confused

[tryin]

what exactly is it that they are asking me, this is a test question that is killing me.

A 120/208V WYE connection for a 800 AMP inductive lighting load with a 200 amp 208V 3phase hermetic refrigerant compressor

the grounded conductor ampacity is?
620
760
820
1000

Now, what I have been doing is GUESSING that if the code requires the minimum amps to be no less than the largest size service conductor, then it would stand to reason that 800 is the answer.But as you can see, thats not an option. Please point me in the right direction, I honestlt do not know what they are asking me.

 

[crossman]

IMHO, there is not enough information given to reach an exact conclusion.

 

[tryin]

I think I'm confused ocver what they are asking for. I have to know, for my own benefit,what the hell the answer is. Do you combine them for a total of 1000 or is it more complex because of the compressor motor

 

[charlie b]

The answer is 620 amps.

First, disregard the 3-phase load, since it will not have any unbalanced current flowing in its (non-existent) ungrounded wire. Next, take 100% of the first 200 amps from the 800 amp lighting load, and add in 70% of the remaining 600 amps. 200 + 0.7*600 = 620.

Reference: 220.61(B).

 

[brian john]

And that is why you have a P.E. after you name right off the cuff.

 

[tryin]

Thanks, Charlie B

Thanks, Charlie B

THANK YOU!!!!!!!!!! THANK YOU!!!!!!!!!!!!
The three phase load was confusing me big time.

 

[charlie b]

You are welcome. And Brian, you are too kind. :smile:

 

[crossman]

The answer is 620 amps.

First, disregard the 3-phase load, since it will not have any unbalanced current flowing in its (non-existent) ungrounded wire. Next, take 100% of the first 200 amps from the 800 amp lighting load, and add in 70% of the remaining 600 amps. 200 + 0.7*600 = 620.

Reference: 220.61(B).

What if all the lighting load is 208 volts?

 

[glene77is]

Please elaborate on how you derive these numbers from 260.61(B).

I understand that the 3 phase motor will not have return neutral currents.

Given the Inductive Lighting requires 800A.
I do not understand why you use 100% of the first 200A,
then add 70% (which I think is a demand factor) on the remaining 600A.

 

[crossman]

Darn it, Charlie, where did you run off to?

What would the answer be if all the lighting was 208 volt luminaires?

 

[wasasparky]

Darn it, Charlie, where did you run off to?

What would the answer be if all the lighting was 208 volt luminaires?

All the lighting is 208V.
22.61(B)2 70% for " all unbalanced load in excess of 200A..."

 

[crossman]

All the lighting is 208V.
22.61(B)2 70% for " all unbalanced load in excess of 200A..."

If all the luminaires are 208 volts, there is no unbalanced load.

 

[glene77is]

Charle B,
I found out whats happening, by re-reading the NEC section.
I was looking from a Physics perspective, not a NEC perspective.
Thanks anyway. Makes me think.

 

[charlie b]

Darn it, Charlie, where did you run off to?

Shucks, can't a guy get in a half hour or so of work for the company without anyone complainin' about being absent from the forum? :-?

What would the answer be if all the lighting was 208 volt luminaires?

I see that someone stepped in during my absence with the simple and correct answer: you don't even pull a neutral wire in that case, so you need not calculate a load, and need not verify that its ampacity exceeds the load. :wink:

 

[LarryFine]

Shucks, can't a guy get in a half hour or so of work for the company without anyone complainin' about being absent from the forum? :-?

Absolutely not! Where are your priorities, man? The forum comes first!

 

[crossman]

I see that someone stepped in during my absence with the simple and correct answer: you don't even pull a neutral wire in that case, so you need not calculate a load, and need not verify that its ampacity exceeds the load. :wink:

The OP's commentary after the question indicates this is a service. If so, we would have to pull a grounded conductor form the neutral point of the utility to the main disconnect, even though there is no neutral load at all. It may not need to have an ampacity of 620 amps.

 

[charlie b]

Absolutely not! Where are your priorities, man? The forum comes first!

I might consider saying something unkind in reply to this statement, but I would not want to surpass you as the forum's most censored member. :grin:

 

[LarryFine]

I might consider saying something unkind in reply to this statement, but I would not want to surpass you as the forum's most censored member. :grin:

Oh, don't let that stop you. Give it your best shot. Besides, I think I earned my nickname on quantity, not quality. :grin:



It makes me wonder, though: Who moderates the moderators? We've always heard that the inmates run the asylum.