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PUMPS AND PUMPINGSTATIONS

i , Civil Engineering DepartmentZerihun Alemayehu

AAiT Water Supply & Urban DrainageBy Zerihun Alemayehu

P UMPING The operation of lifting water or any fluid is calledpumpingPump , a mechanical machine , is used for liftingwateror any fluid to a higher elevations or at higherpressures.

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P URPOSES OF P UMPING

To increase the water pressure at certain points in thedistribution system.To lift treated water to elevated storage tanks flowautomatically under gravity into distribution system.To lift raw river water to carry it to treatment plant.To lift water available from wells to an elevatedstorage tank in stages.To pump water directly into the distribution system.

To take out water from basins, sumps, tanks etc.


T YPES OF PUMPS Classification based on mechanical principle of operation(i) Displacement pumps(ii) Centrifugal pumps(iii) Air lift pumps(iv) Miscellaneous pumps

Classification based on type of power required(i) Steam engine pumps(ii) Diesel engine pumps

(iii) Electrically driven pumps

Classification based on the type of service called for(i) Low lift pumps(ii) High lift pumps(iii) Deep well pumps(iv) Booster pumps

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AAiT

S ELECTION OF PUMP

Capacity of pumpNumber of pump units requiredSuction conditionsLift (total head)Discharge conditions and variations in loadFloor space requirement

Flexibility of operationStarting and priming characteristics

Type of drive requiredInitial costs and running costs.


C ENTRIFUGAL PUMPS Rotodynamic pumps which convert Mechanicalenergy into Hydraulic energy by centripetal force onthe liquid.

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C ENTRIFUGAL PUMPS …

key performance parameters of centrifugal pumpsare capacity, head, BHP (Brake horse power),BEP (Best efficiency point) and specific speed.

pump operating curve


P UMPING TERMS

Head A Centrifugal pump velocity energy to aliquid pressure energy.

60

D RPM vvelocityimpeller peripheral

Head =

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SUCTION LIFT existswhen the source of supplyis below the center line ofthe pump.STATIC SUCTION LIFT isthe vertical distance inmeter from the centerlineof the pump to the free

level of the liquid to bepumped.

P UMPING TERMS …


P UMPING TERMS … STATIC DISCHARGE HEAD isthe vertical distance betweenthe pump centerline and thepoint of free dischargeTOTAL STATIC HEAD is thevertical distance between the

free level of the source of supplyand the point of free dischargeor the free surface of thedischarge liquid.

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P UMPING TERMS …

FRICTION HEAD (h f ) is the head required to overcomethe resistance to flow in the pipe and fittings.TOTAL DYNAMIC SUCTION LIFT (hs) is the staticsuction lift minus the velocity head at the pump suctionflange plus the total friction head in the suction line.TOTAL DYNAMIC SUCTION HEAD (hs) is the staticsuction head plus the velocity head at the pump suctionflange minus the total friction head in the suction line.


P UMPING TERMS …

TOTAL DYNAMIC DISCHARGE HEAD (hd) is the staticdischarge head plus the velocity head at the pumpdischarge flange plus the total friction head in thedischarge line.HEAD (H) or TOTAL Dynamic HEAD (TDH)

TDH = h d + h s with a suction lift)TDH = h d - h s with a suction head)

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P OWER AND E FFICIENCY

The work performed by a pump is a function of the total headand the weight of the liquid pumped in a given time period.Work done by the pump = W x H m-kg/sec = x Q x H m-kg/secPump input or brake horsepower (BHP) is the actualhorsepower delivered to the pump shaft.Pump output or hydraulic horsepower (WHP) is the liquidhorsepower delivered by the pump.

Efficiency PumpGr SpTDH Q

BHP

Gr SpTDH QWHP

75

..75

..

BHP Gr SpTDH Q

BHP WHP Efficiency Pump

75..


E XAMPLE 1Population of a city is 120,000 and rate of water supplyper head per day is 200 liters. Calculate the BHP ofmotor to raise the water to an overhead tank 50 m high.Length and diameter of the rising main is 200 m and 40cm, respectively. Assume motor efficiency 90 % and theof the pump 60 %. Take f = 0.01 and peak hourly

demand as 1.5 times the average demand.

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S OLUTION OF EXAMPLE 1 Average demand = 120,000 x 200

= 24 x 106 L/day= 24000 m3/day (0.28 m3/sec)

Peak hr demand = 1.5 x 0.28 m3/sec = 0.42 m3/secWeight of liquid delivered by the pump

=1000 kg/m3 x 0.42 m3/sec = 420 kg/secTotal static head = 50 m

H = h s + h f = 50 + 2.85 = 52.85 m

η = 0.9 x 0.6 = 0.54


S YSTEM -HEAD CURVE

System Head Curve: therelationship between flowand hydraulic losses in asystem

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S YSTEM -HEAD CURVE


P UMPS OPERATING IN PARALLEL

for a purpose of increasing the total discharge .Pumps should deliver the same head .The total system flow rate is equal to the sum of the flow ratesof contributions from each pump.

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P UMPS OPERATING IN SERIES

for a purpose of increasing the total head .the pumps connected should deliver the same discharge .The total system head is equal to the sum of thecontributions from each pump.


C AVITATION

Cavitation is a phenomenon of cavity formation or theformation and collapse of cavities.Cavities develop when the absolute pressure in aliquid reaches the vapor pressure related to the liquidtemperature.When the net positive suction head (NPSH) is

reduced NPSHmin detrimental cavitationThe minimum static lift is given as

Where p a is atmospheric pressure, p v is vapor pressure of fluidand h ls head loss in the suction pipeNPSH is obtained from manufuctures

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E XAMPLES 2

The pump shown in the figure below has a headcharacteristics that can be expressed by H = 100 6000Q 1.85

Where H = pump head in meters and Q =discharge in m 3/s.

a. Calculate the head and discharge of thepump.

b. Check the potential for cavitation if theanticipated maximum vapor pressure andminimum absolute barometric pressure are0.40 m and 9.70 m, respectively. NPSHrequired for the pump is 3.0 m. Neglect minorhead losses.

600 m P 603 m Elev.

50 m length

250 mm diameter

C = 100

1500 m length

200 mm diameter

C = 100

640 m Elev.


S OLUTION OF E XAMPLE 2

First calculate the TDHTDH = Hs + hld+ hlsHs = 640 – 600 = 40 m

After inserting D and L we getTDH = 40 +8067.66 Q 1.85 Since the TDH and the head delivered by the pump has to be the

same we have:100-6000 Q 1.85 = 40 +8067.66 Q 1.85

Q = 0.0523 m3/sec and H = 74.45 m

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S OLUTION OF E XAMPLE 2

Habs = 9.70 m, Hvap = 0.40 m and NSPH = 3.0 mNPSH = Habs – (Hvap + hls + suction lift)

Suction lift = 603-600 =3 m

NSPH = 9.7 – (0.4 + 0.3824 + 3)= 5.92 > 3no cavitation

AAiT

Project Submission DateJuly 2, 2010 Before 12:00 PM

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