Dwelling Load Calculations Per the National Electrical Code

Courtesy of www.MikeHolt.com. Based on the 2020 NEC.

Sizing the feeder or service load for a dwelling isn’t simply a matter of totaling the loads or breaker ratings. The Code recognizes there’s load diversity and that not all loads run simultaneously. Therefore, it provides “demand factors” that you apply to calculate the “demand load.”

Article 220 allows two methods of calculating the feeder or service load for a dwelling — the standard method [Part III] and the optional method [Part IV].

Using the standard method is almost impossible without software, because of the complexity of the math and the rules. Plus, many judgment calls are needed to decide which demand factors apply to which load. These issues are solved by using the optional method.

Optional method

How do you use the optional method to determine the demand load on the feeder or service conductors for a dwelling unit?

You can use the optional method only for dwellings served by a single 120/240V or 120/208V, 3-wire set of service or feeder conductors with an ampacity of 100A or larger [Sec. 220.82(A)].

Step 1. Determine the total connected load, less HVAC [Sec. 220.82(B)].

General lighting. The total connected load must include a 3VA per sq ft load for general lighting and general-use receptacles. The total square feet is based on the outside dimensions of the dwelling but doesn’t include open porches, garages, or unused or unfinished spaces not adaptable for future use.
Small-appliance and laundry circuits. Two 20A, 120V small-appliance circuits, and one 20A, 120V circuit for the laundry area [Sec. 210.11(C)(1) and Sec. 210.11(C)(2)] are required for each dwelling. The load for each of these circuits is based on 1,500VA. A laundry circuit isn’t required in a dwelling that’s in a multifamily building with laundry facilities for all building occupants [Sec. 210.52(F), Exception No. 1].
Fixed appliances. The nameplate VA rating of appliances fastened in place, permanently connected, or on a specific circuit, examples include dishwashers and waste disposals. Also include the load ratings for dryers, cooking equipment, and special-purpose circuits for ironing stations, elevators, etc.
Motor VA. The VA nameplate rating of motors not part of an appliance used for irrigation, boat lifts, wells, etc. must be included at 100%.

Step 2. Determine the demand load for the Step 1 loads [Sec. 220.82(B)].

Apply a 100% demand load to the first 10kVA of the total connected load, then a 40% demand to the remainder of the total connected load.

Step 3. Determine the air-conditioning versus heating demand load [Sec. 220.82(C)].

Determine the largest of the following loads and add it to the demand load value from Step 2:

Air-conditioning. 100% of the air-conditioning nameplate rating.
Heat-pump compressor without supplemental heating. 100% of the heat-pump nameplate rating.
Heat-pump compressor and supplemental heating. 100% of the nameplate rating of the heat pump and 65% of the supplemental electric heat rating.
Space-heating, three units or less. 65% of the space-electric heating nameplate rating.
Space-heating, four or more units. 40% of the space-electric heating nameplate rating.

Step 4. Determine the service disconnect and conductor sizing.

Determine the total service demand load in amperes for the dwelling by adding the values of Step 2 and Step 3, then divide the sum by the nominal voltage (208V or 240V). Once you know the service demand load, size the service disconnect for a standard size overcurrent protective device per Sec. 240.6(A).

Dwelling service and feeder conductors supplied by single-phase, 120/240V systems can be sized per Sec. 310.12(A) for service conductors and Sec. 310.12(B) for feeder conductors. Dwelling unit feeder conductors supplied by single-phase, 120/208V systems can be sized per Sec. 310.12(B) for feeder conductors.

Example calculation

Even with the optional method, you must consider many factors, as this example shows.

What size feeder/service conductors are required for a 6,000-sq-ft dwelling containing the following loads? (Figure)

Fixed Appliances and Equipment:

Boat Lift Motor, 1/2 hp at 120V (9.80A)
Cooking, Range, 13,800W at 120/240V
Cooking, Cooktop, 6,000W at 240V
Cooking, Oven, two 3,000W at 120/240V
Dishwasher, 12A/120V
Disposal (Waste), 10A/120V
Dryer, 4,000 W, 120/240V
Electric Vehicle Charger, 40A/240V
Elevator, 24A/240V
Hot Tub/Spa Outlet, 48A/240V
Hydromassage Bathtub, 8A/120V
Hydromassage Heater, 12A/120V
Irrigation Motor, 2 hp at 240V (12A)
Microwave, 16A/120V
Pool Heat-Pump Compressor, 40A/240V
Pool Heat-Pump Fan, 3A/240V
Pool Pump Motor, 2 hp at 240V (12A)
RV Receptacle, 30A/120V
Sauna Receptacle, 20A/240V
Trash Compactor, 8A/120V
Water Heater, 4,500W at 240V
Water Heater (Tankless), 40A/240V
Welder Receptacle, 30A/240V
Well Pump Motor, 3 hp at 240V (17A)

Air-Conditioning and Heating Zone 1:

A/C Condenser, 50A/240V
A/C Condenser Fan, 4A/240V
Electric Heating, 9.60kW
Air Handler, 4A/240V

Air-Conditioning and Heating Zone 2:

A/C Condenser, 50A/240V
A/C Condenser Fan, 3A/240V
Electric Heating, 7.50kW
Air Handler, 3A/240V

Even with the optional method to determine the demand load on the feeder or service conductors for a dwelling unit, you must consider many factors.

Step 1. Determine the total connected load, less air-conditioning, and electric heating [Sec. 220.82(B)].

General Lighting ― 6,000 sq ft × 3VA = 18,000VA

Small-Appliance Circuits ― 1,500VA × 2 circuits = 3,000VA

Laundry Circuit ― 1,500VA

Fixed Appliances and Equipment:

Boat Lift Motor, 1/2 hp at 120V ― 120V × 9.80A = 1,176VA
Cooking, Range ― 13,800W = 13,800W
Cooking, Cooktop ― 6,000W = 6,000W
Cooking, Ovens, Two ― 3,000W = 6,000W
Dishwasher ― 12A × 120V = 1,440VA
Disposal (Waste) ― 10A × 120V = 1,200VA
Dryer ― 5,000W = 5,000W
Electric Vehicle Charger ― 40A × 240V = 9,600VA
Elevator ― 24A × 240V = 5,760VA
Hot Tub/Spa Outlet ― 48A × 240V = 11,520VA
Hydromassage Bathtub ― 8A × 120V = 960VA
Hydromassage Heater ― 12A × 120V = 1,440VA
Ironing Center ― 8A × 120V = 960VA
Irrigation Motor, 2 hp at 240V ― 240V × 12A = 2,880VA
Microwave ― 16A × 120V = 1,920VA
Pool Heat-Pump Compressor ― 40A × 240V = 9,600VA
Pool Heat-Pump Fan ― 3A × 240V = 720VA
Pool Motor ― 12A × 240V = 2,880VA
RV Receptacle ― 30A × 120V = 3,600VA
Sauna Receptacle ― 20A × 240V = 4,800VA
Trash Compactor ― 8A × 120V = 960VA
Water Heater ― 4,500W = 4,500W
Water Heater (Tankless) ― 40A × 240V = 9,600VA
Welder Receptacle ― 30A × 240V = 7,200VA
Well Pump Motor, 3 hp at 240V ― 240V × 17A = 4,080VA

Total Connected Load = 140,096VA

Step 2. Determine the demand load for Step 1 loads [Sec. 220.82(B)].

Connected Load = 140,096VA

First 10kW at 100% ― 10,000VA × 100% = 10,000VA

Remainder at 40% ― 130,096VA × 40% = 52,038VA

Demand Load = 10,000VA + 52038VA = 62,038VA

Step 3. Determine the air-conditioning versus electric heating demand load [Sec. 220.82(C)]; air-conditioning at 100% [Sec. 220.82(C)(1)] versus electric heating at 65% [Sec. 220.82(C)(4)].

Air-Conditioning = Volts × Amperes

Zone 1
A/C Condenser = 240V × 50A = 12,000VA
A/C Condenser Fan = 240V × 4A = 960VA
Zone 1 Air-Conditioning Total = 12,000VA + 960VA = 12,960VA

Zone 2
A/C Condenser = 240V × 50A = 12,000VA
A/C Condenser Fan = 240V × 3A = 720VA
Zone 2 Air-Conditioning Total = 12,000VA + 720VA = 12,720VA

Total Air-Conditioning Demand Load (Zone 1 + Zone 2) = 12,960VA + 12,720VA = 25,680VA

Electric Heating at 65%

Electric Heating Zone 1 = 9,600W
Electric Heating Zone 2 = 7,500W
Total Connected Space Heating (Zone 1 + Zone 2) = 9,600W + 7,500W = 17,100W

Total Space Heating Demand Load = 17,100W × 65% = 11,115W (omit) [Sec. 220.82(C)]

Air Handler at 100%

Air Handler = Volts × Amperes

Air Handler, Zone 1 = 240V × 4A = 960VA
Air Handler, Zone 2 = 240V × 3A = 720VA
Total Air Handler Demand Load = 960VA + 720VA = 1,680VA

Air-Conditioning Demand Load = 25,680W + 1,680VA = 27,360VA

Step 4. Determine the service disconnect and conductor sizing.

Demand Load = 62,038VA

Air-Conditioning Load = 27,360VA

Total Demand Load = 62,038VA + 27,360VA = 89,398VA

Service Size in Amperes = VA Demand Load ÷ System Volts

Service Size in Amperes = 89,398VA ÷ 240V = 372A

Using the next size up rule, the service disconnect size should be 400A. Based on Table 310.12, use 400kcmil copper or 600kcmil aluminum wire for this load.

Familiarity breeds accuracy

It’s important to be familiar with the requirements of Art. 220 so you correctly apply the demand factors for a given installation. Time spent studying those requirements will reduce your errors and make you more efficient.

Article 220 presents various tables and specific requirements to follow. Review and apply each of them one at a time, and you’ll be well on the road to mastering these calculations.

These materials are provided to us by Mike Holt Enterprises in Leesburg, Fla. To view Code training materials offered by this company, visit www.mikeholt.com/code.