Mike Holt Enterprises Electrical News Source

NEC Questions and Answers based on 2017 NEC® - January 2020

Figure 01

For EC&M Magazine
By Mike Holt, NEC® Consultant
Here's the follow-up to yesterday's newsletter. This includes the answers to the questions sent, so you can see how you did.

Note: The answers to these questions are based on the 2017 NEC.

Q1. How does the Code allow motor circuit conductors from a feeder to be sized?
A1. Motor circuit conductors tapped from a feeder must have an ampacity in accordance with 430.22, and the tap conductors must terminate in a branch-circuit short-circuit and ground-fault protection device sized in accordance with 430.52. In addition, any of the following requirements must be met [430.28]:
(1) 10-Ft Tap. Tap conductors not over 10 ft long must have an ampacity not less than one-tenth the rating of the feeder overcurrent protection device.
(2) 25-Ft Tap. Tap conductors over 10 ft, but not over 25 ft long, must have an ampacity not less than one-third the ampacity of the feeder conductor.
(3) Ampacity. Tap conductors must have an ampacity not less than the feeder conductors.

Q2. What does the Code require for overload protection of motors?
A2. Part III contains the requirements for overload devices, which are intended to protect motors, motor control equipment, and motor branch-circuit conductors against excessive heating due to motor overloads and failure to start [430.31].

Note 2: An overload is a condition where equipment is operated above its current rating, or where the current exceeds the conductor ampacity. When an overload condition persists for a sufficient length of time, it could result in equipment failure or a fire from damaging or dangerous overheating. A fault, such as a short circuit or ground fault, isn’t an overload [Article 100].

Author’s Comment:
• Motor overload protection sizing is usually accomplished by installing the correct “heater” or setting the overload device in accordance with the controller’s instructions, based on the motor nameplate current rating.
• The intended level of overload protection required in Article 430 Part III is for overload and failure-to-start protection only, in order to protect against the motor becoming a fire hazard.

Overload protection isn’t required if it might introduce additional or increased hazards, as in the case of fire pumps.

Note: See 695.7 for the overcurrent protection requirements for fire pump supply conductors.

Q3. What is the Code requirement for the sizing of motor overload devices?
A3. (A) Motors Rated More Than One Horsepower. Motors rated more than 1 hp, used in a continuous-duty application without integral thermal protection, must have an overload device sized as follows [430.32]:

(1) Separate Overload Device. A separate overload device must be selected to open at no more than the following percent of the motor nameplate full-load current rating:

Service Factor. Motors with a marked service factor (SF) of 1.15 or more on the nameplate must have the overload device sized no more than 125 percent of the motor nameplate current rating.

Author’s Comment:
• Motor service factors are safety factors; they indicate how much the motor capacity can be exceeded for short periods without overheating. For example, a motor with a service factor of 1.15 can operate at 15% more than its rated output without overheating. This is important for motors where loads vary and may peak slightly above the rated torque.

Example: What size dual-element fuse can be used as the overload protection device for a motor with an FLA of 23.50A and a service factor of 1.15?
Solution:
23.50A × 1.25 = 29.38A
Answer: Use a 25A DE fuse because it can’t exceed the nameplate [240.6(A), 430.32(A)(1)].

Temperature Rise. Motors with a nameplate temperature rise of 40°C or less must have the overload device sized no more than 125 percent of the motor nameplate current rating.

Author’s Comment:
• A motor with a nameplate temperature rise of 40°C means the motor is designed to operate so it won’t heat up more than 40°C above its rated ambient temperature when operated at its rated load and voltage. Studies have shown that when the operating temperature of a motor is increased 10°C above its rating, the motor winding insulating material’s anticipated life is reduced by 50 percent.

Example: What size dual-element fuse can be used as the overload protection device for a motor with an FLA of 60A and a temperature rise of 40°C?
Solution:
60A × 1.25 = 75A
Answer: Use a 70A DE fuse because it can’t exceed the nameplate [240.6(A), 430.32(A)(1)].

Other Motors. No more than 115 percent of the motor “nameplate current rating.”

Q4. What is the Code rule for sizing motor branch-circuit short-circuit and ground-fault protection?
A4. (A) General. The motor branch-circuit short-circuit and ground-fault protective device must comply with 430.52(B) and 430.52(C) [430.52].

(B) Motor Starting Current. A motor branch-circuit short-circuit and ground-fault protective device must be capable of carrying the motor’s starting current.

(C) Rating or Setting.

(1) Table 430.52. Each motor branch circuit must be protected against short circuit and ground faults by a protective device sized no greater than the following percentages listed in Table 430.52.

Table 430.52
Motor Type Nontime Delay Dual-Element Fuse Inverse Time Breaker
Wound Rotor 150% 150% 150%
Direct Current 150% 150% 150%
Other Motors 300% 175% 250%

Example: What size conductor and inverse time circuit breaker are required for a 2 hp, 230V, single-phase motor with 60ºC terminals?
Solution:
Step 1: Determine the branch-circuit conductor [Table 310.15(B)(16), 430.22, and Table 430.248]:
12A × 1.25 = 15A, 14 AWG, rated 15A at 60°C
[Table 310.15(B)(16)]
Step 2: Determine the branch-circuit overcurrent protection [240.6(A), 430.52(C)(1), and Table 430.248]:
12A × 2.50 = 30A
Answer: A 14 AWG conductor and a 30A breaker.

Author’s Comment:
• I know it bothers many in the electrical industry to see a 14 AWG conductor protected by a 30A circuit breaker, but branch-circuit conductors are protected against overloads by the overload device, which is sized between 115 and 125 percent of the motor nameplate current rating [430.32]. The small conductor rule contained in 240.4(D) which limits 15A overcurrent protection for 14 AWG doesn’t apply to motor circuit protection. See 240.4(D) and 240.4(G).

Ex 1: If the motor short-circuit and ground-fault protective device values derived from Table 430.52 don’t correspond with the standard overcurrent protection device ratings listed in 240.6(A), the next higher overcurrent protection device rating can be used.

Example: What size conductor and inverse time circuit breaker are required for a 7½ hp, 230V, three-phase motor with 60ºC terminals? Figure 01
Solution:
Step 1: Determine the branch-circuit conductor [Table 310.15(B)(16), 430.22, and Table 430.250]:
22A × 1.25 = 27.50A
Step 2: Determine the branch-circuit overcurrent protection [240.6(A), 430.52(C)(1) Ex 1, and Table 430.250]:
22A × 2.50 = 55A
Answer: A 10 AWG conductor rated 30A at 60°C [110.14(C)(1), Table 310.15(B)(16)] and a 60A breaker (next size up from 55A).

Q5. What does the Code require for the protection of feeder conductors from short circuits and ground faults?
A5. (A) Motors Only. Feeder conductors must be protected against short circuits and ground faults by a protective device sized not more than the largest rating of the branch-circuit short-circuit and ground-fault protective device for any motor, plus the sum of the full-load currents of the other motors in the group [430.62].

Example: What size feeder overcurrent protection (inverse time breakers with 60°C terminals) and conductors are required for the following two motors?
Motor 1—20 hp, 460V, three-phase = 27A FLC [Table 430.250]
Motor 2—10 hp, 460V, three-phase = 14A FLC
Solution:
Step 1: Determine the feeder conductor size [430.24]:
(27A × 1.25) + 14A = 48A
6 AWG rated 55A at 60°C [110.14(C)(1) and
Table 310.15(B)(16)]
A. Feeder overcurrent protection [430.62(A)] isn’t greater than the largest branch-circuit ground-fault and short- circuit protective device plus the other motor’s FLC.
B. Determine the largest branch-circuit ground-fault and short-circuit protective device [430.52(C)(1) Ex]:
20 hp Motor = 27A × 2.50 = 68, next size up = 70A
10 hp Motor = 14A × 2.50 = 35A
Step 2: Determine the size feeder protection:
Not more than 70A + 14A, = 84A,
next size down = 80A [240.6(A)]
Answer: 6 AWG conductors and an 80A breaker.

Author’s Comment:
• The “next size up protection” rule for branch circuits [430.52(C)(1) Ex 1] doesn’t apply to a motor feeder overcurrent protection device rating.

Comments
  • Mike, Could you show the numbers for question 5, step 2.?

    Thanks,

    Steve

    steve  January 16 2020, 2:44 pm EST
    Reply to this comment


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