Motor Output Horsepower and Load from Measurements
Is that 25 hp motor really working at 25 hp? Estimate its actual shaft output and percent load from field readings, using the methods in the U.S. Department of Energy’s motor guide.
Method reviewed October 26, 2026. Version 1.0.0. How it works
This is an estimate. It uses a simplified method and typical assumptions, all listed below. Use it to plan and to check quotes; equipment should be sized from a full room-by-room load calculation (ACCA Manual J) by a qualified professional.
Result
- Estimated output (shaft) horsepower
- 15hp
- Load (share of rated output)
- 60%
About 15 hp — 60 % of the 25 hp rating.
- DOE cautions that the slip method is the least precise: nameplate full-load speed may be off by up to 20 % of the rated slip.
Show the working
- Load = (synchronous − measured) ÷ ((synchronous − full-load) × (V ÷ V_rated)²) (DOE Equations 5–6)30 ÷ (50 × 1) = 60 %
- Output horsepower = load × nameplate horsepower0.6 × 25 = 15 hp
Three methods, in order of reliability
- Input power: measure volts, amps and power factor, compute input kW, and multiply by the motor’s efficiency at that load. Most reliable.
- Current: compare the measured current with the nameplate full-load amps, corrected for voltage. DOE notes current stops tracking load below about 50 % load.
- Slip: compare how far the speed is below synchronous with the nameplate full-load slip. Simplest and safest to measure, but the least precise.
A worked example from DOE
A 25 hp motor with an 1,800 rpm synchronous speed and a 1,750 rpm nameplate speed is measured at 1,770 rpm. Its slip is 30 rpm out of a full-load slip of 50 rpm, so it is at about 60 % load — roughly 15 hp.
Slip also varies inversely with the square of the voltage, so the calculator lets you correct for a supply voltage that differs from the nameplate.
Worked example
These are the example values the calculator opens with, and the result the engine calculates for them.
Inputs
- Method
- Speed (slip method)
- Nameplate horsepower
- 25 hp
- Nameplate voltage
- 460 V
- Synchronous speed
- 1,800 rpm
- Nameplate full-load speed
- 1,750 rpm
- Measured speed
- 1,770 rpm
- Measured voltage (for voltage correction)
- 460 V
Result
- Estimated output (shaft) horsepower
- 15 hp
- Load (share of rated output)
- 60 %
About 15 hp — 60 % of the 25 hp rating.
Assumptions
- DOE field methods: input power × efficiency, current ratio corrected for voltage, or slip ratio corrected for voltage squared.
- 1 hp = 0.7457 kW.
Limitations
- Each method is an estimate; the power method with a measured power factor is the most reliable, and the slip method the least.
- Working on energized equipment is hazardous: measurements must be taken by qualified people with proper protective equipment.
Frequently asked questions
Why estimate output horsepower?
Lightly loaded motors run less efficiently. Knowing the real load tells you whether a smaller or premium-efficiency replacement makes sense.
What efficiency should I enter for the power method?
DOE notes that assuming the nameplate efficiency is reasonable above about 50 % load.
Is it safe to take these measurements?
Only for qualified people with the right meters and protective equipment. The slip method needs only a tachometer on the shaft.
Sources
- Determining Electric Motor Load and Efficiency — U.S. DOE. Equation 1: three-phase input kW = V × I × PF × √3 ÷ 1,000 (worked example 469.7 V, 37 A, PF 0.763 → 22.9 kW); Equation 2: full-load input kW = hp × 0.7457 ÷ efficiency.
- Determining Electric Motor Load and Efficiency — current, slip and efficiency equations — U.S. DOE. Equation 4: load = (I ÷ I_rated) × (V ÷ V_rated), not useful below about 50 % load; Equations 5–6: load = slip ÷ ((S_sync − S_fullload) × (V ÷ V_rated)²), example 1,800/1,750/1,770 rpm → 60 %, 25 hp → 15 hp; Equation 7: η = 0.7457 × hp × load ÷ P_i; efficiency near nameplate is reasonable above 50 % load.
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