图示电路f处发生三相短路,各线路电抗均为0.4Ω/km,长度标在图中,取SB=2

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问题 图示电路f处发生三相短路,各线路电抗均为0.4Ω/km,长度标在图中,取SB=250MVA,f处短路电流周期分量起始值及冲击电流分别为(  )。A.2.677kA,6.815kAB.2.132kA,3.838kAC.4.636kA,6.815kAD.4.636kA,7.786kA

选项 A.2.677kA,6.815kA
B.2.132kA,3.838kA
C.4.636kA,6.815kA
D.4.636kA,7.786kA

答案 A

解析 取SB=250MVA,UB=Uav,计算各元件的参数,形成等值电路如题52解图所示。首先把各元件折算成统一的标幺值:XS*=XS×SB/SN1=1.2×250/1000=0.3Xl1*=Xl1×SB/UB2=50×0.4×250/1152=0.378Xl2*=Xl2×SB/UB2=20×0.4×250/1152=0.151Xl3*=Xl3×SB/UB2=30×0.4×250/1152=0.227Xd*″=Xd″×SB/SN2=0.12×250/250=0.12XT*=US%/100×SB/STN=10.5/100×250/250=0.105则系统的短路阻抗为:X∑*=(XS*+Xl1*)//(Xl2*+XT*+Xd*″)+Xl3*=0.4689。短路电流周期分量起始值为:短路冲击电流为:
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