Audible noise of 735 kV line at 30 m

The audible noise (AN) from each phase is found from the given empirical formula using the distance from that phase to the microphone. The three levels are then added on a power (energy) basis, not arithmetically.

Data

• $N = 4$, sub-conductor diameter $d = 3.05$ cm
• Height $H = 20$ m, phase spacing $S = 14$ m (horizontal)
• Gradients: centre $E = 20$ kV/cm, outer phases $E = 18.4$ kV/cm
• Microphone at ground level, 30 m horizontally from the centre phase
• Formula: $AN_i = 120\log E_i + 55\log d - 11.4\log D_i - 115.4$ dB(A)

 P1(outer)  P2(centre)  P3(outer)
    o--14 m--o--14 m--o        H=20 m
 ///////////////////////////////////[mic]
             |<------- 30 m ------->|

Step 1: Distances from each phase to the microphone

$$\begin{aligned} D_{near\,outer} &= \sqrt{16^2 + 20^2} = 25.61\ \text{m} \\ D_{centre} &= \sqrt{30^2 + 20^2} = 36.06\ \text{m} \\ D_{far\,outer} &= \sqrt{44^2 + 20^2} = 48.33\ \text{m} \end{aligned}$$

Step 2: Common terms

$$\begin{aligned} 55\log 3.05 &= 26.64 \\ 120\log 20 &= 156.12 \\ 120\log 18.4 &= 151.78 \end{aligned}$$

Step 3: AN of each phase

$$\begin{aligned} AN_{centre} &= 156.12 + 26.64 - 11.4\log 36.06 - 115.4 \\ &= 156.12 + 26.64 - 17.75 - 115.4 = 49.61\ \text{dB(A)} \\ AN_{near} &= 151.78 + 26.64 - 11.4\log 25.61 - 115.4 \\ &= 151.78 + 26.64 - 16.06 - 115.4 = 46.96\ \text{dB(A)} \\ AN_{far} &= 151.78 + 26.64 - 11.4\log 48.33 - 115.4 \\ &= 151.78 + 26.64 - 19.20 - 115.4 = 43.81\ \text{dB(A)} \end{aligned}$$

Step 4: Total AN (power addition)

$$\begin{aligned} AN_{total} &= 10\log\left(10^{4.961} + 10^{4.696} + 10^{4.381}\right) \\ &= 10\log\left(91424 + 49640 + 24068\right) \\ &= 10\log(165132) = 52.18\ \text{dB(A)} \end{aligned}$$

The centre phase contributes most because it has the highest surface gradient, even though it is not the nearest phase.

Answer: total AN ≈ 52.2 dB(A) at 30 m from the line centre.