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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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