解
1）计算各元件电抗（取100 ）。 发电机G1 发电机G2
X1???%SBXd100??0.125??0.25100SN50X%S100X2?2?n?0.16??0.32100SN50
变压器
X1???%SBXd100??0.125??0.5100SN25X%S100X2?2?n?0.16??0.64100SN25T1 X1?X2?X0?Uk%?SB?10.5?100?0.175
100SN10060
变压器T2 X1?X2?X0?Uk%?SB100SN?10.5100??0.333 10031.5线路l X1?X2?x1lSB100?0.4?50??0.1522UB115
X0?2X1?2?0.15?0.302）以A相为基准相作出各序网络图，求出等值电抗X1Σ、X2
Σ
、X0Σ。
??j1?j(0.333?0.5)?j1?j(0.25?0.175?0.15)?j1EaΣj(0.25?0.175?0.15?0.333?0.50) X1Σ?j(0.25?0.175?0.15)//j(0.333?0.50)?j0.289
X2Σ?j(0.32?0.175?0.15)//j(0.333?0.64)?j0.388X3Σ?j(0.175?0.3)//j0.333?j0.1963）边界条件。 原始边界条件为 I??0a??U??0 Ubc由对称分量法得出新的边界条件为
??U??U??1U?Ua1a2a0a3??I??I??I??0Iaa1a2a04）据边界条件绘出复合网如图所示。
5）由复合网求各序的电流和电压，得
???Ej1a1Σ??2.385?Ia1?jX?jXj0.388?j0.1962?0??j0.289?jX1???j0.388?j0.196jX2??jX0??X0?0.196 ? ???I??I??2.385??0.80a2a1X2??X0?0.388?0.196??X2?0.388???I??2.385??1.585a0??Ia1?X?X0.388?0.1962?0?????U??U??I? Ua1a2a0a1jX2??jX0?j0.388?j0.196?2.385??j0.311
jX2??jX0?0.388?j0.1966）求各相电流和电压，为
??I??I??I??2.385?0.800?1.585?0Iaa1a2a0??a2I??aI??I??2.375240??(?0.800)120??(?1.585)Iba1a2a0??2.378?j2.758?3.642130.77?
??aI??a2I??I??2.385120??(?0.800)240??(?1.585)Ica1a2a0??2.378?j2.758?3.642130.77???U??U??U??3U??3?j0.311?j0.933Uaa1a2a0a12??aU??aU??U?Uba1a2a0?0.31190??240??0.31190??120??0.31190??0??aU??a2U??U?Uca1a2a0???0.31190?120?0.31190??240??0.31190??0
7）电流电压相量图如图所示。
图（f）