This article was posted 04/28/2009 and is most likely outdated.

Dwelling Unit Calculations
 

 

Topic - NEC
Subject - Dwelling Unit Calculations

April 28, 2009
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Dwelling Unit Calculations

 

By Mike Holt – taken from my NEC Exam Preparation textbook, Unit 9

Apply demand factors for correct load calculations.

A dwelling unit is a single unit that provides complete and independent living facilities, according to the NEC definition found in Article 100.

Dwelling units have special requirements for load calculations. Most of the actual load calculation requirements are in Article 220, but there are many installation requirements scattered throughout the Code that come into play when making these calculations (see Sidebar).

Keep in mind these considerations for dwelling unit calculations:

  • Voltages. Unless other voltages are specified, calculate branch-circuit, feeder, and service loads using the nominal system voltage [220.5(A)]. For a single-family dwelling unit, the nominal voltage is typically 120/240V.
  • Motor VA. Use motor table voltage and current values, such as 115V, 230V, or 460V—not 120V, 240V, or 480V [430.248 and 430.250]. A much more accurate VA rating is obtained by using the motor’s rated voltage and current which were used in developing the Code Tables.
  • Rounding. Where calculations result in a fraction of less than 0.50A, you can drop the fraction [220.5(B)].
  • Receptacles. You can use 15A or 20A receptacles on 20A circuits as long as there is more than one receptacle on the circuit. For these purposes, a duplex receptacle is considered to be two receptacles [210.21(B)(3)].
  • Continuous loads. A continuous load is one in which the maximum current is expected to continue for 3 hours or more, according to the Article 100 definition. Fixed electric heating is one example of a continuous load [424.3(B)]. When sizing branch-circuit conductors and overcurrent devices for a continuous load, multiply the load by 125 percent [210.19(A)(1) and 210.20(A)].
  • Laundry rooms. A laundry area receptacle is required [210.52(F)] and at least one must be within 6 ft of a washing machine [210.50(C)]. Any receptacle within 6 ft of the outside edge of a laundry sink must be GFCI protected [210.8(A)(7)].

Required circuits

In addition to the circuits required for dedicated appliances and the circuits needed to serve the general lighting and receptacle load, a dwelling unit must have the following circuits:

  • A minimum of two 20A, 120V small-appliance branch circuits for receptacles in the kitchen, dining room, breakfast room, pantry, or similar dining areas [220.11(C)(1)]. These circuits must not be used to serve other outlets, such as lighting outlets or receptacles from other areas [210.52(B)(2) Ex]. These circuits are included in the feeder/service calculation at 1,500 VA for each circuit [220.52(A)].
  • One 20A, 120V branch circuit for the laundry receptacle(s). It can’t serve any other outlet(s), such as lighting, and can serve only receptacle outlets in the laundry area [210.52(F) and 210.11(C)(2)]. In your feeder/service load calculation, include 1,500 VA for the 20A laundry receptacle circuit [220.52(B)].

Feeder and Service Calculations

Occupants don’t use all loads simultaneously under normal living conditions, so “demand factors” can be applied to many of the dwelling unit loads in order to size the service. Some demand factors provided in the Code are intended for use in dwellings only, and there are other demand factors which are allowed only in non-dwellings, so be careful to apply demand factors only as allowed by the NEC.

The NEC provides two dwelling service load calculation methods: the Standard Method and the Optional Method.

Standard Method for Feeder and Service Load Calculations

The standard method consists of three calculation steps:

  1. General lighting VA load. When calculating branch circuits and feeder/service loads for dwellings, include a minimum 3 VA 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.
  2. Small appliance and laundry circuits. The 3 VA 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. So, you must calculate those at 1,500 VA per circuit. See 220.14(J) for details.
  3. Number of branch circuits. Determine the number of branch circuits required for general lighting and general use receptacles from the general lighting load and the rating of the circuits [210.11(A)]. You’ll find a useful example in Annex D, Example D1(a) of the NEC, but let’s look at an 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?

(a) 2,100 VA    (b) 4,200 VA    (c) 6,000 VA    (d) 8,400 VA

Answer: (c) 6,000 VA

2,000 sq ft x 3 VA = 6,000 VA. No additional load is required for general-use receptacles and lighting outlets because they are included in the 3 VA 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?

(a) 2 circuits    (b) 3 circuits    (c) 4 circuits    (d) 5 circuits

Answer: (c) 4 circuits

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

General lighting VA = 6,000 VA

Step 2: General lighting amperes:

I = VA/E

I = 6,000 VA/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 [Article 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:

  1. General loads [220.82(B)]
  2. Heating and air-conditioning load [220.82(C)]
  3. Feeder/service conductors [310.15(B)(6)]

Step 1: General loads [220.82(B)]

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

  1. General lighting and receptacles: 3 VA per sq ft
  2. Small-appliance and laundry branch circuits: 1,500 VA for each 20A, 120V small-appliance and laundry branch circuit specified in 220.52.
  3. 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 percent
  • Heat-pump compressor without supplemental heating: 100 percent
  • Heat-pump compressor and supplemental heating: 100 percent of the nameplate rating of the heat-pump compressor and 65 percent 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 percent.
  • Space-heating units (four or more separately controlled units): 40 percent.
  • Thermal storage heating: 100 percent.

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

  • 400A and less. For individual dwelling units of one-family, two-family, and multifamily dwellings, you can 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 file 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: 6,000 VA

Disposal: 900 VA

Dishwasher: 1,200 VA

Dryer: 4,000 VA

Ovens (two each): 3,000 VA

Water heater: 4,500 VA

A/C: 17A, 230V

Electric heating (one control unit): 10 kVA

(a) 6 AWG       (b) 4 AWG       (c) 3 AWG       (d) 2 AWG

Answer: (c) 3 AWG

Step 1: General loads [220.82(B)]

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

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

Laundry circuit: 1,500 VA

Appliances (nameplate):

  • Cooktop: 6,000 VA
  • Disposal: 900 VA
  • Dishwasher: 1,200 VA
  • Dryer: 4,000 VA
  • Ovens (each 3 kW):  6,000 VA
  • Water heater:   4,500 VA

Connected load: 31,600 VA

First 10 kW at 100%: 10,000 VA x 1.00 = 10,000 VA

Remainder at 40%: 21,600 VA x 0.40 = 8,640 VA

Calculated general load: 10,000 VA + 8,640 VA

 

Calculated general load: 18,640 VA

Step 2: Air-Conditioning versus heat [220.82(C)]

Air-conditioning at 100 percent [220.82(C)(1)] versus electric space heating at 65 percent [220.82(C)(4)]

Air conditioner [Table 430.248]:

A/C VA = V x A

A/C VA = 230V x 17A

A/C VA = 3,910 VA (omit)

Electric space heat: 10,000 VA x 0.65 = 6,500 VA

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

Calculated general load (Step 1): 18,640 VA

Heat calculated load (Step 2): 6,500 VA

Total calculated load: 25,140 VA

I = VA/E

I = 25,140 VA/240V

I = 105A

The feeder/service ungrounded conductor is sized to 110A, 3 AWG [310.15(B)(6)].

The Code doesn’t explain how demand factors were derived, and it’s not essential that you understand this in order to apply them correctly. Be sure to work on some practice calculations so you understand how to apply the various demand factors to a dwelling unit calculation.

The standard calculation and the optional calculation methods were both discussed in this article. These are two distinctly different calculation methods, so be careful not to mix them together. Remember that the Standard Method is in Part III of Article 220, the Optional Method is all contained in Part IV. When you are evaluating the necessary loads in either type of calculation method, be sure to follow the requirements for specific loads covered in other Articles outside of Article 220. Which method is better to use? On an exam, you’ll likely be told which method to use on a specific question, but if the question doesn’t specify a method, use the standard calculation. The optional method is usually faster and easier to apply, so it has a natural advantage for daily use on the job.

 

Sidebar: Where to find dwelling unit code requirements outside Article 220

  • Branch Circuits—210
  • Areas supplied by small-appliance circuits [210.52(B)(1)].
  • Feeders—215
  • Services—230
  • Overcurrent Protection—240
  • Wiring Methods—300
  • Conductors—310
  • Appliances—422
  • Electric Space-Heating Equipment—424
  • Motors—430
  • Air-Conditioning Equipment—440.

Mike Holt’s Dwelling Unit Calculations textbook is an extraction from his NEC Exam preparation textbook. To purchase, please click here or call our office at 888.632.2633 for more information.

 

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Comments
  • Something interesting I've encountered is what happens when appliances are rated for one voltage while the service is another voltage. For example, ranges rated 240 volts applied on 120/208 volt systems will effectively derate the range ouput Kw as the square of the voltage. 208 sq'd divided by 240 sq'd is very near .75, A 12 Kw range rated 240 volts, but operated at 208 volts will only deliver .75 x 12 = 9 Kw, so cooking time may be longer. The same reasoning would apply to electric clothes dryers, etc.

    wyatt
    Reply to this comment

  • Nice work. Practical and a useful reference

    Chuck Miller
    Reply to this comment

  • Nice summary. Love the examples. Thanks.

    Malcolm Hebert
    Reply to this comment

  • EXCELLENT ARTICLES. WHERE CAN I GET FOR COMMERCIAL BLDG. THANK YOU. GOD BLESS YOU.

    LUPE JUAREZ
    Reply to this comment

  • Curious, Where are the motor loads; NEC 220.82(B)(4)? Is the blower motor included in the A/C or heat load when they are calculated?

    David Engelhart
    Reply to this comment

  • Thank you mike holt for these e-mails you send there great, and i read every one!

    JIM
    Reply to this comment

  • I use a great software program "Flashworks 2008" in my engineering work for apartments (residential) & commercial buildings. I prints a PERMIT form for inspection authorities and my files.

    I checked it with manual calculations before using. The contact is cepardue@aol.com or google FLASHWORKS. I highly recommend it --produced by an engineer & electrical inspector with lots of experience. these folks are from Florida.

    I am registered in Florida & Alabama.

    Frank Chandler, Jr. PE
    Reply to this comment

  • I got 27,500 watts for the connected load. How did you get 31,600 watts? I tried it twice and got the same answer.

    jim
    Reply to this comment


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