FISEKON GmbH · Fischer Elektrokonstruktion

Calculators

Motor efficiency, losses and cost

A motor costs once to buy and every year to run. At six thousand operating hours the electricity cost usually exceeds the purchase price within the first year. Then efficiency decides, not the quoted price.

Input

From the nameplate or data sheet, at the actual load point.

For example that of a motor one efficiency class higher.

Result

Power drawn
—kW
Losses as heatThis heat has to leave the room.
—kW
Electricity cost per year
—€
Annual saving with the comparison motorEnergy only. Purchase and installation weigh against it.
—€
Energy saved per year
—kWh

A guide, not a design to standard. We give no warranty for the correctness of the results. The applicable standards and case-by-case verification govern.

Result as PDF

Where should we send the PDF?

You receive inputs, result and formula as a PDF by email, to file or pass on.

The grey values are an example. Simply enter your own.

Matching services

Calculating is the easy part.

A formula gives you a number. Designing a plant also demands installation method, grouping, discrimination, the standards in force and a look at the installed base. That is what we take on.

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Method and questions

Motor efficiency

What the motor draws, what stays as heat, and what a better one would save.

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Worked example

The default case the calculator starts with, worked through once. Change the values above and it recalculates immediately.

Inputs

Shaft power delivered
30 kW
Efficiency of the existing motor
91 %
Efficiency for comparison
94 %
Operating hours per year
6,000 h
Electricity price
25 ct/kWh

Result

Power drawn
33 kW
Losses as heat
2.97 kW
Electricity cost per year
49,451 €
Annual saving with the comparison motor
1,578 €
Energy saved per year
6,313 kWh
Formula and standards

Formula

  • P_in = P_out / η
  • P_loss = P_in − P_out
  • W = P_in · t
  • Saving = P_out · t · (1/η_old − 1/η_new)

Assumptions and standards

  • Efficiency applies at a particular load point. At partial load it is lower, markedly so below roughly half rated power. An oversized motor runs permanently in the poor region.
  • The losses are to be treated entirely as heat in the room. For motors in enclosures or confined spaces they belong in the heat balance.
  • Only energy is calculated. Purchase, installation and any changes to coupling or foundation are not included.
Is replacing a running motor worthwhile?

Usually only at high operating hours. Weigh the annual saving against motor, installation and downtime. Under continuous operation, stepping up one efficiency class often pays back in two to four years. At two thousand hours a year it rarely does.

Why is an oversized motor expensive?

Because efficiency collapses at partial load and the power factor with it. A motor running permanently at thirty percent of rated power consumes more than a correctly sized one and additionally loads the network with reactive power.

Where do the losses go?

Entirely into heat, generated in the winding, the core, the bearings and the fan. For a 30 kW motor at 91 percent efficiency that is roughly three kilowatts of continuous heating in the room.

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