ECE 476 Power System Analysis Fall 2014Exam#1,Thursday,October2,2014. 9:30AM-10:50AMName:Problem1(25p)Twobalanced3-phaseloadsareconnectedinparallel. OneisY-connectedanddraws75kW(3-phase)at0.8powerfactorlagging. Theotheris-connectedanddraws60kVA(3-phase)at0.9powerfactorleading. Thesupplyis480V(line-line),60Hz. Determine:(a)Themagnitudeofthesourcelinecurrent.Solution:ForY-connectedload,SY=750.8cos1(0.8)kVA.For-connectedload,S= 60 cos1(0.9)kVA.Totalcomplexpower(3phase):ST= SY+ S= 129 + j30.1 = 13213.33okVAThemagnitudeofthecurrent:|I| =|ST|3|V |= 159A(b)TheamountofcapacitiveVARS(3-phase)neededtomaketheoverallpowerfactortobe1.0.Solution:FromPart(a),withoutcompensation:ST= 129 + j30.1kVA.Withcompensation:ScT= 129 + j(30.1 + Qc)kVA.Tomakethepowerfactortobe1.0,thetotalreactivepowerhastobe0. Therefore,Qc= 30.1kVar(c)Thesourcelinecurrentwhenthecapacitorhavebeenadded.Solution:|I| =|ScT|3|V |=129 1033 480= 155.2A1Problem2(25p)Athree-phasetransmissionlineismountedonthetowerasshownbelow.!# $ The distance between A and B is d meters. The distance between A and C is d meters. The radius of each con-ductor is r meters. The three line currents are dened as positive into the paper and they sum to zero. Determine:(a)Thedistributedux(permeter)linkingconductorAintermsofIA,d,andr.Solution:A=02[IA ln1r

+ IB ln 1d+ IC ln 1d]SinceIA + IB + IC= 0,A=02IA lndr

(b)Thedistributedux(permeter)linkingconductorBintermsofIA,IB,d,andr.Solution:B=02[IA ln 1d+ IB ln1r

+ IC ln12d=02[IA ln 1d+ IB ln1r

(IA + IB) ln12d=02[IA ln2 + IB ln2dr

](c)Thedistributedux(permeter)linkingconductorCintermsofIA,IB,d,andr.Solution:C=02[IA ln 1d+ IB ln12d+ IC ln1r

=02[IA ln 1d+ IB ln12d(IA + IB) ln1r

=02[IA ln r

d+ IB lnr

2d]Problem3(20p)A 3, 300 mile, 345-kV line has series impedance z=j0.48 /mile and shunt admittance y=j6.0106siemens/mile.(a)CalculatethelinescharacteristicimpedanceZcandthepropagationconstant.Solution:Zc=

zy= 282=z y= j1.7 103mile1(b)Ratedlinevoltageisappliedtothesendingendofthisline. Calculatethereceiving-endvoltagewhenthereceivingendisterminatedbyone-halfofthesurge(characteristic)impedance.Solution:Vs= cosh(l)VR + jZc sinh(l)IRandIR=VRZc2. Therefore,VS=cosh(l)VR + jZc sinh(l)IR=[cosh(l) + j2 sinh(l)]VRVR=Vscosh(l) + j2 sinh(l)= 263.4 48.2okVQuestion4(10p)Thevoltageandcurrentofasingle-phaseloadarev(t) =2 sin(2ft)andi(t) =2 sin(2ft),wherefisthestandardvalueoffrequencyintheUS.Denotetheinstantaneouspowerconsumedbytheloadasp(t). Amongthefourdierentpowersnapshotsbelow,circletheonethatcorrespondstop(t). Explainyouranswerwithacoupleofsentences.0 0.01 0.02 0.03 0.04 0.0500.511.52time[s]p(t)[W](a)0 0.01 0.02 0.03 0.04 0.0510.500.51time[s]p(t)[W](b)0 0.01 0.02 0.03 0.04 0.0500.511.52time[s]p(t)[W](c)0 0.01 0.02 0.03 0.04 0.0500.511.52time[s]p(t)[W](d)Explanation:Theforthoptioniscorrect.p(t) = v(t)i(t) = 2 sin(2ft) sin(2ft) = 1 cos(4ft)Therefore,thefrequencyofp(t)istwotimesthestandardvoltagefrequency,i.e.,120Hz.Question5(10p)Theterminalsof athree-phasebalanceddeltaconnectedinductiveloadarelabeleda, b, andc. Theloadissupplied by a three phase abcpositivesequence balanced voltage source. The voltage across terminals a andb of the this three-phase load is vab=2Vcos(t), while the current owing into terminal a of this three-phaseloadis ia(t) =2I cos(t ). Let VabandIadenotethephasors associatedwithvab(t) andia(t). LetS=P+ jQbethecomplexpowerconsumedbytheload. Amongtheoptionsbelow, circlethecorrectone.Explainyourchoiceandwhatiswrongwiththeotheroptions.(1)S=3V IandS=3VabIa(2)S=3V IandP=3V I cos (3)S=3V IandQ =3V I sin( 30)(4)S=3V IandP=3V I cos( + 30)Explanation:(3)iscorrect.Iab=I3( + 30o).S= 3VabIab=3V I( 30o).Therefore,Q =3V I sin( 30)andP=3V I cos( 30)Question6(10p)Athree-phaseY-connectedinductiveloadis suppliedbyathreephaseabc positive sequencebalancedvoltagesource. Theloadisbalancedandtheper-phaseimpedanceof thisloadisZ. Amongthechoicesbelow,picktheonethatiscorrect. Explainyourchoice.V abV bcV caV aV bV cIaIcV ab V bcV caV aV bV cIaIcIbIbV ab V bcV caV aV bV cIaIcIbV abV bcV caV aV bV cIaIb(b) (a)(c) (d)IcExplanation:(a)iscorrect.Sincetheloadisinductive, currentlagsvoltage. Therefore, itshouldbe(a)or(b). Itispositivesequence.Therefore,itshouldbe(a)or(d).