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Most economical span, double circuit line

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.

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.

SpanDmax (m)h1 (m)h2 (m)h3 (m)Ht (m)
250 m3.489.7014.5219.3524.84
300 m5.0911.3016.1320.9626.45
350 m7.0213.2418.0622.8928.39

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.

Method. For each span, the transverse load at the tower comes from two sources:

  • wind on the conductors and earth wires: F=p×d×SF = p \times d \times S, taking the wind span equal to the span SS and using the full projected area;
  • the resultant of the conductor tensions at a line deviation θ\theta: Fd=2Tsin⁡(θ/2)F_d = 2T\sin(\theta/2).

These loads act at the conductor heights. Their overturning moment at ground level is

M=nc (Fc+Fdc)(h1+h2+h3)+ne (Fe+Fde) HtM = n_c\,(F_c + F_{dc})(h_1 + h_2 + h_3) + n_e\,(F_e + F_{de})\,H_t

where ncn_c is the number of conductors at each level and nen_e the number of earth wires. The tower weight is found from Ryle's formula

W=K HtMW = K\,H_t\sqrt{M}

KK 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 h1h_1, h2h_2, h3h_3, so nc=2n_c = 2. Two earth wires at HtH_t, so ne=2n_e = 2.
  • Wind pressure p=100p = 100 kg/m², dc=0.02597d_c = 0.02597 m, de=0.0146d_e = 0.0146 m.
  • All towers are designed for θ=5°\theta = 5°:
Fdc=2×6890×sin⁡2.5°=601.08 kgFde=2×3332×sin⁡2.5°=290.68 kg\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)Fc=p dc SF_c = p\,d_c\,S (kg)Fe=p de SF_e = p\,d_e\,S (kg)
250649.25365.00
300779.10438.00
350908.95511.00

Overturning moments and tower weight index

Sample calculation (250 m):

Mc=2(649.25+601.08)(9.70+14.52+19.35)=108,953.3 kg⋅mMe=2(365.0+290.68)(24.84)=32,574.2 kg⋅mM=141,527.5 kg⋅m,HtM=24.84141527.5=9,344.8\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)McM_c (kg·m)MeM_e (kg·m)MM (kg·m)HtMH_t\sqrt{M}
2505° (all)108,953.332,574.2141,527.59,344.8
3005° (all)133,573.438,547.2172,120.510,973.4
3505° (all)163,656.545,519.4209,175.912,984.4

Tower steel per km

Towers per km =1000/S= 1000/S, so the relative cost per km =(1000/S)×HtM= (1000/S) \times H_t\sqrt{M} (in units of KK):

Span (m)Weighted HtMH_t\sqrt{M}Towers per km =1000/S= 1000/SRelative cost per km
2509,344.84.00037,379.4
30010,973.43.33336,578.1
35012,984.42.85737,098.2

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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