OCPD greater than 800A
- Thread starter Zyb
- Start date
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Using 500 kcm wire on 75 deg terminals means you are limited to 380A per set. This limit does not depend on ambient temperature or number of CCCs. Your 90C rated wire is further limited to 430A per set times any ampacity adjustment (for number of CCCs) or correction (for ambient temperature) factors.
So the answer is 1500A OCPD, as long as the product of the ampacity adjustment and correction factors is at least 1500/1720. And if these conductors and OCPD are carrying a continuous current, it is limited to 1500A / 125% = 1200A unless the OCPD is 100% rated.
Cheers, Wayne
So the answer is 1500A OCPD, as long as the product of the ampacity adjustment and correction factors is at least 1500/1720. And if these conductors and OCPD are carrying a continuous current, it is limited to 1500A / 125% = 1200A unless the OCPD is 100% rated.
Cheers, Wayne
Thank you for your response @wwhitney , just one more thing
Terminal 75°
Wire Rating 90°
Amb. Temp 0.91
No. CCC 0.80
500kcmil @ 75° 380A
500kcmil @ 90° 430A
Current = 1205A (continuous load)
Des. Current = 1205A x 1.25 = 1506.25A
per set (4 sets)
Current = 1205A / 4 = 301.25A
Des. Current = 1506.25A / 4 = 376.6A
500kcmil 75°
380A > 376.6A ; OK
500kcmil 90°
430A x 0.91 x 0.80 > 301.25A
313A > 301.25A; OK
OCPD
1205A x 1.25 = 1506.25A
what OCPD should I select 1500A; 380A(75°) x 4 = 1520A, OCPD < 1520A
or
1600A OCPD and upsize the wire?
Terminal 75°
Wire Rating 90°
Amb. Temp 0.91
No. CCC 0.80
500kcmil @ 75° 380A
500kcmil @ 90° 430A
Current = 1205A (continuous load)
Des. Current = 1205A x 1.25 = 1506.25A
per set (4 sets)
Current = 1205A / 4 = 301.25A
Des. Current = 1506.25A / 4 = 376.6A
500kcmil 75°
380A > 376.6A ; OK
500kcmil 90°
430A x 0.91 x 0.80 > 301.25A
313A > 301.25A; OK
OCPD
1205A x 1.25 = 1506.25A
what OCPD should I select 1500A; 380A(75°) x 4 = 1520A, OCPD < 1520A
or
1600A OCPD and upsize the wire?
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Why the 0.8, do you have two sets of CCCs in each conduit?
Based on the above, if your OCPD is not 100% rated, it has to be at least 1506A, so apparently 1600A if that is the next size up available to you. Because of 240.4, and 240.4(B)'s non-applicability over 800A, that means your ampacity actually has to be at least 1600A, or 400A per set.
400A / 0.91 / 0.8 = 549A. So each set would need to be at least 800 kcmil Cu for 90C rated insulation.
Obviously if you can change some of the variables in your favor, you can get the conductor size down. E.g. more sets, only 3 CCCs, reduce the continuous current to be only 1200A so you can use a 1500A OCPD, or use a 100% rated OCPD.
That check is correct only for 100% rated OCPD. Otherwise this check gets a 125% factor.
That check is OK for the current, but not for the OCPD as per above.
Cheers, Wayne
Thank you Wayne
one set one conduit but i count N as CCCs as there's are comms connecting to Neutral and provision if for future circuit using Neutral on the load center
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
You almost never have to count N as a CCC when it is accompanied by a balanced set of ungrounded conductors. In the 2017 NEC, this is 310.15(B)(5). So you do not need the ampacity adjustment factor of 0.8.
For a brief explanation, consider a set of resistive loads all connected L-N across the 3 different ungrounded conductors (for the case of 3 phase) and suppose the load is a maximum of 100A on any ungrounded conductor. One possible value of the currents (L1,L2,L3,N) is (100,100,100,0), where you have maximum load on each ungrounded conductor. The currents on N add to zero as the contribution from the 3 ungrounded conductors are pairwise 120 degrees out of phase.
This is the worst case heating you can get in the conductors subject to the 100A load limit. If you decrease the current in any one of the ungrounded conductors, current in N will go up accordingly. But as the heating varies as the square of the current, the total heating will actually be less.
Cheers, Wayne
got it. thank you
Carultch
Senior Member
- Location
- Massachusetts
If you can get an OCPD that can be fine-tuned to 1520A or less, than 1520A worth of wire is OK. Otherwise, expect the 1600A with 1600A worth of wire, until proven otherwise.
It's uncommon that this governs. Since the total Ɪ^2 among 4 wires will not be greater than the total Ɪ^2 among just the 3 phases, regardless of the imbalance, the neutral doesn't generate extra heat beyond what the 3 phases would do on their own. Neutrals of insignificant current (instrumentation purposes) also don't need to count.
It's only when harmonic-intensive loads are involved, or when your circuit requires the neutral to carry the full current, even for balanced loads, that neutral would need to count. I find that the code could use more clarity on how to quantify whether non-linear loads are of concern or not for this issue, since harmonic distortion is a lot less than what it used to be.
Thank you
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Presumably you mean "worst case Ɪ^2 among just the 3 phases" given some maximum on any of the currents. But that statement requires some (reasonable and realistic) limitations on the loading. Obviously you can load all 4 wires to full current--just put one load on L1-N and another load on L2-L3.
Cheers, Wayne
Carultch
Senior Member
- Location
- Massachusetts
Yes, that's what I mean. When compared to the maximum possible balanced load, introducing imbalance doesn't generate any more heat. If you load all 3 phases with line-to-neutral loads, the neutral current will add up to zero for the balanced case. It's only for multiples of 180 Hz (triplen harmonics), that neutral current among all three phases is additive.