Question

The single-line diagram of a three-phase power system is shown in Figure 9.17. Equipment ratings are given as follows:

Synchronous generators:

$\begin{array}{l}\mathbf{G}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\mathbf{1000}\mathbf{MVA}\mathbf{15}\mathbf{kV}{X^{\text{'}\text{'}}}_d\mathbf{=}{\mathbf{X}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{18}\mathbf{,}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{07}\mathbf{per}\mathbf{unit}\\ \mathbf{G}\mathbf{2}\mathbf{1000}\mathbf{MVA}\mathbf{15}\mathbf{kV}{X^{\text{'}\text{'}}}_d\mathbf{=}{\mathbf{X}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{,}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{per}\mathbf{unit}\\ \mathbf{G}\mathbf{3}\mathbf{500}\mathbf{MVA}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{kV}{X^{\text{'}\text{'}}}_d\mathbf{=}{\mathbf{X}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1507}\mathbf{,}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{per}\mathbf{unit}\\ \mathbf{G}\mathbf{4}\mathbf{750}\mathbf{MVA}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{kV}{X^{\text{'}\text{'}}}_d\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{,}{\mathbf{X}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{,}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{per}\mathbf{unit}\end{array}$

Transformers:

$\begin{array}{l}\mathbf{T}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1000}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{15}\mathbf{kV}\mathbf{\Delta }/\mathbf{765}\mathbf{kV}\mathbf{Y}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{per}\mathbf{unit}\\ \mathbf{T}\mathbf{2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1000}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{15}\mathbf{kV}\mathbf{\Delta }/\mathbf{765}\mathbf{kV}\mathbf{Y}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{per}\mathbf{unit}\\ \mathbf{T}\mathbf{3}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{500}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{15}\mathbf{kV}\mathbf{\Delta }/\mathbf{765}\mathbf{kV}\mathbf{Y}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{per}\mathbf{unit}\\ \mathbf{T}\mathbf{4}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{750}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{15}\mathbf{kV}\mathbf{\Delta }/\mathbf{765}\mathbf{kV}\mathbf{Y}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{111}\mathbf{per}\mathbf{unit}\end{array}$

Transmission line:

$\begin{array}{l}\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{765}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_1\mathbf{=}\mathbf{50}\mathbf{\Omega }\mathbf{,}{\mathbf{X}}_0\mathbf{=}\mathbf{150}\mathbf{\Omega }\\ \mathbf{1}\mathbf{-}\mathbf{3}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{765}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_1\mathbf{=}\mathbf{40}\mathbf{\Omega }\mathbf{,}{\mathbf{X}}_0\mathbf{=}\mathbf{100}\mathbf{\Omega }\\ \mathbf{2}\mathbf{-}\mathbf{3}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{765}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_1\mathbf{=}\mathbf{40}\mathbf{\Omega }\mathbf{,}{\mathbf{X}}_0\mathbf{=}\mathbf{100}\mathbf{\Omega }\end{array}$

The inductor connected to Generator 3 neutral has a reactance of$\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{per}\mathbf{unit}$ using generator 3 ratings as a base. Draw the zero-, positive-, and negative-sequence reactance diagrams using a$\mathbf{1000}-\mathbf{MVA},\mathbf{765}-\mathbf{kV}$ base in the zone of line 1–2. Neglect the$\mathbf{\Delta }\mathbf{-}\mathbf{Y}$ transformer phase shifts.

Short answer

The per unit zero sequence diagram is as shown below.

Draw the per unit positive sequence diagram as shown below.

Draw the per unit negative sequence diagram 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 $1000\text{\hspace{0.17em}MVA}$, the voltage rating is $15\text{\hspace{0.17em}kV}$, the reactance ${X^{\text{'}\text{'}}}_d$and $X_2$are equal to $0.18\text{\hspace{0.17em}per\hspace{0.17em}unit}$, and the value of $X_0$is $0.07\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of generator $G2$ is $1000\text{\hspace{0.17em}MVA}$, the voltage rating is $15\text{\hspace{0.17em}kV}$, the reactance ${X^{\text{'}\text{'}}}_d$and $X_2$are equal to $0.20\text{\hspace{0.17em}per\hspace{0.17em}unit}$, and the value of $X_0$is $0.10\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of generator $G3$is $500MVA$, the voltage rating is $13.8\text{\hspace{0.17em}kV}$, the reactance ${X^{\text{'}\text{'}}}_d$and $X_2$are equal to $0.15\text{\hspace{0.17em}per\hspace{0.17em}unit}$, and the value of $X_0$ is $0.05\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of generator $G4$is $750MVA$, the voltage rating is $13.8\text{\hspace{0.17em}kV}$, the reactance ${X^{\text{'}\text{'}}}_d$ is $0.30\text{\hspace{0.17em}per\hspace{0.17em}unit}$, $X_2$is $0.40\text{\hspace{0.17em}per\hspace{0.17em}unit}$, and the value of $X_0$is $0.10\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

Write the rating of the transformer.

The power rating of transformer $T1$is $1000MVA$, the voltage rating is $15kV\Delta /765kVY$, and the value of $X_0$is $0.10\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of transformer $T2$ is $1000MVA$, the voltage rating is $15kV\Delta /765kVY$, and the value of $X_0$is $0.10\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of transformer $T3$is $500MVA$, the voltage rating is $15kV\Delta /765kVY$, and the value of $X_0$is $0.12\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of transformer $T4$ is $750MVA$, the voltage rating is $15kV\Delta /765kVY$, and the value of $X_0$ is $0.111\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

Write the rating of transmission.

The voltage rating of transmission line$1-2$ is$765\text{\hspace{0.17em}kV}$ , reactance$X_1$ is $50\Omega$, and$X_0$ is$150\Omega$ .

The voltage rating of transmission line$1-3$ is $765kV$, reactance$X_1$ is $40\Omega$, and $X_0$is $100\Omega$.

The voltage rating of transmission line$2-3$ is$765KV$ , reactance$X_1$ is$40\Omega$ , and$X_0$ is $100\Omega$.

The neutral reactance of generator 3, $Z_{n3}=0.05\text{\hspace{0.17em}pu}$.

The base MVA, $S_{base}=1000\text{\hspace{0.17em}MVA}$.

The base voltage,$V_{base}=15\text{\hspace{0.17em}kV}$ .

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.

${\mathbf{\left(}{\mathbf{X}}_{\mathbf{pu}}\mathbf{\right)}}_{\mathbf{new}}\mathbf{=}{\mathbf{\left(}{\mathbf{X}}_{\mathbf{pu}}\mathbf{\right)}}_{\mathbf{old}}\frac{{\mathbf{\left(}{\mathbf{KV}}_{\mathbf{base}}\mathbf{\right)}}_{\mathbf{old}}}{{\mathbf{\left(}{\mathbf{KV}}_{\mathbf{base}}\mathbf{\right)}}_{\mathbf{new}}}\frac{{\mathbf{\left(}{\mathbf{MVA}}_{\mathbf{base}}\mathbf{\right)}}_{\mathbf{new}}}{{\mathbf{\left(}{\mathbf{MVA}}_{\mathbf{base}}\mathbf{\right)}}_{\mathbf{old}}}$ …… (1)

Here, ${\mathbf{\left(}{\mathbf{X}}_{\mathbf{pu}}\mathbf{\right)}}_{\mathbf{new}}$is the per unit value on new base values, ${\mathbf{\left(}{\mathbf{X}}_{\mathbf{pu}}\mathbf{\right)}}_{\mathbf{old}}$is the per unit value on old base values, ${\mathbf{\left(}{\mathbf{KV}}_{\mathbf{base}}\mathbf{\right)}}_{\mathbf{old}}$it the old voltage rating, ${\mathbf{\left(}{\mathbf{KV}}_{\mathbf{base}}\mathbf{\right)}}_{\mathbf{new}}$is the new voltage rating,${\mathbf{\left(}{\mathbf{MVA}}_{\mathbf{base}}\mathbf{\right)}}_{\mathbf{new}}$is the new power rating, and ${\mathbf{\left(}{\mathbf{MVA}}_{\mathbf{base}}\mathbf{\right)}}_{\mathbf{old}}$is the old power rating.

The equation to calculate the base impedance isrole="math" localid="1656424350222" ${\mathbf{Z}}_{\mathbf{b}\mathbf{a}\mathbf{s}\mathbf{e}}\mathbf{=}\frac{\mathbf{K}{\mathbf{V}}^{\mathbf{2}}}{\mathbf{M}\mathbf{V}\mathbf{A}}$. …… (2)

Here, ${\mathbf{Z}}_{\mathbf{base}}$is the base value, $\mathbf{MVA}$is the power rating, and $\mathbf{KV}$is the voltage rating.

The equation to calculate the new per unit impedance of the transmission line is . ${\mathbf{\left(}\mathbf{X}\mathbf{\right)}}_{\mathbf{new}}\mathbf{=}\frac{{\mathbf{X}}_{\mathbf{old}}}{{\mathbf{Z}}_{\mathbf{base}}}$ …… (3)

Here, ${\mathbf{X}}_{\mathbf{old}}$ is the reactance on old base values and${\mathbf{Z}}_{\mathbf{base}}$ is the base impedance.

Calculate the new value of the reactance and draw the sequence diagram.

As the base old MVA and KV values of generator 1 are same as the new value, the reactance of generator 1 remains the same.

$\begin{array}{l}{X}_1=X_d^{\text{'}\text{'}}\\ X_1=X_2\\ X_1=0.18\text{\hspace{0.17em}pu}\\ {\text{X}}_0=0.07\text{\hspace{0.17em}pu}\end{array}$

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

$\begin{array}{l}{X}_1=X_d^{\text{'}\text{'}}\\ X_1=X_2\\ X_1=0.20\text{\hspace{0.17em}pu}\\ {\text{X}}_0=0.10\text{\hspace{0.17em}pu}\end{array}$

Calculate the new value of the reactance of generator 3.

For positive and negative sequence reactance, substitute $500\text{\hspace{0.17em}MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{old}$, $13.8\text{\hspace{0.17em}KV}$ for ${\left({\text{KV}}_{\text{Base}}\right)}_{old}$, $0.15\text{\hspace{0.17em}pu}$for ${\left({\text{X}}_1\right)}_{old}$, $1000\text{\hspace{0.17em}MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$, $15KV$for ${\left({\text{KV}}_{\text{Base}}\right)}_{new}$into equation (1).

$\begin{array}{l}{\left(X_1\right)}_{new}={\left(X_2\right)}_{new}\\ {\left(X_1\right)}_{new}=0.15\times \frac{{13.8}^2}{{15}^2}\times \frac{1000}{500}\\ {\left(X_1\right)}_{new}=0.2539\text{\hspace{0.17em}pu}\end{array}$

For zero sequence reactance, substitute$500\text{\hspace{0.17em}MVA}$ for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$ ,$13.8KV$ for${\left({\text{KV}}_{\text{Base}}\right)}_{old}$ , role="math" localid="1656427534030" $0.05\text{\hspace{0.17em}pu}$for ${\left({\text{X}}_1\right)}_{old}$, $1000\text{\hspace{0.17em}MVA}$for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$,$15\text{\hspace{0.17em}KV}$ for${\left({\text{KV}}_{\text{Base}}\right)}_{new}$ into equation (1).

$\begin{array}{l}{\left(X_0\right)}_{new}=0.05\times \frac{{13.8}^2}{{15}^2}\times \frac{1000}{500}\\ {\left(X_0\right)}_{new}=0.08464\text{\hspace{0.17em}pu}\end{array}$

Calculate the reactance of neutral of generator 3,$3X_n=3X_0$ .

Substitute$0.05\text{\hspace{0.17em}pu}$ for $X_0$into the equation.

$\begin{array}{l}3X_n=3\times 0.05\\ 3X_n=0.15\text{\hspace{0.17em}pu}\end{array}$

Calculate the per unit reactance for generator 4.

For zero sequence reactance, substitute$750\text{\hspace{0.17em}MVA}$ for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$ ,$13.8\text{\hspace{0.17em}KV}$ for${\left({\text{KV}}_{\text{Base}}\right)}_{old}$ , $0.10\text{\hspace{0.17em}pu}$for ${\left({\text{X}}_0\right)}_{old}$,$1000\text{\hspace{0.17em}MVA}$ for${\left({\text{MVA}}_{\text{base}}\right)}_{new}$ ,$15\text{\hspace{0.17em}KV}$ for${\left({\text{KV}}_{\text{Base}}\right)}_{new}$ into equation (1).

$\begin{array}{l}{\left(X_0\right)}_{new}=0.10\times \frac{{13.8}^2}{{15}^2}\times \frac{1000}{750}\\ {\left(X_0\right)}_{new}=0.1129\text{\hspace{0.17em}pu}\end{array}$

For positive sequence, substitute$750\text{\hspace{0.17em}MVA}$ for ${\left({\text{MVA}}_{\text{base}}\right)}_{old}$,$13.8\text{\hspace{0.17em}KV}$ for ${\left({\text{KV}}_{\text{Base}}\right)}_{old}$, $0.30\mathrm{PU}$for${\left({\text{X}}_1\right)}_{old}$ ,$1000\text{\hspace{0.17em}MVA}$ for${\left({\text{MVA}}_{\text{base}}\right)}_{new}$ , and $15\text{\hspace{0.17em}KV}$for${\left({\text{KV}}_{\text{Base}}\right)}_{new}$ into equation (1).

$\begin{array}{l}{\left(X_1\right)}_{new}=0.30\times \frac{{13.8}^2}{{15}^2}\times \frac{1000}{750}\\ {\left(X_1\right)}_{new}=0.3386\text{\hspace{0.17em}pu}\end{array}$

For negative sequence, substitute $750MVA$ for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$ , $13.8KV$for ${\left({\text{KV}}_{\text{Base}}\right)}_{old}$, $0.40\mathrm{PU}$for${\left({\text{X}}_2\right)}_{old}$ ,$1000\text{\hspace{0.17em}MVA}$ for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$,$15\text{\hspace{0.17em}KV}$ for${\left({\text{KV}}_{\text{Base}}\right)}_{new}$ into equation (1).

$\begin{array}{l}{\left(X_2\right)}_{new}=0.40\times \frac{{13.8}^2}{{15}^2}\times \frac{1000}{750}\\ {\left(X_2\right)}_{new}=0.4514\text{\hspace{0.17em}pu}\end{array}$

Calculate the per unit reactance for transformers.

As the base old MVA and KV values of transformer 1 are same as the new value, the reactance of generator 1 remains the same.

$X_{T1}=0.10\text{\hspace{0.17em}pu}$

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

$X_{T2}=0.10\text{\hspace{0.17em}pu}$

For transformer 3, substitute$750MVA$ for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$ ,$15KV$ for ${\left({\text{KV}}_{\text{Base}}\right)}_{old}$, $0.12\mathrm{pu}$for ${\left({\text{X}}_{T3}\right)}_{old}$,$1000MVA$ for ${\left({\text{MVA}}_{\text{base}}\right)}_{new}$,$15KV$ for${\left({\text{KV}}_{\text{Base}}\right)}_{new}$ into equation (1).

$\begin{array}{l}{\left(X_{T3}\right)}_{new}=0.12\times \frac{{15}^2}{{15}^2}\times \frac{1000}{750}\\ {\left(X_{T3}\right)}_{new}=0.24\text{\hspace{0.17em}pu}\end{array}$

For transformer 4, substitute$750\text{\hspace{0.17em}MVA}$ for${\left({\text{MVA}}_{\text{base}}\right)}_{old}$ ,$15KV$ forrole="math" localid="1656425566783" ${\left({\text{KV}}_{\text{Base}}\right)}_{old}$ , $0.11\mathrm{pu}$for ${\left({\text{X}}_{T4}\right)}_{old}$,$1000MVA$ for${\left({\text{MVA}}_{\text{base}}\right)}_{new}$ ,$15KV$ for${\left({\text{KV}}_{\text{Base}}\right)}_{new}$ into equation (1).

$\begin{array}{l}{\left(X_{T4}\right)}_{new}=0.11\times \frac{{15}^2}{{15}^2}\times \frac{1000}{750}\\ {\left(X_{T4}\right)}_{new}=0.1467\text{\hspace{0.17em}pu}\end{array}$

Calculate the base impedance.

Substitute$1000MVA$ for$S_{Base}$ and $765\text{\hspace{0.17em}kV}$for$V_{base}$ into equation (2).

$\begin{array}{l}{Z}_{base}=\frac{{765}^2}{1000}\\ Z_{base}=585.23\Omega \end{array}$

Calculate the positive sequence reactance for line 1-2.

Substitute$585.23\Omega$ for$Z_{base}$ and$50\Omega$ for $X_{old}$into equation (3).

$\begin{array}{l}{X}_{12}=\frac{50}{585.23}\\ X_{12}=0.08544\text{\hspace{0.17em}pu}\end{array}$

Calculate the positive sequence reactance for lines 1-3 and 2-3.

Substitute$585.23\Omega$ for $Z_{base}$and$40\Omega$ for$X_{old}$ into equation (3).

$\begin{array}{l}{X}_{13}=X_{23}\\ X_{13}=\frac{40}{585.23}\\ X_{13}=0.06835\text{\hspace{0.17em}pu}\end{array}$

Calculate the zero-sequence reactance for line 1-2.

Substitute$585.23\Omega$ for$Z_{base}$ and$150\Omega$ for $X_{old}$into equation (3).

$\begin{array}{l}{X}_{12}=\frac{150}{585.23}\\ X_{12}=0.2563\text{\hspace{0.17em}pu}\end{array}$

Calculate the zero-sequence reactance for lines 1-3 and 2-3.

Substitute $585.23\Omega$for$Z_{base}$ and$100\Omega$ for$X_{old}$ into equation (3).

$\begin{array}{l}{X}_{13}=X_{23}\\ X_{13}=\frac{100}{585.23}\\ X_{13}=0.1709\text{\hspace{0.17em}pu}\end{array}$

Draw the per unit zero sequence diagram as shown below.

Draw the per unit positive sequence diagram as shown below.

Draw the per unit negative sequence diagram as shown below.