Dear Sir,

I would like to ask a simple question, When an induction motor in different loading, why the power factor will be change? also, if the power factor is leading, what is the impact of the motor?

Thanks

Accord

 

Not really sure I understand the question but the power factor changes as the motor is loaded because the power is doing real work which shows up as equivalent to resistance because power is consumed. Unloaded, most of the apparent power is returned to the line out of phase which is equivalant to reactance. In fact the major part is the reactance of the windings.

Regards

cww

 

Responding to Accord. K's, simple question:

Here is a simple answer... and by no means the only one:

Consider the motor an R-L circuit. R produces torque and requires "real" power, kW, from the supply. L produces the magnetizing needs of the motor, and requires "reactive" power, kVAr, from the supply.

The resultant or "apparent" power from the supply, kVA, is the vector addition of kW and kVAr, which are in quadrature. And, power factor, PF, is simply the ratio of kW to kVA, or:

PF = kW / [ kW^2 + kVAr^2 ]

Essentially, kW is proportional to load, while kVAr is constant throughout the operating range of the motor. Thus, at low load the magnetizing kVAr is predominant with respect to kW. The converse is true at full load.

Your question about "leading" pf is unclear. An induction motor, by nature of it being inductive, can never have a "leading" pf. Are you referring to the "displacement" power factor associated with a variable frequency drive?

Regards, Phil Corso, PE {Boca Raton, FL, USA} [[email protected]] ([email protected]) {[email protected]}

 

To List,

Earlier post omitted the sqrt function in the denominator. PF formula should be:

PF = kW / sqrt [ kW^2 + kVAr^2 ]

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected]) {[email protected]}

 

Phil Corso, Thank you for your early reply, I fully understand why the motor will have different PF in different loading.

By the way, the motor became leading because some PF correction device was added such as filter or cap bank.

Actually, I am not sure whether the motor in leading or lagging. Normally, when I used PF meter to measure PF, it will show in negative, (for example -0.9), however, after filter added, the meter show 0.9, which is positive, that's why I guess it is in leading.

 

Phil Corso,
sir ,to avoid the pf lagging we r connecting capacitive bank. if pf is maintained below 0.85 means our TNEB get penulty from us due to over current and transmission losses. if we maintain the pf above 1 means what will happen.
plz explain sir.
thanks

 

I just wonder how you managed to have PF "above 1" ? It is close to inventing the legendary perpetuum mobile.
Meir

 

Responding to Anonymous' Sat, Jul 10, 8:09am request for clarification:

Power Factor correction should be treated as a double edged sword:

A) The Good Edge.
Improvement reduces the apparent power (kVA) drawn from a power company. Therefore, they encourage correction by imposing penalties. But, increasing pf above the level at which the penalty is applied, is uneconomic. Additional correction to change a lagging pf to a leading one will only benefit the power company. Your Benefit-to-Cost Ratio (BCR) will surely decrease.

B) The Bad Edge.
It is unclear if you are applying capacitors at the motor level or at the substation level. The former is called individual compensation, while the latter is called group compensation. In either case voltage regulation now becomes a more serious factor to consider. If applied at the motor, then overcompensation could expose the motor to overvoltages. If applied at substation bus, then as load is reduced the bus voltage may increase to dangerous levels.

BTW, the PF equation had been corrected. It should read:

PF = kW / sqrt[ kW^2 + kVAr^2 ]

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])