T/F: LLF unity for constant power loss

TRUE.

Loss of load factor is defined as

$$LLF = \frac{\text{Average power loss over period}}{\text{Peak power loss}}$$

If the power loss is constant throughout the period, the average loss equals the peak loss, so $LLF = 1$.

Justification:

• Constant (no-load) losses such as transformer core (iron) loss, dielectric loss in cables, corona loss in fair weather and losses in meter potential coils depend on voltage, not on load current. Since voltage is nearly constant, these losses are the same every hour.
• Their energy loss is therefore $E = P_{const} \times 8760$ kWh/year, i.e. the loss acts at its full value for all hours (LLF = 1).
• Example: a 100 kVA transformer with 200 W core loss loses $0.2 \times 8760 = 1752$ kWh/year whether it is lightly or fully loaded.
• Only variable (copper) losses, which vary as $I^2$, need $LLF < 1$, e.g. $LLF = 0.3LF + 0.7LF^2$.

The statement also follows from the formula: a constant load gives $LF = 1$, and $0.3(1) + 0.7(1)^2 = 1$.