Tutorial 1 - Solutions

《Tutorial 1 - Solutions 》由会员上传分享，免费在线阅读，更多相关内容在学术论文-天天文库。

1、ELEN90060PowerSystemAnalysisTutorial1-SolutionsELEN90060PowerSystemAnalysisTutorial1–Solutions:ComplexPowerM.Aldeen,March2011SolutiontoQuestionT3.1Parta):ThepowerfactoriscomputedasP480pf=cosθ===0.7529VI1255.1×rmsrmsor−1θ=cos(0.7529)=41.15Thepowerfactoris“lagging”

2、becausetheangleis(+).Thismeansthattheloadisinductive(comprisingresistorsandinductors)andthereforethecurrent“lags”behindthevoltagebyanangleof41.15degrees.Partb):Thepowersarecalculatedasfollows.Sinceweknowtherealpower,wecancalculatetheapparentpowerfromPS=cosθP480⇒==

3、S=637.5AVpf0.7529Fromthis,wecomputethereactivepowerfromQS=sinθ=637.50.6581=419.53×VARPartc):TheenergystoredintheinductoriscomputedfromQ419.53W===0.5564Joulse2ωπ226××(0)Partd):ThePhasordiagramisadiagramofthevoltageinrelationtothecurrent.Wetakethevoltageasthereferen

4、ce,i.e.VV=/0=V/0=1252/0V.ThuswedrawthecurrentasavectorprmsPage1of21ELEN90060PowerSystemAnalysisTutorial1-Solutionshavingamagnitudeof5.1Aandaphaseangleof41.15degreeslaggingbehindthevoltageasshownbelow.ImRe0V=125Vθ=41.15I=5.1ASolutiontoQuestionT3.2Firstwecomputet

5、heimpedanceofthecircuit,startingwiththeimpedanceofthecapacitor11Xj=−=−j=−j132.63ΩC36−ωC2.513310××310×ωThetotalimpedanceistherefore1Z==(43.78-j16.50)Ω1/R+1/XcorZ=46.7858/20.6−6ΩNowwecomputethermsvoltageasV=IZ=2×=(43.78-j16.50)(87.56-j33)VrmsrmsorV=93.57/20.6

6、6−VrmsNotethatthe(-)signsignifiesthefactthattheloadisacapacitiveload,i.e.comprisesaresistorandacapacitor.Parta):Thecomplex(apparent)powercannowbecalculatedasPage2of21ELEN90060PowerSystemAnalysisTutorial1-Solutions*SVI==93.57/20.66(−Ω=)(2)187.14/2−0.66VAThustheac

7、tive(real,average,effective)powerisPS=cosθ=187.14cos×−=(20.66)175.11WAndthereactivepowerisQS=sinθ=187.14sin×−(20.66)=−66.02VARNotehereagainthatthe(-)signisduetothefactthattheloadcontainsacapacitor,whichisa“reactivepowergenerator”.Inotherwords,thecapacitorwillgener

8、atethisamountofpowerandmakeitavailableforuse,shouldanextrainductiveloadbeconnectedtothecircuit.Partb):ThepowertriangleisdrawnbelowImP=175WRe0θ=20.7Q=66V