Convert VA to Power factor for service panel,

[Breezers612]

I am wiring new restraunt and have 225A panel 3 PH, Load exceeds 80 Percent of rating. Plan checker susgested convert to power factor and it can go to full 225. Do you convert phase by phase or total panel load ? I noticed new code book calls out kVA and not watts as in older books. Is there a formular or do you just use Table
John

 

[rbulsara]

Do you mean the actual current measured exceed the 80%rating of the breaker? (180A for a 225A CB)?

Per NEC you should NOT exceed 180A (80% of 225A) of continuous load (more than 3 hrs), regardless of the power factor.

If this is a calculaion then here are the formulas and you can calculate on your own:

V=line to line Volts, Amp=line current (each phase).

kVA = (1.732*V*A)/1000

kW= kVA*power factor (Assume 0.8 power factor if no other info is available).

If you calculated your loads in kVA, then you can find amps from the first formula. If you calculated your loads in kW, then you need to divide by the p.f. (0.8) to convert kW to kVA and then calculate the amps.

 

[busbar]

Conductor heating is generally a current-based function, so, for instance, using kW-demand history may undersize the proposed equipment. Panelboards don’t care what the load PF is, and voltampere figures consist of real- and reactive-power components.

Voltampere-demand checks will produce larger {or equal, but not smaller} numbers compared to watt-demand measurements, for more safely sizing electrical-distribution equipment.

 

[peebee]

Are you sure "power factor" was the term he used, and not "demand factor"?

 

[jghrist]

Maybe he meant correct the power factor. If you have added up your load in watts at a particular power factor and calculated the current, correcting the power factor to something closer to unity would decrease the current and maybe get it less than 80% of 225A.

Watts = Volts·Amps·Power Factor

You would have to add capacitors to increase the power factor.

Volts·Amps = sqrt(Watts²+(Reactive Load vars - Capacitor vars)²)

With these two equations and some algebra you should be able to figure out how many capacitor vars you need. Keep in mind that you can calculate the Reactive Load vars from the second equation with zero Capacitor vars. I doubt that it would be less expensive than installing a larger panel and conductors, however.

 

[advidana]

No-It would be a lot cheaper to use a 400 amp service. The capactors are expensive to correct pf and this is not the intented application for capactor use.

I have done a lot of restraunts and 225 amps panel is usually just to small an amperage. However use the examples for shown in the NEC code for sizing your load not just 80 percent. If you don't know how to use them hire someone that does to train you- like an master electrician or a professional engineer.