Chapter 6: Short Circuit Studies - Symmetrical Faults

Section III: Symmetrical Fault in a Power System

Calculation of Fault Current Using Impedance Diagram

Let us first illustrate the calculation of the fault current using the impedance diagram with the help of the following examples.

Example 6.1

Consider the power system of Fig. 6.8 in which a synchronous generator supplies a synchronous motor. The motor is operating at rated voltage and rated MVA while drawing a load current at a power factor of 0.9 (lagging) when a three phase symmetrical short circuit occurs at its terminals. We shall calculate the fault current that flow from both the generator and the motor.

We shall choose a base of 50 MVA, 20 kV in the circuit of the generator. Then the motor synchronous reactance is given by

  per unit  

Also the base impedance in the circuit of the transmission line is

  Ω  

 




Fig. 6.8 A generator supplying a motor load though a transmission line.

Therefore the impedance of the transmission line is

  per unit  

The impedance diagram for the circuit is shown in Fig. 6.9 in which the switch S indicates the fault.




Fig. 6.9 Impedance diagram of the circuit of Fig. 6.8.

The motor draws a load current at rated voltage and rated MVA with 0.9 lagging power factor. Therefore

  per unit  

Then the subtransient voltages of the motor and the generator are

  per unit

 per unit

 

Hence the subtransient fault currents fed by the motor and the generator are

  per unit

 per unit

 

and the total current flowing to the fault is

  per unit

 

Note that the base current in the circuit of the motor is

  A

 

Therefore while the load current was 1603.8 A, the fault current is 7124.7 A.

 

Example 6.2

We shall now solve the above problem differently. The Thevenin impedance at the circuit between the terminals A and B of the circuit of Fig. 6.9 is the parallel combination of the impedances j 0.4 and j 0.5148. This is then given as

per unit  

Since voltage at the motor terminals before the fault is 1.0 per unit, the fault current is

per unit  

If we neglect the pre-fault current flowing through the circuit, then fault current fed by the motor and the generator can be determined using the current divider principle, i.e.,

 per unit

 per unit

 

If, on the other hand, the pre-fault current is not neglected, then the fault current supplied by the motor and the generator are

 per unit

 per unit