Question

Equipment ratings for the four-bus power system shown in Figure 7.14 are given as follows:

Generator$\mathbf{G}\mathbf{1}\mathbf{:}\mathbf{500}\mathbf{MVA}\mathbf{,}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{kV}\mathbf{,}{\mathbf{X}\mathbf{\text{'}}\mathbf{\text{'}}}_{\mathbf{d}}\mathbf{=}{\mathbf{X}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{,}{\mathbf{X}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{per}\mathbf{unit}$

Generator$\mathbf{G}\mathbf{2}\mathbf{:}\mathbf{750}\mathbf{MVA}\mathbf{,}\mathbf{18}\mathbf{kV}\mathbf{,}{\mathbf{X}\mathbf{\text{'}}\mathbf{\text{'}}}_{\mathbf{d}}\mathbf{=}{\mathbf{X}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{18}\mathbf{,}{\mathbf{X}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{09}\mathbf{per}\mathbf{unit}$

Generator$\mathbf{G}\mathbf{3}\mathbf{:}\mathbf{1000}\mathbf{MVA}\mathbf{,}\mathbf{20}\mathbf{kV}\mathbf{,}{\mathbf{X}\mathbf{\text{'}}\mathbf{\text{'}}}_{\mathbf{d}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{17}\mathbf{,}{\mathbf{X}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{,}{\mathbf{X}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{09}\mathbf{per}\mathbf{unit}$

Transformer$\mathbf{T}\mathbf{1}\mathbf{:}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{500}\mathbf{}\mathbf{MVA}\mathbf{,}\mathbf{}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{D}\mathbf{/}\mathbf{500}\mathbf{}\mathbf{kV}\mathbf{}\mathbf{Y}\mathbf{,}\mathbf{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$

Transformer$\mathbf{T}\mathbf{2}\mathbf{:}\mathbf{750}\mathbf{MVA}\mathbf{,}\mathbf{18}\mathbf{}\mathbf{∆}\mathbf{/}\mathbf{500}\mathbf{}\mathbf{kV}\mathbf{}\mathbf{Y}\mathbf{,}\mathbf{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$

Transformer$\mathbf{T}\mathbf{3}\mathbf{:}\mathbf{1000}\mathbf{MVA}\mathbf{,}\mathbf{20}\mathbf{}\mathbf{∆}\mathbf{/}\mathbf{500}\mathbf{kV}\mathbf{}\mathbf{Y}\mathbf{,}\mathbf{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$

Each line: ${\mathbf{X}}_{\mathbf{1}}\mathbf{=}\mathbf{50}\mathbf{ohms}\mathbf{,}{\mathbf{X}}_{\mathbf{0}}\mathbf{=}\mathbf{150}\mathbf{ohms}$

The inductor connected to generator neutral has a reactance of . Draw the zero, positive, and negative-sequence reactance diagrams using a base in the zone of generator. Neglect transformer phase shifts.

Short answer

The per unit zero-sequence network is as shown below.

The per unit negative-sequence network is as shown below.

The per unit positive-sequence network is as 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 , $0.17\text{per}\text{unit}$, $X_2$are equal to $0.20\text{per}\text{unit}$, and the value of $X_0$is $0.09\text{per}\text{unit}$.

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$500\text{MVA}$, 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,$V_{base}=20\text{kV}$ .

Determine the formulas to draw the zero, positive and negative sequence diagram of the system.

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,role="math" localid="1656481558527" ${\left(X_{pu}\right)}_{old}$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, role="math" localid="1656481758824" $Z_{base}$is the base value, $MVA$is the power rating, and 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 role="math" localid="1656481944605" $Z_{base}$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, 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).

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

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 value 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.