180) ;

float n5=sin(beta2*3.14/180);

float W1=Cm1/n4;

float W2=Cm2/n5;

cout<<"\nW1="<<W1;

cout<<"\nW2="<<W2;

float Cu2=U2-(Cm2/tan(beta2*3.14/180));

cout<<"\n Cu2="<<Cu2;

62

float C2=sqrt((Cu2*Cu2)+(Cm2*Cm2));

cout<<"\nC2="<<C2;

float r=1-(Cu2/(2*U2));

cout<<"\nr="<<r;

float C3=0.45*sqrt(2*g*H);

cout<<"\n C3="<<C3;

float Cm3=C3*sin(8*3.14/180);

cout<<"\n C3="<<C3;

float B3=1.1*B2;

cout<<"\n B3="<<B3;

float d4=1.45*d2;

cout<<"\n d4="<<d4;

float ds;

if(Nsq>20)

{

ds=0.14;

}

else

{

ds=0.12;

}

getch();

clrscr();

float d3=d2+(2*ds*d2);

float e1=Q/C3;

float e2=10.0/360.0;

float ap=e1*e2;

getch();

clrscr();

cout<<"\n ap="<<ap;

getch();

clrscr();

cout<<"\nINPUT DATA ";

cout<<"\n\tHeadTo ="<<Th<<" m";

63

cout<<"\n\tDischarge Q ="<<Q <<"m cube/s";

cout<<"\n\tRotational speed N="<<N;

cout<<"\n\tShear stress t="<<T;

cout<<"\n\t no. of stages S="<<S;

getch();

clrscr();

//cout<<"\n\t Parameter \t Symbol \t Result ";

cout<<"\n\n\n OUTPUT DATA :- ";

cout<<"\n\n\t H=To/S="<<H<<" m"<<"\n\n";

cout<<"\n\t Parameter"<<setw(18)<< "Symbol"<<setw(15)<<"Result ";

cout<<"\n\n\tSpecific speed"<<"

"<<setw(10)<<"Nsq"<<setw(18)<<setprecision(3)<<Nsq;

cout<<"\n\n\tOverall Efficiency"<<" "<<setw(6)<<"no"<<setw(19)<<no;

cout<<"\n\n\tHydraulic Efficiency"<<" "<<setw(4)<<"nh"<<setw(18)<<nh;

cout<<"\n\n\tVolumetric Efficiency"<<" "<<setw(3)<<"nv"<<setw(19)<<nv;

cout<<"\n\n\tPower"<<setw(21)<<"Pm"<<setw(19)<<Pm;

cout<<"\n\n\tShaft diameter "<<setw(12)<<"dsh"<<setw(18)<<dsh;

cout<<"\n\n\tHub Diameter "<<setw(12)<<"dhi"<<setw(18)<<dhi;

cout<<"\n\n\tinlet diameter "<<setw(11)<<"d1"<<setw(19)<<d1;

cout<<"\n\n\tBreadth at inlet\tB1\t "<<setw(12)<<B1;

cout<<"\n\n\tBlade angle at inlet\tbeta1\t "<<setw(6)<<beta1;

cout<<"\n\n\tImpeller inlet velocity\tCm1\t "<<setw(6)<<Cm1;

cout<<"\n\n\tOutlet diameter "<<setw(10)<<"D2"<<setw(19)<<d2;

cout<<"\n\n\tBreadth at Outlet"<<setw(9)<<"B2"<<setw(19)<<B2;

cout<<"\n\n\tImpeller Outlet Velocity"<<"Cm2"<<setw(18)<<Cm2;

cout<<"\n\n\tOutlet blade angle "<<"beta2"<<setw(15)<<beta2;

cout<<"\n\n\tVolute Width "<<setw(9)<<"B3"<<setw(19)<<B3;

cout<<"\n\n\tVolute Base Circle Diameter "<<setw(5)<<"D3"<<setw(19)<<d3;

cout<<"\n\n\tMean Velocity in the Volute "<<setw(5)<<"C3"<<setw(19)<<C3;

cout<<"\n\n\tNumber of vanes "<<"Z"<<setw(20)<<z;

cout<<"\n\n\tMeridonial Velocity in the Volute "<<"Cm3"<<setw(18)<<Cm3;

getch();

clrscr();

cout<<"\n\n\tArea of Volate at any Cross Section "<<"Ap"<<setw(23)<<ap;

64

getch();

clrscr();

float r1,r2;

r1=d1/2.0;

r2=d2/2.0;

float dr=0.000000;

cout<<"\n Point by Point Method";

int n=10,q=1;

cout<<"\n n="<<n<<",q="<<q<<endl;

cout<<"\n R\t dR\t Cm\t W\t B\t dA\t SdA\t Theta";

float Cm=Cm1;

float W,h1,h2,h3,h11,h12,h13;

r=r1;

W=W1;

float B;

float y0,y1,y2,y3;

y0=pow((Cm/W),2);

y1=1-y0;

y2=r*(Cm/W);

y3=sqrt(y1);

B=y3/y2;

float da=0;

float m1,m2;

m1=B;

float sda=0.0;

float theta;

while(r<=r2)

{

theta=(180/3.14)*sda;

cout<<"\n"<<r<<" "<<dr<<" "<<Cm<<" "<<W<<" "<<B<<" "<<da<<"

"<<sda<<" "<<theta;

dr=(r2-r1)/10.0;

r=r+dr;

h1=(Cm1-Cm2)/(r1-r2);

65

h2=h1*r;

h3=((Cm2*r1)-(Cm1*r2))/(r1-r2);

Cm=h2+h3;

h11=(W1-W2)/(r1-r2);

h12=h11*r;

h13=((W2*r1)-(W1*r2))/(r1-r2);

W=h12+h13;

y0=pow((Cm/W),2);

y1=1-y0;

y2=r*(Cm/W);

y3=sqrt(y1);

B=y3/y2;

m2=B;

da=dr*((m1+m2)/2.0);

m1=B;

sda+=da;

}

getch();

}

66

RESULTS

The results of the design problem with the following input data are tabulated

and results of manual calculation and calculation made by the program are compared,

and the vane shape along the mean streamline is generated.

INPUT DATA: Head To =75 m

Discharge Q =0.015 m3/sec

Rotational speed N=1480 rpm

Shear stress t=2.1e+07 N/m

OUTPUT DATA:

2

No. of stages S=3

H=To/S=25 m Parameter Symbol Result Specific speed Nsq 16.213 Overall Efficiency no 0.85 Hydraulic Efficiency nh 0.922 Volumetric Efficiency nv 0.907 Power Pm 5.197 kw Shaft diameter dsh 0.02 m Hub Diameter dhi 0.027 m Inlet diameter D1 0.103 m Breadth at inlet B1 0.019 m Blade angle at inlet beta1 19.785˚ Impeller inlet velocity Cm1 2.752 m/sec Outlet diameter D2 0.276 m

67

Breadth at Outlet B2 0.009 m Impeller Outlet Velocity Cm2 2.036 m/sec Outlet blade angle beta2 21.555˚ Volute Width B3 0.009 m Volute Base Circle Diameter D3 0.342 m Mean Velocity in the Volute C3 9.966 m/sec Number of vanes Z 8.761 Meridional Velocity in the Volute Cm3 1.386 m/sec Area of Volute at any Cross Section AФ 4.181e-05 m²

68

Point by Point Method Where, n=10, q=1

Table 6.1- Point by Point Method

R dR Cm W B dA SdA Theta

0.051466 0 2.751536 8.13294 54.044975 0 0 0

0.060105 0.008639 2.680012 7.874178 45.964275 0.432004 0.432004 24.764553

0.068745 0.008639 2.608488 7.615417 39.899403 0.3709 0.802904 46.02634

0.077384 0.008639 2.536965 7.356655 35.174023 0.32429 1.127194 64.616219

0.086023 0.008639 2.465441 7.097894 31.383404 0.287504 1.414698 81.097343

0.094662 0.008639 2.393917 6.839132 28.270403 0.257683 1.672381 95.86898

0.103302 0.008639 2.322394 6.58037 25.663776 0.232976 1.905357 109.2243

0.111941 0.008639 2.25087 6.321609 23.444994 0.212132 2.117489 121.3847

0.12058 0.008639 2.179346 6.062847 21.529331 0.194273 2.311762 132.5213

0.12922 0.008639 2.107823 5.804085 19.85453 0.178763 2.490525 142.7689

0.137859 0.008639 2.036299 5.545323 18.373762 0.165132 2.655658 152.2351

POINT BY POINT METHODn=10 , q=1

0

20

40

60

80

100

120

140

160

0.05145 0.060075 0.0687 0.077325 0.08595 0.094575 0.1032 0.111825 0.12045 0.129075 0.1377radius

the

ta

Figure 6.1 – Vane Profile

69

Comparative study

Table 6.2- Comparative Study between Manual calculation and Results

Parameter Symbol

Manual Program

Specific speed N 16.212 q 16.213

Overall Efficiency η 84.95 % o 85.00 %

Hydraulic Efficiency η 92.168 % h 92.20 %

Volumetric Efficiency η 90.727 % v 90.70 %

Power P 5.196 kw 5.197 kw

Shaft diameter D 0.0201 m sh 0.020 m

Hub diameter D 0.02713 m h 0.0270 m

Inlet diameter D 0.1029 m 1 0.1030 m

Breadth at inlet B 0.0187 m 1 0.019 m

Blade angle at inlet β 19.031 19.7850 0

Impeller inlet velocity C 2.75 m/sec m1 2.752 m/sec

Outlet diameter D 0.2754 m 2 0.276 m

Breadth at outlet B 0.00852 m 2 0.009m

Impeller outlet velocity C 2.035 m/sec m2 2.036 m/sec

Outlet blade angle β 21.552 21.550 0

Number of vanes Z 9 8.7641

Volute Width B3 0.00937 m 0.009 m

Volute Base Circle Diameter D3 0.3414 m 0.342 m

Mean Velocity in the Volute C3 9.96 m/sec 9.966 m/sec

Meridional velocity in the

volute

C 1.387 m/sec m3 1.387 m/sec

Area of volute at any cross

section

A 0.000041834 m² Φ 0.000041834 m²

70

CONCLUSION Main dimensions of the multistage radial flow pump impeller have been

calculated using the computer program developed in c++ and the vane shapes have

been plotted by taking value of ‘q’ = 1.and it can be seen from this plots that for

positive values of ‘q’ the variation in relative velocity from inlet to outlet is gradual

but the span angle is large indicating a large length of vane and hence a higher head

loss due to friction and increased hydraulic losses.

SCOPE OF FUTURE WORK Nothing in this universe is perfect everything can be improved, In the same

way the computer program developed in this thesis can also be improved. The

improvements in the program can be done in various ways:

1. Calculation of vane length can be incorporated so that the friction losses can

be calculated.

2. The value of ‘q’, the coefficient in the equation of Relative velocity can be

made user defined and various vane shapes can be generated.

3. The parameters that are given fixed values in the program can be made user

defined so that their effect on pump geometry can be studied.

71

BIBLIOGRAPHY

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72

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6.0”, Department of Civil Engineering, M.A.N.I.T., Bhopal.

16. Vasandani V.P. (1974),” Computerized Design of a centrifugal pump”,

Proceeding of 5th

National Conference on Fluid Mechanics and fluid Power.

17. Yashavant P.Kanetkar, “Let us c++”, First edition-1999, BPB Publication

ISBN 81-7656-106-1.

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Figure A-1Variation of k3 and αś with nq

Figure A-2 Values of ΔS / D2 with nq

74

Figure A-3 Relation between Kcm1 and Kcm2 with N

q