Transformer BIL and maximum arrester distance

The transformer BIL is fixed from the arrester protective level and the transformer protective ratio. The maximum separation is then the distance at which the extra voltage built up between arrester and transformer just uses up the margin between BIL and the arrester protective level.

Assumptions

• Highest system voltage of a 400 kV system = $1.05 \times 400 = 420$ kV (line-to-line, rms).
• Effectively grounded system: coefficient of earthing 0.8, so arrester rating = 80% of highest system voltage.
• "PR = 3" of the arrester is the ratio of its protective level (peak) to its rated voltage (rms).
• The figure gives no surge steepness, so a lightning surge current of 10 kA with 1 μs front is assumed (as in the similar 2070 question). On the 300 Ω line its steepness is $S = 10 \times 300 / 1 = 3000$ kV/μs.
• Velocity $v = 300$ m/μs. The transformer is treated as an open circuit (worst case).

Step 1: Arrester rating and protective level

$$\begin{aligned} V_{rating} &= 0.8 \times 420 = 336\ \text{kV (rms)} \\ V_p &= PR \times V_{rating} = 3 \times 336 = 1008\ \text{kV} \end{aligned}$$

Step 2: BIL of the transformer

$$\text{BIL} = TPR \times V_p = 1.25 \times 1008 = 1260\ \text{kV}$$

The next standard level, 1300 kV, would be chosen in practice.

Step 3: Maximum distance

Voltage at an open-circuited transformer a distance $L$ from the arrester:

$$V_t = V_p + \frac{2SL}{v}$$

Coordination requires $V_t \le \text{BIL}$:

$$\begin{aligned} 1008 + \frac{2 \times 3000 \times L}{300} &\le 1260 \\ 20L &\le 252 \\ L_{max} &= 12.6\ \text{m} \end{aligned}$$

(With the standard BIL of 1300 kV: $L_{max} = 292/20 = 14.6$ m.)

At the 30 m shown in the figure: $V_t = 1008 + 20 \times 30 = 1608$ kV $> 1260$ kV, so the arrester is too far away.

Note on the 350 Ω transformer impedance

If the 350 Ω shown for the transformer is used literally, the coefficients at the transformer are

$$\alpha = \frac{2 \times 350}{300 + 350} = 1.077, \quad \beta = \frac{350 - 300}{650} = 0.077$$

and the transformer voltage cannot exceed about $\alpha V_p = 1085.5$ kV, which is below the BIL at any distance. Real transformer windings have surge impedances of several thousand ohms, so the open-circuit (β ≈ 1) case above is the one used for design.

Answer: BIL = 1260 kV (choose 1300 kV standard); maximum arrester–transformer distance ≈ 12.6 m for the assumed 3000 kV/μs surge (30 m is not acceptable).