- Asked 2/20
- 2075 Bhadra · 10 marks
Most economical span from sag table
Short answer
Using wind load on conductors and Ryle's formula for tower weight, tower steel per km is computed for each span. The 250 m span gives the lowest relative tower cost, making it the most economical span.
For the design of most economical span of a high voltage transmission line, following data are available for a particular conductor. Select the most economical span. Given that: conductor diameter = 28 mm, Cross sectional area = 463 mm², UTS = 15910 kg, Wind pressure = 100 kg/m², Factor of safety = 2.
| Span (m) | Maximum sag (m) |
|---|---|
| 250 | 3.24 |
| 275 | 3.88 |
| 300 | 4.57 |
| 325 | 5.32 |
| 350 | 6.12 |
[Figure: conductor arrangement/air clearance on the tower — three conductors with vertical spacing of 2 m between the top and middle and 2 m between the middle and bottom; the top and bottom conductors on one side and the middle conductor on the other, with a horizontal width of 8.7 m across.] Assume: All towers are straight line towers. The air clearance is shown in figure. Minimum ground clearance is 6 m. Neglect the effect of ground wire.
The tower height is built up from the ground clearance, the sag and the conductor spacing. The tower weight is then found from the overturning moment of the wind load with Ryle's formula, and the total tower steel per km is compared.
Tower heights
With 6 m minimum ground clearance and conductors 2 m apart vertically (from the figure), and the ground wire neglected:
| Span (m) | Max sag (m) | |||
|---|---|---|---|---|
| 250 | 3.24 | 9.24 | 11.24 | 13.24 |
| 275 | 3.88 | 9.88 | 11.88 | 13.88 |
| 300 | 4.57 | 10.57 | 12.57 | 14.57 |
| 325 | 5.32 | 11.32 | 13.32 | 15.32 |
| 350 | 6.12 | 12.12 | 14.12 | 16.12 |
Working tension kg. All towers are straight-line towers, so there is no deviation load. is needed only to check that the given sags are consistent with it. The 8.7 m horizontal width is the same for every span, so it does not affect the comparison.
Method. For each span, the transverse load at the tower comes from two sources:
- wind on the conductors and earth wires: , taking the wind span equal to the span and using the full projected area;
- the resultant of the conductor tensions at a line deviation : .
These loads act at the conductor heights. Their overturning moment at ground level is
where is the number of conductors at each level and the number of earth wires. The tower weight is found from Ryle's formula
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.
Wind load on each conductor ( kg/m², m)
| Span (m) | (kg) | (kg) |
|---|---|---|
| 250 | 700.00 | - |
| 275 | 770.00 | - |
| 300 | 840.00 | - |
| 325 | 910.00 | - |
| 350 | 980.00 | - |
Overturning moment and tower weight index
. For example, at 250 m: kg·m and .
| Span (m) | Tower type (θ) | (kg·m) | (kg·m) | (kg·m) | |
|---|---|---|---|---|---|
| 250 | straight (0°) | 23,604.0 | - | 23,604.0 | 2,034.1 |
| 275 | straight (0°) | 27,442.8 | - | 27,442.8 | 2,299.3 |
| 300 | straight (0°) | 31,676.4 | - | 31,676.4 | 2,593.1 |
| 325 | straight (0°) | 36,363.6 | - | 36,363.6 | 2,921.4 |
| 350 | straight (0°) | 41,512.8 | - | 41,512.8 | 3,284.4 |
Tower steel per km
| Span (m) | Weighted | Towers per km | Relative cost per km |
|---|---|---|---|
| 250 | 2,034.1 | 4.000 | 8,136.6 |
| 275 | 2,299.3 | 3.636 | 8,361.2 |
| 300 | 2,593.1 | 3.333 | 8,643.8 |
| 325 | 2,921.4 | 3.077 | 8,988.9 |
| 350 | 3,284.4 | 2.857 | 9,384.0 |
Here the sag rises quickly with span (3.24 m to 6.12 m), so both tower height and moment grow faster than the number of towers falls.
Answer: the most economical span is 250 m (lowest relative tower cost, 8,136.6 K per km).
✓ Answer checked by Claude Opus 5.5 (high effort)
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