chapter_02 centrifugal pump

7/21/2019 Chapter_02 Centrifugal Pump

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Electric Submersible Pumps Mohamed Dewidar 2013

Chapter 2

1

Centrifugal Pumps

Table of Content

Section Content Page

1 Definition 3

2 Theory 4

2.1 Centrifugal force

2.2 Peripheral velocity and head

2.3 Specific gravity3 General concept 10

4 Energy 10

5 Energy equations for an ideal flow 12

6 Horsepower 12

7 The basic equations for centrifugal pumps 14

8 Characteristics of ideal pump and degree of

reaction 18

9 Impeller with finite number of vanes 21

10 Hydraulic losses in pump and plotting

characteristic curve 24

11 Pump efficiency 26

12 Similarity formulas 28

13 Specific speed and its relation to impeller

geometry 31

13.1 Definition

13.2 Specific speed basics

13.3 Specific speed derivation

14 Net positive suction head & cavitation 38

15 NPSH specific speed 41

16 System curve 42

17 Electric Submersible Pumps 44

17.1 Pump stages

17.2 Types of impellers

17.3 Impeller constructions

18 Recommended operating range 50

18.1 Impeller thrust

18.2 Impeller thrust washers

18.3 Shaft thrust

19 Pump configurations 61


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20 Fluid viscosity effect on centrifugal pumps 64

21 Affinity laws 79

22 Friction loss in pipes 81

23 ESP design examples 83


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Centrifugal Pumps

2.1.DefinitionBy definition, a centrifugal pump is a machine that

imparts energy to a fluid. This energy helpsa liquid to

flow, rise to a higher level, or both.

The centrifugal pump is an extremely simple machine. It is

a member of a family known as rotary machines and consists

of two basic parts:

1. The rotary element or impeller fig (2.1)and

2. The stationary element or diffuser fig (2.2).

Fig (2.1) Impeller


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Fig (2.2) Diffuser

The centrifugal pumps function is as simpleas its design.

It is filled with liquid and the impeller is rotated.

Rotation imparts energy to the liquid causing it to exit

the impellers vanes at a greater velocity than when itentered. This outward flow reduces the pressure at the

impeller eye, allowing more liquid to enter. The liquid that

exits the impeller is collected in the diffuser where its

velocity is converted to pressure before it leaves the

pumps discharge.

2.2.Theory2.2.1. Centrifugal Force

A classic example of the action of centrifugal force

is shown in fig (2.3)below. Here, we see a ball of waterswinging in a circle. The swinging ball generates acentrifugal force that holds the water in the ball. Now,


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if a hole is bored in the ball, water will be thrown out.

The distance the stream carries (tangent to the circle) and thevolume that flows out (per unit time) depends upon the

velocity (in ft/sec) of the rotating ball. The faster theball rotates the greater the centrifugal force and

therefore the greater the volume of water discharged and the

distance it carries.

Fig (2.3)

The description above could be considered that of a simple

centrifugal pump. It demonstrates that the flow and head

(pressure) developed by a centrifugal pump depends upon therotational speed and, more precisely, the peripheral

velocity of its impeller (ball).

In the above figure, the string is in tension and this

means it pulls in both directions. The force pulling the

ball towards the middle is the centripetal force and theforce pulling to outward is the centrifugalforce.

Consider the velocity vector before and after point P has

revolved a small angle d,fig (2.4).

The magnitude of v1and v2are equal so lets denoteit simply

as v. The direction changes over a small period of time dtby dradians.

dsis almost the length of an arc of radius r. If the angle

is small, the length of an arc is radius x angle, so it

follows that,


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Fig (2.4)

Example

Calculate the centrifugal force acting on a small mass of 0.5 kg rotating at

1500 rev/minute on a radius of 300 mm.

Solution

= 2N/60 = 2 x x 1500/60 = 157 rad/s

Centrifugal acc. = 2R = (157)2x 0.3 = 7395 m/s2.

Centrifugal force = Mass x acc. = 0.5 x 7395 = 3697 N

2.2.2. Peripheral Velocity & Head

Gravity is one of the more important forces that acentrifugal pump must overcome. You will find that the

relationship between final velocity, due to gravity, and

initial velocity, due to impeller speed, is a very useful

one.

If a stone is dropped from the top of a building it's

velocity will increase at a rate of 32.2 feet per second

for each second that it falls. This increase in velocity

is known as acceleration due to gravity. Therefore if we

ignore the effect of air resistance on the falling stone, we

can predict the velocity at which it will strike the

ground based upon its initial height and the effect ofacceleration due to gravity.

d

V1

V2r

s


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The equation that describes the relationship of velocity,

height, and gravity as it applies to a falling body is:

v2= 2gh

Where:

v= The velocity of the body in ft/sec

g= The acceleration due to gravity @ 32.2 ft/sec/sec (or

ft/sec2)

h= The distance through which the body falls

For example if a stone is dropped from a building 100 feet high:

v2= 2 x 32.2 ft/sec2 x 100 ft

v2

= 6440 ft2/sec2v = 80.3 ft/sec

The stone, therefore, will strike the ground at a velocity of 80.3 feet per

second.

This same equation allows us to determine the initial

velocity required to throw the stone to a height of 100

feet. This is the case because the final velocity of a

falling body happens to be equal to the initial velocity

required to launch it to height from which it fell. In the

example above, the initial velocity required to throw thestone to a height of 100 feet is 80.3 feet per second, the

same as itsfinal velocity.

The same equation applies when pumping water with a

centrifugal pump. The velocity of the water as it leaves

the impeller determines the head developed. In other wordsthe water is thrown to a certain height. To reach this

height it must start with the same velocity it would attain

if it fell from that height.

If we rearrange the falling body equation we get:

h = v2/2g

Now we can determine the height to which a body (or water)

will rise given a particular initial velocity. For example,

at 10 Ft per Sec:

h = 102ft2/sec2/ (2 x 32.2) ft/sec2

h = 100 ft2/sec2/ 64.4 ft/sec2= 1.55 ft

If you were to try this with several different initial

velocities, you would find out that there is an interesting

relationship between the height achieved by a body and its

initial velocity. This relationship is one of theFundamental laws of centrifugal pumps and we will review


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it in detail a little later. As a finale to this section,

let's apply what we have learned to a practical application.

Example:

For an 1800 RPM pump, find the impeller diameter necessary to develop a

head of 200 feet.

First we must find the initial velocity required to develop a head of 100 feet:

v2= 2 gh

v2= 2 x 32.2 ft/sec2 x 200 ft

v2= 12880 ft2/sec2

v = 113 ft/secWe also need to know the number of rotations the impeller undergoes

each second:

1800 RPM / 60 sec = 30 RPS

Now we can compute the number of feet a point on the impellers rim

travels in a single rotation:

113 ft/sec / 30 rotations/sec = 3.77 ft/rotation

Since feet traveled per rotation are the same as the circumference of the

impeller we can compute the diameter as follows:

Diameter = Circumference /

Diameter = 3.77 ft / 3.1416

Diameter = 1.2 ft or 14.4 inch

Therefore an impeller of approximately 14.4" turning at 1800 RPM will

produce a head of 200 Feet.

2.2.3. Specific GravityWhy head is usually expressed in feet?

The Specific Gravity of a substance is the ratio of the

weight of a given volume of the substance to that of an

equal volume of water at standard temperature and pressure

(STP).

Assuming the viscosity of a liquid is similar to that of

water the following statements will always be true

regardless of the specific gravity:

1. A Centrifugal pump will always develop the same head in

feet regardless of a liquids specific gravity.

2. Pressure will increase or decrease in direct proportion


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to a liquids specific gravity.

3. Brake HP required will vary directly with a liquidsspecific gravity.

The Fig (2.5) below illustrates the relationship between

pressure (in psi) and head (in ft) for three liquids of

differing specific gravity.

Fig (2.5)We can see that the level in each of the three tanks is 100

feet. The resulting pressure at the bottom of each varies

substantially as a result of the varying specific gravity.

If, on the other hand we keep pressure constant as measured

at the bottom of each tank, the fluid levels will vary

similarly.

A centrifugal pump can also develop 100' of head when

pumping water, brine, and kerosene. The resulting

pressures, however, will vary just as those seen in the

above Figure. If that same pump requires 10 HP when pumpingwater, it will require 12 HP when pumping brine and only 8

HP when pumping kerosene.

The preceding discussion of Specific Gravity illustrates

why centrifugal pump head (or pressure) is expressed in

feet. Since pump specialists work with many liquids of

varying specific gravity, head in feet is the most

convenient system of designating head.

When selecting a pump, always remember that factory tests

and curves are based on water at STP. If you are working

with other liquids always correct the HP required for thespecific gravity of the liquid being pumped.


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2.3.General ConceptAs previously mentioned the pump is a machine which

imparts head to a fluid.

From the physical aspect, the work of a pump consists in

transforming the mechanical energy of a motor (drive) into

fluid energy. i.e., imparting power to a flow of fluid

passing through it. The energy imparted to the fluid in the

pump enables the former to overcome hydraulic resistances

and rise to a geometric elevation.

The key idea is that the energy created by the centrifugal

force is kinetic energy. The amount of energy given to theliquid is proportional to the velocityat the edge or vanetip of the impeller. The faster the impeller revolves or the

bigger the impeller is, then the higher will be the velocityof the liquid at the vane tip and the greater the energy

imparted to the liquid.

The pump diffuser catches the liquid and slows it down. In

the discharge nozzle, the liquid further decelerates and its

velocity is converted to pressure according to Bernoullis

principle.

Let's outline the steps a liquid encounters as it moves from

suction to discharge.

1.Rotation of the impeller and the shape of the vaneentrances forces liquid to move from the eye of the

impeller into its vanes.

2.During rotation the curved shape of the vanes causesliquid to continue to flow towards the vane exits.

3.This flow causes a partial vacuum at the eye which allowsatmospheric or some other, outside pressure to force more

liquid into the eye thus regenerating the entire process.

4.As liquid flows through the vanes, it gains velocity andreaches its maximum velocity just as it exits the vanes.

5. Upon exiting the vanes, liquid enters the diffuser wheremost of its kinetic energy of motion is transformed into

pressure energy.

2.4.EnergyLiquid can possess three forms of hydraulic energy:

Potential energy due to elevation.

Kinetic energy due to velocity, and,

Pressure energy due to weight or force.

In physics we say that energy can neither be created nordestroyedit can only change its state or form. Therefore


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these three forms of energy must be able to live in harmony

and their quantities, at a given point in time, will follow

a simple law known as The Conservation of Energy. Now, for

our purposes, we can pretty much eliminate potential energybecause there is little or no elevation change from a pump's

suction to its discharge. But as we saw above, a liquid's

velocity undergoes a large change as it moves through the

impeller and diffuser. Since velocity is, in fact, energy it

must be replaced by some other form as it decreases. That

other form of energy is pressure and it arises of it own

accord as the diffuser volume increases and velocity

decreases.

The fig (2.6) below shows a pipe with water flowing from

left to right at Q gpm. As water reaches the center of the

pipe it encounters a section that has a reduced diameter buta short distance away the pipe returns to its original

diameter. Notice the three pressure gauges the one on the

left points to 12 o'clock while the one in the center is at

10 o'clock.

Fig (2.6)

The gauge on the right displays just a little less than the

one on the far left (due to friction loss). In other words

pressure drops as water enters the constricted area of thepipe but it returns to nearly its original pressure as it

exits the constricted area.

What is happening here? Well, if flow is to remain constant

(Q gpm) the velocity of the water must increase as it

travels through the constricted area of the pipe. We see

this in nature when a slow moving river enters and exits a

narrow gorge. And, it confirms our statement about energy as

one form (velocity) increases, another form (pressure) must

decrease and vice versa. Bernoulli's theorem states thatduring steady flow the energy at any point in a conduit is

the sum of the velocity energy, pressure energy, and thepotential energy due to elevation. It also says the sum will


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remain constant if there are no losses. In our example

above, the small loss in pressure seen in the right hand

portion of the pipe is due to increased friction in the

narrow section. Depending upon the circumstances, pressureloss due to friction may be a result of the generation and

dissipation of heat or it could be due to a small increase

in velocity due to a change in laminar flow. As friction

increases laminar flow becomes less symmetrical which,

essentially, reduces the diameter of the conduit.

2.5.Energy equation for an ideal flowThe pumping head is the sum of increase in pressure

head (static head) and the increase in specific kinetic

energy (dynamic head).

The second term (dynamic head) is usually smaller than the

first term (static head) and if the intake and discharge

pipes have the same diameter (d1=d2, whence v1=v2) and 1=

2, it is zero and

The rate of discharge of a pump is also called its delivery

or capacity, denoted Q.

2.6.HorsepowerThe power of a pump is defined as the energy imparted

by the pump to the fluid flow per second.

Horsepower is the horse which could lift 150 pounds a heightof 220 feet in 1 minute.


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Q = m3/sec

H = m

= kg/m3

Q = m3/sec

H = m

= Specific Gravity

Q = GPM

H = Ft


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= Specific Gravity

Like any other driven machine, a pump consumes more power

than its given off. The ratio of the actual power developed

by the pump (water horsepower) to the power supplies by thepump (shaft horsepower) given the efficiency of the pump:

HP = Water HP

BHP = Brake HP (Shaft HP)

Hence, the shaft horsepower

and, taking into account Eq. (2.1).

This is the equation used in choosing pump drives.

Total or overall, pump efficiency takes into account three

types of energy losses:

Hydraulic losses due to friction and turbulence.

Volume losses due to leakage through internal passages.

Mechanical losses due to mechanical friction in bearings,packing, etc.

2.7.The basic equations for centrifugal pumpsA centrifugal pump operates in the following manner;

the principal working unit is a vaned rotor, called

impeller which is made to revolve at high speed. It

accelerates the incoming fluid, increasing both pressure and

absolute velocity of the latter, driving it to the diffuser.

By interaction between the vanes and fluid, the mechanical

energy of the drive is transformed into the energy of flow.

In the diffuser the kinetic energy of the fluid is partly

converted to pressure energy.

The impeller of a centrifugal pump, fig (2.7),consists oftwo disks like walls, called shrouds, one of which ismounted on the shaft. The other shroud, coupled the former

by the vanes, has a hole in the centre, called eye. Thevanes are curved, cylindrical or have complex surfaces. The

fluid enters the impeller along the axis of rotation through

the eye, flows radially outward between the vanes, and is

discharged around the entire circumference into the

diffuser.


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Fig (2.7)

Fig (2.8) Flow pattern through centrifugal pump impeller

The motion of the fluid through the passages between the

vanes can be regarded as consisting of two motions fig(2.8):

Motion of transport (rotation of the impeller).

Motion relative to the impeller.


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Hence, the absolute velocity Cof the fluid can be found asthe vector sum of the peripheral velocity U and therelative velocity W.

Taking a fluid particle sliding along a vane, one canconstruct a velocity parallelogram for the entrance of the

particle to and discharge form, the vane, assuming the

relative velocity W to be tangent to the vane and theperipheral velocity Utangent to the corresponding circle.A similar velocity parallelogram can be constructed for any

point on the vane. The subscript 1 refers to the entrance

section, and subscript 2 to exit section of the vane.

The angle between the vectors of the peripheral and absolute

velocities is, and the angle between the tangent to thevane and the tangent to the circumference is , with the

corresponding subscripts.

In general case the angle changes with the pump

performance, i.e., the speed of rotation nof the impeller(the velocity U)and the discharge Q(the velocityW).

Angle determines the inclination of a vane at every point

and consequently, does not depend on pump performance.

In order to develop the basic equation of centrifugal pump

theory, we shall accept the following two assumptions:

1.The impeller consists of an infinite number of uniform

vanes of zero thickness (Z = , =0) This means that weassume such flow in the passages between the vanes in

which the geometry of all stream tubes in the relative

motion is identical and corresponds exactly to the vane

geometry, and the velocities depend only on the radius

and are the same on a circle of a given radius. This

possible in the case when each differential stream tube

is guided by its vane.

2.The pump efficiency is unity (=1), i.e., there is noenergy losses in the pump and shaft horsepower is

converted completely into water horsepower.

This is possible in the case of an ideal fluid, no

leakage in the pump and no mechanical friction in packing

and bearings.

Thus, to facilitate our theoretical investigation of a

centrifugal pump we have substantially idealized its

performance. We shall call such a centrifugal pump, in which

Z = and = 1, an ideal pump. After considering thetheory of the idealized pump we shall, naturally, proceed to

deal with real pumps.

Let us develop two equations:


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1.The power equation, it means that the power supplied tothe impeller shaft is equal to the energy imparted every

second to the fluid in the pump,

)5.2(HtQT

2. The equation of momentum, it means that the torque actingon the pump shaft is equal to increase in the angular

momentum of the fluid in the impeller per second.

Denoting by r1the radius of the cylindrical surface on

which the entrance edge of the vanes are located, and by

r2the peripheral radius of the impeller, we have,

From Esq. (2.5) and (2.6) the head developed by an idealized

pump is,

This is the basic equation not only for centrifugal pumps

but also for all rotodynamic machines, such as fans,

compressors, and turbines.

Attention should be paid to the fact that the head developed

by an idealized centrifugal pump in terms of column of

pumped liquid does not depend on type of liquid (i.e. on its

specific weight).

As a rule, a liquid entering the impeller has no whirlcomponent. In the vane passages it moves in radial

direction. This means that the vector C1 is pointed along

the radius and the angle 1 = 90O .

Consequently, the second term in Eq. (2.7) vanishes and the

equation takes the form,

Where U2=r2= peripheral velocity at vane exit;

C2u = projection of absolute velocity on direction of

peripheral velocity, i.e. the tangential component of

absolute velocity C2.


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Equation (2.8) shows that for centrifugal pump to deliver

high head the peripheral velocity must be greater and,

secondary, the vector C2u must be large enough i.e.

sufficient whirl should be imparted to the fluid. The formeris achieved by increasing the speed of rotation and impeller

diameter, the later is attained by providing a sufficient

number of vanes of suitable size and shape.

2.8.Characteristics of ideal pump and degree ofreaction

Equation (2.8) is inconvenient for calculation as does

not contain the rate of discharge Q. Therefore let usrewrite it to express the head Ht as a function ofdischarge Qand impeller radius.

From the velocity triangle for the impeller exit, fig (2.9),

C2m = projection of absolute exit velocity on radius i.e.the radial component of vector C2.

The rate of discharge through the impeller can be expressed

in terms of radial component C2m and impeller radius as

follows:

(Fig 2.9) Velocity triangle at impeller exit

Where b2= width of vane slot at exit, hence,

Substitution of this expression into equation (2.9) yields

Substituting the obtained expression (2.11) for the

tangential velocity component U2u in Eq. (2.8) we obtain

another form of the basic ideal pump equation:


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This equation can be used to plot the theoretical

characteristics of an idealized centrifugal pump, i.e.

curve of the head generated by the pump as a function of

the discharge for constant speed of rotation. It is fromequation (1.12) that the characteristic curve of such a

pump is a straight line the inclination of which depends on

the value of the vane angle 2. The following are threepossible cases:

1.Angle 290O, cot

2 is negative and the head Ht

increases with discharge.

These three theoretical pump characteristics are shown in

Fig(2.10).

Fig (2.10) Ideal centrifugal pump characteristics

The corresponding vane shapes and velocity parallelograms

for the same value of U2and C2mare presented in Fig (11.a,b, and c).

Fig (2.11) Vane design and velocity parallelograms


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It thus follows that the optimum head is produced by a

forward curved vane, when 2>90Oand the head is highest. In

practical, however, the efficiency of such pump is low, and

the performance of backward curved vanes at2


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It will be observed from this expression that the greater

C2u/U2and the smaller angle2, the greater the portion of

the head Ht that produced by pressure increase, i.e. the

higher the degree of reaction of the pump. With angle 2

increasing the portion of the head Ht representing the

increase in the kinetic energy becomes grater. The kinetic

energy, in turn, is associated with higher exit velocity of

the fluid from the impeller, which results in considerable

energy losses and lower pump efficiency. That is why it is

not expedient to use vanes with large values of2, i.e.

forward-bent vanes.

It follows from Eq. (2.15) that for radial vanes (2=90O)

the degree of reaction is and at 2


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vanes than for an infinite number because the less the

number of vanes the less the whirl imparted to the fluid by

the impeller. In the absence of vanes (z=0) the whirl is

zero, i.e.C2u

= 0, and the fluid (in the ideal case) issues

from the impeller in a radial direction.

A reduction of the velocity C2uin passing over to a finite

number of vanes is also accounted for by the prerotation

mentioned before. This relative motion gives rise to an

additional absolute velocity C2uat the outer periphery ofthe impeller, Fig 2.12c, which is directed opposite to C2u

and, hence, is subtracted from the latter.

Fig (2.12) Flow pattern through vane passage

Owing to this velocity triangle at the impeller exit

changes, In Fig (2.13), the solid lines give the velocity

vectors for finite number. The construction was made foridentical values of U2 and C2m, i.e. for identicalrotational speeds and rates of discharge. The primed values

are for the case of a finite number of vanes.

A reduction of the tangential component C2uin transition toa finite number of vanes results in a drop in the pumping

head. The head that would have been generated by a pump if

there were no head losses inside the pump is called

theoretical or ideal head denoted byHtz. From Eq. (2.8), we

have

We shall call the ratio of Htz to Ht the vane number

coefficient ():

So, the head in equation is,


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Fig (2.13) Change of velocity triangle in going over to

finite number of vanes

The problem now is to determine the numerical value of .

Obviously, the coefficient depends first and foremost on

the number of vanes z, through it is also affected by thelength of the vanes, which depends on the ratio (r1/r2) and

on the angle of inclination of the vanes2.

Theoretically, investigations reveal that does not depend

on the operation condition of a pump, i.e. on HQ pump, or n.

It is wholly determined by impeller geometry and is

constant for a given impeller.

Without going into the theory of the effect of the number

of the vanes on the head, here is the conclusion of this

theory as represented by a formula for :

Where

(0.55 to 0.65)+0.6 sin2

Here, for example, is the value of for 2=30O and r1/r2=

0.5

z 4 6 8 10 12 16 24

0.624 0.714 0.768 0.806 0.834 0.870 0.908

Thus at z-->, -->1

As the ratio between Htz and Ht is constant for a given

pump, the theoretical characteristic curve for a finite

number of vanes, like the characteristic curve of an

idealized pump with a uniform speed of rotation

(n=constant) is a straight line. At O90

2 , it parallel to


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Chapter 2

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characteristic curve of an idealized pump, and atO

902 it

intersects the later on the axis of abscissas as Htz=0 and

Ht =0 at the same discharge

2

222

cot

2 ubrQ

2.10. Hydraulic losses in pump and plottingcharacteristic curve

As stated earlier, Htz is the head that would have

been developed by a pump if there were no head losses

inside it. The actual head Hpump is less than the

theoretical head by the total losses inside the pump:

Where hpump= total head losses in the pump (at intake, in

impeller and diffuser).

The ratio of the actual head to the theoretical head for a

finite number of vanes is called hydraulic efficiency,

denoted by hthus,

Hydraulic efficiency is always higher than total efficiencyas it takes into account only head losses inside the pump.

It follows from Esq. (2.18) and (2.21) that,

HHH thtzhpump

Where Ht is given by Esq. (2.7) and (2.12)

The hydraulic losses inside the pump hpumpare conveniently

treated as the sum of the following two components.

1.Ordinary hydraulic losses i.e. losses due to frictionand partly to eddy formation inside the pump. As

turbulent flow is the common regime in a centrifugal

pump, this type of head losses increases approximately

as the square of the discharge and can be expresses by

the equation:

Where k1 is constant depending on hydraulic efficiency and

the dimensions of the pump.

2.Shock losses at impeller and diffuser entrance. if therelative velocity W1of the fluid at the entrance to avane passage is tangent to the vane, the fluid is


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entering the impeller smoothly, without shock or eddy

formation. Shock losses in this case are nil. But this

possible only at some definite rated or normal discharge

QO and corresponding radial entrance velocity (C1m)0,(Fig 2.14).

Fig (2.14) velocity parallelogram at impeller entrance

If the actual discharge Q is more or less than the rateddischarge Q

Oand the radial entrance velocity C1mis more or

less than (C1m)0, the relative velocity Wmakes an angle with the tangent to the vane and the fluid past the vane at

some positive or negative angle of approach. The effect is

that of the fluid impinging on the vane, with eddies formingon the opposite side. Thus, energy is degraded in the impact

and eddy formation. The velocity parallelograms for the same

peripheral velocities corresponding to these non-rated

operating conditions are shown by broken lines in Fig

(2.14).

One of the parallelograms corresponds to the inequality Q >Q0the other, to Q < Q0.

Shock losses can be assumed to vary as the square of the

difference between the actual discharge and the discharge

when they are zero, i.e.,

Shock losses at the diffuser entrance are of the same

nature as it the impeller intake, the minimum being at

about the same rate of discharge Q0 and included in thequantity h2.

The total loss of head inside the pump is the sum of the

two losses considered, i.e.,


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The characteristic curves of a pump at uniform rotation

speed (n = constant) are plotted as following:

First draw for H as a function of Q at n = constant the

theoretical characteristic curves for z=and for a finitenumber of vanes Z.

These are straight lines, Fig (2.15). Below the Q axis,plot curves for the two components h1 and h2 of the headlosses in the pump. Summing the coordinates of the two

curves gives the curve hpumpas a function of the discharge.Now, in accordance with Eq.(2.20) subtract hpump from Htz

which gives the curve Hpump = f(Q), i.e. the actualcharacteristic of the pump for a constant speed.

The curve Hpump = f(Q)in Fig (2.15)typical of a centrifugalpump. The maximum value of head Hpumpis commonly obtainedneither at zero discharge nor at Q = Q0 but at someintermediate value of Q.

Plotting the characteristic curves by the method described

is not very accurate in view of the difficulty of

determining the coefficients k1 and k2 in Esq. (2.22) and

(2.23). Therefore the characteristics are commonly plotted

by direct experiment, i.e., in testing a pump.

Fig (2.15)

2.11. Pump EfficiencyThe energy losses in a pump taken into account in

rating the overall efficiency are:

1.Hydraulic losses, examined in the previous section and

determined by hydraulic efficiency [Eq.(2.21)]:

H

hH

H

H

tz

pumptz

tz

pump

h


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Chapter 2

27

2.Volumetric losses, due to leakage through clearance

spaces between impeller and diffuser. The impeller drives

the fluid from the suction pipe to discharge pipe, but

because of the pressure drop it produces some of thefluid leaks back (Fig.2.16)

Fig (2.16) Leakage in pump impeller

The actual discharge of a pump at outlet is Q, and thenthe discharge through the impeller is equal to,

Whereq

= internal leakage

Volumetric energy losses are evaluated by so called

volumetric efficiency

3.Mechanical losses, which include energy degradation due

to friction in packing and bearings as well as surface

friction of the fluid on the impeller. Denoting loss of

power due to friction by HPmand total shaft horsepowerby HP, the mechanical efficiency of a pump is:

The numerator of this expression represents the so called

hydraulic horsepowerand can be expressed by the formula:

Now let us write the expression of the overall efficiency

of a pump as a ratio of the water horsepowerto the shafthorsepower:


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Chapter 2

28

And multiply the numerator of this expression by HPhand the

denominator by the same quantity, using Eq. (2.28):

After re-arranging:

I.e the overall efficiency of a pump is equal to product of

its hydraulic, volumetric, and mechanical efficiencies.

2.12.Similarity FormulasLet us investigate similar operating conditions of

homologous centrifugal pump. Hydrodynamic similarity is

provided by geometric, kinematic, and dynamic similarity.

As applied to centrifugal pumps, kinematicsimilarity meanssimilarity of the velocity triangles constructed for any

corresponding points of the impeller. Dynamicsimilarity is

insured by equality of the Reynolds' numbersof the flowsthrough the pumps in question.

When operating conditions of centrifugal pumps are similar

a proportionality between heads generated and lost, andbetween actual delivery and leakage, is observed. It can

therefore be assumed that homologous pumps have the same

hydraulic and volumetric efficiencies. The mechanical

efficiencies of homologous pumps will generally vary, but

the total efficiency can nonetheless be assumed equal

without much error.

Let us consider similar operating conditions of two

homologous centrifugal pumps. The values referring to the

first pump are denoted by the additional subscript Iand to

the second, by the subscript II, fig (2.17).

Fig (2.17) Similarity of centrifugal pump


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Chapter 2

29

Taking into account that the peripheral velocities of the

impellers are proportional to the number of rpm times the

respective impeller diameter D, the condition for kinematic

similarity at the impeller exit can be written as follows:

( n; U = r; so, U nD)

From Eq. (2.10),

And from geometric similarity

bb

D

D

II

I

II

I

2

2

We can write from Eq. (2.30)

This means that the rate of discharge of homologous pumps

under similar operating condition conditions is proportional

to the rpmand cube of diameter.

From Eq. (2.8), the theoretical heads for an infinite numberof vanes are proportional to the product of the peripheral

and tangential velocities, while the vane-number coefficient

is the same for homologous impeller. Consequently,

Taking into account (2.30),

The actual head generated by the pump is,

HHH tzhpump

But, as (h)I=(h)IIinstead of Eq. (2.32) we can write,

I.e., the actual heads developed by the homologous pumps

under similar operating conditions are proportional to the


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Chapter 2

30

square of the product of the rpm times the impeller

diameter.

From the expression of the water horsepower of the pump, Eq.

(2.2), and develop equations (2.31) and (2.32), we canwrite the relationship between the power generated by

homologous pumps under similar operation conditions:

If we wish to consider similar operating conditions of the

same pumpat different rotational speed n1and n2, the Eqs(2.31), (2.32), and (2.33) become simpler, as Dand are

the same all through and take the form,

Subscripts 1 and 2 are denoting the different rpm values.

Esq. (2.34) and (2.35) are used to compute pump

characteristics for different relative speeds. If therelationship between Hand Q at n1= constant is given, a

similar curve for n2= constant be obtained by computing the

abscissas of the points of the former curve (the rate of

discharge) proportionally to the rpm ratio, and the

ordinates (the head), proportionally to square of that ratio

(Fig 2.18)

Fig (2.18) Pump characteristics for different relative

speeds


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Chapter 2

31

Thus it is possible to calculate and plot the

characteristics of a pump for any desired rotative speed and

to draw a series of characteristic curves for the pump at

different values of n,fig (2.18).

The points A1, A2, A3, A4, on these curves joined by

coordinate relationship given in (2.34) and (2.35) represent

similar operating conditions. The other rows of points (B1,

B2, B3, B4), (C1, C2, C3, C4)...Etc, give a second, third,

etc, rows of similar operating conditions.

It is easy to develop the equations of the curves joining

the points of similar operating conditions. According to

Esq. (2.34) and (2.35) for any row of points we can write,

Q

H

Q

H

Q

H2

3

3

2

2

2

2

1

1

=Const1

Hence, for a row of similar operation conditions we have,

H = Const1Q2

For another row,

H = Const2Q2

Consequently, the points representing similar operating

conditions in an H-Q coordinate system are located on

parabola of second power through the point of origin shown

in Fig (2.18).

2.13. Specific Speed and its relation to impellergeometry

There is a multitude of pump designs that are

available for any given task. Pump designers have needed a

way to compare the efficiency of their designs across a

large range of pump model and types. Pump users also would

like to know what efficiency can be expected from a

particular pump design. For that purpose pump have been

tested and compared using a number or criteria called thespecific speed (NS) which helps to do these comparisons. The

efficiency of pumps with the same specific speed can be

compared providing the user or the designer a starting point

for comparison or as a benchmark for improving the design

and increase the efficiency.

2.13.1.Definition

Specific speed is defined as the speed of an

imaginary pump geometrically similar in every respect to the

actual pump and capable of delivering unit quantity againsta unit head (gpm and ft in US (English) unitsor m3/s and m


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Chapter 2

32

in SI units). Mathematically, the specific speed for a pump

is given by,

Specific speed has dimensions of L3/4/T3/2. The specific speed

of all pumps, similar in shape, is the same regardless of

size.

Specific speed of a pump can be expressed in the non-

dimensional form as,

Where,

Ns*= the specific speed in non-dimensional form

n = impeller speed (rpm)

Q = the discharge rate m3/s or gpm

H = the head in m or ft

g = gravity acceleration (9.81 m/s2or 32.2 ft /s2)

Pumps are traditionally divided into 3 types, radial flow,mixed flow and axial flow. There is a continuous change from

the radial flow impeller, which develops pressure

principally from the action of centrifugal force, to the

axial flow impeller, which develops most of its head by the

propelling or lifting action of the vanes on the liquid.

2.13.2.Specific Speed basics

The specific speed is largely related to the

impeller discharge angle, relative to the inlet. Pumps in

which the discharge of the impeller is directly radial to

the suction; that is, where the flow transitions rapidly

from one plane to the other, have a low specific speed and

are called radial flow impellers. The N will be in theneighborhood of from 500 to 1700(in US units). These pumpswill usually exhibit a low flow to head ratio. At theother end of the scale, the fluid will be discharged from

the impeller along the same axis as it enters. These

impellers (or propellers) have high specific speeds,

generally above 9000(in US units), and are referred to asaxial flow impellers(or propellers) and have a high flowto head ratio.

Between these two extremes fall:


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Chapter 2

33

Mixed Flow impellers, with an NS range of from around4000 to 9000(in US units) begin to transition away from

the suction axis, but discharge between the axial and

radial angles, and generally exhibit a high flow tomoderate head ratio.

Francis Vaned Impellers, between mixed and radial flow,with NS values from around 1700 to 4000. Francis vanedimpellers are frequently discussed in the industry, and

are simply impellers which have vanes curvature such that

the transition from the inlet axis to radial axis is

completed more gradually (sometimes is considered as high

speed radial flow).

The specific speed determines the general shape or class of

the impeller. As the specific speed increases, the ratio ofthe impeller outlet diameter, D2, to the inlet or eyediameter, D1, decreases. This ratio becomes 1.0 for a true

axial flow impeller.

Radial flow impellers develop head principally through

centrifugal force. Pumps of higher specific speeds develop

head partly by centrifugal force and partly by axial force.

A higher specific speed indicates a pump design with head

generation more by axial forces and less by centrifugal

forces. An axial flow or propeller pump with high specific

speed generates it's head exclusively through axial forces.

Radial impellers are generally low flow high head designs

whereas axial flow impellers are high flow low head designs.

2.13.3.Specific Speed (Ns)derivation

The similarity formula obtained in the previous

section can be used to develop a useful practical factor for

calculating and designing centrifugal pumps which is

commonly known as specific speed.

From Eq. (2.31)

Substituting into Eq. (2.32) yields

Rearranging and raising to the power of we obtain


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Chapter 2

34

This expression is valid no only two homologous pumps I and

II but for any number of homologous pumps operating under

similar conditions.

Suppose that among these homologous pumps we have a standardpump whichdelivers a capacity of one GPM at a head of one

foot, soequation 2.37 becomes,

Where:

Nsis the specific speedn is rpmQ is the pump deliveryH is the pump head

The physical meaning of the quantity Ns is the rpm of a

standard pump homologous with a given pump and generating

under similar operating conditions, a head Hs= 1 ft (in US

units)or 1 m (in SI units)at a rate of discharge of Qs= 1

gpm (in US units) or 1 m3/sec (in SI units).

The hydraulic and volumetric efficiencies of the two pump

are the naturally same.Centrifugal pumps may be classified according to the

specific speed (in SI units) as follows:

1.Radial flow (low speed) Ns~ 20-40 D2/D1= 2.2-1.83.Radial flow (high speed) Ns~> 40-70 D2/D1= 1.8-1.34.Mixed flow Ns~> 70-160 D2/D1= 1.3-1.15.Axial flow, or propeller Ns~> 160 D2/D1= 1

The impeller shapes corresponding to these five types are

presented schematically in fig (2.19)

Fig (2.19)


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Chapter 2

35

The specific speedin U.S units is,

Where:

n in rpm

Q in gpm

H in ft

Centrifugal pumps may be classified according to the

specific speed (in US units) as follows:

1.Radial flow impeller (low speed): Ns~ 1700-40003.Mixed flow impeller: Ns~> 4000-80004.Axial flow impeller, or propeller: Ns~> 8000

Specific speed identifies the approximate acceptable ratio

of the impeller eye diameter (D1) to the impeller maximum

diameter (D2) in designing a good impeller.

Ns: 500 to 4000; D1/D2< 0.5 Radial flow pumpNs: 4000 to 8000; D1/D2> 0.5 Mixed flow pumpNs: 8000 to 12000; D1/D2= 1 Axial flow pump

These figures for Ns and D1/D2 ratio are not restrictive,rather, there is a big amount of overlap in the figures as

pump designers push the envelope of operating range of the

different types of pumps.

Then, the impeller shapes corresponding to these five types

are presented schematically in fig (2.19a)

Fig (2.19a)


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Chapter 2

36

The impeller shapes corresponding to the types are presented

schematically in fig (2.19b)for both SI and Us Units

Fig (2.19b)Example:

It is required to have a pump has a capacity of 1500 US

gal/min (0.0944 m3/s)at 100 ft(30.5 m) of head and is

rotating at 1760 rev/min, what type of impeller has to be

used?

Solution

In US (English) units (gpm, and ft)

Ns = 1760 x (1500)0.5 / 1000.75 = 2156 radial flow, fig

(2.20)

In SI (Metric) units (m3/s and m),

Ns = 1760 x (0.0944)0.5 / 30.50.75 = 42 radial flow, (fig

(2.19)

Note

Dividing the US units by 51.64 will yield the SI equivalentspecific speed.

Ns(SI units) = 5156/51.64 = 41.7 =~ 42

Many pump types have been tested and their efficiencymeasured and plotted in Fig (2.20).

Notice that larger pumps are inherently more efficient.

Efficiency drops rapidly at specific speeds of 1000 or less.

Example:

A pump has a capacity of 500 US gal/min at 97 ftof head and

is rotating at1750 rev/min, calculate:

a. Specific speed. b. Define type of impeller

c. Expected efficiency

Solution


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Chapter 2

37

a.Ns= 1750 * 5000.5/ 970.75= 1266

b.From fig (2.19) it is radial flow impeller pump

c.From fig (2.20 and 2.20') the expected efficiency is ~75-78%

Fig (2.20)

Fig (2.20')

The specific speed determines the general shape or class of

the impeller as depicted in Fig. (2.21). As the specificspeed increases, the ratio of the impeller outlet diameter,

D2, to the inlet or eye diameter, D1, decreases. This ratio

becomes 1.0 for a true axial flow impeller.


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Chapter 2

38

High Ns Low Ns

Fig (2.21)

Under certain conditions the specific speed Nscharacterizesthe ability of a pump to develop head and ensure a certain

delivery.

The higher specific speed the less head (for a given Q and

n) and the greater the capacity (for a given Hand n).

Specific speed depends on impeller design. Pumps with low

specific speed have impellers with small relative width

b2/D2 but a high value of D1/D2 i.e. long vanes, which is

necessary to obtain a higher head. Flow through such an

impeller is in a plane perpendicular to the axis of

rotation.

With Ns increasing the ratio D2/D1 (as well as D2/D0)

decreases, i.e. the vanes are shorter and the relative width

of the impeller b2/D2 is greater. Furthermore, the flow

through the impeller departs from the plane of rotation and

becomes increasingly three dimensional.

In the limit, at maximum value of Ns, the flow is along the

axis of rotation and the impeller is of the axial flow type.

2.14.Net Positive Suction Head (NPSH) andCavitations

The Hydraulic Institute defines NPSH as the total

suction head in feet absolute, determined at the suction

nozzle and corrected to datum, less the vapor pressure of

the liquid in feet absolute. Simply stated, it is an

analysis of energy conditions on the suction side of a pump

to determine if the liquid will vaporize at the lowest

pressure point in the pump.

The pressure which a liquid exerts on its surroundings is

dependent upon its temperature. This pressure, called vapor

pressure, is a unique characteristic of every fluid and

increases with increasing temperature. When the vapor


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Chapter 2

39

pressure within the fluid reaches the pressure of the

surrounding medium, the fluid begins to vaporize or boil.

The temperature at which this vaporization occurs will

decrease the pressure of the surrounding medium decreases.

A liquid increases greatly in volume when it vaporizes. One

cubic foot of water at room temperature becomes 1700 cu. ft.

of vapor at the same temperature.

It is obvious from the above that if we are to pump a fluid

effectively, we must keep it in liquid form. NPSHis simply

a measure of the amount of suction head present to prevent

this excess vaporization at the lowest pressure point in the

pump.

NPSH required is a function of the pump design. As the

liquid passes from the pump suction to the eye of the

impeller, the velocity increases and the pressure decreases.

There are also pressure losses due to shock and turbulence

as the liquid strikes the impeller.

The centrifugal force of the impeller vanes further

increases the velocity and decreases the pressure of the

liquid. The NPSH required is the positive head in feet

absolute required at the pump suction to overcome these

pressure drops in the pump and maintain enough of the liquid

above its vapor pressure to limit the head loss, due to the

blockage of the cavitation vapor bubble, to 3 percent. The3% head drop criteria for NPSHrequired is used worldwideand is based on the ease of determining the exact head drop

off point. Most standard low suction energy pumps can

operate with little or no margin above the NPSH required,

without seriously affecting the service life of the pump.

The NPSHrequired varies with speed and capacity within anyparticular pump. Pump manufacturers curvesnormally provide

this information.

NPSHAvailable is a function of the system in which the pump

operates. It is the excess pressure of the liquid in feetabsolute over its vapor pressure as it arrives at the pump

suction. Fig (2.22)shows four typical suction systems with

the NPSH Available formulas applicable to each.

It is important to correct for the specific gravity of the

liquid and to convert all terms to units of feet absolute

in using the formulas.


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Chapter 2

40

Fig (2.22)

PB = Barometric pressure, in feet absolute.VP = Vapor pressure of the liquid at maximum pumping

temperature, in feet absolute.

p = Pressure on surface of liquid in closed suction tank,

in feet absolute.H = Static suction lift in feet (positive or negative).Hf = Friction loss in feet in suction pipe at required

capacity

Cavitationis a term used to describe the phenomenon, whichoccurs in a pump when there is insufficient NPSH Available.

The pressure of the liquid is reduced to a value equal to or

below its vapor pressure and small vapor bubbles or pockets

begin to form. As these vapor bubbles move along the

impeller vanes to a higher pressure area, they rapidly

collapse.The collapse or implosion is so rapid that it may be heard

as a rumbling noise, as if you were pumping gravel. In high

suction energy pumps, the collapses are generally high

enough to cause minute pockets of fatigue failure on the

impeller vane surfaces. This action may be progressive, and

under severe (very high suction energy) conditions can cause

serious pitting damage to the impeller.

The accompanying noise is the easiest way to recognize

cavitation.

Besides possible impeller damage, excessive cavitationresults in reduced capacity due to the vapor present in the

pump. Also, the head may be reduced and/or be unstable and


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Chapter 2

41

the power consumption may be erratic. Vibration and

mechanical damage such as bearing failure can also occur as

a result of operating in excessive cavitation, with high and

very high suction energy pumps.The way to prevent the undesirable effects of cavitation in

standard low suction energy pumps is to insure that the NPSH

Available in the system is greater than the NPSH required

(NPSHR)by the pump. High suction energy pumps require anadditional NPSH margin, above the NPSH Required. Hydraulic

Institute Standard (ANSI/HI 9.6.1) suggests NPSH margin

ratios of from 1.2 to 2.5times the NPSH required, for highand very high suction energy pumps, when operating in the

allowable operating range.

2.15.NPSH Specific SpeedIn designing a pumping system, it is essential to

provide adequateNPSHavailable for proper pump operation.Insufficient NPSH available may seriously restrict pump

selection, or even force an expensive system redesign. On

the other hand, providing excessive NPSH available mayneedlessly increase system cost.

Suction specific speed may provide help in this situation.

Suction specific speed (Nss) is defined as:

Where:

n = Pump speed RPMgpm = Pump flow at best efficiency point at impeller inletNPSHR= PumpNPSHrequiredat best efficiency point.

Experience has shown that 9000 is a reasonable value of

suction specific speed.

Pumps with a minimum suction specific speed of 9000 arereadily available, and are not normally subject to severe

operating restrictions.

Example:

Pump flows 2,000 gpm; head 600 ft at 3550 rpm. What NPSH

will be required?

9000= 3550 x (2000)0.5/ NPSHR3/4

NPSHR3/4= 17.7

NPSHR= 46 ft

A related problem is in selecting a new pump, especially at

higher flow, for an existing system. Suction specific speed


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Chapter 2

42

will highlight applications where NPSHA may restrict pump

selection.

Example:

Existing system: Flow 2000 GPM; head 600 ft.: NPSHA30 ft.

What is the maximum speed at which a pump can be run without

exceeding NPSH available?

9000= n x (2000)0.5/ 303/4

n = 2580 rpm

Running a pump at this speed would require a gear and at

this speed, the pump might not develop the required head. At

a minimum, existing NPSHAis constraining pump selection.

2.16. System curveFor a specified impeller diameter and speed, a

centrifugal pump has a fixed and predictable performance

curve. The point where the pump operates on its curve is

dependent upon the characteristics of the system in which it

is operating, commonly called the System Head Curve...or,

the relationship between flow and hydraulic losses* in a

system. This representation is in a graphic form and, since

friction losses vary as a square of the flow rate, the

system curve is parabolic in shape.

By plotting the system head curve and pump curve together,it can be determined:

1. Where the pump will operate on its curve.

2. What changes will occur if the system head curve or the

pump performance curve changes.

No static head All friction

As the levels in the suction and discharge are the same, Fig(2.23), there is no static head and, therefore, the systemcurve starts at zero flow and zero head and its shape is

determined solely from pipeline losses. The point ofoperation is at the intersection of the system head curve

and the pump curve. The flow rate may be reduced by

throttling valve.


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Chapter 2

43

Fig (2.23) No Static Head - All Friction

Positive static head

The parabolic shape of the system curve is again determined

by the friction losses through the system. But in this case

there is a positive static head involved. This static head

does not affect the shape of the system curve or its

steepness, but it does dictate the head of the system

curve at zero flow rate.

The operating point is at the intersection of the system

curve and pump curve. Again, the flow rate can be reduced by

throttling the discharge valve, Fig (2.24).

Fig (2.24) Positive Suction Head


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Chapter 2

44

2.17. Electric submersible pumps2.17.1. Pump Stages

Electric submersible pumps (ESP) are multi-stagedcentrifugal pumps. Each stage consists of a rotating

impeller and stationary diffuser.

The pressure energy change is accomplished as the liquid

being pumped around the impeller, and as impeller rotates,

the rotating motion of the impeller imparts a rotating

motion to the liquid. Actually, there are two components to

the motion imparted to the liquid by the impeller. One

motion is in a radial direction outward from the center of

the impeller. The motion is caused by centrifugal force. The

other motion moves in direction tangential to outside

diameter of the impeller. The resultant of these twocomponents is the actual direction of flow (as mentioned

earlier in this chapter).

So, the impellers function is to transfer energy by

rotation to the liquid passing through it, thus raising the

kinetic energy of the fluid.

The diffusers function is to change the high velocity

energy into relatively low velocity energy, while inversing

the pressure energy.

The diffuser section then converts the fluid kinetic energyto potential energy, increasing the fluids potential energy

as it passes through the stage.

Fig (2.25) Impeller

Fig (2.26) Pump Stage

Upthrust washer

Downthrust washer

Eye washer


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Chapter 2

45

Fig (2.27) Fluid path through the vanes

2.17.2.Types of ImpellersAs previously mentioned, there are mainly three

types of impellers, the ones using in ESP are, the radial

flow impeller and the mixed flow impeller.The difference between these two types of designs is

described by the pump impeller vane angles and the size and

shape of the internal flow passages.

Radial flow design

A radial flow impeller has, as previously mentioned,

specific speed of 500-4000 (US units) vane angels at close

to 45 degree, and therefore, is usually found in pumps

designed for lower flow rates and high head.

Fig (2.28) Radial flow impeller

Mixed flow design

A mixed flow impeller has, as previously mentioned, specific

speed of 4000-8000 (US units) vane angels at close to 45

degree, and therefore, is usually found in pumps designed

for higher flow rates.


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Chapter 2

46

Fig (2.29) Mixed flow impeller

For more details see item 2.13 of Specific Speed and its

relation to impeller geometry

2.17.3. Impeller constructionsESP pumps come mainly in two basic varieties:

2.17.3.1. Floater Construction.

Each impeller is free to move up and down the

shaft, so it is said to "float" on the shaft.

Fig (2.30) Radial flow impeller

Why use a floater pump?

Let's look at a floating impeller in detail.

Since a floating impeller is free to move up and down the

shaft, the only thing to stop it is either the upper or

lower diffuser. "Thrust washers" are provided at all mating

surfaces between the impeller and diffuser to absorb any

thrust generated.

There is a small amount of

free play in the coupling such

that the pump shaft can fall

down to where the impellers

ride directly on the lower

diffusers or on the downthrust

washers if available.

Impeller is in full

down position


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Chapter 2

47

Thrust

Washers

Fig (2.31) Washers in impeller

The green area shows, fig (2.32), the "up thrust" washer

between the impeller and upper diffuser.

Force

Upthrust is

absorbed

here

Fig (2.32) up thrust force

The green area shows the "down thrust" washers between the

impeller and lower diffuser, Fig (2.33).

Downthrustis absorbed

here

Force

Fig (2.33) down thrust force

We lose efficiency in the down thrust position because of

the fluid's ability to re-circulate from the high pressure

to low pressure eye area. In addition to loss in efficiency,this can promote erosion in the diffuser in abrasive fluids

fig (2.34).


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Chapter 2

48

Fig (2.34) Fluid leakage

"Floater" Pump is normally used because:

Each stage handles its down thrust; a very large number ofstages can be put in a pump without having to worry about

protector bearing capacity.

Very good with mild abrasives since they prevent materialfrom getting into the radial bearing area.

Easier field assembly - no shimming required.

Fig (2.35) Floater Pump

Motor

Protector

Pump

Impeller

Thrust

Is there

any Thrust

seen here?


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Chapter 2

49

2.17.3.2. Compression Construction.

In a compression pump, all the impellers are

rigidly fixed to the shaft so that if an impeller wants to

move up or down, it will take the shaft with it.

The impeller is normally sitting down on its lower diffuser

during assembly due to gravity. Because of this, the pump

shaft is "raised" with shims in the coupling so that the

impeller is not allowed to touch the diffuser after final

assembly. This allows all thrust developed in the pumpshaft to be transferred to the protector shaft directly.

Every impeller is fixed to the shaft rigidly so that it

cannot move without the shaft moving. All the impellers are

"compressed" together to make one rigid body.

Fig (2.36) Pump shimming Ser.675 or larger

Fig (2.37) Pump shimming Ser.562 or smaller


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Chapter 2

50

Note:

Consult Field Service Manual to determine the appropriate

shimming recommended for every type of pump and stage.

Why Use "Compression" Pumps?

Some stages generate too much thrust to be handled by athrust washer in the stage.

Some fluids (e.g., liquid propane) do not have enoughlubricity to properly lubricate a thrust washer.

If abrasives or corrosives are present, it may bebeneficial to handle the thrust in an area lubricated bymotor oil rather than well fluid.

Occasionally in very gassy wells, the flow volume changesso drastically within the pump that parts of a floater

pump could be in very severe thrust while others are notso a compression pump could be one alternative.

Since all the thrust is handled in the protector, as longas the protector has a great enough capacity, the pump

operating range can be extended over a much wider area

without any increased wear or reduced life.

Fig (2.38) Compression impellers

2.18. Recommended Operating Range (ROR)What exactly is the ROR and why is it important?

ROR is the discharge range which the pump must be worked

within it.

All ThrustCarriedhere

Protector

ThrustBearing

Motor

ThrustBearing


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Chapter 2

51

Most people look at pump operating range as defining

"thrust" limits where the stage goes into down thruston theleftside of the range and it goes into up thruston the

rightside of the range. If the stage is within the range,it is thought to be balanced with no net thrust in either

direction.

Graphically it would look like this:

This is almost always wrong.


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Before we worry too much about how much thrust we have, we

need to know what impeller thrust is?

2.18.1. Impeller ThrustUnder normal operating conditions, fluid re-

circulates on top and underneath the impeller.

Fig (2.39) Operation @ BEP

Accordingly, the fluid creates pressure on the upper and

lower impeller shrouds.

Fig (2.40) Pressure acts on the upper and lower shrouds.


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Since the cross sectional area on the upper shroud is

larger, the net force of the pressure is down.

Fig (2.41) Pressure acts on the upper > lower shrouds.

This causes the impeller to be moved down. This is apositive downward force termed Downthrust

Fig (2.42) impeller moved down

At some point where the volume of fluidgoing up into thepump, i.e. increase the momentum due to increase the fluid

velocity (momentum = mv kg.m/s), will lift the impeller up,

overcoming the Downthrust pressure. This causes the

impeller to be moved up. The downward force is now reversed

(negative); it is termed Upthrust


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Fig (2.43) impeller moved up due to upthrust

This creates a larger cross sectional area on the bottom

shroud which increases the pressure underneath, this causes

the impeller to be held up increases the Upthrust.

In normal operation, the pump is designed to operate inslight to moderate Downthrust.

Fig (2.44) the pump, under normal operating condition

Based on the previous explanation, it is concluded that,

there are three forces acting on the impeller. The sum of

these three forces is the total thrust, they are:

Gravity acting on the buoyed mass of the impeller, its

direction is always downward.

The net force resulting from the differential pressureinthe stage its direction is either downward or zero (zero

occurs at wide open flow - no pressure)

The force from themomentumof the fluid coming into thestage its direction is either upward or zero (zero occurs

at shut-in or no flow condition)

Gravity: An Impeller has a mass which is acted on bygravity which pulls it downward toward the Earth.


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Fig (2.45)

Pressure: The pressure times the area equals force. There

are both a downward force and an upward force.

The downward force is always larger except when the pump

generates no pressure (wide open flow).

Fig (2.46)

Momentum: The fluid entering the bottom of the impeller isforced to change direction. This change in momentum exerts

an upward force on the impeller except when there is no flow

(shut-in).

Direction of Fluid Flow

Fig (2.47)

F = mgWhere g is the acceleration due to gravity

F

An impeller adds pressure to the fluidso that the pressure on the top side isgreater than the pressure on the

Low Pressure

High Pressure


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Pressure: The downward arrows represent the larger force

due to difference in surface areas.

Fig (2.48)

In general, larger diameter impellers will have a higher

down thrust than smaller impellers for the same flow rate.

Why?

Because they have a larger surface area on which the

pressure difference can operate. They also have more mass.

Is it possible to affect the down thrust caused by pressurein any way?

What if we could reduce the pressure on top of the impeller?

If we could lower the pressure on the top of the impeller,

this would reduce the thrust.

By using a "balance ring" between the impeller and diffuserand drilling "balance holes" in the upper impeller skirt,we can re-circulate lower pressure fluid over the majority

of the upper surface.

Fig (2.49)

+ =

The net difference between the two

forces is the down thrust due to

pressure.

Low Pressure Fluid


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Conclusion: Under normal operating conditions, fluid

recirculation on the top and bottom side of the impeller

cause forces to be applied on the upper and lower impeller

shrouds. When the recirculation forces are greater on theupper shroud, the impeller is moved down which is termed

downthrust. When the recirculation forces are greater on the

lower shroud, the impeller is moved up which is termed

upthrust. The magnitude recirculation forces depends upon

the flow rate going thru the impeller vs. its head -

capacity, i.e., its operating range. Downthrust increases as

the flow through the stage decreases (or on the left-hand

side of the pump curve). Upthrust increases as the flow

through the stage increases (or on the right-hand side of

the pump curve).

Fig (2.50) Recommended Operating Range

In ROR, most ES pumps are designed to operate in mild tomoderate downthrust.

2.18.2. Impeller Thrust WashersWashers are fitted to the impeller to prevent wear;

there are three types of washers, upthrust, Downthrust, and

eye washers.

Fig (2.51)


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2.18.3. Shaft thrustWe said that the impellers each handle their own

thrust so why do we have to worry about shaft thrust?

The total thrust is made up of two components:

1.The impeller thrust, already discussed in the previoussection, and

2.The shaft thrust.

Shaft Thrust: Compression Pump

In the compression pump, we could not separate the shaftthrust from the impeller thrust since they were coupled

together.

Shaft Thrust: Floater Pump

In a floater, since the impeller can move on the shaft it

makes sense that the shaft can also move within the

impeller.

Forces acting on shaft:

Pressure acting on top of the shaft

Pump shaft mass


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So will the pump discharge pressure times the shaft crosssectional area give us the shaft thrust?

Force = pressure * area

Note: The force, due to the weight of the shaft, is not

usually significant so we will ignore it for now.

Force = Pressure * Area, or

Thrust = Discharge Pressure x Shaft Cross Section Area

A protector serves to equalize the pressure between the

well fluid and the inside of the motor. Therefore the

motor pressure should be roughly equal to the pump intake

pressure.

This is not exactly true since the motor pressure will be

slightly higher due to the weight of a few feet of motor

oil but this increase will be insignificant compared to the

large pressure differential created by the pump.

It can be shown that all pressures cancel out except for

the pressure on the top of the shaft. It can also be shown

that, regardless of the various diameters, couplings, etc.that the net shaft force can be calculated from the

following:

Shaft Thrust: Floater Pum

Pump Shaft

Protector Shaft

Motor Shaft

Protector ThrustRunner


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Force = (Pd-Pi) *As

From this equation, it can be seen that the larger the

shaft diameter, the greater the force for a given pressure

difference.For two pumps doing the same work, the one with the larger

shaft diameter will generate more shaft thrust.

Reviewing our example of the GN2100 which produced 1850

feet of head, this 1850 feet is a true pump differential so

we can plug it directly into the equation. Or can we?

Let: Specific Gravity = 1.0, and convert to psi:

Pressure = 80031.2

0.11850

xpsi [Pd-Pi]

Force = (Pd-Pi) *As= 800 psi * As

Where do we get the area?

**From the catalog.

For a GN2100, the shaft area is 0.6013 sq. in.

Force = 800 psi *0.6013 = 481 lbs

Remember

We neglected the weight of the shaft, and this is usuallysafe to do unless you are very close to fully loading a

protector bearing.

Larger diameter pumps have larger diameter shafts so, forthe same amount of work, they have higher shaft thrust.

On the other hand, larger diameter pumps can and willgenerally use larger diameter protectors, and larger

protectors can handle much higher thrust loads.

But we must consider either the shaft thrust of a floateror the total thrust of a compression pump when selectinga protector, as we will see later.

Operating outside the range requires a larger pump andmotor to be purchased as well as more electricity to

operate it.

One very real concern with establishing operating rangesis the API RPS2 on pump testing:

+/- 5% head and flow

Less than +8% on HP within the operating range

+/- 10% on efficiency at BEP.


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2.19. Pump configurationsESP pumps come in several different configurations.

Most pumps (especially the smaller diameter ones) come as

"center tandems" (or CT type).

Other types are "upper tandems" (UT), "lower tandems" (LT)

and "single" (S) pumps, fig (2.52).

The actual pump stages are no different regardless of what

"type" it is. The difference in the pumps depends on what

it looks like on the ends.

Fig (2.52)

A "single" pump has an

intake and discharge

head intrinsic to the

pump itself. No other

pumps can be attached toit.

An "upper tandem" pump

has a discharge head

but no intake section.

It can be placed on

top of

another pump or anintake section.

A "center tandem" pump

has no intake or is

charge inherent and must

have these provided in

some fashion either with

another pump or an

intake section and/orbolt-on head.

A "lower tandem" pump

has an integral intake

but no discharge head.

It can go below another

pump or else be

completed with a bolt-ondischarge head.


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Single PumpA "single" pump has an intake and discharge

head intrinsic to the pump itself. No other pumps can be

attached to it, fig (2.53).

Fig (2.53)

Upper tandem

The upper tandem has either another pump below it or else

an intake section to complete the assembly, fig (2.54).

Fig (2.54)

Lower Tandem

The lower tandem has either another pump above it or else

a bolt-on discharge to complete the assembly.

Lower tandems are especially common in the largerdiameter, higher flow rate pumps.

Dischar e Head

Intake

Pump


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This helps to reduce entrance losses associated with

higher flow rates and also, in some cases, allows a gas

separator to be built directly into the intake where a

standard add-on separator could not handle the fluidthroughout, fig (2.55).

Fig (2.55)

Center TandemCenter tandem pumps offer the most flexibility. If the

required number of stages for the well cannot fit into a

single section, more sections can be added until the

stage requirement is met.

Since a CT pump can be either a single (with a bolt-on

discharge and intake added) or a part of a larger pump,inventory requirements are greatly reduced, fig (2.56).

Fig (2.56)


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2.20. Fluid Viscosity Effect on Centrifugal Pumps2.20.1.Viscosity

The viscosity of a fluid is that property whichtends to resist a shearing force. It can be thought of as

the internal friction resulting when one layer of fluid is

made to move in relation to another layer.

Consider the model shown in fig (2.57), which was used byNewton in first defining viscosity. It shows two parallel

planes of fluid of area A separated by a distance dx and

moving in the same direction at different velocities V1andV2.

Fig (2.57)

The velocity distribution will be linear over the distance

dx, and experiments show that the velocity gradient, dv/dx,is directly proportional to the force per unit area, F/A.

F = x dv/dxWhere:

is constant for a given liquid and is called its

viscosity.

The velocity gradient, dv/dx , describes the shearing

experienced

by the intermediate layers as they move with respect to each

other.

Therefore, it can be called the "rate of shear", S. Also,

the force per unit area, F/A, can be simplified and calledthe "shear force" or "shear stress," . With these

simplified terms, viscosity can be defined as follows:

= x S

Viscosity = = /S = shear stress / rate of shear

Newton made the assumption that all materials have, at a

given temperature, a viscosity that is independent of the

rate of shear. In other words, a force twice as large would

be required to move a liquid twice as fast. Fluids which

behave this way are called Newtonian fluids. There are, ofcourse, fluids which do not behave this way, in other words


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Chapter 2

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their viscosity is dependent on the rate of shear. These are

known as Non-Newtonian fluids.There are two related measures of fluid viscosity:

1. Dynamic (or absolute) viscosity.

The absolute viscosity of a liquid is defined as the

resistance to flow and shear under the forces of internal

friction as described above. This internal friction is

caused by the resistance of the liquids molecules moving

relative to each other. The larger the molecules, the higher

the internal resistance and consequently, the higher the

viscosity. The commonly units of absolute viscosity is

described below.

In the SI (international system)system the dynamic viscosity

units are N.s/m2 (Newton.Second/meter2), Pa.s (Pascal.sec)or kg/m.s where,

1 Pa.s = 1 N.s/m2= 1 kg/m.s (N = kg.m/s2)

The dynamic viscosity is also often expressed in the

metric CGS(centimeter-gram-second)system as,

g/cm.s, dyne.s/cm2 or poise (p)

where,

1 poise (p) = dyne.s/cm2= g/cm.s

Pa.s = kg/m.s = 1000g/100cm.s = 10 g/cm.s = 10 poise

Or, N = kg.m/s2= 1000 g.100 cm/s2=105g.cm/s2= 105dyne

where dyne = g.cm/s2

Pa.s = N.s/m2 = 105 dyne.s/104 cm2 = 10 dyne.s/cm2 = 10

poise

Pa.s = 10 Poise, then poise = 0.1 Pa.s

For practical use the Poise is to large and it is usual

divided by 100 into the smaller unit called the

centiPoise (cp)where, 1 p = 100 cp

2. Kinematic Viscosity

Kinematic viscosity is a measure of a liquids resistance

to flow and shear under the forces of gravity.

It is the ratio of absolute or dynamic viscosity to density.

Kinematic viscosity can be obtained by dividing the absolute

viscosity of a fluid with its mass density

= /

Where:

= kinematic viscosity

= absolute or dynamic viscosity


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= density

In the SI system the theoretical unit is m2/sor commonlyused Stoke (St) where 1 St = 10-4 m2/sor 1 m2/s = 10,000

Stoke

Since the Stoke is an unpractical large unit, it is usual

divided by 100 to give the unit called Centistokes (cs)

where

1 St = 100 cs, or 1 cs = 10-6m2/s, m2/s = 1,000,000 cs

Seconds Saybolt Universal (SSU) is used also to measure the

kinematic viscosity. The efflux time is Seconds Saybolt

Universal (SSU) required for 60 milliliters of a petroleum

product to flow through the calibrated orifice of a Saybolt

Universal viscometer, under carefully controlled

temperature.

1cs = 0.220 x SSU 180/SSU

3.20.2.Types of liquidThere are two basic types of liquid that can be

differentiated on basis of their viscosities behavior they

are:

1.NewtonianThese are fluids where viscosity is constant and

independent of shear rate, and where the shear rate islinearly proportional to shear stress. Examples are water

and oil

2.Non- NewtonianThese are liquids where t

chapter_02 centrifugal pump

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