You’re reading an older article from ELECTRICAL CONTRACTOR. Some information, such as code-related information, may be outdated. Find the most up-to-date content in our latest issues.

Provisions in article 220 are few in number compared to the total number of provisions in the National Electrical Code (NEC) book. The number of pages containing requirements in Article 220 makes up approximately 1½ percent of the total number of pages in the NEC that contain requirements. This small percentage is not proportional in size to the importance of this article.

Article 220 contains requirements for calculating branch-circuit, feeder and service loads. Requirements in Article 220 are divided into five parts. General requirements for calculation methods are in Part I. Part II provides calculation methods for branch-circuit loads. Part III contains feeder and service load-calculation methods. Although it is not stated as such in Article 220, Part III is sometimes referred to as the standard method for feeder and service load calculations. Part IV, “Optional Feeder and Service Load Calculations,” also contains load-¬calculation methods for feeders and services. This part contains optional or alternative load-calculation procedures for one-family dwellings, existing dwelling units, multifamily dwellings, two dwelling units supplied by a single feeder, schools, feeder or service loads for existing installations, and new restaurants. Part V provides calculation methods for farms.

Last month’s column concluded by covering permitted reductions or feeder and service neutral loads. This month, the discussion continues with feeder or service neutral loads as specified in 220.61.

Requirements in 220.61 are divided into three sections: basic calculations, permitted reductions and prohibited reductions. Provisions pertaining to prohibited reductions are closely related to 310.15(B)(4)(b) and (c), which pertain to neutral conductors that must be counted as current-carrying conductors when applying the provisions of 310.15(B)(2)(a). As specified in 220.61(C), there shall be no reduction of the neutral or grounded conductor capacity applied to the amount

in 220.61(C)(1), or portion of the amount in (C)(2), from that determined by the basic calculation. Both of these provisions pertain to the same type of electrical system—a three-phase, four-wire, wye-connected system. The prohibited reduction provisions in this section do not apply to other types of electrical systems. It is not permissible to reduce any portion of a three-wire circuit consisting of two ungrounded (hot) conductors and the neutral conductor of a three-phase, four-wire, wye-connected system (see Figure 1).

As discussed last month, it is permissible to apply an additional demand factor of 70 percent to a feeder or service neutral supplying household electric ranges. But, applying the 70 percent demand factor is not permissible if the ranges are supplied by a single-phase panel that is fed from a three-phase, four-wire, wye-connected system. For example, a high school classroom will be converted into a culinary arts classroom. Ten household ranges will be installed. Each range is rated 10.5 kW at 208 volts. A single-phase panelboard having two ungrounded conductors and a neutral will be installed to supply the ranges. This single-phase panelboard will be fed from a three-phase, four-wire, wye-connected system. What load will these ranges add to the feeder neutral load calculation? Start by looking in the left column of Table 220.55 for 10 appliances. The maximum demand for 10 10.5 kW ranges is 25 kW. Because of the system supplying power to this single-phase panelboard, it is not permissible to apply the additional demand of 70 percent to the neutral. The neutral load for 10 10.5 kW household electric ranges is 25 kW (see Figure 2).

When more than three current-carrying conductors are in a raceway or cable, the allowable ampacity of each conductor must be reduced by the adjustment factors in Table 310.15(B)(2)(a). The neutral may or may not be a current-carrying conductor. In accordance with 310.15(B)(4)(a), a neutral conductor that carries only the unbalanced current from other conductors of the same circuit shall not be required to be counted when applying the provisions of 310.15(B)(2)(a). But, if the neutral is part of a three-wire circuit consisting of two phase conductors and the neutral conductor and it is supplied from a four-wire, three-phase, wye-connected system, the neutral must be counted as a current-carrying conductor. [310.15(B)(4)(b)]. For example, a raceway containing eight conductors will supply power to four electric heaters. Two heaters are rated 208-volt, single-phase. Each of these heaters will be on a separate circuit. The other two of the heaters are rated 120-volts and will be on separate circuits. A multiwire branch circuit will supply power to the two 120-volt heaters. There also is an equipment-grounding conductor in the raceway. A three-phase, four-wire, wye-connected panelboard is supplying power to the heaters. What is the Table 310.15(B)(2)(a) adjustment factor for these eight conductors? The six ungrounded (hot) conductors feeding the four heaters count as current-¬carrying conductors. Because of 310.15(B)(5), the grounding conductor does not count. Because of the system supplying power to these two 120-volt heaters, the neutral must be counted as a current-carrying conductor. Because the neutral must be counted, there are seven current-¬carrying conductors. The Table 310.15(B)(2)(a) adjustment factor for seven current-carrying conductors is 70 percent (see Figure 3).

If the single-phase panelboard in the last example was fed from a different system, the adjustment factor would be different. For example, the heaters in the last example will now be supplied from a 120/240 volt, single-phase, three-wire system. The neutral conductor now carries only the unbalanced current from the other conductors. Because of this system, the neutral is not required to be counted as a current-carrying conductor. Now there are six current-carrying conductors. The Table 310.15(B)(2)(a) adjustment factor for six current-carrying conductors is 80 percent (see Figure 4).

Section 310.15(B)(4)(b) also states that in this type of electrical system, a common conductor carries approximately the same current as the line-to-neutral load currents of the other conductors. Although it does not seem possible for the neutral to carry approximately the same current as the ungrounded conductors, it can be verified by the electrical formula for finding neutral current when the system is three-phase, four-wire, wye-connected.

IN= I2A+I2B+I2C–(IA×IB)–(IB×IC)–(IA×IC)

The letter “I” represents current and the subscript letters represent phases A, B and C. The superscript 2 means that the current (or number) must be squared (the square of a number is the product of a number multiplied by itself). For example, each 120-volt heater in Figure 3 draws 15 amperes. The branch-circuit overcurrent devices are in a panelboard that is supplied from a three-phase, four-wire, wye-connected system. Only two of the three phases are needed to supply power to the two 120-volt heaters. The multiwire branch circuit supplying power to the two 120-volt heaters will be terminated on phases B and C. What is the current draw of the neutral in this multiwire branch circuit? Replace the letters in the formula with the known factors and solve for neutral current. Since this multiwire branch circuit does not use phase A, the current on phase A will be 0 amperes. Replace the subscript A with 0. Replace the subscript B with 15. Replace the subscript C with 15. Since there is no current on phase A, the current of A squared is 0. The current of B squared is 225 (15 × 15 = 225). The current of C squared is 225 (15 × 15 = 225). The current of A multiplied by the current of B is 0 (0 × 15 = 0). The current of B multiplied by the current of C is 225 (15 × 15 = 225). The current of A multiplied by the current of C is 0 (0 × 15 = 0). After adding and subtracting, the sum is 225 (0 + 225 + 225 – 0 – 225 – 0 = 225). The square root of 225 is 15. Even with both of these 120-volt heaters energized and drawing 15 amperes each, the current on the neutral is 15 amperes (see Figure 5).

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, [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.