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TOPIC 4: PUMPS AND TURBINES

OBJECTIVES

1. Understand the role of pumps and turbines as energy-conversion devices

and use, appropriately, the terms head, power and efficiency.

2. Be aware of the main types of pumps and turbines and the distinction

between impulse and reaction turbines and between radial, axial and

mixed-flow devices.

3. Match pump characteristics and system characteristics to determine the

duty point.

4. Calculate characteristics for pumps in series and parallel and use the

hydraulic scaling laws to calculate pump characteristics at different

speeds.

5. Select the type of pump or turbine on the basis of specific speed.

6. Understand the mechanics of a centrifugal pump and an impulse turbine.

7. Recognise the problem of cavitation and how it can be avoided.

ENERGY CONVERSION

Pumps: electrical/mechanical energy fluid energy

Turbines: fluid energy electrical/mechanical energy

PUMP

pipeline

TURBINE

total head

total head

pipeline

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ENERGY

Energy per unit weight = head, H

Power = rate of conversion of energy

Efficiency

Turbines:

Pumps:

g

Vz

g

pH

2ρ

2

gQHρpower(fluid)

in

out

power

powerη

gQHρη outpower

inpower

gQHρη

EXAMPLE, PAGE 2

A pump lifts water from a large tank at a rate of 30 L s–1.

If the input power is 10 kW and the pump is operating at

an efficiency of 40%, find:

(a) the head developed across the pump;

(b) the maximum height to which it can raise water if the

delivery pipe is vertical, with diameter 100 mm and

friction factor λ = 0.015.

TYPES OF PUMPS AND TURBINES:

Impulse vs Reaction Turbines

Impulse turbine (Pelton wheel; water wheel)

– change in velocity change in head

Reaction turbine (Francis turbine, Kaplan turbine, windmill)

– change in pressure change in head

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TYPES OF PUMPS AND TURBINES:

Positive-Displacement vs Dynamic Pumps

Positive-displacement (heart; bicycle pump; peristaltic pump)

– change in volume

Dynamic pump

– no change of volume; continuous transfer of energy

– commonest are rotodynamic pumps

TYPES OF PUMPS AND TURBINES:

Radial, Axial and Mixed-Flow Devices

high head,

low flow

low head,

high flow

Radial Axial Mixed

PUMPS

Centrifugal pump

Axial-flow pump

Inward-flow reaction turbine centrifugal pump

(e.g. Francis turbine)

Propeller turbine axial-flow pump

(e.g. Kaplan turbine; windmill)

Volute

'Eye' (intake)

Impeller vane

flow

rotation

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TURBINES

Pelton wheel

– very-high head; hydropower

Francis turbine

– intermediate head; hydropower (pumped storage)

Kaplan turbine

– axial flow

Wells turbine

– wave energy

Bulb turbine

– tidal power

Archimedean screw

– small-scale hydro-power

PELTON WHEEL

PELTON WHEEL

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FRANCIS TURBINE

KAPLAN TURBINE

ARCHIMEDEAN SCREW

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PUMP CHARACTERISTICS

Head: H vs Q

Input power: I vs Q

Efficiency: η vs Q

Q

I

H

Q

IH

centrifugal pump axial-flow pump

PUMP CHARACTERISTICS:

Lab Results

0.0

2.0

4.0

6.0

8.0

10.0

12.0

0 50 100 150

H (m

)

Q (L min-1)

1.1 Head vs Discharge (2800 rpm)

0

50

100

150

200

250

300

350

0 50 100 150

I (W

)

Q (L min-1)

1.2 Input Power vs Discharge (2800 rpm)

0

5

10

15

20

25

30

35

40

45

50

0 50 100 150

(%

)

Q (L min-1)

1.3 Efficiency vs Discharge (2800 rpm)

SYSTEM CHARACTERISTICS

The pump is required to:

lift water through a certain height (static lift, Hs)

overcome frictional (and other flow-related) losses

2aQHH s

Sump

Delivery reservoir

Suction main

Delivery mainStatic liftHs

Suction headPump

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DUTY POINT

The discharge Q is such that … the head provided by the

pump exactly matches the system head (static lift + losses)

The corresponding Q and H are called the duty point.

Ideally, this should be close to the maximum-efficiency point.

Q

HSystemcharacteristic

Pumpcharacteristic

Dutypoint

Hs

EXAMPLE, PAGE 7

A water pump was tested at a rotation rate of 1500 rpm. The following data was

obtained. (Q is quantity of flow, H is head of water, η is efficiency).

It is proposed to used this pump to draw water from an open sump to an

elevation 5.5 m above. The delivery pipe is 20.0 m long and 100 mm diameter,

and has a friction factor of 0.005.

If operating at 1500 rpm, find:

(a) the maximum discharge that the pump can provide;

(b) the pump efficiency at this discharge;

(c) the input power required.

Q (L s–1) 0 10 20 30 40 50

H (m) 10.0 10.5 10.0 8.5 6.0 2.5

η 0.0 0.40 0.64 0.72 0.64 0.40

PUMPS IN PARALLEL AND SERIES

Parallel

Same head: H

Add discharges: Q1 + Q2

Series

Same discharge: Q

Add heads: H1 + H2

Q

H

Single pump

Double the flow

Pumps in parallel

Q

H

Single pump

Pumps in series

Doublethe

head

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PUMPS IN PARALLEL AND SERIES:

Lab Results

0.0

5.0

10.0

15.0

20.0

0 50 100 150

H (m

)

Q (L min-1)

3.1 Head vs Discharge (Series)

Measured

Scaled

EXAMPLE, PAGE 9A rotodynamic pump, having the characteristics tabulated below, delivers water

from a river at elevation 102 m to a reservoir with a water level of 135 m, through a

350 mm diameter cast-iron pipe. The frictional head loss in the pipeline is given by

hf = 550 Q2, where hf is the head loss in m and Q is the discharge in m3 s–1. Minor

head losses from valves and fittings amount to 50 Q2 in the same units.

(a) Calculate the discharge and head in the pipeline (at the duty point).

If the discharge is to be increased by the installation of a second identical pump:

(b) determine the unregulated discharge and head produced by connecting the

pump:

(i) in parallel;

(ii) in series;

(c) determine the power demand at the duty point in the case of parallel operation.

Q (m3 s–1) 0 0.05 0.10 0.15 0.20

H (m) 60 58 52 41 25

η (%) --- 44 65 64 48

DIMENSIONAL ANALYSIS

Variables and dimensions:

discharge Q [L3T­–1]

pressure change ρgH [ML–1T–2]

power P [ML2T–3]

rotor diameter D [L]

rotation rate N [T–1]

fluid density ρ [ML–3]

fluid viscosity μ [ML–1T–1]

# variables = 7

# dimensions = 3 (M,L,T)

# dimensionless Π groups = 4

31ΠND

Q

222ΠDN

gH 533

ρΠ

DN

P Re

μ

ρΠ

2

4 ND

discharge head power viscosity

Scales: D, N, ρ

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DIMENSIONAL ANALYSIS

31ΠND

Q

222ΠDN

gH 533

ρΠ

DN

P Re

μ

ρΠ

2

4 ND

discharge head power viscosity

For fully-turbulent flow there is no significant dependence on Π4 (Re)

Any one of Π1, Π2, Π3 can be replaced by:

or the reciprocal of this for turbines

ηρ

Π

ΠΠ

3

21 P

gQHefficiency (for pumps)

EXAMPLE, PAGE 10

A ¼-scale model centrifugal pump is tested under a

head of 7.5 m at a speed of 500 rpm. It was found that

7.5 kW was needed to drive the model. Assuming similar

mechanical efficiencies, calculate:

(a) the speed and power required by the prototype when

pumping against a head of 44 m;

(b) the ratio of the discharges in the model to that in the

prototype.

HYDRAULIC SCALING LAWS:

(“Affinity Laws”)

2

3

1

3

ND

Q

ND

Q

2

22

1

22

DN

gH

DN

gH

2

53

1

53 ρρ

DN

P

DN

P

Speed

For the same pump/turbine and working fluid (same D, ρ):

1

2

1

2

N

N

Q

Q

2

1

2

1

2

N

N

H

H3

1

2

1

2

N

N

P

P21 ηη

discharge speed head speed2 power speed3

Size

For different-sized, but geometrically similar devices at same speed (same N, ρ):

21 ηη 3

1

2

1

2

D

D

Q

Q2

1

2

1

2

D

D

H

H5

1

2

1

2

D

D

P

P

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FINDING THE DUTY POINT AT A NEW PUMP

SPEED

Hydraulic scaling:1

2

1

2

N

N

Q

Q

2

1

2

1

2

N

N

H

H

Q

System characteristic

New duty point

Hs

N1

N2

H

FINDING THE PUMP SPEED FOR A GIVEN

DUTY POINT

The unknown speed N2 can be found from the ratio of discharges or ratio of heads:

2

22

Q

Q

H

H

Plot a hydraulic-scaling curve back from

the required duty point (Q2,H2) on the

system curve at unknown speed N2:

This will cut the given curve at point (Q1,H1).

1

2

1

2

Q

Q

N

N

1

2

2

1

2

H

H

N

N

or

Q

System characteristic

New duty point

Hs

N1

N2

Scaling curvethroughduty point

H

),( 22 HQ

),( 11 HQ

EXAMPLE, PAGE 12

Water from a well is pumped by a centrifugal pump which delivers water to a

reservoir in which the water level is 15.0 m above that in the sump. When the

pump speed is 1200 rpm its pipework has the following characteristics:

Pipework characteristics:

Discharge (L s–1): 20 30 40 50 60

Head loss in pipework (m): 1.38 3.11 5.52 8.63 12.40

Pump characteristics:

Discharge (L s–1): 0 10 20 30 40

Head (m): 22.0 21.5 20.4 19.0 17.4

A variable-speed motor drives the pump.

(a) Plot the graphs of the system and pump characteristics and determine the

discharge at a speed of 1200 rpm.

(b) Find the pump speed in rpm if the discharge is increased to 40 L s–1.

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SPECIFIC SPEED (PUMPS)

31ΠND

Q

222ΠDN

gH

Find a combination independent of D ...3

42

3

2

2

1

)()Π(

)Π(

gH

NQ

... and proportional to N:4/3

2/14/1

3

2

2

1

)(Π

Π

gH

NQ

Since g is constant, it follows that, as speed changes,

4/3

2/1

H

NQ

is the same at any given (e.g. maximum) efficiency.

Most important parameters are discharge and head:

SPECIFIC SPEED (PUMPS)

4/3

2/1

H

NQNs

Specific speed is that rotation rate which will generate 1 unit of head for 1 unit

of discharge:

Single value, calculated at maximum-efficiency point.

Usual units (UK): N in rpm, Q in m3 s–1, H in m.

Less-common alternative: dimensionless specific speed,

Typical ranges:

4/3

2/1

)(gH

NQKn

Type Ns

Centrifugal 10 – 70 large head

Mixed flow 70 – 170

Axial > 110 small head

EXAMPLE, PAGE 14

A pump is needed to operate at 3000 rpm (i.e.

50 Hz) with a head of 6 m and a discharge of

0.2 m3 s–1. By calculating the specific speed,

determine what sort of pump is required.

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SPECIFIC SPEED (TURBINES)

222ΠDN

gH

Find a combination independent of D ... 5

42

5

2

2

3

)(

)ρ/(

)Π(

)Π(

gH

NP

... and proportional to N:

Since ρ and g are constant, it follows that, as speed changes,

is the same at any given (e.g. maximum) efficiency.

533ρ

ΠDN

P

4/5

2/1

H

NPNS

Most important parameters are head and (output) power:

4/5

2/14/1

5

2

2

3

)(

)ρ/(

Π

Π

gH

NP

SPECIFIC SPEED (TURBINES)

Specific speed is that rotation rate which will generate 1 unit of power for 1 unit

of head:

Single value, calculated at maximum-efficiency point.

Usual units (UK): N in rpm, P in kW, H in m.

Less-common alternative: dimensionless specific speed,

Typical ranges:

4/5

2/1

H

NPN S

4/5

2/1

)(

)ρ/(

gH

NPKn

Type Ns

Pelton wheel (impulse) 12 – 60 very large head

Francis turbine (radial) 60 – 500 large head

Kaplan turbine (axial-flow) 280 – 800 small head

MECHANICS OF CENTRIFUGAL PUMPS

Volute

'Eye' (intake)

Impeller vane

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MECHANICS OF CENTRIFUGAL PUMPS

u impeller velocity (u = rω)

w velocity relative to the impeller

v = u + w absolute velocity

Radial velocity vr determined by flow rate Q

Whirl velocity vt is a combination of impeller (rω) and tangential velocity relative to vanes

2w2β

1w

ω22 ru

ω11 ru

Vane

v,Resultant

w β

ωru

v

tv

rv

MECHANICS OF CENTRIFUGAL PUMPS

Torque = rate of change of angular momentum

)(ρ 1122 rvrvQT tt

Power = torque angular velocity

ω)(ρρ 1122 rvrvQgQH tt

Impeller velocity u = rω

Euler’s turbomachinery equation:

)(1

1122 uvuvg

H tt

Vane

v,Resultant

w β

ωru

v

tv

rv

EFFECT OF BLADE ANGLE

Usually design for small inlet whirl velocity: vt1 0.

22

1uv

gH tThen

βsinwA

Qvr

βcoswuvt

)βcot( 22

A

Qu

g

uH

bQaH

Vane

v,Resultant

w β

ωru

v

tv

rv

Q

H

90β

90β

90β

(backward-facing blades)

(forward-facing blades)

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EXAMPLE, PAGE 17

A centrifugal pump is required to provide a head of

40 m. The impeller has outlet diameter 0.5 m and

inlet diameter 0.25 m and rotates at 1500 rpm. The

flow approaches the impeller radially at 10 m s–1

and the radial velocity falls off as the reciprocal of

the radius. Calculate the required vane angle at the

outlet of the impeller.

MECHANICS OF A PELTON WHEEL

Force (on fluid) = mass flux change in velocity

Change in velocity easiest to establish in the frame of the bucket

)θcos1)((

)(θcos)(

kuv

uvuvkx

velocity- in Change

)θcos1)((ρ kuvQF

Power = force (on bucket) velocity (of bucket) = Fu

)θcos1()(ρ kuuvQP (per jet)

JetBucket

Spear valve

bucket

jet uv

k(v-u)

v-u

MECHANICS OF A PELTON WHEEL

)θcos1()(ρ kuuvQP

2

212

21 )()()( uvvuuv

vu21

Power (per jet)

Theoretical maximum power when

runner speed equals half the jet speed:

If θ = 180 and k = 1 this corresponds

to the absolute velocity on exit being 0

and all kinetic energy converted.

In practice:

θ 165° (to avoid interference with the following bucket)

the jet velocity v is controlled; (turbine usually synchronised to the electricity grid)

the optimal speed ratio u / v is nearer 0.46.

Other considerations:

Head at device, H = (head at reservoir) − (head losses along pipeline)

Maximum jet velocity,

Orifice coefficient, cv 0.97 – 0.99

gHcv v 2

bucket

jet uv

k(v-u)

v-u

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EXAMPLE, PAGE 19

In a Pelton wheel, 6 jets of water, each with a

diameter of 75 mm and carrying a discharge of

0.15 m3 s–1 impinge on buckets arranged around a

1.5 m diameter Pelton wheel rotating at 180 rpm.

The water is turned through 165° by each bucket

and leaves with 90% of the original relative velocity.

Neglecting mechanical and electrical losses within

the turbine, calculate the power output.

CAVITATION

Bubbles of vapour form if pressure is too low; a problem on:

– inlet / suction side of pump

– outlet / draft tube of turbine

Subsequent bubble collapse causes:

– surface pitting

– vibration

– noise

– loss of performance

CAVITATION

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NET POSITIVE SUCTION HEAD

g

pp cavinlet

ρNPSH

losshead sumpinletpump HH

Bernoulli:

fatm

inletinlet h

g

p

g

Vz

g

p

ρ2ρ

2

finletatminlet h

g

Vz

g

p

g

p

2ρρ

2

To prevent cavitation, keep pinlet as large as possible by:

keeping zinlet small (better still, negative – i.e. below level in sump)

keeping V small (use large-diameter pipes)

keeping hf small (use short, large-diameter pipes)

Keeping the pump below the level in the sump is also useful for pump-priming.

Sump

pump inlet

zinletpatm