--- title: "Most economical span, double circuit line" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:economic-span-double-circuit" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-span-double-circuit site: BCT Engineering --- # Most economical span, double circuit line Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2077 Chaitra, 16 marks; 2074 Magh, 10 marks (asked 2 times) - 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 - 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 a double circuit line with a double earth wire is choose for which the following design steps are completed, compute the most economical span. | Span | Dmax (m) | h1 (m) | h2 (m) | h3 (m) | Ht (m) | |---|---|---|---|---|---| | 250 m | 3.48 | 9.70 | 14.52 | 19.35 | 24.84 | | 300 m | 5.09 | 11.30 | 16.13 | 20.96 | 26.45 | | 350 m | 7.02 | 13.24 | 18.06 | 22.89 | 28.39 | where h1, h2 and h3 are height of lower, middle and top power conductor from ground and Ht is total height of tower. The power conductor has maximum working tension of 6890 kg and diameter of 25.97 mm. The earth wire has maximum working tension of 3332 kg and diameter of 14.60 mm. Assume maximum conductor deviation of 5° is permissible for all towers and the wind force is 100 kg/m². ## Short answer Method: Calculate overturning moment from wind and deviation loads, then tower weight index per km using Ryle’s formula. Comparing 250, 300, 350 m spans, the lowest relative cost per km is at 300 m. Most economical span: 300 m. ## Answer **Method.** For each span, the transverse load at the tower comes from two sources: - wind on the conductors and earth wires: $F = p \times d \times S$, taking the wind span equal to the span $S$ and using the full projected area; - the resultant of the conductor tensions at a line deviation $\theta$: $F_d = 2T\sin(\theta/2)$. These loads act at the conductor heights. Their overturning moment at ground level is $$ M = n_c\,(F_c + F_{dc})(h_1 + h_2 + h_3) + n_e\,(F_e + F_{de})\,H_t $$ where $n_c$ is the number of conductors at each level and $n_e$ the number of earth wires. The tower weight is found from **Ryle's formula** $$ W = K\,H_t\sqrt{M} $$ $K$ is the same for all spans, so it cancels in the comparison. Tower cost is proportional to tower weight, so the span that gives the smallest total tower weight is the most economical span (with the conductor already fixed). Wind on the tower body is neglected. #### Data and deviation loads - Double circuit: 2 power conductors at each of $h_1$, $h_2$, $h_3$, so $n_c = 2$. Two earth wires at $H_t$, so $n_e = 2$. - Wind pressure $p = 100$ kg/m², $d_c = 0.02597$ m, $d_e = 0.0146$ m. - All towers are designed for $\theta = 5°$: $$ \begin{aligned} F_{dc} &= 2 \times 6890 \times \sin 2.5° = 601.08\ \text{kg} \\ F_{de} &= 2 \times 3332 \times \sin 2.5° = 290.68\ \text{kg} \end{aligned} $$ #### Wind loads per span | Span (m) | $F_c = p\,d_c\,S$ (kg) | $F_e = p\,d_e\,S$ (kg) | |---|---|---| | 250 | 649.25 | 365.00 | | 300 | 779.10 | 438.00 | | 350 | 908.95 | 511.00 | #### Overturning moments and tower weight index Sample calculation (250 m): $$ \begin{aligned} M_c &= 2(649.25 + 601.08)(9.70 + 14.52 + 19.35) = 108{,}953.3\ \text{kg·m} \\ M_e &= 2(365.0 + 290.68)(24.84) = 32{,}574.2\ \text{kg·m} \\ M &= 141{,}527.5\ \text{kg·m}, \quad H_t\sqrt{M} = 24.84\sqrt{141527.5} = 9{,}344.8 \end{aligned} $$ | Span (m) | Tower type (θ) | $M_c$ (kg·m) | $M_e$ (kg·m) | $M$ (kg·m) | $H_t\sqrt{M}$ | |---|---|---|---|---|---| | 250 | 5° (all) | 108,953.3 | 32,574.2 | 141,527.5 | 9,344.8 | | 300 | 5° (all) | 133,573.4 | 38,547.2 | 172,120.5 | 10,973.4 | | 350 | 5° (all) | 163,656.5 | 45,519.4 | 209,175.9 | 12,984.4 | #### Tower steel per km Towers per km $= 1000/S$, so the relative cost per km $= (1000/S) \times H_t\sqrt{M}$ (in units of $K$): | Span (m) | Weighted $H_t\sqrt{M}$ | Towers per km $= 1000/S$ | Relative cost per km | |---|---|---|---| | 250 | 9,344.8 | 4.000 | 37,379.4 | | 300 | 10,973.4 | 3.333 | 36,578.1 | | 350 | 12,984.4 | 2.857 | 37,098.2 | Moving from 250 m to 300 m saves more on the number of towers than it adds in tower weight. Beyond 300 m, the heavier and taller towers outweigh the saving. **Answer: the most economical span is 300 m**, which has the lowest relative tower cost per km (36,578.1 K). Insulator and foundation costs per km also fall with longer spans, but the spread between 250 m and 350 m is small, so 300 m is a sound choice. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What two sources make up the transverse load on a tower in this method? A. Tower weight and conductor weight B. Wind on conductors/earth wires and the resultant of tensions at line deviation C. Ice load and snow load D. Short-circuit force and vibration Answer: B. F = p d S for wind and Fd = 2T sin(θ/2) for deviation. (Key point: Loads at tower) 2. The deviation load at 5° for a conductor tension T is: A. T sin 5° B. 2T sin 2.5° C. 2T cos 2.5° D. T tan 5° Answer: B. Resultant of two tensions at deviation θ is 2T sin(θ/2). (Key point: Deviation force) 3. What is used as the index of tower weight (cost) to compare spans? A. Ht times M B. Ht times sqrt(M), by Ryle's formula W = K Ht sqrt(M) C. M divided by Ht D. Span divided by Ht Answer: B. Ryle's formula gives weight proportional to Ht sqrt(M); K cancels. (Key point: Tower weight) 4. How is the cost per km found for each span, and which span is the most economical? A. Multiply Ht sqrt(M) by 1000/S; 300 m is lowest B. Multiply by S; 250 m is lowest C. Multiply by 1000/S; 350 m is lowest D. Use Ht sqrt(M) only; 250 m is lowest Answer: A. Towers per km = 1000/S; the lowest relative cost is 36,578 at 300 m. (Key point: Per km cost) ## 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× - [Economical span, double circuit, A/B/C towers](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-double-circuit-abc-towers) — Transmission and Distribution System Design, chapter 4, asked 5× - [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× - [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× ## Related in Transmission and Distribution System Design - [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, 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× - [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 - [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: earth wire choice depends on circuit and voltage](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Atf-earth-wire-choice-circuit-voltage) — 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× - [Design: 250 m head plant, 66 kV double circuit](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Adesign-2071-250m-0-92mh) — Power Plant Design, chapter 4, asked 13×