Question

Consider the one line diagram of a simple power system shown in Figure 9.20. System data in per-unit on a $\mathbf{100}\mathbf{MVA}$base are given as follows:

Synchronous generators:

$\begin{array}{l}\mathbf{}\mathbf{G}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{100}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{20}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_1\mathbf{=}{\mathbf{X}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\\ \mathbf{G}\mathbf{2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{100}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{20}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_1\mathbf{=}{\mathbf{X}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\end{array}$

Transformers:

$\begin{array}{l}\mathbf{T}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{100}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{20}\mathbf{/}\mathbf{220}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_{\mathbf{1}}\mathbf{=}{\mathbf{X}}_{\mathbf{2}}\mathbf{=}{\mathbf{X}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{a}\\ \mathbf{T}\mathbf{2}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{100}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{}\mathbf{20}\mathbf{/}\mathbf{220}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_{\mathbf{1}}\mathbf{=}{\mathbf{X}}_{\mathbf{2}}\mathbf{=}{\mathbf{X}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\end{array}$

Transmission lines:

$\begin{array}{l}\mathbf{L}\mathbf{12}\mathbf{}\mathbf{}\mathbf{100}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{220}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_1\mathbf{=}{\mathbf{X}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{125}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3}\\ \mathbf{L}\mathbf{13}\mathbf{}\mathbf{}\mathbf{100}\mathbf{MVA}\mathbf{220}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_1\mathbf{=}{\mathbf{X}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{35}\\ \mathbf{L}\mathbf{23}\mathbf{}\mathbf{}\mathbf{100}\mathbf{MVA}\mathbf{}\mathbf{}\mathbf{220}\mathbf{kV}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{X}}_1\mathbf{=}{\mathbf{X}}_2\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}{\mathbf{X}}_0\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7125}\end{array}$

The neutral of each generator is grounded through a current-limiting reactor of æ $\mathbf{0}\mathbf{.}\mathbf{08333}$ per unit on a$\mathbf{100}\mathbf{MVA}$ base. All transformer neutrals are solidly grounded. The generators are operating no-load at their rated voltages and rated frequency with their EMFs in phase. Determine the fault current for a balanced three-phase fault at bus 3 through a fault impedance ${\mathbf{Z}}_{\mathbf{F}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$per unit on a$\mathbf{100}\mathbf{MVA}$ base. Neglect $\mathbf{\Delta }\mathbf{-}\mathbf{Y}$phase shifts.

Short answer

The fault current for three phase faults is $820.1\text{\hspace{0.17em}A}$.

Step-by-step solution

Write the given data from the question.

Write the rating of the generators.

The power rating of generator $G1$is $100\mathrm{MVA}$, the voltage rating is $20\mathrm{kV}$, the reactance $X_1$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 $G2$is $100\text{\hspace{0.17em}MVA}$, the voltage rating is $20\text{\hspace{0.17em}kV}$, the reactance $X_1$ and $X_2$are equal to $0.15\text{\hspace{0.17em}per\hspace{0.17em}unit}$and the value of $X_0$is $0.05\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

Write the rating of transformers.

The power rating of transformer $T1$is $100MVA$, the voltage rating is $20/220\mathrm{kV}$, and the value of ${\mathrm{X}}_1,{\mathrm{X}}_2$ and ${\mathrm{X}}_0$is $0.10\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of transformer $\mathrm{T}2$is $100\mathrm{MVA}$, the voltage rating is $20/220\mathrm{kV}$, and the value of ${\mathrm{X}}_1,{\mathrm{X}}_2$ and ${\mathrm{X}}_0$ is $0.10\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

Write the rating for the line.

The power rating of line $L12$ is $100\mathrm{MVA}$, the voltage rating is $220kV$, and the value of ${\mathrm{X}}_1,{\mathrm{X}}_2$is $0.125\text{\hspace{0.17em}per\hspace{0.17em}unit}$and $X_0$is $0.3\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of line $L13$is $100\text{\hspace{0.17em}MVA}$, the voltage rating is $220kV$, and the value of ${\mathrm{X}}_1,{\mathrm{X}}_2$is $0.15\text{\hspace{0.17em}per\hspace{0.17em}unit}$and $X_0$is $0.35\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

The power rating of line $L23$is $100\text{\hspace{0.17em}MVA}$, the voltage rating is $220kV$, and the value of $X_1,X_2$is $0.25\text{\hspace{0.17em}per\hspace{0.17em}unit}$and $X_0$ is $0.7125\text{\hspace{0.17em}per\hspace{0.17em}unit}$.

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

The base voltage, $V_{base}=230\text{\hspace{0.17em}kV}$

The fault impedance,$Z_F=0.1\text{\hspace{0.17em}pu}$

Determine the equation to calculate the fault current

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

${\mathbf{I}}_{\mathbf{base}}\mathbf{=}\frac{{\mathbf{S}}_{\mathbf{base}}}{\sqrt{\mathbf{3}}{\mathbf{V}}_{\mathbf{base}}}$ …… (1)

The equation to calculate per quantities is given as follows.

${\mathrm{I}}_{\mathrm{a}}={\mathrm{I}}_{\mathrm{base}}{\mathrm{I}}_{\mathrm{pu}}$ …… (2)

Here role="math" localid="1656313219030" ${\mathbf{I}}_{\mathbf{pu}}$is per unit current.

Calculate the fault current for three-phase fault.

Since the fault is three phase fault therefore only positive sequence network is existing.

Draw the positive sequence network of the line diagram.

Convert the delta into star as,

The impedance between the line 1 and neutral of the star connection.

$\begin{array}{l}{Z}_{1x}=\frac{\left(j0.125\right)\left(j0.15\right)}{(j0.125+j0.15+j0.25)}\\ Z_{1x}=\frac{\left(j0.125\right)\left(j0.15\right)}{\left(j0.525\right)}\\ Z_{1x}=j0.0357143\end{array}$

The impedance between the line 2 and neutral of the star connection.

$\begin{array}{l}{Z}_{2x}=\frac{\left(j0.125\right)\left(j0.25\right)}{(j0.125+j0.15+j0.25)}\\ Z_{2x}=\frac{\left(j0.125\right)\left(j0.25\right)}{\left(j0.525\right)}\\ Z_{2x}=j0.0595238\end{array}$

The impedance between the line 3 and neutral of the star connection.

$\begin{array}{l}{Z}_{3x}=\frac{\left(j0.15\right)\left(j0.25\right)}{(j0.125+j0.15+j0.25)}\\ Z_{3x}=\frac{\left(j0.15\right)\left(j0.25\right)}{\left(j0.525\right)}\\ Z_{3x}=j0.0714286\end{array}$

The required diagram is shown in Figure below.

Reduce the positive sequence network as,

$\begin{array}{l}{Z}_1=[(j0.15+j0.1)\parallel (j0.15+j0.1)]+(j0.0357143\parallel j0.0595238)+j0.0714286\\ Z_1=[\left(j0.25\right)\parallel \left(j0.25\right)]+j0.0223214+j0.0714286\\ Z_1=j0.125+j0.0223214+j0.0714286\\ Z_1=j0.22\end{array}$

The Thevenin equivalent of positive sequence network is shown below.

The no load generated EMF is$1\angle 0°\text{\hspace{0.17em}pu}$ and fault impedance is$j0.1\text{\hspace{0.17em}pu}$ .

The Thevenin equivalent of positive sequence network with fault impedance.

The positive sequence current from the above circuit.

Calculate the base current as.

Substitute$100MVA$ for$S_{base}$ , and$230Kv$ for$V_{base}$ into equation (1)

Calculate the fault current.

Substitute$262A$ for$I_{base}$ and$-j3.125\text{\hspace{0.17em}pu}$ for$I_a$ into equation (2).

$\begin{array}{l}{I}_a=(-j3.125)\left(262\right)\\ I_a=820.1\text{\hspace{0.17em}A}\end{array}$

Hence, the fault current for three phase faults is$820.1\text{\hspace{0.17em}A}$ .