Audible noise of 765 kV line (18.4 kV/cm outer)

The AN of each phase is found from the given empirical formula and the distance from that phase to the microphone. The three values are then added on an energy basis.

Data

• $N = 4$, $d = 3.05$ cm, $H = 20$ m, $S = 14$ m
• $E_m$ = 20 kV/cm (centre) and 18.4 kV/cm (outer phases)
• Microphone at ground level, 30 m from the centre phase
• Formula:

$$AN_i = 120\log E_i + 55\log d - 11.4\log D_i + 26.4\log N - 128.4\ \ \text{dB(A)}$$

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

Step 1: Distances

The outer phases are 16 m and 44 m away horizontally.

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

Step 2: Constant terms

$$\begin{aligned} 55\log 3.05 &= 26.64 \\ 26.4\log 4 &= 15.89 \\ 120\log 20 &= 156.12 \\ 120\log 18.4 &= 151.78 \end{aligned}$$

Step 3: AN of each phase

$$\begin{aligned} AN_{centre} &= 156.12 + 26.64 - 17.75 + 15.89 - 128.4 = 52.50\ \text{dB(A)} \\ AN_{near} &= 151.78 + 26.64 - 16.06 + 15.89 - 128.4 = 49.85\ \text{dB(A)} \\ AN_{far} &= 151.78 + 26.64 - 19.20 + 15.89 - 128.4 = 46.71\ \text{dB(A)} \end{aligned}$$

Step 4: Total AN (SPL)

$$\begin{aligned} AN &= 10\log\left(10^{5.250} + 10^{4.985} + 10^{4.671}\right) \\ &= 10\log\left[(1.780 + 0.967 + 0.469) \times 10^{5}\right] \\ &= 55.07\ \text{dB(A)} \end{aligned}$$

The centre phase, with the highest gradient, dominates. The total is about 2.6 dB above the loudest single phase.

Answer: total AN ≈ 55.1 dB(A) at 30 m from the centre phase.