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Relation between LLF and LF

Short answer

Using a two-step load curve, LF = (Pm t + P1(T-t))/(Pm T), LLF = (Pm^2 t + P1^2(T-t))/(Pm^2 T). LLF lies between LF^2 and LF; the empirical relation is LLF = k·LF + (1−k)·LF^2, with typical k=0.3 for distribution feeders.

Derive the relation between LLF and LF considering a suitable load curve.

Load factor (LF) is the ratio of average load to peak load. Loss load factor (LLF) is the ratio of average power loss to the loss at peak load. Because loss is proportional to I2I^2, LLF always lies between LF2LF^2 and LFLF. The usual relation is

LLF=k LF+(1−k) LF2,k≈0.2–0.3LLF = k\,LF + (1-k)\,LF^2, \qquad k \approx 0.2\text{–}0.3

Definitions

For a period TT with peak load PmP_m, and assuming constant voltage and power factor:

LF=1T∫0TP dtPm,LLF=1T∫0TP2 dtPm2LF = \frac{\frac{1}{T}\int_0^T P\,dt}{P_m}, \qquad LLF = \frac{\frac{1}{T}\int_0^T P^2\,dt}{P_m^2}

since loss ∝I2∝P2\propto I^2 \propto P^2.

Suitable load curve

Take a two-step daily load curve. The load is at peak PmP_m for time tt, and at a lower load P1P_1 for the rest of the time (T−t)(T - t).

 P
 Pm |-----+
    |     |
 P1 |     +--------------------+
    |                          |
    +-----+--------------------+--> time
    0     t                    T

Then

LF=Pmt+P1(T−t)PmT,LLF=Pm2t+P12(T−t)Pm2TLF = \frac{P_mt + P_1(T-t)}{P_mT}, \qquad LLF = \frac{P_m^2t + P_1^2(T-t)}{P_m^2T}

Limiting cases

Case 1: the off-peak load is zero (P1→0P_1 \to 0, a short sharp peak).

LF=tT,LLF=tT  ⇒  LLF=LFLF = \frac{t}{T}, \qquad LLF = \frac{t}{T} \;\Rightarrow\; LLF = LF

Case 2: the peak lasts a very short time (t→0t \to 0, a nearly flat load at P1P_1).

LF=P1Pm,LLF=P12Pm2  ⇒  LLF=LF2LF = \frac{P_1}{P_m}, \qquad LLF = \frac{P_1^2}{P_m^2} \;\Rightarrow\; LLF = LF^2

Every real load curve lies between these two extremes, so

LF2≤LLF≤LFLF^2 \le LLF \le LF

General relation

LLF is expressed as a weighted combination of the two limits:

LLF=k LF+(1−k) LF2LLF = k\,LF + (1-k)\,LF^2

Here kk is found from measured load curves of the system. For typical distribution feeders k=0.3k = 0.3 (Buller and Woodrow), giving LLF=0.3LF+0.7LF2LLF = 0.3LF + 0.7LF^2. Some utilities use k=0.2k = 0.2 for urban feeders, giving LLF=0.2LF+0.8LF2LLF = 0.2LF + 0.8LF^2.

Example: for LF=0.5LF = 0.5, LLF=0.3(0.5)+0.7(0.25)=0.325LLF = 0.3(0.5) + 0.7(0.25) = 0.325. The energy loss per year is then Ploss,peak×0.325×8760P_{loss,peak}\times0.325\times8760 kWh.

✓ Answer checked by Claude Opus 5.5 (high effort)

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