Restaurant Feeder: Continuous Load, Hot Ceiling, and a 'Close Enough' Conductor
A 208Y/120 V, 3-phase, 4-wire feeder leaves a 400 A switchboard and runs about 70 ft in EMT across a restaurant ceiling to a 225 A panelboard serving kitchen equipment. The load calculation totals 68,000 VA, of which 22,000 VA is continuous equipment that runs more than 3 hours. The ceiling route reaches 108 °F at design conditions. All lugs and breaker terminals in the path are rated 75 °C, and the contractor plans to pull single-rated THWN conductors. The neutral is full size and carries only line-to-neutral imbalance; there are no significant harmonic-producing loads. The contractor proposes 3/0 Cu with a 225 A breaker: '3/0 is good for 200 A at 75 °C, the load is only about 200 A, and a 225 A breaker covers it.'
1. Determine the minimum feeder ampacity required and select a compliant conductor size and overcurrent device. Show every step, including any correction or adjustment factors.
Read the worked answer
1. Continuous-load adder (215.2(A)(1), 215.3):
- 68,000 VA + 0.25 x 22,000 VA = 73,500 VA
2. Feeder current:
- I = 73,500 / (1.732 x 208 V) = 204 A
3. Ambient correction (310.15(B)(2)(a)):
- 108 °F = 42 °C, which falls in the 41–45 °C band.
- Because the conductors are 75 °C-rated and the terminations are 75 °C, the 0.88 factor applies to the 75 °C ampacity column - not the 90 °C column.
4. Number of current-carrying conductors:
- Only the three phase conductors count. The neutral carries just imbalance from linear loads, so it is not a current-carrying conductor (310.15(B)(5)). No 80% adjustment applies.
5. Required uncorrected 75 °C ampacity:
- 204 / 0.88 = 232 A
6. Conductor selection (Table 310.15(B)(16), 75 °C column):
- 4/0 Cu = 230 A → 230 x 0.88 = 202 A < 204 A, fails
- 250 kcmil Cu = 255 A → 255 x 0.88 = 224 A ≥ 204 A, acceptable
7. Overcurrent device:
- 215.3 requires at least 204 A. Use a 225 A device.
- 240.4(B) allows the next standard rating above a non-standard ampacity; 224 A is not a standard size and 225 A does not exceed 800 A, so 225 A protecting 250 kcmil is permitted.
Answer: 250 kcmil Cu THWN with a 225 A breaker. Also verify that both the switchboard lug and the panelboard main lug accept 250 kcmil.
2. The contractor argues the 3/0 proposal should be accepted because the difference is 'only a few amps.' Evaluate the proposal.
Read the worked answer
The proposal fails on two independent grounds:
- Ampacity before any correction: 3/0 Cu at 75 °C is 200 A, which is already below the 204 A required by the 125% continuous-load rule. The load calculation does not round down.
- After ambient correction: 200 x 0.88 = 176 A, roughly 28 A short of the requirement.
A 225 A breaker on 3/0 would violate 240.4, because the conductors would not be protected at their ampacity, and it would violate 215.3, because the device would be sized for a load the conductors cannot carry. 'Close enough' is not a code concept; the arithmetic either satisfies 215.2/215.3/240.4 or it does not. The ambient correction is the trap here - at 30 °C ambient the 3/0 would look marginal, and in a 42 °C ceiling it is clearly inadequate.
3. The owner refuses to buy 250 kcmil. List the compliant options and explain why simply rerouting the conduit through the conditioned dining-room ceiling does not solve the problem.
Read the worked answer
Why rerouting alone fails: Moving the raceway to a space at or below 30 °C removes the 0.88 factor, but 3/0 Cu is still only 200 A at 75 °C, which is less than the 204 A the load requires. Ambient was only part of the problem.
Compliant options:
- Reduce the continuous load below the value where 125% of it plus the noncontinuous load yields 200 A or less - e.g., replace or timer-control some of the 22,000 VA of continuous kitchen equipment. Re-run the calculation; the load calculation must be revised, not just asserted.
- Split the load into two feeders/panelboards so each feeder's calculated ampacity falls within the 3/0 ampacity at the actual ambient.
- Install the larger conductor (250 kcmil with 0.88 correction, or a larger size if the ambient is worse than assumed).
If a different conductor type is proposed, note that a higher-temperature insulation only helps if the termination temperature and the ampacity method used are consistent with 110.14(C); the terminations here are 75 °C, so that avenue is limited. Document whichever option is chosen with a revised load calculation.