Motors/Generators and Power Factor Relationship

Discussion in 'General Electronics Chat' started by Management, Sep 2, 2009.

1. Management Thread Starter Active Member

Sep 18, 2007
306
0
Hi Everyone,

I know that a Motor/Generator, or I think I know, behaves like an inductor because inductors store energy in their magnetic fields.

What I want to make sure of or ask is does the Motor/Generator have a lagging power factor because it behaves like an inductor? Is that the only reason why?

Also, if I am wrong in my assumption then please tell me why it is a leading power factor.

Thank you.

2. GetDeviceInfo AAC Fanatic!

Jun 7, 2009
1,623
240
The power factor is a relationship between the Apparent power and the real power. And yes, motors are built around induction, so you will have a reactive component which affects that relationship (lagging). You can manipulate the reactive component by adding capacitance to the circuit, shifting the angle towards unity. By adding enough capacitance, you can shift the angle to leading.

3. Management Thread Starter Active Member

Sep 18, 2007
306
0
So that is a yes that they have a lagging power factor?

Based on the fact that they are induction type machines.

4. Management Thread Starter Active Member

Sep 18, 2007
306
0
Can someone add how a generator/motor be made to look like a capacitor depending on what is placed at the terminals?

I know the grid is considered inductive and that generators/motors internally behave inductively but my understanding stops there when it comes to manipulating its behavior what it's power factor based on its terminal loads.

5. GetDeviceInfo AAC Fanatic!

Jun 7, 2009
1,623
240
A utility feeding a service, must be able to deliver according to the Apparent power required by the consumer.

Apparent power(S) is the sum of real power(P) and reactive power(Q) such that S = sqrt(P(sq) + Q(sq)).

The power factor is the relationship between S and P such that pf = P/S.
With a unity powerfactor (pf = 1), then the apparent power equals the real power. This is typical of resistive loads.

When you begin to introduce reactive devices into the ciruit, and according to the formula, your apparent power increases, hence your current requirements.

current through an inductor lags it's voltage, current through a capacitor leads it's voltage. If you have both reactive components of equal value, then your supply current and voltage are inphase, as they cancel each other out. If the reactance due to a capacitor is greater than the reactance due to inductance, then the circuit will have leading current, appearing to be capacitive.

In real world situations, a consumer with large inductive loads, such as a manufacturer with many electric motors, will have high currents draws due to the large amount of apparent power, compared to the real power they use. This produces a low power factor and may be penalized for that.

By adding capacitors to offset the inductive reactance, thier power factor will shift toward unity, thereby reducing thier current demands. Note that thier use of real power may not change. Also note that the current between the inductive load and the power correcting capacitors are still higher than the real power used, but in that section of the circuit only, and not the utility infeed.

Typically, because of cost, pf correction is not taken to unity, but closer to unity. If you overdue it, your capacitive reactance will again begin to drive up your current requirements.

Last edited: Sep 8, 2009
6. marktchaos New Member

Sep 7, 2009
5
0
I think that's an over-simplicifaction - for a start there are several types of motor that can act as generators too. In general a motor under load is going to be pretty resistive (at low frequencies) since most of the power is being _dissipated_ from the POV of the circuit - the motor designer wants it to efficiently convert electricity to mechnical energy after all. At high frequencies it will look very inductive though (ignoring RFI suppression components - these might make it capacitive at RF...)

But its more complex as the back EMF depends on the rotation speed, which is not represented in a simple model of complex impedance

7. rvh002@gmail.com AAC Fanatic!

May 15, 2009
119
2
Absolute nonsense, we are talking about normal ac motors which are all inductive.

8. Management Thread Starter Active Member

Sep 18, 2007
306
0
I understand the adding actual capacitors but what about the excitation of the generator. I don't see how increasing the excitation of the coils add reactive current to the "grid" and would make a generator look more inductive or capacitive to whatever it is connected to.

Are there equations that would explain this? Thanks for the discussion.

9. GetDeviceInfo AAC Fanatic!

Jun 7, 2009
1,623
240

If for instance I have a step up converter with an inductor and a smoothing capacitor, can one determine the value of capacitance and induction by looking at the output?

I suppose if one where to draw on the supply and measure switching frequency, power drawn, and ripple, one could calculate the values.

As a source, the presentation of voltage will differ depending on the 'reactance' or method of generation. A flyback inductor or capacitor will provide rapidly decaying voltages, while a conductor sweeping through a magnetic flux will give a sinsodial waveform which may be commutated or rectified.

I don't think your thinking of voltage change in a time domain, so 'reactance' of the supply is mute.

If your talking about a power source in general, remember that a source is just a potential difference with some current carrying capacity. It is up to the load to determine how the current is drawn in relationship to the voltage. The source itself does not affect the power factor, it's in the load.

10. rvh002@gmail.com AAC Fanatic!

May 15, 2009
119
2
Are you talking about AC or DC motors/generators?

11. Management Thread Starter Active Member

Sep 18, 2007
306
0
If that question is for me then I am asking about AC generators.