Finding Power Factor of a 3 phase motor 3

[3phasedaze]

Hello everyone!
I am having a difficult time determining the Power Factor of some older 3 phase motors. The pf is not listed on the data plate. I have looked up formulas about arriving at power factor with KW's and KVA known. I can't figure my KW's correctly because I don't know my power factor, and I can't find the power factor without the KW's. Is there a way I can find all of this out by finding the inductive reactance to actual (measured) resistance ratio of the motor windings? Can the power factor be found by HP or some other listing on the data plate?
I'm really having a tough time of this. It's starting to give me a headache. I would be happy if someone could even direct me to a website that would step me through a formula.
Any help will be greatly appreciated.

Now ya know why I picked 3phasedaze as my user name! HA HA.

 

[bigamp]

If you know hated horsepower (HP), full load current (If) and rated voltage (V) then at full load:

power factor = HP*0.7457/(sqrt(3)*If*V)

The above will give you the power factor at full load. Motor power factor varies with load, power factor tends to get lower with reducing load. If you do not have motor data, you can only determine power factor by measurement.

 

[3phasedaze]

Thank you very much!! Yes, I have all of that data. I knew there had to be a formula. I believe that will give me what I need. Once I figure out the pf on these motors, ya got any suggestions on what type of capacitors to use to correct it? We are going to correct the power factor at each motor, 3 phase, 3 HP or larger. Up to 25 HP. Totalling up to 15 or 20 motors. Anyway, thanks again.

 

[edison123]

bigamp,

the motor efficiency is missing from your calculation.

pf = kw /[sqrt3*V*I*efficieny]

 

[3phasedaze]

Appreciate that. So, I gather, HP to Kw is HP*.7457?
That makes it seem so simple. I like simple.
Thanks for the help.

 

[alehman]

kW = HP * .7475 / efficiency

 

[edison123]

3phasedaze,

If all you need is to know what power factor correction capacitor should be connected to your motors, I suggest
capacitor (in KVAR) = 1/3rd motor HP. For a 30 HP Motor - 10 KVAR, 10 HP motor - 3 KVAR and so on. In most cases, this simple thumb rule will suffice to bring overall pf to near about 0.9.

 

[3phasedaze]

Thanks again. I'll take that under consideration. Most of these motors are not operating under much of a load. That will have some effect on the capacitor I chose. Won't it?
By the way, does it take 3 capacitors for each 3 phase motor, or are there '3 phase' capacitors for one motor?

 

[edison123]

pf correction capacitors do not have anything to do with the actual "load current". They are used to offset the no load magnetizsing current, which is normally 1/3rd of the rated load current. So whether your motor is loaded or not, the no load current remains and these pf capacitors have to be in line (preferably across the motor terminals) to improve your overall pf.

You use 3-phase capacitors (having three terminals)which are standard products.

 

[3phasedaze]

edison,

Great! That makes my task a lot easier, and less time consuming. You've been more than helpful, and very patient.

3pd

 

[jOmega]

alehman,

Allow me to make a couple clarifications to your Aug 18, 2003 post .....
"kW = HP * .7475 / efficiency"

1) The constant is .745699 which is normally rounded up to .746

2) When converting kW (a unit of power) to HP (a unit of power) .... efficiency is not a factor.

Where it sometimes gets muddy in these discussions is the failure to differentiate between input and output terms.

The unit of power on a motor nameplate is that of the motor mechanical shaft power.

Motor electrical input power in kW ÷ 0.745699 = electrical input power in HP units.

Likewise.....motor shaft mechanical power in kW ÷ 0.745699 = motor shaft mechanical power in HP.

Now then....to the issue of efficiency...

When determining the Output Shaft mechanical power from the input elect power, you need to use efficiency in the calculation as you correctly stated.

I raise this distinction because the original question was that of determining Power Factor which is a factor used in determining the input electrical power.

As you correctly stated, the INPUT electrical kW is determined by multiplying the OUTPUT shaft mechanical HP by the conversion constant (0.746) and dividing by motor efficiency.

 

[jbartos]

Suggestion: Voltmeter, Ammeter, and Wattmeter measurements will give you the motor input Power Factor. There is also the power factor meter available on the market.

 

[jbartos]

Suggestion: Visit

http://www.designnotes.com/CIRCUITS/pwrfact.htm

http://www.rudrashakti.com/products.htm

http://www.tpub.com/doeelecscience/electricalscience2169.htm

etc. for more info

 

[alehman]

jOmega,
Thank you for the clarification on this confusing point. You are absolutely correct.

1. My apologies for the dyslexia. '.7475' should have read '.7457'

2. The preceeding discussion was regarding the relationship between nameplate (shaft output) horsepower and input kW which must include the efficiency.