Audible noise of 750 kV line (20 kV/cm gradient)

The audible noise from each phase is calculated with the empirical formula using its own distance to the microphone. The three contributions are then added on an energy basis.

Data and reading of the question

• Flat horizontal line, $H = 20$ m, phase spacing $S = 15$ m
• Bundle of $N = 2$ sub-conductors, diameter given as 0.30 m, so $d = 30$ cm
• All phases: $E_m = 20$ kV/cm
• Microphone at ground level, 20 m horizontally beyond the outermost phase
• The "$-11.4\log N$" term in the formula is read as $-11.4\log D_i$, since the note gives units for $D$ (in m) and the distance must enter:

$$AN_i = 120\log E_m + 55\log d - 11.4\log D_i - 115.4\ \ \text{dB(A)}$$

  P1        P2        P3
  o--15 m---o--15 m---o         H=20 m
 ///////////////////////////////[mic]
                      |<- 20 m ->|

Step 1: Distances to the microphone

Horizontal distances are 20, 35 and 50 m.

$$\begin{aligned} D_3 &= \sqrt{20^2 + 20^2} = 28.28\ \text{m} \\ D_2 &= \sqrt{35^2 + 20^2} = 40.31\ \text{m} \\ D_1 &= \sqrt{50^2 + 20^2} = 53.85\ \text{m} \end{aligned}$$

Step 2: Common part

$$\begin{aligned} &120\log 20 + 55\log 30 - 115.4 \\ &= 156.12 + 81.24 - 115.4 = 121.96 \end{aligned}$$

Step 3: AN from each phase

$$\begin{aligned} AN_3 &= 121.96 - 11.4\log 28.28 = 121.96 - 16.55 = 105.42\ \text{dB(A)} \\ AN_2 &= 121.96 - 11.4\log 40.31 = 121.96 - 18.30 = 103.66\ \text{dB(A)} \\ AN_1 &= 121.96 - 11.4\log 53.85 = 121.96 - 19.74 = 102.23\ \text{dB(A)} \end{aligned}$$

Step 4: Total AN

$$\begin{aligned} AN &= 10\log\left(10^{10.542} + 10^{10.366} + 10^{10.223}\right) \\ &= 10\log\left[(3.481 + 2.325 + 1.671) \times 10^{10}\right] \\ &= 108.74\ \text{dB(A)} \end{aligned}$$

Note on the conductor size

A 30 cm sub-conductor is unrealistically large; the intended value is probably 0.030 m (3 cm). With $d = 3$ cm the term $55\log d$ becomes 26.24 instead of 81.24, so every level drops by exactly 55 dB. The phase levels become 50.42, 48.66 and 47.23 dB(A), and the total is 53.74 dB(A), a typical value for a 750 kV line.

Answer: AN ≈ 108.7 dB(A) with $d = 30$ cm as printed (≈ 53.7 dB(A) if $d = 3$ cm is intended).