Pumps and types of pumps in detail

PUMPSPUMPSPUMPSPUMPSPUMPSPUMPSPUMPSPUMPS

CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER CHAPTER ––––––––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 energy

to 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

POSITIVE DISPLACEMENT PUMPS, (PDP’S)POSITIVE DISPLACEMENT PUMPS, (PDP’S)POSITIVE DISPLACEMENT PUMPS, (PDP’S)POSITIVE DISPLACEMENT PUMPS, (PDP’S)POSITIVE DISPLACEMENT PUMPS, (PDP’S)POSITIVE DISPLACEMENT PUMPS, (PDP’S)POSITIVE DISPLACEMENT PUMPS, (PDP’S)POSITIVE DISPLACEMENT PUMPS, (PDP’S)

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 liquid

5. 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

7. In each cycle the fluid pumped equals the volume of the cavity7. In each cycle the fluid pumped equals the volume of the cavity

8. 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 from

complete shutoff conditions)

PDP’S, contd.PDP’S, contd.PDP’S, contd.PDP’S, contd.PDP’S, contd.PDP’S, contd.PDP’S, contd.PDP’S, contd.RECIPROCATING TYPE PDPS

Diaphragm pumpsPiston OR Plunger pumps

Double acting Simplex pump

Single acting piston pump

Double diaphragm pump

Single diaphragm pump

Double acting Simplex pump Double diaphragm pump

Double acting Duplex pump

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

DYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPSDYNAMIC PUMPS

WORKING PRINCIPLE AND FEATURESWORKING PRINCIPLE AND FEATURES

1. Add somehow momentum to the fluid

(through vanes, impellers or some special design

2. Do not have a fixed closed volume

3. 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 Pumps

Mixed Flow PumpsMixed Flow Pumps

Jet pump or ejector

Electromagnetic pumps for liquid metals

Fluid-actuated: gas-lift or hydraulic-ram

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


Mixed Flow PumpsMixed Flow Pumps

Diffuser PumpDiffuser Pump

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 flow rate for virtually

Constant

RPM

Constant flow rate for virtually

any pressure

OR

Flow rate cannot be changed

without changing RPM

Hence used for metering

Head varies with

flow rate

OR

Flow rate changes with

head for same RPM

Viscosity Virtually no effect Strong effects

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, IMPELLER

2. A stationary element, VOLUTE

3. Filled with fluid & impeller rotated

4. Fluid rotates & leaves with high velocity

Illustration-1

Illustration-2

Impeller-1 Impeller-54. Fluid rotates & leaves with high velocity

5. Outward flow reduces pressure at inlet,

(EYE OF THE IMPELLER), more fluid

comes in.

6. Outward fluid enters an increasing area

region. Velocity converts to pressure

�Impeller Impart Energy/Velocity By Rotating Fluid

�Volute Converts Velocity To Pressure

Impeller-1

Impeller-2

Impeller-3

Impeller-4

Impeller-5

Impeller-6

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 pale

2. Make a bore in hole → water is thrown out

3. Distance the water stream travels tangent to the circle = f(Vr)

4. Volume flow from hole = f(Vr)

5. In centrifugal pumps, flow rate & pressure = f(Vr) (tip velocity)5. In centrifugal pumps, flow rate & pressure = f(Vr) (tip velocity)

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’


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 inches3.77 = circumference of impeller → diameter = 1.2 ft = 14.4 inches

CONCLUSIONCONCLUSION

FLOW THROUGH A CENTRIFUGAL PUMP FOLLOWS THE

SAME RULES OF FREELY FALLING BODIES

DO WE GET

THE SAME DIAMETER OR HEAD OR FLOW RATE

AS PREDICTED BY THESE IDEAL RULES


BASIC PERFORMANCE PARAMETERSBASIC PERFORMANCE PARAMETERS

The Energy Equation for This Case

2 2

1 21 1 1 2 2 2

2 2shaft 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 12 2 1 1

2 2shaft

V VW m h gz h gz

= + + − + +

& &


What would be the difference in ‘z’, can we assume z2-z1≈0

Hence2 2

2 12 1

2 2shaft

V VW m h h

= + − +

& &

2 2

2 2 1 12 1

2 2shaft

p V p VW m u u

ρ ρ

= + + − + +

& &

2 12 2

shaftW m u uρ ρ

= + + − + +

Thermodynamically, u = u(T)

only and Tin ≈ Tout

2 2

2 2 1 1

2 2shaft

p V p VW m

ρ ρ

= + − +

& &

2 2

2 2 1 1

2 2shaft

p V p VW Qρ

ρ ρ

= + − +

&


2 2

2 12 1

2 2w shaft

V VP gHQ W Q p p

ρ ρρ

= = = + − +

&

( )2 2

2 12 1

1

2 2

wP V VH p p

gQ g

ρ ρρ ρ

= = − + −

Where Pw = water power

2 2gQ gρ ρ

Generally V1 and V2 are of same order of magnitude

If the inlet and outlet diameters are same

( )2 1

1wPH p pgQ gρ ρ

= ≅ −


The power required to drive the pump; bhp

The power required to turn the pump shaft at certain RPM

torque required to turn shaftbhp T Tω= =

The actual power required to drive the pump depends upon efficiency

wP gQH

bhp T

ρη

ω= =bhp Tω

Efficiency has three components;

Mechanical

1. Losses in bearings

2. Packing glands etc

Hydraulic

• Shock

• friction,

• re-circulation

Volumetric

• casing leakages

v

L

Q

Q Qη =

+ 1f

m

P

bhpη = − 1

f

v

s

h

hη = −

v h mη η η η=


Torque estimation ⇒ 1D flow assumption

1-D angular momentum balance gives

( )2 2 1 1t tT Q rV rVρ= −

Vt1 and Vt2 absolute circumferential

or tangential velocity components

( ) ( )2 2 1 1 2 2 1 1w t t t tP T Q rV rV Q u V uVω ωρ ρ= = − = −

Torque, Power and Ideal Head depends on,

Impeller tip velocities ‘u’ & abs. tangential velocities Vt

Independent of fluid axial velocity if any

( ) ( )2 2 1 1

2 2 1 1

1t twt t

Q u V uVPH u V uV

gQ gQ g

ρρ ρ

−= = = −

Euler turbo-

machinery

equations;

DO DO

DETAILS DETAILS

IN TUTORIALIN TUTORIAL


Doing some trigonometric and algebraic manipulation

( ) ( ) ( )2 2 2 2 2 2

2 1 2 1 2 1

1

2H V V u u w w

g = − + − + −

2 2 2

2 2

p w rz const

g g g

ωρ

+ + − =

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 uVρ α α= −

2 1

2 2 1 12 2n n

Q QV and V

r b rbπ π= =

With known b1, b2, r1, r2, β1, β2 and ω one can find centrifugal pump’s

ideal power and ideal head as a function of Discharge ‘Q’

DO DO

EX. 11.1 EX. 11.1

IN TUTORIALIN TUTORIAL


EFFECT OF BLADE ANGLES β1, β2 ON PUMP PERFORMANCE

( )2 2 1 1

1wt t

PH u V uV

gQ gρ= = −

Angular

momentum out

Angular

momentum in>>

2

2 22n

QV

r bπ=

momentum out momentum in

2 2 2 2cott nV u V β= −

Doing all this leads to

2

2 2 2

2 2

cot

2

u uH Q

g r b g

βπ

≈ −

if β < 90, backward curve blades, stable op

if β = 90, straight radial blades, stable op

If β > 90, forward curve blades, unstable op

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

experimentally is called “PUMP CHARACTERSTICS” OR “PUMP

extensive testing

1. Centrifugal pumps 2. Axial flow pumps

3. Mixed flow pumps 4. Compressors

experimentally is called “PUMP CHARACTERSTICS” OR “PUMP

CHARACTERSTIC CURVES”

1. This representation is almost always for constant shaft speed ‘N’

2. Q (gpm) discharge is the independent variable

3. H (head developed), P (power), ηηηη (efficiency) and NPSH (net

positive suction head) are the dependent variables

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

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

variables

(LIQUIDS)

(LIQUIDS)

(GASES)

(GASES)

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.

Typical

Characteristic Curves

of Centrifugal Pumps


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

6. η is not an independent parameter → wP gHQρη = =6. η is not an independent parameter →

7. η = ηmax at roughly Q=0.6Qmax to 0.93Qmax

8. η = ηmax is called the BEST EFFICIENCY POINT (BEP)

9. All the parameters corresponding to ηmax are called the design

points, Q*, H*, P*

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

as flat as possible around ηmax

11. ‘P’ rises almost linearly with flow rate

wP gHQ

P P

ρη = =


Typical Characteristic Curves of Commercial Centrifugal Pumps

1. Having same casing size but different impeller diameters

2. Rotating at different rpm

3. For power requirement and efficiency one needs to interpolate

(a ) basic casing with three basic casing with three basic casing with three basic casing with three impeller 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 three

larger impellers at slower speedlarger impellers at slower speedlarger impellers at slower speedlarger impellers at slower speed


Calculate the ideal Head to be developed by the pump

shown in last figure

( ) ( )2 22 2

2

2

1170 2 / 60 / 36.75 / 2 12( ) 1093

32.2 /o

rad s ftrH ideal ft

g ft s

πω × ×= = =

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


IMPORTANT POINTS TO REMEMBER

1. EFFECT OF DENSITY

1. Pump head reported in ‘ft’ or ‘m’ of that fluid → ρ important

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

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

develop the same head in feet of that liquid regardless of the develop the same head in feet of that liquid regardless of the

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 η ↓

2. Viscous liquids tend to increase the pump BHP

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

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)


µµµµ ≥ 300µµµµwor µµµµ > 2000 SSU

PDP’s are preferred

µµµµ ≤≤≤≤ 10µµµµw or µµµµ < 50 SSU

Centrifugal 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

• A centrifugal pump cannot pull or suck liquids

• Suction in centrifugal pump → creation of partial vacuum at pump’s

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

– Increasing the pressure at the other end– Increasing the pressure at the other end

• 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

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 RESERVOIRSCASE 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

• Problem of Cavitation–The lowest pressure occurs at the pump’s inlet

–Pressure at pump inlet < liquid vapor pressure → cavitation occurs

–What are the effects of cavitation

• Lot of noise and vibrations are generated

• Sharp decrease in pump’s ‘H’ and ‘Q’

• Pitting of impeller occurs due to bubble collapse

• May occur before actual boiling in case of dissolved gases / • May occur before actual boiling in case of dissolved gases /

low boiling mixtures of hydrocarbons

• Hence ‘P’ at pump’s inlet should greater than the Pvp

• This extra pressure above Pvp available at pump’s inlet is called

Net Positive Suction Head ‘NPSH’

• Mathematically →→→→2

1

2

vpiPVP

NPSHg gρ ρ

= + −

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.

• NPSH calculated from this equation is the

specified by manufacturer →→→→

• The NPSH actually available at the pump’s inlet is called

•• ‘AVAILABLE NPSH’ must be ‘AVAILABLE NPSH’ must be ≥≥≥≥≥≥≥≥‘REQUIRED NPSH’‘REQUIRED NPSH’

• Rule of thumb for design

ft of liquid

“PUMP’S CHARACTERISTIC”“PUMP’S CHARACTERISTIC”

“SYSTEM’S CHARACTERISTIC”“SYSTEM’S CHARACTERISTIC”

‘REQUIRED NPSH’

‘AVAILABLE NPSH’ →→→→

‘AVAILABLE NPSH’ ‘AVAILABLE NPSH’ ≥≥≥≥≥≥≥≥ (2+‘REQUIRED NPSH’(2+‘REQUIRED 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

ft of liquid

surface vp

available i fi

P PNPSH Z h

g gρ ρ= − − −

‘AVAILABLE NPSH’ ‘AVAILABLE NPSH’ ≥≥≥≥≥≥≥≥ (2+‘REQUIRED NPSH’(2+‘REQUIRED NPSH’)


EFFECT OF VARYING HEIGHT

Given, Psurface, Pvp and hfi , Zi can

be varied to avoid cavitation

The 32-in pump of Fig. 11.7a is to pump 24,000 gpm of water at 1170 rpm from a

reservoir 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) 60°F,

pvp0.26 psia, SG 1.0 and (b) 200°F, pvp 11.52 psia, SG 0.9635?

surface vp

i fi

P PNPSHA Z h NPSHR

g gρ ρ= − − − ≥

An Example

pvp0.26 psia, SG 1.0 and (b) 200°F, pvp 11.52 psia, SG 0.9635?

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

62.4gρ =( )

( ) 1

14.7 0.2640 6

62.4 144

surface vp

i fi i

P PNPSHR Z h Z

g gρ ρ −

−= ≤ − − − = − −

62.4 .9653 60.1gρ = × =

12.7iZ ≤−


TYPICAL EXAMPLE

A pump installed at an altitude of 2500 ft and has a suction lift of 13 ft

while pumping 50 degree water. What is NPSHA? Ignore friction

Actual NPSHA = 17.59 – 2 = 15.59 ft

31 13 0 .41 17.59surface vp

available i fi

P PNPSH Z h ft

g gρ ρ= − − − = − − − =

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 gρ ρ= + = + = − − − = − − − =

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

THREE PERFORMANCE PARAMETERS

1. Head ‘H’ (or pressure difference ∆P-recall that ∆P= ρgH)

2. Volume Flow Rate ‘Q’

3. Power ‘P’

TWO "GEOMETRIC" PARAMETERS:

1. D diameter

EVERY PUMP HASEVERY PUMP HAS

1. D diameter

2. n (or ω) rotational speed

THREE FLUID FLOW PARAMETERS:

1. ρ density

2. µ viscosity

3. ε roughness

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


Buckingham π Theorem suggests 7 -3 = 4 π’s to represent the physical phenomena in a pump.

Any pump’s performance parameters are

1. Head H (or gH ) →2. Power P →

( )1 , , , , ,gH f Q D n ρ µ ε=( )2 , , , , ,P f Q D n ρ µ ε=

Hence The Two π Groups Are

WHERE

D

ε

= relative roughness

( )2 nD DnD ρρµ µ

=

= Re. Number

3 Q

QC

nD

=

= Capacity Coefficient 3 5 P

PC

n Dρ

=

= Power Coefficient

2

12 2 3, ,

gH Q nDg

n D nD D

ρ εµ

=

2

23 5 3, ,

P Q nDg

n D nD D

ρ ερ µ

=

2 2 H

gHC

n D

=

= Head Coefficient


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 functions g1and g2

5. Same is true for ε/D

Hence, we may write:

( )H H QC C C=

( )P P QC C C=

For geometrically similar pumps, For geometrically similar pumps,

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, while Re changes with diameters.


CH, CP and CQ combined to give a coefficient having practical meaning

( )H Q

Q

P

C CC

Cη η= =

Similarly one can also define the CNPSH the NPSH coefficient as

( )g NPSH⋅= = ( )2 2NPSH NPSH Q

g NPSHC C C

n D

⋅= =


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)

•Pump (b) casing 20% > pump

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

•Pump (b) casing 20% > pump

(a) casing.

•Hence same diameter to casing

ratios

DISCRIPENCIES

•A few % in η and CH•pumps not truly dynamically similar

•Larger pump has smaller roughness ratio

•Larger pump has larger Re. number


The BEP lies at η=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

If two pumps are geometrically similar, then

1. Ratio of the corresponding coefficients =1

2. This leads to estimation of performance of one based on the

performance of the other

MATHEMATICALLY THIS CONCEPT LEADS TO

23

QC n D

22 2

gHC n D

23 5

PC n Dρ

2

1

3

2 2

13

1 1

1Q

Q

C n D

QCn D

= =

3

2 2 2

1 1 1

Q n D

Q n D

=

2

1

2 2

2 2

12 2

1 1

1H

H

C n D

gHCn D

= =

2 2

2 2 2

1 1 1

H n D

H n D

=

2

1

3 5

2 2 2

13 5

1 1 1

1P

P

C n D

PCn D

ρ

ρ

= =

3 5

2 2 2 2

1 1 1 1

P n D

P n D

ρρ

=

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


The similarity rules are used to estimate the effect of

1. Changing the fluid

2. Changing the speed

3. Changing the size

VALID ONLY AND ONLY FOR

Geometrically similar family of any dynamic turbo machine

pump/compressor/turbine

Effect of changes in size and speed

on homologous pump performance

(a) 20 percent change

in speed at constant size

(b) 20 percent change in

size at constant speed


For Perfect Geometric Similarity η1 = η2, but

Larger pumps are more efficient due to

1. Higher Reynolds Number

2. Lower roughness ratios

3. Lower clearance ratios

Empirical correlations are available

To estimate efficiencies in geometrically similar family of pumpsTo estimate efficiencies in geometrically similar family of pumps

Moody’s Correlation

Based on size changes

14

2 2

1 1

1

1

D

D

ηη

−≈ −

Anderson’s Correlation

Based on flow rate changes

0.33

2 2

1 1

0.94

0.94

Q

Q

ηη

−≈ −

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

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

deliver 100,000 gal/min of water at 60°F 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 = (CH n2 D2)/g = (5 × n2 D2)/32.2

A confusing example

H = 25 ft = (CH n D )/g = (5 × n D )/32.2

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

Bhp* = Cpρ n3 D5 = 720 hp

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


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

1. High head low flow rate

2. 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 the

three 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 the

centrifugal force to the fluid.

PHYSICS FOR OUR RESCUE


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

7. More will be the clearance between the impeller and casing 7. More will be the clearance between the impeller and casing

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


• We represent the performance of a family of geometrically similar

pumps 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 of impeller and running at certain rpmimpeller, shape of impeller and running at certain rpm

• 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 of π’s is also a π, giving the same information in a different form


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=s sN N

Points to remember

1. Ns refers only to BEP

2. Directly related to most efficient

pump design

3. Low N means low Q, High H( ) ( )( )

12

34,

=s

RPM GPMN

H 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 cavitation

If Nss ≥ 8100


GEOMETRICAL

VARIATION OF SPECIFIC

SPEED

Detailed shapes


Specific speed is an indicator of

Pump performance

Pump efficiency

The Q is a rough indicator of

Pump size

Pump Reynolds Number THE PUMP CURVES


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

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,ηmax= 0.84.

Ns = 12000

D = 14 in, n = 690 rpm, Q* = 4400 gpm.D = 14 in, n = 690 rpm, Q* = 4400 gpm.

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

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

discharge

Since we know the Ns and Dimensionless coefficients then using

similarity rules let us calculate the Diameter and RPM

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

head which the selected pump would have to supply

1. Static head due to elevation

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

3. Minor head losses

( )2 1sysH z z a= − =, min 4

128f la ar

LQh

gD

µπρ

=2 1sys

Mathematically,

3 possibilities

( )2

2

2 12

sys

V fLH z z K a cQ

g D

= − + + = + ∑ ∑

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

, min 4f la argDπρ

, 'f turbulenth 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

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

•In industrial situation the resistance often varies for various

reasons

•If the resistance factor increases, the slope of the system

curve (Resistance vs flow) increases & intersect the

characteristic curve at a lower flow.

•The designed operating points are chosen as close to the •The designed operating points are chosen as close to the

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

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

•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 parallel in parallel

•To increase the head at a given flow

1. 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

PUMPS IN SERIESPUMPS IN SERIESPUMPS IN SERIESPUMPS IN SERIES

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

If the characteristic is

such that the system

finds two flow rates for

a given head it cannot

decide where to stay. decide where to stay.

The pump could

oscillate between

points. It is called

hunting.

TableTableTableTableTableTableTableTable--------11111111



Axial flow pump cross section

Radial flow pump cross section

Mixed flow pump cross section

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Pumps and types of pumps in detail