Lighting panel feeder sizing

[Rouse]

I am studying for an exam that I have already taken. This may be a silly question but here goes. When you are asked to size a feeder conductor for a 225a panel used only for lighting with a total connected load of 90,000va, do you size it to the panel or the connected load? I don't remember the voltage or the type of building it is in.

 

[jumper]

I am studying for an exam that I have already taken. This may be a silly question but here goes. When you are asked to size a feeder conductor for a 225a panel used only for lighting with a total connected load of 90,000va, do you size it to the panel or the connected load? I don't remember the voltage or the type of building it is in.

You would size it per the connected load, unless speced otherwise.

I believe voltage is irrelevant.

Lighting loads are calculated per T220.42.

 

[don_resqcapt19]

It has to be sized to supply the load and to be protected by the supply side OCPD.

 

[Rouse]

It has to be sized to supply the load and to be protected by the supply side OCPD.

So you guys are saying the 225a is just extra info I don't need and just size it to the connected load?

 

[bob]

I am studying for an exam that I have already taken. This may be a silly question but here goes. When you are asked to size a feeder conductor for a 225a panel used only for lighting with a total connected load of 90,000va, do you size it to the panel or the connected load? I don't remember the voltage or the type of building it is in.

You size the feeder for the load. If you assume the voltage is 120/208 volts
3 phase 90000 va = 250 amps. Lighting is usually considered a continuous load so you would need to multiply 250 x 1.25 = 313 amps. Going back to your origional question, you can size the conductor to match the ampacity of the panel if it is protected by a 225 amp breaker.

 

[Smart $]

You would size it per the connected load, unless speced otherwise.

I believe voltage is irrelevant.

Lighting loads are calculated per T220.42.

You size the feeder for the load. If you assume the voltage is 120/208 volts
3 phase 90000 va = 250 amps. Lighting is usually considered a continuous load so you would need to multiply 250 x 1.25 = 313 amps. Going back to your origional question, you can size the conductor to match the ampacity of the panel if it is protected by a 225 amp breaker.

Voltage is very relevant! A 225A panel with 90kVA lighting load would be an overloaded panel (by 88 amperes) if supplied by a 208Y/120 source and an occupancy which does not qualify for any lighting load demand, regardless of the feeder being sized to carry the load. Supplied with 480Y/277 would make a world of difference: 90kVA?(480?√3)=108A

 

[Rouse]

You size the feeder for the load. If you assume the voltage is 120/208 volts
3 phase 90000 va = 250 amps. Lighting is usually considered a continuous load so you would need to multiply 250 x 1.25 = 313 amps. Going back to your origional question, you can size the conductor to match the ampacity of the panel if it is protected by a 225 amp breaker.

I believe it was 277v because several lighting questions were 277v. I think I know were my problem was. Tell me if you think this is right, 90,000va @ 277v = 188 amps. What I think I did wrong was not multiply this amount by 125% for it being a continious load. At that point I was not sure to size it by the load of 188 amps or the panel at 225 amps. But If I had multiplied the lighting load by 125% that would make it 235 amps so I would want to size it by 235 amps? (which if THW would be a 250 kcmil?)

 

[Dennis Alwon]

If the total connected load is 188 amps and the load is continuous then you have a load of 235 amps --as you stated. Now article 215. 3 states the feeder OCPD must be not less than the sum of the continuous plus the non continuous load. This is a problem since you would have to protect this 225 amp panel with a 250 amp OCPD. I don't believe you can do this. 408.36

There is an excep.- if the ocpd is rated for 100% of its rating then you do not have to figure the load as continuous.

Not given any more info then I would have to say the feeder could be 188 amps or 3/0 copper at 75C

Now do your formula with 90000/ 480*1.73 Sorry I just caught that mistake.

 

[charlie b]

Tell me if you think this is right, 90,000va @ 277v = 188 amps.

It is not right, as Dennis has pointed out. You are dealing with a three phase system. Current flowing out on any wire will return on one or both of the other two wires. The voltage driving this current is the line-to-line voltage of 480. So the correct calculation of current is 90,000 divided by (480 x 1.732), or about 108 amps. You then multiply that value by 1.25, and get 135 amps. So the MINIMUM feeder size (which is what a test question is likely to be asking for) is based on 135 amps.

That said, in a real world situation, I would at least consider sizing the feeder for the full 225 amps of the panel's rating. I might not choose that feeder size, but I would consider it. Tests, however, are asking for specific things, and that is what you must base your answer upon.

 

[jumper]

Voltage is very relevant! A 225A panel with 90kVA lighting load would be an overloaded panel (by 88 amperes) if supplied by a 208Y/120 source and an occupancy which does not qualify for any lighting load demand, regardless of the feeder being sized to carry the load. Supplied with 480Y/277 would make a world of difference: 90kVA?(480?√3)=108A

Let me rephrase what what I meant.

"Voltage is irrelevant in applying the demand factors of T220.42. A higher voltage will result in a lower amperage, yet the demand factor for the occupancy will stay the same."

I did not do the math for the OPs question, I was only trying to make that point.

For some reason I have heard guys around this area think that if the commercial lighting is 277v instead of 120v, they will not have to apply the 100% demand factor. Where this comes from, I do not have a clue.