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Motor Efficiency Savings Calculator

Convert the same mechanical shaft output through two entered efficiencies, then compare annual energy charges and simple payback.

115 / COMPARE MATCHED-LOAD EFFICIENCIES

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Starting values are illustrative. Use your own costs, rates and assumptions.

INPUT GROUP 01

Matched actual operating load

INPUT GROUP 02

Energy and project costs

Check the assumptions before acting on a result.

Read the method and example

YOUR CALCULATION

The alternative efficiency reduces modeled annual energy charges at equal shaft output.

Existing electrical input · kW

45

Alternative electrical input · kW

40

Existing minus alternative annual electricity · kWh

10,000

Existing minus alternative annual energy charge

$1,500.00

Incremental cost / positive annual saving · years

2

Keep the calculation

Saving is optional and stays on this device. Unsaved edits are not kept when you leave. CSV includes every input and the current results.

Illustrative inputs loaded. Replace them with your own assumptions.

Put power ratings in matching units

Convert known power units before comparing the same mechanical output. Unit conversion does not apply motor efficiency. Use the power conversion calculator.

How the motor efficiency savings calculator works

Electrical input kW = required mechanical shaft output ÷ motor efficiency fraction. Both scenarios provide the same constant shaft output. Efficiency must be positive and no more than 100%, and should be determined at the actual matching load point.

Annual electricity is unrounded input kW × annual operating hours. Each annual energy charge extends that energy at the same rate and rounds half-up to cents. Saving subtracts the rounded alternative charge from the rounded existing charge. Lower alternative efficiency can produce negative savings and remains visible.

Simple payback divides rounded incremental project cost by positive annual charge saving; zero or negative saving has no calculated payback. The model excludes power factor, demand charges, drives, variable load, starts, process changes, financing and asset life. It does not select a motor or establish electrical capacity. DOE’s motor-system guidance supports comparing system performance at an appropriate load rather than assuming a nameplate efficiency is universal.

Worked example

36 kW shaft output requires 45 kW electrical input at 80% efficiency and 40 kW at 90%. Over 2,000 hours, the difference is 10,000 kWh. At an illustrative 0.15/kWh, annual charge saving is 1,500. An incremental project cost of 3,000 has two-year simple payback. Changing the alternative efficiency to 80% makes the modeled charges equal and payback unavailable.

Sources and assumptions

The source material explains the underlying calculation or pricing structure. Entered prices, costs, schedules and scenario choices are your assumptions; this tool does not obtain supplier quotes or predict actual results.

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