Exam Question Help

1055 Views 6 Replies 4 Participants Last post by  sraistrick
Just wondering if someone out there could assist me with this question...

A three phase motor with a FLA of 14A , service factor of 1.15 and PF of .8 lag, is connected to a 460V, 60Hz supply. A capacitor bank of 2386 VAR is connected to the load side of the motor controller at the motor terminals. Calculate the maximum setting of the overload device in this controller.

I want to say 17.5A (14x1.25) but that is not the answer and I can't figure out how to include the capacitor bank in the formula as it should be decreasing the current draw and increasing the PF, any help would be great, thanks.
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What is the maximum allowable per article 430?, you have to read the question & understand what they are asking, not what you think they are asking.....
What is the maximum allowable per article 430?, you have to read the question & understand what they are asking, not what you think they are asking.....
Right, but this doesn't get me any farther ahead.
Duty rating?
Service factor

They don't alway make motors in the size you need, so they use a service factor which is how many % over the rated horsepower it is good for. Try 1.15 times the current first and see if that works
They don't alway make motors in the size you need, so they use a service factor which is how many % over the rated horsepower it is good for. Try 1.15 times the current first and see if that works
****, forgot to mention that I am working off of the CEC, not sure how much different our books are, I should have posted this in the CEC forum, sorry.
Hey maybe a bit late but maybe this is it

460v x 14a= 6440VA

make your power triangle with that as the hypotenuse then

6440 x .8pf =5152Watts put this on the bottom of the same triangle then

Pythagoras: 6440(2) -5152(2) = square root the answer = 3864 VARS.. add this to vertical or triangle

those are Vars of inductance so if you subtract the added vars of capacitance:

3864-2386=1478 VARS

if you do pythagoras again with the same VA and the new VARS then:

6440(2) - 1478(2) = sqr rt = 6268 Watts

6268/6440=.97pf

hope it helps a bit, I'm doing 4th year right now, IP in a week
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