Article 220 of the National Electrical Code (NEC) provides requirements for calculating branch-circuit, feeder and service loads. Article 220 is divided into five parts. Part I contains general requirements for calculation methods. Part II provides calculation methods for branch-circuit loads. Calculation methods for branch circuits are not only useful for calculating branch-circuit loads, they are also useful for calculating feeder and service loads. For example, results from Part II can be used to size branch-circuit conductors and branch-circuit overcurrent devices. Results from Part II can also be incorporated with results from other sections in Article 220 to size conductors and overcurrent devices for feeders and services. Part III contains feeder and service load calculation provisions. Optional feeder and service load calculation methods are covered in Part IV. This part contains optional or alternative load calculation methods for one-family dwellings, existing dwelling units, multifamily dwellings, two-family dwelling units (where supplied by a single feeder), schools, existing installations, and new restaurants. The last part in Article 220, Part IV, contains calculation methods for farms.

Last month’s column concluded by covering the first three types of heating and air conditioning loads in 220.82(C). This month, the discussion continues with optional feeder or service load calculations as specified in 220.82(C)(4).

When calculating feeder and service loads (in a one-family dwelling) by the optional method, heating and air conditioning loads are calculated separately from the general loads. The heating and air conditioning load in kilovolt-amperes (kVA) is the largest of the selections in 220.82(C)(1) through (6). Where there are one, two or three separately controlled, electric space heating units, multiply the load by 65 percent [220.82(C)(4)]. For example, a total of three electric wall heaters will be installed in a one-family dwelling. Two of the units are rated 3,000 watts (W) at 240 volts (V), and one unit is rated 4,800W at 240V. Each wall heater is separately controlled. This dwelling does not have an air conditioning load. Calculating by the optional method and after applying the appropriate demand factors, what is the heating load for these three electric space heaters? Start by calculating the total load in volt-amperes (VA). The total rating for all three heaters is 10,800 VA (3,000 + 3,000 + 4,800 = 10,800). Because there are less than four separately controlled space heating units, multiply the total rating by 65 percent. After applying 65 percent to the total, the load is 7,020 VA (10,800 65% = 7,020). When calculating these three heaters by the optional method, add 7,020 VA (7,020 ÷ 1,000 = 7.02 kVA) to the general loads (see Figure 1).

If there are four or more separately controlled, electric space heating units, it is permissible to multiply the nameplate ratings by 40 percent [220.82(C)(5)]. For example, six separately controlled electric wall heaters will be installed in a one-family dwelling. Two heaters are rated 1,500W at 120V, one heater is rated 2,000W at 240V, two heaters are rated 3,000W at 240V and one heater is rated 4,000W at 240V. This dwelling does not have an air conditioning load. Calculating by the optional method and after applying the appropriate demand factors, what is the heating load for these six electric space heaters? The total rating for all six heaters is 15,000 VA (1,500 + 1,500 + 2,000 + 3,000 + 3,000 + 4,000 = 15,000). Because there are at least four separately controlled space heating units, multiply the total rating by 40 percent. After applying 40 percent to the total, the load is 6,000 VA (15,000 40% = 6,000). The total heating load is 6,000 VA or 6.0 kVA (see Figure 2).

The last selection, the electric thermal storage system (ETS), is not as well known as the other five selections in 220.82(C). Some electric utilities offer a discounted rate for kilowatt-hours used during certain hours of the day and night. During off-peak hours, customers pay rates that are less than rates during peak hours. During off-peak hours, electric elements are used to heat ceramic bricks. The stored heat is then released throughout the day when rates are higher. Many installations include multiple units placed throughout the house. In accordance with 220.82(C)(6), in electric thermal storage and other heating systems, where the usual load is expected to be continuous at the full nameplate value, the nameplate ratings must be calculated at 100 percent. Systems qualifying under this selection shall not be calculated under any other selection in 220.82(C). Since all of the electric thermal storage units could be heating bricks at the same time, no demand factor can be applied to the units. For example, four electric thermal storage units will be installed in a one-family dwelling. Two units are rated 5.4 kilowatts (kW) at 240V and two units are rated 7.2 kW at 240V. This dwelling does not have an air conditioning load. Calculating by the optional method and after applying the appropriate demand factors, what is the heating load for this dwelling? The total rating for all four units is 25.2 kVA or 25,200 VA (5.4 + 5.4 + 7.2 + 7.2 = 25.2 kVA). Because electric thermal storage system loads are calculated at 100 percent, 25.2 kVA will be added to the general loads (see Figure 3).

Up to this point, each of the six selections or types of systems in 220.82(C) has been covered one at a time, and, therefore, it was not necessary to compare the loads and select the largest. But in many installations, air conditioning loads and heating loads are not energized at the same time. Where more than one selection in 220.82(C)(1) through (6) is used to calculate heating and air conditioning loads, calculate each selection separately. After applying demand factors, if any, compare the loads and select the largest load in kilovolt-amperes. For example, three window air conditioners and five separately controlled electric wall heaters will be installed in a one-family dwelling. One window unit is rated 11.5 amperes at 240V and the other two are rated 9 amperes at 240V. Two wall heaters are rated 2,000W at 240V, two heaters are rated 3,000W at 240V and one heater is rated 4,000W at 240V. Calculating by the optional method, how much heating and air conditioning load will be added to the general loads? The larger air conditioner has a rating of 2,760 VA (11.5 240 = 2,760). Each smaller window unit has a rating of 2,160 VA (9 240 = 2,160). The total load for all three units is 7,080 VA (2,760 + 2,160 + 2,160 = 7,080). Since the air conditioning load is calculated at 100 percent, the total air conditioning load is 7,080 VA (7,080 100% = 7,080). The total rating for all five heaters is 14,000 VA (2,000 + 2,000 + 3,000 + 3,000 + 4,000 = 14,000). Because there are at least four separately controlled space heating units, multiply the total rating by 40 percent. After applying 40 percent to the total, the heating load is 5,600 VA (14,000 40% = 5,600). Because the heating and air conditioning systems in this example will not run simultaneously, it is only required to include the larger of the two loads. The larger load, after applying demand factors, is the air conditioning load. Add 7,080 VA or 7.08 kVA (7,080 ÷ 1,000 = 7.08) to the general loads (see Figure 4).

Next month’s Code in Focus continues the discussion of feeder and service load calculations.

MILLER, owner of Lighthouse Educational Services, teaches classes and seminars on the electrical industry. He is the author of “Illustrated Guide to the National Electrical Code” and “The Electrician’s Exam Prep Manual.” He can be reached at 615.333.3336, [email protected] and www.charlesRmiller.com.