Calculate electric motor heat loss (heat dissipation) in W and Btu/hr from power and efficiency or volts and amps, with typical efficiencies by motor size.
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Electric Motor Heat Loss Formula
The calculator uses one of three formulas, depending on the selected power or supply method.
Heat loss from rated power and efficiency:
Room heat gain (depends on what sits in the room):
Heat loss from measured amps:
- Qloss = motor heat loss (kW)
- Qroom = heat dumped into the room (kW)
- Pinput = electrical power drawn by the motor (kW)
- Poutput = shaft or nameplate output power (kW)
- eta = motor efficiency (decimal, e.g. 0.85)
- V = line voltage (V)
- I = line current (A)
- PF = power factor (0 to 1)
Select shaft output, electrical input, three-phase AC, single-phase AC or DC. Optional room-boundary settings distinguish motor loss from shaft-work heat gain. Full electrical input becomes room heat only when the motor and all shaft work dissipate in that space; energy carried away by fluid, air or products must be accounted for separately.
Typical Values
These broad illustrative ranges are not motor ratings. Use manufacturer data or measurements at the actual operating point; efficiency and power factor depend on load, speed, design and supply.
| Motor size | Typical efficiency | Typical power factor |
|---|---|---|
| Under 1 kW (under ~1.3 hp) | 70 to 80% | 0.70 to 0.80 |
| 1 to 3 kW | 75 to 85% | 0.78 to 0.85 |
| 3 to 15 kW | 85 to 90% | 0.82 to 0.88 |
| 15 to 150 kW | 90 to 94% | 0.85 to 0.90 |
| Over 150 kW | 93 to 96% | 0.87 to 0.92 |
| Output | Heat loss at 85% eff | Heat loss at 90% eff | Heat loss at 94% eff |
|---|---|---|---|
| 1 kW | 176 W (601 Btu/hr) | 111 W (379 Btu/hr) | 64 W (218 Btu/hr) |
| 5 kW | 882 W (3,010 Btu/hr) | 556 W (1,896 Btu/hr) | 319 W (1,089 Btu/hr) |
| 15 kW | 2,647 W (9,032 Btu/hr) | 1,667 W (5,687 Btu/hr) | 957 W (3,267 Btu/hr) |
| 50 kW | 8,824 W (30,107 Btu/hr) | 5,556 W (18,956 Btu/hr) | 3,191 W (10,889 Btu/hr) |
| 100 kW | 17,647 W (60,214 Btu/hr) | 11,111 W (37,913 Btu/hr) | 6,383 W (21,778 Btu/hr) |
Worked Examples
Example 1: 7.5 kW pump motor. Nameplate output is 7.5 kW. Efficiency is 88%. Input power is 7.5 / 0.88 = 8.523 kW. Heat loss is 8.523 − 7.5 = 1.023 kW, or about 3,490 Btu/hr. If the pump and motor both sit in a plant room, the room sees the full 8.523 kW of heat once the fluid energy dissipates back. If only the motor is inside, the room gain is 1.023 kW.
Example 2: From amps. A three-phase motor runs at 400 V, draws 18 A, with PF 0.85 and efficiency 90%. Input power is 1.732 × 400 × 18 × 0.85 / 1000 = 10.6 kW. Heat loss is 10.6 × (1 − 0.90) = 1.06 kW, or about 3,617 Btu/hr.
FAQ
Why does the room gain sometimes equal the full input power? If both the motor and all shaft work ultimately dissipate in the room, the steady heat gain equals electrical input. Equipment location alone is insufficient: energy may leave with fluid, air, products or mechanical work.
Should I use nameplate power or measured power? Nameplate output is a rated operating point, not necessarily actual shaft load. Prefer measured real input power with an operating efficiency estimate; current alone cannot establish load, efficiency or power factor, especially at light load.
Does power factor change the heat loss? Power factor affects the apparent power and the current, but the real power (kW) is what becomes heat. The calculator already uses real power in the amps tab.
How do I convert the result to cooling tons? One ton of cooling equals 12,000 Btu/hr. The result panel shows a thermal-power equivalent. A complete cooling-load calculation, operating conditions and equipment performance are still required; this number alone does not size an AC unit or fan.
