0 10 20 30 40 50 60 70 80 90 100 120 130 140 150 0 10 20 30 40 50 60 70 80 90 100 SkUP=26.7 MVA Ik=36.5 kA 1SDC010054F0001以下示例演示了某些不同类型安装中短路电流的计算。示例1上游网络：Ur=20000 V Sknet=500 MVA变压器：Sr=1600 kVA uk%=6%U1r/U2r=20000/400电机：Pr=220 kW I kmot/I r=6.6昂贵的r=0.9η=0.917一般负载：I rL=1443.4 A昂贵的=0.9计算不同元件的短路功率网络：Sknet=500 MVA变压器：电机：Skmot=6.6。Srmot=1.76 MVA对于前5-6个周期（50 Hz约100 ms）短路电流的计算用于选择断路器CB1对于断路器CB1，当故障发生在断路器自身的正下游时，出现坏情况。在上游发生故障的情况下，断路器将仅受到来自电机的故障电流的影响，该电流明显小于网络贡献。

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 0 10 20 30 40 50 60 70 80 90 100 SkUP = 26.7 MVA Ik = 36.5 kA 1SDC010054F0001 Thefollowing examples demonstrate the calculation of the short-circuit current in some different types of installation. Example 1 Upstream network:Ur = 20000 V Sknet = 500 MVA Transformer: Sr = 1600 kVA uk % = 6% U1r / U2r =20000/400 Motor: Pr = 220 kW I kmot/Ir = 6.6 cosϕr = 0.9 η = 0.917 Generic load: I rL= 1443.4 A cosϕr = 0.9 Calculation of the short-circuit power of different elements Network: Sknet= 500 MVA Transformer: Motor: Skmot = 6.6.Srmot = 1.76 MVA for the first 5-6 periods (at 50 Hz about 100 ms) Calculation of the short-circuit current for the selection of circuitbreakers Selection of CB1 For circuit-breaker CB1, the worst condition arises when the fault occurs right downstream of the circuit-breaker itself. In the case of a fault right upstream, the circuit-breaker would be involved only by the fault current flowing from the motor, which is remarkably smaller than the network contribution.