Arrester with 1000 kV BIL on 350 Ω line

When the travelling wave reaches the arrester at the end of the line, the arrester conducts and holds the voltage at its rated (protective) level of 1000 kV. It acts as a resistance $R_a$ terminating the line, so a large negative voltage wave is reflected.

Data

• Incident voltage, $V_i = 4200$ kV
• Arrester voltage (BIL rating / protective level), $V_a = 1000$ kV
• Surge impedance, $Z = 350\ \Omega$

Incident current

$$I_i = \frac{V_i}{Z} = \frac{4200}{350} = 12\ \text{kA}$$

(i) Reflected voltage and current

$$\begin{aligned} V_r &= V_a - V_i = 1000 - 4200 = -3200\ \text{kV} \\ I_r &= -\frac{V_r}{Z} = \frac{3200}{350} = 9.14\ \text{kA} \end{aligned}$$

(ii) Arrester discharge current and resistance

$$\begin{aligned} I_a &= I_i + I_r = \frac{2V_i - V_a}{Z} \\ &= \frac{8400 - 1000}{350} = 21.14\ \text{kA} \end{aligned}$$ $$R_a = \frac{V_a}{I_a} = \frac{1000}{21.143} = 47.30\ \Omega$$

Check

$$\beta = \frac{R_a - Z}{R_a + Z} = \frac{47.30 - 350}{397.30} = -0.762, \quad \beta V_i = -3200\ \text{kV}$$

Compared with a 300 Ω line, the higher surge impedance gives a smaller arrester current and a larger equivalent resistance.

Answer: $V_r = -3200$ kV, $I_r = 9.14$ kA, $I_a = 21.14$ kA, $R_a = 47.30\ \Omega$.