Dwelling Load Calculations

Figure 01

Note: This article is based on the 2020 NEC.

The standard method for sizing a dwelling feeder or service load presents several problems that are all solved by the optional method.

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 [220.82(A)].

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

• General lighting. The total connected load must include a 3 VA 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 [210.11(C)(1) and 210.11(C)(2)] are required for each dwelling. The load for each of these circuits is based on 1,500 VA.
• A laundry circuit isn't required in a dwelling that's in a multifamily building with laundry facilities for all building occupants [210.52(F) Ex 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 [220.82(B)].

Apply a 100% demand load to the first 10 kVA 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 [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 and 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 dividing 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 240.6(A).

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

Example
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 01
(a) 400 kcmil copper     (b) 600 kcmil aluminum     (c) a or b     (d) 750 kcmil
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
Ironing Center, 8A/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, 9.60 kW
Air Handler, 4A/240V

Air-Conditioning and Heating Zone 2:
A/C Condenser, 50A/240V
A/C Condenser Fan, 3A/240V
Electric Heating, 7.50 kW
Air Handler, 3A/240V

Step 1. Determine the total connected load, less air-conditioning and electric heating [220.82(B)].
General Lighting 6,000 sq ft × 3 VA = 18,000 VA
Small-Appliance Circuits 1,500 VA × 2 circuits = 3,000 VA
Laundry Circuit 1,500 VA
Fixed Appliances and Equipment
Boat Lift Motor, 1/2 hp at 120V 120V × 9.80A = 1,176 VA
Cooking, Range 13,800 W = 13,800 W
Cooking, Cooktop 6,000 W = 6,000 W
Cooking, Ovens, Two 3,000 W= 6,000 W
Dishwasher 12A × 120V = 1,440 VA
Disposal (Waste) 10A × 120V = 1,200 VA
Dryer 4,000 W= 5,000 W
Electric Vehicle Charger 40A × 240V = 9,600 VA
Elevator 24A × 240V = 5,760 VA
Hot Tub/Spa Outlet 48A × 240V = 11,520 VA
Hydromassage Bathtub 8A × 120V = 960 VA
Hydromassage Heater 12A × 120V = 1,440 VA
Ironing Center 8A × 120V = 960 VA
Irrigation Motor, 2 hp at 240V 240V × 12A = 2,880 VA
Microwave 16A × 120V = 1,920 VA
Pool Heat-Pump Compressor 40A × 240V = 9,600 VA
Pool Heat-Pump Fan 3A × 240V = 720 VA
Pool Motor 12A × 240V = 2,880 VA
RV Receptacle 30A × 120V = 3,600 VA
Sauna Receptacle 20A × 240V = 4,800 VA
Trash Compactor 8A × 120V = 960 VA
Water Heater 4,500 W= 4,500 W
Water Heater (Tankless) 40A × 240V = 9,600 VA
Welder Receptacle 30A × 240V = 7,200 VA
Well Pump Motor, 3 hp at 240V 240V × 17A = + 4,080 VA
Total Connected Load 140,096 VA

Step 2. Determine the demand load for Step 1 loads [220.82(B)].
Connected Load 140,096 VA
First 10 kW at 100% “10,000 VA × 100% 10,000 VA
Remainder at 40% 130,096 VA × 40% + 52,038 VA
Demand Load 62,038 VA

Step 3. Determine the air-conditioning versus electric heating demand load [220.82(C)]; air-conditioning at 100% [220.82(C)(1)] versus electric heating at 65% [220.82(C)(4)].
Air-Conditioning = Volts × Amperes.

Zone 1
A/C Condenser = 240V × 50A
A/C Condenser = 12,000 VA
A/C Condenser Fan = 240V × 4A
A/C Condenser Fan = 960 VA
Zone 1 Air-Conditioning Total = 7,200 VA + 960 VA
Zone 1 Air-Conditioning Total = 12,960 VA

Zone 2
A/C Condenser = 240V × 50A
A/C Condenser = 12,000 VA
A/C Condenser Fan = 240V × 3A
A/C Condenser Fan = 720 VA
Air-Conditioning Total = 4,800 VA + 720 VA
Air-Conditioning Total = 12,720 VA
Total Air-Conditioning Demand Load = 12,960 VA + 12,720 VA
Total Air-Conditioning Demand Load = 25,680 VA

Electric Heating at 65%
Electric Heating Zone 1 = 9,600 W
Electric Heating Zone 2 = 7,500 W
Total Connected Space Heating = 17,100 W
Space Heating Demand Load = 17,100 W × 65%
Space Heating Demand Load = 11,115 W (omit) [220.82(C)]

Air Handler at 100%
Air Handler = Volts × Amperes
Air Handler, Zone 1 = 240V × 4A
Air Handler, Zone 1 = 960 VA
Air Handler, Zone 2 = 240V × 3A
Air Handler, Zone 2 = 720 VA
Air Handler Demand Load = 1,680 VA
Air-Conditioning Demand Load = 25,680 W + 1,680 VA
Air-Conditioning Demand Load = 27,360 VA

Step 4. Determine the service disconnect and conductor sizing.
Step 1: Demand Load 62,038 VA
Step 2: Air-Conditioning Load + 27,360 VA
Total Demand Load 89,398 VA

Service Size in Amperes = VA Demand Load/System Volts
Service Size in Amperes = 89,398 VA/240V
Service Size in Amperes = 372A
Service Disconnect Size = 400A
Use 400 kcmil copper or 600 kcmil aluminum [Table 310.12]

Answer: (c) a or b.

Familiarity breeds accuracy
It's important to be familiar with the requirements of Article 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.