Conductor selection, 220 kV line for 200 MW

The conductor must pass all four technical checks. The cheapest arrangement that passes is chosen.

Assumptions: pf 0.9 lagging at the receiving end, 220 kV at the receiving end; standard ACSR table (IOE); the tower geometry from the air-clearance design (adjacent phases 11.03 m apart, outer phases 9.77 m apart, so GMD = 10.59 m); medium line with the nominal-π model; $\delta = 1$, $m = 0.85$; $GMR = 0.7788r$; twin bundle spacing $s = 0.4$ m.

i) Thermal rating

$$I = \frac{200\times10^6}{\sqrt3\times220\times10^3\times0.9} = 583.2\ \text{A}$$

• Single conductor: Sheep (592 A) or larger.
• Twin bundle: each sub-conductor carries 291.6 A, so Coyote/Tiger (about 311 A) or larger.

Checks on single conductors

Using $x = 2\pi f\times2\times10^{-4}\ln(GMD/GMR)$ Ω/km and $V_s = AV_r + BI_r$:

Every single conductor fails the 12% regulation limit. The reactance (about 0.43 Ω/km) is too high. A twin bundle is needed to cut the reactance.

Checks on twin bundles

2 × Lion only just passes, at 11.94%. With resistance at the operating temperature (about 10% higher), its regulation rises to about 12.2% and it fails. 2 × Bear is the smallest bundle with a safe margin.

Detailed check: twin ACSR Bear

Data: $r = 23.45/2 = 11.725$ mm, $R_{20} = 0.1093$ Ω/km per sub-conductor.

$$\begin{aligned} GMR_b &= \sqrt{0.7788r\times s} = \sqrt{0.009131\times0.4} = 0.06044\ \text{m}\\ x &= 0.06283\ln\frac{10.592}{0.06044} = 0.3246\ \Omega/\text{km}\\ r_{eq} &= \sqrt{rs} = 0.06848\ \text{m}, \quad b = \frac{2\pi f\times2\pi\varepsilon_0}{\ln(GMD/r_{eq})} = 3.467\ \mu\text{S/km}\\ Z &= (0.1093/2)\times120 + j0.3246\times120 = 6.558 + j38.95\ \Omega\\ Y &= j4.160\times10^{-4}\ \text{S}, \quad A = 1 + \frac{ZY}{2} = 0.99190\angle0.08^\circ\\ V_r &= 127.0\angle0^\circ\ \text{kV}, \quad I_r = 583.2\angle{-25.84^\circ}\ \text{A}\\ V_s &= AV_r + ZI_r = 139.33 + j18.95\ \text{kV} = 140.61\angle7.75^\circ\ \text{kV} \end{aligned}$$

i) Thermal: 2 × 480 = 960 A, which exceeds 583 A. Passes.

ii) Corona: for the bundle,

$$V_c \approx \frac{21.1\,m\,\delta\,n\,r\ln(GMD/r_{eq})}{1 + (n-1)r/s} = 206.0\ \text{kV (phase)}$$

This exceeds the 141.5 kV maximum phase voltage, so there is no fair-weather corona.

iii) Power angle: δ = 7.75°, which is below 30°.

iv) Regulation:

$$\frac{|V_s|/|A| - V_r}{V_r} = \frac{140.61/0.9919 - 127.02}{127.02} = 11.61\%$$

This is below 12%. The sending-end voltage is 243.6 kV (line), within the 245 kV maximum.

Answer: use ACSR "Bear" (30/7/3.35 mm) in a twin bundle (2 sub-conductors per phase, 400 mm apart). This gives δ = 7.75°, regulation 11.6%, corona voltage 206 kV per phase and a 960 A rating. No single conductor can meet the 12% regulation limit at 0.9 pf.