Pump Sand Types of Pumps

7/29/2019 Pump Sand Types of Pumps

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PUMPSPUMPSPUMPSPUMPSPUMPSPUMPSPUMPSPUMPS

CHAPTERCHAPTERCHAPTERCHAPTERCHAPTERCHAPTERCHAPTERCHAPTER 1111111111111111

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INTRODUCTIONINTRODUCTIONINTRODUCTIONINTRODUCTIONINTRODUCTIONINTRODUCTIONINTRODUCTIONINTRODUCTION

DESIGNING OF ANY FLUID FLOWING SYSTEM REQUIRES;

1. Design of system through which fluid will flow

2. Calculation of losses that will occur when the fluid flows

3. Selection of suitable device which will deliver enough energyto the fluid to overcome these losses

Devices: Deliver Energy To Liquids/Gases: Pumps/CompressorsPumps/Compressors

TYPES OF PUMPSTYPES OF PUMPS

POSITIVE DISPLACEMENT PUMPSPOSITIVE DISPLACEMENT PUMPS DYNAMIC PUMPSDYNAMIC PUMPS

ROTARY PUMPSROTARY PUMPS

RECIPROCATING PUMPSRECIPROCATING PUMPSCENTRIFUGALCENTRIFUGAL

PUMPSPUMPS

Devices: Extracts Energy From Fluids: TurbinesTurbines


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POSITIVE DISPLACEMENT PUMPS, (PDPS)POSITIVE DISPLACEMENT PUMPS, (PDPS)POSITIVE DISPLACEMENT PUMPS, (PDPS)POSITIVE DISPLACEMENT PUMPS, (PDPS)POSITIVE DISPLACEMENT PUMPS, (PDPS)POSITIVE DISPLACEMENT PUMPS, (PDPS)POSITIVE DISPLACEMENT PUMPS, (PDPS)POSITIVE DISPLACEMENT PUMPS, (PDPS)

WORKING PRINCIPLE AND FEATURES;WORKING PRINCIPLE AND FEATURES;

1. Fixed volume cavity opens

2. Fluid trapped in the cavity through an inlet

3. Cavity closes, fluid squeezed through an outlet

4. A direct force is applied to the confined liquid5. Flow rate is related to the speed of the moving parts of the pump

6. The fluid flow rates are controlled by the drive speed of the pump

. n eac cyc e t e u pumpe equa s t e vo ume o t e cav ty8. Pulsating or Periodic flow

9. Allows transport of highly viscous fluids

10. Performance almost independent of fluid viscosity

11.Develop immense pressures if outlet is shut for any reason,

HENCE

1. Sturdy construction is required

2. Pressure-relief valves are required (avoid damage fromcomplete shutoff conditions)


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PDPS, contd.PDPS, contd.PDPS, contd.PDPS, contd.PDPS, contd.PDPS, contd.PDPS, contd.PDPS, contd.RECIPROCATING TYPE PDPS

Diaphragm pumpsPiston OR Plunger pumps

Single acting piston pump

Single diaphragm pump

Double acting Duplex pump


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ROTARY TYPE PDPSROTARY TYPE PDPSROTARY TYPE PDPSROTARY TYPE PDPSROTARY TYPE PDPSROTARY TYPE PDPSROTARY TYPE PDPSROTARY TYPE PDPS

SINGLE ROTOR MULTIPLE ROTORS

Flexible tube or lining

Gear PumpSliding vane pump

2 Lobe Pump

AND MANY MOREAND MANY MOREAND MANY MOREAND MANY MOREAND MANY MOREAND MANY MOREAND MANY MOREAND MANY MORE

3 Lobe PumpScrew pump

Radial Pump


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DYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPS

WORKING PRINCIPLE AND FEATURESWORKING PRINCIPLE AND FEATURES1. Add somehow momentum to the fluid

(through vanes, impellers or some special design

2. Do not have a fixed closed volume3. Fluid with high momentum passes through open passages and

converts its high velocity into pressure

TYPES OF DYNAMIC PUMPSTYPES OF DYNAMIC PUMPS

ROTARY PUMPSROTARY PUMPS SPECIAL PUMPSSPECIAL PUMPS

Centrifugal PumpsCentrifugal Pumps

Axial Flow PumpsAxial Flow PumpsMixed Flow PumpsMixed Flow Pumps

Jet pump or ejector

Electromagnetic pumps for liquid metalsFluid-actuated: gas-lift or hydraulic-ram


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DYNAMIC PUMPS, contd.DYNAMIC PUMPS, contd.DYNAMIC PUMPS, contd.DYNAMIC PUMPS, contd.DYNAMIC PUMPS, contd.DYNAMIC PUMPS, contd.DYNAMIC PUMPS, contd.DYNAMIC PUMPS, contd.

Jet pump or ejector

Centrifugal PumpsCentrifugal Pumps

hydraulic-ram

1 vane Pump1 vane Pump

Ax a F ow PumpsAx a F ow PumpsMixed Flow PumpsMixed Flow Pumps

Diffuser PumpDiffuser Pump


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COMPARISON OF PDPS AND DYNAMIC PUMPSCOMPARISON OF PDPS AND DYNAMIC PUMPSCOMPARISON OF PDPS AND DYNAMIC PUMPSCOMPARISON OF PDPS AND DYNAMIC PUMPSCOMPARISON OF PDPS AND DYNAMIC PUMPSCOMPARISON OF PDPS AND DYNAMIC PUMPSCOMPARISON OF PDPS AND DYNAMIC PUMPSCOMPARISON OF PDPS AND DYNAMIC PUMPS

CRITERIA PDPS DYNAMIC PUMPS

Flow rate Low, typically 100 gpm As high as 300,000 gpm

Pressure As high as 300 atm Moderate, few atm

Priming Very rarely Always

Flow Type Pulsating Steady

Constant

RPM

any pressure

OR

Flow rate cannot be changed

without changing RPMHence used for metering

Head varies withflow rate

OR

Flow rate changes with

head for same RPM

Viscosity Virtually no effect Strong effects


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CENTRIFUGAL PUMPSCENTRIFUGAL PUMPSCENTRIFUGAL PUMPSCENTRIFUGAL PUMPSCENTRIFUGAL PUMPSCENTRIFUGAL PUMPSCENTRIFUGAL PUMPSCENTRIFUGAL PUMPS

Centrifugal Pumps: Construction Details and Working

1. A very simple machine

2. Two main parts

1. A rotary element, IMPELLER2. A stationary element, VOLUTE

3. Filled with fluid & impeller rotated

Illustration-1

Illustration-2

.

5. Outward flow reduces pressure at inlet,

(EYE OF THE IMPELLER), more fluid

comes in.

6. Outward fluid enters an increasing arearegion. Velocity converts to pressure

Impeller Impart Energy/Velocity By Rotating FluidVolute Converts Velocity To Pressure

Impeller-1

Impeller-2

Impeller-3

Impeller-4

Impeller-5

Impeller-6


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CENTRIFUGAL PUMPS, contd.CENTRIFUGAL PUMPS, contd.CENTRIFUGAL PUMPS, contd.CENTRIFUGAL PUMPS, contd.CENTRIFUGAL PUMPS, contd.CENTRIFUGAL PUMPS, contd.CENTRIFUGAL PUMPS, contd.CENTRIFUGAL PUMPS, contd.

Centrifugal Pumps: Working Principal

1. Swinging pale generates centrifugal force holds water in pale2. Make a bore in hole water is thrown out3. Distance the water stream travels tangent to the circle =f(Vr)

4. Volume flow from hole =f(Vr)

=. , r

A freely falling body achieves a velocity V = (2gh)1/2

A body will move a distance h = V2/2g, having an initial velocity V

OR

Find diameter that will generate V to get required h for given N


11/65


Q. FOR AN 1800 RPM PUMP FIND THE DIAMETER

OF IMPELLER TO GENERATE A HEAD OF 200 FT.

Find first initial velocity V = (2gh)1/2 = 113 ft/sec

Convert RPM to linear distance per rotation

1800 RPM = 30 RPS V/RPS = 113/30 = 3.77 ft/rotation

3.77 = circumference of impeller diameter = 1.2 ft = 14.4 inches

CONCLUSIONCONCLUSION

FLOW THROUGH A CENTRIFUGAL PUMP FOLLOWS THESAME RULES OF FREELY FALLING BODIES

DO WE GET

THE SAME DIAMETER OR HEAD OR FLOW RATEAS PREDICTED BY THESE IDEAL RULES


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BASIC PERFORMANCE PARAMETERSBASIC PERFORMANCE PARAMETERS

The Energy Equation for This Case

2 2

1 21 1 1 2 2 2

2 2

sh aft vis

V VQ W W m h gz m h gz

= + + + + +

& & & & &

Assumptions:

No heat generation

No viscous work. Mass in = mass out

2 2

2 1

2 2 1 12 2sh aft

V V

W m h gz h gz

= + + + +

& &


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What would be the difference in z, can we assume z2-z

10

Hence2 2

2 12 1

2 2sh aft

V VW m h h

= + +

& &

2 2

2 2 1 12 1ha t

p V p VW m u u

= + + + +& &

Thermodynamically, u = u(T)

only and Tin Tout

2 2

2 2 1 1

2 2sh aft

p V p VW m

= + +

& &

2 2

2 2 1 1

2 2sh aft

p V p V

W Q

= + +

&


14/65


2 2

2 12 1

2 2w shaft V VP gHQ W Q p p

= = = + +

&

( )2 2

2 12 1

1

2 2

wP V VH p p

= = +

Where Pw

= water power

Generally V1 and V2 are of same order of magnitude

If the inlet and outlet diameters are same

( )2 11wP p p

gQ g

=


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The power required to drive the pump; bhp

The power required to turn the pump shaft at certain RPM

orque required to turn shaftbhp T T = =

The actual power required to drive the pump depends upon efficiency

wP QH

bh T

= =

Efficiency has three components;

Mechanical1. Losses in bearings

2. Packing glands etc

Hydraulic

Shock

friction,

re-circulation

Volumetric casing leakages

vL

Q

Q Q = + 1 fmbhp

= 1f

vhh

=

v h m =

CENT IF GAL P MPS tdCENT IF AL P MPS tdCENT IF AL P MPS tdCENT IF GAL P MPS tdCENT IF GAL P MPS tdCENT IF AL P MPS tdCENT IF AL P MPS tdCENT IF GAL P MPS td


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Torque estimation 1D flow assumption

1-D angular momentum balance gives

( )2 2 1 1t tT Q r V rV =

Vt1 and Vt2 absolute circumferential

or tangential velocity components

( ) ( )2 2 1 1 2 2 1 1w t t t t T Q r V rV Q u V u V = = =

Torque, Power and Ideal Head depends on,

Impeller tip velocities u & abs. tangential velocities VtIndependent of fluid axial velocity if any

( )( )2 2 1 1 2 2 1 1

1t twt t

Q u V u V PH u V u V

gQ gQ g

= = =

Euler turbo-

machineryequations;

DODO

DETAILSDETAILSIN TUTORIALIN TUTORIAL

dddddddd


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Doing some trigonometric and algebraic manipulation

( ) ( ) ( )2 2 2 2 2 22 1 2 1 2 11

2H V V u u w w

g = + +

2 2 2

2 2

p w rz const

g g g

+ + =

BERNOULLI EQUATION IN ROTATING COORDINATES

Applicable to 1, 2 and 3D Ideal Incompressible Fluids

One Can Also Relate the Pump Power With Fluid Radial Velocity

( )2 2 2 1 1 1cot cotw n nP Q u V u V =

2 1

2 2 1 12 2n n

Q QV and V

r b r b = =

With known b1, b2, r1, r2, 1, 2 and one can find centrifugal pumpsideal power and ideal head as a function of Discharge Q

DODO

EX. 11.1EX. 11.1

IN TUTORIALIN TUTORIAL

A tdA tdA tdA tdA tdA tdA tdA td


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EFFECT OF BLADE ANGLES 1, 2 ON PUMP PERFORMANCE

( )2 2 1 11w

t t

PH u V u V

gQ g= =

Angular Angular>>

2

2 22

n

QV

r b=

2 2 2 2cott nV u V =

Doing all this leads to

2

2 2 2

2 2

cot

2

u u

Qg r b g

if < 90, backward curve blades, stable opif = 90, straight radial blades, stable opIf > 90, forward curve blades, unstable op

CENTRIFUGAL PUMPS CHARACTERISTICSCENTRIFUGAL PUMPS CHARACTERISTICSCENTRIFUGAL PUMPS CHARACTERISTICSCENTRIFUGAL PUMPS CHARACTERISTICSCENTRIFUGAL PUMPS CHARACTERISTICSCENTRIFUGAL PUMPS CHARACTERISTICSCENTRIFUGAL PUMPS CHARACTERISTICSCENTRIFUGAL PUMPS CHARACTERISTICS


19/65

CENTRIFUGAL PUMPS, CHARACTERISTICSCENTRIFUGAL PUMPS, CHARACTERISTICSCENTRIFUGAL PUMPS, CHARACTERISTICSCENTRIFUGAL PUMPS, CHARACTERISTICSCENTRIFUGAL PUMPS, CHARACTERISTICSCENTRIFUGAL PUMPS, CHARACTERISTICSCENTRIFUGAL PUMPS, CHARACTERISTICSCENTRIFUGAL PUMPS, CHARACTERISTICS

1. Whatever discussed earlier is qualitative due to assumptions.

2. Actual performance of centrifugal pump 3. The presentation of performance data is exactly same for

4. The graphical representation of pumps performance data obtained

ex erimentall is called PUMP CHARACTERSTICS OR PUMP

extensive testing

1. Centrifugal pumps 2. Axial flow pumps

3. Mixed flow pumps 4. Compressors

CHARACTERSTIC CURVES1. This representation is almost always for constant shaft speed N

2. Q (gpm) discharge is the independent variable

3. H (head developed), P (power), (efficiency) and NPSH (netpositive suction head) are the dependent variables

4. Q (ft3/m3/min), discharge is the independent variable

5. H (head developed), P (power), (efficiency) are the dependentvariables

(LIQUIDS)

(LIQUIDS)

(GASES)

(GASES)

CENTRIFUGAL PUMPS CHARACTERISTICS co tdCENTRIFUGAL PUMPS CHARACTERISTICS c tdCENTRIFUGAL PUMPS CHARACTERISTICS c tdCENTRIFUGAL PUMPS CHARACTERISTICS co tdCENTRIFUGAL PUMPS CHARACTERISTICS co tdCENTRIFUGAL PUMPS CHARACTERISTICS c tdCENTRIFUGAL PUMPS CHARACTERISTICS c tdCENTRIFUGAL PUMPS CHARACTERISTICS co td


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CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.CENTRIFUGAL PUMPS, CHARACTERISTICS, contd.

TypicalCharacteristic Curves

of Centrifugal Pumps

CENTRIFUGAL PUMPS CHARACTERISTICS contdCENTRIFUGAL PUMPS CHARACTERISTICS contdCENTRIFUGAL PUMPS CHARACTERISTICS contdCENTRIFUGAL PUMPS CHARACTERISTICS contdCENTRIFUGAL PUMPS CHARACTERISTICS contdCENTRIFUGAL PUMPS CHARACTERISTICS contdCENTRIFUGAL PUMPS CHARACTERISTICS contdCENTRIFUGAL PUMPS CHARACTERISTICS contd


21/65


General Features of Characteristic Curves of Centrifugal Pumps

1. H is almost constant at low flow rates

2. Maximum H(shut off head) is at zero flow rate

3. Head drops to zero at Qmax

4. Q is not greater than Qmax N and/or impeller size is changed

5. Efficiency is always zero at Q = 0 and Q = Qmax

H.

7. = max at roughly Q=0.6Qmax to 0.93Qmax8. =max is called the BEST EFFICIENCY POINT (BEP)

9. All the parameters corresponding to max are called the designpoints, Q*, H*, P*

10. Pumps design should be such that the efficiency curve should be

as flat as possible around max11. P rises almost linearly with flow rate

P = =



22/65


Typical Characteristic Curves of Commercial Centrifugal Pumps

1. Having same casing size but different impeller diameters2. Rotating at different rpm

3. For power requirement and efficiency one needs to interpolate

(a ) basic casing with threebasic casing with threebasic casing with threebasic casing with threeimpeller sizesimpeller sizesimpeller sizesimpeller sizes

(b) 20 percent larger casing with three20 percent larger casing with three20 percent larger casing with three20 percent larger casing with threelarger impellers at slower speedlarger impellers at slower speedlarger impellers at slower speedlarger impellers at slower speed

CENTRIFUGAL PUMPS CHARACTERISTICS tdCENTRIFUGAL PUMPS CHARACTERISTICS tdCENTRIFUGAL PUMPS CHARACTERISTICS tdCENTRIFUGAL PUMPS CHARACTERISTICS tdCENTRIFUGAL PUMPS CHARACTERISTICS tdCENTRIFUGAL PUMPS CHARACTERISTICS tdCENTRIFUGAL PUMPS CHARACTERISTICS tdCENTRIFUGAL PUMPS CHARACTERISTICS td


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Calculate the ideal Head to be developed by the pump

shown in last figure

( ) ( )2 2

2 22

21170 2 / 60 / 36.75 / 2 12( ) 1093

32.2 /o

rad s ft rH ideal ftg ft s

= = =

Actual Head = 670 ft or 61% of Ho(ideal) at Q=0

Differences are due to

1. Impeller recirculations, important at low flow rates

2. Frictional losses

3. Shock losses due to mismatch of blade angle and flow

inlet important at high flow rates



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IMPORTANT POINTS TO REMEMBER

1. EFFECT OF DENSITY

1. Pump head reported in ft or m of that fluid important

2. These characteristic curves, valid only for the liquid reported

3. Same pump used to pump a different liquid H and would be almost same. OR. A centrifugal pump will always

fluid density

4. However P will change. Brake HP will vary directly with the

liquid density

2. EFFECT OF VISCOSITY

1. Viscous liquids tend to decrease the pump Head, Discharge

and efficiency tends to steepen the H-Q curve with

2. Viscous liquids tend to increase the pump BHP


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CentiPoise

cP)

centiStokes

(cSt)

Saybolt Second

Universal (SSU) Typical liquid

Specific

Gravity

1 1 31 Water 1

3.2 4 40 Milk -

12.6 15.7 80 No. 4 fuel oil 0.82 - 0.95

16.5 20.6 100 Cream -

34.6 43.2 200 Vegetable oil 0.91 - 0.95

88 110 500 SAE 10 oil 0.88 - 0.94

176 220 1000 Tomato Juice -352 440 2000 SAE 30 oil 0.88 - 0.94

820 650 5000 Glycerine 1.26

1561 1735 8000 SAE 50 oil 0.88 - 0.94

1760 2200 10,000 Honey -

5000 6250 28,000 Mayonnaise -

15,200 19,000 86,000 Sour cream -

17,640 19,600 90,000 SAE 70 oil 0.88 - 0.94


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Viscosity Scales

CentiPoises (cp) = CentiStokes (cSt) / SG (Specific Gravity)

SSU = Centistokes (cSt) 4.55

Degree Engler 7.45 = Centistokes (cSt)

Seconds Redwood 0.2469 = Centistokes (cSt)



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300wor > 2000 SSUPDPs are preferred

10w or < 50 SSUCentrifugal pumps are preferred

SUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFT


28/65

A centrifugal pump cannot pull or suck liquids

Suction in centrifugal pump creation of partial vacuum at pumps

inlet as compared to the pressure at the other end of liquid

Hence, pressure difference in liquid drives liquid through pump How one can increase this pressure difference

ncreas ng e pressure a e o er en

Equal to 1 atm for reservoirs open to atmosphere

> or < 1 atm for closed vessels

Decreasing the pressure at the pump inlet

Must be > liquid vapor pressure

By increasing the capacity

temperature very important

Bernoulli's equation

SUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFTSUCTION HEAD AND SUCTION LIFT


29/65

MAXIMUM SUCTION DEPENDS UPON

Pressure applied at liquid surface at liquid source, hence Maximum suction decreases as this pressure decreases

Vapor pressure of liquid at pumping temperature

Maximum suction decreases as vapor pressure increases

Capacity at which the pump is operating

CASE OF OPEN RESERVOIRS Maximum suction varies inversely with altitude Table-1

CASE OF HOT LIQUIDS

Maximum suction varies inversely with temp. Table-2

CASE OF INCREASING CAPACITY

Maximum suction varies inversely with capacity Table-3

NET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEADNET POSITIVE SUCTION HEAD


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Problem of Cavitation

The lowest pressure occurs at the pumps inletPressure at pump inlet < liquid vapor pressure cavitation occursWhat are the effects of cavitation

Lot of noise and vibrations are generated Sharp decrease in pumps H and Q

Pitting of impeller occurs due to bubble collapse

ay occur e ore ac ua o ng n case o sso ve gases

low boiling mixtures of hydrocarbons

Hence P at pumps inlet should greater than the Pvp

This extra pressure above Pvp available at pumps inlet is calledNet Positive Suction Head NPSH

Mathematically 2

1

2

vpiVPNPSH

g g

= +

NET POSITIVE SUCTION HEAD, contd.NET POSITIVE SUCTION HEAD, contd.NET POSITIVE SUCTION HEAD, contd.NET POSITIVE SUCTION HEAD, contd.NET POSITIVE SUCTION HEAD, contd.NET POSITIVE SUCTION HEAD, contd.NET POSITIVE SUCTION HEAD, contd.NET POSITIVE SUCTION HEAD, contd.


31/65

,,,,,,,,

NPSH calculated from this equation is the

specified by manufacturer The NPSH actually available at the pumps inlet is called

AVAILABLE NPSH must beAVAILABLE NPSH must be REQUIRED NPSHREQUIRED NPSH

Rule of thumb for design

PUMPS CHARACTERISTICPUMPS CHARACTERISTIC

SYSTEMS CHARACTERISTICSYSTEMS CHARACTERISTIC

REQUIRED NPSH

AVAILABLE NPSH

HOW TO CALCULATE AVAILABLE NPSH

Write Energy Equation between the free surface of fluid reservoir

and pump inlet

Thus Zi can be important parameter in designers hand to ensure that

cavitation does not occur for a given Psurface and temperature

o qu

surface vp

available i fi

PNPSH Z hg g

=



32/65

EFFECT OF VARYING HEIGHT

Given, Psurface, Pvp and hfi, Zi canbe varied to avoid cavitation

The 32-in pump of Fig. 11.7a is to pump 24,000 gpm of water at 1170 rpm from areservoir whose surface is at 14.7 psia. If head loss from reservoir to pump inlet is 6

ft, where should the pump inlet be placed to avoid cavitation for water at (a) 60F,

p 0.26 psia, SG 1.0 and (b) 200F,p 11.52 psia, SG 0.9635?

surface vp

i fi

P PNPSHA Z h NPSHR

g g =

An Example

Pump must be placed at least 12.7 ft below the reservoir surface to

avoid cavitation.

38.4iZ

Pump must now be placed at least 38.4 ft below the reservoir surface,to avoid cavitation

62.4g =( )

( )1

14.7 0.2640 6

62.4 144

surface vp

i fi i

P PNPSHR Z h Z

g g

= =

62.4 .9653 60.1g = =

12.7i

Z



33/65

TYPICAL EXAMPLE

A pump installed at an altitude of 2500 ft and has a suction lift of 13 ftwhile pumping 50 degree water. What is NPSHA? Ignore friction

Actual NPSHA = 17.59 2 = 15.59 ft

31 13 0 .41 17.59

surface vp

available i fi

P P

NPSH Z h ftg g = = =

TYPICAL EXAMPLE

We have a pump that requires 8 ft of NPSH at I20 gpm. If the pump is

installed at an altitude of 5000 ft and is pumping cold water at 60oF,what is the maximum suction lift it can attain? Ignore friction

2 8 2 28.2 0 .59 17.59surface vp

i fi i

P PNPSHA NPSHR Z h Z ft

g g = + = + = = =

DIMENSIONLESS PUMP PERFORMANCEDIMENSIONLESS PUMP PERFORMANCEDIMENSIONLESS PUMP PERFORMANCEDIMENSIONLESS PUMP PERFORMANCEDIMENSIONLESS PUMP PERFORMANCEDIMENSIONLESS PUMP PERFORMANCEDIMENSIONLESS PUMP PERFORMANCEDIMENSIONLESS PUMP PERFORMANCE--------11111111


34/65

THREE PERFORMANCE PARAMETERS

1. Head H(or pressure difference P-recall that P= gH)

2. Volume Flow Rate Q

3. Power P

TWO "GEOMETRIC" PARAMETERS:

EVERY PUMP HASEVERY PUMP HAS

1. D diameter2. n (or ) rotational speed

THREE FLUID FLOW PARAMETERS:

1. density2. viscosity

3. roughness

Above parameters involve only three dimensions, M-L-T



35/65

Buckingham Theorem suggests

7 -3 = 4 sto represent the physical phenomena in a pump.

Any pumps performance parameters are

1. HeadH(orgH) 2. Power P

( )1 , , , , ,gH f Q D n =

( )2 , , , , ,P f Q D n =

Hence The Two Groups Are

WHERE

= relative roughness

( )2 nD DnD

=

= Re. Number

3 QQ C

nD =

= Capacity Coefficient 3 5 PP Cn D = = Power Coefficient

2

12 2 3, ,

gH Q nDg

n D nD D

=

2

23 5 3, ,

P Q nDg

n D nD D

=

2 2 H

gHC

n D

=

= Head Coefficient



36/65

Reynolds number inside a centrifugal pump

1. 0.80 to 1.5x107)2. Flow always turbulent

3. Effect of Re, almost constant

4. May take it out of the functionsg1andg

25. Same is true for /D

Hence, we may write:

( )H H QC C C=

( )P QC C C=

For eometricall similar um s

Head and Power coefficients should be (almost)unique functions of the capacity coefficients.

In real life, however:

-manufacturers use the same case for different rotors

(violating geometrical similarity)

-larger pumps have smaller ratios of roughness and clearances

-the fluid viscosity is the same, whileRe changes with diameters.



37/65

CH, CPand CQ combined to give a coefficient having practical meaning

( )H Q QP

C CC

C = =

Similarly one can also define the CNPSHthe NPSH coefficient as

2 2NPSH NPSH QC C Cn D= =



38/65

Representing the pump performance data in dimensionless form

Pump data

Choose two geometrically

similar pumps

32 in impeller in pump (a) & 38

in in pump (b)

Pum (b) casin 20% > um

Results in graphical formResults in graphical formResults in graphical formResults in graphical form

(a) casing.Hence same diameter to casing

ratios

DISCRIPENCIESA few % in and CHpumps not truly dynamically similar

Larger pump has smaller roughness ratio

Larger pump has larger Re. number



39/65

The BEP lies at =0.88, corresponding to,

CQ* 0.115 CP* 0.65 CH* 5.0 CNPSH* 0.37

A unique set of values

Valid for all pumps of this geometrically similar family

Used to estimate the performance of this family pumps at BEP

Comparison of Values

D, ft n, r/s

Discharge

nD3, ft3/s

Head

n2D2/g, ft

Power

n3D5/550, hp

Fig. 11.7a 32/12 1170/60 370 84 3527

Fig. 11.7b 38/12 710/60 376 44 1861

Ratio - - 1.02 0.52 0.53

SIMILARITY RULES/AFFINITY LAWSSIMILARITY RULES/AFFINITY LAWSSIMILARITY RULES/AFFINITY LAWSSIMILARITY RULES/AFFINITY LAWSSIMILARITY RULES/AFFINITY LAWSSIMILARITY RULES/AFFINITY LAWSSIMILARITY RULES/AFFINITY LAWSSIMILARITY RULES/AFFINITY LAWS--------11111111


40/65

If two pumps are geometrically similar, then

1. Ratio of the corresponding coefficients =12. This leads to estimation of performance of one based on the

performance of the other

MATHEMATICALLY THIS CONCEPT LEADS TO

2Q 2gH 2P

2

1

2 2

13

1 1

1Q

Q

n D

QCn D

= =

3

2 2 2

1 1 1

Q n D

Q n D

=

2

1

2 2

12 2

1 1

1H

H

n D

gHCn D

= =

2 2

2 2 2

1 1 1

H n D

H n D

=

2

1

2 2 2

13 5

1 1 1

1P

P

n D

PCn D

= =

3 5

2 2 2 2

1 1 1 1

P n D

P n D

=

THESE ARE CALLLED SIMILARITY RULESTHESE ARE CALLLED SIMILARITY RULESTHESE ARE CALLLED SIMILARITY RULESTHESE ARE CALLLED SIMILARITY RULESTHESE ARE CALLLED SIMILARITY RULESTHESE ARE CALLLED SIMILARITY RULESTHESE ARE CALLLED SIMILARITY RULESTHESE ARE CALLLED SIMILARITY RULES



41/65

The similarity rules are used to estimate the effect of

1. Changing the fluid2. Changing the speed

3. Changing the size

VALID ONLY AND ONLY FORGeometrically similar family of any dynamic turbo machine

pump/compressor/turbine

Effect of changes in size and speedon homologous pump performance

(a) 20 percent change

in speed at constant size

(b) 20 percent change in

size at constant speed



42/65

For Perfect Geometric Similarity 1 = 2,but

Larger pumps are more efficient due to1. Higher Reynolds Number

2. Lower roughness ratios

3. Lower clearance ratios

Empirical correlations are available

Moodys Correlation

Based on size changes

14

2 2

1 1

1

1

D

D

Andersons Correlation

Based on flow rate changes

0.33

2 2

1 1

0.94

0.94

Q

Q

Concept of Specific SpeedConcept of Specific SpeedConcept of Specific SpeedConcept of Specific SpeedConcept of Specific SpeedConcept of Specific SpeedConcept of Specific SpeedConcept of Specific Speed--------11111111


43/65

We want to use a centrifugal pump from the family of Fig. 11.8 to

deliver 100,000 gal/min of water at 60F with a head of 25 ft. What

should be (a) the pump size and speed and (b) brake horsepower,

assuming operation at best efficiency?

H* = 25 ft = C n2 D2 / = 5 n2 D2 /32.2

A confusing example

Q* = 100000 gpm = 222.8 ft3/m = CQ n D3 = 0.115 n D3

Bhp* = Cp n3 D5 = 720 hp

Solving simultaneously gives, D = 12.4 ft, n = 62 rpm



44/65

The type of applications for which centrifugal pumps are required are;

1. High head low flow rate2. Moderate head and moderate flow rate

3. Low head and high flow rate

Q. Would a general design of the centrifugal pump will do all thethree jobs?

Ans. No

Q. What should be the design features to accomplish the three

specified jobs?

1. Answer to this question lies in the basic concept of centrifugal

pump working principle.

2. Vanes are used to impart momentum to the fluid by applying thecentrifugal force to the fluid.

PHYSICS FOR OUR RESCUE



45/65

3. More the diameter of the vane more will be the centrifugal force

4. More will be the diameter more will be the radial component of

velocity and lesser will be the axial component

5. More will be the radial velocity more will be the head developed

6. Hence to get more head you need longer vanes and vice versa

. ore w e e c earance e ween e mpe er an cas ng

more will the flow rate & also more will be the axial component

8. These simple physics principles lead us to the variation in

impeller design to accomplish the three jobs mentioned


POINT TO PONDER


46/65

We represent the performance of a family of geometrically similarpumps by a single set of dimensionless curves

Can we use even a smaller amount of information or even a single

number to represent the same information?

POINT TO PONDER

We have a huge variety of pumps each with a different diameter

,

Impeller shape ultimately dictates the type of application

RPM is not related to the pump design however it effects its

performance

Hence the biggest problem is to avoid diameter in the pump

performance information

Again dimensional analysis comes to rescue, a combination ofs isalso a , giving the same information in a different form



47/65

REARRANGE THE THREE COEFFICIENTS INTO A NEW

COEFFICIENT SUCH THAT DIAMETER IS ELIMINATED

( )

( )

1 122

3 344

/ Q

s

H

C n QN

C gH= =

Rigorous form, dimensionless

/17182= sN

Points to remember

1. Ns refers only to BEP

2. Directly related to most efficient

pump design

( ) ( )( )

12

34,

=s RPM GPMNH ft

Lazy but common form,Not dimensionless

3. Low Ns means low Q, High H

4. High Ns means High Q, Low H

5. Ns leads to specific pump

applications

6. Low Ns means high head pump

7. High Ns means high Q pump

Similarly one can define Nss, based on NPSH

Experimental data suggests, pump is in

danger of cavitationIf Nss 8100



48/65

GEOMETRICALVARIATION OF SPECIFIC

SPEED

Detailed shapes



49/65

Specific speed is an indicator of

Pump performancePump efficiency

The Q is a rough indicator of

Pump sizePump Reynolds Number THE PUMP CURVES



50/65

Note How The Head, Power and Efficiency curves change as

specific speed changes

Revisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing Example--------11111111


51/65

Dimensionless performance curves for a

typical axial- flow pump. Ns = 12.000.Constructed from data for a 14-in pump

at 690 rpm.

CQ* =0.55, CH*=1.07, Cp*=0.70,max= 0.84.Ns = 12000

D = 14 in n = 690 r m * = 4400 m.

Revisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing ExampleRevisit of Confusing Example--------22222222


52/65

Can this propeller pump family provide a 25-ft head & 100,000 gpm

discharge

Since we know the Ns and Dimensionless coefficients then usingsimilarity rules let us calculate the Diameter and RPM

D = 48 in and n = 430 r/min with bh = 750:

a much more reasonable design solution

Pump vs System CharacteristicsPump vs System CharacteristicsPump vs System CharacteristicsPump vs System CharacteristicsPump vs System CharacteristicsPump vs System CharacteristicsPump vs System CharacteristicsPump vs System Characteristics

Any piping systems has the following components in its total


53/65

Any piping systems has the following components in its total

head which the selected pump would have to supply1. Static head due to elevation

2. The head due to velocity head, the fictional head loss

3. Minor head losses

( )2 1ys z z a= = , min 4128

f la ar

LQh

=

Mathematically,

3 possibilities

( )2

2

2 1

2

sys

V fLz z K a cQ

g D

= + + = +

( )2 1 , minsys f la arz z h a bQ= + = +

,'

f turbulent

h Through Moody s Method =

Pump vs System Characteristics, contdPump vs System Characteristics, contdPump vs System Characteristics, contdPump vs System Characteristics, contdPump vs System Characteristics, contdPump vs System Characteristics, contdPump vs System Characteristics, contdPump vs System Characteristics, contd

Graphical Representation Of The Three Curves


54/65

Graphical Representation Of The Three Curves

Match between pump & systemMatch between pump & systemMatch between pump & systemMatch between pump & system


55/65

Match between pu p & systeMatch between pump & systemMatch between pump & systemMatch between pu p & syste

In industrial situation the resistance often varies for various

reasons

If the resistance factor increases, the slope of the systemcurve (Resistance vs flow) increases & intersect the

characteristic curve at a lower flow.

e es gne opera ng po n s are c osen as c ose o e

highest efficiency point as possible.Large industrial systems requiring different flow rates often

change the flow rate by changing the characteristic curve with

change in blade pitch or RPM

If K changes system curve shiftsIf K changes system curve shiftsIf K changes system curve shiftsIf K changes system curve shifts


56/65

Pump in Parallel or SeriesPump in Parallel or SeriesPump in Parallel or SeriesPump in Parallel or Series


57/65

pppp

To increase flow at a given head

1. Reduce system resistance factor with valve

2. Use small capacity fan/pumps in parallel.Some loss in flow rate may occur when operating

in arallel

To increase the head at a given flow1. Reduce system resistance by valve

2. Use two smaller head pumps/fans in series.

But some head loss may occur.

PUMPS IN PARALLELPUMPS IN PARALLELPUMPS IN PARALLELPUMPS IN PARALLEL


58/65

PUMPS IN SERIESPUMPS IN SERIESPUMPS IN SERIESPUMPS IN SERIES


59/65

UUUUnstable operation (Huntingnstable operation (Huntingnstable operation (Huntingnstable operation (Hunting)


60/65

If the characteristic is

such that the system

finds two flow rates fora given head it cannot

decide where to sta .

The pump could

oscillate between

points. It is called

hunting.

TableTableTableTableTableTableTableTable--------11111111


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Axial flow pump cross section

Radial flow pump cross section

Mixed flow pump cross section


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