2005-2006-学年第1-学期电力系统分析考试试题参考答案

2005-2006学年第1学期电力系统分析考试试题参考答案一、简答题（1）表示在节点j取出单位有功功率而将其注入节点i而其它节点注入有功功率为零时各节点电压相位。（2）线路两端的电压损耗和电压降落概念不同。电压损耗为12||||UU，电压降落为12UU。（3）G:10.5kVT1:10.5kV/121kVT2:110kV/38.5kV/11kVT3:35kV/6.6kVT4:10kV/420VM:6kV（4）110kV以上系统使用的是大电流接地方式以降低绝缘造价，60kV及以下使用的是小电流接地方式以提高供电可靠性。（5）UIGEdjxUIMEdjx发电机次暂态模型异步电动机次暂态模型二（1）计算等值电路参数，选110kV级为基本级对于线路116410.12000.420020802/22.6610200/22.6610SZrljxljjYjbljj对于变压器2212222160221502211901102.3171000100031.5%10.511040.3310010031.531.052.56610S10001000110%0.731.51.822310S100100110kNTNkNTNmNNmNPURSUUXSPGUISBUmmGjBTZZ2Y2YBCSBCSABSABSABC（2）设B、C节点电压均为额定电压110kV，则变压器消耗的功率为222222010()(2.3240.33)0.0961.67MVA110CTTTCSSRjXjjU2652(2.566101.822310)1100.0310.22MVAmmBSYUjj2422(2)2.66101103.22MVAYBSYUjj20100.0961.6722.09611.67MVABCCTSSSjjj222.09611.67MVA0.0310.223.2222.1278.67MVAABBCmYSSSSjjjj2222222.1278.67(2080)0.712.84MVA110BZBSSZjjU22.1278.670.712.8422.83711.51MVAABABZSSSjjj2422()(2)2.66101153.52MVAYAASYUjj2()22.83711.513.5222.8378.0MVAAABYASSSjjj计算电压损失22.8372011.518011.97kV115ABABABAPRQXUU11511.97103.1kVBAABUUU25.0962.31711.6740.335.1kV103.1BCBCBCBPRQXUU103.15.198.0kVCBBCUUU折算到低压侧9.80kV110/11CCUU三、（1）因122221222TUUkkZIAUBIIkICUDI故1221111eTeTeTZBkZDkYBkZAkYBkZeTZkZ11eTkYkZ221eTkYkZ（2）设新增加的节点编号为j则节点i的自导纳发生变化，增加的量为11TiiTTyΔYy()ykk节点j的自导纳为2211TTjjTyyYy()kkkk则节点i和节点j的互导纳为TijjiyYYk导纳矩阵增加1阶，节点j和其它节点间的互导纳为零。其它节点的自导纳及其间的互导纳均不变。四画出零序等值电路图3TjX40ljX3TjX4TjX3TjX3NjX30ljX20TjX10ljX1TjX1TjX1TjX0kU五（1）节点1：平衡节点节点2：PV节点节点3：PQ节点（2）可写出节点导纳矩阵为52.52.52.552.52.52.55jjjYjjjjjj设2310UU，可得52.52.55B故2333(0)22222213(0)33333313(0)3333331(cossin)0.8(cossin)1.0(sincos)0.6mjjjjjjmjjjjjjmjjjjjjPPUUGBPPUUGBPQUUGB(0)2222(0)3333PUUBUPU得230.080.16从而230.080.16(0)333QBUU得30.11U从而310.110.89U故214.98U，30.899.17U六取100BSMVA，BavUU（1）则对于变压器112132321223133132312%0.5(%%%)15%0.5(%%%)9%0.5(%%%)0kkkkkkkkkkkkUUUUUUUUUUUU则112233%0.125100%0.075100%0100kBTNkBTNkBTNUSXSUSXSUSXS对于发电机111000.1830.1830.156/cos100/0.851000.2230.2230.190/cos100/0.85BdNNBdNNSXPSXP对于电力线路21200.4160/()0.1202.150.1200.26BllBlUXXSX设1210EE，则121325.84()//()dTlTTGEIjjXjXjXjXjX有名值为1005.845.842.93kA33115BdBSIjjU（2）21GjXljX1GE2GE1TjX2TjX3TjX1jX1E1dU1dI1dU正序等值网络及其戴为南等值电路22GjXljX1TjX2TjX3TjX2jX2dU2dI2dU负序等值网络及其戴为南等值电路0ljX10TjX20TjX30TjX0dU0SjX零序等值网络及其戴为南等值电路（3）k点发生单相接地故障时，复合序网由各序网络串联而得到，故112012012012.04(0.1710.1760.152)akkEIIIjjXjXjXjjXjXjX1120212010()0.670.360.31dadadaUIjXjXUIjXUIjX故短路点0122012201201.01117.531.01117.53dAddddBddddCdddUUUUUUaUaUUUaUaU由发电机G2提供的短路电流为2112210.2451.25900.2450.1560.2451.15900.2450.19GaGaIIII303021222303021223023021222.0798902.07991jjGaGGjjGbGGjjGcGGIeIeIIaeIaeIIaeIaeI故有名值为1202.07911.43kA310.50kAGaGcGbIII