--- title: "Conductor selection, 220 kV line for 200 MW" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:conductor-selection-220kv-200mw-120km" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/conductor-selection-220kv-200mw-120km site: BCT Engineering --- # Conductor selection, 220 kV line for 200 MW Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2067 Chaitra (old course), 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/conductor-selection-220kv-200mw-120km - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question For a transmission line to deliver 200 MW of power over 120 km, it has been decided to design a 220 kV transmission line of single circuit. Select the best suitable conductor and conductor composition which meets the following technical criterion. i) Satisfy the thermal rating ii) No corona in fair weather condition iii) Power angle between sending end and receiving end must be within 30° iv) Initial voltage regulation within 12%. [ACSR conductor table to be provided; not reproduced with this paper.] ## Short answer Use twin ACSR Bear (30/7/3.35 mm), two sub‑conductors per phase at 400 mm spacing. Thermal rating 960 A > 583 A, corona voltage 206 kV > 141.5 kV, power angle 7.75° < 30°, regulation 11.6% < 12%. Single conductors fail the regulation limit. ## Answer 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 $V_s = AV_r + BI_r$: | Conductor | X total (Ω) | δ (°) | Regulation (%) | $V_c$ phase (kV) | |---|---|---|---|---| | Sheep | 51.88 | 9.87 | 16.04 | 166.1 | | Deer | 51.37 | 9.93 | 15.46 | 175.9 | | Moose | 50.91 | 10.03 | 14.76 | 185.3 | 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 | Bundle (2 per phase) | X (Ω) | R (Ω) | δ (°) | Regulation (%) | $V_c$ (kV) | |---|---|---|---|---|---| | 2 × Tiger | 40.27 | 13.21 | 7.21 | 14.54 | 151.4 | | 2 × Lion | 39.15 | 7.28 | 7.70 | 11.94 | 196.8 | | 2 × Bear | 38.95 | 6.56 | 7.75 | 11.61 | 206.0 | | 2 × Goat | 38.57 | 5.35 | 7.82 | 11.04 | 225.1 | 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the line current found for thermal rating? A. I = P/V B. I = P/(3 V cos phi) C. I = P/(sqrt3 V cos phi) = 583 A D. I = P/(sqrt3 V) = 525 A Answer: C. I = 200e6 / (sqrt3 x 220e3 x 0.9) = 583.2 A. (Key point: Thermal current) 2. Why can no single conductor be used? A. It cannot carry 583 A B. Its reactance is too high, so regulation exceeds 12% C. It has fewer corona losses D. It gives a power angle above 30° Answer: B. Single Sheep, Deer and Moose gave regulation of 14.8-16%; a bundle lowers the reactance. (Key point: Why twin bundle) 3. Why is 2 x Bear chosen over 2 x Lion, which gives 11.94%? A. Lion fails the thermal limit B. Lion causes corona C. Lion's regulation is only just within 12% and would fail with resistance at operating temperature D. Lion has a larger power angle than 30° Answer: C. With about 10% higher resistance, Lion's regulation rises to about 12.2%; Bear (11.61%) has a safe margin. (Key point: Why Bear not Lion) 4. What is the corona check for 2 x Bear? A. Vc of 206 kV per phase exceeds the maximum phase voltage of 141.5 kV, so no fair-weather corona B. Vc must be below the phase voltage C. Vc is 141.5 kV D. Corona is ignored for bundles Answer: A. No fair-weather corona requires the critical disruptive voltage to exceed the maximum operating phase voltage. 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