Dependency of energy loss on load factor

Energy loss in a line depends on how the load varies with time, not just on the peak. Since copper loss is proportional to the square of current, the energy loss is linked to the load factor (LF) through the loss of load factor (LLF).

Definitions

For a line of resistance $R$ (per phase) carrying load $P(t)$ at voltage $V$ and power factor $\cos\phi$:

$$\begin{aligned} P_{loss}(t) &= 3 I^2 R = \frac{P(t)^2 R}{V^2 \cos^2\phi} = k\,P(t)^2 \\ E_{loss} &= \int_0^T k P(t)^2 dt = k P_{max}^2 \, T \cdot \underbrace{\frac{1}{T}\int_0^T \left(\frac{P}{P_{max}}\right)^2 dt}_{LLF} \end{aligned}$$

So $E_{loss} = P_{loss,peak} \times LLF \times T$, and $LF = \dfrac{1}{T}\int_0^T \dfrac{P}{P_{max}} dt$.

Limits of LLF in terms of LF

Consider a two-step load: $P_{max}$ for time $t$ and $P_{min}$ for the rest $(T - t)$.

• Case 1, $P_{min} = 0$: $LF = t/T$ and $LLF = t/T$, so $LLF = LF$ (upper limit).
• Case 2, very short peak ($t \to 0$), off-peak $P_{min}$ for nearly all the time: $LF \approx P_{min}/P_{max}$ and $LLF \approx (P_{min}/P_{max})^2$, so $LLF = LF^2$ (lower limit).

Hence

$$LF^2 \le LLF \le LF$$

Real load curves lie between these, giving the empirical relation (Buller–Woodrow):

$$LLF = 0.3\,LF + 0.7\,LF^2$$

Dependency of % energy loss on LF

$$\%E_{loss} = \frac{P_{loss,peak} \times LLF \times T}{P_{max} \times LF \times T} \times 100 = \%P_{loss,peak} \times \frac{LLF}{LF}$$

With the empirical formula, $\dfrac{LLF}{LF} = 0.3 + 0.7\,LF$.

Example: peak loss 8% of peak demand.

Conclusions:

• For a given peak, a lower LF means fewer kWh lost but also fewer kWh sold; the % energy loss is always ≤ % peak power loss.
• For a given energy delivered, a higher LF lowers the peak, and since loss ∝ peak², the absolute energy loss falls.
• Energy loss can never be found by multiplying peak loss by hours; LLF must be used.
• Constant losses (transformer core loss, corona) do not depend on load, so for them LLF = 1.