You're reading an older article from ELECTRICAL CONTRACTOR. Some content, such as code-related information, may be outdated. Visit our homepage to view the most up-to-date articles.

Article 220 of the National Electrical Code (NEC) contains requirements for calculating branch-circuit, feeder and service loads. While understanding load calculation requirements is essential to the professional electrician, knowing what to do with the results from the load calculations is just as important.

Article 220 is not a stand-alone article; results from load calculations are used in conjunction with provisions from other articles. For example, a panelboard will be installed in a new addition to a church. The new load will be calculated per Part III of Article 220. With the result of the load calculation, the size of the feeder conductors can be determined. In accordance with 215.2(A)(1), feeder conductors must have an ampacity not less than required to supply the load as calculated in Parts III, IV and V of Article 220. The minimum feeder-circuit conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load. Load calculation results are also used to size the overcurrent device for the feeder. Feeders shall be protected against overcurrent in accordance with the provisions of Part I of Article 240. Where a feeder supplies continuous loads or any combination of continuous and noncontinuous loads, the rating of the overcurrent device shall not be less than the noncontinuous load plus 125 percent of the continuous load [215.3]. It is also required to use Article 220 to calculate loads for outside branch circuits and feeders for systems rated 600 volts (V) nominal or less. The load on outdoor branch circuits shall be as determined by 220.10 [225.3(A)]. The load on outdoor feeders shall be as determined by Part III of Article 220 [225.3(B)].

Optional method load calculations for multifamily dwellings are covered in 220.84. There are five categories or types of connected loads listed in 220.84(C). The fourth type of load to which the demand factors in Table 220.84 apply is the nameplate ampere (A) or kilovolt-ampere (kVA) rating of all permanently connected motors not included in item (3) of 220.84(C). If, at this point in the multifamily dwelling load calculation, there are any permanently connected motors that have not been included, add the nameplate ampere or kilovolt-ampere rating of those motors to the calculation. Do not include motors that are part of the air conditioning system or motors that are part of the fixed electric space-heating system. These motors will be included as 220.84(C)(5) requires. Include only permanently connected motors that are supplied from dwelling units. Do not include permanently connected motors that are supplied by house power. For example, the load of a multifamily dwelling will be calculated by the optional method. Two swimming pool pump motors and two automatic power attic roof ventilators will be powered by the house panel. Can these permanently connected motors be included with the connected loads to which Table 220.84 apply? Because these permanently connected motors are supplied by house power, do not add them to the loads that will be calculated in accordance with Table 220.84. These motors must be calculated in accordance with Part III of Article 220 (see Figure 1).

The electric space heating load is not calculated the same way in multifamily dwellings as it is in one-family dwellings. With the optional method one-family dwelling load calculation, it is permissible to apply a demand factor to space heating units. The demand factor depends on the number of units. With the optional method multifamily dwelling load calculation, the load for space heating units must be added to the calculation at the nameplate rating. The air conditioning load is calculated the same: at 100 percent of the nameplate rating. When calculating a multifamily dwelling by the optional method, use the larger of the air conditioning loads or the fixed electric space-heating load. For example, what is the optional method service load calculation (before applying the Table 220.84 demand factor) for a 30-unit multifamily dwelling that will have electric space heaters and window air conditioners? The heating for each unit will consist of two separately controlled wall heaters rated 2,250W each and one heater rated 1,500W. The air conditioning for each unit will consist of two window air conditioning units rated 11.5A at 240V. Assume space-heating watts are equivalent to volt-amperes (VA). The heat load for each unit is 6,000 VA (2,250 + 2,250 + 1,500 = 6,000). The total heat load for 30 units is 180,000 VA (6,000 30 = 180,000). Multiply volts, amperes and the number of air conditioners in each unit to find the air conditioning load for each unit. The air conditioning load for each unit is 5,520 VA (240 11.5 2 = 5,520). The total air conditioning load for 30 units is 165,600 VA (5,520 30 = 165,600). Since the heating load is larger than the air conditioning load, omit the air conditioning load. The optional method service load calculation (before applying the Table 220.84 demand factor) for the heating and air conditioning loads in this 30-unit multifamily dwelling is 180,000 VA (see Figure 2).

Do not assume the heating load will always be larger than the air conditioning load. The heating system could be gas or oil. The dwelling could also be located in a warm climate where the air conditioning load is larger than the heating load. For example, what is the optional method service load calculation (before applying the Table 220.84 demand factor) for a 12-unit multifamily dwelling that will have an identical central heating and air unit for each dwelling unit? Each package unit will contain a compressor, a blower motor, a condenser fan motor and electric heat. The compressor draws 23A at 240V, the blower motor draws 5A at 240V, and the condenser fan motor draws 2A at 240V. The rating of the electric heat is 5 kW. Assume the heating kilowatt rating is equivalent to kilovolt-amperes. The load of each air conditioner compressor is 5,520 VA (23 240 = 5,520). The load of each blower motor is 1,200 VA (5 240 = 1,200). The load of each condenser fan motor is 480 VA (2 240 = 480). The air conditioning load for each unit is 7,200 VA (5,520 + 1,200 + 480 = 7,200). The total air conditioning load for 12 units is 86,400 VA (7,200 12 = 86,400). Since the blower motor also works with the heat, add the load of the blower motor to the heat load of 5,000 VA (5 1,000 = 5,000). The heating load for each unit is 6,200 VA (5,000 + 1,200 = 6,200). The total heating load for 12 units is 74,400 VA (6,200 12 = 74,400). In this example, the air conditioning load is larger than the heating load; therefore, omit the heating load. The optional method service load calculation (before applying the Table 220.84 demand factor) for the heating and air conditioning system in this 12-unit multifamily dwelling is 86,400 VA (see Figure 3).

Fixed electric space heating is not limited to space heaters and electric strip heat that is part of a package unit or furnace. Fixed electric space heating could also be a heat pump with supplementary heat. With a heat pump, the compressor (and accompanying motors) and some or all of the electric heat can be energized at the same time. The load contribution of a heat pump is the air conditioning system load plus the maximum amount of heat that can be on while the air conditioner compressor is energized. For example, what is the optional method service load calculation (before applying the Table 220.84 demand factor) for a six-unit multifamily dwelling that will have an identical heat pump for each dwelling unit? Each unit’s compressor draws 26.8A at 240V, the blower motor draws 5.8A at 240V, and the condenser fan motor draws 2.6A at 240V. The electric heat in this unit has a rating of 15 kW. The compressor and all of the heat in this heat pump can be energized at the same time. Assume the heating kilowatt rating is equivalent to kilovolt-amperes. The load of the air conditioner compressor is 6,432 VA (26.8 240 = 6,432). The load of the blower motor is 1,392 VA (5.8 240 = 1,392). The load of the condenser fan motor is 624 VA (2.6 240 = 624). The air conditioning load for each unit is 8,448 VA (6,432 + 1,392 + 624 = 8,448). Since the air conditioning system and all of the heat can be on at the same time, add the two together (8,448 + 15,000 = 23,448 VA). The total load for six units is 140,688 VA (23,448 6 = 140,688). The optional method service load calculation (before applying the Table 220.84 demand factor) for the heating and air conditioning system in this six-unit multifamily dwelling is 140,688 VA (see Figure 4).

**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.

### About The Author

Charles R. Miller, owner of Lighthouse Educational Services, teaches custom-tailored seminars on the National Electrical Code and NFPA 70E. He is the author of “Illustrated Guide to the National Electrical Code” and “Electrician's Exam Prep Manual.” He can be reached at 615.333.3336 and [email protected]. Connect with him on LinkedIn.