The Line-Neutral voltage on the secondary of the transformer is 0.4/√3 = 0.230 kV, giving: Selecting P base as 20 MVA and V base as 11 kV and using the above equations: We want to find the fault level at the transformer secondary. Per-unit values for these elements can be quickly derived from:Ĭonsider a system of source impedance 4.48 Ω connected to a 20 MVA transformer (11/0.4 kV) at 6% impedance. In this instance it is possible to convert the unit:įault calculation problems typically deal wit power sources, generators, transformers and system impedances. Some times per-unit values are available for a given base kV, but the problem being solved is using a different base. Having selected a base power and voltage, the base per unit values of impedance, admittance and current can be calculated from: Per Unitĭividing a system element by it’s per-unit base value gives the per-unit value of the element, for example Tip: while the base power and voltage be any value, typically it would make sense to select values related to the system under construction (for example 11 kV and 20 MVA may be appropriate for a distribution type system) Per Unit Method In applying the per unit method, the first step is to select an arbitrary voltage (V base) and power (P base) base. From this fault level can be readily determined. Any per unit impedance will have the same value on both the primary and secondary of a transformer and is independent of voltage level.Ī network of per unit impedances can then be solved using standard network analysis (see the example). These normalised impedances are know as per unit impedances. To use the per unit method, we normalise all the system impedances (and admittances) within the network under consideration to a common base. By removing the impact of varying voltages, the necessary calculations are simplified. Per unit fault calculations is a method whereby system impedances and quantities are normalised across different voltage levels to a common base.