T/F: % power loss equals % energy loss

FALSE.

The percentage power loss is measured at peak load, while the percentage energy loss is averaged over time. Because losses vary as the square of the load, they are not equal.

Justification:

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

Since $LF^2 \le LLF \le LF$, we get $LLF/LF \le 1$, so % energy loss is less than % peak power loss.

Example: % peak power loss = 10%, LF = 0.4. Using $LLF = 0.3LF + 0.7LF^2$:

$$\begin{aligned} LLF &= 0.3(0.4) + 0.7(0.16) = 0.232 \\ \%E_{loss} &= 10 \times \frac{0.232}{0.4} = 5.8\% \end{aligned}$$

So 10% power loss corresponds to only 5.8% energy loss.

The two are equal only when $LF = LLF = 1$, i.e. a perfectly constant load, which never occurs in practical lines. (If constant losses such as corona or transformer core loss are included, those parts have LLF = 1, but the overall figures still differ.)