电力系统分析课后解答.ppt

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1、Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.9-15静态稳定性分析—小扰动分析法（2）p248（9-35）根据p320（9-39），系统静态稳定解：（1）网络参数：Eval

2、uationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.9-16暂态稳定分析—等面积定则解：（1）网络参数（2）正常时的功率特性当发生两相接地短时，负序和零序网络如下图所示：Evaluatio

3、nonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.负序和零序等值电抗两相短路接地时，附加电抗为故障时系统的转移阻抗和功率特性为：Evaluationonly.CreatedwithAspose.S

4、lidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.切除故障后系统转移电抗及功率特性为由等面积定则得极限切除角(P273(9-123))：当时，所以系统能够保持暂态稳定。P343(8-111)Evaluationonly.CreatedwithAs

5、pose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.8-9各序网络的制订（1）正序（2）负序Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Co

6、pyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.（3）零序Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspo

7、se.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.8-11简单不对称短路计算（正序等效定则）（1）计算元件各序电抗，制订各序网络，取SB=100MVA,VB=Vav0.20.1670.1510.1670.5240.20.1670.1510.1670.5240.1670.4530.1670.524（a）正序（b）负序（c）零序Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientPr

8、ofile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011Aspose