Formula
\(P_{in} = \frac{P_{out}}{\eta} \qquad S = \frac{P_{in}}{\cos\varphi} \qquad I = \frac{S}{\sqrt{3}\,U} = \frac{P_{out}}{\sqrt{3}\,U\,\eta\,\cos\varphi} \qquad Q = \sqrt{S^2 - P_{in}^2} \qquad P_{loss} = P_{in} - P_{out}\)
Variables
| Symbol | Meaning | SI unit |
|---|---|---|
| \(U\) | line-to-line voltage | V |
| \(P_{out}\) | rated mechanical output at the shaft | W |
| \(\eta\) | efficiency at full load | – |
| \(\cos\varphi\) | power factor at full load | – |
| \(P_{in}\) | active electrical input | W |
| \(I\) | line current | A |
| \(S\) | apparent power | VA |
| \(Q\) | reactive power | var |
| \(P_{loss}\) | losses dissipated as heat | W |
Method and standards
Balanced three-phase power relations, with efficiency and power factor as defined in IEC 60034-1. The output may be entered in kW or horsepower (1 hp = 745.70 W mechanical; 1 metric hp = 735.50 W).
Assumptions and limits
- Balanced three-phase supply at rated voltage and frequency.
- Full-load values of efficiency and power factor; both change with load.
References
- IEC 60034-1 (rating and performance definitions)
- IEC 60034-30-1 (efficiency classes)
Questions and answers
Why is the current higher than P/(√3·U)?
The shaft output is only part of what the motor draws. Losses raise the active power to P/η, and the power factor means the current carries reactive power as well, so I = P/(√3·U·η·cosφ).
Which efficiency and power factor should I use?
The nameplate or data-sheet values at full load. Both fall at part load; for a motor running at 50 % load use the part-load values from the manufacturer's curve.
Is this the starting current?
No. Direct-on-line starting current is typically 6 to 8 times the full-load current for a few seconds.