--- title: "Most economical span, 180 km single circuit" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:economic-span-180km-single-circuit" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-span-180km-single-circuit site: BCT Engineering --- # Most economical span, 180 km single circuit Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2078 Chaitra, 10 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-180km-single-circuit - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question Calculate most economical span for 180 km long single circuit transmission system with single earth wire chosen, considering the following design step data. | Span (m) | Dmax (m) | H1 (m) | H2 (m) | H3 (m) | Ht (m) | |---|---|---|---|---|---| | 250 | 3.15 | 10.22 | 13.60 | 16.98 | 28.84 | | 300 | 4.32 | 11.40 | 14.72 | 18.04 | 29.98 | | 350 | 5.48 | 12.58 | 15.95 | 19.32 | 31.22 | 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 of tower. The power conductor has maximum working tension of 6860 kg and diameter of 25.2 mm. The earth wire has maximum working tension of 2840 kg and diameter of 16.4 mm. Assume the towers consisted to be 80% type A, and other remaining are of type B in the system and the wind force of 100 kg/m². ## 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. #### Assumptions - Single circuit: one conductor at each of $H_1$, $H_2$, $H_3$ ($n_c = 1$), and one earth wire at $H_t$ ($n_e = 1$). - Standard tower types (IS 802 practice): type A is a suspension tower for 0°–2° deviation and type B is a small-angle tower for 2°–15°. Each type is designed for its maximum angle. - Number of towers for 180 km: $N = \lceil 180000/S \rceil + 1$, with 80% type A and 20% type B. #### Deviation loads | Type | $F_{dc} = 2 \times 6860 \sin(\theta/2)$ | $F_{de} = 2 \times 2840 \sin(\theta/2)$ | |---|---|---| | A (2°) | 239.45 kg | 99.13 kg | | B (15°) | 1,790.82 kg | 741.39 kg | #### Wind loads ($p = 100$ kg/m²) | Span (m) | $F_c = p\,d_c\,S$ (kg) | $F_e = p\,d_e\,S$ (kg) | |---|---|---| | 250 | 630.00 | 410.00 | | 300 | 756.00 | 492.00 | | 350 | 882.00 | 574.00 | #### Moments and tower weight index Sample calculation (250 m, type A): $$ \begin{aligned} M_c &= (630 + 239.45)(10.22 + 13.60 + 16.98) = 35{,}473.4\ \text{kg·m} \\ M_e &= (410 + 99.13)(28.84) = 14{,}683.3\ \text{kg·m} \\ H_t\sqrt{M} &= 28.84\sqrt{50156.7} = 6{,}458.9 \end{aligned} $$ | Span (m) | Tower type (θ) | $M_c$ (kg·m) | $M_e$ (kg·m) | $M$ (kg·m) | $H_t\sqrt{M}$ | |---|---|---|---|---|---| | 250 | A (2°) | 35,473.4 | 14,683.3 | 50,156.7 | 6,458.9 | | 250 | B (15°) | 98,769.4 | 33,206.1 | 131,975.5 | 10,477.1 | | 300 | A (2°) | 43,958.9 | 17,722.1 | 61,681.0 | 7,445.7 | | 300 | B (15°) | 112,467.5 | 36,977.0 | 149,444.5 | 11,589.7 | | 350 | A (2°) | 53,661.2 | 21,015.1 | 74,676.3 | 8,531.5 | | 350 | B (15°) | 127,894.4 | 41,066.4 | 168,960.8 | 12,832.9 | #### Total tower steel for the line Weighted index $= 0.8(H_t\sqrt{M})_A + 0.2(H_t\sqrt{M})_B$, multiplied by the number of towers $N$: | Span (m) | Weighted $H_t\sqrt{M}$ | Towers $N$ | Relative tower steel $N\times H_t\sqrt{M}$ | |---|---|---|---| | 250 | 7,262.6 | 721 | 5.236 × 10⁶ | | 300 | 8,274.5 | 601 | 4.973 × 10⁶ | | 350 | 9,391.8 | 516 | 4.846 × 10⁶ | The total tower weight keeps falling up to the largest span tabulated. **Answer: the most economical span is 350 m** (lowest total tower weight index, 4.846 × 10⁶ K, with 516 towers). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are type A and type B towers distinguished here? A. By voltage B. By deviation angle: A is 0-2°, B is 2-15° C. By height D. By conductor material Answer: B. Each type is designed for its maximum deviation angle. (Key point: Tower types) 2. The weighted index for each span is: A. 0.8 times A plus 0.2 times B of Ht sqrt(M) B. 0.2 times A plus 0.8 times B C. Average of A and B D. Sum of A and B Answer: A. 80% are type A and 20% type B. (Key point: Weighted index) 3. Total steel for the line is obtained by: A. Multiplying the weighted Ht sqrt(M) by the number of towers N B. Dividing by N C. Adding N D. Using the span only Answer: A. N x weighted Ht sqrt(M) gives relative tower steel. (Key point: Total steel) 4. Which span is most economical? A. 250 m B. 300 m C. 350 m D. None, cost is equal Answer: C. 350 m gives the lowest total index, 4.846 x 10^6. 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