Question

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

Generator G1:$\mathbf{500}\mathbf{}\mathbf{MVA}\mathbf{,}\mathbf{}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{kV}\mathbf{,}\mathbf{}\mathbf{X}\mathbf{\text{'}\text{'}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$

Generator G2:$\mathbf{750}\mathbf{}\mathbf{MVA}\mathbf{,}\mathbf{}\mathbf{18}\mathbf{}\mathbf{kV}\mathbf{,}\mathbf{}\mathbf{X}\mathbf{\text{'}\text{'}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{18}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$

Generator G3:$\mathbf{1000}\mathbf{}\mathbf{MVA}\mathbf{,}\mathbf{}\mathbf{20}\mathbf{}\mathbf{kV}\mathbf{,}\mathbf{}\mathbf{X}\mathbf{\text{'}\text{'}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{17}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$

Transformer T1:$\mathbf{500}\mathbf{}\mathbf{MVA}\mathbf{,}\mathbf{}\mathbf{13}\mathbf{.}\mathbf{8}\mathbf{∆}\mathbf{/}\mathbf{500}\mathbf{Y}\mathbf{}\mathbf{kV}\mathbf{,}\mathbf{}\mathbf{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$

Transformer T2:$\mathbf{750}\mathbf{}\mathbf{MVA}\mathbf{,}\mathbf{}\mathbf{18}\mathbf{∆}\mathbf{/}\mathbf{500}\mathbf{Y}\mathbf{}\mathbf{kV}\mathbf{,}\mathbf{}\mathbf{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$

Transformer T3:$\mathbf{1000}\mathbf{}\mathbf{MVA}\mathbf{,}\mathbf{}\mathbf{20}\mathbf{∆}\mathbf{/}\mathbf{500}\mathbf{Y}\mathbf{}\mathbf{kV}\mathbf{,}\mathbf{}\mathbf{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}$

Each 500 kV line:.${\mathbf{X}}_{\mathbf{1}}\mathbf{=}\mathbf{50}\mathbf{}\mathbf{\Omega }$

A three-phase short circuit occurs at bus 1, where the prefault voltage is 525 kV . Prefault load current is neglected. Draw the positive-sequence reactance diagram in per unit on a $\mathbf{1000}\mathbf{-}\mathbf{MVA}\mathbf{,}\mathbf{}\mathbf{20}\mathbf{-}\mathbf{kV}$, base in the zone of generatorG3. Determine (a) the Thévenin reactance in per unit at the fault, (b) the subtransient fault current in per unit and in kA rms , and (c) contributions to the fault current from generator GI and from line $\mathbf{1}\mathbf{-}\mathbf{2}$.

Short answer

(a) The Thévenin reactance per unit fault is 0.27 p.u.

(b) The sub transient fault current in per unit and kA are $-j5.33\mathrm{p}.\mathrm{u}.\mathrm{and}-j6.18\mathrm{kA}$.

(c) The current contributed from GI and1-2 line is $-j1.89\mathrm{kA}\mathrm{and}-j2.62\mathrm{kA}$.

Step-by-step solution

Write the given data from the question.

Consider a single-line diagram of four-bus power system.

Consider the following data:

GeneratorGI : Power rating: 500 MVA

Voltage rating: 13.8 kV

Reactance: 0.2 p.u.

GeneratorG2: Power rating: 750 MVA

Voltage rating: 18 kV

Reactance: 0.18 p.u.

Generator: Power rating: 1000 MVA

Voltage rating: 20 kV

Reactance: 0.17 p.u.

TransformerT1: Power rating: 500 MVA

Voltage rating:$13.8∆/500\mathrm{YkV}$

Reactance: 0.12 p.u.

TransformerT2: Power rating: 750 MVA

Voltage rating: $18∆/500\mathrm{YkV}$

Reactance: 0.10 p.u.

TransformerT3: Power rating:1000 MVA

Voltage rating: $20∆/500\mathrm{YkV}$

Reactance:0.10 p.u.

Prefault Voltage: 525 kV

Base: Power rating: 1000 MVA

For G3 Voltage rating: 20 kV

Determine the required formula.

Write the formula ofThévenin reactance per unit fault.

${\mathbf{X}}_{\mathbf{thp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{=}\left(\begin{array}{c}{X}_{T1p.u.\mathrm{new}}+\\ +X_{G1p.u.\mathrm{new}}\end{array}\right)\mathbf{֏}\left(\begin{array}{c}\left(X_{\mathrm{line}12p.u.}+X_{\mathrm{line}24p.u.}+X_{G2p.u.\mathrm{new}}\right)֏\\ \left(X_{\mathrm{line}23}+X_{T3p.u.\mathrm{new}}+X_{G3p.u.\mathrm{new}}\right)\end{array}\right)$ …… (1)

Here, ${\mathbf{X}}_{\mathbf{T}\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$is per unit reactance of $\mathbf{T}\mathbf{1}$transformer according to new base, ${\mathbf{X}}_{\mathbf{G}\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$is per unit reactance of generator $\mathbf{G}\mathbf{1}$according to new base, ${\mathbf{X}}_{\mathbf{line}\mathbf{12}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unit line reactance${\mathbf{X}}_{\mathbf{line}\mathbf{12}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}$ is per unit line reactance,${\mathbf{X}}_{\mathbf{12}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$ is per unit reactance, ${\mathbf{X}}_{\mathbf{G}\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$is per unit reactance of generator $\mathbf{G}\mathbf{2}$according to new base, ${\mathbf{X}}_{\mathbf{line}\mathbf{23}}$is line reactance, ${\mathbf{X}}_{\mathbf{T}\mathbf{3}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$is per unit reactance of transformer$\mathbf{T}\mathbf{3}$ according to new base and ${\mathbf{X}}_{\mathbf{G}\mathbf{3}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$is per unit reactance of generator $\mathbf{G}\mathbf{3}$according to new base.

Write the formula ofsub transient fault current in per unit.

${\mathbf{I}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}}{{\mathbf{X}}_{\mathbf{thp}\mathbf{.}\mathbf{u}\mathbf{.}}}$ …… (2)

Here, ${\mathbf{V}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unit voltage at the fault and ${\mathbf{X}}_{\mathbf{thp}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unit Thévenin reactance.

Write the formula ofsub transient fault current in .

${\mathbf{I}}_{\mathbf{F}}\mathbf{=}{\mathbf{I}}_{\mathbf{base}\mathbf{4}}\mathbf{\times }{\mathbf{I}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$ …… (3)

Here, ${\mathbf{I}}_{\mathbf{base}\mathbf{4}}$is base current and ${\mathbf{I}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unit fault current.

Write the formula ofcurrent contributed from.

${\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{F}}\mathbf{=}{\mathbf{I}}_{\mathbf{base}\mathbf{4}}\mathbf{\times }{\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$ …… (4)

Here,${\mathbf{I}}_{\mathbf{base}\mathbf{4}}$ is base current and${\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$ is per unit GI fault current.

Write the formula ofcurrent contributed from 1-2 line.

${\mathbf{I}}_{\mathbf{12}}\mathbf{=}{\mathbf{I}}_{\mathbf{F}}\mathbf{-}{\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{F}}$ …… (5)

Here, ${\mathbf{I}}_{\mathbf{F}}$is fault current and ${\mathbf{I}}_{\mathbf{G}\mathbf{1}\mathbf{F}}$is fault current.

Determine the Thévenin reactance per unit fault.

(a)

First determine the per unit reactance of generatorG3.

$\frac{{\mathrm{X}}_{\mathrm{G}3\mathrm{p}.\mathrm{u}.\mathrm{new}}}{{\mathrm{X}}_{\mathrm{G}3\mathrm{p}.\mathrm{u}.}}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{S}}_{\mathrm{G}3\mathrm{rated}}}\times {\left(\frac{{\mathrm{V}}_{\mathrm{G}3\mathrm{rated}}}{{\mathrm{V}}_{\mathrm{baseG}3}}\right)}^2$...... (6)

Here, localid="1655443161508" $X_{\mathrm{G}3\mathrm{p}.\mathrm{u}.\mathrm{new}}$isper unit reactance of generator G3, $X_{\mathrm{G}3\mathrm{p}.\mathrm{u}.}$is per unit reactance of the rated generator,$S_{base}$is given power base,$S_{\mathrm{G}3\mathrm{rated}}$is rated power as base, $V_{\mathrm{G}3\mathrm{rated}}$is rated voltage base and $V_{\mathrm{baseG}3}$is new voltage base for G3 .

Substitute 0.17 for $X_{\mathrm{G}3\mathrm{p}.\mathrm{u}.}$, 1000 for $S_{base}$, 1000 for $S_{\mathrm{G}3\mathrm{rated}}$, 20 for $V_{\mathrm{G}3\mathrm{rated}}$and 20 for $V_{\mathrm{baseG}3}$into equation (6).

$$\begin{gathered} X_{G3p.u.new}=0.17\left(\frac{1000}{1000}\right){\left(\frac{20}{20}\right)}^2 \\ =0.17\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the per unit reactance of generatorG2.

localid="1655444118138" $\frac{{\mathrm{X}}_{\mathrm{G}3\mathrm{p}.\mathrm{u}.\mathrm{new}}}{{\mathrm{X}}_{\mathrm{G}2\mathrm{p}.\mathrm{u}.}}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{S}}_{\mathrm{G}2\mathrm{rated}}}\times {\left(\frac{{\mathrm{V}}_{\mathrm{G}2\mathrm{rated}}}{{\mathrm{V}}_{\mathrm{baseG}2}}\right)}^2$...... (7)

Here, ${\mathrm{X}}_{\mathrm{G}3\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of generatorG2 , $X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.}$is per unit reactance of the rated generator, localid="1655443738182" $S_{base}$is given power base, $S_{\mathrm{G}2\mathrm{rated}}$is rated power as base, $V_{\mathrm{G}2\mathrm{rated}}$is rated voltage base and $V_{\mathrm{baseG}2}$is new voltage base for G2.

Substitute 0.18 for $X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.}$, 1000 for $S_{base}$, 750 for $S_{\mathrm{G}2\mathrm{rated}}$, 18 for $V_{\mathrm{G}2\mathrm{rated}}$and 18 for $V_{\mathrm{baseG}2}$into equation (7).

$$\begin{gathered} X_{G2p.u.new}=0.18\left(\frac{1000}{750}\right){\left(\frac{18}{18}\right)}^2 \\ =0.24\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the per unit reactance of generatorG1.

localid="1655444634253" $\frac{{\mathrm{X}}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}}{{\mathrm{X}}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.}}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{S}}_{\mathrm{G}1\mathrm{rated}}}\times {\left(\frac{{\mathrm{V}}_{\mathrm{G}1\mathrm{rated}}}{{\mathrm{V}}_{\mathrm{baseG}1}}\right)}^2$...... (8)

Here,$X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of generatorG2,$X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.}$is per unit reactance of the rated generator,$S_{base}$is given power base, $S_{\mathrm{G}1\mathrm{rated}}$is rated power as base,$V_{\mathrm{G}1\mathrm{rated}}$is rated voltage base and $V_{\mathrm{baseG}1}$is new voltage base for G1.

Substitute 0.2 for $X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.}$, 1000 for $S_{base}$, 500 for $S_{\mathrm{G}1\mathrm{rated}}$, 13.8 for$V_{\mathrm{G}1\mathrm{rated}}$and 13.8 for $V_{\mathrm{baseG}1}$into equation (8).

$$\begin{gathered} X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}=0.2\left(\frac{1000}{500}\right){\left(\frac{13.8}{13.8}\right)}^2 \\ =0.4\mathrm{p}.\mathrm{u} \end{gathered}$$

For transformerT1 determine the new per unit reactance of transformer according to new base.

$\frac{{\mathrm{X}}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}}{{\mathrm{X}}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.}}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{S}}_{\mathrm{T}1\mathrm{rated}}}\times {\left(\frac{{\mathrm{V}}_{\mathrm{T}1\mathrm{rated}}}{{\mathrm{V}}_{\mathrm{baseT}1}}\right)}^2$ …… (9)

Here, $X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of transformer T1 according to new base, $X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.}$is per unit reactance on the rated base, $S_{\mathrm{base}}$is given power base, $S_{\mathrm{T}1\mathrm{rated}}$is rated power as base, $V_{\mathrm{T}1\mathrm{rated}}$is rated voltage base and $V_{\mathrm{baseT}1}$is new voltage base for T1.

Substitute 0.12 for $X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.}$, 100 for $S_{\mathrm{base}}$, 500 for $S_{\mathrm{T}1\mathrm{rated}}$, 13.8 for$V_{\mathrm{T}1\mathrm{rated}}$and 13.8 for $V_{\mathrm{baseT}1}$into equation (9).

localid="1655445300189" $$\begin{gathered} X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.new}=\left(0.12\right)\left(\frac{1000}{500}\right){\left(\frac{13.8}{13.8}\right)}^2 \\ =0.24\mathrm{p}.\mathrm{u}. \end{gathered}$$

For transformer determine the new per unit reactance of transformer according to new base.

$\frac{{\mathrm{X}}_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}}{{\mathrm{X}}_{\mathrm{T}2\mathrm{p}.\mathrm{u}.}}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{S}}_{\mathrm{T}2\mathrm{rated}}}\times {\left(\frac{{\mathrm{V}}_{\mathrm{T}2\mathrm{rated}}}{{\mathrm{V}}_{\mathrm{baseT}2}}\right)}^2$ …… (10)

Here, $X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of transformer T2 according to new base, $X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.}$is per unit reactance on the rated base,$S_{base}$is given power base,$S_{\mathrm{T}2\mathrm{rated}}$is rated power as base, $V_{\mathrm{T}2\mathrm{rated}}$is rated voltage base and$V_{\mathrm{baseT}2}$is new voltage base for T2.

Substitute 0.10 for $X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.}$, 1000 for $S_{\mathrm{base}}$, 750 for $S_{\mathrm{T}2\mathrm{rated}}$, 18 for $V_{\mathrm{baseT}2}$and for into equation (10).

localid="1655446317104" $$\begin{gathered} X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}=\left(0.10\right)\left(\frac{1000}{750}\right){\left(\frac{18}{18}\right)}^2 \\ =0.13\mathrm{p}.\mathrm{u}. \end{gathered}$$

For transformer T3 determine the new per unit reactance of transformer according to new base.

$\frac{{\mathrm{X}}_{\mathrm{T}3\mathrm{p}.\mathrm{u}.\mathrm{new}}}{{\mathrm{X}}_{\mathrm{T}3\mathrm{p}.\mathrm{u}.}}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{S}}_{\mathrm{T}3\mathrm{rated}}}\times {\left(\frac{{\mathrm{V}}_{\mathrm{T}3\mathrm{rated}}}{{\mathrm{V}}_{\mathrm{baseT}3}}\right)}^2$ …… (11)

Here, $X_{\mathrm{T}3\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of transformer T3 according to new base, $X_{\mathrm{T}3\mathrm{p}.\mathrm{u}.}$is per unit reactance on the rated base, $S_{\mathrm{base}}$is given power base, $S_{\mathrm{T}3\mathrm{rated}}$is rated power as base, $V_{\mathrm{T}3\mathrm{rated}}$is rated voltage base and$V_{\mathrm{baseT}3}$is new voltage base for T3.

Substitute 0.10 for $X_{\mathrm{T}3\mathrm{p}.\mathrm{u}.}$, 1000 for$S_{\mathrm{base}}$, 1000 for $S_{\mathrm{T}3\mathrm{rated}}$, 20 for $V_{\mathrm{T}3\mathrm{rated}}$and 20 for$V_{\mathrm{baseT}3}$into equation (11).

$$\begin{gathered} X_{\mathrm{T}3\mathrm{p}.\mathrm{u}.\mathrm{new}}=\left(0.10\right)\left(\frac{1000}{1000}\right){\left(\frac{20}{20}\right)}^2 \\ =0.13\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the base impedance for the transmission lines.

${\mathrm{Z}}_{\mathrm{base}}=\frac{{\mathrm{V}}_{\mathrm{base}}^2}{{\mathrm{S}}_{\mathrm{base}}}$ …… (12)

Here,${\mathrm{V}}_{\mathrm{base}}^2$is base voltage and$S_{\mathrm{base}}$is given power base.

Substitute $500\times {10}^3$for ${\mathrm{V}}_{\mathrm{base}}^2$and $1000\times {10}^6$for $S_{\mathrm{base}}$into equation (12).

$$\begin{gathered} {\mathrm{Z}}_{\mathrm{base}}=\frac{{\left(500\times {10}^3\right)}^2}{1000\times {10}^6} \\ =250\mathrm{\Omega } \end{gathered}$$

Determine the per unit reactance of the line.

$X_{\mathrm{linep}.\mathrm{u}.}=\frac{{\mathrm{X}}_{\mathrm{line}}}{{\mathrm{Z}}_{\mathrm{base}}}$ …… (13)

Here, $X_{\mathrm{line}}$is line reactance and$Z_{\mathrm{base}}$ is base impedance.

Substitute 50 for$X_{\mathrm{line}}$ and 250 for $Z_{\mathrm{base}}$into equation (13)

$$\begin{gathered} {\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.}=\frac{50}{250} \\ =0.2\mathrm{p}.\mathrm{u}. \end{gathered}$$

Solve further as,

$$\begin{gathered} {\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.12}={\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.23} \\ ={\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.24} \\ =0.2\mathrm{p}.\mathrm{u}. \end{gathered}$$

Here, fault occurs at bus 2.

Thévenin reactance at the fault will be the equivalent reactance of the circuit with respect to the fault point and ground.

Determine the per unit Thévenin reactance at the fault point.

Substitute 0.2 for $X_{\mathrm{line}12\mathrm{p}.\mathrm{u}.}$, 0.24 for role="math" localid="1655447421985" $X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}$, 0.4 for role="math" localid="1655447433574" $X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}$, 0.2 for role="math" localid="1655447447800" $X_{\mathrm{line}24\mathrm{p}.\mathrm{u}.}$, 0.13 for role="math" localid="1655447460824" $X_{12\mathrm{p}.\mathrm{u}.\mathrm{new}}$, 0.24 for role="math" localid="1655447473115" $X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}}$, 0.2 for role="math" localid="1655447487258" $X_{\mathrm{line}23}$, 0.1 for $X_{\mathrm{T}3\mathrm{p}.\mathrm{u}.\mathrm{new}}$and 0.17 for $X_{\mathrm{G}3\mathrm{p}.\mathrm{u}.\mathrm{new}}$into equation (1).

role="math" localid="1655447671921" $$\begin{gathered} X_{\mathrm{thp}.\mathrm{u}.}=\left(0.24+0.4\right)\mathrm{֏}\left(0.2\left(\left(0.2+0.13+0.24\right)\mathrm{֏}\left(0.2+0.1+0.17\right)\right)\right) \\ =0.64\mathrm{֏}\left(0.2+0.57\mathrm{֏}0.47\right) \\ =0.27\mathrm{p}.\mathrm{u} \end{gathered}$$

Determine thesub transient fault current in per unit and kA.

(b)

First determine the per unit fault voltage.

$V_{\mathrm{Fp}.\mathrm{u}.}=\frac{V_{\mathrm{F}}}{V_{\mathrm{base}4}}$ …… (14)

Here,$V_{\mathrm{F}}$ is fault voltage and $V_{\mathrm{base}4}$is base voltage for the zone 4

Substitute 525 for $V_{\mathrm{F}}$and 500 for$V_{\mathrm{base}4}$ into equation (14).

$$\begin{gathered} V_{\mathrm{Fp}.\mathrm{u}.}=\frac{525}{500} \\ =1.05\mathrm{p}.\mathrm{u}. \end{gathered}$$

For the actual current, determine the base current.

$I_{\mathrm{base}4}=\frac{S_{\mathrm{base}}}{V_{\mathrm{baseLL}}}$ …… (15)

Here, $S_{\mathrm{base}}$is given power base, $V_{\mathrm{baseLL}}$is base line to line voltage.

Substitute $1000\times {10}^6$for $S_{\mathrm{base}}$and$\sqrt{3}\left(500\times {10}^3\right)$ for $V_{\mathrm{baseLL}}$into equation (15).

$$\begin{gathered} I_{\mathrm{base}4}=\frac{1000\times {10}^6}{\sqrt{3}{\mathrm{V}}_{\mathrm{baseLN}}} \\ =\frac{1000\times {10}^6}{\sqrt{3}\left(500\times {10}^3\right)} \\ =1.16\mathrm{kA} \end{gathered}$$

Determine the per unit current at the fault

Substitute 1.05 for $V_{\mathrm{Fp}.\mathrm{u}.}$and$j0.27$ for $X_{\mathrm{thp}.\mathrm{u}.}$into equation (2).

$$\begin{gathered} I_{\mathrm{Fp}.\mathrm{u}.}=\frac{1.05}{j0.27} \\ =-j3.89\mathrm{p}.\mathrm{u}. \end{gathered}$$

Now determine the actual current.

Substitute 1.16 for $I_{\mathrm{base}4}$and $-j3.89\times {10}^3$for $I_{\mathrm{Fp}.\mathrm{u}.}$into equation (3).

$$\begin{gathered} I_{\mathrm{F}}=\left(1.16\right)\left(-j3.89\times {10}^3\right) \\ =-j4.51\mathrm{kA} \end{gathered}$$

Determine the current contributed from GI and 1-2 line is.

(c)

Determine the current contributed from the generator GI by using the current division rule between the generator circuit and the circuit of 1-2-3.

Substitute 1.16 for $I_{\mathrm{base}4}$and $-j3.89$for $I_{\mathrm{Fp}.\mathrm{u}.}$into equation (4).

$$\begin{gathered} I_{12F}=\left(1.16\right)\left(-j3.89\right)\left(\frac{X_{123}}{\left(X_{\mathrm{G}1}+X_{\mathrm{T}1}\right)+X_{123}}\right) \\ =\left(1.16\right)\left(-j3.89\right)\left(\frac{0.2+0.57֏0.47}{0.4+0.24+\left(0.2+0.57֏0.47\right)}\right) \\ =\left(1.16\right)\left(-j3.89\right)\left(0.42\right) \\ =-j1.89\mathrm{kA} \end{gathered}$$

Determine the remaining current will be contributed from the line 1-2 is.

Substitute $-j4.51\mathrm{kA}$for ${\mathrm{I}}_{\mathrm{F}}$and $j1.89\mathrm{kA}$for ${\mathrm{I}}_{\mathrm{G}1\mathrm{F}}$into equation (5).

$$\begin{gathered} I_{12}=-j4.51\mathrm{kA}+j1.89\mathrm{kA} \\ =-\mathrm{j}2.62\mathrm{kA} \end{gathered}$$