Skip to main content

Economical span, double circuit, A/B/C towers

Short answer

Using wind and deviation loads, bending moment, and Ryle's formula, the weighted average tower steel per km is lowest at 250 m (29.82 t/km). 275 m gives 29.83 t/km, but 250 m is the most economical.

To transmit a given amount of power to a given distance, double circuit line with a double earth wire is chosen for which the following design step are completed, compute the most economical span.

Span (m)Dmax (m)H1 (m)H2 (m)H3 (m)Ht (m)
2505.2712.17814.99817.81828.444
2756.2413.14815.96818.78829.414
3007.314.20817.02819.84830.474
3258.4315.33818.15820.97831.604
3509.6516.55819.37822.19832.824

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. The power conductor has maximum working tension of 6000 kg and diameter of 4 cm. The earth wire has maximum tension of 2000 kg and diameter of 2 cm. Assume 80% as tower A, 15% as tower B and 5% tower C and the wind force is 100 kg/m².

The most economical span gives the least average tower weight (cost) per km. Here the line has a mix of tower types, so a weighted average tower weight is used.

Tower types assumed (usual IS 802 classification):

TypeUseDesign deviation θShare
ATangent / small angle0–2° (use 2°)80%
BMedium angle2–15° (use 15°)15%
CLarge angle15–30° (use 30°)5%

Tower weight is taken from Ryle's formula (as given with IOE papers):

Wt=0.0016 HtM×FS  tonneW_t = 0.0016\,H_t\sqrt{M\times FS}\ \ \text{tonne}

Here HtH_t is in ft, MM 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/L1000/L. So the span with the least tower weight per km is the most economical.

Loads and bending moment

For each span LL 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 h1,h2,h3h_1, h_2, h_3):
Fc=p dc L+2Tcsin⁡θ2=100×0.04 L+2×6000sin⁡θ2 kgF_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 HtH_t):
Fe=p de L+2Tesin⁡θ2=100×0.02 L+2×2000sin⁡θ2 kgF_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=2Fc(h1+h2+h3)+2FeHtM = 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.

The deviation factor 2sin⁡(θ/2)2\sin(\theta/2) is 0.0349 for 2°, 0.2611 for 15° and 0.5176 for 30°.

Sample (L = 250 m, tower A, θ = 2°):

Fc=1000+6000×0.0349=1209.4 kgFe=500+2000×0.0349=569.8 kgMA=2(1209.4)(44.994)+2(569.8)(28.444)=141,249 kg⋅mWA=0.0016(93.32)141249×0.007233×2=6.749 t\begin{aligned} F_c &= 1000 + 6000\times0.0349 = 1209.4\ \text{kg}\\ F_e &= 500 + 2000\times0.0349 = 569.8\ \text{kg}\\ M_A &= 2(1209.4)(44.994) + 2(569.8)(28.444) = 141{,}249\ \text{kg·m}\\ W_A &= 0.0016(93.32)\sqrt{141249\times0.007233\times2} = 6.749\ \text{t} \end{aligned}

The weighted weight per tower is W=0.8WA+0.15WB+0.05WCW = 0.8W_A + 0.15W_B + 0.05W_C, and the weight per km is W×1000/LW\times1000/L.

Results

Span L (m)MAM_A (kg·m)MBM_B (kg·m)MCM_C (kg·m)
250141,249289,083456,814
275161,916318,524496,211
300184,822351,019539,585
325209,947386,366586,529
350237,662425,116637,801
Span L (m)WAW_A (t)WBW_B (t)WCW_C (t)Weighted WW (t)Weight/km (t)
2506.7499.65612.1387.45529.819
2757.47310.48113.0828.20429.834
3008.27211.39914.1339.03430.113
3259.14312.40315.2829.93930.581
35010.10313.51216.55110.93731.248

The weight per km is lowest at 250 m (29.819 t/km). 275 m is almost equal (29.834 t/km, only about 0.05% more). From 300 m onwards the weight rises clearly.

Answer: the most economical span is 250 m, with an average tower weight of 7.455 t and 29.82 t of tower steel per km. 275 m is practically as good and could be chosen if tower count or terrain favours it.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Transmission and Distribution System Design chapter 4 questions →