Arrester coordination for 8000 kV strike

The arrester is at the end of the line, next to the transformer. When it conducts, the line's Thevenin equivalent is $2V_w$ behind $Z$, so

$$2V_w = V_d + I_dZ, \qquad V_d = kI_d^{\beta}$$

The surge impedance of the line is not given. A typical 132 kV overhead line value, $Z = 400\ \Omega$, is assumed.

Arrester characteristic $V_d = kI_d^{\beta}$

From the two table points:

$$\begin{aligned} \frac{360}{300} &= \left(\frac{20}{10}\right)^{\beta} \;\Rightarrow\; \beta = \frac{\ln 1.2}{\ln 2} = 0.2630 \\ k &= \frac{300}{10^{0.2630}} = 163.71 \end{aligned}$$

So $V_d = 163.71\,I_d^{0.263}$ (kV, kA).

Discharge current and residual voltage

$$2 \times 8000 = 163.71\,I_d^{0.263} + 400\,I_d$$

Solving by trial (or iteration):

So $I_d = 38.93$ kA and $V_d = 428.9$ kV.

Protective ratio

$$PR = \frac{\text{BIL}}{V_d} = \frac{500}{428.9} = 1.166$$

Required $PR = 1.2$, which means $V_d \le 500/1.2 = 416.7$ kV. Since $428.9 > 416.7$ kV, i.e. $1.166 < 1.2$:

The transformer is NOT properly coordinated with this arrester for an 8000 kV incident surge.

Notes

• For $V_d = 416.7$ kV the arrester current must be at most $I_d = (416.7/163.71)^{1/0.263} = 34.87$ kA. That needs $Z \ge (16000 - 416.7)/34.87 = 447\ \Omega$. So for any usual line surge impedance (about 300–400 Ω, i.e. below 447 Ω) the result is the same: not coordinated. With $Z = 300\ \Omega$, $I_d = 51.8$ kA and $PR = 1.08$.
• Remedy: choose a transformer BIL of at least $1.2 \times 428.9 = 515$ kV (standard 550 kV), or use an arrester with a lower residual voltage.