Stumped on exam practice question..need help!

[johnnyelectric369]

Question is as written..What's the required feeder inverse time circuit breaker required for one 25 hp,460v,three phase squirrel cage motor with a name plate full- load current rating of 32a and a design B service factor of 1.15;two 30 hp,460v,three- phase wound rotor motors with a name plate primary full-load current rating of 38a and a name plate secondary full load current rating of 65a with a 40 degree Celsius rise?
(A)150A
(B)175A
(C)200A
(D)225A
I see the answer is annex D8 example but unless this question is worded wrong cannot figure out how they arrived at this conclusion.Thanks for the help!

 

[Dennis Alwon]

What specifically don't you understand?

This is for feeders

Feeder Short-Circuit and Ground-Fault Protection

(a) Example using nontime-delay fuse. The rating of the feeder protective device is based on the sum of the largest branch-circuit protective device for the specific type of device protecting the feeder. In the previous step above, the calculation for the 25 hp squirrel-cage motor results in the largest branch-circuit protective device: 300% × 34 A = 102 A (therefore the next largest standard size, 110 A, would be used) plus the sum of the full-load currents of the other motors, or 110 A + 40 A + 40 A = 190 A. The nearest standard fuse that does not exceed this value is 175 A [see 240.6, Table 430.52, and 430.62(A)].
(b) Example using inverse time circuit breaker. The largest branch-circuit protective device for the specific type of device protecting the feeder. The calculation for the 25-hp squirrel-cage motor results in the largest branch-circuit protective device, 250% × 34 A = 85. The next larger standard size is 90 A, plus the sum of the full-load currents of the other motors, or 90 A + 40 A + 40 A = 170 A. The nearest standard inverse time circuit breaker that does not exceed this value is 150 A [see 240.6, Table 430.52, and 430.62(A)].

 

[johnnyelectric369]

What specifically don't you understand?

This is for feeders

Dang...you're good and thank you for explaining.now I understand more clear.i guess I got confused when I went to the the annex D example D8.they figured it out by multiplying 34a by 125% and then added the sum.i multiplied by 250% and then added the sum and my answer came out higher.that will teach me to rush.thank you very much for the response!

 

[Tulsa Electrician]

Did some reading this morning on this. Found it interesting.