Lesson Motors 2 Lesson 8
Determine the maximum overload protection by using separate overload protective devices to protect a 20-horsepower, 460-volt, 3-phase continuous-duty motor marked with a temperature rise of 50°C and a motor nameplate full-load current rating of 13 amperes.
430.32(A)(1), Separate Overload Device All other motors, Max. overload rating = 115%Max. overload = motor nameplate FLC rating × 115%= 13 × 1.15= 14.95 A The correct answer is: 14.95 A
Determine the maximum overload protection, using overload relays, for a 25-horsepower, 240-volt, 3-phase continuous-duty motor with a motor nameplate full-load current rating of 65 amperes, a temperature rise of 40°C, and a service factor of 1.15.
430.32(A)(1), Separate Overload Device40°C rise, 1.15 service factor Max. overload rating = 125% Max. overload = motor nameplate FLC rating × 125%= 65 × 1.25= 81.25 A The correct answer is: 81.25 A
Overload relays are used for the overload protection of a 100-horsepower, 440-volt, 3-phase, continuous-duty motor marked with a service factor of 1.15, a temperature rise of 40°C, and a motor nameplate full-load current rating of 132 amperes. Calculate the maximum overload protection.
430.32(A)(1), Separate Overload DeviceMax. overload rating = 125%Max. overload = motor nameplate FLC rating × 125%= 132 × 1.25= 165 A The correct answer is: 165 A
Overload relays are used for the overload protection of a 7.5-horsepower, 208-volt, single-phase, continuous-duty motor with a motor nameplate full-load current rating of 46 amperes, a service factor of 1.10, and a temperature rise of 50°C. Calculate the maximum overload protection.
430.32(A)(1), Separate overload device All other motors, Max. overload rating = 115% Max. overload = motor nameplate FLC rating × 115%= 46 × 1.15= 52.9 A The correct answer is: 52.9 A
Overload relays are used for the overload protection of a five-horsepower, 240-volt, single-phase motor with a 1.15 service factor and a motor nameplate full-load current rating of 31.5 amperes. Calculate the maximum overload protection.
430.32(A)(1), Separate overload device Max. overload rating = 125% Max. overload = motor nameplate FLC rating × 125% = 31.5 × 1.25 = 39.375 A The correct answer is: 39.375 A
Determine the maximum overload protection permitted for this 75-horsepower motor when starting current is not a problem.
430.32(A)(1)Max. overload rating = 125%Max. overload = motor nameplate FLC rating × 125%= 188 × 1.25= 235 A The correct answer is: 235 A
What is the maximum rating of a thermal protector, built integral with the motor, when used for the motor overload protection of a 20-horsepower, 460-volt, 3-phase, continuous-duty motor with a motor nameplate full-load current rating of 30 amperes?
430.32(A)(2), Thermal Protector, Use NEC table value for FLC Table 430.250, 3-phase, 20 hp, 460 V = 27 A FLC 430.32(A)(2), FLC greater than 20 A Max. thermal protector rating = 140% Max. thermal protector = FLC × 140%= 27 × 1.40= 37.8 A Note: Review Informational Note following 430.32(C). The correct answer is: 37.8 A
Determine the maximum overload protection when starting current is a problem by using separate overload protective devices to protect a 20-horsepower, 460-volt, 3-phase, continuous-duty motor marked with a temperature rise of 50°C and a motor nameplate full-load current rating of 13 amperes.
430.32(C), Starting current a problem 50°C temperature rise, Max. overload rating = 130% Max. overload = motor nameplate FLC rating × 130%= 13 × 1.30= 16.9 A The correct answer is: 16.9 A
What is the maximum size time-delay fuse permitted to be used for the overload protection for a 5-horsepower, 240-volt, single-phase motor with a motor nameplate full-load current rating of 30 amperes, a service factor of 1.15, and a temperature rise of 40°C, when starting current is a problem? Fuse sizes must be selected from the NEC standard sizes.
430.32(C), Starting current a problem Max. overload rating = 140% Max. overload = motor nameplate FLC rating × 140% = 30 × 1.40 = 42 A Next smaller size fuse, use 240.6(A) standard sizes42 A → 40 A fuse selection The correct answer is: 40 A
Overload relays are used for the overload protection of a five-horsepower, 240-volt, single-phase motor with a 1.15 service factor and a motor nameplate full-load current rating of 31.5 amperes. Calculate the maximum overload protection when starting current is a problem.
430.32(C), Starting current a problem Max. overload rating = 140% Max. overload = motor nameplate FLC rating × 140% = 31.5 × 1.40 = 44.1 A The correct answer is: 44.1 A
Overload relays are used for the overload protection of a 100-horsepower, 440-volt, 3-phase, continuous-duty motor marked with a service factor of 1.15, a temperature rise of 40°C, and a motor nameplate full-load current rating of 132 amperes. Calculate the maximum overload protection when starting current is a problem.
430.32(C), Starting current a problem Max. overload rating = 140% Max. overload = motor nameplate FLC rating × 140%= 132 × 1.40= 184.8 A The correct answer is: 184.8 A
Overload relays are used for the overload protection of a 7 1/2-horsepower, 208-volt, single-phase, continuous-duty motor with a motor nameplate full-load current rating of 46 amperes, a service factor of 1.10, and a temperature rise of 50°C. Calculate the maximum overload protection when starting current is a problem.
430.32(C), Starting current is a problem All other motors, Max. overload rating = 130%Max. overload = motor nameplate FLC rating × 130% = 46 × 1.30 = 59.8 A The correct answer is: 59.8 A
A 40-horsepower, 3-phase, 208-volt, continuous-duty squirrel cage motor is installed using an inverse-time circuit breaker for branch-circuit short-circuit and ground-fault protection. The motor nameplate full load current rating is 120 amperes and the overload protection is provided by overload relays. Starting current is a problem for this motor. Calculate the maximum size overload protection.
430.32(C), Starting current is a problem Max. overload rating = 130% Max. overload = motor nameplate FLC rating × 130% = 120 × 1.30 = 156 A The correct answer is: 156 A
Determine the maximum overload protection, using overload relays, when starting current is a problem, for a 50-horsepower, 208-volt, continuous-duty motor with a motor nameplate full-load current rating of 148 amperes and a service factor of 1.15.
430.32(C), Starting current is a problem Service factor = 1.15 Max. overload rating = 140% Max. overload = motor nameplate FLC rating × 140%= 148 × 1.40= 207.20 A The correct answer is: 207.2 A
Determine the maximum rating of motor branch-circuit short-circuit and ground-fault protective device using time-delay fuses when starting current is not a problem for this 75-horsepower motor.
430.52(C)(1) Exception No.1Max time-delay fuse = 175%Max. rating = FLC × 175%= 192 × 1.75= 336 ANext higher OCPD permitted240.6(A) standard sizes336 A → 350 A time-delay fuses The correct answer is: 350 A
What is the maximum overload protection rating of a separate overload device for a 3/4-horsepower, 120-volt, single-phase motor, not automatically started, not permanently installed, not within sight from the controller location, with a motor nameplate full-load current rating of 14.5 amperes and no other nameplate markings? Starting current for this motor is not a problem.
Solution:430.32(D)(2)(b) refers to 430.32(B)(1), and then to 430.32(A)(1) Max. overload rating = 115% Max. overload = motor nameplate FLC rating × 115% = 14.5 × 1.15 = 16.675 A Note: The Code requirement path to arrive at the 115% value involves the following key stipulations:(1) less than 1 hp; 2) non-automatically started; (3) not permanently installed; (4) not within sight from controller; (5) separate overload devices; (6) no service factor on nameplate; and (7) no temperature rise on nameplate. The correct answer is: 16.675 A
Determine the appropriate minimum THWN copper wire size for this 75-horsepower motor based on the ampacity of the branch-circuit conductors.
Table 310.16240 A = 250 kcmil The correct answer is: 250 kcmil
A 40-horsepower, 3-phase, 208-volt, continuous-duty squirrel-cage motor is installed using an inverse-time circuit breaker for branch-circuit short-circuit and ground-fault protection. The motor nameplate full load current rating is 120 amperes, and the overload protection is provided by overload relays. Starting current is a problem. Using an inverse-time circuit breaker, calculate the maximum branch-circuit short-circuit and ground-fault protection size permitted for this application.
Table 430.250, 3-phase40 hp, 208 V = 114 A FLC 430.52(C)(1), Exception No. 2(c) Max. inverse-time circuit breaker = 300% Max. rating = FLC × 300%= 114 × 3.00 = 342 A Next lower standard OCPD size, 240.6(A) standard sizes 342 A→ 300 A The correct answer is: 300 A
A 3-phase, 75-horsepower, 240-volt, continuous-duty motor is installed with a motor nameplate full-load current rating of 188 amperes, a temperature rise of 40°C, and a marked service factor of 1.15. Starting current is not an issue. Determine the ampacity of the branch-circuit conductors for this 75-horsepower motor.
Table 430.250, 3-phase75 hp, 240 V = 192 A FLC430.22(A)Branch circuit ampacity = FLC × 125%= 192 × 1.25= 240 A The correct answer is: 240 A
Thermal protective devices for overload protection shall not be used with fractional horsepower motors.
The correct answer is: False The correct answer is: 430.32(B)(2)
A 1/2-horsepower, 120-volt, continuous-duty motor is not permanently installed, is not automatically started, is within sight of the controller, and utilizes a 20-ampere branch-circuit short-circuit and ground-fault protective device as the overload protection. According to the NEC, this is an acceptable installation.
The correct answer is: True The correct answer is: 430.32(D)(2)(a) Exception