Motor Run Hours for an Energy Budget
The run hours an annual energy budget buys.
Example
You enter
- Rated horsepower 25
- Full-load efficiency (%) 93
- Average load (% of rated) 100
- Energy rate ($/kWh) 0.12
- Annual energy budget ($) 5000
You get
- Input power 20.05 kW
- Max run hours per year 2077.7
No published worked example is on file for this calculation. It is checked against a hand-worked derivation of its method and an independent reference test.
Details, formula, and sources
Scopehours = budget / (input_kW x rate), input_kW = HP x 0.746 x load / efficiency. A 25 HP motor at 93% and $0.12/kWh draws 20.05 kW, so a $5,000 budget covers about 2,078 hours. Answers 'how long can I run it' instead of the cost from a set duty. Energy charge only; the utility tariff (demand, time-of-use, power-factor) governs the full bill, so the real hours are fewer.
Formula or tableinput_kW = HP x 0.746 x (load_factor/100) / (efficiency/100); max_hours = cost_budget / (input_kW x rate); annual_kWh = input_kW x max_hours. The inverse of annual_cost = input_kW x run_hours x rate.
Source and editionFirst-principles electrical-input power and the 0.746 kW-per-HP identity, solved for the run hours.
Free accessFirst-principles physics; no licensed source required.
What governsEstimate. AHJ and licensed professional govern.
API field nameshp, efficiency_pct, load_factor_pct, rate_usd_per_kwh, cost_budget_usd, input_kw, max_hours_per_year
- Energy charge only the result is the energy-charge component only and excludes demand charges, time-of-use rates, and power-factor penalties, so the real hours before a bill target are fewerutility tariff governs the full bill
- Efficiency basis efficiency is the full-load efficiency; partial-load efficiency differs and the load factor scales the input power linearly as a first approximationmotor nameplate / manufacturer curve