Help Understanding Ampacities and Parallels, table 310.15(B)(16)

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Koplinsky

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I have a 1000A/3phase feeder that's called out to be 4 sets of [(3)#300, 2/0g, in 3" EMT)], copper.

My question is, according to table 310.15(B)(16), (1)#300 in column 75C has ampacity rating of 285. Why must I have 4sets of (3)300s, a total of 12, if (4)#300s provides 1140A. I am clearly reading this table wrong. Please help me understand.
 
Perhaps there is something in details that are missing:
Conductor insulation type ?
Environment (wet, dry, ambient) ?
Number of sets in each conduit ?
Is voltage drop an issue ?
 
What is the Actual Load? The Phrase "1000A 3 Phase Feeder" just tell me the equipment the feeder is fed from is rated at 1000 amps. Outdoor install or indoor install? (ambient temp? ) Conductor length?

From what you have provided, you are reading the table correctly. But remember, the ampacities given in the table are the ampacity of the conductors while they are on the shelf at big blue. :)
 
It appears you are mixing up the single phase and three phase requirements with paralleling. This is a three phase feeder, so it needs three conductors as a base (phase A, B, and C). It does not have a neutral, or you would need four conductors as your base (phase A, B, C, and the neutral). As you noted, 300 kcmil has an ampacity of 285A. One set of (3) 300 kcmil is a 285A, three phase feeder. The conductors are not paralleled. Since you need 1000A, you need four sets of (3) 300 kcmil. That will give you a 1000A, three phase feeder, which has 1000A of conductor on each phase (A, B, and C).

So, to answer your question, you need (12) 300 kcmil conductors because you need 4 paralleled conductors for each of your three phases (A, B, and C).
 
300 kcm is rated 285 amps as you stated. If you parallel then each phase must be paralleled also. so you have 4 sets. Each phase will have 4 x 300kcm = 1140 amps. Each phase must be cpable of the 1140 amps.
One could pull 4 sets of 250 kcm as that is rated 255 x 4 = 1020 amps but you have to do what is speced.
 
Koplinsky said:
I have a 1000A/3phase feeder that's called out to be 4 sets of [(3)#300, 2/0g, in 3" EMT)], copper.

My question is, according to table 310.15(B)(16), (1)#300 in column 75C has ampacity rating of 285. Why must I have 4sets of (3)300s, a total of 12, if (4)#300s provides 1140A. I am clearly reading this table wrong. Please help me understand.
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Each "set" is rated for 285 amps, 4 sets equals 1140 amps as others have stated. Also as Dennis stated since you could use 250 kcmil the #2/0 EGC in each raceway is too small and needs to be up-sized according to 250.122(B).
 
I think there are several possible misunderstandings here.

1) A '1000A feeder' is 1000A for each phase, not 1000A total. Since this is a 3 phase without neutral feeder, you need 3 phase conductors each rated 1000A. 4 x 285A = 1140, which is greater than 1000 and fits the bill. 3 phases * 4 wires = 12

2) The 285A value for 300 kcmil Cu is based on specific conditions of ambient temperature, number of conductors in each raceway, etc. It is possible that the actual ampacity is reduced because of not standard conditions.

3) Ampacity is not the only reason for selecting a particular conductor size. Another important consideration is voltage drop. In general for long feeders you need larger conductors to compensate for the resistance of the wire.

-Jon
 
Another thing to consider is that if the OCPD rating is >800A, the derated ampacity of the conductors has to be equal to or greater than the rating of the OCPD, i.e., no "next size up"for OCP.
 
winnie said:
I think there are several possible misunderstandings here.

1) A '1000A feeder' is 1000A for each phase, not 1000A total.
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One common misunderstanding is not getting that in a balanced three phase system the current per phase and the total current are the same.
 
ggunn said:
One common misunderstanding is not getting that in a balanced three phase system the current per phase and the total current are the same.
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I don't really like to say 'total current'. In a balanced 3 phase system you have 3 conductors, and _each_ conductor is carrying the same current.

With 3 wires each carrying 1000A, how many 'total amps' do you have?? 3000A? 1732A? 0A? In different situations one could argue for each of these numbers.

The OP needs to understand that in his system he needs 3 separate conductors, _each_ capable of carrying 1000A.

-Jon
 
Koplinsky said:
I have a 1000A/3phase feeder that's called out to be 4 sets of [(3)#300, 2/0g, in 3" EMT)], copper.

My question is, according to table 310.15(B)(16), (1)#300 in column 75C has ampacity rating of 285. Why must I have 4sets of (3)300s, a total of 12, if (4)#300s provides 1140A. I am clearly reading this table wrong. Please help me understand.
Click to expand...

Think of each of the three phases, as if they were each polarities of a battery, in order to understand why you don't add up ampacities among the three phases. If you had a DC battery bank that needed a 1000A circuit to connect it to the load, both the positive and the negative polarities would need to be built from a group of conductors in parallel that add up to 1000A. Because 1000A will be flowing on both of them, in opposite directions. You wouldn't get to take credit for the fact that there are two polarities, and install an ampacity that adds up to 500A on the positive, and 500A on the negative, thinking that will add up to 1000A. Both the positive and negative are needed to complete the full closed path of the current from the source and back to the source, and will both be required to carry 1000A.

With 3-phase, each of the three phases do not each draw their peak instantaneous current simultaneously. They are each time-delayed from each other, by a third of a cycle each. At t=0, suppose phase A has its peak outbound current from the source. At this instant, both phases B and C will carry an inbound current that adds up to the total current on A. B's current would be increasing, and C's current would be decreasing. The three phases each are a component of what is needed to deliver the power, and complete the round trip path. As a side note: a waveform with a nominal current of 1000A isn't really peaking at 1000A, but is instead peaking at 1414 A. Look up "root mean square" if you are curious to learn more about this.

The take-away:
Ampacities add up for conductors in parallel on the same phase, which is why you can multiply the 285A value by 4 in this example.
Ampacities don't add up for different phases. Each phase needs to be rated for the required ampacity independently of all the other phases being present.
 
A 1000 amp 3 phase 60 Hz. feeder is required to deliver 1000 amps to the 1000 amp load. Each phase must be sized to carry 1000 amps. Not 1000 / 3 = 334 amps. I am confused. 334 x 3 = 1000 amps. This would require 3 parallel runs in 3 separate conduits. Each conduit with 3 - 400 Kcmil + Equipment ground. Each conduit with ( A , B & C phases ) = ( 334 A ) x 3 conduit runs. = 1000 Amps
My calculation is 3 runs of conduit each with 3 - 1 / C # 400 Kcmil + 2 / 0 EG. Makes a 1000 Amp power supply to a 1000 Amp load.
1000 A load x 1.25 = 1250 Amps required copper conductor size. = 1250 / 3. = 417 amps per conductor. Three conduit runs each with
 
Is there confusion here with what is meant as a 1000 amp feeder. My circuit for a 3 phase / 3 wire 1000 amp load would be 1000 x 1.25 = 1250 amps. equals 1600 A breaker, and 3 runs of conduit each with ( 3 - 1 / C # 600 Kcmil + 4 / 0 EG. ) 420 A x 3 = 1260 Amps
 
It's really kind of an academic question as to why. Contractually it would appear he is obligated to supply what the drawings ask for as long as they meet code and they seem to.
 
LarryFine said:
Since nobody mentioned it, the EGC can be eliminated.
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True can eliminate the EGC if using EMT, but there also was no mention of the neutral (grounded conductor). Again not enough info provided by OP to make absolute determination that the suggested sizing can be reduced or even might need to be upsized.
 
Since he is running 4 conduits anyway, why not run (6) 1/0s in each conduit?

I did some looking once and found that 1/0 gives you the most ampacity for the amount of copper.
 
winnie said:
With 3 wires each carrying 1000A, how many 'total amps' do you have?? 3000A? 1732A? 0A? In different situations one could argue for each of these numbers.
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ggunn said:
1000A. Current going out on one conductor is returning on the other two.
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Okay, with 2 wires each carrying 1000A in a single phase circuit, how many 'total amps' do you have?

And before you say the answer is the same, consider that at the same line-neutral voltage the 3 phase circuit is delivering more power with the 'same' current. This is why I don't like the idea of saying 'total amps'.

-Jon
 
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