How many times has it been said that the National Electrical Code (NEC) is a minimum set of electrical standards or that the Code contains minimum electrical installation requirements? While these statements are true, wording such as this is not in the Code book, that is, not in the Code book anymore.
The first sentence in 90-1(B) of the 1968 edition states that the NEC contains basic minimum provisions considered necessary for safety. In the 1971 edition, the sentence was changed by deleting the words “basic” and “minimum.” As stated in the 1971 edition, this Code contains provisions considered necessary for safety [90.1(B).] Other than a small revision in the 1999 edition that added the words “that are,” the sentence is exactly the same today. Except for a small grammatical change in the 2005 edition, the rest of this section has also remained the same. Compliance with the Code and proper maintenance results in an installation that is essentially free from hazards but not necessarily efficient, convenient or adequate for good service or future expansion of electrical use [90.1(B)]. The fine print note (FPN) under this section has also remained the same except for a small grammatical change. The FPN provides a warning of what could happen by not providing for future increases in the use of electricity. In accordance with the FPN under 90.1(B), hazards often occur because of overloading of wiring systems by methods or usage not in conformity with the NEC. This occurs because initial wiring did not provide for increases in the use of electricity.
Last month’s column concluded by covering the general loads when calculating feeders and services by the optional method [220.82(B)]. This month, the discussion continues with optional feeder or service load calculations as specified in 220.82.
Heating and air conditioning loads in a one-family dwelling are calculated separately from the general loads when calculating feeder and service loads by the optional method. Heating and air conditioning loads are divided into six sections. The heating and air conditioning load in kilovolt-amperes (kVA) is the largest of the selections in 220.82(C)(1) through (6). First multiply the load by the percentage that is applicable to the type of equipment and then select the largest load in kilovolt-amperes. If a dwelling has some type of heat other than electric, calculate the air conditioning load only. For example, a central heating and air conditioning unit will be installed in a one-family dwelling. The air conditioning system is electric, and the heating system is gas. The air conditioner compressor has a rating of 16.6 amperes (A) at 230 volts (V). The condenser fan motor has a rating of 2A at 115V and the air-handler (blower motor) has a rating of 3.2A at 115V. Calculating by the optional method, how much air conditioning load will be added to the general loads? Start by calculating the volt-ampere (VA) load for each motor. The compressor load is 3,818 VA (16.6 230 = 3,818). The condenser fan load is 230 VA (2 115 = 230). The air handler load is 368 VA (3.2 115 = 368). Now, add the air conditioning loads to find the total (3,818 + 230 + 368 = 4,416 VA). Because air conditioning loads are calculated at 100 percent, 4,416 VA or 4.416 kVA (4,416 ÷ 1,000 = 4.416) will be added to the general loads (see Figure 1).
Where there is more than one air conditioning unit, the calculation method is the same as it is for one air conditioning unit. For example, four window air conditioners will be installed in a one-family dwelling. The heating system is not electric. Two units are rated 12.5A at 230V and the other two are rated 8.5 amperes at 230 volts. Calculating by the optional method, how much air conditioning load will be added to the general loads? Each of the larger air conditioners has a rating of 2,875 VA (12.5 230 = 2,875). Each of the smaller window units has a rating of 1,955 VA (8.5 230 = 1,955). The total load for all four units is 9,660 VA (2,875 + 2,875 + 1,955 + 1,955 = 9,660). The multiplication factor for four air conditioning units is the same as it is for one unit (9,660 100% = 9,660). When calculating these four window units by the optional, add 9,660VA or 9.66 kVA (9,660 ÷ 1,000 = 9.66) to the general loads (see Figure 2).
Heat pumps are either equipped with or without electric supplemental heat. Supplemental heat is sometimes referred to as auxiliary, backup or even emergency heat. A dual-fuel heat pump is an electric heat pump and a gas furnace all in one. Dual fuel heat pumps can be fueled with natural gas or propane. Because geothermal heat pumps (sometimes referred to as geoexchange, earth-coupled, ground-source or water-source heat pumps) do not depend on the temperature of the outside air, they may or may not be equipped with supplemental heat. Heat pumps not equipped with supplemental electric heat are calculated exactly the same as the air conditioning equipment specified in 220.82(C)(1). When a heat pump is used without any supplemental electric heat, multiply the nameplate rating(s) by 100 percent and then add the result to the general loads [220.82(C)(2)].
When a heat pump is used with supplemental electric heat, multiply the nameplate rating(s) of the heat pump compressor by 100 percent and multiply the supplemental electric heating for central electric space-heating systems by 65 percent [220.82(C)(3)]. For example, a heat pump with supplemental electric heat will be installed in a one-family dwelling. The-21/2 ton heat pump system has nameplate rating of 24A at 240V. This heat pump is equipped with 15 kilowatts (kW) of backup heat. Calculating by the optional method, how much heating and air conditioning load will be added to the general loads? The heat pump load is 5,760 VA (24 240 = 5,760). The backup or supplemental heat is 15 kVA or 15,000 VA (15 1,000 = 15,000). Multiply the backup heat by 65 percent. After applying 65 percent to the backup heat, the load is 9,750 VA (15,000 65% = 9,750). The total heating and air conditioning load is 15,510 VA or 15.51 kVA (5,760 + 9,750 = 15,510) (see Figure 3).
Some heat pumps are designed so that only part of the electric heat will operate while the compressor is in operation. If the heat pump compressor is prevented from operating at the same time as the supplementary heat, it does not need to be added to the supplementary heat for the total central space-heating load [220.82(C)(3)]. For example, a heat pump with supplemental electric heat will be installed in a one-family dwelling. The 3-ton heat pump system has nameplate rating of 30A at 240V. Although this heat pump is equipped with 15 kW of backup heat, it is designed so that only 10 kW can operate while the compressor is running. The other 5 kW of electric heat will only operate when the compressor is in the off position. Calculating by the optional method, how much heating and air conditioning load will be added to the general loads? The heat pump load is 7,200 VA (30 240 = 7,200). Because only part of the heat can operate while the compressor is running, multiply 10 kW (or 10,000 VA) by 65 percent. After applying 65 percent to the supplemental heat that operates with the compressor, the load is 6,500 VA (10,000 65% = 6,500). The total heating and air conditioning load is 13,700 VA or 13.7 kVA (7,200 + 6,500 = 13,700) (see Figure 4).
Next month’s column 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, charles@charlesRmiller.com and www.charlesRmiller.com.