Minimum service size conductor

[tdotboys]

Hi guys. Need some help with this question. i am using canadian electrical code book 2006 edition, TABLE 2.

According to me for 1200 a service , 4 conductors per phase in PVC Conduit.
1200 amps / number of conductors per phase
1200 amps / 4
300 amps per conductor

Now if i look in TABLE 2, RW90(XLPE) , i will pick 350 kcmil. but according to the question in attachment its something else. Help will be appreciated. Thanks guys.

[Ohm1]

In this case, it has a lot to do with the conductor being able to protect at its ampacity.

Looking at your list of answers, the only one that will fulfill this is 600kcmil.

Lets look at the one that is closest to being correct [250kcmil]. So: 265(4)=1060A, yet it can only be loaded to the 75 degree column. So: 255(4)= 1020A<---regardless 250kcmil want work. Per our code, if you are over 800 amps you cannot go to the next higher size, unless the conductor ampacity is equal to or higher than the breaker size of 1200 amps. As you can see--per our code--you cannot use 250Kcmil.

Sorry, if what I explained is unclear. I'm on the road.

[suemarkp]

So why do you say 600 KCMil? The 350 KCMil is rated just at 310A per conductor at 75C. So 4 in parallel should be good for over 1240 amps which is over the 1200A requirement. So 350 Copper should be the answer.

Is there some hidden derating in the question -- bundled conductors with a harmonic load, high ambient temperatures, ??? Did they ask for aluminum instead of copper?

Most of us are NEC people. Don't know if there is some special Canadian rule that applies that we're not familiar with.

[Ohm1]

So why do you say 600 KCMil? The 350 KCMil is rated just at 310A per conductor at 75C. So 4 in parallel should be good for over 1240 amps which is over the 1200A requirement. So 350 Copper should be the answer.

Is there some hidden derating in the question -- bundled conductors with a harmonic load, high ambient temperatures, ??? Did they ask for aluminum instead of copper?

Most of us are NEC people. Don't know if there is some special Canadian rule that applies that we're not familiar with.

I don't see an answer for 350KCMIL, Nor do I see an answer that says "none of these", so you choose the best answer in this question.
It's basically seeing if you know how to use code.
Yet, we aren't talking NEC. The CEC may not prohibit upsizing. But I think it does---based on the anwsers listed. 600Kcmil is the best answer for this question.

[suemarkp]

Didn't even see the first attachment with the question and answers....doh!

[Roger]

Whats that rule say in the test attachment? It appears it has something to do with doubling the conductor size in order to make 600 kcmil the minimum ampacity conductor. If not... the answer is ridiculous. And if this is about test writing and using the table to derive the answer from the list of answers this teaches you nothing in my opinion. So in the field I'll just say "well I had a test question once that told me to use 600 kcmil" using 4 conductors per phase so now you just say "ok 350 is my minimum so lets just double the size and we're home free...jeeez.

This is a prime example of why so many apprentices are confused at tests cause the test material is out in left field...focusing on whether or not they can trick you not if you know how to use your training. It really is just a game for them. Not so for the guy trying to make a living. I got tired of trying to "pass lets play games" tests....

Using a conductor as the correct answer that would never be correct in the real application is plain ignorant....or you had a test writer that wasn't sure himself/herself.

[Ohm1]

Whats that rule say in the test attachment? It appears it has something to do with doubling the conductor size in order to make 600 kcmil the minimum ampacity conductor. If not... the answer is ridiculous. And if this is about test writing and using the table to derive the answer from the list of answers this teaches you nothing in my opinion. So in the field I'll just say "well I had a test question once that told me to use 600 kcmil" using 4 conductors per phase so now you just say "ok 350 is my minimum so lets just double the size and we're home free...jeeez.

This is a prime example of why so many apprentices are confused at tests cause the test material is out in left field...focusing on whether or not they can trick you not if you know how to use your training. It really is just a game for them. Not so for the guy trying to make a living. I got tired of trying to "pass lets play games" tests....

Using a conductor as the correct answer that would never be correct in the real application is plain ignorant....or you had a test writer that wasn't sure himself/herself.I agree for the most part!
All you can do, is do the best you can do--appling the rules you know to be right, then use common sense where needed!!

[Roger]

May be but if you apply the rules and what you know you come up with what Mark did. Now you sit there going jeeez the correct answer is 350 kcmil but only one I have is 600 kcmil that works for enough ampacity... that can't be right... no way I'm that far off. And you miss the question even though you had it right.....sorry.. down right stupid question. If I met the pin head that wrote that and if I missed my masters by a tad cause of a bs question like that I wouldn't want to meet the test writer.... I'd ask him what hospital he wants the ambulance to take him to.....:evil:

[tdotboys]

I don't see an answer for 350KCMIL, Nor do I see an answer that says "none of these", so you choose the best answer in this question.
It's basically seeing if you know how to use code.
Yet, we aren't talking NEC. The CEC may not prohibit upsizing. But I think it does---based on the anwsers listed. 600Kcmil is the best answer for this question.


Thanks guys. i got it now. From my calculation my answere was coming up minimum 350 kcmil. As Ohm1 explained, we do not have 350 kcmil in answere and there is no way we can use 250 kcmil for this, so i shud pick a next higher ampacity cable. This question is just to confuse a guy like me :D Thanks Ohm1, i am trying to learn the tricks day by day :p to answere these bs questions.

[Ohm1]

Thanks guys. i got it now. From my calculation my answere was coming up minimum 350 kcmil. As Ohm1 explained, we do not have 350 kcmil in answere and there is no way we can use 250 kcmil for this, so i shud pick a next higher ampacity cable. This question is just to confuse a guy like me :D Thanks Ohm1, i am trying to learn the tricks day by day :p to answere these bs questions.You are welcome!!! Thank you for returning!!! :smile:

[Ohm1]

May be but if you apply the rules and what you know you come up with what Mark did. Now you sit there going jeeez the correct answer is 350 kcmil but only one I have is 600 kcmil that works for enough ampacity... that can't be right... no way I'm that far off. And you miss the question even though you had it right.....sorry.. down right stupid question. If I met the pin head that wrote that and if I missed my masters by a tad cause of a bs question like that I wouldn't want to meet the test writer.... I'd ask him what hospital he wants the ambulance to take him to.....:evil: LOL!!!!! :laugh: You are starting to sound like me...........LOL!!

[Wgoodrich]

Many electrical tests are written by local inspectors that are not professional test writers. Many inspectors are not really qualified to be the inspector let alone a test writer. However they write the test to the best of the knowledge not knowing what they don't know. To me it is very depressing and frustrating to find this type test with inaccurate questions.

Wg

[scuba_dave]

It's not just electrical tests
They do the same thing for computer certs

[transister1]

1200/4=300A
300/0.8=375A(see table 5c)
#500(see table 2)

[transister1]

1200/4=300A
300/0.8=375A(four conductors per phase(four hot wireses) in four seperate runs of PVC conduit, so, use T-5c)
#500AWG (see T-2)

is it correct?