Transformer BIL and arrester resistance for 500 kV line

The transformer BIL is obtained from the arrester protective level and the required margin. The arrester resistance is then found from the travelling-wave current through the arrester during the stroke.

Assumptions

• Highest system voltage = $1.05 \times 500 = 525$ kV (rms, line-to-line).
• Effectively grounded: arrester rating = 80% of highest system voltage.
• Arrester protective ratio = protective level (peak) / rated voltage (rms).
• The 5 kA stroke current travels on the line, producing a voltage wave $V_w = I Z$.

Step 1: Arrester rating and protective level

$$\begin{aligned} V_{rating} &= 0.8 \times 525 = 420\ \text{kV (rms)} \\ V_p &= 2.83 \times 420 = 1188.6\ \text{kV} \end{aligned}$$

Step 2: BIL of the transformer

$$\text{BIL} = 1.4 \times 1188.6 = 1664.0\ \text{kV}$$

The next standard BIL, 1675 kV, would be chosen.

Step 3: Arrester resistance for a 5 kA stroke

Voltage wave on the line:

$$V_w = I Z = 5 \times 360 = 1800\ \text{kV}$$

Arrester current (line terminated by the arrester at voltage $V_p$):

$$\begin{aligned} I_a &= \frac{2V_w - V_p}{Z} \\ &= \frac{2 \times 1800 - 1188.6}{360} \\ &= 6.70\ \text{kA} \end{aligned}$$

Arrester resistance:

$$R_a = \frac{V_p}{I_a} = \frac{1188.6}{6.698} = 177.4\ \Omega$$

(If 5 kA is taken directly as the arrester discharge current, $R_a = 1188.6/5 = 237.7\ \Omega$.)

Answer: BIL ≈ 1664 kV (use 1675 kV); arrester resistance ≈ 177.4 Ω (237.7 Ω if 5 kA is the arrester current itself).