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电力系统继电保护课程设计参考

上传者:苏堤漫步 |  格式:doc  |  页数:49 |  大小:588KB

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1=1.05/2×1.2718=0.4128If(2)*=1.732Ifa(1)(2)=1.732×0.4128=0.7150If(2)=If(2)*IB=0.7150×0.502=0.3589KA(3)最大运行方式两相短路零序短路电流.图3.2.1(c)零序短路电路图XTT(0)=XT1(0)/2=0.1525XT(0)=0.1525×0.1601/(0.1525+0.0.1601)=0.0781XTBL(0)=(0.4536+0.0781)×0.2401/(0.4536+0.0781+0.2401)=0.167XTL(0)=(0.2268+0.6405)×0.2401/(0.2268+0.6405+0.2401)=0.188XTLB(0)=(0.363+0.167)×0.188/(0.363+0.167+0.188)=0.1388Xff1(0)=XL3(0)+XTLB(0)=0.8166+0.1388=0.9554Iff1(0)*=E(0)/Xff1(0)=1.05/0.9554=1.099Iff1(0)=Iff1(0)*IB=1.099×0.502=0.552KA2.3.2.2d2发生短路时流过断路2(1)最大运行方式正序短路电流Xff2=XdT+XL3=0.18255+0.2722=0.45475Id2·max*=E/Xff2=1.05/0.45475≈2.309Id2·max=Id2·max*IB=2.309×0.502≈1.159KA(2)最小运行方式两相短路正序短路电流?X2=(XL1+XL2)×XL3/(XL1+XL2+XL3)=XL3/2=0.1361Xff2=Xd1+X2=0.18255+0.1361=0.31865If(3)*=E/2Xff1=1.05/2×0.31865=1.6475If(3)=If(3)*IB=3.295×0.502=0.827KA

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