Course Hero uses AI to attempt to automatically extract content from documents to surface to you and others so you can study better, e.g., in search results, to enrich docs, and more. This preview shows page 1 out of 1 page. Show
Note: This article is based on the 2008 NEC. A dwelling unit is a single structure that provides complete and independent living facilities, according to the NEC definition found in Art. 100 (Fig. 1 ). Dwelling units have special requirements for load calculations. Although most of the actual load calculation requirements are in Art. 220, others are scattered throughout the Code and still come into play when making certain calculations (see SIDEBAR: Where to Find Dwelling Unit Code Requirements Outside Art. 220 at end of article). Keep the following considerations in mind when making dwelling unit calculations:
Required circuitsIn addition to the circuits required for dedicated appliances and those needed to serve the general lighting and receptacle load, a dwelling unit must have the following circuits:
Feeder and service calculationsOccupants 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; others are allowed only in non-dwellings. Therefore, 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 calculationsThe standard method consists of three calculation steps:
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 calculationsYou 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:
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:
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):
Step 3: Feeder/service conductors [310.15(B)(6)]
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 circuits = 3,000VA Laundry circuit: 1,500VA Appliances (nameplate): Cooktop: 6,000VA Disposal: 900VA Dishwasher: 1,200VA Dryer: 4,000VA Ovens (each 3 kW): 6,000VA Water heater: 4,500VA Total connected load: 31,600VA First 10kW at 100%: 10,000VA x 1.00 = 10,000VA Remainder at 40%: 21,600VA x 0.40 = 8,640VA Calculated general load: 10,000VA + 8,640VA Calculated general load: 18,640VA Step 2: Air-Conditioning versus heat [220.82(C)] Air-conditioning at 100% [220.82(C)(1)] vs. electric space heating at 65% [220.82(C)(4)] Air conditioner [Table 430.248]: A/C VA = V x A A/C VA = 230V x 17A A/C VA = 3,910VA (omit) Electric space heat: 10,000VA x 0.65 = 6,500VA Step 3: Feeder/service conductors [310.15(B)(6)] Calculated general load (Step 1): 18,640VA Heat calculated load (Step 2): 6,500VA Total calculated load = 18,640VA + 6,500VA = 25,140VA I = VA ÷ E I = 25,140VA ÷ 240V = 105A Therefore, the feeder/service ungrounded conductor is sized to 110A, 3 AWG [310.15(B)(6)]. The NEC 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. Remember that the standard method is in Part III of Art. 220, and the optional method is contained in Part IV. When you are evaluating the necessary loads in either type of calculation method, follow the requirements for specific loads covered in other Articles outside of Art. 220. Which method is better to use? On an exam, you’ll likely be told which method to use on a specific question. However, 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 Art. 220Branch circuits — Art. 210 Areas supplied by small appliance circuits — 210.52(B)(1) Feeders — Art. 215 Services — Art. 230 Overcurrent protection — Art. 240 Wiring methods — Art. 300 Conductors — Art. 310 Appliances — Art. 422 Electric space-heating equipment — Art. 424 Motors — Art. 430 Air-conditioning equipment — Art. 440 For more information, read "Load Calculations -- Part 1." |