Audible noise of 750 kV line (formula with Di)

Audible noise (AN) of each phase is found from the given empirical formula, and the noise from the three phases is added on an energy (logarithmic) basis.

Data and assumptions

• $N = 2$ sub-conductors, average height $H = 20$ m, phase spacing $S = 15$ m, $E_m = 20$ kV/cm for all three phases.
• Sub-conductor diameter is read as 3.0 cm (0.030 m). A diameter of 0.30 m (30 cm) is not a realistic conductor size and would give about 116.7 dB(A), which is unrealistic, so the "0.30" is taken as a misprint.
• The microphone is at ground level, 20 m horizontally from the nearest outer phase.

  A(near)      B(centre)     C(far)     height 20 m
    o------15 m----o-----15 m---o
    |
    |  20 m
    |
  ground ---20 m---> P

Horizontal distances from P: $20$ m, $35$ m and $50$ m. Radial distance $D_i = \sqrt{H^2 + x_i^2}$.

Common part of the formula

$$\begin{aligned} 120\log_{10} 20 &= 156.124 \\ 55\log_{10} 3.0 &= 26.242 \\ 26.4\log_{10} 2 &= 7.947 \\ AN_i &= 156.124 + 26.242 + 7.947 - 115.4 - 11.4\log_{10} D_i \\ &= 74.913 - 11.4\log_{10} D_i \end{aligned}$$

Noise from each phase

Total noise

$$\begin{aligned} AN &= 10\log_{10}\sum_{i=1}^{3} 10^{AN_i/10} \\ &= 10\log_{10}\left(10^{5.837} + 10^{5.661} + 10^{5.518}\right) \\ &= 10\log_{10}\left[(6.86 + 4.58 + 3.30)\times 10^{5}\right] \\ &= 61.68\ \text{dB(A)} \end{aligned}$$

Answer: Audible noise at the point $\approx$ 61.7 dB(A) (phase contributions 58.4, 56.6 and 55.2 dB(A)). With $d = 30$ cm taken literally, every value rises by exactly 55 dB, giving about 116.7 dB(A).

This is above the usual design limit of about 52–55 dB(A) at the right-of-way edge, so a larger bundle (more or bigger sub-conductors, which lowers $E_m$) would be needed in practice.