Cable to limit transformer surge voltage

Instead of an arrester, a short length of cable of low surge impedance $Z_c$ is inserted between the overhead line ($Z_0 = 400\ \Omega$) and the transformer ($Z_t = 1000\ \Omega$). At the line–cable junction most of the surge is reflected back, so only a small voltage enters the cable; it then doubles (nearly) at the transformer. $Z_c$ is chosen so that the transformer voltage is 800 kV.

 Overhead line     Cable        Transformer
 Z0 = 400 ohm ---- Zc ---------- Zt = 1000 ohm
   1500 kV ->   J1           J2

Assumption: only the first transmitted wave is considered (the cable is long enough that successive reflections inside it arrive after the surge crest).

Voltages

At junction J1 (line to cable):

$$V_1 = \frac{2Z_c}{Z_0 + Z_c}\times 1500$$

At junction J2 (cable to transformer):

$$V_2 = \frac{2Z_t}{Z_c + Z_t}V_1 = 1500 \times \frac{2Z_c}{400 + Z_c}\times\frac{2000}{Z_c + 1000}$$

Condition $V_2 = 800$ kV

$$\begin{aligned} 800(400 + Z_c)(1000 + Z_c) &= 1500 \times 4000\,Z_c \\ (400 + Z_c)(1000 + Z_c) &= 7500\,Z_c \\ Z_c^2 + 1400Z_c + 4\times10^5 &= 7500Z_c \\ Z_c^2 - 6100Z_c + 4\times10^5 &= 0 \end{aligned}$$ $$Z_c = \frac{6100 \pm \sqrt{6100^2 - 1.6\times10^6}}{2} = \frac{6100 \pm 5967.4}{2}$$ $$Z_c = 66.3\ \Omega \quad \text{or} \quad 6034\ \Omega$$

A cable has low surge impedance (30–70 Ω typical), so $Z_c = 66.3\ \Omega$ is taken. Any cable with $Z_c \le 66.3\ \Omega$ keeps the transformer at or below 800 kV.

Voltage rating of the cable

$$V_1 = \frac{2 \times 66.3}{400 + 66.3}\times 1500 = 426.5\ \text{kV}$$

Check: $V_2 = \frac{2000}{66.3 + 1000}\times 426.5 = 800$ kV.

Without the cable, the transformer would see $\frac{2 \times 1000}{1400}\times1500 = 2143$ kV.

Answer: use a cable of surge impedance about 66.3 Ω with a surge (impulse) voltage rating of about 427 kV (peak); the transformer voltage is then limited to 800 kV.