Premium Motor Upgrade Energy Savings
The annual energy and dollars a premium-efficiency motor saves over the one it replaces.
Example
You enter
- Motor horsepower 50
- Load fraction (0-1) 0.75
- Existing motor efficiency (0-1) 0.9
- Premium motor efficiency (0-1) 0.945
- Annual run hours 4000
- Electricity rate ($/kWh) 0.12
You get
- Kw standard 31.08
- Kw premium 29.6
- Annual saving 710
Checked against published worked examples from: U.S. Department of Energy (AMO).
Details, formula, and sources
ScopeInput kW = HP x 0.746 x load / efficiency, so the saving = (kW at the old efficiency - kW at the new) x run hours x rate. A 50 hp motor at 75% load going 90% -> 94.5% over 4,000 hr at $0.12 saves about $710/yr; at full load the same swap saves about $947, because the fixed efficiency gap acts on more power. Energy charge only; the utility tariff and any rebate change the payback. A screening estimate, not a metered M&V.
Formula or tablekw = HP x 0.746 x load / efficiency; annual_saving = (kw_standard - kw_premium) x hours x rate.
Source and editionPremium-motor energy-saving comparison, standard motor-efficiency retrofit practice (NEMA Premium / DOE motor efficiency levels) by name; first-principles input-power relation.
Free accessThe input-power relation (HP x 0.746 / efficiency) and the saving arithmetic are first-principles; NEMA Premium efficiency levels are published free.
What governsEstimate. AHJ and licensed professional govern.
API field nameshp, load, eff_standard, eff_premium, hours, rate_kwh, kw_standard, kw_premium, annual_saving
- Input power input kW = HP x 0.746 x load / efficiencyfirst-principles motor relation
- Saving (kW at old eff - kW at new eff) x run hours x rateenergy-charge arithmetic
- Energy only excludes demand / TOU / power-factor charges and rebatesscope of this tile