Add up every load on a standby panel and you get a number that will never happen. The lift is not running while the boiler is firing while every pump is at full flow. Sizing a generator to that total buys a machine that spends its life lightly loaded, wet-stacking and burning fuel for nothing.

$P_g = P_c\, D_f (1 + m)$ — P-g equals P-c D-f times one plus m. $P_c$ is the connected load, the honest sum of every nameplate on the panel, in kilowatts. $D_f$ is the demand factor, the fraction of that total genuinely running at once — a bare decimal between 0 and 1, and 0.7 is a common engineering judgement rather than a measurement. $m$ is the spare-capacity margin, also bare: 0.25 means twenty-five percent left for growth, and 0 means none. $P_g$ is the continuous rating the set must carry, in kilowatts.

The measured route to $P_c$ is the honest one. Clamp the feeder and use $P = \sqrt{3}\, V_L I_L\,\mathrm{PF}$ — $V_L$ the line-to-line voltage, $I_L$ the line current, PF the logged power factor. Multiply amps by volts and $\sqrt{3}$ and you have kVA; multiply by the power factor as well and you have kW. Quoting the first as though it were the second is the classic overspend in this whole chapter.

Two cautions from the field. Enter $D_f$ as 0.7, not 70 — a demand factor above 1 says more runs than is connected, which cannot happen. And on a motor-heavy site the running load is often not what sizes the set at all: the starting inrush from the largest motor is, because an alternator has to hold its voltage through that inrush. The kW answer is where you start, not where you stop.