Economical span, double circuit, five spans

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

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.

 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.