Medium Voltage technical guide - Schneider Electric

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Design rules Short-circuit currents Here is the solution to the problem with the calculation method. Solving the exercise b Determining the various short-circuit currents The three sources which could supply power to the short-circuit are the two transformers and the alternator. We are supposing that there can be no feedback of power through D4, D5, D6 and D7. In the case of a short-circuit downstream of a circuit breaker ( D4, D5, D6, D7), then the short-circuit current flowing through it is supplied by T1, T2 and G1. b Equivalent diagram Each component comprises a resistance and an inductance. We have to calculate the values for each component. The network can be shown as follows: DE59015EN Za = alternator impedance different according to state (transient or subtransient) Z15 = transformer impedance 15 MVA Busbars Zr = network impedance Z20 = transformer impedance 20 MVA Experience shows that the resistance is generally low compared with, reactance, so we can therefore deduce that the reactance is equal to the impedance (X = Z). b To determine the short-circuit power, we have to calculate the various values of resistances and inductances, then separately calculate the arithmetic sum: Rt = R 22 AMTED300014EN.indd Xt = X b Knowing Rt and Xt, we can deduce the value of Zt by applying the equation: Z = (∑ R 2 + ∑ X 2) N.B.: since R is negligible compared with X, we can say that Z = X.
Design rules Short-circuit currents And now here are the results! N.B.: a circuit breaker is defined for a certain breaking capacity of an r ms value in a steady state, and as a percentage of the aperiodic component which depends on the circuit breaker’s opening time and on R X of the network (about 30%). For alternators the aperiodic component is very high; the calculations must be validated by laboratory tests. The breaking capacity is defined at the transient state. Subtransient period is very short (10 ms) and approximatively is the necessary duration for the protection relay to analyse the fault and give the trip order. AMTED300014EN.indd Component Calculation Z = X (ohms) Network Ssc = 2000 MVA U op. = 10 kV 15 MVA transformer (usc = 10%) U op. = 10 kV 20 MVA transformer (usc = 10%) U op. = 10 kV 15 MVA alternator U op. = 10 kV Transient state (Xsc = 20%) Subtransient state (Xsc = 15%) Busbars Parallel-mounted with the transformers Series-mounted with the network and the transformer impedance Parallel-mounting of the generator set Transient state Subtransient state Zr = U2 Ssc Z15 = U2 Sr Z20 = U2 Sr Za = U2 Sr Zat = 102 15 Zas = 102 15 102 = 0.05 2000 = • Usc = • Usc • Xsc • 20 100 • 15 100 10 2 15 10 2 20 • 10 100 • 10 100 Z15 • Z20 0.67 • 0.5 Z15 // Z20 = = Z15 + Z20 0.67 + 0.5 Zr + Zet = 0.05 + 0.29 Zer • Zat Zer // Zat = Zer + Zat Zer • Zat Zer // Zat = Zer + Zat 0.34 • 1.33 = 0.34 + 1.33 0.34 • 1 = 0.34 + 1 0.67 0.5 Zat = 1.33 Zas = 1 Zet = 0.29 Zer = 0.34 ≈ 0.27 ≈ 0.25 Circuit breaker Equivalent circuit Breaking capacity Closing capacity Z (ohm) in kA r ms U 10 1 2 Isc = = 3 • Zsc • 3 Zsc 2.5 Isc (in kA peak) D4 to D7 Zr Transient state Z = 0.27 Za Z15 Z20 Subtransient state 21.4 21.4 • 2.5 = 53.5 Z = 0.25 Zt = [ Zr + (Z15//Z20) ] //Za D3 alternator Z15 Zr Z20 Zt = Zr + (Z15//Z20) D1 15 MVA transformer Za Zr Z20 Zt = (Zr + Z20)//Za D2 20 MVA transformer Za Zr Z20 Zt = (Zr + Z15)//Za Z = 0.34 17 17 • 2.5 = 42.5 Transient state Z = 0.39 Subtransient state Z = 0.35 Transient state Z = 0.47 Subtransient state Z = 0.42 14.8 14.8 • 2.5 = 37 12.3 12.3 • 2.5 = 30.7 23
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Units of measure Names and symbols
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Units of measure Names and symbols
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Standards The standards mentioned i
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DE59063EN Standards IEC - ANSI/IEEE
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Standards IEC - ANSI/IEEE compariso
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Medium Voltage technical guide Refe
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Medium Voltage technical guide - Schneider Electric

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