- Asked 5/20
- 2072 Asoj · 15 marks
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) |
|---|---|---|---|---|---|
| 250 | 5.27 | 12.178 | 14.998 | 17.818 | 28.444 |
| 275 | 6.24 | 13.148 | 15.968 | 18.788 | 29.414 |
| 300 | 7.3 | 14.208 | 17.028 | 19.848 | 30.474 |
| 325 | 8.43 | 15.338 | 18.158 | 20.978 | 31.604 |
| 350 | 9.65 | 16.558 | 19.378 | 22.198 | 32.824 |
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):
| Type | Use | Design deviation θ | Share |
|---|---|---|---|
| A | Tangent / small angle | 0–2° (use 2°) | 80% |
| B | Medium angle | 2–15° (use 15°) | 15% |
| C | Large angle | 15–30° (use 30°) | 5% |
Tower weight is taken from Ryle's formula (as given with IOE papers):
Here is in ft, 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 . So the span with the least tower weight per km is the most economical.
Loads and bending moment
For each span 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 ):
- Each earth wire (2 of them, at height ):
- Bending moment at the base:
Wind on the tower body is neglected because it is about the same for all spans.
The deviation factor is 0.0349 for 2°, 0.2611 for 15° and 0.5176 for 30°.
Sample (L = 250 m, tower A, θ = 2°):
The weighted weight per tower is , and the weight per km is .
Results
| Span L (m) | (kg·m) | (kg·m) | (kg·m) |
|---|---|---|---|
| 250 | 141,249 | 289,083 | 456,814 |
| 275 | 161,916 | 318,524 | 496,211 |
| 300 | 184,822 | 351,019 | 539,585 |
| 325 | 209,947 | 386,366 | 586,529 |
| 350 | 237,662 | 425,116 | 637,801 |
| Span L (m) | (t) | (t) | (t) | Weighted (t) | Weight/km (t) |
|---|---|---|---|---|---|
| 250 | 6.749 | 9.656 | 12.138 | 7.455 | 29.819 |
| 275 | 7.473 | 10.481 | 13.082 | 8.204 | 29.834 |
| 300 | 8.272 | 11.399 | 14.133 | 9.034 | 30.113 |
| 325 | 9.143 | 12.403 | 15.282 | 9.939 | 30.581 |
| 350 | 10.103 | 13.512 | 16.551 | 10.937 | 31.248 |
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)
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