why power factor is increased as increasing load,...? 1

[polkadot]

1.why the power factor of induction motor increased as increasing load...?
2.as i know the inductive-reactive is incresed as increasing load(XL= 2*Pi*f*L).
XL = reactive-inductive
pi= 3.14
f= rotor frequecy
L= inductancies
increasing inductive-reactive cause decreasing power factor.
how we can understand this opposite,..? which is the dominant,...?

3. why the current is increased as increasing load,..?

thanks for the respons....

 

[electricpete]

I think the question is best answered from the equivalent circuit.

As a very rough approximation neglecting leakage reactance, there are two parallel branches:
1 - One has the magnetizing reactance Xm which draws a constant inductive magnetizing current, regardless of load.
2 - The other branch has a load resistance R2/s, which draws a load-dependent resistive current which supplies the motor real output power (and rotor losses).

As load increases, the 2nd branch resistive current increases, inductive current stays the same. The sum of these two increases in magnitude and becomes more resistive (higher power factor).

What this simplified approximatte version of the equivalent circuit is telling us is that the inductive current necessary to establish the fields is constant, and the resistive current necessary to supply real output power increases with load.

 

[jbartos]

Suggestion: Reference:
1. Gordon R. Slemon "Magnetoelectric Devices, Transducers, Transformers, and Machines," John Wiley and Sons, 1966,
par 5.4.1 Current, Torque and Speed Relations (for example)
Zr'=Rr'/s + jwsyn x LLr'
PF~(Rr'/s)/[(Rr'/s)**2 + (wsyn x LLr')**2]**(.5)
if you increase the load, the slip s decreases; therefore, PF (Power Factor) increases.

 

[Marke]

Good explanation there electricpete. One of those problems that is easiest explaned using diagrams, and certainly an eye on the equivilent circuit is the way to go. Draw it out first, then the real and reactive current vectors and it should become clear. The secret is that the magnetising current is essentially constant, drops very slightly with load, and the load current changes with shaft load. There is a small reactive component that increases with load, but this is small. (Had someone recently trying to tell me that the iron loss quadrupled under no load conditions, and that was why an energy saver worked so well!!)
I have a page on my web site that may also be of interest,

http://www.lmphotonics.com/pwrfact.htm

Best regards. Mark Empson

http://www.lmphotonics.com

 

[infomaniac]

Electricpete, good explanation....just they way I would have explained it.

 

[infomaniac]

Electricpete, good explanation....just they way I would have explained it (along with some vector example).

 

[jbartos]

Suggestion: Visit

http://www.jmpco.com/tech350.htm

http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm

http://www.fpl.com/savings/energy_advisor/EA-30.html

etc. for more info