RI levels of 400 kV line at 1 MHz

RI of each phase is calculated with the CIGRE formula and combined by the CIGRE rule. Rain and frequency corrections are then added.

Data

• $H = 14$ m, $S = 15$ m, bundle $4 \times 3.18$ cm, so $d = 3.18$ cm
• $g_m$ = 16.2 kV/cm (outer) and 17.3 kV/cm (centre)
• Point at ground level, 30 m horizontally from an outer phase (outside the line)
• $RI_i = 3.5g_m + 6d - 33\log_{10}(D_i/20) - 30$ dB (0.5 MHz, average fair weather)

(i) Fair weather at 0.5 MHz

Horizontal distances are 30, 45 and 60 m.

$$\begin{aligned} D_{near} &= \sqrt{30^2 + 14^2} = 33.11\ \text{m} \\ D_{centre} &= \sqrt{45^2 + 14^2} = 47.13\ \text{m} \\ D_{far} &= \sqrt{60^2 + 14^2} = 61.61\ \text{m} \end{aligned}$$

With $6d = 19.08$, $3.5 \times 16.2 = 56.7$ and $3.5 \times 17.3 = 60.55$:

$$\begin{aligned} RI_{near} &= 56.7 + 19.08 - 7.22 - 30 = 38.56\ \text{dB} \\ RI_{centre} &= 60.55 + 19.08 - 12.28 - 30 = 37.35\ \text{dB} \\ RI_{far} &= 56.7 + 19.08 - 16.12 - 30 = 29.66\ \text{dB} \end{aligned}$$

The two highest (38.56 and 37.35) differ by less than 3 dB, so

$$RI_{fair} = \frac{38.56 + 37.35}{2} + 1.5 = 39.45\ \text{dB}$$

Rainy weather. Average heavy-rain RI is about 17 dB above average fair weather (CIGRE):

(ii) Values at 1 MHz

CIGRE frequency correction ($f$ in MHz):

$$\begin{aligned} \Delta RI &= 5\left[1 - 2(\log_{10} 10f)^2\right] \\ f = 1:\ \Delta RI &= 5[1 - 2(1)^2] = -5\ \text{dB} \end{aligned}$$

(At 0.5 MHz the correction is only +0.11 dB, so the formula is taken as the 0.5 MHz value.)

RI falls at higher frequency because the corona pulse spectrum falls off and the line attenuates higher frequencies more.

Answer: fair weather ≈ 39.5 dB and rain ≈ 56.5 dB at 0.5 MHz; at 1 MHz ≈ 34.5 dB (fair) and 51.5 dB (rain), all above 1 μV/m.