Optional Method Load Calculation for a 200 A Dwelling Service
An 1,800 ft² single-family dwelling is being rewired with a 120/240 V, single-phase service. The owner wants a 200 A service. The calculated general lighting load is 3 VA/ft²; there are two 1,500 VA small-appliance branch circuits, one 1,500 VA laundry circuit, a 12 kW range, a 5.5 kW dryer, a 4.5 kW water heater, a 1.0 kW disposal, a 1.5 kW dishwasher, a 4.0 kW air conditioner, and 8.0 kW of electric resistance heat. The heat and air conditioner are not intended to operate at the same time. The plan reviewer used the optional method in 220.82 and came up with a demand of about 116 A, then approved the 200 A service without checking the neutral. The contractor asks whether the 200 A service is adequate and whether the neutral can be reduced.
1. Using the optional method in 220.82, calculate the service demand and determine whether the 200 A service is adequate.
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Under 220.82(B), the general lighting, small-appliance, and laundry loads are added first: 1,800 ft² × 3 VA/ft² = 5,400 VA; two small-appliance circuits = 3,000 VA; laundry = 1,500 VA; total = 9,900 VA. Add the named appliances at nameplate rating: range 12,000 VA, dryer 5,500 VA, water heater 4,500 VA, disposal 1,000 VA, and dishwasher 1,500 VA, for 24,500 VA. The combined total is 34,400 VA. Apply the optional-method demand factors: first 10,000 VA at 100% = 10,000 VA; remaining 24,400 VA at 40% = 9,760 VA; subtotal = 19,760 VA. Under 220.82(C), add the larger of the heating or air-conditioning load, not both. The 8,000 VA heat is larger than the 4,000 VA air conditioner, so add 8,000 VA. Total demand = 27,760 VA. At 240 V, that is 27,760 ÷ 240 = 115.7 A. A 200 A service is more than adequate for the calculated demand, even after allowing for the service rating rules.
2. How should the heating and air-conditioning loads be handled, and what if the heat pump and resistance heat can operate together?
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The optional method does not simply add all heating and cooling equipment. Under 220.82(C), the calculation uses the larger of the heating or air-conditioning demand, with specific rules for heat pumps and supplemental resistance heat. In this case the 8 kW resistance heat is larger than the 4 kW air conditioner, so only the 8 kW is added. If the heat pump and supplemental resistance heat can operate simultaneously, the calculation must reflect that combination rather than treating the resistance heat as a separate either/or load. The same principle applies if the air conditioner and heat are interlocked or controlled so that only one can run at a time; the load calculation may use the larger. The plan reviewer's 115.7 A result is consistent with the optional method as long as the heating and cooling equipment cannot operate together.
3. Can the service neutral be reduced? Explain the code basis and the limit.
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The neutral is not sized by simply matching the ungrounded service conductors. The service neutral demand is calculated under 220.61, which permits certain demand factors and allows 70% of the neutral load for electric ranges and dryers under 220.61(B). The calculated neutral load may therefore be less than the ungrounded load, and the neutral may be reduced accordingly. However, the reduced neutral must still comply with 250.24(C) for service grounded conductors and with 310.12 for dwelling service conductors. It cannot be reduced below the minimum required for the service rating or for fault-current return. The 200 A service does not require a full-size neutral if the calculated neutral load is lower, but the contractor must document the calculation and verify termination temperature ratings. The neutral must also be large enough for the available fault current and the grounding electrode conductor connection.