发布时间 : 星期六 文章电力系统分析试题及参考答案修改1更新完毕开始阅读
100?0.52520100X2*?0.4?10?2?0.292 37100X3*?X4*?0.07??2.193.2E*?1X1*?0.105?等值网络如图所示。
X?*?0.525?0.292?周期分量电流的有效值 I*?有名值电流为
I?I*?IB?0.523?若取KM=1.8，则冲击电流
iM?2?1.8?I?2.55I?2.55?2.88?7.34(kA) 短路功率为
Sk?S*?SB?I*?SB?0.523?100?52.3(MVA)
五、已知某系统接线如图四所示，各元件电抗均已知，当k点发生BC两相接地短路时，求
短路点各序电流、电压及各相电流、电压，并绘出短路点的电流、电压相量图。
1003?10.5?2.88(kA)
11??0.523 X?*1.9121?2.19?1.912 2
图四 题五图
解
1）计算各元件电抗（取SB=100MVA UB=Uav）。 发电机G1
X1???%SBXd100??0.125??0.25100SN50
X%S100X2?2?n?0.16??0.32100SN50
发电机G2
X1?
X2%Sn100X2???0.16??0.64100SN25U%S10.5100变压器T1 X1?X2?X0?k?B???0.175
100SN10060变压器T2 X1?X2?X0?线路l X1?X2?x1lSB100?0.4?50??0.152UB1152
??%SBXd100??0.125??0.5100SN25Uk%SB10.5100????0.333 100SN10031.5 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、X2Σ、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。
??0原始边界条件为 Ia??U??0 Ubc由对称分量法得出新的边界条件为 ??I??I??I??0Iaa1a2a0 1? ???Ua1?Ua2?Ua0?Ua34）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?jX2??jX0??2.385?j0.388?j0.196?j0.311 Ua1a2a0a1jX2??jX0?0.388?j0.1966）求各相电流和电压，为
??I??I??I??2.385?0.800?1.585?0Iaa1a2a02??aI??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?j0.933Uaa1a2a0a1??a2U??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）