Radio interference of 765 kV line by CIGRE

Radio interference (RI) from each phase is found with the CIGRE formula using its distance to the measuring point. The line RI is then obtained by the CIGRE addition rule.

Data

• $N = 4$, sub-conductor diameter $d = 3.05$ cm, $B = 45.72$ cm
• Height $H = 20$ m, phase spacing $S = 14$ m (horizontal)
• $g_m = 20$ kV/cm (centre) and 18.4 kV/cm (outer phases)
• Measuring point at ground level, 15 m horizontally outside one outer phase
• CIGRE formula (0.5 MHz, average fair weather), $D$ in m, $d$ in cm:

$$RI_i = 3.5\,g_m + 6d - 33\log_{10}\frac{D_i}{20} - 30\ \ \text{dB above 1 μV/m}$$

  P1        P2        P3
  o--14 m---o--14 m---o          H=20 m
 ///////////////////////////////[P]
                      |<- 15 m ->|

Step 1: Distances

Horizontal distances are 15, 29 and 43 m.

$$\begin{aligned} D_3 &= \sqrt{15^2 + 20^2} = 25.00\ \text{m} \\ D_2 &= \sqrt{29^2 + 20^2} = 35.23\ \text{m} \\ D_1 &= \sqrt{43^2 + 20^2} = 47.42\ \text{m} \end{aligned}$$

Step 2: RI of each phase

$6d = 6 \times 3.05 = 18.3$; $3.5 \times 20 = 70$; $3.5 \times 18.4 = 64.4$.

$$\begin{aligned} RI_3 &= 64.4 + 18.3 - 33\log\frac{25}{20} - 30 = 64.4 + 18.3 - 3.20 - 30 = 49.50\ \text{dB} \\ RI_2 &= 70 + 18.3 - 33\log\frac{35.23}{20} - 30 = 70 + 18.3 - 8.11 - 30 = 50.19\ \text{dB} \\ RI_1 &= 64.4 + 18.3 - 33\log\frac{47.42}{20} - 30 = 64.4 + 18.3 - 12.37 - 30 = 40.33\ \text{dB} \end{aligned}$$

Step 3: RI of the line (CIGRE rule)

• If the highest phase value exceeds the others by 3 dB or more, the line RI equals that highest value.
• Otherwise, line RI = (average of the two highest) + 1.5 dB.

Here the two highest values (50.19 and 49.50 dB) differ by only 0.69 dB, so

$$RI = \frac{50.19 + 49.50}{2} + 1.5 = 51.34\ \text{dB}$$

This is the average fair-weather value at 0.5 MHz. In heavy rain it would be about 17 dB higher (≈ 68.3 dB).

Answer: phase RI = 49.50, 50.19, 40.33 dB; line RI ≈ 51.3 dB above 1 μV/m (fair weather, 0.5 MHz).