--- title: "Economical span, double circuit, 5° deviation" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:economic-span-double-circuit-6000kg" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-span-double-circuit-6000kg site: BCT Engineering --- # Economical span, double circuit, 5° deviation Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2072 Magh, 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-double-circuit-6000kg - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question The design data for a double circuit transmission line with a double earth wire are as follows: | Span | Dmax (m) | h1 (m) | h2 (m) | h3 (m) | Ht (m) | |---|---|---|---|---|---| | 250 m | 5.27 | 12.18 | 15.00 | 17.82 | 28.44 | | 300 m | 7.30 | 14.21 | 17.03 | 19.85 | 30.47 | | 350 m | 9.65 | 16.56 | 19.38 | 22.20 | 32.82 | where h1, h2 and h3 are heights of lower, middle and top power conductor from ground and Ht is the total height of tower. The power conductor has maximum working tension of 6000 kg and diameter of 4 cm. The earth wire has maximum working tension of 2000 kg and diameter of 2 cm. Assume maximum conductor deviation of 5° is permissible for all towers and the wind force is 100 kg/m². Compute the most economical span. ## Short answer All towers designed for 5° deviation; using wind and deviation loads, bending moment, and Ryle's formula, tower weight per km is 30.09 t at 250 m, 30.30 t at 300 m, and 31.39 t at 350 m. The most economical span is 250 m. ## Answer The most economical span is the one that gives the **least total tower weight (cost) per km**. Each tower's bending moment is computed from the wind load plus the 5° deviation load, and the tower weight is then found from Ryle's formula. 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. All towers are designed for the permissible deviation θ = 5°. **Sample calculation for L = 250 m, θ = 5°:** $$ \begin{aligned} F_c &= 100\times0.04\times250 + 12000\sin2.5^\circ = 1000+523.4 = 1523.4\ \text{kg}\\ F_e &= 100\times0.02\times250 + 4000\sin2.5^\circ = 500+174.5 = 674.5\ \text{kg}\\ M &= 2(1523.4)(12.18+15.00+17.82) + 2(674.5)(28.44) = 175{,}473\ \text{kg·m}\\ &= 1269.2\ \text{klb-ft}\\ W_t &= 0.0016\times(28.44\times3.281)\sqrt{1269.2\times2} = 7.522\ \text{t}\\ \text{Weight/km} &= 7.522\times\frac{1000}{250} = 30.09\ \text{t/km} \end{aligned} $$ #### Results | Span L (m) | $F_c$ (kg) | $F_e$ (kg) | M (kg·m) | $W_t$ (t) | Towers/km | Weight/km (t) | |---|---|---|---|---|---|---| | 250 | 1523.4 | 674.5 | 175,473 | 7.522 | 4.000 | 30.09 | | 300 | 1723.4 | 774.5 | 223,297 | 9.091 | 3.333 | 30.30 | | 350 | 1923.4 | 874.5 | 281,057 | 10.986 | 2.857 | 31.39 | The tower weight per km rises steadily from 250 m to 350 m. The extra height and moment of the longer spans outweigh the saving from having fewer towers. **Answer: of the three spans given, the most economical is 250 m, with about 30.09 t of tower steel per km, against 30.30 t/km for 300 m and 31.39 t/km for 350 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The most economical span is the one with the least what? A. Tower weight per tower B. Sag C. Tower weight per km D. Conductor tension Answer: C. Cost is taken proportional to weight, and 1000/L towers are needed per km. (Key point: Criterion) 2. How is the 5 degree deviation included in a conductor load? A. Add T sin(5 deg) only B. Multiply wind load by sin(5 deg) C. Ignore it since towers are tangent D. Add 2T sin(2.5 deg) to the wind load p d L Answer: D. Fc = 1000 + 12000 sin 2.5 deg = 1523.4 kg at 250 m. (Key point: Deviation load) 3. How is weight per km found from the tower weight? A. Multiply by 1000/L B. Multiply by L/1000 C. Divide by 1000/L D. Multiply by L only Answer: A. 7.522 t x 4 towers/km = 30.09 t/km at 250 m. (Key point: Towers per km) 4. What is the result comparing 250, 300 and 350 m? A. 350 m is least because it needs the fewest towers B. 250 m is least at 30.09 t/km; 300 m gives 30.31 and 350 m gives 31.39 C. 300 m is least at 30.31 t/km D. All three are equal Answer: B. Extra height and moment of longer spans outweigh the saving from fewer towers. 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