CENTRIFUGALPUMP

BASICEQUATIONRotationalspeed

= = ?@ =ABC

60

= = linearvelocityinm/s? = radiusinm@ = angularvelocityinrad/sB = diameterinmC = rotationperminutePowerOPQR? = S ∙ U = O ∙ V ∙ U = WXℎ ∙ VU = WXℎ ∙ Z

OPQR? = WXZ ∙ ℎ

[P?\ = Torque×AngularvelocityWork

[P?\ = 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.

rs = r + ℎt +Uuv

2X=Ov − OxWX

+Uvv − Ux

v

2X

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

rs =Ov − OxWX

VELOCITYTRIANGLE

Legend:Atinlet(1)=x = ?x@ = Tangetialvelocityofimpeller{x = Absolutevelocityat|xtotangent{}x = {x − =x = RelataivevelocitytoimpellerbladeComponentvelocityfor{x:{~x = Whirlvelocity{tx = Radialflowvelocity�x = InletbladeangleAtoutlet(2)=v = ?v@ = Tangetialvelocityofimpeller{v = Absolutevelocityat|vtotangent{}v = {v − =v = RelataivevelocitytoimpellerbladeComponentvelocityfor{v:{~v = Whirlvelocity{tv = Radialflowvelocity�v = 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:

rÇuÉÑÖ = V − ÜZZ = volumeflowrateattheimpelleroutletV, Ü = constantforagivenimpellerrunningatagivenspeed

Additionally,Ü ∝ cot �v .

Donotethatthevalueoftheactualheaddevelopedbythepumpwillbe lower than this idealvalueowing toshocks

ràâäãå = çx Zu − Zv

Zu = designvolumeflowrateZ = actualvolumeflowrate

Frictioncanbecalculatedby:

ℎt = çvZv

whichtogetherconstitutehydrauliclosses.

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

O = WXZ ∙ rÇuÉÑÖTherepresentativecurvesaregivenbelow.

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 = è{~x?x

Momentumsudutkeluar = è{~v?v

è = kadarjisimmengalirsesaatKadarperubahanmomentumsudut:

ê = è{~v?v − è{~x?x

è = WVU = WZ

ê = WZ {~v?v − {~x?x

Diketahuipowerialah:

O = ê@

O = WZ {~v?v − {~x?x @Diketahui:

= = ?@

=x = ?x@

=v = ?v@

?x =ëíìand?v =

ëîì

Diketahuipowerialah:

O = WZ {~v?v − {~x?x @

= WZ {~v=v@− {~x

=v@

@

= WZ {~v=v − {~x=x

Powerjugabolehditulissebagai:

O = WXZ ∙ ℎ

Jika power adalah maksimum, nilai h ialah nilaimaksimum, iaitu niai power dalam keadaan tiadakehilangantenaga(losses,friction,etc).Nilaihbolehditulissebagairñ(Eulerhead)

O = WXZ ∙ rñ = WZ {~v=v − {~x=x

rñ =1

X{~v=v − {~x=x

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

PUMPEFFICIENCY(kecekapanpam)Manometricefficiency

ósÑòä =Kuasaairyangdihasilkan

Kuasaimpeller

=WXZ ∙ rsWXZ ∙ rñ

=

WXZ ∙ rs

WXZ ∙1X {~v=v − {~x=x

ósÑòä =Xrs

{~v=v − {~x=x

Mechanicalefficiency

ósÉãâ =Kuasaimpeller

Kuasayangdiberikankepadasyaf

ósÉãâ =

1X {~v=v − {~x=x

OÇòôëö

Overallefficiency

óä =Kuasaairyangdihasilkan

Kuasayangdiberikankepadasyaf

óä =WXZ ∙ rsOÇòôëö

CENTRIFUGALPUMPTUTORIALO1

QUESTION1A centrifugal pump is driven by an electricmotor at 1450 rpm.Outletdiameterofblade,outletbladewidthandoutletbladeangleare600mm,400mmand30o,respectively.Inletdiameterofblade,inletbladewidthandinletbladeangleare300mm,80mmand20ᵒrespectively. Pressure at suction pipe and delivery are positive13.5 bar and negative 0.5 bar, respectively. Assume that thediameter for suction and delivery pipes is equal. The flowrateinsidethepumpis0.3m3/s.Determine:i. Themonometrichead,Hmii. Themanometricefficiency,WXiii. Powerrequiredbyelectricmotorifoverallefficiencyis98%.QUESTION2Acentrifugalpumphasinletandoutletdiameterof30cmand60cm,respectively.Impellerwidthatoutletis12cm.Bladethicknessoccupied10percentofthecircumference.Bladeisbackwardwithinlet and outlet blade angle is 30ᵒ and 40ᵒ, respectively. Theflowrateis0.5m3/s.Assumetherearenowhirlatinletandvelocityofflowisconstantdetermine:i. Therotationofpumpinrpmii. Theoutputpowerifmanometricefficiencyis85%iii. Thepressuredifferenceacrosstheimpeller

QUESTION3Centrifugalpumpsupplieswaterattherateof400liter/sandthepressuredifferenceacrossthepumpis200kN/m3.Outletdiameterandoutletwidthare40cmand10cm,respectively.Bladethicknessoccupied10%ofthecircumference.Impellerinletdiameterishalfof theoutletdiameter.Assume losses in casingand impellerarenegligible and zero whirl at inlet. The diameter of suction anddeliverypipesisequal.Ifthebladesareradial,determine:i. Thepumppowerinputinhorsepowerifoverallefficiencyis

80%ii. Theimpellerspeedinrpmiii. TheinletbladeangleifvelocityofflowisconstantQUESTION4Theinletandoutletimpellerdiameterofcentrifugalpumpare200mmand400m,respectively.Impellerwidthatinletandoutletare15mmand8mm,respectively.Bladesarebackwardwithangleof38ᵒ. Pump operates at 1500 rpm. The flowrate is 15 liter/s.Determinethepressurechangesintheimpeller.Assumenoenergylosses.

QUESTION01

QUESTION02

QUESTION03

QUESTION04