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RI of 400 kV line at 0.5 MHz

Short answer

Fair-weather RI per phase: near outer 38.56 dB, centre 38.55 dB, far outer 31.58 dB. Using the CIGRE addition rule (mean of two highest +1.5 dB if within 3 dB), total line RI is 40.1 dB fair. Rainy weather adds 17 dB, giving 57.1 dB.

A 400 kV line has conductors in horizontal configuration at average height H = 14 m and phase spacing S = 11 m. The conductors of each phase are 2 × 0.0318 m diameter at B = 0.4572 spacing. Calculate the RI level of each phase at a distance of 30 m from one of the outer phase at ground level at 0.5 MHz using the CIGRE formula. Take gm = 17.3 kV/cm on the center phase and 16.2 kV/cm in the two outer phases. What are the values for rainy weather? What is the total value of the RI in fair and rainy weathers?

RI of each phase is calculated with the CIGRE formula at 0.5 MHz, then the line RI is obtained by the CIGRE addition rule. Rainy-weather values are found by adding the CIGRE rain increment.

Data

  • H=14H = 14 m, S=11S = 11 m, bundle 2×3.182 \times 3.18 cm, so d=3.18d = 3.18 cm
  • gmg_m = 17.3 kV/cm (centre) and 16.2 kV/cm (outer phases)
  • Point at ground level, 30 m horizontally from one outer phase
  • RIi=3.5gm+6d−33log⁡10(Di/20)−30RI_i = 3.5g_m + 6d - 33\log_{10}(D_i/20) - 30 dB above 1 μV/m (average fair weather, 0.5 MHz)
  P1        P2        P3
  o--11 m---o--11 m---o           H=14 m
 ////////////////////////////////[P]
                      |<- 30 m ->|

Step 1: Distances (horizontal 30, 41, 52 m)

D3=302+142=33.11 mD2=412+142=43.32 mD1=522+142=53.85 m\begin{aligned} D_3 &= \sqrt{30^2 + 14^2} = 33.11\ \text{m} \\ D_2 &= \sqrt{41^2 + 14^2} = 43.32\ \text{m} \\ D_1 &= \sqrt{52^2 + 14^2} = 53.85\ \text{m} \end{aligned}

Step 2: Fair-weather RI of each phase

With 6d=19.086d = 19.08, 3.5×16.2=56.703.5 \times 16.2 = 56.70 and 3.5×17.3=60.553.5 \times 17.3 = 60.55:

RI3=56.70+19.08−33log⁡33.1120−30=56.70+19.08−7.22−30=38.56 dBRI2=60.55+19.08−33log⁡43.3220−30=60.55+19.08−11.08−30=38.55 dBRI1=56.70+19.08−33log⁡53.8520−30=56.70+19.08−14.20−30=31.58 dB\begin{aligned} RI_3 &= 56.70 + 19.08 - 33\log\frac{33.11}{20} - 30 = 56.70 + 19.08 - 7.22 - 30 = 38.56\ \text{dB} \\ RI_2 &= 60.55 + 19.08 - 33\log\frac{43.32}{20} - 30 = 60.55 + 19.08 - 11.08 - 30 = 38.55\ \text{dB} \\ RI_1 &= 56.70 + 19.08 - 33\log\frac{53.85}{20} - 30 = 56.70 + 19.08 - 14.20 - 30 = 31.58\ \text{dB} \end{aligned}

Step 3: Rainy weather

Average heavy-rain RI is about 17 dB above average fair-weather RI.

PhaseDD (m)Fair (dB)Rain (dB)
Near outer33.1138.5655.56
Centre43.3238.5555.55
Far outer53.8531.5848.58

Step 4: Total RI of the line (CIGRE rule)

If one phase exceeds the others by ≥ 3 dB, the line RI is that value. Otherwise it is the mean of the two highest plus 1.5 dB. Here the two highest are almost equal:

RIfair=38.56+38.552+1.5=40.05 dBRIrain=40.05+17=57.05 dB\begin{aligned} RI_{fair} &= \frac{38.56 + 38.55}{2} + 1.5 = 40.05\ \text{dB} \\ RI_{rain} &= 40.05 + 17 = 57.05\ \text{dB} \end{aligned}

The near outer phase and the centre phase contribute almost equally: the centre phase has a higher gradient but is farther away.

Answer: phase RI (fair) = 38.56, 38.55, 31.58 dB; rain = 55.56, 55.55, 48.58 dB; line RI ≈ 40.1 dB (fair) and 57.1 dB (rain) above 1 μV/m at 0.5 MHz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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