Why are the results of Short-Citcuit calculation depending on Max. Voltage Tolerance?

Category:
Short-Circuit
Summary

For IEC method in the Short-Circuit calculation, when the Calculate option is set for Min. Short-Circuit Currents and the cmin factor is 0.95, the calculation results are different depending which value for Max. Voltage Tolerance for LV-System is used 6% or 10%.

Answer

The new VDE/IEC standard distinguishes between low-voltage networks with 6% and 10% of voltage tolerance.

For networks with a tolerance of 6% cmin and cmax are set according with IEC 60909-0:2001 Page 41 Table 1 - Voltage factor c , Section 2.3.1 (cmax=1.05 and cmin=0.95). For networks with voltage tolerance of 10% cmin is set to 0.95 and cmax to 1.1.

This means that for Min. Short-Circuit Calculation you would get the same result, but not for Max. Short-Circuit Calculation. However, if there is any transformer or generator in the network, according with the new standard, a correction factor has to be considered.

For example, for transformers (refer to IEC 60909-0: 2001 section 3.3.3):

Kt = 0.95 * cmax / (1 + 0.6 * xt); then Kt = Ct ZTK

Where xt is the Short-Circuit voltage of the transformer in p.u., cmax the maximum voltage factor, Nt the transformer impedance and ZTK the corrected impedance.

Despite computing a minimum Short-Circuit Calculation the correction factor for a maximum voltage factor is used. For example there are obtained in dependence on the tolerance of the network different correction factors for the transformers and also different Short-Circuit results.

Back