CHAPTER 10 Motor Installation What is the minimum conductor ampacity, overload p
ID: 2267249 • Letter: C
Question
CHAPTER 10 Motor Installation What is the minimum conductor ampacity, overload protection, and dual-element fuse rating for a circuit supplying a 560-volt, 3-phase, code J, 40°C, 125-horsepower motor with a nameplate full-load current of 115 amperes? What is the minimum conductor size, overload protection, and inverse-time circuit- breaker rating for a circuit supplying a 230-volt, 3-phase, code A, 40°C, 75-horse 8. ower synchronous motor operating at an 80% power factor with a nameplate full load current of 185 amperes? 9. What size conductor would be required to supply three motors connected to a 440- volt, 3-phase branch circuit? Motor #1 is a 50-horsepower, code B induction motor, motor #2 is a 40-horsepower, code H induction motor, and motor #3 is a 50-horse- power, code J induction motor. circuit serving the three motors in the previous question? What would be the rating of a dual-element fuse? 10. What would be the rating of an inverse-time circuit breaker required for the branch Inverse-time circuit breaker: Dual-element fuse:Explanation / Answer
Answer for Qustion no. 7 is
So minimum Ampacity of conductor is 125% of Full Load Current = 115*1.25=143.75A
so 150A fuse rating is sufficient.
Over Load Protection is 115*1.25=143.75A so next fuse rating available is 150A .
Dual Element Fuse Rating is 125% to 130% of Full Load Current so 115*1.25=143.75A
Or 115*1.3=149.5A so 150A fuse rating is sufficient.
Answer for Qustion no. 8 is
So minimum Ampacity of conductor is 125% of Full Load Current = 185*1.25=231.25A so 225A fuse rating is sufficient.
Over Load Protection is 185*1.25=231.25A so 225A fuse rating is sufficient.
Inverse Time Circuit Breaker Rating is 2.5 times the Rated Full Load Current i.e
185*2.50=462.5A but next available Inverse Time Circuit Breaker Rating is 500A
Answer for Qustion no. 9 is
Circuit conductors that supply several motors must not be smaller than the minimum ampacity found by adding
· 125% of the Full Load Current of the highest rated motor.
· The Full Load Currents of other motors
Full load current of Highest Rated motor is (50*746)/(1.732*440)=48.94A
So 125% of Full load current of Highest Rated motor is 1.25*48.94=61.17A
Full load Current of other motors are 39.15A,48.94A respectively so when we add all these currents we get 61.17+39.15+48.94=149.26A
so 150A fuse rating is sufficient.
Answer for Qustion no. 10 is
For the above Question Inverse Time Circuit Breaker Rating is 2.5 times the Rated Full Load Current i.e
149.26*2.50=373.15A but next Available Inverse Time Circuit Breaker Rating is 400A
Dual Element Fuse Rating is 125% to 130% of Full Load Current so 149.26*1.25=186.57A
or 149.26*1.3=194.03 so 200A fuse rating is sufficient.