电力系统暂态分析的课件以及习题答案全集

1１－２发电机G1和G2具有相同的容量，他们的额定电压分别为6.3kV和10.5kV。若以其额定条件为基准的发电机电抗标幺值相等，这两台发电机电抗的欧姆数的比值是多少？解：SG1=SG2UG1=6.3kVUG2=10.5kVxG1*=xG2*2111*1GGGGUSxx2222*2GGGGUSxx36.05.103.622222122212121GGGGGGGGUUUUSSxx１－３如图所示的电力网，图中已标明各元件的参数。要求：⑴准确计算各元件电抗的标幺值（采用变压器实际变比），基本级为I段，UBI=10.5kV。⑵近似计算各元件电抗的标幺值（采用变压器平均额定变比）。SB取100MVA。G100kmIIIX”d=0.1510.5/121kV50MVAIIILT110.5kV110/6.6kV60MVA30MVAUk%=10.50.4Ω/kmT2Uk%=10.5解：(1)SB=100MVAUBI=10.5kVUBII=121kVkVUBIII26.76.61101213.05.105.105010015.02222*BINNBddUUSSxx175.05.105.10601001005.10100%2222*1BINNBkTUUSSux2732.01211001004.0221*BIIBLUSlxx2892.0121110301001005.10100%2222*2BIINNBkTUUSSux(2)SB=100MVAUB=Uav3.05010015.0*NBddSSxx175.0601001005.10100%*1NBkTSSux3024.01151001004.0221*avBLUSlxx35.0301001005.10100%*2NBkTSSux１－４已知网络接线如图，图中已标明各元件的参数。试求：⑴以SB=100MVA，UB=Uav为基值的电抗标幺值。⑵各元件归算到短路点所在电压等级为基本段的电抗有名值。2G65kmX”d=0.1310.5/121kV50MWT1E”=11kV110/38.5kV60MVA31.5MVAUk%=10.50.4Ω/km(单回)T2Uk%=10.5L1cosφ=0.8f(3)L20.4Ω/km40km解：(1)SB=100MVA，UB=Uav208.08.0/5010013.0*NBddSSxx175.0601001005.10100%*1NBkTSSux0983.01151006524.02221*1avBLUSlxx3333.05.311001005.10100%*2NBkTSSux169.137100404.0221*2avBLUSlxx(2)短路点所在电压等级的Uav=37kV69.131003722BavBSUZ8475.269.13208.0*BddZxxxT1=xT1*×ZB=0.175×13.69=2.3958ΩxL1=xL1*×ZB=0.0983×13.69=1.3457ΩxT2=xT2*×ZB=0.3333×13.69=4.5629ΩxL2=xL2*×ZB=1.169×13.69=16Ω１－５如图所示系统，电源为无限大功率系统。当取SB=100MVA，UB=Uav，冲击系数Kch=1.8时，在f点发生三相短路时的冲击电流是多少？短路电流的最大有效值是多少？G115kV0.4Ω/km40kmf(3)T1L1L2R6.3kV6.3kV30MVAUk%=10.5XR%=40.3kA0.08Ω/km0.5km解：121.0115100404.02211avBLUSlxx35.0301001005.10100%1NBkTSSux222.13.03.6310010043100%NNBRRIUSxx1.03.61005.008.02212avBLUSlxxx∑=1.793If=1/x∑=1/1.793=0.557734197.15577.08.12MiIM=1.52If=0.8477有名值：kAiM01.133.631004197.1kAIM769.73.631008477.0３－２电力系统接线如图所示，元件参数标于图中，当f点发生三相短路时，试计算：⑴次暂态电流初始有效值I；⑵冲击电流值ich。60kmX”d=0.1252×50MWL1T1110kV2×60MVASs=100MWUk%=10.5T2XS=0.1G2G1cosφ=0.8f(3)L26.3kV0.4Ω/km(单回)Gs解：取SB=100MVA，UB=Uav2.08.0/50100125.021NBdGGSSxxx175.0601001005.10100%21NBkTTSSuxx1815.0115100604.022121avBLLUSlxxx1.01001001.01.0NBSSSx0946.0)1.021815.0//(2175.02.0])////[()]//()[(212211SLLTGTGxxxxxxxx(1)57.100946.011xI有名值：)(307.5115310057.10kA(2)91.2657.1028.1chi有名值：)(51.13115310091.26kA３－３如图所示的网络中，各元件参数在同一基准条件下的标幺值如下：E1=1.44，E2=1.54，X1=0.67，X2=0.2，X3=0.25，X4=0.2，X=2.4(负荷)。试求：电源对短路点f的等值电抗。4X1G2G1f(3)E1E2X2XX4X3解：x5=x1//x2+x3=0.67//0.2+0.25=0.404(517.12.02.0//67.054.167.02.0//67.044.1////22121211xxxExxxEE)x、x4、x5的Y形变换为三角形：取Ef支路得：6377.04.2404.02.0404.02.05454xxxxxxEf３－４如图所示系统，当f点发生三相短路时，试计算各电源对f点的转移阻抗。已知图中110kV母线a处短路时，由系统S供给的短路电流为16.6kA。30.4kmX”d=0.1252×15MVALT16kV2×75MVAXR%=10T2Sf(3)115kV0.4Ω/km(单回)GsG2G1R0.6kA115/6.3kVUk%=10.51365472a解：取SB=100MVA，UB=Uav)(164.93.63100)(502.01153100)6()110(kAIkAIBB8333.015100125.021NBdSSxxxI”3=16.6kA，0302.007.331107.33502.06.16*33*3IxI14.0751001005.10100%54NBkSSuxx046.011510024.34.022216avBUSlxx527.16.0164.910010100%7BNNBRUUIIxx将4、5、7的三角形变换为8、9、10的Y形：0108.0527.114.014.014.014.0754548xxxxxx51183.0527.114.014.0527.114.0754749xxxxxx1183.0527.114.014.0527.114.07547510xxxxxxx11=x3+x6+x8=0.087x12=x2+x10=0.9513x1f=x1=0.833363.2087.09513.01183.09513.01183.0111291292xxxxxxf2161.09513.0087.01183.0087.01183.012119119xxxxxxSf第四章4-6-14-7-1(1)在题中，若有两台升压变压器和一台自偶变压器高压侧星形中性点直接接地，330KV系统中有中性点接地。计及330KV两回路的互感，试绘制f1点接地短路时以下两种情况的零序网络。（1）中性点接地的自偶变压器与f1点在同一回线路上（2）中性点接地的自偶变压器与f1点不在同一回线路上解1（6(2)第五章５－２在如图所示的电力系统中，各元件参数的标幺值为：发电机G1：Ea=1,Xd=0.2,X2=0.2,X0=0.1；发电机G2：Ea=1,Xd=0.2,X2=0.18,X0=0.1；变压器T-1：XT1=0.12；变压器T-2：XT2=0.1；线路L1、L2：X1=X2=0.3，X0=1.5。试制订在f点发生不对称故障时的各序网络，并求出各序网的等值电抗。解：187.0)2.01.023.0//()12.02.0(])////[()(2221111GTLLTGxxxxxxx1835.0)18.01.023.0//()12.02.0(2x1052.0)1.025.1//(12.0])////[(22110TLLTxxxxx５－３电力系统如图所示。设A、B发电机暂态电势E'*=1.0，在f点发生两相短路接地故障，试计算故障处的A、B、C三相电流。解：72283.0)25.01.023.0//()12.03.0(])////[()(22111GBTLLTGAxxxxxxx1778.0)15.01.023.0//()12.02.0(2x1234.0)1.027.0//()12.005.0(])////[()(22110TLLTGAxxxxxxf(1,1)时三序网并联：321.31234.0//1778.02283.01)//(10211jjxxxjIf361.11234.01778.01234.0)321.3(02012jjxxxIIff96.11234.01778.01778.0)321.3(02210jjxxxIIff0faI000021206.144594.2055.496.190120361.190240321.3jjIIaIaIffffb000022194.35594.2055.496.190240361.190120321.3jjIIaIaIffffc５－４计算图中f点发生A相短路接地时，流经发电机G1的负序电流，变压器T1为YN，d11接线（Yo/△-11)。8解：取SB=100MVA，UB=UavxG1(1)=0.13，xG1(2)=0.20833.01201001.0100%1NBkTSSux1512.02301002004.0221)2()1(BBLLUSlxxxxL(0)=3xL(1)=0.45361667.0601001.0100%2NBkTSSux2167.06010013.0)1(2Gx3333.0601002.0)1(2Gx1869.0)2167.01667.0//()1512.00833.013.0(1x2325.0)3333.01667.0//()1512.00833.02.0(2x1272.01667.0//)4536.00833.0(0xf(1)时三序网串联：829.11272.02325.01869.01)(1021021jjxxxjIIIfff9786.0333.01667.01512.00833.02.0)3333.01667.0(829.1)(2211222')2(1jjxxxxxxxIIGTLTGGTfG