--- title: "Economical span, double circuit, A/B/C towers" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:economic-span-double-circuit-abc-towers" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-span-double-circuit-abc-towers site: BCT Engineering --- # Economical span, double circuit, A/B/C towers Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2072 Asoj, 15 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/economic-span-double-circuit-abc-towers - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question To transmit a given amount of power to a given distance, double circuit line with a double earth wire is chosen for which the following design step are completed, compute the most economical span. | Span (m) | Dmax (m) | H1 (m) | H2 (m) | H3 (m) | Ht (m) | |---|---|---|---|---|---| | 250 | 5.27 | 12.178 | 14.998 | 17.818 | 28.444 | | 275 | 6.24 | 13.148 | 15.968 | 18.788 | 29.414 | | 300 | 7.3 | 14.208 | 17.028 | 19.848 | 30.474 | | 325 | 8.43 | 15.338 | 18.158 | 20.978 | 31.604 | | 350 | 9.65 | 16.558 | 19.378 | 22.198 | 32.824 | Where, H1 → height of lower conductor from ground, H2 → height of middle conductor from ground, H3 → height of top conductor from ground, Ht → total height. The power conductor has maximum working tension of 6000 kg and diameter of 4 cm. The earth wire has maximum tension of 2000 kg and diameter of 2 cm. Assume 80% as tower A, 15% as tower B and 5% tower C and the wind force is 100 kg/m². ## Short answer Using wind and deviation loads, bending moment, and Ryle's formula, the weighted average tower steel per km is lowest at 250 m (29.82 t/km). 275 m gives 29.83 t/km, but 250 m is the most economical. ## Answer The most economical span gives the least **average tower weight (cost) per km**. Here the line has a mix of tower types, so a weighted average tower weight is used. **Tower types assumed** (usual IS 802 classification): | Type | Use | Design deviation θ | Share | |---|---|---|---| | A | Tangent / small angle | 0–2° (use 2°) | 80% | | B | Medium angle | 2–15° (use 15°) | 15% | | C | Large angle | 15–30° (use 30°) | 5% | Tower weight is taken from **Ryle's formula** (as given with IOE papers): $$ W_t = 0.0016\,H_t\sqrt{M\times FS}\ \ \text{tonne} $$ Here $H_t$ is in ft, $M$ is the bending moment at the base in klb-ft, and FS = 2 (assumed). The conversions are 1 m = 3.281 ft and 1 kg·m = 0.007233 klb-ft. Tower cost is taken as proportional to tower weight, and the number of towers per km is about $1000/L$. So the span with the **least tower weight per km** is the most economical. #### Loads and bending moment For each span $L$ the loads on a tower are found as follows. The wind span is taken equal to the span. - Each power conductor (6 of them, 2 at each height $h_1, h_2, h_3$): $$ F_c = p\,d_c\,L + 2T_c\sin\frac{\theta}{2} = 100\times0.04\,L + 2\times6000\sin\frac{\theta}{2}\ \text{kg} $$ - Each earth wire (2 of them, at height $H_t$): $$ F_e = p\,d_e\,L + 2T_e\sin\frac{\theta}{2} = 100\times0.02\,L + 2\times2000\sin\frac{\theta}{2}\ \text{kg} $$ - Bending moment at the base: $$ M = 2F_c(h_1+h_2+h_3) + 2F_e H_t $$ Wind on the tower body is neglected because it is about the same for all spans. The deviation factor $2\sin(\theta/2)$ is 0.0349 for 2°, 0.2611 for 15° and 0.5176 for 30°. **Sample (L = 250 m, tower A, θ = 2°):** $$ \begin{aligned} F_c &= 1000 + 6000\times0.0349 = 1209.4\ \text{kg}\\ F_e &= 500 + 2000\times0.0349 = 569.8\ \text{kg}\\ M_A &= 2(1209.4)(44.994) + 2(569.8)(28.444) = 141{,}249\ \text{kg·m}\\ W_A &= 0.0016(93.32)\sqrt{141249\times0.007233\times2} = 6.749\ \text{t} \end{aligned} $$ The weighted weight per tower is $W = 0.8W_A + 0.15W_B + 0.05W_C$, and the weight per km is $W\times1000/L$. #### Results | Span L (m) | $M_A$ (kg·m) | $M_B$ (kg·m) | $M_C$ (kg·m) | |---|---|---|---| | 250 | 141,249 | 289,083 | 456,814 | | 275 | 161,916 | 318,524 | 496,211 | | 300 | 184,822 | 351,019 | 539,585 | | 325 | 209,947 | 386,366 | 586,529 | | 350 | 237,662 | 425,116 | 637,801 | | Span L (m) | $W_A$ (t) | $W_B$ (t) | $W_C$ (t) | Weighted $W$ (t) | Weight/km (t) | |---|---|---|---|---|---| | 250 | 6.749 | 9.656 | 12.138 | 7.455 | 29.819 | | 275 | 7.473 | 10.481 | 13.082 | 8.204 | 29.834 | | 300 | 8.272 | 11.399 | 14.133 | 9.034 | 30.113 | | 325 | 9.143 | 12.403 | 15.282 | 9.939 | 30.581 | | 350 | 10.103 | 13.512 | 16.551 | 10.937 | 31.248 | The weight per km is lowest at 250 m (29.819 t/km). 275 m is almost equal (29.834 t/km, only about 0.05% more). From 300 m onwards the weight rises clearly. **Answer: the most economical span is 250 m, with an average tower weight of 7.455 t and 29.82 t of tower steel per km. 275 m is practically as good and could be chosen if tower count or terrain favours it.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What criterion picks the most economical span here? A. Least weighted tower weight (cost) per km B. Least sag at the span C. Least tower weight per tower D. Largest span that meets ground clearance Answer: A. Longer spans mean heavier towers but fewer per km, so weight per km is compared. (Key point: Economic span criterion) 2. How is the weight per tower combined for A, B and C towers? A. Simple average of WA, WB and WC B. W = 0.8WA + 0.15WB + 0.05WC, then multiplied by 1000/L for weight per km C. Only WA is used since it is the most common D. W = WA + WB + WC divided by the span Answer: B. Each type is weighted by its share of the line (80%, 15%, 5%). (Key point: Weighted tower weight) 3. What is the transverse load on each power conductor at an angle tower? A. Wind load p d L only B. Tension T only C. Wind load p d L plus the angle load 2T sin(theta/2) D. Wind load p d L times 2 sin(theta/2) Answer: C. Fc = 100 x 0.04 L + 2 x 6000 sin(theta/2) kg. (Key point: Conductor load) 4. How is the base bending moment M calculated? A. M = Fc(h1+h2+h3) + Fe Ht, ignoring the second circuit B. M = 6Fc Ht + 2Fe Ht C. M = 2Fc(h1+h2+h3) + 2Fe(h1+h2+h3) D. M = 2Fc(h1+h2+h3) + 2Fe Ht, with 6 conductors and 2 earth wires Answer: D. Two conductors sit at each of three heights and two earth wires at Ht. (Key point: Bending moment) 5. Which span is the most economical and why not 300 m or more? A. 250 m (29.82 t/km); weight per km rises from 300 m onwards B. 350 m, because fewer towers are needed C. 300 m, because tower weight per tower is lowest D. 275 m, because its tower weight is lowest Answer: A. Weight/km is 29.819 at 250 m, 29.834 at 275 m and then rises. (Key point: Result) ## Close variants set in other exams - [Economical span, double circuit, five spans](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-double-circuit-five-spans) — Transmission and Distribution System Design, chapter 4, asked 6× - [Most economical span, double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-double-circuit) — Transmission and Distribution System Design, chapter 4, asked 6× - [Economical span, 200 km, A/B/C tower types](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-200km-three-tower-types) — Transmission and Distribution System Design, chapter 4, asked 6× ## Related in Transmission and Distribution System Design - [Economical span, 200 km, 10% angle towers](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-200km-angle-towers) — Transmission and Distribution System Design, chapter 4, asked 2× - [Most economical span, 180 km single circuit](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-180km-single-circuit) — Transmission and Distribution System Design, chapter 4 - [Economical span, double circuit, 5° deviation](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-double-circuit-6000kg) — Transmission and Distribution System Design, chapter 4, asked 4× - [Most economical span from sag table](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-sag-table-28mm) — Transmission and Distribution System Design, chapter 4, asked 2× - [Most economical EHV span from sag table](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-sag-table-21mm) — Transmission and Distribution System Design, chapter 4, asked 2× - [T/F: double earth wire for 400 kV single circuit](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atf-400kv-single-circuit-double-earth) — Transmission and Distribution System Design, chapter 3 ## Related in other subjects - [Block diagram reduction: G1–G4 with H1, H2, H3](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2h3-2073) — Control Systems, chapter 3, asked 3× - [Block diagram reduction: G1–G4 with H1, H2, H3 loops](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2h3-2070r) — Control Systems, chapter 3, asked 2×