---
title: "Economical span, 200 km, 10% angle towers"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:economic-span-200km-angle-towers"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-span-200km-angle-towers
site: BCT Engineering
---

# Economical span, 200 km, 10% angle towers

Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection

- Asked in: 2073 Bhadra, 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-200km-angle-towers
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

To transmit the given amount of power to a given distance a single circuit with single earth wire is chosen for which the following design steps are completed, compute the most economical span if the transmission length is 200 km. Assume 10% of towers have been used to take care of maximum angle deviation of 15° and rest are straight line towers.

| Span (m) | Dmax (m) | H1 (m) | H2 (m) | H3 (m) | Ht (m) |
|---|---|---|---|---|---|
| 250 | 3.14 | 10.245 | 13.555 | 16.865 | 28.815 |
| 275 | 3.69 | 10.801 | 14.111 | 17.421 | 29.371 |
| 300 | 4.27 | 11.383 | 14.693 | 18.003 | 29.953 |
| 325 | 4.88 | 11.965 | 15.298 | 18.608 | 30.558 |
| 350 | 5.50 | 12.548 | 15.923 | 19.233 | 31.183 |

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 UTS of 8000 kg and diameter of 20 mm. The earth wire has maximum working tension of 4000 kg and diameter of 16 mm. Wind force: 80 kg/m² and factor of safety for tension is 2.

## Short answer

Using conductor working tension 4000 kg (UTS 8000 kg, safety factor 2) and wind 80 kg/m², overturning moments are computed for straight and 15° angle towers. A weighted index (90% straight, 10% angle) gives total tower steel over 200 km; minimum occurs at 350 m with 3.870×10⁶ relative steel.

## 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
- Conductor working tension $T_c = \text{UTS}/\text{FOS} = 8000/2 = 4000$ kg, $d_c = 0.020$ m.
- Earth wire $T_e = 4000$ kg, $d_e = 0.016$ m. Wind pressure $p = 80$ kg/m².
- Single circuit with single earth wire: $n_c = 1$, $n_e = 1$.
- 90% straight towers (no deviation load) and 10% angle towers designed for 15°.
- $N = \lceil 200000/S \rceil + 1$.

Deviation load on the angle towers:

$$
F_{dc} = F_{de} = 2 \times 4000 \times \sin 7.5° = 1,044.21\ \text{kg}
$$

#### Wind loads
| Span (m) | $F_c = p\,d_c\,S$ (kg) | $F_e = p\,d_e\,S$ (kg) |
|---|---|---|
| 250 | 400.00 | 320.00 |
| 275 | 440.00 | 352.00 |
| 300 | 480.00 | 384.00 |
| 325 | 520.00 | 416.00 |
| 350 | 560.00 | 448.00 |

#### Moments and tower weight index
Sample calculation (250 m, angle tower):

$$
\begin{aligned} M_c &= (400 + 1044.21)(10.245 + 13.555 + 16.865) = 58{,}728.8\ \text{kg·m} \\ M_e &= (320 + 1044.21)(28.815) = 39{,}309.7\ \text{kg·m} \\ H_t\sqrt{M} &= 28.815\sqrt{98038.5} = 9{,}022.3 \end{aligned}
$$

| Span (m) | Tower type (θ) | $M_c$ (kg·m) | $M_e$ (kg·m) | $M$ (kg·m) | $H_t\sqrt{M}$ |
|---|---|---|---|---|---|
| 250 | straight (0°) | 16,266.0 | 9,220.8 | 25,486.8 | 4,600.2 |
| 250 | angle (15°) | 58,728.8 | 39,309.7 | 98,038.5 | 9,022.3 |
| 275 | straight (0°) | 18,626.5 | 10,338.6 | 28,965.1 | 4,998.7 |
| 275 | angle (15°) | 62,831.0 | 41,008.1 | 103,839.1 | 9,464.5 |
| 300 | straight (0°) | 21,157.9 | 11,502.0 | 32,659.9 | 5,413.1 |
| 300 | angle (15°) | 67,185.6 | 42,779.2 | 109,964.8 | 9,932.7 |
| 325 | straight (0°) | 23,852.9 | 12,712.1 | 36,565.0 | 5,843.3 |
| 325 | angle (15°) | 71,751.9 | 44,621.1 | 116,372.9 | 10,424.4 |
| 350 | straight (0°) | 26,714.2 | 13,970.0 | 40,684.2 | 6,289.7 |
| 350 | angle (15°) | 76,527.2 | 46,531.6 | 123,058.8 | 10,938.9 |

#### Total tower steel for 200 km
Weighted index $= 0.9\,(\text{straight}) + 0.1\,(\text{angle})$:

| Span (m) | Weighted $H_t\sqrt{M}$ | Towers $N$ | Relative tower steel $N\times H_t\sqrt{M}$ |
|---|---|---|---|
| 250 | 5,042.4 | 801 | 4.039 × 10⁶ |
| 275 | 5,445.3 | 729 | 3.970 × 10⁶ |
| 300 | 5,865.1 | 668 | 3.918 × 10⁶ |
| 325 | 6,301.4 | 617 | 3.888 × 10⁶ |
| 350 | 6,754.6 | 573 | 3.870 × 10⁶ |

**Answer: the most economical span is 350 m** (573 towers: about 516 straight and 57 angle towers). The curve is flat beyond 300 m, so 325–350 m spans are nearly equally economical.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the conductor working tension found?
   A. UTS x FOS
   B. UTS / FOS = 8000/2 = 4000 kg
   C. UTS - FOS
   D. Wind load x span
   Answer: B. Working tension is the ultimate strength divided by the factor of safety. (Key point: Working tension)

2. For a 15° angle tower, the deviation load per wire is:
   A. 2 x 4000 x sin 15° = 2070 kg
   B. 2 x 4000 x sin 7.5° = 1044 kg
   C. 4000 x sin 7.5° = 522 kg
   D. 4000 kg
   Answer: B. F = 2T sin(θ/2) with θ = 15° gives 1044.21 kg. (Key point: Deviation load)

3. What loads act on the 90% straight towers?
   A. Wind on conductors and earth wire only, no deviation load
   B. Deviation load only
   C. Both wind and 15° deviation
   D. No load
   Answer: A. At 0° deviation the tension components cancel, leaving wind only. (Key point: Straight tower loading)

4. Why is the 325-350 m range nearly equally economical?
   A. The total steel curve flattens beyond 300 m (350 m is lowest, saving small)
   B. Both have identical tower counts
   C. Sag is equal
   D. Angle towers are not used
   Answer: A. Total relative steel falls from 3.888 to 3.870 x 10^6, so 350 m is lowest with only a tiny gain. (Key point: Result)

## Close variants set in other exams

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