Total audible noise of 735 kV line

The AN of each phase is calculated with the empirical relation using the distance from that phase to the measuring point on the ground. The levels are then added on an energy (power) basis.

Data

• $N = 4$, $d = 3.05$ cm, bundle spacing 45.72 cm (not needed in the formula)
• $H = 20$ m, $S = 14$ m (horizontal)
• $E_m$ = 20 kV/cm (centre) and 18.4 kV/cm (outer phases)
• Measuring point on the ground, 30 m from the centre phase

$$AN_i = 120\log E_i + 55\log d - 11.4\log D_i + 26.4\log N - 128.4$$

Step 1: Distances

The outer phases are $30 - 14 = 16$ m and $30 + 14 = 44$ 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: AN of each phase

Constant part: $55\log 3.05 + 26.4\log 4 - 128.4 = 26.64 + 15.89 - 128.4 = -85.87$.

$$\begin{aligned} AN_{centre} &= 120\log 20 - 11.4\log 36.06 - 85.87 \\ &= 156.12 - 17.75 - 85.87 = 52.50\ \text{dB(A)} \\ AN_{near} &= 120\log 18.4 - 11.4\log 25.61 - 85.87 \\ &= 151.78 - 16.06 - 85.87 = 49.85\ \text{dB(A)} \\ AN_{far} &= 151.78 - 19.20 - 85.87 = 46.71\ \text{dB(A)} \end{aligned}$$

Step 3: Total AN

$$\begin{aligned} AN_{total} &= 10\log\sum 10^{AN_i/10} \\ &= 10\log\left(178033 + 96672 + 46871\right) \\ &= 10\log(321576) = 55.07\ \text{dB(A)} \end{aligned}$$

Sound levels in dB cannot be added arithmetically. Two equal sources add only 3 dB, so the total (55.1) is only about 2.6 dB above the loudest phase.

Answer: total AN ≈ 55.1 dB(A).