Power factor and motors

[JHZR2]

I'm working a system for humanitarian aid, which is being requested for fielding in Haiti. Trying to verify generator suitability for full-load.

I understand for the most part the mathematics and reason for power factor, but it is not necessarily clear why for determining current requirement for a motor, why power factor and motor efficiency are used in the calculation. Note I'm talking 480v 3phase motors...

Say I'm calculating motor power, iirc, in doing
V x I x 1.732 x pf x motor efficiency.

This would yield shaftwork, not electric power. To determine power or current requirements, the calculation is the same without pump efficiency terms, right?

Also, what is the rough effect on pf due to a 208-480 upconvert transformer, say 150-250kva?

Thanks!

 

[hertzer]

I can't answer your question, but you could post your question over at the garage journal forum and get an answer. A few Electrical Engineers frequently post over there.

http://www.garagejournal.com/forum/forumdisplay.php?f=30

 

[tomcat27]

you could also call your local Grainger branch, and ask for their tech dept.

 

[tom slick]

You may already know this but;
Power factor is a "back force" (back emf) to flow due to an inductor. Perfect power factor is 1.0. Lower than 1.0 means the power supplied in volt-amps needs to be higher. i.e. a pf of .8 means you need to supply 1.2 times the power to overcome the back emf. Capacitors can be used correct for a poor power factor. The wattage of the motor remains the same but because of this back force more power (volt-amps) needs to be supplied to overcome it.

 

[severach]

Back EMF is found in all inductive loads.

http://en.wikipedia.org/wiki/Power_factor

Power factor is the multiplicative correction to real power as the amount of imaginary power the load draws increases. Real power is worth paying for since all of it you get performs work. Imaginary power performs no work but it does waste energy and requires that the conductors be larger to transport it which is why the power company meters are designed to charge you for imaginary power even though they know it performs no work. Correct the power factor at the load and everyone is happy.

Paddle ball is an example of a mechanical system with a power factor of almost zero if you're doing it right.

Proper motor loading is often discussed at DSL Reports.

http://www.dslreports.com/forum/homerepair

 

[XS650]

Originally Posted By: severach
Back EMF is found in all inductive loads.

http://en.wikipedia.org/wiki/Power_factor

Power factor is the multiplicative correction to real power as the amount of imaginary power the load draws increases. Real power is worth paying for since all of it you get performs work. Imaginary power performs no work but it does waste energy and requires that the conductors be larger to transport it which is why the power company meters are designed to charge you for imaginary power even though they know it performs no work.




Private residence power meters (nearly all of then in the US) measure power, not VA. Home owners pay for actual power used, not VA. Industrial meters often charge extra if the PF is below a certain amount, but AFAIK, even they don't charge by VA

 

[Shannow]

When considering AC, a power factor of one means that the volts and the current are in phase , the peaks happening at the same time.

When they get out of phase, the volt and current peaks don't match, with peak voltage occuring either before or after the peak current. The power that is available to do real work is the integral of all of the "slices" of the VxI that make up the two curves, i.e. less than voltsxcurrent... the power factor is the percentage of real power compared to the "apparent" power that VxI would suggest.

So your shaft power is VxIxPFxnx(1.72??).

The motor and cables will feel the apparent power, which will be basically VxI when it comes to heating etc.

Best way of thinking of it is to put in real power, and imagine reactive power as being perpendicular to it. Then the apparent power is the hypotenuse of the right triangle so formed.

We get payed to import or export VARs (Volts Amps Reactive) at the power stations, and adust the generator excitation to move the power factor around to try to keep the grid at PF=1 over as much area as possible.

 

[JHZR2]

well, I think my shaft power is that, with the 1.732 due to the 120 degree shift between phases (thus giving me max). I assume if I take motor efficiency out, and devide to solve for current, this IS the way to determine it, right?

 

[Papa Bear]

"We get payed to import or export VARs (Volts Amps Reactive) at the power stations, and adust the generator excitation to move the power factor around to try to keep the grid at PF=1 over as much area as possible."

Are you operating as a synchronous capacitor at that time ??

 

[Shannow]

Originally Posted By: JHZR2
well, I think my shaft power is that, with the 1.732 due to the 120 degree shift between phases (thus giving me max). I assume if I take motor efficiency out, and devide to solve for current, this IS the way to determine it, right?


1.732 is square root of 3.

Cable Current will be shaft power/(nxPFx1.732xV).

Heating, voltage drops, and losses will relate to that value.

 

[JHZR2]

right.. sqrt of 3. Thanks!