# Power factor meaning on induced motor

Let say

A 440-V 60Hz three-phase star-connected four-pole induction motor developed 10 hp at 1745. The input power is 10kw at 0.85 lagging power factor. the no load input power is 0.6kw

My question is that what does the power factor mean in this sentence? What is this power factor value derived from and the meaning of it ?

## 3 Answers

First things first there is something called Displacement Power Factor This is basically related to the phase difference between the voltage applied and the current drawn $$DPF = \cos( \Theta )$$ What this means for your case is if 10kW of real power is being consumed then there is 11.76kVA of apparent power (industry are charged based upon their VA consumption as oppose to W consumption hence why PF/DPF is important)

For linear loads and an induction machine will appear as a linear load: PowerFactor == DisplacementPowerFactor

For non-linear loads (rectifier-based systems etc) the disctinction between DPF and PF becomes important. PowerFactor takes into account harmonic distortion

$$PowerFactor = DisplacementPowerFactor * DistortionPowerFactor$$ $$Power Factor = \cos( \Theta ) * \tfrac{1}{\sqrt{1 + THD_{i}^{2}}}$$

So you can see for a linear load the THD = 0 and thus the DistortionPowerFactor = 1 and thus PowerFactor = DisplacementPowerFactor and is the cosine of the phase difference between the voltage and current

If you apply a sinewave to an inductor or capacitor, current flows but no power is taken from the sinewave generator because, for inductive components, current leads or lags voltage by 90º. Power factor is cos(phase angle) and so for a resistor (phase angle is zero), power factor = 1. For an inductor PF = 0 because cos(90º)=0.

For an induction motor the power factor is somewhere between 0 and 1; the zero end of the scale is due to leakage inductances in the machine and the unity end of the scale is because the motor needs real power to turn the shaft and pump water or turn a wheel.

Power Factor is defined as the Real Power divided by the Apparent Power, where
$$\mathrm{Real\,Power} = \frac{1}{T} \int_T v(t) i(t) \mathrm{d}t\\ \mathrm{Apparent\,Power} = V_{rms} \times I_{rms}$$ and T is one period. For sinusoidal voltages and currents, this calculation reduces to the cosine of the phase angle between the voltage and current. Lagging power factor means the current is lagging the voltage, as would be the case for an inductive load.