Question

A 345-kV, three-phase transmission line has a 2.2-kA continuous current rating and a 2.5-kA maximum short-time overload rating with a 356-kV maximum operating voltage. The maximum symmetrical fault current on the line is 30 kA. Select a circuit breaker for this line from Table 7.10.

Short answer

The specification for the circuit breaker to be selected from the table is$40\mathrm{kA}$rated short circuit current.

Step-by-step solution

Select a circuit breaker from Table 7.10

The circuit breaker which is selected must have the maximum operating voltage greater than or equal to $35\mathrm{kV}$. Hence, select the circuit breakerof$362\mathrm{KV}$.

Also the continuous current rating should be greater than or equal to$2.2\mathrm{kA}$. So the current rating of the circuit breaker which should be selected from the table is$3\mathrm{kA}$.

Calculate the minimum voltage

The formula to calculate the minimum voltage,

${\mathbf{V}}_{\mathbf{min}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{max}}}{\mathbf{K}}$

Here,${\mathrm{V}}_{\max }$is the maximum voltage and $K$is the voltage range factor.

Substitute $362\mathrm{kV}\mathrm{for}{\mathrm{V}}_{\max }$ and 1 for K in the equation.

$$\begin{gathered} {\mathrm{V}}_{\min }=\frac{362\mathrm{kV}}{1} \\ 362\mathrm{kV} \end{gathered}$$

Thus,$362\mathrm{kV}$is selected for$V_{min}$.

Calculate the symmetrical interrupting current

The formula to calculate the symmetrical interrupting current is

$\mathrm{Symmetrical}\mathrm{interrupting}\mathrm{current}=\left(\frac{\left(\mathrm{Rated}\mathrm{short}\mathrm{circuit}\mathrm{current}\right)\left({\mathrm{V}}_{\max }\right)}{\mathrm{Operating}\mathrm{voltage}}\right)$...... (1)

Substitute 1 for $\left(\frac{V_{max}}{Operatingvoltage}\right)$in equation (1)

$$\begin{gathered} Symmetricalinterruptingcurrent=\left(Ratedshortcircuitcurrent\right)\left(1\right) \\ =\left(Ratedshortcircuitcurrent\right) \end{gathered}$$

Now, the given value of the maximum symmetrical fault current on the line is$30\mathrm{kA}$

So, the value of rated short-circuit current should be chosen in such a way that it exceeds the value of the maximum symmetrical fault that can occur on the line, only then the circuit breaker will be able to protect. So, choose the value which is just next to the value of rated short-circuit current.

Hence, the value chosen from the table is$40\mathrm{kA}$.

Calculate the maximum rated short circuit current.

The formula to calculate the maximum rated short circuit current is

$I_{\max }=\mathrm{K}\times \mathrm{Rated}\mathrm{short}\mathrm{circuit}\mathrm{current}$

Substitute 1 for $K$and $40\mathrm{kA}$for in the above equation.

$$\begin{gathered} I_{max}=1\times 40\mathrm{kA} \\ =40\mathrm{kA} \end{gathered}$$

The specifications for the circuit breaker to be selected from the table is$40\mathrm{kA}$rated short circuit current.