Question

Faults at bus n in Problem 9.5 are of interest (the instructor selects or ). Determine the Thevenin equivalent of each sequence network as viewed from the fault bus. Prefault voltage is per unit. Prefault load currents and phase shifts are neglected.

Short answer

The Thevenin’s equivalent circuit viewed from bus 1 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 $500\text{MVA}$, the voltage rating is$13.8\text{kV}$, 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 $750\text{MVA}$, the voltage rating is $18\text{kV}$, the reactance $X\text{'}{\text{'}}_d$and $X_2$are equal to $0.18\text{per}\text{unit}$, and the value of $X_0$is$0.09\text{per}\text{unit}$.

The power rating of generator $G3$is $1000\text{MVA}$, the voltage rating is$20\text{kV}$, the reactance$X\text{'}{\text{'}}_d$is ,$0.17\text{per}\text{unit}$,$X_2$is equal to$0.20\text{per}\text{unit}$, and the value of$X_0$is .

The neutral has reactance of $0.09\text{per}\text{unit}$.

The neutral has reactance of $X_n=0.028\text{per}\text{unit}$.

Write the rating of the transformer.

The power rating of transformer $T1$is$750\text{MVA}$, the voltage rating is$13.8kV∆/500kVY$, and the reactance value $X$is$0.12\text{per}\text{unit}$.

The power rating of transformer $T2$is $750\text{MVA}$, the voltage rating is$18kV∆/500kVY$, and the reactance value$X$is$0.10\text{per}\text{unit}$.

The power rating of transformer $T3$is$1000\text{MVA}$, the voltage rating is$20kV∆/500kVY$, and the reactance value $X$is $0.10\text{per}\text{unit}$.

Write the reactance of the line.

The positive sequence impedance of line,$X_1=50\text{ohms}$.

The zero-sequence impedance of line,$X_0=150\text{ohms}$.

The base MVA,$S_{base}=1000\text{MVA}$ .

The base voltage,$V_{base}=20\text{kV}$ .

The prefault value of voltage,$V_F=1.0\text{per}\text{unit}$.

Determine the Thevenin’s equivalent for each sequence network.

The equation to calculate the reactance on the new base value can be mathematically presented as shown below.

${\left(X_{pu}\right)}_{new}={\left(X_{pu}\right)}_{old}\frac{{\left(MV{A}_{base}\right)}_{new}}{{\left(MV{A}_{base}\right)}_{old}}$ …… (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, ${\left(MV{A}_{base}\right)}_{new}$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 .

$Z_{base}=\frac{K{V}^2}{MVA}$ …… (2)

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

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

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

Here,$X_{old}$ is the reactance on old base values and$Z_{base}$ is the base impedance.

Draw the zero, positive and negative sequence diagram of the system.

Consider the line diagram for the system.

Calculate the new value of the reactance for generator 1.

Calculate the zero-sequence reactance.

Substitute $500\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$,$0.10\text{pu}$for${\left({\text{X}}_0\right)}_{old}$, and $1000\text{MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$, into equation (1).

$$\begin{gathered} {\left(X_0\right)}_{new}=0.10\times \frac{1000}{500} \\ {\left(X_0\right)}_{new}=0.2\text{pu} \end{gathered}$$

Calculate the positive and negative sequence reactance.

Substitute $500\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$,$0.20\text{pu}$for ${\left({\text{X}}_1\right)}_{old}$, and $1000\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{new}$, into equation (1).

$$\begin{gathered} {\left(X_1\right)}_{new}={X\text{'}\text{'}}_d \\ {\left(X_1\right)}_{new}={\left(X_2\right)}_{new} \\ {\left(X_1\right)}_{new}=0.20\times \frac{1000}{500} \\ {\left(X_1\right)}_{new}=0.4\text{pu} \end{gathered}$$

Calculate the new value of the reactance for generator 2.

Calculate the zero-sequence reactance.

Substitute$750\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$, $0.09\text{pu}$for ${\left({\text{X}}_0\right)}_{old}$, and $1000\text{MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$, into equation (1).

$$\begin{gathered} {\left(X_0\right)}_{new}=0.09\times \frac{1000}{750} \\ {\left(X_0\right)}_{new}=0.12\text{pu} \end{gathered}$$

Calculate the positive and negative sequence reactance.

Substitute $750\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$,$0.18\text{pu}$for${\left({\text{X}}_1\right)}_{old}$, and $1000\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{new}$, into equation (1).

$$\begin{gathered} {\left(X_1\right)}_{new}={X\text{'}\text{'}}_d \\ {\left(X_1\right)}_{new}={\left(X_2\right)}_{new} \\ {\left(X_1\right)}_{new}=0.18\times \frac{1000}{500} \\ {\left(X_1\right)}_{new}=0.24\text{pu} \end{gathered}$$

The new values of the reactance for generator 3 is given below.

$$\begin{gathered} X_0=0.09\text{pu} \\ {\text{X}}_1=0.17\text{pu} \\ {\text{X}}_2=0.20\text{pu} \\ {X}_n=0.028\text{pu} \end{gathered}$$

Calculate the new value of the reactance fortransformer1.

Substitute $500\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$, $0.12\text{pu}$for${\left({\text{X}}_{T1}\right)}_{old}$, and$1000\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{new}$into equation (1).

$$\begin{gathered} {\left(X_{T1}\right)}_{new}=0.12\times \frac{1000}{750} \\ {\left(X_{T1}\right)}_{new}=0.24\text{pu} \end{gathered}$$

Calculate the new value of the reactance fortransformer2.

Substitute $750\text{MVA}$for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$, $0.10\text{pu}$for${\left({\text{X}}_{T1}\right)}_{old}$, and$1000\text{MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$into equation (1).

$$\begin{gathered} {\left(X_{T2}\right)}_{new}=0.10\times \frac{1000}{750} \\ {\left(X_{T2}\right)}_{new}=0.133\text{pu} \end{gathered}$$

Calculate the new value of the reactance fortransformer3.

${\left(X_{T3}\right)}_{new}=0.10\text{pu}$

Calculate the base impedance.

Substitute$1000\text{MVA}$for$S_{Base}$and $500\text{kV}$for$V_{base}$into equation (2).

$$\begin{gathered} Z_{base}=\frac{{500}^2}{1000} \\ {Z}_{base}=250\Omega \end{gathered}$$

Calculate the positive and negative sequence reactance for line.

Substitute$250\Omega$for$Z_{base}$and $50\Omega$for $X_{old}$into equation (3).

$$\begin{gathered} X_1=X_2 \\ {X}_1=\frac{50}{250} \\ {X}_1=0.20\text{pu} \end{gathered}$$

Calculate the zero-sequence reactance for line.

Substitute $250\Omega$for$Z_{base}$and$150\Omega$for $X_{old}$into equation (3).

$$\begin{gathered} X_0=\frac{150}{250} \\ {X}_0=0.60\text{pu} \end{gathered}$$

Draw the per unit zero-sequence network as shown below.

Draw the per unit negative-sequence network as shown below.

Draw the per unit positive-sequence network as shown below.

Reduce the zero-sequence network as shown below.

Calculate the zero-sequence equivalent impedance.

$$\begin{gathered} X_0=0.24\parallel \left(0.6+\left(0.7333\parallel 0.7\right)\right) \\ {X}_0=0.24\parallel 0.9581 \\ {X}_0=0.1919\text{pu} \end{gathered}$$

Reduce the negative-sequence network as shown below.

Calculate the negative-sequence equivalent impedance.

$$\begin{gathered} X_2=0.64\parallel \left(0.2+\left(0.5733\parallel 0.47\right)\right) \\ {X}_2=0.64\parallel 0.4671 \\ {X}_2=0.270\text{pu} \end{gathered}$$

Similarly, calculate the positive-sequence equivalent impedance.

$$\begin{gathered} X_1=0.64\parallel 0.4583 \\ {X}_1=0.2670\text{pu} \end{gathered}$$

The Thevenin’s equivalent circuit viewed from bus 1 is shown below.