Question

Faults at bus n in Problem 9.8 are of interest (the instructor selects , or ). Determine the Thevenin equivalent of each sequence network as viewed from the fault bus. Prefault voltage is 1.0 per unit. Prefault load currents and D–Y phase shifts are neglected.

Short answer

The zero sequence Thevenin equivalent circuit is shown below.

The positive sequence Thevenin equivalent circuit is shown below.

The negative sequence Thevenin equivalent circuit is shown below.

Step-by-step solution

Write the given data from the question:

Write the rating of the generators.

The power rating of generator $G1$is$50\text{MVA}$, the voltage rating is , $12\text{kV}$the reactance $X\text{'}{\text{'}}_d$and $X_2$are equal to$0.20\text{per}\text{unit}$and the value of $X_0$is $0.10\text{per}\text{unit}$.

The power rating of generator $G2$is$100\text{MVA}$, the voltage rating is$15\text{kV}$, the reactance$X\text{'}{\text{'}}_d$is $0.20\text{per}\text{unit}$$X_2$,is $0.23\text{per}\text{unit}$and the value of$X_0$is $0.10\text{per}\text{unit}$.

Write the rating of transformers.

The power rating of transformer $T1$is $50\text{MVA}$, the voltage rating is$10kVY/138kVY$ , and the value of $X_0$is$0.10\text{per}\text{unit}$ .

The power rating of transformer $T2$is$100\text{MVA}$ , the voltage rating is$15kVY/138kVY$, and the value of $X_0$is$0.10\text{per}\text{unit}$ .

The reactance of each line,$X_1$ is $40\Omega$and$X_0$ is $100\Omega$.

The base MVA,$S_{base}=100\text{MVA}$

The base voltage for generator side,$V_{base}=15\text{kV}$

The prefault fault voltage,$V_F=1\text{per unit}$

 Step 2: Determine the formulas to calculate the reactance of the zero, positive and negative sequence network. 

The equation to calculate the reactance on the new base value is given as follows.

${\left(X_{pu}\right)}_{new}={\left(X_{pu}\right)}_{old}\frac{{\left(K{V}_{base}\right)}_{old}}{{\left(K{V}_{base}\right)}_{new}}\frac{{\left(MV{A}_{base}\right)}_{new}}{{\left(MV{A}_{base}\right)}_{old}}$ …… (1)

Here ${\left(X_{pu}\right)}_{new}$is the per unit value on new base values, ${\left(X_{pu}\right)}_{old}$is the per unit value on old base values, ${\left(K{V}_{base}\right)}_{old}$it the old voltage rating, ${\left(K{V}_{base}\right)}_{new}$is the new voltage rating, ${\left(MV{A}_{base}\right)}_{new}$is the new power rating and ${\left(MV{A}_{base}\right)}_{old}$is the old power rating.

The equation to calculate the base impedance is given as follows.

$Z_{base}=\frac{K{V}^2}{MVA}$ …… (2)

Here,$Z_{base}$ is the base value,$MVA$ is the power rating,$KV$ is the voltage rating.

The equation to calculate the new per unit impedance of the transmission line is given as follows.

${\left(X\right)}_{new}=\frac{X_{old}}{Z_{base}}$ …… (3)

Calculate the reactance of the zero, positive and negative sequence network.

Calculate the new value of the reactance of generator 1.

For positive and negative sequence reactance.

Substitute $50\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$, $12\text{KV}$for${\left({\text{KV}}_{\text{Base}}\right)}_{old}$,$0.20\text{pu}$for${\left({\text{X}}_1\right)}_{old}$, $100\text{MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$,$10\text{KV}$for ${\left({\text{KV}}_{\text{Base}}\right)}_{new}$into equation (1).

$$\begin{gathered} {\left(X_1\right)}_{new}={\left(X_2\right)}_{new} \\ {\left(X_1\right)}_{new}={\left({X\text{'}\text{'}}_d\right)}_{new} \\ {\left(X_1\right)}_{new}=0.20\times \frac{{12}^2}{{10}^2}\times \frac{100}{50} \\ {\left(X_1\right)}_{new}=0.576\text{pu} \end{gathered}$$

For zero-sequence reactance.

Substitute$50\text{MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{old}$, $12\text{KV}$for ${\left({\text{KV}}_{\text{Base}}\right)}_{old}$,$0.10\text{pu}$for ${\left({\text{X}}_1\right)}_{old}$,$100\text{MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$, $10\text{KV}$for${\left({\text{KV}}_{\text{Base}}\right)}_{new}$into equation (1).

$$\begin{gathered} {\left(X_0\right)}_{new}=0.10\times \frac{{12}^2}{{10}^2}\times \frac{100}{50} \\ {\left(X_0\right)}_{new}=0.288\text{pu} \end{gathered}$$

As the base old MVA and KV values of generator 2 is same as the new value, therefore the reactance of generator remains the same.

$$\begin{gathered} X_1={X\text{'}\text{'}}_d \\ {X}_1=0.2\text{pu} \\ {\text{X}}_2=0.23\text{pu} \\ {\text{X}}_0=0.1\text{pu} \end{gathered}$$

Calculate the new value of the reactance of transformer1.

Substitute $50\text{MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{old}$, $10\text{KV}$for ${\left({\text{KV}}_{\text{Base}}\right)}_{old}$,$0.10\text{pu}$for${\left({\text{X}}_1\right)}_{old}$, $100\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{new}$,$10\text{KV}$for${\left({\text{KV}}_{\text{Base}}\right)}_{new}$into equation (1).

$$\begin{gathered} {\left(X_{T1}\right)}_{new}=0.10\times \frac{{10}^2}{{10}^2}\times \frac{100}{50} \\ {\left(X_{T1}\right)}_{new}=0.20\text{pu} \end{gathered}$$

Calculate the new value of the reactance of transformer2.

Substitute $100\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$, width="44">10KVfor${\left({\text{KV}}_{\text{Base}}\right)}_{old}$,$0.10\text{pu}$for${\left({\text{X}}_1\right)}_{old}$,$100\text{MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$,$10\text{KV}$for${\left({\text{KV}}_{\text{Base}}\right)}_{new}$into equation (1).

$$\begin{gathered} {\left(X_{T2}\right)}_{new}=0.10\times \frac{{10}^2}{{10}^2}\times \frac{100}{100} \\ {\left(X_{T2}\right)}_{new}=0.10\text{pu} \end{gathered}$$

Calculate the base impedance.

Substitute localid="1656593416892" $1000\text{MVA}$for localid="1656593439738" $S_{Base}$andlocalid="1656593447267" $138\text{kV}$for localid="1656593454308" $V_{base}$into equation (2).

localid="1656593461674" $$\begin{gathered} Z_{base}=\frac{{138}^2}{1000} \\ {Z}_{base}=190.44\Omega \end{gathered}$$

Calculate the positive and negative sequence reactance for line.

Substitute localid="1656593468806" $190.44\Omega$for localid="1656593475727" $Z_{base}$and localid="1656593483904" $40\Omega$for localid="1656593490377" $X_{old}$into equation (3)

localid="1656593496481" $$\begin{gathered} X_1=X_2 \\ {X}_1=\frac{40}{190.44} \\ {X}_1=0.210\text{pu} \end{gathered}$$

Calculate the zero-sequence reactance for line.

Substitutelocalid="1656593502999" $190.44\Omega$forlocalid="1656593509006" $Z_{base}$andlocalid="1656593514879" $100\Omega$for localid="1656593521416" $X_{old}$into equation (3)

localid="1656593527536" $$\begin{gathered} X_0=\frac{100}{190.44} \\ {X}_0=0.5251\text{pu} \end{gathered}$$

Draw the zero-sequence network.

Draw the positive sequence network.

Draw the negative network.

Calculate the zero-sequence equivalent impedance.

localid="1656593534563" $$\begin{gathered} X_0=0.2+0.5251+0.5251+0.1 \\ {X}_0=1.3502\text{pu} \end{gathered}$$

The zero sequence Thevenin equivalent circuit is shown below.

Calculate the positive-sequence equivalent impedance.

localid="1656593541576" $$\begin{gathered} X_1=\left(0.576\right)\parallel \left(0.2+0.21+0.1+0.2\right) \\ {X}_1=\left(0.576\right)\parallel \left(0.92\right) \\ {X}_1=0.354\text{pu} \end{gathered}$$

The positive sequence Thevenin equivalent circuit is shown below.

Calculate the negative-sequence equivalent impedance.

localid="1656593428874" $$\begin{gathered} X_2=\left(0.576\right)\parallel \left(0.2+0.21+0.1+0.23\right) \\ {X}_2=\left(0.576\right)\parallel \left(0.94\right) \\ {X}_2=0.3586\text{pu} \end{gathered}$$

The negative sequence Thevenin equivalent circuit is shown below.