无穷大系统中,求发生三相短路时短路点的短路电流I〞、I∞ 、ish ,线路电抗均为每公里0.4Ω,参数见图

1个回答

  • 等效图(略)X1* X2* X3* X4 *

    标么值:

    X1*= × = × =0.18(Ω)

    X2*=X =0.4×80× =0.198(Ω)

    X3 *= × = × =0.833(Ω)

    X4*=X =0.4×4× =1.199(Ω)

    X*=X1*+X2*+X3*+X4*=0.18+0.198+0.833+1.199=2.41(Ω)

    Ij= = =5(KA)

    I"= = =2.08(KA)

    根据DL/T5222-2005附录F第F.2.2条:当供电电源为无穷大时:I"=I∞=2.08(KA)

    根据DL/T5222-2005附录F第F.4.1条:冲击电流ich=√2Kch I"=√2×1.8×2.08=5.29(KA)

    从你提供的电压等级看,似乎不对,应该是220/110KV和110/10KV,则:

    X*=X1*+X2*+X3*+X4*=0.18+0.242+0.833+1.451=2.71(Ω)

    Ij==5.5(KA)

    I"=I∞=2.03(KA)

    ich=5.17(KA)

    另外,10KV线路电抗太高. (我复制后公式变了,你下载后看变没变)