Network Reduction Technique: For simple power system networks, the analysis of the symmetrical faults can be made easy by the network reduction technique. For modern complex problems, computers are used for these computations.

Three symmetrical faults is a balanced fault occurring in all three phases. Due to this balance nature of the fault, we can reduce the problem of the fault analysis to a single-phase problem. As any condition happening in one phase equally applies to the other phases.

In-network reduction technique, the equivalent impedance up to the fault point can be determined or obtained by network reduction. The network reduction involves series-parallel combinations and star to delta or delta to star conversion of the reactance or impedance.

Contents

## Steps of Network Reduction Technique

The steps involved in solving a problem on symmetrical fault analysis using the network reduction technique are as follows.

1. Form a single line diagram (SLD) of the given power system network. The single line diagram should include the component along with its details such as its rating, voltage, resistance, and reactance or impedance.
2. Covert all the resistance and the reactance to per-unit values by choosing a suitable base kVA (or MVA) and base kV values.
3. Draw a reactance diagram from the single line diagram. Use the values of reactance or impedance per unit as obtained from step 2.
4. Now use the network reduction methods such as series-parallel combinations and star to delta or delta to star conversion of the reactance or impedance. In doing so, keep the identity of the fault point intact (unchanged). Now find the reactance of the system as seen from the fault point. This will give the value of the equivalent reactance of the system up to the fault point.
5. By using the mentioned mathematical relations, calculate the fault current or short circuit current and fault level.

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 1. To calculate the voltage and current distribution throughout the network, retrace the step.

## Examples of Network Reduction Technique

Let’s see some examples for the analysis of symmetrical faults using the network reduction technique.

### Example-1:

Here a three-phase symmetrical fault F1 occurs at the middle of the line between transformer T1 and T2 having a reactance of 0.848 Ω/km. Figure: Power System Network

At first, consider the base MVA of the system to be 20 MVA, the base kV of the generator side be 11 kV and for the transmission, side is 66 kV.

Secondly converting the reactance into per-unit values. Furthermore, we draw the figure showing the reactance diagram of the system with the per-unit reactance value. Figure: Reactance Diagram

Then we reduce this network as  Figure: Reactance Diagram

Also, the equivalent reactance of the system is Now finally the network is reduced to Figure: Reactance Diagram ### Example-2:

Here a three-phase symmetrical fault F occurs at the receiving bus. The length of the transmission line is 200 km long. Ω Figure: Power System Network

At first, consider the base MVA of the system to be 20 MVA, the base kV of the generator side be 11 kV and the transmission side is 132 kV.

Secondly converting the reactance into per-unit values. Furthermore, the reactance diagram of the system with the per-unit reactance value is shown below Figure: Reactance Diagram

Then network is further reduced as (j0.25 || j0.36) + j 0.115 i.e. Figure: Reactance Diagram  Figure: Simplified Reactance Diagram

Finally, the total fault current supplied by the two generators

= total fault current on the feeder side x transformation ratio of the transformer Moreover, the fault current supplied by the generator G1 Also, the fault current supplied by the generator G2 FOR MORE ARTICLES RELATED TO ELECTRICAL ENGINEERING VISIT

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