Dwelling unit feeder/service conductor calculations

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Whether they're for journeymen, master electricians, or contractors, most electrical licensing exams require you to calculate residential loads and service or feeders sizes using one of two methods. The standard method, which is contained in Art. 220, Part II, involves more steps, but many people use it exclusively to avoid using the wrong method. However, most residential construction qualifies for the optional method in Art. 220, Part III, so it's prudent to understand both methods. In either case, you're free to exceed the NEC requirements — these are minimum requirements, not design specifications.

The standard method is where we'll start. It requires six sets of calculations for general lighting and receptacles, small-appliance, and laundry; air conditioning versus heat; appliances; clothes dryer; cooking equipment; and conductor size.

The following example should help illustrate how to apply these steps.

What size service conductor does a 1,500 sq ft dwelling unit need, if it contains the following loads? The service is 120/240V.

Disposal (940VA)

Dishwasher (1,250VA)

Trash compactor (1,100VA)

Water heater (4,500VA)

Dryer (4,000VA)

Cooktop (6,000VA)

Two ovens (each 3,000VA)

Air conditioning (5 hp, 230V)

Three electric space heating units (each 3,000W)

General lighting and receptacles, small-appliance, and laundry. The NEC recognizes these circuits won't all be on (loaded) simultaneously. Thus, you may apply a demand factor to the total connected general lighting and receptacle load (220.16). To determine the service/feeder demand load, refer to Table 220.11 and follow these steps:

Fig. 1. In calculating conductor sizes for a dwelling unit, the NEC recognizes a homeowner won’t run all the appliances, turn on all the lights, and load all receptacles at the same time.

First, determine the total connected load for general lighting and receptacles (3VA per sq ft) [Table 220.3(A)], two small-appliance circuits each at 1,500VA, and one laundry circuit at 1,500VA (220.16) (Fig.1).

Second, apply Table 220.11 demand factors to the total connected general lighting and receptacle load. Calculate the first 3,000VA at 100% demand and the remaining VA at 35% demand.

General lighting/receptacles: 1,500 sq ft× 3VA=4,500VA

Small-appliance circuits: 1,500VA×2 =3,000VA

Laundry circuit: 1,500VA×1= 1,500VA

Total connected load: 4,500VA+ 3,000VA+1,500VA=9,000VA

First 3,000VA at 100%=3,000VA× 1.00=3,000VA

Remainder at 35%=(9,000VA- 3,000VA)×0.35=2,100VA

Total demand load=5,100VA

Fig. 2. It should be obvious that the heating and cooling systems won’t be generated at the same time. Therefore, your calculations should be based on the larger of the two loads.

Air-conditioning versus heat. Because air-conditioning and heating loads aren't on simultaneously, you may omit the smaller of the two loads (220.21). Calculate each of these at 100% (220.15) (Fig. 2).

Air conditioning: 5 hp, 230V

VA=E×I (Table 430.148)

28 FL×230V=6,440 VA

Heat: 3,000W×3 units=9,000W

The air conditioning load is smaller than the heat load, therefore it can be omitted.

Appliances. Per 220.17, you can use a 75% demand factor when four or more “fastened in place” appliances, such as a dishwasher or waste disposal, are on the same feeder. Don't include clothes dryers, cooking equipment, air conditioning, or heat in this category (Fig. 3).

Waste disposal: 940VA

Dishwasher: 1,250VA

Trash compactor: 1,100VA

Water heater: 4,500VA

Total connected appliance load: 7,790VA× 0.75=5,843VA

Fig. 3. When determining load for appliances, the Code allows you to use a 75% demand factor when four or more fastened-in-place appliances are on the same feeder.

Fig. 4. Electric clothes dryer demand must be accounted for as a separate load item.

Clothes dryer. Per 220.18, the feeder or service demand load for electric clothes dryers in a dwelling unit shall not be less than 5,000W. However, if the nameplate rating exceeds 5,000W, use that rating as the load. You can omit this calculation if the unit has no electric dryer provision. However, it's common to provide both gas and electric sources. If you see gas on the plans, verify electric will be omitted (Fig. 4).

The service and feeder demand load for a 4kW dryer is 5,000W.

Cooking equipment. For household-cooking appliances rated higher than 1.75kW, you can use the demand factors listed in 220.19, Table and Notes 1, 2, and 3.

All three of the units in the example are rated higher than 1.75kW and not higher than 8.75kW, so follow the instructions in Note 3. The two ovens are rated less than 3.5kW, so Table 220.19 Column A demand factor applies. The cooktop is 6kW, so Column B demand factor applies (Fig. 5).

Column A demand: 3kW×2 units× 0.75 demand factor=4.5kW

Column B demand: 6kW×1 unit× 0.8=4.8kW

Demand load=4.5kW+4.8kW= 9.3kW=9,300W

Fig. 5. See Table 220.19 to determine demand load for cooling equipment rated higher than 1.75kW.

Feeder and service conductor size. 400A and less: For 3-wire, 120/240V, single-phase systems, size the feeder or service conductors according to Table 310.15(B)(6). For all others, use Table 310.16. Size the grounded (neutral) conductor to the maximum unbalanced load (220.22) per Table 310.16.

Over 400A: Size the ungrounded and grounded (neutral) conductors per Table 310.16.

Now we can total up the demand loads from steps 1 through 5.

Step 1: 5,100VA

Step 2: 9,000VA

Step 3: 5,843VA

Step 4: 5,000VA

Step 5: 9,300W

Step 6: 34,243VA total demand load To determine the amperes need for the service use the formula: I5VA÷E.

I=34,243VA÷240V=143A

We can use 310.15(B)(6) for a 120/240V single-phase dwelling service and feeder up to 400A. This Table allows a smaller conductor size than Table 310.16.

A 143A demand load means this house requires at least a 150A service with 1 AWG conductors.

Optional method. You can use the easier optional method found in 220.30 only when the total connected load is served by a single 3-wire, 120/240V or 208Y/120V set of service or feeder conductors with an ampacity of 100A or greater. Because this condition describes the typical residential service, the optional method is likely to apply. Using it can simplify the design process and save you time because you have so many fewer sets of calculations.

General loads. The calculated load shall not be less than 100% for the first 10kW, plus 40% of the remainder of the following loads:

Small-appliance and laundry branch circuits: 1,500VA for each 20A circuit.

General lighting and receptacles: 3VA per sq ft

Appliances: The nameplate VA rating of all appliances and motors fastened in place (permanently connected) or on a specific circuit. Be sure to use the range and dryer at nameplate rating.

HVAC. Include the largest of the following:

100% of the nameplate rating of the air-conditioning equipment.

100% of the heat-pump compressors and supplemental heating, unless the controller prevents simultaneous operation of the compressor and supplemental heating.

100% of the nameplate ratings of electric thermal storage and other heating systems where you expect the usual load to be continuous at the full nameplate value. Don't configure such systems under any other selection in this table.

65% of the nameplate rating(s) of the central electric space heating, including integral supplemental heating in heat pumps where the controller prevents simultaneous operation of the compressor and supplemental heating.

65% of the nameplate rating(s) of electric space heating, if there are less than four separately controlled units.

40% of the nameplate rating(s) of electric space heating of four or more separately controlled units.

Sizing service/feeder conductors. Now that we've seen how to determine residential loads, let's size the service/feeder conductors. We'll use the same specifications that we used for the standard method so we can compare apples to apples.

Step 1: Determine general loads [230.30(B)].

Small appliance: 1,500VA×2 circuits = 3,000VA

General lighting: 1,500 sq ft×3VA= 4,500VA

Laundry circuit=1,500VA

Now add up the appliance ratings. Disposal (940VA) Dishwasher (1,250VA) Trash compactor (1,100VA) Water heater (4,500VA) Dryer (4,000VA) Ovens (3,000VA×2 units=6,000VA) Cooktop (6,000VA)

The total connected load=32,790VA

Calculate the first 10,000VA at 100%=10,000VA× 1.00=10,000VA

Calculate the remainder at 40%= 22,790VA×0.40= 9,116VA

Demand load=10,000VA+9,116VA= 19,116VA

Step 2: Compare air conditioner at 100% vs. heat at 65% [220.30(C)].

Air conditioner: 230V×28A=6,440VA

Heat [220.30(C)(5)]: 9,000 W×0.65 = 5,850 W (omit)

Step 3: Calculate service/feeder conductors per 310.15(B)(6).

General loads=19,116VA

Air conditioning=6,440VA

Total demand load=25,556VA

I=VA÷E=25,556VA ÷ 240V = 106.5A

310.15(B)(6) requires at least a 110A service with 3 AWG conductors.

As you can see, in this case the optional method permitted a smaller service than the standard method of calculating a service for a dwelling.

Now that we've walked through the process of calculating residential services and feeders, you can see that doing so is fairly easy. You need to calculate the loads first, and then move on to the service and feeder size. The NEC provides the requirements in Art. 220 and 230. Doing these calculations correctly can save you money during design and construction, while providing safe homes for the families who occupy them.

Multifamily Dwelling Unit Service and Feeder Calculations

The NEC defines a dwelling unit as a single unit that provides complete and independent living facilities for one or more persons that must include permanent provisions for living, sleeping, cooking, and sanitation (Fig. 1). A dwelling becomes "multifamily" when it contains three or more dwelling units [Art. 100 Definitions] (Fig. 2).

Fig. 1. A dwelling unit is defined as a single unit that provides permanent provisions for living, sleeping, cooking, and sanitation.

When you size the service for a single-family dwelling, you calculate the load and apply the appropriate demand factors. For a multifamily dwelling, you do the same thing, except you apply the appropriate demand factors to the sum of the individual dwelling units of that multifamily dwelling. You are allowed to use the standard method from Part III of Art. 220 or the optional method from Part IV of Art. 220.

Standard method. The same method used for single dwellings can be applied to multifamily dwellings. The NEC allows some additional demand factors for multifamily dwellings, on the presumption that there will be diversity of usage between the various units. For example, it's very unlikely that four families will run their clothes dryers, ranges, and small appliances at exactly the same time.

The following steps can be used to determine feeder and service sizes for a multifamily dwelling using the standard method contained in Art. 220, Part III:

Step 1: General Lighting, Small Appliance, and Laundry Demand [Table 220.42] 3VA per sq ft for general lighting and general-use receptacles [Table 220.12]. 1,500VA for each small-appliance circuit (minimum of 2 circuits) [220.52(A)]. 1,500VA for each laundry circuit [220.52(B)].

Step 2: Air-Conditioning versus Heat [220.51]

The larger of the air-conditioning load or the space-heating load.

Step 3: Appliance Calculated Load [220.53]

Fig. 2. A multifamily dwelling is defined as a building with three or more dwelling units. Examples include apartment buildings, condominiums, some hotels and motels.

Nameplate ratings of all appliances (except heating, air-conditioning, cooking equipment, and dryers) are taken times a 75% multiplier if there are four or more on the feeder.

Step 4: Household Dryer Calculated Load [220.54]

Dryers are allowed the demand factors of Table 220.54, but this table does not allow less than 100% demand until there are five units or more. The 5kW minimum per dryer applies to all dwelling units [220.54]. A laundry circuit isn't required for an individual dwelling unit if the multifamily unit has common laundry facilities.

Step 5: Household Cooking Equipment Calculated Load [220.55]

Perhaps one of the most confusing tables in the NEC is Table 220.55 for household ranges. This table is confusing because the first two columns are percentage multipliers, while the third column is a final kVA value. The notes to this table further complicate matters. Be sure you study this table carefully and pay close attention to how to properly apply each column.

Step 6: Service Conductor Size [Table 310.16]

When sizing the service or feeder conductors for a single-family dwelling, you can use Table 310.15(B)(6), but that is not the case when sizing conductors for the service or feeder to a two-family or multifamily dwelling. For sizing those conductors, use Table 310.16 instead. Use Table 310.15(B)(6) for the feeders to an individual dwelling unit within the building.

Optional method. When should you use the optional method instead of the standard one? If you have the necessary information, you'll probably want to use the optional method, because it's faster and easier to calculate.

Fig. 3. Sample problem of how to size the ungrounded conductors for a 12-unit multifamily building.

The optional method for multifamily dwellings is different from the one for single-family dwellings. That's because with multifamily dwellings, you apply demand factors in recognition of the diversity of usage of all the loads in all the separate units.

Let's make sure this is clear. In a single family unit, you have diversity among the various types of loads. Although you have that in multifamily units as well, you have diversity among the units that make up the multifamily dwelling — all of the families in a multifamily dwelling aren't using identical loads at identical times.

You can use the optional method [220.84] for multifamily dwelling unit feeder and service calculations only if each dwelling unit is equipped with electric cooking equipment and electric heating and/or air-conditioning, and is supplied by no more than one feeder. Follow these rules:1. Use the demand factors of Table 220.84, based on the number of dwelling units.2. Determine the feeder/service neutral calculated load per 220.61.3. Calculate house loads for common areas per Art. 220, Part III and then add them to the Table 220.84 calculated load.

House loads are those not directly associated with the individual dwelling units of a multifamily dwelling. Some examples include landscape and parking lot lighting, hall and stairway lighting, common recreation areas, and common laundry facilities.

Follow these steps:1. Determine total connected load.

2. Calculate the load.3. Size feeder and service conductors.

Let's look at these three steps in a bit more detail by walking through an example. In practice, you may see NEC-compliant variations of executing these steps.

Step 1: Determine total connected load [220.84(C)].

Add the following loads (from all the dwelling units) together, then apply the Table 220.84 demand factor: 3VA per sq ft for general lighting and general-use receptacles. 1,500VA for each small-appliance circuit (minimum of two circuits). 1,500VA for each laundry circuit. The nameplate rating of all appliances. The nameplate rating of all motors. The larger of the air-conditioning load or the space-heating load.

A laundry circuit isn't required for an individual dwelling unit if the multifamily unit has common laundry facilities.

Step 2: Calculate the load.

Apply the demand factor from Table 220.84 to the total connected load (Step 1). You can convert the calculated load (kVA) to amperes by:

Single-Phase Formula: I = VA ÷ E

Three-Phase Formula: I = VA ÷ (1.732 x E)

Step 3: Size feeder and service conductors.

Size the ungrounded conductors per Table 310.16, based on the calculated load.

Fig. 4. Sample calculation of how to determine service conductor size on a multifamily dwelling

Apply the demand factor from Table 220.84 to the total connected load (Step 1). You can convert the calculated load (kVA) to amperes by:

Single-Phase Formula: I = VA ÷ E

Three-Phase Formula: I = VA ÷ (1.732 x E)

Step 3: Size feeder and service conductors.

Size the ungrounded conductors per Table 310.16, based on the calculated load.

Example ProblemA 120/240V, single-phase system supplies a 12-unit multifamily building (Fig. 3). Each 1,500 sq ft unit contains:

Dishwasher — 1.5kVA

Water Heater — 4.0kVA

Washing Machine — 1.2kVA

Dryer — 4.5kVA

Range — 14.4kVA

A/C (230V x 17A) — 3.91kVA

Electric Space Heating — 5.0kVA

Question: What size conductor is required if the service is rated 120/240V, single-phase and the conductors are installed in parallel in two separate raceways?

Step 1: Total Connected Load

Step 1a: Determine the General Lighting Load:

General Lighting(1,500 sq ft x 3VA = 4,500VA)

Small-Appliance Circuits(2 circuits x 1,500VA = 3,000VA)

Laundry Circuit (1,500VA)

4,500 + 3,000 + 1,500 = 9,000

9,000VA x 12 units = 108,000VA

Step 1b: Determine the Appliance Calculated Load:

Dishwasher (1,500VA)

Water Heater (4,000VA)

Dryer [nameplate] (4,500VA)

Range [nameplate] (14,400VA)

1,500 + 4,000 +4,500 + 14,400 = 24,400

24,400VA x 12 units = 292,800VA

Step 1c: Compare the Air-Conditioning versus Heat Load:

A/C = 3,910VA (omit)

Heat = 5,000VA x 12 units = 60,000VA

Step 2: Total Connected Loads

General Lighting, Receptacles (108,000VA)

Appliances Connected Load (292,800VA)

Heat (60,000VA)

108,000 + 292,800 + 60,000 = 460,800

Total Connected Load = 460,800VA

Total Calculated Load = Total Connected Load x Demand Factor [Table 220.84]

Total Calculated Load = 460,800VA x 0.41 [Table 220.84]

Total Calculated Load = 188,928VA

Step 3: Service Conductor Size (Fig. 4 on page xx)

I = VA ÷ E

I = 188,928VA ÷ 240V

I = 787A

Conductor size if paralleled in two raceways [240.4(B)]:

787A ÷ 2 raceways = 393A per conductor

Feeder/Service Conductors:

Parallel 600kcmil conductors rated 420A at 75ºC [Table 310.16] would meet these load requirements.

Grounding electrode conductor sizing. What size grounding electrode conductor is required if the service ungrounded conductors are 600kcmil with two conductors in parallel in two separate raceways?

600kcmil x 2 = 1,200kcmil (equivalent area of the ungrounded conductors) [Table 250.66, Note 1].

Over 1,100kcmil for ungrounded conductors requires a grounding electrode conductor of 3/0 AWG [Table 250.66].

Feeder installation. When installing feeders, include an equipment grounding conductor in each raceway. Section 250.118 lists allowable equipment grounding conductors.

To size this conductor using a wire-type equipment grounding conductor, go to Table 250.122 and select the equipment grounding conductor based on the overcurrent device protecting the conductors in the raceway [250.122(F)]. For instance, the equipment grounding conductor in each raceway of an 800A feeder, which is paralleled using two 600kcmil conductors per phase, will require a 1/0 AWG equipment grounding conductor in each raceway.

Two more samples. Working these two additional sample problems will reinforce what we've learned thus far.

Size the grounding electrode conductor.

Question: What size grounding electrode conductor is required if the service ungrounded conductors are 300kcmil with three conductors in parallel in three separate raceways?

300kcmil x 3 = 900kcmil equivalent area of the ungrounded conductors [Table 250.66, note 1]

900kcmil for ungrounded conductors requires a grounding electrode conductor of 2/0 AWG [Table 250.66]

Size the parallel service conductor.

Question: What size conductor is required if the service with a calculated load of 787A is rated 120/240V, single-phase and the conductors are installed in parallel in four separate raceways?

787A ÷ 4 raceways = 196.75A

3/0 AWG copper is rated 200A at 75ºC [Table 310.16], so four 3/0 AWG conductors can be paralleled for this service.

A word about two-family dwellings. The feeder for a two-family dwelling unit is calculated using the standard method in Part III of Art. 220. When that calculated load exceeds the calculation for three identical units using the optional method of 220.84, the lesser of the two calculations is permitted to be used [220.85].

Avoiding confusion. The sizing of branch circuits, feeders, and service conductors for multifamily dwellings is similar to the sizing for single-family dwellings. You size the feeders to individual dwelling units in the same manner, whether that dwelling unit is a single-family dwelling or an individual unit of an apartment building.

The NEC allows the use of Table 310.15(B)(6) for sizing the feeders or service conductors to an individual dwelling unit. However, to size the conductors that provide the service to a two-family or multifamily dwelling you must use Table 310.16.

Whether calculating the service for a single-family or a multifamily dwelling, be sure to follow the Code rules for the specific calculation you are working on and do not intermix the standard method with the optional method. Follow the steps outlined in this article and apply the demand factors allowed for each method carefully, and you will be successful.