We get a fair number of questions on calculating the loads for an existing dwelling and often the students get lost in the small details of the process. In an effort to simplify the prospect of being successful on an electrical exam that may have an existing load calculation question the below information will help reduce that test-taking stress.
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When it comes to doing a load calculation on existing dwellings it can become quite confusing for the electrician much less an apprentice or helper studying for an electrical exam. We direct the electrician or student to Section 220.83 which pertains to an existing dwelling that will be getting some type of additional electrical load. The additional load could be from building a new addition onto an existing dwelling, replacing the existing heating and air conditioning system with a system that draws more volt-amperes (VA), adding a swimming pool, etc. Regardless of the type of load added, it shall be permitted, not required, to use the calculations in this section to determine if the existing size of service or feeder will be of sufficient capacity to ultimately serve the additional loads being proposed. Make a note, this section only applies to an existing dwelling unit.
Optional method calculations for two dwelling units are in 220.85 and calculations for multifamily dwellings are in 220.84. This optional calculation for an existing dwelling unit can only be used where the dwelling is supplied by a single-phase service. The service can be fed from a 120/240Vor 208Y/120V system but as it states in the second sentence, it must be a 3-wire system.
The calculation process found in 220.83 is segregated into two groups. The first calculation shall be used for existing and additional new loads where additional air conditioning equipment or electric space-heating equipment is not to be installed. This calculation method is almost identical to the calculation method in 220.82(B). The only difference is the amount of load that is rated at 100 percent. For example in 220.82(B) the first 10,000 VA is at 100% and in 220.83 the first 8,000 VA is at 100% and Includes all of the existing loads and the new loads when using this formula. For example, an addition will be built onto an existing one-family dwelling; the existing kitchen will also be renovated. The area, as determined by 220.11, of the existing dwelling is 1,500 square feet. The existing dwelling has the following loads: two small-appliance branch circuits; one laundry branch circuit; two attic roof ventilators rated 506 VA each; one motor rated 1,176 VA; a range rated 8,000 VA; a clothes dryer rated 5,000 VA, and a water heater rated 4,500 VA. The service supplying this existing dwelling is 120/240V, single-phase, and is rated 100 amperes (A).
The addition built onto this dwelling will add 600 square feet (as determined by 220.11) of floor area. Therefore, the total calculated floor area will be 2,100 square feet. The existing range will be replaced by a new 12,000 VA range. The kitchen renovation will also include two additional appliance loads: a dishwasher rated 1,380 VA and a kitchen waste disposer rated 864 VA. Two more small-appliance branch circuits will be installed, thus making a total of four small-appliance branch circuits. No additional air conditioning equipment or electric space-heating equipment will be installed.
Is this 100A service of sufficient capacity to serve these additional loads? If not, what size of service is required?
When calculating an existing dwelling unit, include all the existing loads and all the additional new loads. Start by calculating the general lighting and general--use receptacles at 3 VA per square foot [220.83(A)(1)]. The general lighting and general-use receptacle load is 6,300 VA (2,100 sq. ft. 3 VA per sq. ft. = 6,300). The small-appliance branch circuit (4) and laundry branch circuit (1) load is 7,500 VA (1,500 x 5 = 7,500). The appliance load, as specified in 220.83(A)(3)(a), is 4,432 VA (506 + 506 + 1,176 + 1,380 + 864 = 4,432). Because of the new range, the range load is 12,000 VA. The clothes dryer load is 5,000 VA and the water heater load is 4,500 VA. Before applying the demand factor, the total for the existing loads and the new loads is 39,732 VA (6,300 + 7,500 + 4,432 + 12,000 + 5,000 + 4,500 = 39,732).
In accordance with 220.83(A), calculate the first 8 kVA (8,000 VA) of load at 100 percent and the remainder of load at 40 percent. The first 8,000 VA calculated at 100 percent is 8,000 VA (8,000 100% = 8,000). After deducting 8,000, the remainder is 31,732 VA (39,732 – 8,000 = 31,732). The remainder calculated at 40 percent is 12,693 VA (rounded to the nearest whole number) (31,732 x 40% = 12,692.8 = 12,693). The total load after applying demand factors is 20,693 VA (12,693 + 8,000 = 20,693). Find the minimum ampere rating by dividing volt-amperes by voltage. The minimum ampere rating required for this service is 86A (20,693 ÷ 240 = 86.2 = 86). Since the service supplying this dwelling is rated 100A, it is of sufficient capacity to serve these additional loads.
Oh Oh - Before the job starts, the homeowners change their mind. Like that never happens.
The homeowners decide have installed a central air conditioning system that is rated 7,200 VA. Now, you have to ask yourself this question. Is this 100A service of sufficient capacity to serve these additional loads plus the central air conditioning unit? If not, what size of service is required?
Use the second formula, 220.83(B), for existing and additional new loads where additional air conditioning equipment or electric space-heating equipment is to be installed. The calculations in 220.83(A) and (B) are identical except for the demand factors for air conditioning and heating equipment. The larger connected load of air conditioning or space heating, but not both, shall be used. The following loads must be added to the calculation at 100 percent of nameplate rating: air conditioning equipment, central electric space heating, and less than four separately controlled space-heating units. Do not apply the 40 percent demand factor to heating and air conditioning equipment, this equipment must be added to the calculation at 100 percent.
Since this existing dwelling unit has already been calculated with everything except the air conditioning system, add the total (100 percent) load of the air conditioning system to the calculated load of 20,693 VA. The total load including the air conditioning equipment is 27,893 VA (20,693 + 7,200 = 27,893).
The minimum ampere rating required for this service is 116A (27,893 ÷ 240 = 116.2 = 116). In accordance with 240.6(A), the next standard ampere rating above 116 is 125 amperes. Therefore, the minimum size service for this dwelling is 125 amperes.
The above was an extract from Mr. Charles Miller, who happens to be a good friend of mine and the reason I actually wrote my first book. When I was just starting to think about writing a technical publication in 2005 I spoke with him on the phone and Charlie said "Just do it Paul" and well the rest was history. Since the NEC is technical and he did such a great job at explaining it I figured I would pay homage and post it above with my own small artistic adaptations for the 2020 NEC.
Original Author: Charles Miller - Lighthouse Educational Services
Adaptation Edits: Paul W Abernathy
* Full Credit To Charles Miller - Lighthouse Educational Services
Original Article - https://www.ecmag.com/section/codes-standards/branch-circuit-feeder-and-service-calculations-part-lii
So you are replying to an EXISTING dwelling calculation method. The washing machines is covered by the Laundry Circuit and 1,500 VA.