Ampacity of 3 or 4 conductor cable

[Lars Holm]

Hello forum

I just need to be sure that I have understood the NEC correctly, as I am a European citizen!

A 3 conductor feeder cable - Size 4/0 - Temp 90°C/194°F - feeds, for example, a Single Phase installation.
The cable can handle 260A (NEC 310.16)- is this correct?

A similar feeder cable with 4 conductors - for example for use of 3PH + Neutral - must the ampacity be reduced 20 % due to the extra conductor.
So max 208A (NEC 310.15(C)(1))- is this also correct?

We disregard other factors such as ambient temperature etc.

Thanks to all,
Lars Holm

 

[LarryFine]

Welcome to the forum.

No, the neutral of a 1ph/3w or a 3ph/4w circuit, feeder, etc., need not be counted as a CCC.

 

[infinity]

A 3 conductor feeder cable - Size 4/0 - Temp 90°C/194°F - feeds, for example, a Single Phase installation.
The cable can handle 260A (NEC 310.16)- is this correct?

The terminals and equipment connected to the cable will have a 75° C rating so you need to use the 75° C ampacity.

 

[don_resqcapt19]

Hello forum

I just need to be sure that I have understood the NEC correctly, as I am a European citizen!

A 3 conductor feeder cable - Size 4/0 - Temp 90°C/194°F - feeds, for example, a Single Phase installation.
The cable can handle 260A (NEC 310.16)- is this correct?

A similar feeder cable with 4 conductors - for example for use of 3PH + Neutral - must the ampacity be reduced 20 % due to the extra conductor.
So max 208A (NEC 310.15(C)(1))- is this also correct?

We disregard other factors such as ambient temperature etc.

Thanks to all,
Lars Holm

If you have more than 3 current carrying conductors in the cable you must adjust the conductor ampacity. If the ambient temperature is higher than 30°C for the ampacities shown in Table 310.16, you will have to correct the ampacity for the ambient temperature.

Where both conditions exist you will have to do both.

If you have 4 current carrying conductors and an ambient of 41°C you correct the ampacity by 0.87 for the temperature and then adjust the ampacity by 0.8 for 4 current carrying conductors. If you had a 4/0 THHN conductor, the ampacity of that 4/0 conductor under those conditions would be 260 x 0.87 x 0.8 or 181 amps.

 

[Todd0x1]

Welcome to the forum.

No, the neutral of a 1ph/3w or a 3ph/4w circuit, feeder, etc., need not be counted as a CCC.

unless the 1ph/3w is 2 phases of a 3 phase service yes?

 

[LarryFine]

unless the 1ph/3w is 2 phases of a 3 phase service yes?

Correctamundo.

 

[LarryFine]

A similar feeder cable with 4 conductors - for example for use of 3PH + Neutral - must the ampacity be reduced 20 % due to the extra conductor.

Just to reiterate, the answer to this question as a stand-alone question is no under almost all conditions.

 

[Julius Right]

According to NEC [2023 ed.] Art.310.15 Ampacity Tables.

Par.(E) Neutral Conductor

(3) On a 4-wire, 3-phase wye circuit where the major portion

of the load consists of nonlinear load- harmonic currents

are present in the neutral conductor- the neutral conductor

shall therefore be considered a current-carrying conductor.

According to Table 310.15(C) ( 1 ) for 4 loaded wires the ampacity shall be reduced to 80%.However, no referring is noted about percent of harmonics load.

However, IEC 60364-5-52 it is more detailed [see Annex D for instance].

 

[infinity]

According to NEC [2023 ed.] Art.310.15 Ampacity Tables.

Par.(E) Neutral Conductor

(3) On a 4-wire, 3-phase wye circuit where the major portion

of the load consists of nonlinear load- harmonic currents

are present in the neutral conductor- the neutral conductor

shall therefore be considered a current-carrying conductor.

According to Table 310.15(C) ( 1 ) for 4 loaded wires the ampacity shall be reduced to 80%.However, no referring is noted about percent of harmonics load.

However, IEC 60364-5-52 it is more detailed [see Annex D for instance].

The percentage of harmonics is more than 50% due to the words major portion. Anything less than 50% is a minor portion so the neutral is not considered a CCC.

 

[Lars Holm]

All,
Thank you for your time to answer!
I just want to try simple dimensioning, so please correct if necessary...

One Feeder (AWG 2/0-4C) feeds a Sub Panel.
The Feeder run close in Cable Tray with one (1) AWG12/4C, and one (1) AWG12/3C, the temp. = 30°C
The last part of the Cable Tray the Feeder run alone, temp. = 41°C.
The Cables, AWG 2/0-4C, 90°, current max 195A
The Breaker is molded type used for back-feed the Sub-Panel

The task;

• What size Breaker must be placed to protect the Feeder?
• What is the max Ampacity load for the Sub Panel?

Calc:
(TERMINALS) The Current is derated to 75°C due to the terminals from 195 to 175A
(CCC 11 EA WIRES) The Current is derated 50% due to 11 ea. CCC, (I use the value 195A as the terminals are not relevant here) = 195*0.5 = 97.5A
(CCC 4 EA WIRES + TEMP) The Current is derated 20% due to 4 ea. CCC and factor 0.87 due to temp. 41°C = 195*0.8*0.87 = 135.7A


Result:
The lowest value of the above is 97.5A, which is therefore the maximum Current for the Cable.

• I choose a molded Breaker, QO390, 3 Pole, 90A, 240V
• As a Breaker must be loaded with a maximum of 80% of the rated current, the current used in Sub Panel must not exceed 90*0.8 = 72A.

Is this calculation really correct???

Thanks and best regards
Lars

 

[wwhitney]

Is this calculation really correct???

I'm not super familiar with cable trays, but I suggest you read NEC 392.80(A) and tell us which subsection covers your installation. Seems like most of those section says "The adjustment factors of 310.15(C)(1) shall not apply to the ampacity of cables in cable trays" or the like. So it's unlikely that the portion of your calculation that considers 11 CCCs is correct.

Cheers, Wayne

 

[Julius Right]

You have to check the allowable width against actual width, according to art.392.22 A 1) c).
For instance, if the cable tray width is 6"[minimum according to NEMA V1], then according to table 392.22 the allowable width for all the cables of diameter less than 4/0 will be 7-1.2*0.528[overall diameter of 4*4/0 cable] =6.3664 inches. The actual remaining width it is 1.39+0.47+0.43=2.29 inches.
Then you can use Table 310.16 [260*0.8=208A].
However, even at the last part 2/0 cable run alone, for 41oC ambient, another reducing factor of 0.87 has to be employed. But, depending on the total length of the cable and the last part length it is permissible to use 208 A instead of 208*0.87 some time (see 310.14 A 2).