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Conductor selection, 100 MW 80 km, no corona

Short answer

ACSR "Goat" (30/7/3.71 mm, 400 mm²) meets all criteria. It carries 486 A within its 543 A rating, gives line efficiency of 95.2% (>94%), and has a critical disruptive voltage of 148 kV per phase against 83.7 kV operating, so no fair-weather corona.

For a 3-phase 50 Hz transmission line to deliver 100 MW of power over 80 km, 132 kV single circuit line with following air clearances of phase conductors has been decided to design. Suggest the best suitable conductor which meets following technical criterion: (i) Satisfy the thermal limit (ii) Efficiency of line not to be less than 94% (iii) No corona in fair weather condition.

[Figure: conductor arrangement — three phase conductors, top and bottom on the right and the middle one on the left; vertical spacing 2.5 m from top to middle level and 2.5 m from middle to bottom level; horizontal distance between the left (middle) conductor and the right conductors is 9 m.] [The conductor table referred to as attached is not reproduced with this paper.]

The conductor is chosen as the smallest standard ACSR that passes all three checks.

Assumptions: power factor 0.9 lagging; the standard ACSR table supplied with IOE papers (DC resistance at 20°C, current rating at 80°C conductor temperature); fair weather with δ=1\delta = 1 and irregularity factor m=0.85m = 0.85; maximum system voltage for 132 kV is 145 kV.

Spacing (GMD) from the figure

The middle conductor is 9 m horizontally and 2.5 m vertically from the top and bottom conductors. The top and bottom conductors are 5 m apart vertically.

D12=D23=92+2.52=9.341 m,D13=5 mGMD=9.341×9.341×53=7.584 m\begin{aligned} D_{12} &= D_{23} = \sqrt{9^2+2.5^2} = 9.341\ \text{m}, \quad D_{13} = 5\ \text{m}\\ GMD &= \sqrt[3]{9.341\times 9.341\times 5} = 7.584\ \text{m} \end{aligned}

(i) Thermal limit

I=P3 Vcos⁡ϕ=100×1063×132×103×0.9=485.99 AI = \frac{P}{\sqrt3\, V\cos\phi} = \frac{100\times10^6}{\sqrt3\times132\times10^3\times0.9} = 485.99\ \text{A}
  • Bear is rated 480 A, which is less than 486 A, so Bear is not acceptable.
  • Goat is rated 543 A, which is more than 486 A, so Goat passes. Goat is the smallest conductor that does.

(ii) Efficiency of at least 94%

Ploss,max=P(10.94−1)=6.383 MWRmax=6.383×1063×485.992=9.009 Ω  ⇒  0.1126 Ω/km\begin{aligned} P_{loss,max} &= P\left(\frac{1}{0.94}-1\right) = 6.383\ \text{MW}\\ R_{max} &= \frac{6.383\times10^6}{3\times485.99^2} = 9.009\ \Omega \;\Rightarrow\; 0.1126\ \Omega/\text{km} \end{aligned}
ConductorR (Ω/km)R for 80 km (Ω)Loss (MW)η (%)
Bear0.10938.7446.19694.17
Goat0.08917.1285.05195.19
Sheep0.07716.1684.37095.81

Here η=P/(P+Ploss)\eta = P/(P+P_{loss}) and Ploss=3I2RP_{loss} = 3I^2R. Goat meets the target with a margin. Even if its resistance is raised by 10% for temperature and skin effect, η is still about 94.7%.

(iii) No corona in fair weather

The disruptive critical voltage (phase, rms) for Goat, with r=25.97/2=1.2985r = 25.97/2 = 1.2985 cm, is:

Vc=21.1 m δ rln⁡GMDr=21.1×0.85×1×1.2985×ln⁡758.41.2985=148.3 kV (phase)\begin{aligned} V_c &= 21.1\, m\,\delta\, r \ln\frac{GMD}{r}\\ &= 21.1\times0.85\times1\times1.2985\times\ln\frac{758.4}{1.2985}\\ &= 148.3\ \text{kV (phase)} \end{aligned}

The highest operating phase voltage is 145/3=83.7145/\sqrt3 = 83.7 kV. Since 148.3 kV is greater than 83.7 kV, there is no corona in fair weather. This also holds for Bear, whose Vc=136.1V_c = 136.1 kV.

Selection

CheckBearGoat
Thermal (486 A)Fails (480 A)Passes (543 A)
Efficiency ≥ 94%94.17%95.19%
No coronaPassesPasses

Answer: ACSR "Goat" (30/7/3.71 mm, 400 mm² total area) is the most suitable conductor. It carries 486 A within its 543 A rating, gives an efficiency of 95.2%, and has a corona voltage of 148 kV per phase against 83.7 kV operating.

✓ Answer checked by Claude Opus 5.5 (high effort)

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