Arrester current and resistance for 750 kV line

Note on the data: a surge impedance of 3000 Ω is not realistic for an overhead line (typical values are 300–500 Ω). The same problem in other papers uses $Z = 300\ \Omega$, so it is solved here with $Z = 300\ \Omega$. Results for 3000 Ω are given at the end. The reflection and refraction coefficients are the same in both cases.

Data: $V_w = 3000$ kV, $V_p = 1700$ kV (arrester protective level), $Z = 300\ \Omega$.

When the wave reaches the arrester (line end), the Thevenin equivalent is $2V_w$ behind $Z$, and the arrester holds the voltage at $V_p$.

(i) Current in the line before reaching the arrester

$$I_w = \frac{V_w}{Z} = \frac{3000}{300} = 10\ \text{kA}$$

(ii) Current through the arrester

$$\begin{aligned} I_a &= \frac{2V_w - V_p}{Z} = \frac{2 \times 3000 - 1700}{300} \\ &= \frac{4300}{300} = 14.33\ \text{kA} \end{aligned}$$

(iii) Arrester resistance

$$R = \frac{V_p}{I_a} = \frac{1700}{14.33} = 118.6\ \Omega$$

(iv) Reflection and refraction coefficients

Treat the arrester as a termination $R = 118.6\ \Omega$:

$$\begin{aligned} \text{Voltage reflection: } \Gamma_v &= \frac{R - Z}{R + Z} = \frac{118.6 - 300}{418.6} = -0.4333 \\ \text{Voltage refraction: } T_v &= \frac{2R}{R + Z} = \frac{V_p}{V_w} = 0.5667 \\ \text{Current reflection: } \Gamma_i &= -\Gamma_v = +0.4333 \\ \text{Current refraction: } T_i &= \frac{2Z}{R + Z} = 1.4333 \end{aligned}$$

Check: reflected voltage $= -0.4333 \times 3000 = -1300$ kV, and $3000 - 1300 = 1700$ kV ✓. Reflected current $= 0.4333 \times 10 = 4.33$ kA, and $10 + 4.33 = 14.33$ kA ✓.

Discussion

• The arrester clamps the voltage at 1700 kV instead of letting it double to 6000 kV at the open end. A negative voltage reflection of −1300 kV travels back along the line.
• The arrester current (14.3 kA) is larger than the line current (10 kA), because current reflection at a low-resistance termination is positive.
• The arrester must be rated to carry this current and to absorb the energy.