T/F: feeder average consumption near peak

FALSE.

The ratio of average demand to peak demand of a feeder is its load factor:

$$LF = \frac{\text{Average demand}}{\text{Peak demand}} = \frac{\text{Energy in period } T}{P_{peak} \times T}$$

If average consumption were almost equal to the peak, the load factor would be close to 1, which happens only for a flat, round-the-clock load (e.g. a continuous-process industry).

Justification:

• A distribution feeder mainly supplies domestic and commercial consumers whose use varies strongly through the day: low at night, moderate in the day, and a sharp evening peak (lighting, cooking, TV).
• Typical feeder load factors are about 0.3–0.5 for residential/rural feeders and 0.5–0.7 for mixed urban feeders. So the average demand is usually only 30–60% of the peak.
• Example: a feeder with a 1 MW peak and 9,600 kWh/day has an average of 400 kW, i.e. LF = 0.4.
• This is why the feeder and transformer must be sized for the peak, not the average, and why energy loss is computed using the loss of load factor ($LLF = 0.3LF + 0.7LF^2$) rather than assuming peak loss all day.

A high load factor is desirable (better use of equipment, lower loss per kWh), and utilities try to raise it by demand-side management, but in practice the average is well below the peak.