Question

Consider the system shown in Figure 9.18. (a) As viewed from the fault at F, determine the Thevenin equivalent of each sequence network. Neglect phase shifts. (b) Compute the fault currents for a balanced three phase fault at fault point F through three fault impedances $Z_{FA}=Z_{FB}=Z_{FC}=j0.5$per unit. Equipment data in per-unit on the same base are given as follows:

Synchronous generators:

$$\begin{gathered} G1 X_1=0.2 X_2=0.12 X_0=0.06 \\ G2 X_1=0.33 X_2=0.22 X_0=0.066 \end{gathered}$$

Transformers:

$$\begin{gathered} T1 X_1=X_2=X_0=0.2 \\ T2 X_1=X_2=X_0=0.225 \\ T3 X_1=X_2=X_0=0.27 \\ T4 X_1=X_2=X_0=0.16 \end{gathered}$$

Transmission lines:

$$\begin{gathered} L1 X_1=X_2=0.14 X_0=0.3 \\ L1 X_1=X_2=0.35 X_0=0.6 \end{gathered}$$

Short answer

(a) The Thevenin equivalent for positive sequence network.

The Thevenin equivalent circuit for negative sequence network.

The Thevenin equivalent for zero-sequence network.

(b) The fault current for balance three phase fault is $1.315\angle -90°\text{pu}$.

Step-by-step solution

Write the given data from the question:

Write the reactance of the generators.

For generator$G1$

The negative sequence reactance$X_2$is $0.20\text{per}\text{unit}$, the positive sequence $X_1=0.12\text{per}\text{unit}$and zero sequence reactance$X_0$is $0.06\text{per}\text{unit}$.

For generator$G2$

The negative sequence reactance$X_2$is$0.33\text{per}\text{unit}$, the positive sequence$X_1=0.22\text{per}\text{unit}$and zero sequence reactance $X_0$is$0.066\text{per}\text{unit}$.

Write the reactance of the transformer.

For transformer$T1$

The negative sequencereactance $X_2$, the positive sequence reactance$X_1$and zero sequence reactance$X_0$are$0.2\text{per}\text{unit}$.

For transformer$T2$

The negative sequencereactance $X_2$, the positive sequence reactance $X_1$and zero sequence reactance $X_0$are $0.225\text{per}\text{unit}$.

For transformer$T3$

The negative sequencereactance $X_2$, the positive sequence reactance $X_1$and zero sequence reactance $X_0$are $0.27\text{per}\text{unit}$.

For transformer$T4$

The negative sequencereactance , $X_2$,the positive sequence reactance$X_1$and zero sequence reactance$X_0$are $0.16\text{per}\text{unit}$.

Write the reactance of the transmission lines.

For line $L1$,

The negative sequencereactance , $X_2$the positive sequence reactance $X_1$are$0.14\text{per}\text{unit}$and zero sequence reactance $X_0$is$0.3\text{per}\text{unit}$.

For line $L2$,

The negative sequencereactance $X_2$, the positive sequence reactance $X_1$are$0.35\text{per}\text{unit}$ and zero sequence reactance $X_0$is$0.6\text{per}\text{unit}$ .

The faulty impedances,

$Z_{FA}=Z_{FB}=Z_{FC}=j0.5$

 Step 2: Determine the Thevenin equivalent of each sequence network. 

Draw the per unit positive sequence network.

Reduce the circuit as,

Reduce the circuit as,

Reduce further circuit as,

Reduce further circuit as,

Draw the Thevenin equivalent for positive sequence network.

Draw the per unit negative sequence network.

Reduce the circuit as,

Reduce the circuit as,

Reduce the circuit as,

Determine the negative sequence equivalent impedance.

$$\begin{gathered} X_2=\left(j0.1872\right)\parallel \left(j0.48\right)+j0.0736 \\ {X}_2=j0.1346+j0.0736 \\ {X}_2=j0.2084 \end{gathered}$$

Draw the Thevenin equivalent circuit for negative sequence network.

Draw the Thevenin equivalent circuit for negative sequence network.

Draw the per unit zero-sequence network.

Reduce the circuit as,

Reduce the circuit further as,

Draw the Thevenin equivalent for zero-sequence network.

Calculate the fault currents for a balanced three phase fault at fault point F through three fault impedances.

Since the fault is symmetrical, therefore only positive sequence network exist.

Draw the positive sequence network.

Calculate the fault current,

$$\begin{gathered} I_{sc}=I_{a1} \\ {I}_{sc}=\frac{V_F}{Z_1+Z_F} \end{gathered}$$

Substitute$1\angle 0°\text{V}$ for$V_F$ , $j0.26$for $X_1$and $j0.5$for$Z_F$ into above equation.

$$\begin{gathered} I_{sc}=\frac{1\angle 0°}{j0.26+j0.5} \\ {I}_{sc}=\frac{1\angle 0°}{j0.76} \\ {I}_{sc}=-j1.315\text{pu} \\ {I}_{sc}=1.315\angle -90°\text{pu} \end{gathered}$$

Hence, the fault current for balance three phase fault is$1.315\angle -90°\text{pu}$.