6 circuits in 1 conduit. still not sure

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collier

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Can you install 6 circuits (9 wires) of #12 thhn in one conduit under the following conditions:

Given:
? 20 amp single pole breakers (not concerned with handle tie code issue for now)
? 6 circuits in one conduit ? phase A,B,C, and N, Phase A,B,C and N, 1 equipment ground)
? Assume a continuous load of 16 amps or less per circuit of T8 -32 watt fluorescent lights with electronic ballasts
? Assume voltage at panel is 120/208 3 phase 4 wire
? Assume ambient temperature is 110 degrees F

Not sure if the below is correct:
? Do you derate for 4-6 conductors @ 80% or
? for 7-9 conductors @ 70% because the fluorescent lighting makes it a non-linear load with harmonics where the 2 neutrals are carrying more than the unbalance?

Step 1--------#12 thhn. 90 degree column = 30 amps x .87 for ambient temperture = 26.1 amp

Step 2--------26.1 amp x 80% (4-6 conductors ? neutral only carrying imbalance) = 20.88 amps or

or

26.1 amp x 70% (7-9 conductors ? neutral carrying more than unbalance because nonlinear/harmonics
present) = 18.27 amps

1. Is the ampacity of the #12 now 18.27 amps?
2. Does the installation still meet NEC code because?.

A continuous load on a 20 amp breaker can be no more than 80% or 16 amps??and the newly derated #12 wire now has an ampacity of 18.27, which is greater than the 16 amp load?
 
70% for derating. Eight CCCs (grounding only counts for 'fill).

You are correct in your calculations, but you can't use a 20a breaker. You can either install 10s, or a 15a breaker and break it up into 2 circuits.
 
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you lost me there. i can't use a 20 amp breaker on 16 amp (or less) continuous loaded circuit? why not?

there are 6 circuits. each pulling 16 amps or less
 
collier said:
you lost me there. i can't use a 20 amp breaker on 16 amp (or less) continuous loaded circuit? why not?

there are 6 circuits. each pulling 16 amps or less
Click to expand...

In this case (lighting circuits) you could use 240.4(B) and round up to the 20 amp breaker. If the circuit was feeding receptacles you could not use 240.4(B) and the 18.27 rating of the conductor would hold you back to a 15 amp breaker.
 
iwire said:
In this case (lighting circuits) you could use 240.4(B) and round up to the 20 amp breaker. If the circuit was feeding receptacles you could not use 240.4(B) and the 18.27 rating of the conductor would hold you back to a 15 amp breaker.
Click to expand...

I agree, the fact that the load is only lighting 240.4(B) will allow you to go up to the next standard size OCPD which would be 20 amps.
 
Volta said:
If the circuit is pulling 16 amps for 3 hours, 16 x 1.25 = 20a ampacity needed.

Collier showed that the ampacity of the #12 @ 110 deg @ 70 % (I didn't check the math) is 18.27 a.

So 18.27a x .8 = 14.616 max load continuous.

That sound right?
Click to expand...

Reading 210.19(A)(1), I understand the ampacity rating of the conductor needs to be 125% of the continuous load before adjustment or correction factors, not after.
 
hardworkingstiff said:
Reading 210.19(A)(1), I understand the ampacity rating of the conductor needs to be 125% of the continuous load before adjustment or correction factors, not after.
Click to expand...

I think that's what is happening.

If the load was 14.6 a, x 1.25 continuous, then 18.25 amp minimum conductor size.

Then look for a wire worth 18.25 or more after derating and ambient adjustment.

#12 THHN
20 a breaker
14.6 a load max
 
Volta said:
I think that's what is happening.

If the load was 14.6 a, x 1.25 continuous, then 18.25 amp minimum conductor size.

Then look for a wire worth 18.25 or more after derating and ambient adjustment.

#12 THHN
20 a breaker
14.6 a load max
Click to expand...

Load = 16-amps. Since it's continuous, 16 * 1.25 = 20-amp conductor required. Before any adjustments, #12 is rated for 20-amps so it is OK.

Now you go through the adjustments as posted earlier, and after the adjustments the #12 is good for 18.25-amps, which is more than the 16-amp load. Since this is lighting, you may install a 20-amp breaker.
 
Volta said:
So 18.27a x .8 = 14.616 max load continuous
Click to expand...

(Volta) Ballast-listed amps divides into 14.616

(Lou) Ballast-listed amps * 1.25 divides into 18.27

The same continuous load * ccc's * ambient-adjustment factors multiply in any order, so both results are NEC perfect --if all equipment is listed for 75?C--.

The NEC finds ballasts highly inductive for switches 600.6(A)(2)(B), but not for derating. NFPA-70 derating ignores these inductive loads and power factors for motors, see Table 430.250 *Note Amps adjustment x 1.25 ignored when derating.

See how equipment labels not adjusted for extra load/pwrFactor will result in conductors that exceed temperature limits of equipment & terminations 110.14(C)(1), perhaps perpetuating the cottege industries for thermal-camera operators, Electrical-fire expert litigators, and insurers disqualifying claims & contractors that lacked engineering supervision.

The UL General Information Directory (White Book,page 3) clearly indicates that the 60?C and 75?C provisions for equipment have been determined using conductors from Table 310.16.

The problem is the NFPA, publishing NEC tables to avoid International Standards & formulas.
 
iwire said:
I have no idea
Click to expand...

IMO, lighting-ballast conductors should be adjusted for inductive load?

The NEC specifically adjusts inductive-motor loads in Table 430.250 *Note x 1.25, and finds ballasts highly inductive for switches 600.6(A)(2)(B); but this inductive-load adjustment (higher-current/power-factor) gets ignored when derating from NEC 310.15?

Using these NFPA Table adjustments, missing this inductive adjustment, does anyone see those #12 conductor temperatures around ~75?c, violating 110.14(C)(1) with a false sense of safety?
 
For electronic ballasts, I would utilize a #10 neutral conductor on the main multi-wire feeders.

Gotta consider the extra loading of the non-linear characteristics of fully-loaded circuits here.
 
iwire said:
In this case (lighting circuits) you could use 240.4(B) and round up to the 20 amp breaker. If the circuit was feeding receptacles you could not use 240.4(B) and the 18.27 rating of the conductor would hold you back to a 15 amp breaker.
Click to expand...
Can you load a multi-outlet circuit at a point above what would be dereated at 7-9 conductors? Unless the ambient temp was pretty high, or the load continuous.

Started to do the math and realized the limit was much closer than I thought..... ;) 96+
 
hardworkingstiff said:
Load = 16-amps. Since it's continuous, 16 * 1.25 = 20-amp conductor required. Before any adjustments, #12 is rated for 20-amps so it is OK.
Click to expand...
Ok

Now you go through the adjustments as posted earlier, and after the adjustments the #12 is good for 18.25-amps, which is more than the 16-amp load.
Click to expand...
So we don't need the conductor feeding a continuous load to be 125% of load after all is said and done? Only before adjustments.
I think you are right, just never looked at it that way.
 
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