Question

In the system shown in Figure 7.16, a three-phase short circuit occurs at point. Assume that prefault currents are zero and that the generators are operating at rated voltage. Determine the fault current.

Short answer

The fault current is$1609.56\angle -75.3^{\circ }\mathrm{A}$

Step-by-step solution

Write the given data from the question.

In the system shown in figure 7.16.

Consider the following data:

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

Voltage rating:$11\mathrm{kV}$

Generator$\mathrm{G}2$: Power rating:$10\mathrm{MVA}$

Voltage rating:$11\mathrm{kV}$

Transformer$\mathrm{T}1$: Power rating:$30\mathrm{MVA}5\%$

Voltage rating:$1:3$

Base Voltage:$33\mathrm{kV}$

Prefault current: $0$

Determine the required formula.

Write the formula offault current.

${\mathit{I}}_{\mathbf{F}}\mathbf{}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{F}}}{{\mathbf{Z}}_{\mathbf{e}\mathbf{q}}}$ …… (1)

Here,${\mathit{V}}_{\mathbf{F}}$is fault voltage and ${\mathit{Z}}_{\mathbf{eq}}$is equivalent impedance.

Determine current.the fault

Consider a schematic diagram of the system just before the fault occurs is shown:

Figure 1

Select the base power or the base MVA as$S_{base}=30\mathrm{MVA}$and the base voltage a$V_{base}=33\mathrm{kV}$

Determine the new values of per unit reactance.

$X_{p.u.}=X_{\mathrm{actual}}\frac{S_{base}}{S_{actual}}$ …… (2)

Here, $X_{\mathrm{actual}}$is actual reactance,$S_{\mathrm{base}}$is given power and$S_{\mathrm{actual}}$is actual power.

For generator$\mathrm{G}1$,

Substitute$0.15\mathrm{for}{\mathrm{X}}_{\mathrm{actual}},30\mathrm{for}{\mathrm{S}}_{\mathrm{base}}\mathrm{and}20\mathrm{for}{\mathrm{S}}_{\mathrm{actual}}$into equation (2).

$$\begin{gathered} X_{\mathrm{G}1}=\left(0.15\right)\left(\frac{30}{20}\right) \\ =\frac{0.45}{2} \\ =0.{225}_{\mathrm{perunit}} \end{gathered}$$

For generator$\mathrm{G}2$,

Substitute$0.10\mathrm{for}{\mathrm{X}}_{\mathrm{actual}},30\mathrm{for}{\mathrm{S}}_{\mathrm{base}}\mathrm{and}10\mathrm{for}{\mathrm{S}}_{\mathrm{actual}}$into equation (2).

$$\begin{gathered} X_{\mathrm{G}2}=\left(0.10\right)\left(\frac{30}{10}\right) \\ =\frac{3}{10} \\ =0.3\mathrm{perunit} \end{gathered}$$

For the transformer,

Substitute$0.05\mathrm{for}{\mathrm{X}}_{\mathrm{actual}},30\mathrm{for}{\mathrm{S}}_{\mathrm{base}}\mathrm{and}30{\mathrm{S}}_{\mathrm{actual}}$into equation (2).

$$\begin{gathered} X_{TR}=0.05\left(\frac{30}{30}\right) \\ =0.05\mathrm{perunit} \end{gathered}$$

Determine the new value of per unit line impedance.

$Z_L=\frac{\left({\mathrm{Z}}_{\mathrm{actual}}\right)\left({\mathrm{S}}_{\mathrm{base}}\right)}{{\left({\mathrm{V}}_{\mathrm{base}}\right)}^2}$ …… (3)

Here,$Z_{actual}$is actual impedance,$S_{base}$is given power and $V_{base}$is base voltage.

Substitute$3+J5\mathrm{FOR}{Z}_{actual},30\mathrm{for}{\mathrm{S}}_{\mathrm{base}}\mathrm{and}33\mathrm{for}{\mathrm{V}}_{\mathrm{base}}$into equation (3).

$$\begin{gathered} Z_L=\frac{90+J150}{1089} \\ =0.0826+J0.1377\mathrm{perunit} \end{gathered}$$

Determine the equivalent impedance of the system in per unit.

$Z_{eq}=\left(j{X}_{G2}𝆏j{X}_{G2}\right)+j{X}_{TR}+Z_L$ …… (4)

Here,$X_{G1}$is reactance on the rated base generator$G1$, $X_{G2}$is reactance on the rated base generator$\mathrm{G}2,{\mathrm{X}}_{\mathrm{TR}}$is transformer reactance.

Substitute$j0.225\mathrm{for}j{\mathrm{X}}_{\mathrm{G}1},j0.3\mathrm{for}j{\mathrm{X}}_{\mathrm{G}2},j0.05\mathrm{for}j{X}_{TR}and0.0826+j0.1377for{Z}_L$into equation (4).

$$\begin{gathered} Z_{eq}=\left(j0.225𝆏j0.3\right)+j0.05+0.0826+j0.1377 \\ =j0.128+0.05+0.0826+j0.1377 \\ =0.0826+j0.3157+j0.1377 \\ =0.0826+j0.3157\mathrm{perunit} \\ =0.326\angle 75.3^{\circ }\mathrm{perunit} \end{gathered}$$

Determine the base current

$I_{base}=\frac{S_{base}}{\sqrt{3}{V}_{base}}$ .......(5)

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

Substitute$30\mathrm{MVA}\mathrm{for}{\mathrm{S}}_{\mathrm{base}}\mathrm{and}33\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\mathrm{base}}$into equation (5).

Determine the fault current.

Substitute$1.0\mathrm{V}\mathrm{for}{\mathrm{V}}_{\mathrm{f}}\mathrm{and}0.326\angle -75.3^{\circ }\mathrm{for}{\mathrm{V}}_{\mathrm{base}}$into equation (1).

$$\begin{gathered} {\mathrm{I}}_{\mathrm{f}}=\frac{1.0}{0.326\angle 75.3^{\circ }} \\ =3.067\angle -75.3^{\circ }\mathrm{perunit} \\ =3.067\angle ‐75.3^{\circ }\times {\mathrm{I}}_{\mathrm{base}} \\ =(3.067\angle -75.3^{\circ })\left(0.5248\mathrm{kA}\right) \end{gathered}$$

Solve further as,

${\mathrm{I}}_{\mathrm{f}}=1609.56\angle -75.3^{\circ }\mathrm{A}$