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Efficiency of a pump

Volumetric Hydraulic Mechanical

• QL is the loss of fluid due to leakage in the impeller-casing • hf is the summation of shock loss (imperfect match between inlet flow and blade entrances);

friction loss (in blades); and circulation loss (imperfect matching at the exit side) • Pf is power loss due to mechanical friction in bearings, packing and similar contact points

In pump design improving all three is needed as overall efficiency is given by

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η = 𝜂𝑣𝜂ℎ𝜂𝑚 = 𝜂𝑝

Also,

𝜂𝑝 =𝜌𝑔𝑄𝐻

𝜏𝜔

𝜂𝑀 =𝜏𝜔

𝑉𝐼 Motor

Pump

Overall efficiency of the pump can be calculated from above two equations. Then the efficiency of the motor can be found by;

Product of overall efficiency of the pump and efficiency of the motor will give the efficiency of the complete system (i.e. the system consisting of the motor and the pump)

𝜂𝑠𝑦𝑠𝑡𝑒𝑚/𝑜𝑣𝑒𝑟𝑎𝑙𝑙 = 𝜂𝑀 × 𝜂𝑝

Pump (efficiency)- examples

• Example 1: A pump delivers 1500 L/min of water at 20°C against a pressure rise of 270 kPa. Kinetic and potential energy changes are negligible. If the driving motor supplies 9 kW, what is the overall efficiency?

• Example 2: A centrifugal pump delivers 2.1 m3/min of water at 20°C (ρ =1000 kg/m3) when the brake horsepower is 16.5 kW and the efficiency is 71%. (a) Estimate the head rise in ft and the pressure rise in kPa. (b) Also estimate the horsepower if instead the delivery is 2.1 m3/min of gasoline at same efficiency (ρ =680 kg/m3) at 20°C.

• Example 3: A pump must move a discharge of 0.031 m3/s against a head of 9.1 m. Total head loss in the pipelines is 1.2 m. If the pump is driven by a 5 KW motor with an efficiency of 75%, calculate the pump efficiency and break horse power?

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Elementary pump theory

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• Simple one-dimensional flow formula • Complex computer aided models that account

three-dimensionality and viscosity

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By angular momentum theory

Vt1 and Vt2 are absolute circumferential velocity components of the flow; then Power delivered to fluid can be calculated as

Referred as Euler equations for pumps. It shows torque, power and head are functions of rotor-tip velocities (u) and absolute fluid velocities (v) only; in other words independent of any axial velocities through the machine

Bernoulli equation in rotating coordinates

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Example: Given are the following data for a commercial centrifugal water pump: r1= 4 in, r2 = 7

in, β1= 30°, β2= 20°, speed = 1440 r/min. Estimate (a) the design-point discharge, (b) the water horse power, and (c) the head if b1 = b2 = 1.75 in.

• As radial velocity Vn can be expressed as 𝑉𝑛 = 𝑉𝑡𝑡𝑎𝑛 ∝

• Therefore, by continuity equation power of the pump can be expressed as : • b1 and b2 are inlet and outlet blade widths. This

equation also can be used to calculate the design flow rate and usually estimated assuming the flow enters exactly normal to the impeller (i.e. α1 is 90°)

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Effect of blade angle on pump head

Effect of blade angle on pump head

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• If inlet angular momentum is neglected

• You already know by application of angular momentum theory, that

• Where Vt2 is

• Then H becomes

𝑦 = 𝑐 − 𝑚𝑥

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Pump performance and characteristics

Use of characteristic pump curves

• Pump performance most commonly shown by means of plotted curves

• Even though a pump is designed for specific conditions (e.g. specified speed and flow rate), in reality it may operate under different conditions

• In such cases to predict the behavior (output) under varying conditions, the pump curves are important.

• These curves are derived after running several pumps of the same design

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Hea

d/P

ow

er/E

ffic

ien

cy

Pump performance

Shut off head

Actually only two curves are needed as head, power and efficiency is related through the following equation

# above plot is derived for a pump operating at a constant impeller speed

will be discussed later

Pump efficiency vs discharge

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Overall pump efficiency (refer slide 21)

Effi

cie

ncy

Discharge

Head vs discharge

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Shut off head

He

ad

Discharge

• Head developed by the pump at zero discharge is called shut-off head

• Represents rise in pressure head across the pump when the discharge valve is closed

• As there is no discharge (flow), the efficiency is zero, and the energy supplied to the pump (i.e. bhp at Q=0) will be dissipated as heat

• Discharge valve of centrifugal pumps can be kept closed for a short period.

• Prolonged closed valves will damage the pump due to overheating and mechanical stress

Horse power vs discharge

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Ho

rse

po

wer

Discharge

Ho

rse

po

wer

Discharge

or

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Effi

cie

ncy

Discharge

He

ad

Discharge

He

ad

Discharge

Q optimum

Efficiency maximum

Q optimum Q optimum

#The points on different curves that corresponds to maximum efficiency is refereed as best efficiency points #It is advisable to operate a pump near its maximum efficiency

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Best efficiency points

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Measured performance curves Impellers with different diameters may be used in a given casing.

He

ad

Discharge

Decreasing diameter

Relationship is same for

decreased speed

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Iso-efficiency lines

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It should be noted that pumps has to supply power (head) to overcome losses in pipes (e.g. friction loss) Therefore, the head required is elevation the water needed to pump + losses

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Net Positive Suction Head (NPSH) • NPSH is the head required at the pump inlet to keep the

fluid from cavitation ! (a noisy, vibrating pump with a sharp drop of pump head and discharge)

• The pump inlet (suction side) is the low pressure point where cavitation first occurs

• NPSH is defined as (pv is the vapour pressure)

Needs to be greater than NPSH to avoid cavitation

• If the pump inlet is at a height z1 above the reservoir (; i.e. free water surface is zi m below the pump), whose free water surface is at pressure pa NPSH can be written as (hf is the friction loss);

Therefore, NPSH is the difference between suction head and liquids vapour head. Vapour pressure can be obtained from known charts/books. Depending on the Zi, the required NPSH can be worked out

NPSH(approximate) = 𝑝𝑎

𝜌𝑔−

𝑝𝑣

𝜌𝑔

A few more details of cavitation

• Formation of vapour cavities (bubbles/voids) in a liquid (air bubbles/voids means cavities have low pressure!)

• Occurs when a liquid pressure at a given location reduces to the vapour pressure of the liquid

• Once subjected to higher pressure, these cavities will crash/explode and can generate an intense shockwave.

• Collapsing voids that implode near to a metal surface cause surface fatigue of the metal of impeller and/or casing

• Also a sharp drop of discharge and a higher power (electrical) consumption take place when cavitation occurs

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Cavitation- quantitative explanation

• A pump delivers 1500 L/min of water at 20 °C against a pressure rise of 270 kPa, to a 10 m elevation. Inlet diameter is 150 mm and the delivery diameter is 100 mm. • (a) Estimate the useful energy supplied by the pump if its locates on the ground level? • (b)What would be the pressure at suction inlet (friction loss in the suction pipe is given by

2v2/2g; v is the velocity of the suction pipe) (b) What would be the suction inlet’s pressure if the reservoir is 10 m below the level of (a)

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