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RI of 330 kV line at point M

Short answer

RI per phase: 31.43 dB (far), 38.77 dB (centre), 39.29 dB (near) using CIGRE formula with distances 48.08, 37.70, 27.78 m and gradients 16.0, 17.1, 16.0 kV/cm. Line RI = (38.77+39.29)/2 + 1.5 = 40.5 dB above 1 μV/m.

With the help of transmission line and data given below, calculate the RI level of each phase at point M at 500 kHz in fair weather condition using CIGRE formula. Also calculate the RI level of the line. [Figure: three phase conductors in flat horizontal configuration spaced 11 m apart at a height of 14 m above ground; point M on the ground 46 m horizontally from the farthest outer conductor] Nominal voltage = 330 kV Bundle spacing = 45.72 cm Conductor diameter = 30 mm Number of conductors per phase = 2 Maximum surface gradient at central and outer phases at 330 kV are 1710 kV/m and 1600 kV/m respectively. The CIGRE formula is given by RIi(dB) = 3.5Em + 6d − 33 log10(Di/20) − 30, where Em in kV/cm, d in cm, Di in m.

RI from each phase is computed with the CIGRE formula using its distance to point M. The line RI is then obtained by the CIGRE rule for combining phase values.

Data

  • Flat horizontal, S=11S = 11 m, H=14H = 14 m; M on the ground, 46 m horizontally from the farthest outer conductor
  • N=2N = 2, d=30d = 30 mm =3.0= 3.0 cm, B=45.72B = 45.72 cm
  • gmg_m centre = 1710 kV/m = 17.1 kV/cm; outer = 1600 kV/m = 16.0 kV/cm
  • 500 kHz (0.5 MHz), fair weather:
RIi=3.5Em+6d−33log⁡10Di20−30  dB above 1 μV/mRI_i = 3.5E_m + 6d - 33\log_{10}\frac{D_i}{20} - 30\ \ \text{dB above 1 μV/m} 
 far      centre    near
  o--11 m---o--11 m---o           H=14 m
 /////////////////////////////////[M]
  |<------------- 46 m ----------->|

Step 1: Horizontal and direct distances to M

Horizontal distances are 46 m (far outer), 35 m (centre) and 24 m (near outer).

Dfar=462+142=48.08 mDcentre=352+142=37.70 mDnear=242+142=27.78 m\begin{aligned} D_{far} &= \sqrt{46^2 + 14^2} = 48.08\ \text{m} \\ D_{centre} &= \sqrt{35^2 + 14^2} = 37.70\ \text{m} \\ D_{near} &= \sqrt{24^2 + 14^2} = 27.78\ \text{m} \end{aligned}

Step 2: RI of each phase

6d=186d = 18; 3.5×16.0=56.03.5 \times 16.0 = 56.0; 3.5×17.1=59.853.5 \times 17.1 = 59.85.

RInear=56.0+18−33log⁡27.7820−30=56.0+18−4.71−30=39.29 dBRIcentre=59.85+18−33log⁡37.7020−30=59.85+18−9.08−30=38.77 dBRIfar=56.0+18−33log⁡48.0820−30=56.0+18−12.57−30=31.43 dB\begin{aligned} RI_{near} &= 56.0 + 18 - 33\log\frac{27.78}{20} - 30 = 56.0 + 18 - 4.71 - 30 = 39.29\ \text{dB} \\ RI_{centre} &= 59.85 + 18 - 33\log\frac{37.70}{20} - 30 = 59.85 + 18 - 9.08 - 30 = 38.77\ \text{dB} \\ RI_{far} &= 56.0 + 18 - 33\log\frac{48.08}{20} - 30 = 56.0 + 18 - 12.57 - 30 = 31.43\ \text{dB} \end{aligned}
PhaseEmE_m (kV/cm)DD (m)RI (dB)
Near outer16.027.7839.29
Centre17.137.7038.77
Far outer16.048.0831.43

Step 3: RI level of the line

CIGRE rule: if the highest phase value exceeds the others by 3 dB or more, the line RI equals it; otherwise take the average of the two highest plus 1.5 dB. Here 39.29 and 38.77 differ by only 0.52 dB, so

RIline=39.29+38.772+1.5=40.53 dBRI_{line} = \frac{39.29 + 38.77}{2} + 1.5 = 40.53\ \text{dB}

Answer: phase RI at M = 39.29 dB (near outer), 38.77 dB (centre), 31.43 dB (far outer); line RI ≈ 40.5 dB above 1 μV/m at 500 kHz in fair weather.

✓ Answer checked by Claude Opus 5.5 (high effort)

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