250.122
- Thread starter harriz
- Start date
- Status
- Location
- Massachusetts
How many 500s per phase?
- Location
- Illinois
- Occupation
- retired electrician
This should be fun...I will say you need a 350kcmil for the EGC.
bphgravity
Senior Member
- Location
- Florida
The code is not clear exactly how to figure this out. For one, there are several wiring configurations in which a 500A circuit can be supplied. Here's one way to do it:
Calculation:
Assume you really only needed 800 kcmil of ungrounded conductor for the 500A circuit. You are installing 2,500 kcmil. This is a 3.125 increase. (2,500,00 / 800,000).
A typcial 500A circuit requires a #2 AWG EGC. #2 = 66,360 cm X 3.125 = 207,375 cm = 250 kcmil EGC.
So, the cable tray may serve as the EGC (250.118 / 392.3 / 392.7) or 1 - 250 kcmil copper EGC (250.122(B).
Calculation:
Assume you really only needed 800 kcmil of ungrounded conductor for the 500A circuit. You are installing 2,500 kcmil. This is a 3.125 increase. (2,500,00 / 800,000).
A typcial 500A circuit requires a #2 AWG EGC. #2 = 66,360 cm X 3.125 = 207,375 cm = 250 kcmil EGC.
So, the cable tray may serve as the EGC (250.118 / 392.3 / 392.7) or 1 - 250 kcmil copper EGC (250.122(B).
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
This could be a matter of opinion. The starting point for an OCPD of 500 amps is a #2 copper. The starting point for a conductor ampacity of 500 amps would be either a single 900 MCM, or a pair of 250 MCM (total cross sectional area of 500 MCM). The final configuration has a total cross sectional area of 2500 MCM. That is a ratio of either 2.8 or 5, depending on what you consider to be the ?correct? starting point. A factor of 2.8 times the area of a #2 would lead you to select a 4/0 EGC. A factor of 5 times the area of a #2 would lead you to select a 350 MCM EGC. I can think of at least one other possible way to look at the situation, and that would give another answer. :huh:
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
That would only work if the calculated load were less than 490 amps. We were not given any information on that subject.
inspector141
Senior Member
- Location
- Westminster, MD
Since the design calls for 5 sets of upsized parallel conductors, we must use 5 sets to find the minimum conductor size required for a 500 amp feeder. But to follow the parallel rule requirement in 310.4, we are forced to use the minimum size of 1/0 in this case.
5x105,600mcm=528kcmil
2500kcmil/528kcmil=4.7. Therefore the minimum size equipment grounding conductor(#2) must be increased 4.7.
#2(66360) x 4.7=311,892mcm or 350kcmil
The reason why we should follow the number of sets in the design is because the lowest kcmil would more than likely be used with paralleling to derive the lowest possible mcm.
for example: a 1200 ocpd could use 8 sets of 1/0 or 3 sets of 600. The mcm would be far less with the 8 sets of 1/0, therefore the increase of the egc would be significantly less. So whatever the design is, then thats what should be used to derive the increase in the egc, IMHO of course. In this case, 5 sets.
5x105,600mcm=528kcmil
2500kcmil/528kcmil=4.7. Therefore the minimum size equipment grounding conductor(#2) must be increased 4.7.
#2(66360) x 4.7=311,892mcm or 350kcmil
The reason why we should follow the number of sets in the design is because the lowest kcmil would more than likely be used with paralleling to derive the lowest possible mcm.
for example: a 1200 ocpd could use 8 sets of 1/0 or 3 sets of 600. The mcm would be far less with the 8 sets of 1/0, therefore the increase of the egc would be significantly less. So whatever the design is, then thats what should be used to derive the increase in the egc, IMHO of course. In this case, 5 sets.
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