Six Conductors in a Hot Ceiling Void
A commercial kitchen has three 20 A, 230 V branch circuits wired with 3.5 mm² THHN copper, all pulled through one 25 mm PVC conduit across a ceiling void. Under full load the circuits draw 16 A, 14 A, and 12 A respectively, and all three are classed as continuous loads. A temperature logger in the void recorded 46 °C at 2 p.m. The designer took the 30 A ampacity of 3.5 mm² THHN from the 90 °C column, applied no adjustment factors, and installed 20 A breakers on all three circuits.
1. Compute the actual permitted ampacity of the 3.5 mm² conductors in this installation.
Read the worked answer
Step 1 - count current-carrying conductors. Three 2-wire circuits = 6 current-carrying conductors in one raceway. Adjustment factor 0.80.
Step 2 - ambient correction. 46 °C falls in the 46–50 °C row: factor 0.82 for a 90 °C-rated insulation.
Step 3 - apply to the 90 °C ampacity.
30 A × 0.80 × 0.82 = 19.7 A
Step 4 - cap at the termination rating. 3.5 mm² at 75 °C ≈ 25 A, which is not the binding limit here. Governing ampacity = 19.7 A.
The design used 30 A. The installation permits 19.7 A. That is a 34 % overstatement, and the 20 A breakers are now one size larger than the conductor can carry.
2. The installer argues that the next-size-up rule lets a 20 A breaker protect a 19.7 A conductor. Is he right?
Read the worked answer
No. The next-size-up allowance is expressly unavailable for conductors that are part of a multioutlet branch circuit supplying receptacles for cord-and-plug-connected portable loads - which is exactly what a kitchen branch circuit is. Even setting that aside, the allowance is a rounding convenience of a fraction of an amp, not a mechanism for a 34 % derating error.
There is a second, independent failure the installer has missed. Continuous loads are limited to 80 % of the breaker rating: 20 A × 0.80 = 16 A. The 16 A circuit is sitting exactly at that limit, so there is zero margin - and the conductor feeding it is only good for 19.7 A. Both the conductor and the continuous-load rule are violated, and the 20 A breaker will not protect the insulation if the ambient climbs a few more degrees on a July afternoon.
3. Compare the practical fixes and state which you would choose.
Read the worked answer
Option A - Upsize the conductors to 5.5 mm². 90 °C ampacity ≈ 40 A; 40 × 0.80 × 0.82 = 26.2 A; 75 °C limit ≈ 35 A; governing 26.2 A ≥ 20 A. The 20 A breakers stay, and the continuous 16 A load is comfortably within 80 % of 20 A. Cost: pulling new conductors.
Option B - Split into two conduits. Two raceways of 3 conductors each: factor 1.00, so 30 × 0.82 = 24.6 A, capped at the 75 °C value of 25 A. 24.6 A is adequate for 20 A breakers. This is often cheaper than repulling if there is a spare route, but it doubles the penetrations and hangers.
Option C - Drop the breakers to 16 A. This does not work: the 16 A continuous circuit needs 125 % of 16 A = 20 A of conductor, and 16 A × 0.80 capacity gives only 12.8 A of continuous allowance. It converts an overload problem into a nuisance-trip problem.
My choice: Option A, unless the route makes Option B trivial. Upsizing also buys margin against the void running hotter than the logger recorded - a 46 °C reading at 2 p.m. in a kitchen ceiling is a snapshot, not a maximum, and the ambient factor steps down again at 51 °C. Verify the final selection against the actual conduit fill and the codebook tables in use.