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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類型安裝中短路電流的計算。示例1上游網絡：Ur=20000 V Sknet=500 MVA變壓器：Sr=1600 kVA uk%=6%U1r/U2r=20000/400電機：Pr=220 kW I kmot/I r=6.6昂貴的r=0.9η=0.917一般負載：I rL=1443.4 A昂貴的=0.9計算不同元件的短路功率網絡：Sknet=500 MVA變壓器：電機：Skmot=6.6。Srmot=1.76 MVA對于前5-6個周期（50 Hz約100 ms）短路電流的計算用于選擇斷路器CB1對于斷路器CB1，當故障發(fā)生在斷路器自身的正下游時，出現壞情況。在上游發(fā)生故障的情況下，斷路器將僅受到來自電機的故障電流的影響，該電流明顯小于網絡貢獻。

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 cos?r = 0.9 η = 0.917 Generic load: I rL= 1443.4 A cos?r = 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.