Rooftop Raceway: Ambient Correction, Derating, Voltage Drop, and Fill in One Decision
A 208Y/120 V, 3-phase, 4-wire circuit supplies a 24 A continuous load on a rooftop mechanical platform. It leaves a panel 185 ft away and runs in 3/4 in EMT across the roof, where the measured ambient inside the raceway area is 118 F (48 C). Conductors are THHN, terminations are rated 75 C, and the design calls for three phase conductors plus a neutral plus an equipment grounding conductor. The inspector has already told the crew that the roof ambient is real, not theoretical, and that he will look at both ampacity and voltage drop. A spare 20 A GFCI receptacle circuit for servicing equipment is also being requested on the same run.
1. Determine the minimum conductor size for the 24 A continuous load, showing ambient correction and conductor adjustment.
Read the worked answer
Step 1 - circuit rating. 24 A continuous requires a 30 A branch circuit (24 x 125%). Conductors must have a corrected ampacity of at least that 30 A so a 30 A device protects them.
Step 2 - adjustment for current-carrying conductors. 4 current-carrying conductors (3 phase + neutral, since the neutral carries the unbalanced load of line-to-neutral loads) falls in the 4-to-6 range: 0.80 per 310.15(C)(1).
Step 3 - ambient correction. 118 F = 48 C, which is the 46-to-50 C band; 90 C column factor is 0.82 per 310.15(B)(1). Use the 90 C column for correction and adjustment because the conductor is THHN, then compare the result to the 75 C ampacity for the termination check.
Step 4 - try #10 Cu THHN
40 A x 0.80 x 0.82 = 26.2 A
26.2 A is below the 30 A circuit rating, so #10 fails. Do not lean on 240.4(B) here; the corrected ampacity is under the device rating.
Step 5 - #8 Cu THHN
55 A x 0.80 x 0.82 = 36.1 A - acceptable, and #8 at 75 C (50 A) satisfies the termination check.
Minimum is #8 Cu, protected at 30 A.
2. Check voltage drop on the 185 ft run and decide whether to upsize beyond the ampacity minimum.
Read the worked answer
Three-phase voltage drop: VD = 1.732 x K x I x L / cmil, with K = 12.9 for copper.
#8 Cu (16,510 cmil):
VD = 1.732 x 12.9 x 24 x 185 / 16,510 = 99,197 / 16,510 = 6.0 V, which is 2.9% of 208 V.
#10 Cu (10,380 cmil):
VD = 99,197 / 10,380 = 9.6 V, or 4.6% - over the 3% branch-circuit recommendation.
So #8 lands just under the 3% target, and it is already the ampacity minimum, which makes it an easy approval. Two cautions before you sign off:
- The 3% branch-circuit figure is an informational recommendation in 210.19(A) Informational Note, not a mandatory rule. If the equipment is sensitive or the utility voltage is low, say so in writing.
- The margin is 0.1%. Any field change that adds length or load pushes it over. If the run grows past roughly 190 ft, go to #6 Cu and re-run the correction (75 A x 0.80 x 0.82 = 49.2 A).
3. The GFCI receptacle circuit is added to the same 3/4 in EMT. Check conduit fill and determine the conductor size for that circuit.
Read the worked answer
Fill check, Chapter 9 Table 1 and Table 4. 3/4 in EMT at 40% fill = 0.213 sq in.
Existing: 4 x #8 THHN (0.0366 each) = 0.1464; plus 1 x #10 THHN EGC (0.0211) = 0.1675 sq in.
Adding 2 x #12 THHN (0.0133 each) = 0.0266: total 0.1941 sq in, which is 91% of the allowable 0.213. It fits - barely - and the pull will be stiff with five conductors on a long rooftop run. Note that if the neutral were counted as a current-carrying conductor for a line-to-neutral load, you now have 6 current-carrying conductors and the adjustment stays at 0.80, but the circuit mix still has to be evaluated.
Conductor size for the GFCI circuit. This is where the fill answer gets interesting. With 6 current-carrying conductors at 48 C ambient:
- #12 THHN: 30 A x 0.80 x 0.82 = 19.7 A - below a 20 A device rating, so #12 Cu cannot be used for a 20 A circuit here
- #10 THHN: 40 A x 0.80 x 0.82 = 26.2 A - acceptable, protects at 20 A, and drops the fill back down because #10 is the same size class as the existing conductors
Recommendation: run the receptacle circuit in #10 Cu on a 20 A GFCI breaker, and re-verify fill with the actual conductor list (2 x #10 instead of 2 x #12 adds 0.0156 sq in, total 0.210 sq in - still inside 0.213, but that is a hard stop). If the receptacle circuit is not essential, a separate small raceway is cleaner than a run at 98% fill.