A feeder neutral with a load of 400 A would be permitted the demand factor applied to of the load

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Characteristics of the Neutral Conductor

In any electrical system, the neutral is a grounded conductor that you must size and treat differently from ungrounded phase conductors.

Do you know how to properly size a neutral conductor? Do you know the rules surrounding its proper application? If not, read on. This article discusses how to calculate the neutral current for various circuit configurations to meet the requirements set forth in the Code.

Sizing the neutral: Sec. 220-22. You must size the neutral conductor to carry the maximum unbalanced current in the circuit (i.e. the largest load between the neutral and any one ungrounded phase conductor). You calculate the first 200A of neutral current at 100%. For all resistive loads on the neutral exceeding 200A, you must apply a demand factor of 70%. Then, you add this value to the first 200A, which we calculated at 100%.

You calculate all inductive neutral current at 100% with no demand factor applied. When working with cooking equipment or a dryer load, the feeder neutral load shall also be 70% of the demand load. You must use a multiplier of 140% when calculating the neutral current for a 3-wire, 2-phase or 5-wire, 2-phase system. The neutral conductors do not become overloaded because 120V loads switch in and out on the circuits at different intervals of time.

Using the neutral: Sec. 310-15(b)(4). This section has three subdivisions explaining the loading conditions and use of the neutral conductor. Let’s take a closer look at each of these sections to help you fully understand their application.

Part (a). The Code considers the neutral conductor a current-carrying conductor only when it carries the unbalanced current from other ungrounded phase conductors. When circuits are properly balanced, the neutral carries very little current. When sizing the load for a 2-wire circuit, the grounded neutral conductor carries the same amount of current as the ungrounded phase conductor. This type of installation has no unbalanced load; therefore, the neutral conductor carries full current.

Example: What is the neutral load for a single-phase, 120V, 2-wire circuit supplying a load of 14A?

Step 1: Find amperage per Sec. 220-22 and Sec. 310-15(b)(4)(a).

Ungrounded conductor = 14A

Grounded neutral conductor = 14A

Solution: Size the neutral conductor to carry a load of 14A.

When sizing the load for a 3-wire circuit, the grounded neutral conductor must carry the unbalanced load of the two ungrounded phase conductors. This type of installation has an unbalanced load – unless both ungrounded conductors pull the same amount of current on each ungrounded phase conductor.

Example: What is the unbalanced neutral load for a 3-wire circuit carrying 64A and 52A on the ungrounded phase conductors?

Step 1: Find amperage per Sec. 220-22 and Sec. 310-15(b)(4)(a).

Ungrounded phase conductor: Phase A = 64A

Ungrounded phase conductor: Phase B = 52A

Unbalanced load = 12A

Solution: The grounded neutral conductor load is 12A for the unbalanced condition.

You must use a specific formula to calculate the neutral current for 3-phase feeder-circuits. Where currents on Phases A, B, and C are of different values, you can compute the neutral current

Part (b). The Code requires the grounded neutral conductor of a 3-wire, 120/208V feeder-circuit to be the same size as the ungrounded phase conductors for a feeder-circuit derived from a 4-wire, 120/208V system.

This is because the grounded neutral of a 3-wire circuit (consisting of 2-phase conductors) carries approximately the same amount of current as the ungrounded phase conductor. Therefore, the Code does not allow a reduction in ampacity.

Example: What is the grounded neutral conductor load for a 120/208V, single-phase circuit taken from a 4-wire wye, 3-phase system with 190A on phase A, 170A on phase B, and 90A for the neutral?

Solution: You must size the grounded (neutral) conductor based on the largest ungrounded phase conductor. Therefore, you must size the grounded conductor to carry 190A.

Part (c). The grounded neutral conductor of a 4-wire, 3-phase system supplying nonlinear loads must be the same size as the ungrounded phase conductors. The Code considers the grounded neutral conductor a current-carrying conductor due to the harmonic currents generated by these loads.

A demand factor of 70% applies to neutral loads exceeding 200A for nonlinear loads. You shall calculate nonlinear related loads at 100%.

Example: What is the load for the neutral if it exceeds 200A and has more than 50% of its load affected by harmonics? The ungrounded phase conductors carry a total neutral load of 275A respectively.

Step 1: Find amperage per Sec. 310-15(b)(4)(c). Phases4275A

Step 2: Calculate amperage per Sec. 220-22.

First, 200A x 100% = 200A

Next, 75A x 100% = 75A

Therefore, the total = 275A

Solution: You must size the neutral conductor to carry 275A.

The Code considers the grounded neutral conductor a current-carrying conductor because of the harmonic currents generated by these loads. You must apply Sec. 310-15(b)(2)(a) for four or more current-carrying conductors in a conduit, cable, etc.

Example: What is the neutral load for 120V loads having harmonic currents of 400A per phase?

Step 1: Find amperage per Sec. 310-15(b)(4)(c). Ungrounded conductors = 400A

Step 2: Calculate amperage per Sec. 220-22. 400A2 x 100% = 400A

Solution: The neutral load is 400A.

Note: The Code does not permit reduction of ampacity due to harmonic currents.

You must determine the size of the neutral conductor (based on its use with ungrounded circuit conductors) carefully. For example, the manner in which it shares loads between the other conductors determines if you can reduce its ampacity rating. Likewise, you must consider the number of current-carrying conductors to see if you must derate the neutral’s ampacity. The neutral conductor is special; therefore, you must size it accordingly.

Whether they’re for journeymen, master electricians, or contractors, most electrical licensing exams require you to calculate residential loads and service or feeders sizes using one of two methods. The standard method, which is contained in Art. 220, Part II, involves more steps, but many people use it exclusively to avoid using the wrong method. However, most residential construction qualifies for the optional method in Art. 220, Part III, so it’s prudent to understand both methods. In either case, you’re free to exceed the NEC requirements — these are minimum requirements, not design specifications.

The standard method is where we’ll start. It requires six sets of calculations for general lighting and receptacles, small-appliance, and laundry; air conditioning versus heat; appliances; clothes dryer; cooking equipment; and conductor size.

The following example should help illustrate how to apply these steps.

What size service conductor does a 1,500 sq ft dwelling unit need, if it contains the following loads? The service is 120/240V.

  • Disposal (940VA)

  • Dishwasher (1,250VA)

  • Trash compactor (1,100VA)

  • Water heater (4,500VA)

  • Dryer (4,000VA)

  • Cooktop (6,000VA)

  • Two ovens (each 3,000VA)

  • Air conditioning (5 hp, 230V)

  • Three electric space heating units (each 3,000W)

General lighting and receptacles, small-appliance, and laundry. The NEC recognizes these circuits won’t all be on (loaded) simultaneously. Thus, you may apply a demand factor to the total connected general lighting and receptacle load (220.16). To determine the service/feeder demand load, refer to Table 220.11 and follow these steps:

First, determine the total connected load for general lighting and receptacles (3VA per sq ft) [Table 220.3(A)], two small-appliance circuits each at 1,500VA, and one laundry circuit at 1,500VA (220.16) (Fig. 1).

Second, apply Table 220.11 demand factors to the total connected general lighting and receptacle load. Calculate the first 3,000VA at 100% demand and the remaining VA at 35% demand.

General lighting/receptacles: 1,500 sq ft× 3VA=4,500VA

Small-appliance circuits: 1,500VA×2 =3,000VA

Laundry circuit: 1,500VA×1= 1,500VA

Total connected load: 4,500VA+ 3,000VA+1,500VA=9,000VA

First 3,000VA at 100%=3,000VA× 1.00=3,000VA

Remainder at 35%=(9,000VA- 3,000VA)×0.35=2,100VA

Total demand load=5,100VA

Air-conditioning versus heat. Because air-conditioning and heating loads aren’t on simultaneously, you may omit the smaller of the two loads (220.21). Calculate each of these at 100% (220.15) (Fig. 2).

Air conditioning: 5 hp, 230V

VA=E×I (Table 430.148)

28 FL×230V=6,440 VA

Heat: 3,000W×3 units=9,000W

The air conditioning load is smaller than the heat load, therefore it can be omitted.

Appliances. Per 220.17, you can use a 75% demand factor when four or more “fastened in place” appliances, such as a dishwasher or waste disposal, are on the same feeder. Don’t include clothes dryers, cooking equipment, air conditioning, or heat in this category (Fig. 3).

Waste disposal: 940VA

Dishwasher: 1,250VA

Trash compactor: 1,100VA

Water heater: 4,500VA

Total connected appliance load: 7,790VA× 0.75=5,843VA

Clothes dryer. Per 220.18, the feeder or service demand load for electric clothes dryers in a dwelling unit shall not be less than 5,000W. However, if the nameplate rating exceeds 5,000W, use that rating as the load. You can omit this calculation if the unit has no electric dryer provision. However, it’s common to provide both gas and electric sources. If you see gas on the plans, verify electric will be omitted (Fig. 4).

The service and feeder demand load for a 4kW dryer is 5,000W.

Cooking equipment. For household-cooking appliances rated higher than 1.75kW, you can use the demand factors listed in 220.19, Table and Notes 1, 2, and 3.

All three of the units in the example are rated higher than 1.75kW and not higher than 8.75kW, so follow the instructions in Note 3. The two ovens are rated less than 3.5kW, so Table 220.19 Column A demand factor applies. The cooktop is 6kW, so Column B demand factor applies (Fig. 5).

Column A demand: 3kW×2 units× 0.75 demand factor=4.5kW

Column B demand: 6kW×1 unit× 0.8=4.8kW

Demand load=4.5kW+4.8kW= 9.3kW=9,300W

Feeder and service conductor size. 400A and less: For 3-wire, 120/240V, single-phase systems, size the feeder or service conductors according to Table 310.15(B)(6). For all others, use Table 310.16. Size the grounded (neutral) conductor to the maximum unbalanced load (220.22) per Table 310.16.

Over 400A: Size the ungrounded and grounded (neutral) conductors per Table 310.16.

Now we can total up the demand loads from steps 1 through 5.

Step 1: 5,100VA

Step 2: 9,000VA

Step 3: 5,843VA

Step 4: 5,000VA

Step 5: 9,300W

Step 6: 34,243VA total demand load To determine the amperes need for the service use the formula: I5VA÷E.

I=34,243VA÷240V=143A

We can use 310.15(B)(6) for a 120/240V single-phase dwelling service and feeder up to 400A. This Table allows a smaller conductor size than Table 310.16.

A 143A demand load means this house requires at least a 150A service with 1 AWG conductors.

Optional method. You can use the easier optional method found in 220.30 only when the total connected load is served by a single 3-wire, 120/240V or 208Y/120V set of service or feeder conductors with an ampacity of 100A or greater. Because this condition describes the typical residential service, the optional method is likely to apply. Using it can simplify the design process and save you time because you have so many fewer sets of calculations.

General loads. The calculated load shall not be less than 100% for the first 10kW, plus 40% of the remainder of the following loads:

  • Small-appliance and laundry branch circuits: 1,500VA for each 20A circuit.

  • General lighting and receptacles: 3VA per sq ft

  • Appliances: The nameplate VA rating of all appliances and motors fastened in place (permanently connected) or on a specific circuit. Be sure to use the range and dryer at nameplate rating.

HVAC. Include the largest of the following:

  • 100% of the nameplate rating of the air-conditioning equipment.

  • 100% of the heat-pump compressors and supplemental heating, unless the controller prevents simultaneous operation of the compressor and supplemental heating.

  • 100% of the nameplate ratings of electric thermal storage and other heating systems where you expect the usual load to be continuous at the full nameplate value. Don’t configure such systems under any other selection in this table.

  • 65% of the nameplate rating(s) of the central electric space heating, including integral supplemental heating in heat pumps where the controller prevents simultaneous operation of the compressor and supplemental heating.

  • 65% of the nameplate rating(s) of electric space heating, if there are less than four separately controlled units.

  • 40% of the nameplate rating(s) of electric space heating of four or more separately controlled units.

Sizing service/feeder conductors. Now that we’ve seen how to determine residential loads, let’s size the service/feeder conductors. We’ll use the same specifications that we used for the standard method so we can compare apples to apples.

Step 1: Determine general loads [230.30(B)].

Small appliance: 1,500VA×2 circuits = 3,000VA

General lighting: 1,500 sq ft×3VA= 4,500VA

Laundry circuit=1,500VA

Now add up the appliance ratings.

  • Disposal (940VA)
  • Dishwasher (1,250VA)
  • Trash compactor (1,100VA)
  • Water heater (4,500VA)
  • Dryer (4,000VA)
  • Ovens (3,000VA×2 units=6,000VA)
  • Cooktop (6,000VA)

The total connected load=32,790VA

Calculate the first 10,000VA at 100%=10,000VA× 1.00=10,000VA

Calculate the remainder at 40%= 22,790VA×0.40= 9,116VA

Demand load=10,000VA+9,116VA= 19,116VA

Step 2: Compare air conditioner at 100% vs. heat at 65% [220.30(C)].

Air conditioner: 230V×28A=6,440VA

Heat [220.30(C)(5)]: 9,000 W×0.65 = 5,850 W (omit)

Step 3: Calculate service/feeder conductors per 310.15(B)(6).

General loads=19,116VA

Air conditioning=6,440VA

Total demand load=25,556VA

I=VA÷E=25,556VA ÷ 240V = 106.5A

310.15(B)(6) requires at least a 110A service with 3 AWG conductors.

As you can see, in this case the optional method permitted a smaller service than the standard method of calculating a service for a dwelling.

Now that we’ve walked through the process of calculating residential services and feeders, you can see that doing so is fairly easy. You need to calculate the loads first, and then move on to the service and feeder size. The NEC provides the requirements in Art. 220 and 230. Doing these calculations correctly can save you money during design and construction, while providing safe homes for the families who occupy them.

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