Problem#1 : A 2200 square foot residence with the following loads lighting load

Question

Problem#1 : A 2200 square foot residence with the following loads lighting load 3 small appliance branch circuits laundry circuit 1500 W natural gas heating air conditioner 7000 VA . electric range 10,000 W hot tub 8000 W (2 hp motor) Level II electric vehicle charger 6000 W electric dryer 6000 W garbage disposal 900 W microwave 1400 W dishwasher 1300 W electric water heater 5000 W . . (1) Calculate the unit load using the standard method (use a table) (2) Calculate the unit load using the optional method (use a table) (3) Size the feeder (4) Size the neutral

Explanation / Answer

General lighting

VA load. When calculating branch circuits and feeder/service loads for dwellings, include a minimum 3VA per sq ft for general lighting and general-use receptacles [220.12]. When determining the area, use the outside dimensions of the dwelling. Don’t include open porches, garages, or spaces not adaptable for future use.

Small appliance and laundry circuits.

The 3VA per sq ft rule includes general lighting and all 15A and 20A, 125V general-use receptacles, but doesn’t include small-appliance or laundry circuit receptacles. Therefore, you must calculate those at 1,500VA per circuit. See 220.14(J) for details.

Number of branch circuits

Determine the number of branch circuits required for general lighting and general-use receptacles from the general lighting load and rating of the circuits [210.11(A)]. Although this is explained in Annex D, Example D1(a) of the NEC, let’s look at an another example.

Question:What’s the general lighting and receptacle load for a 2,000-sq-ft dwelling unit that has 34 convenience receptacles and 12 luminaires rated 100W each (Fig. 3)?

The calculation is pretty simple.

2,000 sq ft x 3VA = 6,000VA.

No additional load is required for general-use receptacles and lighting outlets because they are included in the 3VA per sq ft load specified by Table 220.12 for dwelling units. See 220.14(J).

Now let’s work through an example to determine the number of circuits required.

Question: How many 15A circuits are required for a 2,000-sq-ft dwelling unit?

Step 1: General lighting VA = 2,000 sq ft x 3VA = 6,000VA

Step 2: General lighting amperes:

I = VA ÷ E

I = 6,000VA ÷ 120V*

I = 50A

*Use 120V, single-phase unless specified otherwise.

Step 3: Determine the number of circuits:

Number of circuits = General lighting amperes ÷ circuit amperes

Number of circuits = 50A ÷ 15A

Number of circuits = 3.30, or 4 circuits. Any fraction of a circuit must be rounded up.

Optional method for feeder and service load calculations

You can use the optional method [Art. 220, Part IV] only for dwelling units 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]. The optional method consists of three calculation steps:

General loads [220.82(B)]

Heating and air-conditioning load [220.82(C)]

Feeder/service conductors [310.15(B)(6)]

Step 1: General loads [220.82(B)]

The general calculated load must be at least 100% for the first 10kVA, plus 40% of the remainder of the following loads:

General lighting and receptacles: 3VA per sq ft

Small-appliance and laundry branch circuits: 1,500VA for each 20A, 120V small-appliance and laundry branch circuit specified in 220.52.

Appliances: The nameplate VA rating of all appliances and motors that are fastened in place (permanently connected) or located on a specific circuit, not including heating or air-conditioning.

Be sure to calculate the range and dryer at their nameplate ratings.

Step 2: Heating and air-conditioning load [220.82(C)]

Include the larger of (1) through (6):

Air-conditioning equipment: 100%

Heat-pump compressor without supplemental heating: 100%

Heat-pump compressor and supplemental heating: 100% of the nameplate rating of the heat-pump compressor and 65% of the supplemental electric heating for central electric space-heating systems. If the control circuit is designed so that the heat-pump compressor can’t run at the same time as the supplementary heat, omit the compressor from the calculation.

Space-heating units (three or fewer separately controlled units): 65%.

Space-heating units (four or more separately controlled units): 40%.

Thermal storage heating: 100%.

Step 3: Feeder/service conductors [310.15(B)(6)]

400A and less. For individual dwelling units of one-family, two-family, and multi-family dwellings, use Table 310.15(B)(6) to size 3-wire, single-phase, 120/240V service or feeder conductors (including neutral conductors) that serve as the main power feeder. Feeder conductors aren’t required to have an ampacity rating greater than the service conductors [215.2(A)(3)]. Size the neutral conductor to carry the unbalanced load per Table 310.15(B)(6). Table 310.15(B)(6) can’t be used for sizing the feeder or service conductors that supply more than a single dwelling unit.

Over 400A. Size ungrounded conductors and the neutral conductor using Table 310.16 for feeder/services over 400A and those that do not fill all of the requirements for using Table 310.15(B)(6). Let’s try a calculation example.

Question: What size service conductor is required for a 1,500-sq-ft dwelling unit containing the following loads?

Cooktop: 6,000VA

Disposal: 900VA

Dishwasher: 1,200VA

Dryer: 4,000VA

Ovens (two each): 3,000VA

Water heater: 4,500VA

A/C: 17A, 230V

Electric heating (one control unit): 10kVA

Step 1: General loads [220.82(B)]

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

Small-appliance circuits: 1,500VA x 2 ci

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