Transmission Line Formulas: Drop and Losses
Short transmission line formulas: voltage drop and regulation from line resistance and reactance, three-phase line loss and efficiency, with a worked example.
The formulas
| To find | Formula |
|---|---|
| Phase voltage drop | ΔVphase ≈ I × (R × cos φ + X × sin φ) |
| Line-to-line drop | ΔVLL = √3 × ΔVphase |
| Regulation | ΔVLL / VLL × 100 |
| Three-phase line loss | Ploss = 3 × I² × R / 1000 (kW) |
| Efficiency | η = Pdelivered / (Pdelivered + Ploss) × 100 |
R and X are the resistance and reactance of one conductor over the whole length, in ohms, I the line current and φ the load angle for a lagging load. This is the short-line model, which ignores the line capacitance, and it suits lines up to a few tens of kilometres at distribution voltages. Longer lines need the medium or long-line model.
Worked example: 11 kV line, 100 A at power factor 0.9 lagging
Each conductor has R = 0.5 Ω and X = 0.8 Ω over the length of the line.
- sin φ = √(1 − 0.9²) = 0.4359
- ΔVphase = 100 × (0.5 × 0.9 + 0.8 × 0.4359) = 79.87 V, so ΔVLL = √3 × 79.87 = 138.3 V, which is 1.258% of 11 kV
- Delivered power = √3 × 11,000 × 100 × 0.9 / 1000 = 1,715 kW
- Ploss = 3 × 100² × 0.5 / 1000 = 15 kW, so the efficiency is 99.13%
Drop and loss by current
| Current | Line-to-line drop | Regulation | Line loss |
|---|---|---|---|
| 50 A | 69.17 V | 0.6288% | 3.75 kW |
| 100 A | 138.3 V | 1.258% | 15 kW |
| 150 A | 207.5 V | 1.886% | 33.75 kW |
Frequently asked questions
Why does reactance matter as well as resistance?
The reactive part of the drop depends on the load angle. At a low power factor the X term is large, and a line with high reactance drops more voltage.
Is the loss only in the resistance?
Yes. Reactance stores and returns energy and does not dissipate power, so only R appears in the loss.
How can I reduce the drop?
Use a larger conductor, raise the voltage, shorten the line or improve the power factor with capacitors.
Is the approximate formula exact?
It is an approximation, accurate when the drop is small against the voltage. The exact method uses the phasor relation between sending and receiving voltages.
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