KVA vs. KW for lighting

[audioaddict]

Since studying for and taking the Electrical Power PE exam recently, I've been more in tune with power factor, kva, kw, kvars.

Was wondering...when calculating lighting and other loads given in real power (kw), should these kw figures be converted to kva to be "pure"?

Assuming a .9 pf for fluorescent light fixtures, one should technically divide the kw by .9 to get kva. It might not add much but for large loads it looks like it will increase the load enough to consider the next higher OCPD and wire size.

Thankfully for motor loads you can get the FLA from the tables in section 430, and multiply by the voltage to get VA directly. If given watts instead, I'd be inclined to convert with a .8 pf or whatever was appropriate.

Most EEs seem to keep this stuff simple either because they forgot about these relationships, don't care, or are ignorant (mostly non EE graduates...designers). It's sad that MEP Electrical Engineering has been reduced to just knowing the NEC and Ohm's law...and knowing about sqrt(3) for 3 phase power.

Watts and KVA are interchangeable to most, yet you cannot even begin to pass the PE exam if you don't understand these relationships WELL.

 

[HornTootinEE]

Does flourescent lighting actually have a .9 pf? I am just curious. I don't do any lighting design. I have always assumed lighting to be pf =1 when doing utility transformer sizing for buildings and such. Seems to hold in that case, but if they truly are .9 pf I'd like to know to adjust. Incandescent I know would be nearly 1, since it's just a really hot resistor.

 

[pelaw]

I have to disagree with your assessment of the EE professionals.

In Building, or systems, design field, we deal exclusively with VA. On the plans, when doing load calcs, such as panelboards, the loads are shown in VA, or kVA. The reason is simple: we ultimately want to get correct amperage to size the components, and VA / V = A gives us that.

In the case of purely resistive loads kW = kVA. This is the case for incadescent lighting, or heaters. In some cases reactive (inductive, capacitive) loads are miniscule compared to the resistive loads within equipment and can be disregarded, such as heaters with digital controls.

But on the plans, it is always kVA, even if only consisting of kW, in which case it would be an equivalent of kVA =sqrt( X kW^2 + j0 kVAR^2) -> kVA = kW.

 

[wilheldp_PE]

In my limited time in the A/E world, we almost always simply interchanged kW and kVA. This was for any lighting (fluor, incand, LED, etc.) and other loads (hand dryers, water heaters, etc.) unless kVA loads were known for those appliances. There were many a 5 kW instantaneous water heater that went on the panel schedule as a 5 kVA load.

 

[Machiavelli999]

I agree with pelaw. KVA should be used because using KW would give you an innaccurate ampere rating. Usually though you won't get in to too much trouble just taking KW - > KVA. But for large loads its very important to use KVA. However, for large loads FLA is usually given.

And yes new fluorescent lighting does have a power factor. It's not a purely resistive load like an incandescent light bulb.

 

[Kuku]

In most cases the power factor will be negligible, so in the grand scheme of things if you are sizing your distribution equipment correctly, it will not matter if you use VA or W.

 

[USFishin]

When you're doing your load calcs (either in a panel or for the service) you multiply your lighting load by 1.25 anyways. So I doubt the PF really makes much of a difference in the long run.

 

[cdcengineer]

See the NEC handbook 220.5 (Blue Text). It states that kVA and kW can be considered equivalent. I do consider PF when looking at large loads or when given kVA