Question

A three-phase short circuit occurs at the generator bus (bus 1) for the system shown in Figure 7.17. Neglecting prefault currents and assuming that the generator is operating at its rated voltage, determine the subtransient fault current using superposition.

Short answer

The sub transient fault current is$8393.5\mathrm{A}$.

Step-by-step solution

Write the given data from the question.

In the system shown in figure 7.17.

Consider the following data:

Generator$\mathrm{G}1$: Power rating:$25\mathrm{MVA}$

Voltage rating:$13.8\mathrm{kV}$

Reactance:$15\%$

MotorPower rating:$15\mathrm{MVA}$

Voltage rating:$13\mathrm{kV}$

Reactance:$15\%$

Transformer$\mathrm{T}1$: Power rating:$25\mathrm{MVA}$

Voltage rating:$13.2/69\mathrm{kV}$

Reactance:$15\%$

Transformer$\mathrm{T}2$: Power rating:$25\mathrm{MVA}$

Voltage rating:$69/13.2\mathrm{kV}$

Reactance:$11\%$

Transmission line impedance is$j65\Omega$

Determine the required formula.

Write the formula of subtransientfault current.

${\mathbf{I}}_{\mathbf{FAULT}}\mathbf{=}{\mathbf{I}}_{\mathbf{F}}^{{\mathbf{*}}^{\mathbf{-}}}\mathbf{\times }{\mathbf{I}}_{\mathbf{base}}$ …… (1)

Here,${\mathbf{I}}_{\mathbf{F}}^{{\mathbf{*}}^{\mathbf{-}}}$ is fault current and ${\mathbf{I}}_{\mathbf{base}}$is base current.

Determine the subtransient fault current.

Consider a single line diagram of the given system is shown below:

Consider line-to-line base voltage and base power for generator as:

$$\begin{gathered} {\mathrm{V}}_{\mathrm{base}}=13.8\mathrm{kV} \\ {\mathrm{S}}_{\mathrm{base}}=25\mathrm{MVA} \\ {\mathrm{V}}_{\mathrm{RATING}}=69/13.2\mathrm{kV} \end{gathered}$$

Determine base voltage at transmission line is

$V_{base}\left(trans\right)=V_{base}\times V_{RATING}$ …… (2)

Here, $V_{base}$is base voltage and $V_{RATING}$is voltage rating.

Substitute,$13.8\mathrm{for}{\mathrm{V}}_{\mathrm{base}}\mathrm{and}69/13.2\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{RATING}}$into equation (2).

$$\begin{gathered} V_{base}\left(trans\right)=\left(13.8\right)\left(\frac{69}{13.2}\right) \\ =72.136\mathrm{kV} \end{gathered}$$

Determine the per unit reactance using the below formulate.

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

$\frac{X_{\mathrm{G}1}}{X_{rated}}=\frac{S_{base}}{S_{rated}}\times {\left(\frac{V_{rated}}{V_{base}}\right)}^2$ ...... (3)

Here, $X_{\mathrm{G}1}$is reactance of generator$\mathrm{G}1,{\mathrm{X}}_{\mathrm{rated}}$is reactance of the rated generator, $S_{base}$is given power base, $S_{\mathrm{G}2\mathrm{rated}}$is rated power as base, $V_{\mathrm{rated}}$is rated voltage base and$V_{\mathrm{base}}$is voltage base.

Substitute$0.15\mathrm{for}{\mathrm{X}}_{\mathrm{rated}},25\mathrm{for}{\mathrm{S}}_{\mathrm{base}},25\mathrm{for}{\mathrm{S}}_{\mathrm{rated}},13.8\mathrm{for}{\mathrm{V}}_{\mathrm{base}}\mathrm{and}13.8\mathrm{for}{\mathrm{V}}_{\mathrm{base}}$into equation (3).

$$\begin{gathered} X_{G1}=0.15\left(\frac{25}{25}\right){\left(\frac{13.8}{13.8}\right)}^2 \\ =0.101\mathrm{p}.\mathrm{u}. \end{gathered}$$

For transformer $\mathrm{T}1$determine the reactance of transformer.

$\frac{X_{TRANS}}{X_{rated}}=\frac{S_{base}}{S_{rated}}\times {\left(\frac{V_{rated}}{V_{base}}\right)}^2$ …… (4)

Here,$X_{TRANS}$is transformer reactance of transformer $\mathrm{T}1,{\mathrm{X}}_{\mathrm{rated}}$,is reactance of the rated transformer, $S_{base}$is given power base,$S_{\mathrm{rated}}$is rated power, $V_{\mathrm{rated}}$is rated voltage for $\mathrm{T}1$and$V_{\mathrm{base}}$is base voltage for$\mathrm{T}1$.

Substitute$0.11\mathrm{for}{\mathrm{X}}_{\mathrm{rated}},25\mathrm{for}{\mathrm{S}}_{\mathrm{base}},25\mathrm{for}{\mathrm{S}}_{\mathrm{rated}},13.2\mathrm{for}{\mathrm{V}}_{\mathrm{rated}}\mathrm{and}13.8\mathrm{for}{\mathrm{V}}_{\mathrm{base}}$into equation (4).

$$\begin{gathered} X_{TRANS}=\left(0.11\right)\left(\frac{25}{25}\right){\left(\frac{13.2}{13.8}\right)}^2 \\ =0.101\mathrm{p}.\mathrm{u}. \end{gathered}$$

Determine the impedance of transmission line.

$Z_{LINE}=Z_L\times \frac{S_{base}}{V_{base}^2\left(trans\right)}$ …… (5)

Here,localid="1655400632407" $Z_L$is transmission line impedance,localid="1655400609828" $S_{base}$is given power base,localid="1655400623505" $V_{base}^2$is base voltage for transmission line.

Substitutelocalid="1655400565700" $j65\Omega \mathrm{for}{\mathrm{Z}}_{\mathrm{L}},25\mathrm{for}{\mathrm{S}}_{\mathrm{base}}\mathrm{and}72.136\mathrm{for}{\mathrm{V}}_{\mathrm{base}}^2\left(\mathrm{trans}\right)$into equation (5).

localid="1655400642947" $$\begin{gathered} Z_{LINE}=\left(j65\right)\frac{25}{{\left(72.136\right)}^2} \\ =0.132\mathrm{p}.\mathrm{u}. \end{gathered}$$

For motor determine the motor reactance.

localid="1655400648407" $\frac{X_M}{X_{rated}}=\frac{S_{base}}{S_{rated}}\times {\left(\frac{V_{rated}}{V_{base}}\right)}^2$ …… (6)

Here,localid="1655400656330" $X_M$is motor reactance of motorlocalid="1655400666055" $\mathrm{M}1,{\mathrm{X}}_{\mathrm{rated}}$is reactance of the rated motor,localid="1655400675732" $S_{base}$is given power base,localid="1655400687487" $S_{rated}$is rated power,localid="1655400697815" $V_{rated}$is rated voltage forlocalid="1655400706601" $\mathrm{M}1$andlocalid="1655400714016" $V_{base}$is base voltage for.

Substitutelocalid="1655400722592" $0.15\mathrm{for}{\mathrm{X}}_{\mathrm{rated}},25\mathrm{for}{\mathrm{S}}_{\mathrm{base}},15\mathrm{for}{\mathrm{S}}_{\mathrm{rated}},13\mathrm{for}{\mathrm{V}}_{\mathrm{rated}}\mathrm{and}13.8\mathrm{for}{\mathrm{V}}_{\mathrm{base}}$intoequation (6).

localid="1655400733096" $$\begin{gathered} X_M=\left(0.15\right)\left(\frac{25}{15}\right){\left(\frac{13}{13.8}\right)}^2 \\ =0.222\mathrm{p}.\mathrm{u}. \end{gathered}$$

Consider a system which reduces with per unit values as shown below in the reactance diagram:

The pre-fault voltage islocalid="1655400751775" $\stackrel{-}{V_F}$.

The machine pre-fault internal voltages are

Generator:localid="1655400740135" $\stackrel{—}{E_g^{*}}$

Motor:localid="1655400759758" $\stackrel{—}{E_g^{*}}$

The superposed networks for the above network are shown below:

Select the source voltagelocalid="1655400768074" $\stackrel{—}{V_F}$as equal to the voltage prior to the fault. So,

localid="1655400775417" $$\begin{gathered} \stackrel{—}{E_g^{*}}=\stackrel{—}{E_m^{*}} \\ =\stackrel{—}{V_F} \\ =1.0\mathrm{p}.\mathrm{u} \end{gathered}$$

So, the fault current component is zero.

localid="1655400784092" $\stackrel{—}{I_{F2}^{*}}=0$

Redraw the superposed networks as

The impedance between points a and b is:

$$\begin{gathered} Z=\frac{\left(j0.15\right)\left(j0.736\right)}{j0.15+j0.736} \\ =j0.1246 \end{gathered}$$

Determine the fault current.

$\stackrel{—}{I_F^{*}}=\frac{V_F}{Z}$ …… (7)

Here,$V_F$is fault voltage and$Z$is impedance.

Substitute$1.0for{V}_{F}andj0.1246forZ$into equation (7).

$$\begin{gathered} \stackrel{—}{I_F^{*}}=\frac{1.0}{j0.1246} \\ =-j8.025p.u. \end{gathered}$$

Determine the base current.

$I_{base}=\frac{S_{base}}{\sqrt{3}{V}_{base}}$ …… (8)

Here, $S_{base}$is given power base, $V_{base}$is base voltage.

Substitute$25\times {10}^{6}for{S}_{base}and\sqrt{3}\left(13.8\times {10}^3\right)for{V}_{baseLL}$into equation (8).

$$\begin{gathered} I_{base}=\frac{25\times {10}^6}{\sqrt{3}{V}_{base}} \\ =\frac{25\times {10}^6}{\sqrt{3}\left(13.8\times {10}^3\right)} \\ =1045.92\mathrm{A} \end{gathered}$$

Determine the fault current in amperes.

Substitute$-j8.025\mathrm{for}\stackrel{—}{{\mathrm{I}}_{\mathrm{F}}^{*}}\mathrm{and}1045.92\mathrm{for}{I}_{base}$into equation (1).

$$\begin{gathered} I_{FAULT}=\left(-j8.025\right)\left(1045.92\right)\mathrm{A} \\ =-j8393.5\mathrm{A} \end{gathered}$$

Therefore, the value of the fault current is$8393.5\mathrm{A}$.