Question

A $\mathbf{69}\mathbf{}\mathbf{kV}$circuit breaker has a voltage range factor$\mathbf{K}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{25}$, a continuous current rating of$\mathbf{1200}\mathbf{}\mathbf{A}$, and a rated short-circuit current of$\mathbf{19000}\mathbf{}\mathbf{A}$at the maximum rated voltage of$\mathbf{72}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{kV}$. Determine the maximum symmetrical interrupting capability of the breaker. Also, explain its significance at lower operating voltages.

Short answer

The value of the maximum symmetrical interrupting capability of the breaker is$23750\mathrm{A}$.The minimum voltage is and below that the current value remains constant at as derived i.e.$23750\mathrm{A}$. If the voltage value is more than$V_{min}$and less than$V_{max}$, then the interrupting current also varies between rated short circuit current to$I_{max}$.

Step-by-step solution

Given data

Circuit breaker of with a voltage range factor of$k=1.25$continuous current rating of$1200A$, and a rated short-circuit current of$19000A$ at the maximum rated voltage of$72.5kV$.

Determine the maximum symmetrical interrupting capability of the breaker

The general formula to calculate the maximum symmetrical interrupting capability,

$$\begin{gathered} I_{max}=Voltagerangefactor\times Ratedshortcircuitcurrent \\ =KI \end{gathered}$$...... (1)

Substitute the value $1.25\mathrm{fo}rK19000A$for $Ratedshortcircuitcurrent$in the above equation,

$$\begin{gathered} {\mathrm{I}}_{\max }=\left(1.25\right)\times \left(19000\mathrm{A}\right) \\ =23750\mathrm{A} \end{gathered}$$

Thus, the value of the maximum symmetrical capability is$23750\mathrm{A}$.

Determine the significance of lower operating voltage Vmin

The general formula to calculate the value of ${\mathrm{V}}_{\min }$,

$V_{min}=\frac{V_{max}}{K}......\left(2\right)$

Here, $V_{max}$is the maximum rated voltage and K is the voltage range factor.

Substitute the value $1.25\mathrm{for}\mathrm{K}$, $72.5kV\mathrm{for}{V}_{max}$in equation (2),

localid="1655375027981" $$\begin{gathered} V_{\min }=\frac{72.5\mathrm{kV}}{1.25} \\ =58\mathrm{kV} \end{gathered}$$

As derived above the minimum voltage is$58\mathrm{kV}$and below that the current value remains constant at as derived i.e. $23750A$. If the voltage value is more than$V_{min}$and less than$V_{max}$, then the interrupting current also varies between rated short circuit current to$I_{max}$.