Dwelling Unit Service Load Calculations – Optional Method

In the last article, we covered Part III in article 220 of the 2020 edition of the National Electrical codebook. Part III is known as the Standard Method. In this article, we will cover Part IV, The Optional Method for calculating the size of service and feeder of dwelling units. Part IV starts at 220.80 and states in 220.82(A) that it is permissible to calculate the feeder and service in accordance with this section instead of the standard method in Part III. There are some differences between the two methods – the major difference has to do with the demand factors.

A demand factor is a multiplier that we use to help us to get the calculation as close as possible to the actual demand from the loads – in this case, a dwelling unit. Dwelling units are different from other occupancies. In dwelling units, there are a lot of receptacles that are often not used because they are behind furniture. There are receptacles in places such as hallways that aren’t used that often but are required by code. Also, receptacle outlets in dwelling units aren’t usually used at the same time. 210.52 covers where receptacle outlets are to be placed. Receptacle outlets in occupancies other than dwelling units don’t really have any code requirements for placement. However, the National Electrical Code is written to keep people, equipment, and structures safe, as stated in 90.1(A). If there were no requirements for receptacle placement in a dwelling unit, there would be a greater risk of hazards because homeowners would be tempted to string extension cords throughout the house. 

In Part IV, The Optional Method, we simply take the nameplate value of all the loads and add them up, then do one calculation for the demand factor. 220.82(B)(1) through (4) gives all the requirements for the optional method. The optional method does have some similarities to the standard method, and one of those similarities is that we use three VA per square foot for the general lighting and general use receptacles. This can be found in 220.82(B)(1). Unlike the standard method, there is no demand factor specifically for general lighting. 220.82(B)(2) gives us the information we need for the two small-appliance branch circuits and the laundry circuit that is required by 210.11(C)(1) and (2). This is also similar to the standard method – each small-appliance branch circuit is required to have a calculated load of 1,500 VA. 210.11(C)(1) says that each dwelling is required to have two or more small-appliance branch circuits. This gives us 3,000 VA for the two small appliances. 210.11(C)(2) says that we need to have one or more laundry circuits. This gives us 1,500 VA for the laundry circuit. Both the 1,500 VA for each of the two small appliances and the 1,500 VA for the laundry can be found in 202.82(B)(2). There is no demand factor for these specific loads. 220.82(B)(3) tells us to take the nameplate values of all appliances, ranges, and other cooking equipment, as well as the nameplate rating of the dryer, and add them up. There is no demand factor for these at this point. 220.82(B)(4) says to take the nameplate rating for all permanently connected motors –  but not the one that falls under (3), which covers appliances. We would now add all this up and do ONE BIG demand factor. 220.82(B) says to take the first 10,000 VA at 100% and the remainder at 40%.

This is different from the Standard Method in Part III of Article 220, where each load gets a demand factor of its own as you do the calculations. Because the optional method has fewer steps, electrical exams often have you size a dwelling unit using the optional method. Sizing the feeder or service for a full dwelling unit using the standard method takes too much time. In exams, a test question will often say, “using the optional method, size the feeder or service for this dwelling”. Exams have questions from Part III the standard method, but the questions are based on the individual load types. When studying make sure not to mix up Part III and Part IV, as most times the answer will offer both options. 

As an example, 220.54 of Part III tells us that the load for a clothes dryer is to be calculated using the nameplate of the dryer or 5,000 Watts, whichever one is larger. This isn’t the case for the optional method. If you had an optional method exam question that asked you to size the feeder or service load for a 4,500 Watt dryer, you wouldn’t round up to 5,000 Watts like you would for the standard method. Instead, you would just use the nameplate rating.

We will now do a calculation using the optional method and will be using Example D2(a) found in the back of the codebook. In this example, we will be calculating a single-family dwelling where the heater load is larger than that of the air conditioning. 

All the codes pertaining to a single-family dwelling can be found in 220.82(A), (B), and (C). (A) gives us general information about the optional method. (B) gives us information about the general loads. It is broken into four different loads, which are as follows:

1. General lighting and general use receptacles: three VA per square foot of the dwelling. The floor area can be calculated from the outside dimensions of the dwelling unit.
2. 1,500 VA for each 20-amp small-appliance branch circuit that is required by 210.11(C)(1) and 1,500 VA for each laundry circuit required by 210.11(C)(2). The small-appliance branch circuits in 210.11(C)(1) states that there are to be at least two small-appliance branch circuits in a dwelling. That puts us at 3,000 VA for the two. In example D2(a) there is only one laundry room, so that would give us 1,500 VA. That puts the loads for this dwelling unit at 4,500 VA. 
3. The Nameplate rating of the following: 
   1. All appliances that are fastened in place
   2. Ranges, wall-mounted ovens, counter-mounted cooling units
   3. Clothes dryer that aren’t connected to the laundry
4. The nameplate rating of permanently connected motors

220.82(C) gives us information on heating and cooling loads. This is similar to 220.60 Noncoincident Loads, where it states that you choose whichever is the larger of the air-conditioning and heating; this is because these two loads will not be operating at the same time. The same goes with 220.82(C) – it tells us to use the largest six different loads that all pertain to heating and cooling.

The dwelling unit in example D2 (a) is a 1,500 square foot, single-family dwelling. It has a 12,000-Watt range, a 2,500-Watt water heater, a 1,200-Watt dishwasher, 9,000-Watts of electric space heating installed in five rooms, a 5,000-Watt clothes dryer, and a 6-Amp, 230 Volt room air-conditioning unit. 

Now for the demand factor. In 220.82(B) it states that the load shall not be less than 100% of the first 10,000 VA plus 40% of the remainder.

The calculated load for the dwelling in example D2(a) is 90 Amps according to Table 240.6 there is a breaker that is sized at 90 amps but we can’t size the service for a dwelling at 90 amps because of a couple of codes 230.42 (B) for conductors and 230.79 (C ) for the service disconnecting means, it is stated that a one-family dwelling shall not have a disconnecting means rated less than 100 amps. 

Table 310.12 in the 2020 codebook is the Table one would use to size the main service or feeder to a dwelling unit. Some refer to it as the 83% rule. This table has moved around over the past few code cycles. In the 2008 National Electrical codebook the table was 310.15(B)(6), in 2011 it moved to 310.15(B)(7), in 2014 it was taken out and 83% was added. The original Table had the conductors sized at 83% so you just had to do the math. In the 2017 codebook, the table could be found in annex D and the 83% could still be found in 310.15(B)(7). I personally hope that it has found its home in 310.12. The wording in 310.12(A) states that services rated 100 amperes through 400 amperes the conductors supplying the entire load associated with an individual dwelling unit in a two-family or multifamily dwelling shall be permitted to have an ampacity not less than 83% of the service rating. If we go to Table 310.12 we can see that at 100 amps we would have to use 4 AWG copper or 2 AWG aluminum.