Switching resistance Rs for 1.8 pu limit

Assumptions: the line is represented by the given single-phase lumped series $R$–$L$–$C$ circuit; the breaker closes at the peak of the source voltage $E$ (worst case), the line has no trapped charge, and 1 pu = peak of source phase voltage. When the auxiliary breaker closes first, $R_s$ is in series with the line resistance.

Given: $R = 10\ \Omega$, $L = 0.5$ H, $C = 5\ \mu$F.

Formula

For energisation of a series $RLC$ circuit by a step $E$ with $V_0 = 0$:

$$V_{max} = E\left(1 + e^{-\alpha\pi/\omega}\right), \quad \alpha = \frac{R_T}{2L}, \quad \omega = \sqrt{\frac{1}{LC} - \alpha^2}$$

where $R_T = R_s + 10$. Also

$$\frac{\alpha}{\omega} = \frac{R_T}{\sqrt{4L/C - R_T^2}}$$

Overvoltage without $R_s$ (check)

$\alpha = 10/(2 \times 0.5) = 10\ \text{s}^{-1}$, $\omega = \sqrt{1/(0.5 \times 5\times10^{-6}) - 10^2} = 632.4$ rad/s

$$V_{max} = E(1 + e^{-10\pi/632.4}) = 1.952\ \text{pu}$$

which exceeds 1.8 pu, so a resistor is needed.

Required total resistance

$$\begin{aligned} 1 + e^{-\alpha\pi/\omega} &= 1.8 \\ e^{-\alpha\pi/\omega} &= 0.8 \\ \frac{\alpha}{\omega} &= \frac{-\ln 0.8}{\pi} = \frac{0.22314}{\pi} = 0.07103 \end{aligned}$$

Let $k = 0.07103$. From $\frac{R_T}{\sqrt{4L/C - R_T^2}} = k$:

$$R_T = \frac{k\sqrt{4L/C}}{\sqrt{1+k^2}}$$ $$\sqrt{\frac{4L}{C}} = \sqrt{\frac{4 \times 0.5}{5\times10^{-6}}} = \sqrt{4\times10^5} = 632.46\ \Omega$$ $$R_T = \frac{0.07103 \times 632.46}{\sqrt{1 + 0.07103^2}} = \frac{44.92}{1.00252} = 44.81\ \Omega$$

Switching resistance

$$R_s = R_T - R = 44.81 - 10 = 34.81\ \Omega$$

Check: $\alpha = 44.81/(2\times0.5) = 44.81\ \text{s}^{-1}$, $\omega = \sqrt{4\times10^5 - 44.81^2} = 630.87$ rad/s, $1 + e^{-44.81\pi/630.87} = 1.800$ pu.

Answer: $R_s \approx 34.8\ \Omega$ (total series resistance 44.8 Ω) limits the maximum switching overvoltage to 1.8 pu. After a short insertion time (about 8–10 ms) the main CB closes and shorts $R_s$.