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Quizzes on the Code

Mar 1, 2000 12:00 PM, By James Stallcup, Jr., NEC and OSHA Consultant

Motors -- Art. 430

Choose the best answer:

1. What minimum size inverse time circuit breaker is required to start and run a 40-hp, 460V, 3-phase, Design B motor?

(a) 100A
(b) 125A
(c) 150A
(d) 200A

2. What is the next larger size inverse time circuit breaker required to start and run a 40-hp, 460V, 3-phase, Design B motor?

(a) 100A
(b) 125A
(c) 150A
(d) 200A

3. What is the maximum size inverse time circuit breaker required to start and run a 40-hp, 460V, 3-phase, Design B motor?

(a) 100A
(b) 125A
(c) 150A
(d) 200A

4. What size THWN copper conductors are required for a 40-hp, 460V, 3-phase, Design B motor?

(a) No. 8
(b) No. 6
(c) No. 4
(d) No. 2

5. What is the locked-rotor current for a 40-hp, 230V, 3-phase, Design G motor?

(a) 580A
(b) 632A
(c) 734A
(d) 824A

6. What is the locked-rotor current for a 40-hp, 230V, 3-phase, Design B motor?

(a) 580A
(b) 632A
(c) 734A
(d) 824A

Answers and Discussion

1. (b). Based on the rules in Sec. 430-6(a)(1), the full-load current rating of the motor is first obtained from Table 430-150. The motor’s protective device is then sized based on the rules in Sec. 430-52(c)(1) and the values (percentages) in Table 430-152.

40 hp = 52A
52A x 250% = 130A
130A requires a 125A OCPD

Stallcup’s Code Loop: Sec. 430-6(a)(1), Table 430-150, Sec. 430-52(c)(1), Table 430-152, Sec.240-3(g), and Sec.240-6(a).

2. (c). The same rules in question 1 apply, but Exception No. 1 of Sec. 430-52(c)(1) allows the motor’s protective device to be increased to the next higher standard size.

40 hp = 52A
52A x 250% = 130A
130A requires 150A OCPD

Stallcup’s Code Loop: Sec. 430-6(a)(1), Table 430-150, Sec. 430-52(c)(1), Ex. 1, Table 430-152, Sec.240-3(g), and Sec.240-6(a).

3. (d). The same rules in question 1 apply, but Exception No. 2(c) of Sec. 430-52(c)(1) allows the motor’s protective device to be increased by 400%, as long as the rating or ting does not exceed the calculated value.

40 hp = 52A
52A x 400% = 208A
208A requires 200A OCPD

Stallcup’s Code Loop: Sec. 430-6(a)(1), Table 430-150, Sec. 430-52(c)(1), Ex. 2 (c), Table 430-152, Sec.240-3(g), and Sec.240-6(a).

4. (b). The same rules in question 1 apply, but Sec. 430-22(a) requires the branch-circuit conductors to have an ampacity of no less than 125%. The conductor size is then determined based on the information in Table 310-16.

40 hp = 52A
52A x 125% = 65A
65A requires No. 6

Stallcup’s Code Loop: Sec. 430-6(a)(1), Table 430-150, Sec.430-22(a), and Table 310-16.

5. (b). You can find Code letters on the nameplate of the motor. Locked-rotor current for a variety of Code letters is listed in Table 430-7(b) in kVA (kilo-volt amps) per horsepower

Code letter G = 6.29 kVA
A4kVA per hp x 1,000/V x 1.732
A = 6.29kVA x 40 hp x 1,000/230V x 1.732
LRC = 632A

6. (a). You can find the locked-rotor current of a motor in Tables 430-151(A) and (B). The locked-rotor current for single-phase and 3-phase motors are selected from these Tables based on the number of phases, voltage level, and horsepower rating of the motor. Motors will be marked either as Design B, C, D, or E.

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