故障處短路電流和電壓計算二

第十章不對稱故障的分析和計算第十章不對稱故障的分析和計算第二節(jié)非故障處電流和電壓的計算第十章第十章不對稱故障的分析和計第一節(jié)故障點(diǎn)短路電流和電壓的計四、正序等效定兩相相間短G~Lf邊界條.Ia= Ib+Ic= Ub=.. . Ib+Ic= Ub=Ia=序分量表 Ia1+Ia2+Ia0=a2Ia1+aIa2+Ia0+aIa1+a2Ia2+Ia0=. . aUa1+aUa2+Ua0=aUa1+aUa2+Ua 2..................................................................2.Ua1=Ua Ia1+Ia2=.Ia0=復(fù)合序jX.I+E+-U.jX I+-UjX.I.IjX+E.+.U-U例G-例G-~X"=fLG-~.E"=jX=X=X=X=X"=X=0.33X=.E"=jj1jjjjjj12 I45X=(X+X)//(X+X+X=(0.2+0.18)//(0.12+0.33+0.4jX1=j.Ia=+E1=j+-Ua例G-~X"=fLG-~.E"=jjX=X=X=X"=X=X=0.33X=.E"=jjjjj12 I45X=(X+X)//(X+X+X=(0.26+0.18)//(0.12+0.33+0.51)jX=j=.I+U-..合序jX=j+E=jI.I=+. E jX+jXjj0.26+j.-U=jX=j.I.I=-+U U=E-jXI=j1-j0.26=j U=-jX2Ia=-j0.30(-1.79=j.解各相電流.Ia=Ib=aIa1+aIa2+Ia0=12401.79+1120(-.=-j3.10Ic=j 2 各相電 Ua=Ua1+Ua2+Ua0=j0.54+j0.54+0=jUb=aUa1+aUa2+Ua=1240j0.54+1120j0.54+ 2 =-Uc=-.解相量解相量...Ic.Ic Ua Ua.Ia.Ia.Ib..Ub.Uc.Uc.Ub..Uc兩相接地短G~Lf(1,邊界條.Ia= Ub=Uc=.... UaUa1=Ua2=Ua Ia1+Ia2+Ia0=邊界條件Ia=.Ia1+Ia2+Ia0= Ub=Uc=0aUa1+aUa2+Ua0=aUa1+aUa2+Ua 2 2 =jX.I+E+-U.jX I+Ua1=Ua2=Ua -UIa1+Ia2+Ia0=. jX.I+U-復(fù)合復(fù)合序.jXjX+E.I.UI+I-.U jXU2=Ua+Ia0=..例G-~X"=f(1,L.E"=jX=X=X=X=G-~X"=X=0.33X=.E"=jj1jjjjjj12.I345jX=j.IX=(X+X)//(X+X+X)+=(0.2+0.18)//(0.12+0.33+0E4)=j=+U-例例G-~X"=f(1,LG-~X=X=X"=.E"=jjX=X=X=0.33X=.E"=jjjjj12.I345jX=j.IX=(X+X)//(X+X+X=(0.26+0.18)//(0.12+0.33+0.51=+U-例G-~X"=f(1,LG-~X=.E"=jX=X=X=X"=X=0.33X=.E"=jj2.Ij3j4X=X//(X+X=0.18//(0.24+0.33=jX=j.I+U-解j解jX=j+E=j.I.I+-U. E =jX+jX//jXj=j0.26+j0.30//j= X .(-IjX=j.II+ X+X(-2.850.30+-UjX0=j.I=-.I=-+U U=U=U=-jXI=-j0.13(-=j 解各相電流.Ia= Ib=a2Ia1+aIa2+Ia=12402.85+1120(-0.86)+(-=-2.99-j3.21=4.39Ic=aIa1+aIa2+Ia=-2.99+j3.21=4.39 Ua=Ua1+