暂态分析—第五次作业答案和常见错误

1、如图所示输电系统，在f点发牛接地短路，试绘出各序网络，并计算:(1) 电源的等值电势；(2) 短路点的各序输入电抗；(3) 计算变压器I的d侧的各相电压和各相电流。T-26.3kV4W各元件的参数如下：发电机:SN=I2OMVA，VN=IO.5kV 巳二 1.67 X(I)=O.9，x(2)=0.45变压器 T-I : SN= 60MVA Vs% =10.5 kT, =10.5/115 ;变压器 T-2 : Sn=60MVA Vs%=10.5 kT2=115/6.3 ;线路 L 每回路:I = 105km X(l)=0.41 / km X(O)= 3x(1);负荷 LD-I : S = 60M

2、VA, X(l)=l .2，XG)= 0.35 ;负荷 LD-2 : SN 40MVA, X=1 .2 X(2)=0.35。解:基准功率 気=12OMVA基准电压UB =UaV各序网络图如下，图中标出各元件电抗标 幺值。正序网：E| XG(I)167怙42.4 0.91.21Ua( i)%侧等效电抗为A xgo)/xldi(i)xti(D xl(i 严 0.66 0.21 0.19 =1.062巳二+XLDI xGUa(】)右侧等效电抗 为X2 二 xt2(i)xld2(i)=o.21 3.6 =3.81则正序网的等效电抗为-w=O.83 厶3.81+1.06等值电势Ee厂亠二远炎1 =0.9

3、5qx,+ X2 3.81 +1 .06网：丄丄1丄X 1 2 - (X G( 2)/XlD 1(2) XT 1( 2)-X L(2)/(Xt2(2)X LD 2(2)=0.442负序零序网络：X O-Xti(O) 1 XL(OrO.78假设f点发生了单相接地故障故障相为A相j(xVX，；)j(0.83 1.22)魏)=j(X2+X o)lfa(i)=j(0.440.78) *(-j0.46) =0.56fa(2)-二x 2 I fa(i)二0.44*( -j )0.46= J .2fa(O)jX H tad)= j0-78*(- j0.46)= -0.36TlEl-U fa(l) 1r21

4、0.56 沖.060.61XiT1(2)XG(2)/XLD1 (2) XTI (2) X L (2)A44fa(2)*( JO.46) =JO.310.67eq 0.95lfa(l) =lfa (2)=1 fa(O)=E= j 0.46则变压器d侧的电流为j30j30ITId(l)E1 ITI(l)eU-j0.61eTld(2) =，TI(2re-J3 二一朋耶。ITIda 二 ITId(l)+1 Tid(2)A0.61ej3-j0.3eT3022j30Tldb -a I Tld(l) a I Tid(2)-ja *0.61 e - ja *0.3 e-j 302ITIdC =alTid(D

5、a ITId(2)=-ja*0.61 ej 302-j30-ja 0.3e变压器Tly侧的电压为1UM)=%j(Xti 八)56 0 461 t4804U 叶汨爲 j (XgX L(2) Tl =.2 0.4*0.3=0.08则变压器d侧的电压为U Tld(l)=IJ Tly (l)A=0.804eU AeO6eT1d(2)b k U T1y(2)u uTldaTld(l) *,zHd(2)j 30c-j300 804e 0 08eUTIdAa2UTid(i) -al)Tid(2)=a2O.8O4ej30-aO.O8e*j30UTIdC=all Tld(l) aJ Tld=aO.8O4d30

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