Coordination with residual voltage 200 Id^0.25 (PR 1.25)

The transformer is properly coordinated only if the arrester residual voltage at the given discharge current is low enough that $\dfrac{\text{BIL}}{V_{res}} \ge$ the required protective ratio (1.25).

Data

• BIL of transformer = 500 kV
• Required protective ratio, $PR = 1.25$
• Arrester discharge current, $I_d = 24$ kA
• Residual voltage characteristic: $V_{res} = 200\, I_d^{0.25}$ kV ($I_d$ in kA)

Step 1: Residual voltage of the arrester

$$\begin{aligned} V_{res} &= 200 \times (24)^{0.25} \\ &= 200 \times 2.2134 \\ &= 442.67\ \text{kV} \end{aligned}$$

Step 2: Protective ratio actually obtained

$$PR_{actual} = \frac{\text{BIL}}{V_{res}} = \frac{500}{442.67} = 1.13$$

Step 3: Compare with the requirement

The maximum residual voltage allowed for $PR = 1.25$ is

$$V_{res,max} = \frac{\text{BIL}}{PR} = \frac{500}{1.25} = 400\ \text{kV}$$

Since $442.67\ \text{kV} > 400\ \text{kV}$ (or $1.13 < 1.25$), the protective margin is not enough.

Step 4: Largest discharge current the arrester can handle with PR = 1.25

$$\begin{aligned} 200\, I_d^{0.25} &= 400 \\ I_d &= \left(\frac{400}{200}\right)^4 = 16\ \text{kA} \end{aligned}$$

So the arrester keeps the required margin only for discharge currents up to 16 kA; a 24 kA discharge exceeds this.

Remedies

• Use an arrester with a lower protective level (lower rating or better non-linear blocks, e.g. ZnO).
• Choose a transformer with a higher BIL: at least $1.25 \times 442.67 = 553.3$ kV, i.e. the next standard level above it, 650 kV.
• Reduce the stroke current reaching the arrester by better shielding of the line near the substation.

Answer: $V_{res} = 442.67$ kV, giving $PR = 1.13 < 1.25$, so the transformer is NOT properly coordinated (the required margin holds only up to $I_d = 16$ kA).