Question

Equipment ratings for the five-bus power system shown in Figure 7.15 are as follows:

Generator G1:$\mathbf{50}\mathbf{MVA}\mathbf{,}\mathbf{12}\mathbf{kV}\mathbf{,}\mathbf{X}\mathbf{\text{'}\text{'}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{.}$

Generator G2:$\mathbf{100}\mathbf{MVA}\mathbf{}\mathbf{,}\mathbf{15}\mathbf{kV}\mathbf{,}\mathbf{X}\mathbf{\text{'}\text{'}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{.}$

Transformer T1:$\mathbf{50}\mathbf{MVA}\mathbf{,}\mathbf{10}\mathbf{kVY}\mathbf{/}\mathbf{138}\mathbf{kVY}\mathbf{,}\mathbf{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{.}$

Transformer T2:$\mathbf{50}\mathbf{MVA}\mathbf{,}\mathbf{10}\mathbf{kV}\mathbf{△}\mathbf{/}\mathbf{138}\mathbf{kVY}\mathbf{,}\mathbf{X}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{per}\mathbf{}\mathbf{unit}\mathbf{.}$

Each $\mathbf{138}\mathbf{-}\mathbf{kV}$line: localid="1655450324555" ${\mathbf{X}}_{\mathbf{1}}\mathbf{=}\mathbf{40}\mathbf{\Omega }\mathbf{.}$

A three-phase short circuit occurs at bus 5, where the prefault voltage is 15 kV . Prefault load current is neglected. (a) Draw the positive-sequence reactance diagram in per unit on a $\mathbf{100}\mathbf{‐}\mathbf{MVA}\mathbf{,}\mathbf{15}\mathbf{‐}\mathbf{kV}$, base in the zone of generatorG2. Determine (b) the Thévenin equivalent at the fault, (c) the subtransient fault current in per unit and inkA rms , and (d) contributions to the fault from generator G2 and from transformer T2 .

Short answer

(a)The positive-sequence reactance diagram in per unit on a 100-MVA is,

(b)The Thevenin equivalent at the fault is $0.1733\mathrm{p}.\mathrm{u}.$

(c)The sub transient fault current in per unit andkA rms is $-j5.772\mathrm{p}.\mathrm{u}.$and $-j22.22\mathrm{kA}$respectively.

(d)The current contributions to the fault from generator G2 and from transformer T2 is $-j19.24\mathrm{kA}\mathrm{and}-j2.98\mathrm{kA}$and respectively.

Step-by-step solution

Write the given data from the question.

Consider the following data:

GeneratorG1: Power rating: 50 MVA

Voltage rating: 12 kV

Reactance: 0.2 p.u.

GeneratorG2: Power rating: 100 MVA

Voltage rating: 15 kV

Reactance: 0.2 p.u.

TransformerT1: Power rating: 50 MVA

Voltage rating: 10 kV Y/138 kV Y

Reactance: 0.10 p.u.

TransformerT2: Power rating: 100 MVA

Voltage rating: $15\mathrm{kV}∆/138\mathrm{kV}\mathrm{Y}$

Reactance:0.10 p.u.

Each 138-kV line: Reactance$40\mathrm{\Omega }$

Prefault Voltage: 15 kV

Base: Power rating: 100 MVA

For G2 Voltage rating: 15 kV

Determine the required formula.

Write the formula of per unit Thévenin equivalent at the fault.

${\mathbf{X}}_{\mathbf{thp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{=}\left(X_{G2p.u.\mathrm{new}}\right)\mathbf{}\mathbf{֏}\mathbf{}\left(X_{T1p.u.\mathrm{new}}+X_{T1p.u.\mathrm{new}}+X_{G1p.u.}+X_{\mathrm{line}23p.u.}+X_{\mathrm{line}34}\right)$ …… (1)

Here, ${\mathbf{X}}_{\mathbf{G}\mathbf{2}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$is per unit reactance of generator $\mathbf{G}\mathbf{2}$according to new base, ${\mathbf{X}}_{\mathbf{T}\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$is per unit reactance of transformer $\mathbf{T}\mathbf{1}$according to new base,${\mathbf{X}}_{\mathbf{T}\mathbf{2}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$is per unit reactance of transformer role="math" localid="1655452247801" $\mathbf{T}\mathbf{2}$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, ${\mathbf{X}}_{\mathbf{line}\mathbf{23}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unit line reactance and role="math" localid="1655452310749" ${\mathbf{X}}_{\mathbf{line}\mathbf{23}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unit line reactance.

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 .

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

Here, ${\mathbf{I}}_{\mathbf{base}\mathbf{2}}$is base current and ${\mathbf{I}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unit fault current.

Write the formula of current contributions to the fault from generator.

${\mathbf{I}}_{\mathbf{G}\mathbf{2}\mathbf{F}}\mathbf{=}{\mathbf{I}}_{\mathbf{base}\mathbf{2}}{\mathbf{I}}_{\mathbf{G}\mathbf{2}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$…… (4)

Here, ${\mathbf{I}}_{\mathbf{base}\mathbf{2}}$is base current and ${\mathbf{I}}_{\mathbf{G}\mathbf{2}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unitfault current from generator.

Write the formula of current contributions to the fault from transformerT2.

${\mathbf{I}}_{\mathbf{T}\mathbf{2}}\mathbf{=}{\mathbf{I}}_{\mathbf{F}}\mathbf{‐}{\mathbf{I}}_{\mathbf{G}\mathbf{2}\mathbf{F}}$ …… (5)

Here, ${\mathbf{I}}_{\mathbf{F}}$is fault current and${\mathbf{I}}_{\mathbf{G}\mathbf{2}\mathbf{F}}$ is fault current from generator$\mathbf{G}\mathbf{2}$.

Determine the positive-sequence reactance diagram.

(a)

Refer to single-line diagram of five-bus power system in Figure 7.14 in the textbook.

Determine the per unit reactance of generatorG2.

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

Here, $X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of generator G2 , $X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.}$is per unit reactance of the rated generator, $S_{base}$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 0.2 for $X_{G2p.u.},100\times {10}^6\mathrm{for}{S}_{base},100\times {10}^6$for $S_{\mathrm{G}2\mathrm{rated}},15\times {10}^3$for $V_{\mathrm{G}2\mathrm{rated}}$and $15\times {10}^3$for $V_{\mathrm{baseG}2}$into equation (6).

localid="1655454043862" $$\begin{gathered} X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}}=0.2\left(\frac{100\times {10}^6}{100\times {10}^6}\right){\left(\frac{15\times {10}^3}{10\times {10}^3}\right)}^2 \\ =0.2\mathrm{p}.\mathrm{u} \end{gathered}$$

Determine the per unit reactance of generatorG1.

$\frac{{\mathrm{X}}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}}{{\mathrm{X}}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.}}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{S}}_{\mathrm{G}1\mathrm{rated}}}\times {\left(\frac{{\mathrm{V}}_{\mathrm{G}1\mathrm{rated}}}{{\mathrm{V}}_{\mathrm{baseG}1}}\right)}^2$...... (7)

Here, $X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of generator G2, $X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.}$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 localid="1655454606059" $X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.},100\times {10}^6$for $S_{base}$, localid="1655454703244" $50\times {10}^6$for $S_{\mathrm{G}1\mathrm{rated}},12\times {10}^3$for $V_{\mathrm{G}1\mathrm{rated}}$and $10\times {10}^3$for $V_{\mathrm{baseG}1}$into equation (7).

$$\begin{gathered} X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}=0.2\left(\frac{100\times {10}^6}{50\times {10}^6}\right){\left(\frac{12\times {10}^3}{10\times {10}^3}\right)}^2 \\ =0.576\mathrm{p}.\mathrm{u}. \end{gathered}$$

For transformer T1 determine the new per unit reactance of transformer according to new base.

$\frac{{\mathrm{X}}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}}{{\mathrm{X}}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.}}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{S}}_{\mathrm{T}1\mathrm{rated}}}\times {\left(\frac{{\mathrm{V}}_{\mathrm{T}1\mathrm{rated}}}{{\mathrm{V}}_{\mathrm{baseT}1}}\right)}^2$ …… (8)

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 0.10 for $X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.},100\times {10}^6$for localid="1655455309014" $S_{\mathrm{base}},50\times {10}^6$for localid="1655455357340" $S_{\mathrm{T}1\mathrm{rated}},10\times {10}^3$for $V_{\mathrm{T}1\mathrm{rated}}$and $10\times {10}^3\mathrm{for}{V}_{\mathrm{baseT}1}$for into equation (8).

$$\begin{gathered} X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}=\left(0.10\right)\left(\frac{100\times {10}^6}{50\times {10}^6}\right){\left(\frac{10\times {10}^3}{10\times {10}^3}\right)}^2 \\ =0.20\mathrm{p}.\mathrm{u}. \end{gathered}$$

For transformer T2 determine the new per unit reactance of transformer according to new base.

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

Here, localid="1655457753653" $X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of transformer T2 according to new base, localid="1655457762276" $X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.}$is per unit reactance on the rated base, localid="1655457770737" $S_{\mathrm{base}}$is given power base, localid="1655457778675" $S_{\mathrm{T}2\mathrm{rated}}$is rated power as base, localid="1655457785895" $V_{\mathrm{T}2\mathrm{rated}}$is rated voltage base and localid="1655457792364" $V_{\mathrm{baseT}2}$is new voltage base for T2

Substitute 0.1 for localid="1655457800601" $X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.},100\times {10}^6\mathrm{for}{S}_{\mathrm{base}},100\times {10}^6$for localid="1655457806927" $S_{\mathrm{T}2\mathrm{rated}},15\times {10}^3$for localid="1655457813422" $V_{\mathrm{T}2\mathrm{rated}}$andlocalid="1655457819606" $15\times {10}^3\mathrm{for}{V}_{\mathrm{baseT}2}$into equation (9).

localid="1655457826755" $$\begin{gathered} X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}=\left(0.1\right)\left(\frac{100\times {10}^6}{100\times {10}^6}\right){\left(\frac{15\times {10}^3}{15\times {10}^3}\right)}^2 \\ =0.1\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the base impedance for the transmission lines.

localid="1655457833837" ${\mathrm{Z}}_{\mathrm{base}}=\frac{{\mathrm{V}}_{\mathrm{base}}^2}{{\mathrm{S}}_{\mathrm{base}}}$ …… (10)

Here, localid="1655457841411" $V_{\mathrm{base}}^2$is base voltage andlocalid="1655457848657" $S_{\mathrm{base}}$is given power base.

Substitutelocalid="1655457855122" $138\times {10}^3\mathrm{for}{V}_{\mathrm{base}}^2\mathrm{and}100\times {10}^6\mathrm{for}{S}_{\mathrm{base}}$for and for into equation (10).

localid="1655457862073" $$\begin{gathered} Z_{base}=\frac{{\left(138\times {10}^3\right)}^2}{100\times {10}^6} \\ =190.44\mathrm{\Omega } \end{gathered}$$

Determine the per unit reactance of the line.

localid="1655457871812" $X_{\mathrm{linep}.\mathrm{u}.}=\frac{X_{\mathrm{line}}}{Z_{\mathrm{base}}}$ …… (11)

Here,localid="1655457878444" $X_{\mathrm{line}}$is line reactance and localid="1655457885468" $Z_{\mathrm{base}}$is base impedance.

Substitute 40 forlocalid="1655457898625" $X_{\mathrm{line}}$and 190.44 forlocalid="1655457906269" $Z_{\mathrm{base}}$into equation (11)

localid="1655457913049" $$\begin{gathered} X_{\mathrm{linep}.\mathrm{u}.}=\frac{40}{190.44} \\ =0.21\mathrm{p}.\mathrm{u}. \end{gathered}$$

Hence,

localid="1655457924506" $$\begin{gathered} X_{\mathrm{linep}.\mathrm{u}.12}=X_{\mathrm{linep}.\mathrm{u}.23} \\ =X_{\mathrm{linep}.\mathrm{u}.24} \\ =0.21\mathrm{p}.\mathrm{u}. \end{gathered}$$

The per unit positive-sequence reactance diagram is shown in figure below.

Determine theThévenin equivalent at the fault.

(b)

Determine the Thevenin equivalent reactance for the fault at bus 5.

Substitute 0.2 for $X_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}}$,0.2 for $X_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}}$, 0.1 for $X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}$,0.576 for $X_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}}$, 0.21 for $X_{\mathrm{line}23\mathrm{p}.\mathrm{u}.}$and 0.21 for $X_{\mathrm{line}34}$into equation (1).

$$\begin{gathered} X_{\mathrm{thp}.\mathrm{u}.}=\left(0.2\right)\mathrm{֏}\left(0.2+0.1+0.576+0.21+0.21\right) \\ =\frac{\left(0.2\right)\left(1.296\right)}{0.2+1.296} \\ =0.1733\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine thesub transient fault current in per unit and.

(c)

The fault current can be calculated by the Thévenin equivalent parameters.

First determine the per unit fault voltage.

$V_{\mathrm{Fp}.\mathrm{u}.}=\frac{V_{\mathrm{F}}}{V_{\mathrm{base}2}}$ …… (12)

Here,$V_{\mathrm{F}}$is fault voltage and$V_{\mathrm{base}2}$is base voltage for the zone 4

Substitute 15 for $V_{\mathrm{F}}$and 15 for $V_{\mathrm{base}2}$into equation (12).

$$\begin{gathered} V_{\mathrm{Fp}.\mathrm{u}.}=\frac{15}{15} \\ =1\mathrm{p}.\mathrm{u}. \end{gathered}$$

For the actual current, determine the base current.

$I_{\mathrm{base}2}=\frac{S_{\mathrm{base}}}{V_{\mathrm{baseLL}}}$ …… (13)

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

Substitute $100\times {10}^6\mathrm{for}{S}_{\mathrm{base}}$and $\sqrt{3}\left(15\times {10}^3\right)\mathrm{for}{V}_{\mathrm{baseLL}}$into equation (13).

$$\begin{gathered} I_{\mathrm{base}4}=\frac{100\times {10}^6}{\sqrt{3}{V}_{\mathrm{baseLN}}} \\ =\frac{100\times {10}^6}{\sqrt{3}{V}_{\mathrm{baseLN}}} \\ =0.85\mathrm{kA} \end{gathered}$$

Determine the per unit current at the fault

Substitute $1\mathrm{for}{V}_{\mathrm{Fp}.\mathrm{u}.}\mathrm{and}j0.1772\mathrm{for}{X}_{\mathrm{thp}.\mathrm{u}.}$into equation (2).

$$\begin{gathered} I_{\mathrm{Fp}.\mathrm{u}.}=\frac{1}{j0.1733} \\ =-j5.772\mathrm{p}.\mathrm{u}. \end{gathered}$$

Now determine the actual current.

Substitute $3.85\mathrm{for}{I}_{\mathrm{base}2}\mathrm{and}-j5.772\mathrm{for}{I}_{\mathrm{Fp}.\mathrm{u}.}$into equation (3).

$$\begin{gathered} I_{\mathrm{F}}=\left(3.85\mathrm{kA}\right)\left(-j5.772\right) \\ =-j22.22\mathrm{kA} \end{gathered}$$

Determine the current contributions to the fault from generator and from transformer.

(d)

Determine the current contributed from the generator G2 by using the current division rule between the generator circuit and the circuit of ${\mathrm{T}}_2$.

Substitute $3.85\times {10}^3\mathrm{for}{I}_{\mathrm{base}2}\mathrm{and}-j5.772\mathrm{for}{I}_{\mathrm{Fp}.\mathrm{u}.}$for and for into equation (4).

$$\begin{gathered} I_{12\mathrm{F}}=\left(3.85\times {10}^3\right)\left(-j5.772\right)\left(\frac{X_{\mathrm{T}2}+X_{\mathrm{line}34}+X_{\mathrm{line}23}+X_{\mathrm{T}1}+X_{\mathrm{G}1}}{X_{\mathrm{T}2}+X_{\mathrm{line}34}+X_{\mathrm{line}23}+X_{\mathrm{T}1}+X_{\mathrm{G}1}+X_{\mathrm{G}2}}\right) \\ =\left(3.85\times {10}^3\right)\left(-j5.772\right)\left(\frac{1.296}{1.296+0.2}\right) \\ =\left(3.85\times {10}^3\right)\left(-j5.772\right)\left(0.866\right) \\ =-J19.24\mathrm{kA} \end{gathered}$$

Determine the remaining current will be contributed from the transformerT2.

Substitute $-j22.22\mathrm{kA}\mathrm{for}{\mathrm{I}}_{\mathrm{F}}\mathrm{and}j19.24\mathrm{kA}\mathrm{for}{\mathrm{I}}_{\mathrm{G}2\mathrm{F}}$into equation (5).

$$\begin{gathered} I_{\mathrm{T}2}=-j22.22\mathrm{kA}+j19.24\mathrm{kA} \\ =-\mathrm{j}2.98\mathrm{kA} \end{gathered}$$