Nine Conductors in a Hot Mechanical Room
You are pulling circuits in a mechanical room feeding four rooftop units. One 27 mm EMT, 40 m long, leaves a 120/208 V panel and carries nine 12 AWG RW90 current-carrying conductors - three 20 A, 120 V circuits for RTU control transformers and condensate pumps - plus one 12 AWG bonding conductor. Each circuit is protected at 20 A by a breaker with 75 °C terminals, and the panel bus terminations are also rated 75 °C. A thermometer hung on the conduit run reads 42 °C at 2 p.m. in July; the room has no ventilation and the boiler is out of service. Conduit fill is under 40 % for the run. The foreman says: "12 AWG RW90 is 30 amps, you're fine at 20 - stop overthinking it." The home runs are already pulled and the panel is scheduled to be trimmed out tomorrow.
1. Calculate the corrected ampacity of these 12 AWG conductors and state whether the 20 A breakers comply.
Read the worked answer
Take the derating factors in the right order: start from the 90 °C column, then apply ambient, then apply conductor count.
- Base ampacity: 12 AWG RW90 in Table 2, 90 °C column = 30 A. The foreman stopped here.
- Ambient correction (Table 5A, 90 °C column): the room is 42 °C. Correction factors are tabulated in 5 °C steps and you do not interpolate in the safe direction - use the next higher ambient, 45 °C, factor 0.87. (At 40 °C it would be 0.91; measure and document 40 °C if you want that number.)
- Conductor count (Table 5C): nine current-carrying conductors in one raceway falls in the 7-9 band, factor 0.70.
Corrected ampacity = 30 x 0.87 x 0.70 = 18.3 A.
Does the 20 A breaker comply? No. Rule 14-104 requires the overcurrent device not to exceed the ampacity of the conductors it protects, so a 20 A breaker on 12 AWG needs at least 20 A of ampacity. 18.3 A fails, and the load cannot be counted on to stay under 18.3 A either, since these are 20 A circuits.
Do not forget the termination limit. Rule 4-006 limits the usable ampacity to the lowest temperature rating in the circuit. The breaker and panel are 75 °C, so the 75 °C column (25 A for 12 AWG) is the ceiling for the final connection. Applying the same factors: 25 x 0.87 x 0.70 = 15.2 A. The conductors are thermally comfortable; the terminals are the tighter constraint and the circuit is non-compliant on both counts.
2. Give three compliant fixes with the numbers for each, and say which you would actually choose given tomorrow's trim-out.
Read the worked answer
Option 1 - split the raceway. Three conduits of three current-carrying conductors each removes the Table 5C factor entirely (no correction for three or fewer). Ampacity = 30 x 0.87 = 26.1 A, terminal-limited to 25 x 0.87 = 21.8 A. Both exceed 20 A. Cost: two extra runs of EMT and re-pulling; the existing conduit is inside 40 % fill, so the wire fits, but the conduits do not exist yet.
Option 2 - upsize the conductors to 10 AWG RW90. 10 AWG 90 °C = 40 A; 40 x 0.87 x 0.70 = 24.4 A, terminal-limited 35 x 0.87 x 0.70 = 21.3 A. Both clear 20 A. Cost: re-pull four home runs and confirm the 20 A breakers still suit the load. This is the least disruptive to the trim-out because the raceway and fill stay as built.
Option 3 - fix the environment. Ventilate or cool the room so the measured ambient at the conduit is 30 °C or less: 30 x 1.00 x 0.70 = 21.0 A and the terminal check 25 x 1.00 x 0.70 = 17.5 A, which still fails the terminal test. So ambient alone is not enough with nine conductors - a useful reminder that the conductor-count factor, not the heat, is the dominant problem here.
What I would choose: Option 2 tomorrow (re-pull with 10 AWG, keep the raceway, keep the schedule) and recommend Option 1 at the next available shutdown. Note the decision and the calculation on the as-built so the next person does not re-derive it.
3. The installer asks whether the bonding conductor counts in the nine, and whether the neutral of a 120/240 V multiwire branch circuit would count. Answer both.
Read the worked answer
Bonding conductor: does not count. Table 5C applies to current-carrying conductors. A bonding conductor carries current only during a fault, so it is excluded from the count. That is why the run is described as nine current-carrying conductors plus one bonding conductor, not ten.
Neutral of a multiwire branch circuit: it depends, and this is where people get it wrong.
- On a 120/240 V or 120/208 V multiwire circuit where the two ungrounded conductors are on opposite legs and the neutral carries only the unbalanced current, the neutral is not counted as current-carrying for Table 5C purposes.
- If the neutral is connected to a nonlinear load that produces triplen harmonics (electronic ballasts, dimmers, switch-mode power supplies - common on RTU controls), the neutral can carry harmonic current even when the phase currents balance, and good practice is to count it. If the circuit consists of only two phase conductors and a neutral from a 3-phase wye system, the neutral is counted as current-carrying regardless, because it carries the full phase current.
Practical consequence for this job: if any of the three 120 V circuits were reworked as a shared-neutral multiwire circuit with the neutral not counted, the count could drop into a lower Table 5C band and the ampacity would improve. That is a legitimate design route, but only if the two phases are on different legs, a common-trip two-pole breaker is used, and there is no harmonic concern. Given RTU control transformers, assume harmonics and do not take the credit without a load study.