Question

If the source impedance at a $\mathbf{13}\mathbf{.}\mathbf{2}\mathbf{kV}$distribution substation bus is $\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{+}\mathbf{j}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{)}\mathbf{\Omega }$per phase, compute the rms and maximum peak instantaneous value of the fault current for a balanced three-phase fault. For the system $\left(\frac{X}{R}\right)$ratio of $\mathbf{3}\mathbf{.}\mathbf{0}$, the asymmetrical factor is $\mathbf{1}\mathbf{.}\mathbf{9495}$and the time of peak is $\mathbf{7}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{ms}$(see Problem 7.4). Comment on the withstanding peak current capability to which all substation electrical equipment need to be designed.

Short answer

The value of maximum instantaneous current is $9403.56A$.The equipment installed in the sub-station should be able to withstand it.

Step-by-step solution

Write the given data from the question.

$$\begin{gathered} \mathrm{The}\mathrm{line}\mathrm{voltage},{\mathrm{V}}_{\mathrm{LL}}=13.2\mathrm{KV} \\ \mathrm{The}\mathrm{impedance},\mathrm{Z}=0.5+\mathrm{j}1.5\mathrm{\Omega } \\ \mathrm{The}\mathrm{peak}\mathrm{time},{\mathrm{t}}_{\mathrm{P}}=7.1\mathrm{ms} \\ \mathrm{The}\mathrm{reactance}\mathrm{to}\mathrm{resistance}\mathrm{ratio},\frac{\mathrm{X}}{\mathrm{R}}=3 \\ \mathrm{Asymmetrical}\mathrm{current}\mathrm{factor}{\mathrm{l}}_{\mathrm{a}\mathrm{sy},\mathrm{f}}=1.9495 \end{gathered}$$

Determine the formulas to calculate the maximum peak instantaneous current.

The expression to calculate the phase current is given as follows.

${\mathrm{V}}_{\mathrm{LN}}=\frac{{\mathrm{V}}_{\mathrm{LL}}}{\sqrt{3}}$ …… (1)

The equation to calculate the RMS peak instantaneous fault current is given as follows.

${\mathrm{I}}_{\mathrm{RMS}}=\frac{{\mathrm{V}}_{\mathrm{LN}}}{\left|\mathrm{Z}\right|}$ …… (2)

The equation to calculate the peak value of maximum fault current is given as follows

$({\mathrm{I}}_{\max )\mathrm{peak}}={\mathrm{l}}_{\mathrm{RMS}}\times {\mathrm{l}}_{\mathrm{asy},\mathrm{f}}$

Calculate the formulas to calculate the maximum peak instantaneous current.

Calculate the magnitude of the impedance.

$$\begin{gathered} \left|Z\right|=\sqrt{0.5^2+1.5^2} \\ \left|Z\right|=1.58\Omega \end{gathered}$$

Calculate the phase voltage.

Substitute$13.2\mathrm{kV}$for $V_{LL}$ into equation (1).

$$\begin{gathered} {\mathrm{V}}_{\mathrm{LN}}=\frac{13.2\times {10}^3}{\sqrt{3}} \\ {\mathrm{V}}_{\mathrm{LN}}=7621.024\mathrm{V} \end{gathered}$$

Calculate the RMS peak instantaneous fault current.

Substitute $7621.024\mathrm{V}$for ${\mathrm{V}}_{\mathrm{LN}}$and $1.58\Omega$ for $\left|Z\right|$ into equation (2).

$$\begin{gathered} {\mathrm{l}}_{\mathrm{RMS}}=\frac{7621.024}{1.58} \\ {\mathrm{l}}_{\mathrm{RMS}}=4823.43\mathrm{A} \end{gathered}$$

Calculate the peak value of maximum instantaneous current.

Substitute $1.9495$for $l_{ays,f}$, $4823.43A$for $l_{RMS}$into equation (3).

$$\begin{gathered} {\left(l_{max}\right)}_{peak}=4823.43\times 1.9495 \\ {\left(l_{max}\right)}_{peak}=9403.56A \end{gathered}$$

Hence the value of maximum instantaneous current is $9403.56A$. The equipment installed in the sub-station should be able to withstand it.