chapter11 ac power analysis.ppt

chapter11acpoweranalysis,11.1instantaneousandaveragepower11.2apparentpowerandpowerfactor11.3maximumaveragepowertransfer11.4complexpower11.5conservationofacpower11.6powerfactorcorrection,1,introduction,inthischapter,ourgoalsandobjectivesincludedeterminingtheinstantaneouspowerdeliveredtoanelementdefiningtheaveragepowersuppliedbyasinusoidalsourceusingcomplexpowertoidentifyaverageandreactivepoweridentifyingthepowerfactorofagivenload,andlearningmeansofimprovingit,2,instantaneousandaveragepower,1.instantaneouspower（瞬时功率）,3,theaveragepoweristheaverageoftheinstantaneouspoweroveroneperiod.,2.averagepower,realpower(平均功率、有功功率),4,instantaneouspower：瞬时功率,averagepower：平均功率/有功功率,poweranalysistor:,5,poweranalysistol:,instantaneouspower：,averagepower：,storedmagneticenergy：,qisameasureoftheenergyexchangebetweenthesourceandthereactivepartoftheload,6,poweranalysistoc:,instantaneouspower：,averagepower：,storedelectricenergy：,7,inductor=90,capacitor=-90,p=0,8,maximumaveragepowertransfer,thecurrentthroughtheloadis,9,averagepowerdeliveredtoloadis:,ourobjectiveistoadjusttheloadparametersrlandxlsothatpismaximum.todothiswesetp/rlandp/xlequaltozero.weobtain,p/rl=0,p/xl=0,10,formaximumaveragepowertransfer,theloadimpedancezlmustbeequaltothecomplexconjugateofthetheveninimpedancezth.,settingrl=rthandxl=-xthineq.(1)givesusthemaximumaveragepoweras,11,inasituationinwhichtheloadispurelyreal,theconditionformaximumpowertransferisobtainedfromeq.(2)bysettingxl=0;thatis,12,example：giventhatv(t)=120cos(377t+450）vandi(t)=10cos(377t-100)afindtheinstantaneouspowerandtheaveragepowerabsorbedbythepassivelinearnetworkoffig.11.1.,13,solution:theinstantaneouspowerisgivenbyp=vi=1200cos(377t+450)cos(377t-100)applyingthetrigonometricidentity,orp(t)=344.2+600cos(754t+350)w,theaveragepoweris,p(t)=600cos(754t+350)+cos550,14,example：forthecircuitshowninfig.,findtheaveragepowersuppliedbythesourceandtheaveragepowerabsorbedbytheresistor.,15,solution:,thecurrentthroughtheresistoris,theaveragepowersuppliedbythevoltagesourceis,andthevoltageacrossitis,16,whichisthesameastheaveragepowersupplied.zeroaveragepowerisabsorbedbythecapacitor.,theaveragepowerabsorbedbytheresistoris,example:determinethepowergeneratedbyeachsourceandtheaveragepowerabsorbedbyeachpassiveelementinthecircuitfig11.3.,17,solution:weapplymeshanalysisasshowninfig11.3.formesh1,formesh2,18,forthevoltagesource,thecurrentflowingfromitisi2=10.5879.10a.andthevoltageacrossitis60300v,sothattheaveragepoweris,v1=20i1+j10(i1-i2)=80+j10(4-4-j10.39)=183.9+j20=184.9846.210v,19,theaveragepowersuppliedbythecurrentsourceis,p1+p2+p3+p4+p5=-735.6+320+0+0+415.6=0,v3=-j5i2=(5-900)(10.5879.10)=52.9(79.10-900),20,11.4effectiveorrmsvalue,theeffectivevalueofaperiodiccurrentisthedccurrentthatdeliversthesameaveragepowertoaresistorastheperiodiccurrent.,theeffectivevalueofaperiodicsignalisitsrootmeansquare(rms)value.,21,thepowerfactoristheratiooftheaveragepowertotheapparentpower,whichisdimensionless.,apparentpowerandpowerfactor,1.apparentpower（视在功率）,theapparentpower(inva)istheproductofthermsvaluesofvoltageandcurrent.,1)pfisthecosineofthephasedifferencebetweenvoltageandcurrent.2)pfisalsothecosineoftheangleoftheloadimpedance.,22,generallyspeaking,0cosj1,x0,j0,inductive,currentlagsvoltage,x0,j0,pc=uicos=uicos(-90)=0,qc=uisin=uisin(-90)=-ui=u2/xc=i2xc0,weknowzlisinductiveload.,zl=r+jxl,(1),weset,becausezlisinductiveload,(2),(1),(2),analyzethecircuitoffig.11.46tofindthecomplexpowerabsorbedbyeachofthefivecircuitelements.,example11.6,solution：,example11.8:theloadinfig.11.44draws10kvaatpf=0.8lagging.if|il|=40arms,whatmustbethevalueofctocausethesourcetooperateatpf=0.9lagging?,solution:,example:considerthecircuitoffig.11.45.specifythevalueofcapacitancerequiredtoraisethepfofthetotalloadconnectedtothesourceto0.92laggingifthecapacitanceisadded(a)inserieswiththe100-mhinductor;(b)inparallelwiththe100-mhinductor.,zeff=j100+j300|200=23754.25o.=pf=cos54.25o=0.5843lagging.,(a)raisepfto0.92laggingwithseriescapacitance,znew=j100+jxc+j300|200=138.5+j(192.3+xc)w,solution:,solving,wefindthatxc=133.3=1/wc,sothatc=7.501mf,(b