Analysis of Symmetrical Faults: In power system networks, the symmetrical fault occurs infrequently. A three-phase line-to-line fault is an asymmetrical fault and the occurrence of such fault in the power system network is rare. Here we will discuss the analysis of symmetrical faults in the power system network.

The analysis of symmetrical fault includes the determination of voltage at any point in the network, the value of current in the network branch, and the value of reactance necessary to limit the fault current to the desired level.

Analysis of symmetrical fault is an easy calculation and the MVA breaking capacity of the circuit breaker is based upon this three-phase fault MVA level.

Contents

## Simplifying Assumptions for Analysis of Symmetrical Faults

We undertake the following assumptions for symmetrical fault analysis.

1. In comparison to the fault currents, we consider the fault currents to be negligible.
2. The iron loss and the core loss associated with the transformer are neglected. The transformer is represented by a series reactance because the series resistance of the transformer is low compared to the reactance.
3. Transmission lines shunt capacitance is neglected.
4. We only take the inductive reactance of the system into account while neglecting the resistance of the system. In the case of the overhead lines and the underground cables of considerable lengths, this assumption cannot be made.
5. We take generators emf to be 1∠0. This assumption clarifies that the system voltage is at its nominal value and is operating at no load during the occurrence of the fault. And 0-degree phase means that the source is the reference.
6. To account for the dc component use correction factors. Such as for a circuit breaker whose breaking capacity is to be determined then for a two-cycle breaker a factor of 1.4 is suitable and for eight cycle breaker a factor of 1 is sufficient.
[Always take the sub transient reactance of the generator. If the transient current is to be determined then use the transient reactance.]

## Fault MVA and Fault Current

Per unit fault current or short circuit current Per unit fault level (MVA) Or, Fault level (MVA) Actual value of fault current ## Analysis Problem Example

Here in the following case, we will calculate the fault current and fault level of the three-phase symmetrical short circuit fault occurring at 22 kV bus bar indicated by ‘F’ Figure: Power System Network

At first, consider a base value of voltage and power for the entire system.

Then let the base MVA of the system be 100 MVA.

Furthermore let the base kV for the generator side be 11 kV, for the feeder side be 132 kV and for the load side be 22 kV.

Then converting the reactance of the power system components into per unit (PU) value. Furthermore assuming the generator voltage to be 1∠0. The figure below shows the reactance diagram of the system. Figure: Reactance Diagram

Therefore Base current IB Then per unit fault current ISC PU As a result, the actual value of fault current ISC Finally, fault Level (MVA) FOR MORE ARTICLES RELATED TO ELECTRICAL ENGINEERING VISIT

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