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Audible noise of 750 kV line using two formulas

Short answer

For N=2, formula (ii) without the 26.4logN term is appropriate. Using the given conductor diameter 0.30 m (30 cm), the total AN at 20 m from the outer phase is 108.7 dB(A). If the more realistic 3 cm diameter were used, it would be 53.7 dB(A).

A 750 kV flat horizontal configured transmission line has average ground clearance of 20 m and phase spacing of 15 m. The line uses bundle conductor of 2 × 0.30 m diameter. Determine the audible noise at the horizontal distance of 20 m from the outer most conductor. Assume empirical formula from below. (i) 120 log10(Em(i)) + 55 log10(d) − 11.4 log10(D(i)) + 26.4 log10(N) − 128.4 (ii) 120 log10(Em(i)) + 55 log10(d) − 11.4 log10(D(i)) − 115.4 Em = kV/cm, D = m, d = cm

The AN of each phase is computed from the empirical formula and the phase-to-microphone distance, then added on an energy basis. Both given formulas are evaluated. Formula (i), with the 26.4log⁡N26.4\log N term, is meant for bundles with N≥3N \ge 3. Formula (ii) is meant for N<3N < 3, so (ii) is the appropriate one here (N=2N = 2).

Data

  • H=20H = 20 m, S=15S = 15 m, N=2N = 2
  • Sub-conductor diameter 0.30 m as printed, so d=30d = 30 cm
  • Gradient EmE_m not given; assume 20 kV/cm on all phases (the value used in the companion question)
  • Microphone at ground, 20 m horizontally beyond the outer phase

Step 1: Distances (horizontal 20, 35, 50 m)

D1=202+202=28.28 mD2=352+202=40.31 mD3=502+202=53.85 m\begin{aligned} D_1 &= \sqrt{20^2 + 20^2} = 28.28\ \text{m} \\ D_2 &= \sqrt{35^2 + 20^2} = 40.31\ \text{m} \\ D_3 &= \sqrt{50^2 + 20^2} = 53.85\ \text{m} \end{aligned}

Step 2: Common terms

120log⁡20=156.1255log⁡30=81.2426.4log⁡2=7.9511.4log⁡D=16.55, 18.30, 19.74\begin{aligned} 120\log 20 &= 156.12 \\ 55\log 30 &= 81.24 \\ 26.4\log 2 &= 7.95 \\ 11.4\log D &= 16.55,\ 18.30,\ 19.74 \end{aligned}

Formula (i)

ANi=156.12+81.24−11.4log⁡Di+7.95−128.4=116.91−11.4log⁡Di\begin{aligned} AN_i &= 156.12 + 81.24 - 11.4\log D_i + 7.95 - 128.4 \\ &= 116.91 - 11.4\log D_i \end{aligned}

This gives 100.37, 98.61 and 97.18 dB(A).

AN=10log⁡(1010.037+109.861+109.718)=103.68 dB(A)\begin{aligned} AN &= 10\log\left(10^{10.037} + 10^{9.861} + 10^{9.718}\right) \\ &= 103.68\ \text{dB(A)} \end{aligned}

Formula (ii)

ANi=156.12+81.24−11.4log⁡Di−115.4=121.96−11.4log⁡DiAN_i = 156.12 + 81.24 - 11.4\log D_i - 115.4 = 121.96 - 11.4\log D_i

This gives 105.42, 103.66 and 102.23 dB(A).

AN=10log⁡(1010.542+1010.366+1010.223)=108.74 dB(A)\begin{aligned} AN &= 10\log\left(10^{10.542} + 10^{10.366} + 10^{10.223}\right) \\ &= 108.74\ \text{dB(A)} \end{aligned}
Phase distance(i) dB(A)(ii) dB(A)
28.28 m100.37105.42
40.31 m98.61103.66
53.85 m97.18102.23
Total103.68108.74

The two formulas differ by a constant 26.4log⁡2−128.4+115.4=−5.0526.4\log 2 - 128.4 + 115.4 = -5.05 dB per phase.

Note on conductor size

A 30 cm sub-conductor is unrealistic; 0.030 m (3 cm) is probably intended. With d=3d = 3 cm every value falls by exactly 55 dB, giving 48.68 dB(A) by (i) and 53.74 dB(A) by (ii). These are typical levels for a 750 kV line.

Answer: with d = 30 cm, AN = 103.7 dB(A) by (i) and 108.7 dB(A) by (ii); formula (ii) applies for N = 2. With d = 3 cm, 48.7 and 53.7 dB(A).

✓ Answer checked by Claude Opus 5.5 (high effort)

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