# Network Reduction Technique

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

Three symmetrical fault is a balanced fault occurring in all the three phases. Due to this balanced nature of the fault, the problem of the fault analysis can be reduced 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.

## Steps of Network Reduction Technique

The steps involve in solving a problem related to symmetrical fault analysis using the network reduction technique are given below.

- 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.
- Covert all the resistance and the reactance to per unit values by choosing a suitable base kVA (or MVA) and base kV values.
- Draw a reactance diagram from the single line diagram. Use the values of reactance or impedance in per unit as obtained from the step 2.
- 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. On 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.
- 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

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

**Examples**

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

__Example-1:__

Here a three phase symmetrical fault F_{1} occurs at the middle of the line between transformer T_{1} and T_{2} having reactance 0.848 Ω/km.

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

Now converting the reactance into per unit values.

The reactance diagram of the system with the per unit reactance value is shown below

This network can be more reduced as following.

As the equivalent reactance of the system is given by

Now finally the network is reduced to

**Example-2:**

Here a three phase symmetrical fault F occurs at the receiving bus. The length of the transmission line is 200 km long.

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

Now converting the reactance into per unit values.

The reactance diagram of the system with the per unit reactance value is shown below

Network is further reduced as (j0.25 || j0.36) + j 0.115 i.e.

Now the total fault current supplied by the two generators

= total fault current on the feeder side x transformation ratio of the transformer

Fault current supplied by the generator G_{1 }

Fault current supplied by the generator G_{2 }

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**FAULTS IN ELCECTRICAL POWER SYSTEM**

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Good presentation!aditionally&I

hope the bests will came!

Thank you