Calculating number of circuits required for lighting please help asap

[Matt Mckenzie]

When calculating number of circuits required for a continuous lighting load do we multiple by 125% then divide by 80% of the amperage of the circuit? For example how many 20amp 277volt circuits are required to feed a 14,625.5Va continuous lighting load? 125%x14,625.5Va= 18,282Va/277v = 66amps/16amps(80% of 20amps) = 4.125. At least 5 circuits are required?

 

[kwired]

I think the simplest method here is if you know you are going to use 20 amp circuits then simply don't put more then 80% (16 amps) on a circuit - that would be 4432 VA if these are 277 volt circuits.

Otherwise once you have added an additional 25% to the load, you are done with adjustments unless something else requiring conductor ampacity adjustments comes up like ambient temp or number of conductors in raceway. By doing so you are essentially making 16 amps of actual load the limit for a 20 amp branch circuit.

So in your example you have come up with 66 amps of load - divide by 20 amps per circuit (because 125% factor is already in that 66 amps) and you get 3.3 - round up to 4 circuits needed. What you did ultimately de-rated the same load two times

 

[Smart $]

When calculating number of circuits required for a continuous lighting load do we multiple by 125% then divide by 80% of the amperage of the circuit? For example how many 20amp 277volt circuits are required to feed a 14,625.5Va continuous lighting load? 125%x14,625.5Va= 18,282Va/277v = 66amps/16amps(80% of 20amps) = 4.125. At least 5 circuits are required?

Like so...

14,625.5VA x 125% / 277V / (20A/ckt) = 3.3ckts... and round up to 4ckts.

14,625.5VA / 277V / 4ckts = 13.2A (per ckt)

 

[Dennis Alwon]

When calculating number of circuits required for a continuous lighting load do we multiple by 125% then divide by 80% of the amperage of the circuit? For example how many 20amp 277volt circuits are required to feed a 14,625.5Va continuous lighting load? 125%x14,625.5Va= 18,282Va/277v = 66amps/16amps(80% of 20amps) = 4.125. At least 5 circuits are required?

The problem with your method is you multiplied the load by 125% AND divided by 16 amps. If you had divided by 20 amps you would have gotten 3.3 cir== 4.

If you didn't use the 125% then you would divide by 16.

14625.5 / 277 =52.8

Now divide by 16 amps = 3.3 cir= 4

 

[samblv]

When calculating number of circuits required for a continuous lighting load do we multiple by 125% then divide by 80% of the amperage of the circuit? For example how many 20amp 277volt circuits are required to feed a 14,625.5Va continuous lighting load? 125%x14,625.5Va= 18,282Va/277v = 66amps/16amps(80% of 20amps) = 4.125. At least 5 circuits are required?

Divide 14,625.5 by 4432 and that's your minimum number of circuits. But then you need to look at voltage drop issues and may have to add a circuit or two to compensate for that? And if you're in California, you may have to check Title 24 to see what your "allowed" watts are, if applicable?

 

[kwired]

Divide 14,625.5 by 4432 and that's your minimum number of circuits. But then you need to look at voltage drop issues and may have to add a circuit or two to compensate for that? And if you're in California, you may have to check Title 24 to see what your "allowed" watts are, if applicable?

Voltage drop is a different issue. Sure splitting the load into more circuits will help with voltage drop - assuming you continue to use same size conductors, but by adding one more circuit you have now thinned the load enough that 15 amp circuits are possible - but if you decrease conductor size just because you now have 15 amp circuits you still may have similar voltage drop issues.

 

[samblv]

Voltage drop is a different issue. Sure splitting the load into more circuits will help with voltage drop - assuming you continue to use same size conductors, but by adding one more circuit you have now thinned the load enough that 15 amp circuits are possible - but if you decrease conductor size just because you now have 15 amp circuits you still may have similar voltage drop issues.

It's not really a different issue. It's one of the considerations you think about when solving the problem of how many circuits are required. So is conductor sizing. They're both part of the process. Load, circuit length, circuit voltage, related code requirements, site conditions, cost, all potentially related factors in determining the number of circuits required for a particular lighting load.

You know, I used to hate doing algebra word problems. I took algebra one semester and got a "D". I took it the very next semester and got another "D". I finally got an "A", but doing those word problems taught me the mechanics of determining "related factors" to a problem and its eventual solution. I'm sure I'm preaching to the choir here when I say, "In this business, not having those skills, can have drastic monetary consequences."

 

[kwired]

It's not really a different issue. It's one of the considerations you think about when solving the problem of how many circuits are required. So is conductor sizing. They're both part of the process. Load, circuit length, circuit voltage, related code requirements, site conditions, cost, all potentially related factors in determining the number of circuits required for a particular lighting load.

You know, I used to hate doing algebra word problems. I took algebra one semester and got a "D". I took it the very next semester and got another "D". I finally got an "A", but doing those word problems taught me the mechanics of determining "related factors" to a problem and its eventual solution. I'm sure I'm preaching to the choir here when I say, "In this business, not having those skills, can have drastic monetary consequences."

Sorry dude but I don't see it that way. Voltage drop is fixed by increasing conductor size, redistributing the load just means recalculating everything, including voltage drop. It also still takes the same KVA to operate the same load even if you split it up into more circuits.

Fixing voltage drop doesn't mean you need to increase conductor size for the entire circuit either, maybe just for the first run of the circuit and then branch in to more then one direction with a smaller conductor. With lighting you are typically dropping off some load at each junction point so it is not like you are still seeing a full 16 amps on a 20 amp circuit by the time you get to the last luminaire at the longest point from the starting of the circuit.