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Loss-factor approximation (LSF ≈ 0.3·LF + 0.7·LF²)8 min read

Worked Example: Estimating a Feeder's Annual I²R Loss Cost from Load Factor

Converting a feeder's peak load and load factor into an annual energy-loss estimate — without needing a full year of interval load data.

Scenario

Peak load (individual, pre-diversity)500 kW
Diversity factor1.2
Load factor60%
Line resistance (per phase)0.5 Ω
System voltage11 kV
Power factor0.9
Energy cost$0.12 / kWh

Step-by-step calculation

Step 1: Apply the diversity factor to find coincident peak demand

Coincident peak = individual peak / diversity factor
500 / 1.2
Coincident peak = 416.7 kW

Step 2: Convert load factor to loss factor

Loss factor estimates how loss (which scales with current squared) averages over time, given only the load factor (which is a simple linear average) — using the widely referenced empirical approximation.

LSF ≈ 0.3 x LF + 0.7 x LF²
0.3 x 0.6 + 0.7 x 0.6²
Loss factor = 0.432

Step 3: Find peak current and peak I²R loss

I = P / (√3 x V x PF) Ploss = 3 x I² x R
I = 416,667 / (1.732 x 11,000 x 0.9) = 24.3 A Ploss = 3 x 24.3² x 0.5
Peak current = 24.3 A, peak loss = 0.886 kW

Step 4: Scale peak loss to an annual energy total and cost

Annual loss = peak loss x loss factor x 8760 hours
0.886 x 0.432 x 8760
Annual energy loss = 3352 kWh -> annual cost = $402.20

Result summary

CheckRequirementActualStatus
Peak I²R lossn/a (informational)0.886 kW✓ PASS
Annual energy lossn/a (informational)3352 kWh/year✓ PASS
Annual loss costn/a (informational)$402.20/year✓ PASS
This feeder's technical losses cost an estimated $402 per year — a small figure at 500 kW peak load, but the same method scales directly to much larger feeders where loss cost becomes a real factor in conductor upsizing decisions.

Key insight: Loss factor (0.432) is always less than load factor (0.6) whenever load factor is below 1.0, because loss scales with the square of current — a feeder that runs at partial load most of the time wastes proportionally less energy to resistive loss than one that runs flat-out constantly, even at the same average load factor.

Try it with your own numbers

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Frequently asked questions

Why not just use load factor directly to estimate energy loss?

Because I²R loss is proportional to the square of current, not current itself, so simply averaging loss using the linear load factor would overstate annual losses whenever load varies over time — the loss-factor approximation exists specifically to correct for this squaring effect without requiring a full year of interval (load-duration) data, which most sites don't have readily available.

How accurate is the 0.3·LF + 0.7·LF² approximation?

It's a widely cited empirical fit, not an exact relationship — the true loss factor for a given load factor depends on the actual shape of the load curve over time, which this approximation doesn't know. Where real interval/AMI data is available, computing loss factor directly from the actual load curve will always be more accurate than this formula.

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