motor power factor

[ShaunE]

Can anybody give a brief practical explanation of motor power factors / phase diagrams etc.

This is my understanding so far: for a purely inductive load the current peaks 90 degrees after the voltage peak on the sin wave and this would give a power factor of 1. The problem is that motors aren't purely inductive loads and so the 90 degrees goes out the window. This affects the power wave and means that more work goes into the motor in terms of V & I than you get out of it in KW.

Can anybody explain further?
What are the practical implications?

 

[MerleH]

You are close. Actually, motors are not close to pure inductive loads, but are close to pure resitive load. A pure resistive load has a power factor of zero. That means that all of the current is in phase with the voltage. Motors however have some inductive loading. The inductive loading draws current 90 degrees out of phase from the voltage. The plant power system needs to be designed to deliver the in-phase load and the inductive load, even though only the in-phase load is doing work.

The power factor is the ratio of the in-phase current to the total current, which is equal to the tangent of the phase angle.

The electric utilities frequently charge a penalty for low power factor, since it puts increased loads on their lines.

Sometimes capacitor banks are added, which draw power 180 degrees out of phase from the inductive motor loads. This balances out the inductive current, increasing the power factor. This results in a more efficient power distribution.

See

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

 

[electricpete]

Good comments. I expect a typo:
"A pure resistive load has a power factor of zero"
Should be 1.

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[rbulsara]

Also in the original question:

Purely inductive (or capactive) loads will have power facor of 0 not 1 (Cosine of +90 or -90 degrees =0).

Purely restive load will have pf of 1 as electricpete indicated. (Cosine of 0 degree=1).

Real Power (watts) =VI*Cos(phase angle). = "work"

A purely inductive (or capacitive) load will consume zero real power as cosine of 90 is always zero but it draws current in the line causing I2R losses in conductors and tranforers etc feeding the motor loads.

The line power factor does not affect the 'work' done (kW used) by the motor but whatever inductacen it has causes extra current in the line as stated above.

Again the PFC does not correct the pf of the motor but that of the system on the line side of the connection of the PFC and thus reducing the copper losses in the lines and reducing kVA demand of the source.

 

[MerleH]

Yes, of course, the power factor of a resitive load is one.

Also I said the tangent of the phase angle when I should have said the cosine.

 

[aolalde]

Power Factor: A measurement of the time phase difference between the voltage and current in an A.C. circuit. It is represented by the cosine of the angle of this phase difference. For an angle of 0 degrees, the power factor is 100% and the volt/amperes of the circuit are equal to the watts.

 

[jbartos]

Comment on rbulsara (Electrical) Apr 2, 2004 marked ///\\The line power factor does not affect the 'work' done (kW used) by the motor but whatever inductacen it has causes extra current in the line as stated above.
///The line power factor affects the line voltage drop vector. This in turn causes the motor terminal voltage to be affected. The higher voltage at the motor terminals causes the motor output kW increase and vice-versa.\\

 

[ShaunE]

Thanks Guys

I think I have it now:

For purely resistive loads the power factor is 1. This means that VI in = KW out.

For motors there is some inductive load (which is work done to maintain magnetic field)

The inductive load pulls the voltage and current out of phase so for any instantaneous value of VI you dont get peak V * peak I hence power output is reduced.

therefore Work out = V*I*PF, where pf = cos phase angle.

for loads with an inductive component the voltage leads the current and for capacitive loads voltage lags.

I have also found out that the power factor depends on the % load because the 'wasted' emf work is constant and so has less impact at higher loads. I am guessing that the motor manufacturers state pf at full load.

 

[jbartos]

Comment on the previous posting marked ///\\
I have also found out that the power factor depends on the % load because the 'wasted' emf work is constant and so has less impact at higher loads. I am guessing that the motor manufacturers state pf at full load.
///Please, would you clarify "...the 'wasted' emf work is constant..." in terms of the emf work waste in the rotor reactance through which is the rotor current, dependent on the motor load, passing.\\\

 

[ShaunE]

Refer to:

http://www.lima-int.co.uk/whatispfc.htm

The web page above refers to the 'wasted' emf production power as KVAR or Reactive Power:-

"The KVARs required for an induction motor are the same whether the motor is loaded or not. Therefore when the motor is not loaded the power factor is low, and as the motor load increases the power factor improves..."

 

[rbulsara]

Shaune:
your previous statement that because of a lower p.f., power output is 'reduced' is not accurate.

Motor will put out kW as required by the load as long as it is within its rating. By your own admisison the kVAR remains virtually constant. Therefore the actual kW drawn and kVAR will determine the final p.f of the motor, p.f.=kW/kvAR, and not other way around, that is p.f. does not dictate how much kW a motor will produce once its design (kvAR) is fixed.

 

[ShaunE]

to Rbulsara

Yes I agree, my statement wasn't correct, I should have said efficiency is reduced not power output.

Thanks

Shaun

 

[jbartos]

Comment on rbulsara (Electrical) Apr 5, 2004 marked ///\\Therefore the actual kW drawn and kVAR will determine the final p.f of the motor, p.f.=kW/kvAR, and not other way around, that is p.f. does not dictate how much kW a motor will produce once its design (kvAR) is fixed.
///Please notice that:
PF=kW/kVA=(Active Power)/(Apparent Power)
or
PF=cos(arctan(kVAR/kW))
Reference:
IEEE Std 100 Dictionary\\\