Audible noise of 765 kV line (18 kV/cm outer)

The audible noise of each phase is found from the given empirical formula and the phase-to-microphone distance. The three levels 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 kV/cm (outer phases)
• Microphone at ground, 30 m from the centre phase
• Formula (as given):

$$AN_i = 120\log E_i + 55\log d - 10.4\log D_i + 26.4\log N - 128.4$$

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

$$\begin{aligned} 55\log 3.05 &= 26.64 \\ 26.4\log 4 &= 15.89 \\ 120\log 20 &= 156.12 \\ 120\log 18 &= 150.63 \end{aligned}$$

Step 3: AN of each phase

$$\begin{aligned} AN_{centre} &= 156.12 + 26.64 - 10.4\log 36.06 + 15.89 - 128.4 \\ &= 156.12 + 26.64 - 16.19 + 15.89 - 128.4 = 54.06\ \text{dB(A)} \\ AN_{near} &= 150.63 + 26.64 - 14.65 + 15.89 - 128.4 = 50.12\ \text{dB(A)} \\ AN_{far} &= 150.63 + 26.64 - 17.52 + 15.89 - 128.4 = 47.25\ \text{dB(A)} \end{aligned}$$

Step 4: Total AN

$$\begin{aligned} AN &= 10\log\left(10^{5.406} + 10^{5.012} + 10^{4.725}\right) \\ &= 10\log\left[(2.547 + 1.027 + 0.531) \times 10^{5}\right] \\ &= 56.13\ \text{dB(A)} \end{aligned}$$

The centre phase dominates because of its higher gradient. (The standard BPA/Begamudre formula has $-11.4\log D_i$; the printed $-10.4$ has been used as given. With $-11.4$ and the same data, the total is about 54.6 dB(A).)

Answer: total AN ≈ 56.1 dB(A) at 30 m from the centre phase.