Ecmweb 6506 Nec Code Quandaries July 2014
Ecmweb 6506 Nec Code Quandaries July 2014
Ecmweb 6506 Nec Code Quandaries July 2014
Ecmweb 6506 Nec Code Quandaries July 2014
Ecmweb 6506 Nec Code Quandaries July 2014

Stumped by the Code? Disconnects for Motors

July 16, 2014
Your most pressing National Electrical Code (NEC) questions answered

All questions and answers are based on the 2011 NEC.

Q. What is the NEC requirement for disconnects for a motor?

A. A disconnecting means is required for each motor controller, and it must be located within sight from the controller [430.102(A)]. According to Art. 100, “within sight” means that it’s visible and not more than 50 ft from one to the other.

A motor disconnect must be provided in accordance with either one of the following [430.102(B)], as shown in Fig. 1.

Fig. 1. The NEC defines “within sight” as being visible and located no more than 50 ft away.

• A disconnecting means is required for each motor, and it must be located in sight from the motor location and the driven machinery location [430.102(B)(1)].

• The controller disconnecting means [430.102(A)] can serve as the disconnecting means for the motor, if the disconnect is located in sight from the motor location [430.102(B)(2)].

Exception: A motor disconnecting means isn’t required under either condition (a) or (b) if the controller disconnecting means [430.102(A)] is capable of being locked in the open position. The provision for locking or adding a lock to the disconnecting means must be installed on or at the switch or circuit breaker, and it must remain in place with or without the lock installed.

(a)  If locating the disconnecting means is impracticable or introduces additional or increased hazards to persons or property.

(b) In industrial installations, with written safety procedures, where conditions of maintenance and supervision ensure only qualified persons will service the equipment.

Informational Note 2: For information on lockout/tagout procedures, see NFPA 70E, Standard for Electrical Safety in the Workplace.

Q. How do I size the grounding electrode conductor for a service that is fed by parallel service conductors?

A. Except as permitted in (A) through (C), a grounding electrode conductor must be sized in accordance with Table 250.66 [250.66]. If the grounding electrode conductor is connected to a ground rod as permitted in 250.52(A)(5), that portion of the grounding electrode conductor that’s the sole connection to the ground rod isn’t required to be larger than 6 AWG copper [250.66(A)].

If the grounding electrode conductor is connected to a concrete-encased electrode, the portion of the grounding electrode conductor that’s the sole connection to the concrete-encased electrode isn’t required to be larger than 4 AWG copper [250.66(B)].

If the grounding electrode conductor is connected to a ground ring, the portion of the conductor that’s the sole connection to the ground ring isn’t required to be larger than the conductor used for the ground ring [250.66(C)].

Here are two additional comments to consider:

• A ground ring encircling the building/structure in direct contact with the earth must consist of at least 20 ft of bare copper conductor not smaller than 2 AWG [250.52(A)(4)]. See 250.53(F) for the installation requirements for a ground ring.

• Table 250.66 is used to size the ground-ing electrode conductor when the conditions of 250.66(A), (B), or (C) don’t apply.

Q. What are the rules on conductor box fill, and can you give me an example?

A. Boxes containing 6 AWG and smaller conductors must be sized to provide sufficient free space for all conductors, devices, and fittings. In no case can the volume of the box, as calculated in 314.16(A), be less than the volume requirement as calculated in 314.16(B). Conduit bodies must be sized in accordance with 314.16(C).

Note: The requirements for sizing boxes and conduit bodies containing conductors 4 AWG and larger are contained in 314.28. The requirements for sizing handhole enclosures are contained in 314.30(A).

The volume of a box includes the total volume of its assembled parts, including plaster rings, extension rings, and domed covers that are either marked with their volume in cubic inches or are made from boxes listed in Table 314.16(A).

The calculated conductor volume determined by (1) through (5) and Table 314.16(B) are added together to determine the total volume of the conductors, devices, and fittings. Raceway and cable fittings, including locknuts and bushings, aren’t counted for box fill calculations.

Fig. 2. When counting the number of conductors in a box, remember that length plays a key factor in your calculation.

Each unbroken conductor that runs through a box and each conductor that terminates in a box is counted as a single conductor volume in accordance with Table 314.16(B) [314.16(B)(1)]. Each loop or coil of unbroken conductor having a length of at least twice the minimum length required for free conductors in 300.14 must be counted as two conductor volumes. Conductors that originate and terminate within the box, such as pigtails, aren’t counted at all (Fig. 2).

Note: According to 300.14, at least 6 in. of free conductor, measured from the point in the box where the conductors enter the enclosure, must be left at each outlet, junction, and switch point for splices or terminations of luminaires or devices.

Exception: Equipment grounding conductors and up to four 16 AWG and smaller fixture wires can be omitted from box fill calculations if they enter the box from a domed luminaire or similar canopy, such as a ceiling paddle fan canopy.

One or more internal cable clamps count as a single conductor volume in accordance with Table 314.16(B), based on the largest conductor that enters the box. Cable connectors that have their clamping mechanism outside of the box aren’t counted [314.16(B)(2)].

Each luminaire stud or luminaire hickey counts as a single conductor volume in accordance with Table 314.16(B), based on the largest conductor that enters the box [314.16(B)(3)]. Luminaire stems don’t need to be counted as a conductor volume.

Each single-gang device yoke (regardless of the ampere rating of the device) counts as two conductor volumes, based on the largest conductor that terminates on the device in accordance with Table 314.16(B) [314.16(B)(4)]. Each multigang-device yoke counts as two conductor volumes for each gang, based on the largest conductor that terminates on the device in accordance with Table 314.16(B).

Note: A device that’s too wide for mounting in a single-gang box, as described in Table 314.16(A), is counted based on the number of gangs required for the device.

All equipment grounding conductors in a box count as a single conductor volume in accordance with Table 314.16(B), based on the largest equipment grounding conductor that enters the box. Insulated equipment grounding conductors for receptacles having insulated grounding terminals (isolated ground receptacles) [250.146(D)] count as a single conductor volume in accordance with Table 314.16(B) [314.16(B)(5)]. Conductor insulation isn’t a factor that’s considered when determining box volume calculations.

Let’s take a look at a sample calculation to hit these points home.

How many 14 AWG conductors can be pulled through a 4-in.-square × 2½-in.-deep box with a plaster ring with a marking of 3.60 cu in.? The box contains two receptacles, five 12 AWG conductors, and two 12 AWG equipment grounding conductors.

Step 1 – Determine the volume of the box assembly [314.16(A)]:

Box (30.30 cu in.) + 3.60 cu in. plaster ring = 33.90 cu in.

Note: A 4-in. × 4-in. × 218-in. box will have a gross volume of 34 cu in., but the interior volume is 30.30 cu in., as listed in Table 314.16(A).

Step 2 – Determine the volume of the devices and conductors in the box:

Two — receptacles

4 — 12 AWG

Five — 12 AWG

5 — 12 AWG

Two — 12 AWG grounds

1 — 12 AWG

Total 10 — 12 AWG × 2.25 cu in. = 22.50 cu in.

Step 3 – Determine the remaining volume permitted for the 14 AWG conductors:
33.90 cu in. - 22.50 cu in. = 11.40 cu in.

Step 4 – Determine the number of 14 AWG conductors permitted in the remaining volume:

14 AWG = 2.00 cu in. each [Table 314.16(B)

Final calculation – 11.40 cu in. ÷ 2.00 cu in. = 5 conductors.   

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