3-01-CP note 03 - fkm.utm.mysyahruls/resources/SKMM2323/3-01-CP-note-03(1).… · motor is rated at...
Transcript of 3-01-CP note 03 - fkm.utm.mysyahruls/resources/SKMM2323/3-01-CP-note-03(1).… · motor is rated at...
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.
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
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
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�����