--- title: "Economical span, double circuit, five spans" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:economic-span-double-circuit-five-spans" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-span-double-circuit-five-spans site: BCT Engineering --- # Economical span, double circuit, five spans Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2070 Bhadra, 16 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-five-spans - 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 chosen 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 | 5.27 | 12.178 | 14.998 | 17.818 | 28.444 | | 275 m | 6.24 | 13.148 | 15.968 | 18.788 | 29.414 | | 300 m | 7.30 | 14.208 | 17.028 | 19.848 | 30.474 | | 325 m | 8.43 | 15.338 | 18.158 | 20.978 | 31.604 | | 350 m | 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 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. ## Short answer The most economical span is 275 m, yielding the minimum tower weight per km of 30.06 t/km. It is determined by computing base bending moment and tower weight via Ryle's formula for each span and selecting the least weight per km. ## Answer The most economical span gives the **minimum tower weight (cost) per km of line**. For each span, compute the base bending moment of a 5° deviation tower, then the tower weight from Ryle's formula, then the weight per km. 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 maximum permissible deviation θ = 5°, so $2\sin2.5^\circ = 0.08724$. **Sample calculation for L = 250 m, θ = 5°:** $$ \begin{aligned} F_c &= 100\times0.04\times250 + 12000\sin2.5^\circ = 1000+523.4 = 1523.4\ \text{kg}\\ F_e &= 100\times0.02\times250 + 4000\sin2.5^\circ = 500+174.5 = 674.5\ \text{kg}\\ M &= 2(1523.4)(12.178+14.998+17.818) + 2(674.5)(28.444) = 175{,}460\ \text{kg·m}\\ &= 1269.1\ \text{klb-ft}\\ W_t &= 0.0016\times(28.444\times3.281)\sqrt{1269.1\times2} = 7.522\ \text{t}\\ \text{Weight/km} &= 7.522\times\frac{1000}{250} = 30.090\ \text{t/km} \end{aligned} $$ #### Results for all spans | 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 | 175,460 | 7.522 | 4.000 | 30.090 | | 275 | 1623.4 | 724.5 | 198,157 | 8.267 | 3.636 | 30.061 | | 300 | 1723.4 | 774.5 | 223,283 | 9.092 | 3.333 | 30.305 | | 325 | 1823.4 | 824.5 | 250,773 | 9.992 | 3.077 | 30.745 | | 350 | 1923.4 | 874.5 | 281,041 | 10.986 | 2.857 | 31.390 | Steps followed: 1. $F_c$ and $F_e$ increase with span because the wind span increases. 2. $M$ increases because both the loads and the heights ($h_1,h_2,h_3,H_t$, which include the sag) increase. 3. $W_t$ increases, but the number of towers per km falls. 4. The product, weight per km, is lowest at 275 m (30.061 t/km). It is only slightly higher at 250 m (30.090 t/km) and then rises clearly from 300 m onwards. ```text t/km 31.4 | * 30.7 | * 30.3 | * 30.1 | * * +--+------+------+------+------+-- 250 275 300 325 350 span (m) ``` **Answer: the most economical span is 275 m, with a tower weight of 8.267 t per tower and 30.06 t per km.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What decides the economical span? A. Maximum tower weight per tower B. Minimum tower weight per km C. Minimum sag D. Minimum number of towers Answer: B. Cost is proportional to weight. (Key point: Criterion) 2. Why does M rise with span? A. Tension decreases B. Only the conductor weight increases C. Wind load rises with wind span and heights include larger sag D. Angle increases Answer: C. Both Fc and heights increase. (Key point: Moment) 3. What is 2T sin(theta/2) with theta = 5 deg for conductor? A. About 1000 kg B. About 1523 kg C. About 87 kg D. About 523 kg (12000 x sin2.5) Answer: D. 12000 x 0.04362 = 523.4 kg. (Key point: Deviation term) 4. Which span is most economical? A. 275 m, 30.06 t/km B. 250 m, 30.09 t/km C. 300 m, 30.31 t/km D. 350 m, 31.39 t/km Answer: A. 275 m is just below 250 m, then it rises. 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