Motor Input Power, Annual Energy, and Cost
The input kilowatts a motor draws at its efficiency (HP x 0.746 x load / efficiency).
Example
You enter
- Rated horsepower 25
- Full-load efficiency (%) 93
- Average load (% of rated) 100
- Run hours per year 4000
- Energy rate ($/kWh) 0.12
You get
- Input power 20.05 kW
- Annual energy 80215 kWh/yr
- Annual energy cost $9626/yr
Checked against published worked examples from: U.S. Department of Energy (AMO).
Details, formula, and sources
ScopeThe annual kilowatt-hours at a given duty, and the energy cost; run it twice at two efficiencies to price a premium-motor retrofit. The energy charge only; the utility tariff governs demand, time-of-use, and power-factor penalties.
Formula or tableinput_kW = HP x 0.746 x (load_factor/100) / (efficiency/100); annual_kWh = input_kW x run_hours; annual_cost = annual_kWh x rate. The 0.746 kW/HP is the mechanical-to-electrical conversion.
Source and editionFirst-principles electrical-input power and the 0.746 kW-per-HP identity.
Free accessFirst-principles physics; no licensed source required.
What governsEstimate. AHJ and licensed professional govern.
API field nameshp, efficiency_pct, load_factor_pct, hours_per_year, rate_usd_per_kwh, input_kw, annual_kwh, annual_cost
- Energy charge only the result is the energy-charge component only and excludes demand charges, time-of-use rates, and power-factor penaltiesutility 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