Looking for someone to help me with a code question and a power question as I prepare to write my IP!

Lucas Organ

Member
Location
NL CANADA
Occupation
Industrial Electrician
Hey, Im preparing to write my IP in about a months time. I am in desperate need of someone who is very strong with code and theory. I simply have 2 questions in particular that are troubling me. Please if anyone out there is strong in with the code and can seek me help with with it would be extremely appreciated.
looking forward to responses.
One stressed out apprentice
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Welcome to the forum.

Post your questions here. You'll get much more response than waiting for someone to message you.
 

Lucas Organ

Member
Location
NL CANADA
Occupation
Industrial Electrician
This question pertains to the 2018 CEC.
I have A 400A splitter that feeds a 100A panel board temp specified 60/75 degrees C, that is 1.5M away from the splitter what size feeder cable is to be used?
To begin, I understand that code 4-006(2) tells me that for 100A and below I would use the 60 degree column out of table 2. Which is straight forward and I get that, however would sub rule (4) change this? As it states that this rule only applies to the first 1.2M. And on another note, I have always been told by instructors that when terminating on a breaker to use the 75 degree Column, I struggle to find a code for this though, however if this is proven true then it would again change my answer.

A Second question I have does not pertain to the CEC.
If I have a 3 phase system with PTs and CTs, and my PT is 600:20 and my CT is 250:5, the PT reads 112.5V and the PT reads 3A along with a power factor of 90%, what is my total power in Kw?


Thanks you all so much in advance who take the time to try and help me in my time of need.

Looking forward to response
 

Eddy_Current

Member
Location
Canada
For the first question concerning 4-006

You need to use the termination temperature on the equipment. Your instructor told you to always use 75 because 99% of equipment is rated for 75 degrees. There is no code that tells you that, it’s just trade practice.
Sub rule 4 does state that it only applies to the first 1.2m but that can only be used if there is a transition to another conductor. For instance if you had a 1.2 m long conductor connected to a 75 degree rated breaker, then in a junction box you spliced to another conductor, 4-006 would only apply to the 1.2 m of conductor connected to the equipment. It does not apply for the question you posted.

You need to look at 14-100 (tap conductor code) as that is the code they are asking you about in the question.
 

Lucas Organ

Member
Location
NL CANADA
Occupation
Industrial Electrician
Thank you so much. The most clarification I’ve had all year. So, would you go with 60 degrees or 75. The question did state that the is specified for 60/75 degrees. Thanks
 

Eddy_Current

Member
Location
Canada
Thank you so much. The most clarification I’ve had all year. So, would you go with 60 degrees or 75. The question did state that the is specified for 60/75 degrees. Thanks
Where are you getting this question from? I assume it is from an instructor that does not understand 4-006 entirely.

Ontario has a bulletin that says for a slash rated breaker we can use the higher one, but there is nothing about that in the CEC, that is Ontario only.

Also, you can not use 4-006(2) because that is for equipment that is not marked with a termination temperature.

You will not get a question on the Red Seal exam that specifies a slash rated breaker. It will be labelled with a termination temp and given to you in the question, or it won’t have one so you use sub rule 2
 

Lucas Organ

Member
Location
NL CANADA
Occupation
Industrial Electrician
Yeah, your exactly right. I think it’s basically questions he’s came up with. Me personally I would’ve went with the 75 degrees. But he rambles on about 4-006.
I cannot thank you enough for helping clear this up. Hats off to you
 

Rock86

Senior Member
Location
new york
Occupation
Electrical Engineer / Electrician
A Second question I have does not pertain to the CEC.
If I have a 3 phase system with PTs and CTs, and my PT is 600:20 and my CT is 250:5, the PT reads 112.5V and the PT reads 3A along with a power factor of 90%, what is my total power in Kw?


Looking forward to response

P(kW) = (VL * I * 1.732 * PF) / 1000
 

Rock86

Senior Member
Location
new york
Occupation
Electrical Engineer / Electrician
Thanks a million for the formula. Still having trouble getting the answer. I am reaching a answer of 131kw
131kW!!?? Maybe to help you understand your answer... you have 112.5V, and 3A.... you're not going to be near 1000W (1kW). VL=112.5V, I = 3A, PF = .9 (90%). Just at a quick glance, you could say 3A* 100V = 300A, so you know you should see an answer somewhere around there before included the 1.723 and PF.
 

Lucas Organ

Member
Location
NL CANADA
Occupation
Industrial Electrician
Okay sorry for the confusion. This is really not my strong point. So I have
112.5x3x1.732x90%= 526.095/1000
0.53
does this look right
 

Rock86

Senior Member
Location
new york
Occupation
Electrical Engineer / Electrician
Actually... if we are taking the PT & CT readings to be the adjusted output numbers, than your calculation is correct. I was just thinking, if this is an exam question and they are expecting you calculate the primary voltage and current, than it may require more work.

Don't be sorry for being confused. I didn't mean to come off harsh. I'd suggest looking at the question and just doing the simple math first to determine you approximate range. If you know P = V*I at the basic level, than you can extend the calculation for 3-phase to include the PF. Learning is a journey.
 

Rock86

Senior Member
Location
new york
Occupation
Electrical Engineer / Electrician
First I worked my V by 112.5/120x600 and my I at 3/5x250.
Or should I leave my values at 112.5 and 3a
PT is 600:20 and the that makes the primary side 30 times larger than the secondary. CT is 250:5, so primary is 50 times larger.

If our secondary is 112.5V and 3A, than our primaries should be 3,375 V, and 150A. Therefore we end up with (3375 * 150 * 1.723 * .9) / 1000 = 78.5kW
 

Rock86

Senior Member
Location
new york
Occupation
Electrical Engineer / Electrician
If you have an instructor, I would run that by them to clarify if the information given is the secondary side or the calculated primary on the meter.
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Retired Electrical Contractor
131kW!!?? Maybe to help you understand your answer... you have 112.5V, and 3A.... you're not going to be near 1000W (1kW). VL=112.5V, I = 3A, PF = .9 (90%). Just at a quick glance, you could say 3A* 100V = 300A, so you know you should see an answer somewhere around there before included the 1.723 and PF.


I assume you meant 300 watts not amps
 

synchro

Senior Member
Location
Chicago, IL
Occupation
EE
PT is 600:20 and the that makes the primary side 30 times larger than the secondary. CT is 250:5, so primary is 50 times larger.

If our secondary is 112.5V and 3A, than our primaries should be 3,375 V, and 150A. Therefore we end up with (3375 * 150 * 1.723 * .9) / 1000 = 78.5kW
That would be if the PT's were connected in delta or open delta to measure the L-L voltages. If the PT's were connected as wye-wye, then the power delivered would be 1.732 times higher.
 

Lucas Organ

Member
Location
NL CANADA
Occupation
Industrial Electrician
I Am wondering why you said the PT is 600:20. Is that on purpose or did you mean to do 600:120
Thank you again so much, i cannot believe the head way I am making today.
 
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