电力系统分析英文课件PowerFlowAnalysis

1Instructor:KaiSunFall2014ECE421/599ElectricEnergySystems6–PowerFlowAnalysis2Introduction•Powerflow(loadflow)analysis–Steady-stateanalysisofaninterconnectedpowersystem–TosolvethepowerflowequationsforVi(i.e.|Vi|andδi)andSi(i.e.PiandQi)–Basisofpowersystemanalysisanddesign•Assumptionsforapowersystemtobestudied–Balancedconditions–Representedbyasingle-phasenetwork–Modeledbynodes(buses)andbranches–Allimpedances/admittancesareinperunitonacommonMVAbase(typically100MVA)3BusAdmittanceMatrix•Branchadmittance•ApplyKCLtonodes(buses)1-4Ii:currentinjectedbyanequivalentcurrentsource11ijijijijyzrjx==+1101121213132202122123232332133134343443()()()()0()()()0()IyVyVVyVVIyVyVVyVVyVVyVVyVVyVV=+−+−=+−+−=−+−+−=−z12=z10=z13=z20=z23=z34=11012131122133212120122322331312321323343344343344()()0()0IyyyVyVyVIyVyyyVyVyVyVyyyVyVyVyV=++−−=−+++−=−−+++−=−+1110121322201223331323344434YyyyYyyyYyyyYy=++=++=++=122112133113233223344334YYyYYyYYyYYy==−==−==−==−14410YY==42240YY==0111101101101110()(0)(0)IEVyEyVyIVy∆=−=+−=+−E1V1E2I010,1,=≠=≠=−∑∑nniiijijjjjijjiIVyyV4•Ibus:currentsinjectedbyexternalcurrentsources•Vbus:busvoltagesrelativetoareference(notincluded),usuallytheground•Ybus:busadmittancematrix(symmetricandsparse)–Diagonalelements:calledself-admittancesordrivingpointadmittances–Off-diagonalelements:calledmutualadmittancesortransferadmittances•Zbus=Y-1bus:busimpedancematrix1111122133144221122223324433113223333444411422433444IYVYVYVYVIYVYVYVYVIYVYVYVYVIYVYVYVYV=+++=+++=+++=+++11111211222122221212............::::::......::::::......ininiiiiiniinninnnnnIVYYYYYYYYIVYYYYIVYYYYIV=busbusbusI=YV0,niiijjjiYy=≠=∑ijjiijYYy==−-1busbusbusbusbus=V=YIZIji≠0,1,1=≠=≠==−=∑∑∑nnniiijijjijjjjijjijIVyyVYV5•IfitisknownthatE1=1.1∠0opuandE2=1.0∠0opu8.502.505.0002.508.755.0005.005.0022.5012.500012.5012.50jjjjjjjjjjjj−−=−−busY10.500.400.450.450.400.480.440.440.450.440.5450.5450.450.440.5450.625jjjjjjjjjjjjjjjj−==busbusZYHowmanynon-zeroelementsforasystemwithNbuses(notincludingtheground)andMbranches?111022201.11.1pu1.01.01.25pu0.8EIjzjEIjzj===−===−12341.11.0501.251.04001.04501.045VjVjVV−−=bus=Zz12=z10=z13=z20=z23=z34=2M+NE1E2Whyaretheysame?6Amoregeneralpowerflowstudy•Howtosolveallbusvoltagesandlinerealandreactivepowerflows?z12=z10=z13=z20=z23=j0.2z34=|V1|=1|V2|=1.05puP4+jQ4=1+j0.2puP2=0.5pu7PowerFlowEquation•Consideratypicalbusofann-bussystem–Alllinesrepresentedbyequivalentπmodels–AdmittancesareinpuonacommonMVAbase•ApplyKCL•Solve|Vi|,δi,Pi,QiandthencalculatePij,Qij011220121122()()...()...()(...)......iiiiiiiijijininiiiiniiiijjinnIyVyVVyVVyVVyVVyyyyVyVyVyVyVji=+−+−++−++−=+++−−−−−−≠01nniiijijjjjIVyyVji===−≠∑∑*j=+=iiiiiSPQVI*j−=iiiiPQIV*01j==−=−≠∑∑nniiiijijjjjiPQVyyVjiV01||||||jδδδ==−=∠−∠≠∠−∑∑nniiiiijijjjjjiiPQVyyVjiV−ncomplexnonlinearalgebraicequations(2×nrealequations)with4×nrealquantities−canbesolvedbyiterativetechniques8PowerFlowSolution•Determining:–|Vi|andδi(magnitudeandphaseangleofeachbusvoltage)–PijandQij(realandreactivepowerflowsineachline)•Thesystemisassumedtobeoperatingunderbalancedconditionsandasingle-phasemodelisused•4quantities,i.e.|Vi|,δi,PiandQi,areassociatedwitheachbus•SystembusesareusuallyclassifiedintothreetypesSlackbus(swingbusorV-δbus)•Selectedasthereferencehaving|Vi|andδifixed•PiandQiareusuallyunlimitedandcantakeanyvaluestomakeupthegapbetweensystemgenerationandloadLoadbuses(P-Qbuses)•PiandQiarespecifiedRegulatedbuses(generatorbusesorP-Vbuses)•PiandViarespecified.•LimitsofQiarealsospecified9MoreThinkingonTypesofBuses|Vi|δiPiQiV-δXXP-QXXP-VXXQ-VXXP-δXXQ-δXX•Needtoknowtwoof|Vi|,δi,PiandQi(eitherconstantorobserved)•Relaxtheothertwowithinupperandlowerlimits•Mayassumemoretypesofbusesforavarietyofnaturesofbuses10Amoregeneralpowerflowstudyz12=z10=z13=z20=z23=j0.2z34=|V1|=1|V2|=1.05puP4+jQ4=1+j0.2puP2=0.5puδ1=0(slackbus)11SolutionofNonlinearAlgebraicEquations•Gauss-SeidelMethod•Newton-RaphsonMethod12Gauss-SeidelMethod:Example6.232()6940fxxxx=−+−=32164()999xxxgx=−++=3roots:x1,2=1andx3=4(0)2x=(0)32164()(2)(2)2.2222999gx=−++=x(2)=g(x(1))=2.5173|x(2)-x(1)|=0.2951x(3)=g(x(2))=2.8966|x(3)-x(2)|=0.3793x(4)=g(x(3))=3.3376|x(4)-x(3)|=0.4410x(5)=g(x(4))=3.7398|x(5)-x(4)|=0.4022x(6)=g(x(5))=3.9568|x(6)-x(5)|=0.2170x(7)=g(x(6))=3.9988|x(7)-x(6)|=0.0420x(8)=g(x(7))=4.0000|x(8)-x(7)|=0.0012x(9)=g(x(8))=4.0000|x(9)-x(8)|=0.0001(1)x=|x(1)-x(0)|=0.222213Gauss-SeidelMethod•Tosolvenonlinearequationf(x)=0•Re-writex=g(x)•Startiterationfromaninitialestimatex(0)x(1)=g(x(0))x(2)=g(x(1))…x(k+1)=g(x(k))•Stopwhen|x(k+1)-x(k)|≤ε.Solution:x=x(k+1)14x(0)g(x(0))x(1)x(5)•“Zigzag”graphicalillustration–Cantheiterationbefaster?–Howtofindallroots?–Doestheiterationalwaysconvergetoaroot?x=g(x)32164()999==−++ygxxx=yxg(x(1))15Fasteriteration?•Usinganaccelerationfactorwhenupdatingx(k)(1)()()()[()]kkkkxxgxxα+=+−(1)()()()()()[()]kkkkkxgxxgxx+==+−Adjustmentonxx(1)x(0)g(x(0))x(4)α=1.2532164()999==−++ygxxx=yxg(x(1))16-10-8-6-4-20246810-10-8-6-4-20246810xg(x)=-1/9x3+6/9x2+4/9xdivergentInitial