310.15(B)(16) wire size requirement vs. breaker lug size....

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EEEC

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So I've got a 600A rated panelboard being fed by a 600A main breaker (all specified in the architectural drawings). According to 310.15(B)(16), I would need to use 1000 kcmil. So why does the MB only accept up to 750 kcmil? And secondly when I approached the supplier for a cost on the 1000 kcmil, he acted as if I was crazy and said they don't carry that size and would have to hunt around. Am I missing something here or figuring wrong? :huh:
 
The issue is no sane person is going to run that large of wire for a 600 amp circuit. 2 or even 3 parallel conductors are much more economical to purchase and install.
 
I went and looked at the MB specs and it has two 750 kcmil lugs per phase which means I'd need to run two wires in parallel for each phase and the neutral (3-wire single phase).

Based on this, can someone confirm the following:


600A / 2 = 300A each


Derate (80%) based on 310.15(B)(3)(a) for six current-carrying conductors in one raceway.


300A + 20% = 360A


Size according to 310.15(B)(16) since there are no other factors (ambient is correct, short run, no motors, etc.)


So I'd need to run 400 kcmil conductors, yeah?
 
EEEC said:
I went and looked at the MB specs and it has two 750 kcmil lugs per phase which means I'd need to run two wires in parallel for each phase and the neutral (3-wire single phase).

Based on this, can someone confirm the following:


600A / 2 = 300A each


Derate (80%) based on 310.15(B)(3)(a) for six current-carrying conductors in one raceway.


300A + 20% = 360A


Size according to 310.15(B)(16) since there are no other factors (ambient is correct, short run, no motors, etc.)


So I'd need to run 400 kcmil conductors, yeah?
Click to expand...

That should be +25%.

Yes, 400 kcmil Cu THHN is 380 amps@ 90° C.

380 * 80% = 304 amps * 2 sets = 608 amps.

We never run cables larger than 750 kcmil.
 
EEEC said:
I went and looked at the MB specs and it has two 750 kcmil lugs per phase which means I'd need to run two wires in parallel for each phase and the neutral (3-wire single phase).

Based on this, can someone confirm the following:


600A / 2 = 300A each


Derate (80%) based on 310.15(B)(3)(a) for six current-carrying conductors in one raceway.


300A + 20% = 360A


Size according to 310.15(B)(16) since there are no other factors (ambient is correct, short run, no motors, etc.)


So I'd need to run 400 kcmil conductors, yeah?
Click to expand...

Even 750 is rarely used. 750 if the largest I have ever seen in the wild, and I have only used it once. The ampacity per cmil really starts to tank and get inefficient when you get into those larger wire sizes, plus working with it can be a hassle. Look into more sets of smaller conductors, usually in separate pipes then you wont have derating.
 
EEEC said:
So I've got a 600A rated panelboard being fed by a 600A main breaker (all specified in the architectural drawings). According to 310.15(B)(16), I would need to use 1000 kcmil.
Click to expand...

If you were running a single conductor per phase, you'd need to use 1500 kcmil for a full 600A ampacity. The 90C ampacity can be used as a starting point for derating, but for connecting to 75C equipment (the breaker), you are limited to the 75C ampacity.

Cheers, Wayne
 
infinity said:
That should be +25%.

Yes, 400 kcmil Cu THHN is 380 amps@ 90° C.

380 * 80% = 304 amps * 2 sets = 608 amps.

We never run cables larger than 750 kcmil.
Click to expand...

Your Derating process found Ampacity, but left your overload-termination temperatures @ 90°C.

Ampacity Table shows 304 amps within 75°C on #400, because table assumes 3 ccc's.

Conduit is cheep, get your termination temperature back in line with more conduit, for max of 3 ccc's.
 
ramsy said:
Your Derating process found Ampacity, but left your overload-termination temperatures @ 90°C.

Ampacity Table shows 304 amps within 75°C on #400, because table assumes 3 ccc's.

Conduit is cheep, get your termination temperature back in line with more conduit, for max of 3 ccc's.
Click to expand...

You lost me. My calculation is correct can you explain further what I'm missing? 400 kcmil is a 335 amp conductor at 75° C. Two sets in a single raceway brings each set down to 304 amps.
 
ramsy said:
Your Derating process found Ampacity, but left your overload-termination temperatures @ 90°C.

Ampacity Table shows 304 amps within 75°C on #400, because table assumes 3 ccc's.

Conduit is cheep, get your termination temperature back in line with more conduit, for max of 3 ccc's.
Click to expand...

Derates apply to wire, not terminations. See 110.14(C). The 75C termination temperature, not corrected for anything, is the starting point (generally). Using 90C wire allows flexibility for some derate calculations, which is the primary advantage of 90C wire.
 
infinity said:
You lost me. My calculation is correct..
Click to expand...


1) NEC Table Methods
When Adjusting to 6 ccc's, #400 wire is Derated to 304A. Temperature not known.
Continuous loads not assumed.

2) NEC Formula Methods
With Note #2 formula, in Chap.9 Table 8: 6 ccc's of #400 @ 304A ea approaches 90°C.
Formula assumes continuous loads.


A continuous-load adjustment to the Table calc, 304A x 1.25 = 380A approaches the 90°C column.

Both methods show the same result when the assumptions match.
 
ramsy said:
1) NEC Table Methods
When Adjusting to 6 ccc's, #400 wire is Derated to 304A. Temperature not known.
Continuous loads not assumed.

2) NEC Formula Methods
With Note #2 formula, in Chap.9 Table 8: 6 ccc's of #400 @ 304A ea approaches 90°C.
Formula assumes continuous loads.


A continuous-load adjustment to the Table calc, 304A x 1.25 = 380A approaches the 90°C column.

Both methods show the same result when the assumptions match.
Click to expand...

The continuous load presence or absence is already accounted for, in the sizing of the 600A breaker. Either it is up to 600A noncontinuous, 480A continuous, or a mixture of the two that "adds up" to 600A with the appropriate factors for each. It is just a coincidence that the 6 ccc's has a derate factor of 0.8, which is the reciprocal of the 125% continuous load factor.

Here's the way I think through the rules:
1. OCPD not continuous rated >= 1.25*continuous+noncontinuous
2. Terminal rating >= 1.25*continuous + noncontinuous
3. Wire rating >= (continuous + noncontinous)/(total derate)
4. Wire rating * total derate > previous OCPD and not be an exact match, if next-size-up rule can apply
5. Terminal rating > previous OCPD and not be an exact match, if next-size-up rule can apply
6. Otherwise, if NSU rule cannot apply, for 4 and 5, you need at least as much amps of terminals and derated wire as OCPD
 
Since the 75 deg rating of #350 is 310A, I would be able to use it with two separate conduits without having to derate anything, correct? It is a 20ft run to a 600A panelboard with about 200A worth of distributed load that includes LED lighting, a handful of courtesy receptacles, and four 35A HVAC units, and another 200A subpanel that includes LED lighting, a handful of courtesy receptacles, and two 35A HVAC units.

Also with regard to grounding, according to 250.122(F) "...the equipment grounding conductors, where used, shall be installed in parallel in each raceway or cable." and "Each equipment grounding conductor shall be sized in compliance with 250.122". According to 250.122 the size of the grounding conductor is not be sized below that of the "Rating or Setting of Automatic Overcurrent Device in Circuit Ahead of Equipment, Conduit, etc.,". So the way I interpret this is that each grounding conductor must meet the 600A rating of the OCPD, and therefore must be sized at #1 for each conduit run. Agree or disagree?
 
EEEC said:
Also with regard to grounding, according to 250.122(F) "...the equipment grounding conductors, where used, shall be installed in parallel in each raceway or cable." and "Each equipment grounding conductor shall be sized in compliance with 250.122". According to 250.122 the size of the grounding conductor is not be sized below that of the "Rating or Setting of Automatic Overcurrent Device in Circuit Ahead of Equipment, Conduit, etc.,". So the way I interpret this is that each grounding conductor must meet the 600A rating of the OCPD, and therefore must be sized at #1 for each conduit run. Agree or disagree?
Click to expand...

Correct or use a metallic raceway that qualifies as an EGC and a wire conductor EGC would not be needed. Design choice.
 
EEEC said:
Since the 75 deg rating of #350 is 310A, I would be able to use it with two separate conduits without having to derate..
Click to expand...

Or, #500 Cal-XHHx with optional 2/0 EGC, if Aluminum is that much cheaper than Copper.
 
Carultch said:
Here's the way I think through the rules:
1. OCPD not continuous rated >= 1.25*continuous+noncontinuous
2. Terminal rating >= 1.25*continuous + noncontinuous
3. Wire rating >= (continuous + noncontinous)/(total derate)
4. Wire rating * total derate > previous OCPD and not be an exact match, if next-size-up rule can apply
5. Terminal rating > previous OCPD and not be an exact match, if next-size-up rule can apply
6. Otherwise, if NSU rule cannot apply, for 4 and 5, you need at least as much amps of terminals and derated wire as OCPD
Click to expand...
I think I understand the rules, but I find the relevant NEC texts fairly opaque on the issue. Any chance you could provide section references for each of the above?

Also, here's one example that I'm unclear on: Is it NEC-compliant to feed a 48A continuous load with #6 NM cable protected by a 60A breaker? NM cable is limited to its 60C ampacity, or 55A, which is greater than the continuous current, but not greater than 125% of the continuous current. And is the answer the same for a wiring method with actual 60C insulation?

Thanks, Wayne
 
wwhitney said:
I think I understand the rules, but I find the relevant NEC texts fairly opaque on the issue. Any chance you could provide section references for each of the above?

Also, here's one example that I'm unclear on: Is it NEC-compliant to feed a 48A continuous load with #6 NM cable protected by a 60A breaker? NM cable is limited to its 60C ampacity, or 55A, which is greater than the continuous current, but not greater than 125% of the continuous current. And is the answer the same for a wiring method with actual 60C insulation?

Thanks, Wayne
Click to expand...

No, it is not NEC compliant to supply a 48A continuous load with #6NM cable, per 210.19(A)(1).
 
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