Derating Parallel Conductors with Conduit Size

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Electricalhelp

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I have a store with it's incoming Electrical Service to be 400 Amp/208V/3P. But I don't have the actual demand load, so I assumed the store needs exactly 400AMP.

The drawing calls out for 2 sets (4# 3/0 + 1 #1/0 Ground in 4" Conduit)

Based on Table 310.15(B)(16), 3/0 is good for 200AMP

So 2 Sets of (4# 3/0 + 1 #1/0 Ground in 4"C) is exactly at 400 AMP

If I use Table 310.15(B)(3)(a) for more than Three Current Carrying Conductors, since there are 8 Current Carry Conductors, the percent factor is 70%.

So 2 Sets 2 sets (4# 3/0 + 1 #1/0 Ground) with a 70% factor is only good for 280AMP

But when I use Cerrowire Ampacity Calculator, (default temp), they said it's good for 315A.

So based on Cerrowire Ampacity Calculator, I would need to use at least 250kcmil, which would come out to 406A.
My calculation came out as 250kcmil at 255 AMP. So (2*255*.8) = 357 AMP

How do I fix my canulation so I can get similar answers to the calculator.

Do I need to include the neutral and ground for my calculation for deratings and Conduit Size.

Based on Conduit Fill Calculator, I believe I can fit about 17 conductors for 3/0 or about 11 Conductors for 4" for 250kcmil in a 4" Conduit.

So instead of running 2 separate 4" Conduit, can I ran everything in 1 4" Conduit.

I believe based on the calculation, will a 2 Sets of 4#250kcmil+1#1/0G in a 4" Conduit work for 400AMP
 
So 2 Sets 2 sets (4# 3/0 + 1 #1/0 Ground) with a 70% factor is only good for 280AMP
Click to expand...


Use the 90° C ampacity for the calculation. Also there should not be an EGC is that raceway.
 
Electricalhelp said:
If I use Table 310.15(B)(3)(a) for more than Three Current Carrying Conductors, since there are 8 Current Carry Conductors, the percent factor is 70%.So 2 Sets 2 sets (4# 3/0 + 1 #1/0 Ground) with a 70% factor is only good for 280AMP
Click to expand...

I looked at this again. There are two problems with this calculation, one you would use the 90° C ampacity of #3/0 for derating so that number is 225 amps.

Secondly unless you're feeding something that will have a non-linear load of more than 50% you only have 6 CCC's (the neutrals do not count as a CCC) which is 80% derating factor not 70%.

So here's the actual calculation:
225 amps*2*80%=360 amps after adjustment. Next size up rule 400 amp OCPD (calculated load would need to be 360 amps or less).

One other note as mentioned this is on the line side of the service disconnect so no EGC's are used.
 
infinity said:
I looked at this again. There are two problems with this calculation, one you would use the 90° C ampacity of #3/0 for derating so that number is 225 amps.

Secondly unless you're feeding something that will have a non-linear load of more than 50% you only have 6 CCC's (the neutrals do not count as a CCC) which is 80% derating factor not 70%.

So here's the actual calculation:
225 amps*2*80%=360 amps after adjustment. Next size up rule 400 amp OCPD (calculated load would need to be 360 amps or less).

One other note as mentioned this is on the line side of the service disconnect so no EGC's are used.
Click to expand...
Why does the neutral not count as a CCC since it's 4#3/0

Based on 310.15(B)(5)(c), it notes that the neutral wire is considered a CCC
 
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What is considered a nonlinear and linear loads?

Since it's a restaurant, wouldn't it be consider a nonlinear load since it varies from day to day
 
Electricalhelp said:
What is considered a nonlinear and linear loads?

Since it's a restaurant, wouldn't it be consider a nonlinear load since it varies from day to day
Click to expand...
No a restaurant would not be considered a non-linear load. Today there are less and less buildings that would have more than 50% (major portion) of their loads as non-linear. Here's some info on neutral conductors and CCC's:

Neutral Conductors:
Click to expand...

Here's some examples of when to count and not count the neutral as a current carrying conductor or CCC:
3Ø- 208Y/120 or 480Y/277 volt system-different circuit types:
A) 2 wire circuit w/ 1 ungrounded, 1 neutral = 2 CCC's
B) 3 wire circuit w/ 2 ungrounded, 1 neutral = 3 CCC's
C) 4 wire circuit w/ 3 ungrounded, 1 neutral = 3 CCC's*
Notes:
A) A normal 2 wire circuit has equal current flowing in each of the circuit conductors so they both count as CCC's.
B) In this circuit the neutral current will be nearly equal to the current in the ungrounded conductors so the neutral counts as a CCC
C) In this circuit the neutral will only carry the imbalance of the current between the three ungrounded conductors so it is not counted as a CCC, with an exception,
*if the current is more than 50% nonlinear (see below for NEC article 100 definition) then the neutral would count as a CCC.

1Ø- 120/240 volt system-different circuit types:
D) 2 wire circuit w/ 1 ungrounded, 1 neutral = 2 CCC's
E) 3 wire circuit w/ 2 ungrounded, 1 neutral = 2 CCC's
Notes:
D) A normal 2 wire circuit has equal current flowing in each of the circuit conductors so they both count as CCC's.
E) In this circuit the neutral will only carry the imbalance between the two ungrounded conductors so the neutral is not counted as a CCC.
Nonlinear Load. A load where the wave shape of the steady-state current does not follow the wave shape of the applied voltage.
Informational Note: Electronic equipment, electronic/electric-discharge lighting, adjustable-speed drive systems, and similar equipment may be nonlinear loads.
Click to expand...
 
Also, single phase L-L loads and 3-phase 3-wire loads can be nonlinear without affecting the current in the neutral at all.
 
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