Question

For the power system given in Problem 7.14, a three-phase short circuit occurs at bus 4, where the prefault voltage is$\mathbf{138}\mathbf{}\mathbf{kV}$. Prefault load current is neglected. Determine (a) the Thévenin equivalent at the fault, (b) the subtransient fault current in per unit and in$\mathbf{kA}\mathbf{}\mathbf{rms}$, and (c) contributions to the fault from transformer $\mathbf{T}\mathbf{2}$and from line 3–4.

Short answer

(a)TheThévenin equivalent at the fault is$0.2398\mathrm{p}.\mathrm{u}.$

(b)The sub transient fault current in per unit and $\mathrm{kA}$are $-j4.1701\mathrm{p}.\mathrm{u}.$and.

(c)The current contributed to the fault from transformer $\mathrm{T}2$and$3-4$line is $-j1.40\mathrm{A}$and$-j0.35\mathrm{A}$.

Step-by-step solution

Write the given data from the question.

Reference problem 7.14.

Consider the following data:

Generator$\mathrm{G}1$: Power rating:$500\mathrm{MVA}$

Voltage rating:$13.8\mathrm{kV}$

Reactance:$0.2\mathrm{p}.\mathrm{u}.$

Generator$\mathrm{G}2$: Power rating:$750\mathrm{MVA}$

Voltage rating:$18\mathrm{kV}$$18\mathrm{kV}$

Reactance:$018\mathrm{p}.\mathrm{u}.$

Generator$\mathrm{G}3$: Power rating:$1000\mathrm{MVA}$

Voltage rating:$20\mathrm{kV}$

Reactance:$0.17\mathrm{p}.\mathrm{u}.$

Transformer$\mathrm{T}1$: Power rating:$500\mathrm{MVA}$

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

Reactance:$0.12\mathrm{p}.\mathrm{u}.$

Transformer$\mathrm{T}2$: Power rating:$750\mathrm{MVA}$

Voltage rating:$18∆/500\mathrm{YkV}$

Reactance:$0.10\mathrm{p}.\mathrm{u}.$

Transformer$\mathrm{T}3$: Power rating:$1000\mathrm{MVA}$

Voltage rating:$20∆/500\mathrm{YkV}$

Reactance:$0.10\mathrm{p}.\mathrm{u}.$

Prefault Voltage:$138\mathrm{kV}$

Base: Power rating:$1000\mathrm{MVA}$

For$\mathrm{G}3$Voltage rating:$20\mathrm{kV}$

Determine the required formulas.

Write the formula of Thévenin equivalent at the fault.

${\mathbf{X}}_{\mathbf{thp}\mathbf{.}\mathbf{u}\mathbf{.}}\mathbf{=}\left(X_{G2p.u.\mathrm{new}}+X_{T2p.u.\mathrm{new}}\right)\mathbf{𝆏}\left(X_{T1p.u.\mathrm{new}}+X_{G1p.u.\mathrm{new}}+X_{\mathrm{line}23p.p.}+X_{\mathrm{line}34}\right)$

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 $\mathbf{T}\mathbf{2}$according to new base, ${\mathbf{X}}_{\mathbf{G}\mathbf{1}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}\mathbf{new}}$is per unit reactance of transformer $\mathbf{G}\mathbf{1}$according to new base, ${\mathbf{X}}_{\mathbf{line}\mathbf{23}\mathbf{p}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unit reactance of generator according to new base, is per unit line reactance and${\mathbf{X}}_{\mathbf{line}\mathbf{34}\mathbf{}}$is 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 Thevenin reactance.

Write the formula of sub transient fault current in $\mathrm{kA}$.

${\mathbf{i}}_{\mathbf{F}}\mathbf{=}{\mathbf{I}}_{\mathbf{base}\mathbf{2}\mathbf{}}\mathbf{\times }{\mathbf{I}}_{\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$ …… (3)

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

Write the formula of fault current contributed from transformer.

${\mathbf{I}}_{\mathbf{T}\mathbf{2}\mathbf{F}}\mathbf{}\mathbf{=}{\mathbf{I}}_{\mathbf{base}\mathbf{3}}\mathbf{}\mathbf{\times }{\mathbf{I}}_{\mathbf{T}\mathbf{2}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$ …… (4)

Here,${\mathbf{I}}_{\mathbf{base}\mathbf{3}}$is base current and ${\mathbf{I}}_{\mathbf{T}\mathbf{2}\mathbf{Fp}\mathbf{.}\mathbf{u}\mathbf{.}}$is per unit fault current.

Write the formula ofcurrent contributed from line.

${\mathbf{I}}_{\mathbf{line}\mathbf{34}}\mathbf{=}{\mathbf{I}}_{\mathbf{F}}\mathbf{-}{\mathbf{I}}_{\mathbf{T}\mathbf{2}\mathbf{F}}$ …… (5)

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

Determine the Thévenin reactance per unit fault.

(a)

First determine the per unit reactance of generator$\mathrm{G}2$.

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

Here,${\mathrm{X}}_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of generator$\mathrm{G}2,{\mathrm{X}}_{\mathrm{G}3\mathrm{p}.\mathrm{u}.}$, is per unit reactance of the rated generator,${\mathrm{S}}_{\mathrm{base}}$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 $V_{\mathrm{baseG}2}$is new voltage base for $\mathrm{G}2$.

Substitute $0.2\mathrm{fo}{\mathrm{X}}_{\mathrm{G}2\mathrm{p}.\mathrm{u}.},100\mathrm{for}{\mathrm{S}}_{\mathrm{base}},100\mathrm{for}{\mathrm{S}}_{\mathrm{G}2\mathrm{rated}},15\mathrm{for}{\mathrm{V}}_{\mathrm{G}2\mathrm{rated}}\mathrm{and}15\mathrm{for}{\mathrm{V}}_{\mathrm{baseG}2}$into equation (6).

$$\begin{gathered} {\mathrm{X}}_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}}=0.2\left(\frac{100}{100}\right){\left(\frac{15}{15}\right)}^2 \\ =0.2\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the per unit reactance of generator$\mathrm{G}1$.

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

Here, $X_{G1p.u.new}$is per unit reactance of generator$\mathrm{G}1,{\mathrm{X}}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.}$is per unit reactance of the rated generator, ${\mathrm{S}}_{\mathrm{base}}$is given power base, ${\mathrm{S}}_{\mathrm{G}2\mathrm{rated}}$is rated power as base, $V_{G1rated}$is rated voltage base and $V_{\mathrm{baseG}1}$is new voltage base for $\mathrm{G}1$.

Substitute$0.2\mathrm{for}{\mathrm{X}}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.},100\mathrm{for}{\mathrm{S}}_{\mathrm{base}},50\mathrm{for}{\mathrm{S}}_{\mathrm{G}1\mathrm{rated}},12\mathrm{for}{\mathrm{V}}_{\mathrm{G}1\mathrm{rated}}\mathrm{and}10\mathrm{for}{\mathrm{V}}_{\mathrm{baseG}1}$into equation (7).

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

For transformer $\mathrm{T}1$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$\mathrm{T}1$according to new base,${\mathrm{X}}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.}$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 $\mathrm{T}1$.

Substitute$0.10\mathrm{for}{\mathrm{X}}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.},100\mathrm{for}{\mathrm{S}}_{\mathrm{base}},50\mathrm{for}{\mathrm{S}}_{\mathrm{T}1\mathrm{rated}},10\mathrm{for}{\mathrm{V}}_{\mathrm{T}1\mathrm{rated}}$into equation (8).

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

For transformer $\mathrm{T}2$determine the new per unit reactance of transformer according to new base.

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

Here,$X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}$is per unit reactance of transformer $\mathrm{T}2$according to new base,$X_{\mathrm{T}2\mathrm{p}.\mathrm{u}.}$is per unit reactance on the rated base,$S_{\mathrm{base}}$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 $\mathrm{T}2$.

Substitute$0.10\mathrm{for}{\mathrm{X}}_{\mathrm{T}2\mathrm{p}.\mathrm{u}.},100\mathrm{for}{\mathrm{S}}_{\mathrm{base}},100\mathrm{for}\mathrm{S},15\mathrm{for}{\mathrm{V}}_{\mathrm{T}2\mathrm{rated}}\mathrm{and}15\mathrm{for}{\mathrm{V}}_{\mathrm{baseT}2}$into equation (9).

$$\begin{gathered} {\mathrm{X}}_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}}=\left(0.10\right)\left(\frac{100}{100}\right){\left(\frac{15}{15}\right)}^2 \\ =0.10\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the base impedance for the transmission lines.

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

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

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

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

Determine the per unit reactance of the line.

${\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.}=\frac{{\mathrm{X}}_{\mathrm{line}}}{{\mathrm{Z}}_{\mathrm{base}}}$ …… (11)

Substitute$40\mathrm{for}{\mathrm{X}}_{\mathrm{line}}\mathrm{and}190.44\mathrm{for}{\mathrm{Z}}_{\mathrm{base}}$into equation (11)

$$\begin{gathered} {\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.}=\frac{40}{190.44} \\ =0.21\mathrm{p}.\mathrm{u}. \end{gathered}$$

Hence,

$$\begin{gathered} {}_{\mathrm{linep}.\mathrm{u}.}23={\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.34} \\ ={\mathrm{X}}_{\mathrm{linep}.\mathrm{u}.34} \\ =0.21\mathrm{p}.\mathrm{u}. \end{gathered}$$

Here, fault occurs at bus 4.

Thevenin 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$$\begin{gathered} 0.2\mathrm{for}{\mathrm{X}}_{\mathrm{G}2\mathrm{p}.\mathrm{u}.\mathrm{new}},0.1\mathrm{for}{\mathrm{X}}_{\mathrm{T}2\mathrm{p}.\mathrm{u}.\mathrm{new}},0.2\mathrm{for}{\mathrm{X}}_{\mathrm{T}1\mathrm{p}.\mathrm{u}.\mathrm{new}},0.576\mathrm{for}{\mathrm{X}}_{\mathrm{G}1\mathrm{p}.\mathrm{u}.\mathrm{new}},0.21 \\ \mathrm{for}{\mathrm{X}}_{\mathrm{line}23\mathrm{p}.\mathrm{u}.},0.21\mathrm{for}{\mathrm{X}}_{\mathrm{line}34} \end{gathered}$$into equation (1).

$$\begin{gathered} {\mathrm{X}}_{\mathrm{thp}.\mathrm{u}.}=\left(0.2+0.1\right)𝆏\left(0.2+0.576+0.21+0.21\right) \\ =0.3𝆏\left(1.196\right) \\ =\frac{\left(0.3\right)\left(1.196\right)}{0.3+1.196} \\ =0.2398\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the sub transient fault current in per unit and kA.

(b)

First determine the per unit fault voltage.

${\text{V}}_{Fp.u.}=\frac{V_F}{V_{base3}}$ …… (12)

Here, ${\mathrm{V}}_{\mathrm{F}}$is fault voltage and$V_{base3}$is base voltage for the zone 4

Substitute$138\mathrm{for}{\mathrm{V}}_{\mathrm{F}}\mathrm{and}138\mathrm{for}{\mathrm{V}}_{\mathrm{base}3}$into equation (12).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{Fp}.\mathrm{u}.}=\frac{138}{138} \\ =1\mathrm{p}.\mathrm{u}. \end{gathered}$$

For the actual current, determine the base current.

${\mathrm{I}}_{\mathrm{base}4}=\frac{{\mathrm{S}}_{\mathrm{base}}}{{\mathrm{V}}_{\mathrm{baseLL}}}$ …… (13)

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

Substitute$100\times {10}^6\mathrm{for}{\mathrm{S}}_{\mathrm{base}}\mathrm{and}\sqrt{3}\left(138\times {10}^3\right)\mathrm{for}{\mathrm{V}}_{\mathrm{baseLL}}$into equation (13).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{base}}=\frac{1000\times {10}^6}{{\sqrt{3\mathrm{V}}}_{\mathrm{baseLN}}} \\ =\frac{100\times {10}^6}{\sqrt{3}\left(138\times {10}^3\right)} \\ =0.4184\mathrm{kA} \end{gathered}$$

Determine the per unit current at the fault

Substitute$1\mathrm{for}{\mathrm{V}}_{\mathrm{Fp}.\mathrm{u}.}\mathrm{and}j0.2398\mathrm{for}{\mathrm{X}}_{\mathrm{thp}.\mathrm{u}.}$into equation (2).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{Fp}.\mathrm{u}.}=\frac{1}{\mathrm{j}0.2398} \\ =j4.1701\mathrm{p}.\mathrm{u}. \end{gathered}$$

Now determine the actual current.

Substitute$0.4184\mathrm{for}{I}_{\mathrm{base}3}\mathrm{and}-j401701\mathrm{for}{I}_{Fp.u.}$into equation (3).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{F}}=\left(0.4184\right)\left(-\mathrm{j}4.1701\right) \\ =-\mathrm{j}1.7448\mathrm{kA} \end{gathered}$$

Determine the current contributed to the fault from transformer T2and 3-4line.

(c)

Determine the current contributed from the transformer$\mathrm{T}2$by using the current division rule between the generator circuit and the circuit of line$3-4.$

Substitute$1.16\mathrm{for}{I}_{\mathrm{base}3}and-j3.89\mathrm{for}{I}_{\mathrm{Fp}.\mathrm{u}.}$into equation (4).

$$\begin{gathered} I_{\mathrm{T}2\mathrm{F}}=\left(0.4184\right)\left(-j4.1701\right)\left(\frac{{\mathrm{X}}_{\mathrm{line}34}+{\mathrm{X}}_{\mathrm{line}23}+{\mathrm{X}}_{\mathrm{T}1}+{\mathrm{X}}_{\mathrm{G}1}}{{\mathrm{X}}_{\mathrm{G}1}+{\mathrm{X}}_{\mathrm{line}34}+{\mathrm{X}}_{\mathrm{line}23}+{\mathrm{X}}_{\mathrm{T}1}+{\mathrm{X}}_{\mathrm{G}2}+{\mathrm{X}}_{\mathrm{T}2}}\right) \\ =\left(0.4184\right)\left(-j4.1701\right)\left(\frac{1.196}{1.296+0.3}\right) \\ =\left(0.4184\right)\left(-j4.1701\right)\left(0.7995\right) \\ =-j1.40A \end{gathered}$$

Determine the remaining current will be contributed from the line$3-4$is.

Substitute$-j1.7448\mathrm{A}\mathrm{for}{I}_F\mathrm{and}j1.40\mathrm{A}\mathrm{for}{I}_{T2F}$nto equation (5).

$$\begin{gathered} I_{line34}=-j1.7448\mathrm{A}+j1.40\mathrm{A} \\ =-j0.35\mathrm{A} \end{gathered}$$