Coordination with residual voltage 200 Id^0.25 (PR 1.2)

The transformer is properly coordinated if the arrester residual voltage at the given discharge current is low enough to give the required protective ratio:

$$PR = \frac{\text{BIL}}{V_d} \ge 1.2$$

Residual voltage at 24 kA

$$\begin{aligned} V_d &= 200\,I_d^{0.25} = 200 \times 24^{0.25} \\ 24^{0.25} &= \sqrt{\sqrt{24}} = \sqrt{4.899} = 2.2134 \\ V_d &= 200 \times 2.2134 = 442.7\ \text{kV} \end{aligned}$$

Protective ratio

$$PR = \frac{500}{442.7} = 1.129$$

Check

• Maximum allowed residual voltage $= 500/1.2 = 416.7$ kV.
• Actual $V_d = 442.7$ kV > 416.7 kV.
• Achieved $PR = 1.13 < 1.2$.

The transformer is NOT properly coordinated.

The largest discharge current for which coordination holds is

$$I_d = \left(\frac{416.7}{200}\right)^4 = 18.8\ \text{kA}$$

Remedies

• Raise the BIL to at least $1.2 \times 442.7 = 531$ kV (standard 550 kV).
• Use an arrester with a lower residual voltage.
• Reduce the stroke current reaching the arrester with better shielding and footing resistance.

Answer: $V_d$ = 442.7 kV and PR = 1.13 < 1.2, so the transformer is not properly coordinated.