12 AWG in a Hot Mechanical Room: Ambient, Fill, and Derating
A 20-ampere branch circuit supplies sales-floor lighting in a warehouse. From the panel, the circuit runs 200 ft through a mechanical room where the ceiling ambient is measured at 113°F, then continues in the same 3/4-inch EMT to the sales floor. The conduit now contains nine 12 AWG THHN copper current-carrying conductors, and the connected lighting load is 16 A continuous on the circuit in question. The breaker is a standard 20-ampere thermal-magnetic type. Nothing in the run is bundled or covered with insulation, and terminations at both ends are rated 75°C. The owner wants to add one more lighting circuit to the same conduit next month.
1. Is the 20 A breaker with 12 AWG THHN acceptable for this circuit? Show the ampacity check and explain which factor actually governs.
Read the worked answer
Start from the 90°C ampacity of 12 AWG THHN (30 A), because adjustment and correction factors may be applied to that column; 110.14(C) then limits the final number to the 75°C termination rating (25 A for 12 AWG), which is not the binding constraint here.
- Ambient correction at 113°F (45°C), 90°C column, Table 310.15(B)(1): 0.82
- Nine current-carrying conductors in one raceway, 310.15(C)(1): 70%
30 A × 0.82 × 0.70 = 17.2 A
That is below the 20 A breaker and below the 20 A required for a 16 A continuous load (125% under 210.19(A)(1)). Because 240.4 requires the conductors to be protected at their ampacity, the circuit is noncompliant.
Note what is not the problem: conduit fill. Nine 12 AWG THHN take about 0.12 in² against roughly 0.213 in² permitted at 40% fill in 3/4-inch EMT (Chapter 9, Table 1). The heat, not the space, is the issue.
2. What is the smallest copper conductor that works in this raceway with the 20 A breaker, and does adding the owner's fourth lighting circuit change that?
Read the worked answer
Now: try 10 AWG. 40 A × 0.82 × 0.70 = 22.96 A, which exceeds 20 A, so 10 AWG THHN is the smallest that works. Fill still passes: nine 10 AWG THHN ≈ 0.19 in² versus 0.213 in² permitted.
After the addition: a fourth circuit means 12 current-carrying conductors, which moves the adjustment from 70% to 50% (310.15(C)(1)).
- 10 AWG: 40 × 0.82 × 0.50 = 16.4 A - fails a 20 A circuit.
- 8 AWG: 55 × 0.82 × 0.50 = 22.6 A - works.
So the practical answer is either upsize now to 8 AWG or plan a second raceway; pulling 10 AWG and adding a circuit next month means re-pulling. Once 12 conductors are in the pipe, fill also becomes tight, so splitting the conduit is usually the better judgment call.
3. The circuit is 200 ft long. Even with the code-minimum conductor, is the installation a good design? Support your answer with numbers.
Read the worked answer
Do a voltage-drop check at 120 V, 16 A, 200 ft one-way (Chapter 9, Table 9 AC resistance, 75°C):
- 10 AWG (1.21 Ω/kft): 2 × 1.21 × 0.2 × 16 ≈ 7.7 V, about 6.5%
- 8 AWG (0.764 Ω/kft): ≈ 4.9 V, about 4.1%
- 6 AWG (0.491 Ω/kft): ≈ 3.1 V, about 2.6%
The NEC does not mandate 3% for branch circuits (see the informational notes to 210.19(A) and 215.2(A)); it is a design and specification target. But a 6.5% drop on a lighting circuit means noticeably dim lamps at the far end and possible driver/ballast problems.
Judgment: meet the code minimum by calculation, then sell the customer 6 AWG (or a closer panel) if the run stays at 200 ft. Document the choice on the proposal so the inspector's code-minimum pass and the customer's performance expectation are both satisfied.