CENTRIFUGALPUMP

BASICEQUATIONRotationalspeed

𝑢 = 𝑟𝜔 =𝜋𝐷𝑁60

𝑢 = linearvelocityinm/s𝑟 = radiusinm𝜔 = angularvelocityinrad/s𝐷 = diameterinm𝑁 = rotationperminutePower𝑃𝑜𝑤𝑒𝑟 = 𝐹 ∙ 𝑉 = 𝑃 ∙ 𝐴 ∙ 𝑉 = 𝜌𝑔ℎ ∙ 𝐴𝑉 = 𝜌𝑔ℎ ∙ 𝑄

𝑃𝑜𝑤𝑒𝑟 = 𝜌𝑔𝑄 ∙ ℎ

𝑊𝑜𝑟𝑘 = Torque×AngularvelocityWork

𝑊𝑜𝑟𝑘 = Force×Distance

TURBOMACHINESTurbomachines are the commonly employed devicesthateithersupplyorextractenergyfromaflowingfluidbymeansofrotatingpropellersorvanes.PUMP:Pumpaddsenergytoasystem,withtheresultthatthepressureisincreased.Italsocausesflowtooccuroritincreasestherateofflow.TURBINE:Aturbineextractsenergyfromasystemandconvertsittosomeotherusefulform,typically,toelectricpower.Hydroturbine:isamachinethatgeneratespowerfromhigh-pressure water; relatively large conduits ortunnels deliver fluid to closed turbines in order togeneratepower.Anotherexample:steamturbineandairturbine.

PUMPCLASSIFICATION

CENTRIFUGALPUMPAcentrifugalpumpconsistsoftwoprincipalparts:(1) Impeller: which imparts a rotarymotion to the

liquid.(2) Housing or casing:which directs the liquid into

theimpellerregionandtransportsitawayunderahighpressure.

Theimpellerismountedonashaftandisoftendrivenbyanelectricmotor.The casing includes the suction anddischargenozzlesand houses the impeller assembly. The portion of thecasingsurroundingtheimpelleristermedthevolute.Liquidentersthroughthesuctionnozzletotheimpellereye and travels along the shroud, developing a rotarymotionduetotheimpellervanes.

It leaves the volute casing peripherally at a higherpressurethroughthedischargingnozzle.Somesingle-suctionimpellersareopen,withthefrontshroudremoved.Double-suctionimpellershaveliquidenteringfrombothsides.

HEADOFPUMP(Manometrichead)This isdefinedbyBritishStandardsas the sumof theactuallift(H)+thefrictionlossesinthepipes(hf)+thedischargevelocityhead.

𝐻s = 𝐻 + ℎt +𝑉uv

2𝑔=𝑃v − 𝑃x𝜌𝑔

+𝑉vv − 𝑉xv

2𝑔

However, for special pumps allowance must also bemadeforthevelocityofflowtowardsthesuctionintakeandanypressuredifferencesatthewatersurfacesinthesupplyandreceivingtanks.Commonly thesuctionanddeliverypipesareofequaldiameter.Inwhichcase:

𝐻s =𝑃v − 𝑃x𝜌𝑔

VELOCITYTRIANGLE

Legend:Atinlet(1)𝑢x = 𝑟x𝜔 = Tangetialvelocityofimpeller𝑣x = Absolutevelocityat𝛼xtotangent𝑣}x = 𝑣x − 𝑢x = RelataivevelocitytoimpellerbladeComponentvelocityfor𝑣x:𝑣~x = Whirlvelocity𝑣tx = Radialflowvelocity𝛽x = InletbladeangleAtoutlet(2)𝑢v = 𝑟v𝜔 = Tangetialvelocityofimpeller𝑣v = Absolutevelocityat𝛼vtotangent𝑣}v = 𝑣v − 𝑢v = RelataivevelocitytoimpellerbladeComponentvelocityfor𝑣v:𝑣~v = Whirlvelocity𝑣tv = Radialflowvelocity𝛽v = Inletbladeangle

BLADETYPE1. Forwardblade2. Radialblade3. Backwardblade

FORWARDBLADE

RADIALBLADE

BACKWARDBLADE

THEEFFECTOFBLADETYPECentrifugalpumpsdonotalwayshavebackwardcurvedvanes. Butwhen they do, it ismostly for fluids in theincompressibleregimeofoperationsuchaswater.Forcompressible operation of fluids such as air, forwardcurve-vanedcentrifugalpumpsareused.Thenetidealheaddevelopedbyacentrifugalpumpisgivenby:

𝐻�u��� = 𝐴 − 𝐵𝑄𝑄 = volumeflowrateattheimpelleroutlet𝐴, 𝐵 = constantforagivenimpellerrunningatagivenspeedAdditionally,𝐵 ∝ cot 𝛽v .

Donotethatthevalueoftheactualheaddevelopedbythepumpwillbe lower than this idealvalueowing toshocks

𝐻����� = 𝐾x 𝑄u − 𝑄 v𝑄u = designvolumeflowrate𝑄 = actualvolumeflowrateFrictioncanbecalculatedby:

ℎt = 𝐾v𝑄vwhichtogetherconstitutehydrauliclosses.

The power required to drive the pump to provide agivenflow-rateisgivenas:

𝑃 = 𝜌𝑔𝑄 ∙ 𝐻�u���Therepresentativecurvesaregivenbelow.

Asisevidentfromthepower-dischargecharacteristicsof the radialand forwardvanedcentrifugalpump, thepower requirement increases monotonically with anincreaseindischarge.Hence,ifthepumpmotorisratedformaximumpower,thenitwillremainunder-utilizedformostoftheoperatingtime,andresultinanincreasedcost due to its higher rating. On the other hand, if amotor is rated at the design point, and due to somereasontheflow-rateexceedsthedesignflowrate,thenthepowerrequirementwillshootup(incaseofforwardandradialvanesonly),causingoverloadingandmotorfailure.However,forbackwardcurve-vanedcentrifugalpumps,if the flow-rate exceeds the design flow rate (occursquiteclosetothemaximaofthepower-dischargecurve),thencontrarytotheearliercase,thepowerrequirementdropsdownasevidentfromthecurves.Thisenablesthemotorwhichisratedatthedesignpowertohandletheentire range of flow-rates without any problems. Theactualdesignpointislocatedcorrespondingtotheflow-rateatwhichmaximumefficiencyoccurs.

EULERHEADTorque=kadarperubahanmomentumsudutMomentumsudut=(jisim)×(halajutangen)×(jejari)

Momentumsudutmasuk = 𝑚𝑣~x𝑟x

Momentumsudutkeluar = 𝑚𝑣~v𝑟v

𝑚 = kadarjisimmengalirsesaatKadarperubahanmomentumsudut:

𝑇 = 𝑚𝑣~v𝑟v − 𝑚𝑣~x𝑟x𝑚 = 𝜌𝐴𝑉 = 𝜌𝑄

𝑇 = 𝜌𝑄 𝑣~v𝑟v − 𝑣~x𝑟x Diketahuipowerialah:

𝑃 = 𝑇𝜔𝑃 = 𝜌𝑄 𝑣~v𝑟v − 𝑣~x𝑟x 𝜔

Diketahui:

𝑢 = 𝑟𝜔𝑢x = 𝑟x𝜔𝑢v = 𝑟v𝜔

𝑟x =���and𝑟v =

���

Diketahuipowerialah:𝑃 = 𝜌𝑄 𝑣~v𝑟v − 𝑣~x𝑟x 𝜔

= 𝜌𝑄 𝑣~v𝑢v𝜔− 𝑣~x

𝑢v𝜔

𝜔

= 𝜌𝑄 𝑣~v𝑢v − 𝑣~x𝑢x Powerjugabolehditulissebagai: 𝑃 = 𝜌𝑔𝑄 ∙ ℎJika power adalah maksimum, nilai h ialah nilaimaksimum, iaitu niai power dalam keadaan tiadakehilangantenaga(losses,friction,etc).Nilaihbolehditulissebagai𝐻�(Eulerhead)

𝑃 = 𝜌𝑔𝑄 ∙ 𝐻� = 𝜌𝑄 𝑣~v𝑢v − 𝑣~x𝑢x

𝐻� =1𝑔𝑣~v𝑢v − 𝑣~x𝑢x

IakenalisebagaiEulerhead(turusEuler).Unitnyadalammeter(m).Ia adalah turus ideal yang dihasilkan oleh impeller(pendesak)dalamsystempam.

PUMPEFFICIENCY(kecekapanpam)Manometricefficiency

𝜂s��� =Kuasaairyangdihasilkan

Kuasaimpeller

=𝜌𝑔𝑄 ∙ 𝐻s𝜌𝑔𝑄 ∙ 𝐻�

=

𝜌𝑔𝑄 ∙ 𝐻s

𝜌𝑔𝑄 ∙ 1𝑔 𝑣~v𝑢v − 𝑣~x𝑢x

𝜂s��� =𝑔𝐻s

𝑣~v𝑢v − 𝑣~x𝑢x

Mechanicalefficiency

𝜂s��� =Kuasaimpeller

Kuasayangdiberikankepadasyaf

𝜂s��� =

1𝑔 𝑣~v𝑢v − 𝑣~x𝑢x

𝑃�����

Overallefficiency

𝜂� =Kuasaairyangdihasilkan

Kuasayangdiberikankepadasyaf

𝜂� =𝜌𝑔𝑄 ∙ 𝐻s𝑃�����