Fault in a Power System

· Calculation of Fault Current Using Impedance Diagram

· Calculation of Fault Current Using Z_bus Matrix 

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

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.

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.

Calculation of Fault Current Using Z_bus Matrix

Consider the circuit of Fig. 3.3 which is redrawn as shown in Fig. 6.10.

Fig. 6.10 Network depicting a symmetrical fault at bus-4.

We assume that a symmetrical fault has occurred in bus-4 such that it is now connected to the reference bus. Let us assume that the pre-fault voltage at this bus is V_f . To denote that bus-4 is short circuit, we add two voltage sources V_f and - V_f together in series between bus-4 and the reference bus. Also note that the subtransient fault current I²_f flows from bus-4 to the reference bus. This implies that a current that is equal to - I²_f is injected into bus-4. This current, which is due to the source - V_f will flow through the various branches of the network and will cause a change in the bus voltages. Assuming that the two sources and V_f are short circuited. Then - V_f is the only source left in the network that injects a current - I²_f into bus-4. The voltages of the different nodes that are caused by the voltage - V_f  and the current - I²_f are then given by