I need a little clarification on 3phase vs single phase. If I have a 3 phase 277V service and am using it to run 16 fixtures that each indivually use 6.3 amps. How many fixtures can be on a 10awg neutral wire if the load is distributed evenly across the 3 phases. Right now I've got 3 10-5 cables for the 16 fixtures but not sure if that shouldn't be 4 cables instead. I think I've been oversizing the wire in the past. I know in 3 phase 480V that you take the number of fixtures x amps per fixture and divide by 1.73 to get total amps, but I'm not sure if that formula applies to all 3 phase calculations. Any insight would be appreciated.
3 phase calculation
- Thread starter aklock
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I am a bit confused by your question.
Welcome to the forum.
I don’t know what you mean by this. I know of 3-phase services that have 480 volts phase to phase, and 277 volts phase to neutral. Is this what you mean? For the rest of my response, I will assume it is.aklock said:
So you have 6.3 amps per fixture, times 277 volts, gives you 1,745 watts per fixture. Multiply by 16 to get about 27,900 watts. Divide by the line to line voltage of 480, and divide again by 1.732 (i.e., the square root of 3), and you get a phase current of 33.6 amps.aklock said:
If the load is distributed evenly, the neutral current will be zero. So the question does not make sense to me.aklock said:
You lost me here, I fear. What do the 3, 10, 5, and 4 mean?aklock said:
I showed the calculation above. Let's start there, and tell us what you are trying to figure out.aklock said:
Welcome to the forum.
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aklock said:
Yes it applies to each conductor. If you calculate the load at 20 amps then all three phases would also be 20 amps if the load is distributed evenly. The neutral as charlie says will draw zero so it only needs to be as large as the phase conductors.
I am assuming 5 conductor cables means 3 phases , neutral and ground.
That is correct, the 10-5 cable would be 3 phases, neutral and ground. I was confusing myself between 3 phase and single phase neutral load. I kept thinking what ever was on the phase conductor had to return on the neutral conductor regarless of single or 3 phase. So if I understand correctly any time it is a 3phase system the total watts/volts/1.73 would equal the total current of all 3 phases and if equally distributed across each phase then the total current can be divided by 3 to get each legs current.
LarryFine
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Just a side note, 16 is not evenly divisible by 3.
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No!aklock said:
This concept of "total 3 phase current" is one that is best forgotten as early in one's career as possible. There is no such thing as "total current" in the context you are using it. If you add the currents in phase A, phase B, phase C, and the neutral, you get a total of zero. If the phases are balanced, the the neutral current is zero, and the sum of A, B, and C is also zero.
Currents flow in the shape of a sine wave: positive for one half cycle, and negative in the other half cycle. Let's talk about a balanced system. As the current in phase A is at its positive peak, the currents in phases B and C are negative, and not quite yet at their peak values. What is actually happening is that for a short part of the cycle the current leaves the source along the phase A wire and returns to the source along the phase B and C wires. Somewhat later in the cycle, the current leaves the source along the phase B wire and returns to the source along the phase A and C wires. But the sum of the currents is always zero.
If the loads are not perfectly balanced, then some of the current that leaves the source along the phase A wire will returns to the source along the phase B and C wires, and the rest will return along the neutral wire.
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aklock said:
Actually that would be a 10-4 cable not a 10-5 cable.
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Yes. If you calculate the load based on watts/volts/1.73 then that would give the load for each phase. Thus if the calculated load was 45 amps for each phase then you would only need a 60 amp diconnect
I have to caution you in doing it this way, for it is only good for an ideally balanced 3? system. To size OCPDs, feeders, disconnects for even slightly unbalanced systems, you need to determine the current on each Line and use the highest of the three.aklock said:
For example, your 16 fixtures @ 277V cannot be balanced on 3 phases of a 480Y/277 system. You can balance 15 at 5 per Line for a current of 31.5A per Line. Including the 16th fixture will make one Line's current 37.8A.
In addition to Smart $'s caution on the unbalanced load, you also need to consider that you really want to divide total VA by volts/1.73 to get amps, not watts. Charles' example works because he actually calculated VA not watts by using amps per fixture. If the fixtures are incandescent, then the power factor is 1 and VA=watts.
real example
real example
Here is an example of one I'm doing now. It is 18 fixtures the way the diagram is drawn up. They are not balanced for some reason, instead it's 7-A, 6-B and 5 on C. They are HID's that and are 1650 watts. So even though they are not completely balanced the 2 questions I have concerning the drawing is they show 3 #10's for the the neutrals, 1 per phase, and show an 80A disconnect for the 18 fixtures. Will the unbalance have any negative effect, are the #10's large enough and is the 80A disconnect big enough? This is all on a 277V 3phase system. After the back and forth responses here I think it should be ok.
real example
Here is an example of one I'm doing now. It is 18 fixtures the way the diagram is drawn up. They are not balanced for some reason, instead it's 7-A, 6-B and 5 on C. They are HID's that and are 1650 watts. So even though they are not completely balanced the 2 questions I have concerning the drawing is they show 3 #10's for the the neutrals, 1 per phase, and show an 80A disconnect for the 18 fixtures. Will the unbalance have any negative effect, are the #10's large enough and is the 80A disconnect big enough? This is all on a 277V 3phase system. After the back and forth responses here I think it should be ok.
kingpb
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Smart $ said:
I'm glad we can always count on someone providing a rational answer.
I'll break it down for you...aklock said:
Assumptions:
- Fixtures operate at a .85 power factor.
- OCP equipment is not rated for continuous operation.
- Conductors are Cu.
?A: 7 fixtures X 1650W ? 277 ? .85 = 49.1A
?B: 6 fixtures X 1650W ? 277 ? .85 = 42.0A
?C: 5 fixtures X 1650W ? 277 ? .85 = 35.0A
#10 is too small for all three circuits and an 80A disconnect (rated 100A, fused at 80A) will not protect #10 conductors. The NEC limits the max' ampacity rating of #10's to 30A, and further limits this size conductor to a max' continuous load of 24A (1/125% or 80%).
Are you sure there is not something in between the disconnect and the branch circuit conductors?
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