Something I ran into...

Merry Christmas
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iwire said:
Yes, all over the place, the NEC has been changing the wording of the section to try to clear up the confusion.

Only the conductors that supply 100% of the dwelling units power can be sized using table 310.15(B)(6).
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It's gonna be odd buying 90 amp breakers.
 
chris kennedy said:
1000' of #8 has .8 ohms resistance
1000' of #6 has .5 ohms resistance

.8-.5=.3
.3?.8=.375 or 37.5% difference in resistance.

That sound right?
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Not really sure??? :-? :roll:
But still less than an ohm at 1000' - at 100' add a zero after the decimal? .03 difference?
 
e57 said:
Not really sure??? :-? :roll:
But still less than an ohm at 1000' - at 100' add a zero after the decimal? .03 difference?
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I don't think that it is the actual difference in resistance, it is the ratio between the 2.
 
e57 said:
But still less than an ohm at 1000' - at 100' add a zero after the decimal? .03 difference?
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Wrong way. If theres .3 ohm difference at 1000', then theres 3 ohm difference at 100'

120?.8=150

120?.5=240

240-150=90

90?240=.375 or 37.5 percent difference in Amps.

So ohms law backs up my first attempt at the math.

Therefore if A? has a load of 80A the #6 would carry 50A and the #8 would carry 30A.

Now does that sound right?(I'm just guessing here)
 
If the difference in impedances quoted by Chris is right, and assuming that the only difference is impedance then, the current flow will be in the ratio of 0.5:0.8, so at 60A the heavier wire will carry about 37A and the lighter about 23A.

Edit: Bugger! Beaten to it.

The way I did it: The ratio is 0.5/0.8, and ratios stay constant so make it easy, shift the decimal point, and say 5/8.

Add the ratio parts together, so 5 + 8 = 13.

Current A = total current * 5 / 13

Current B = total current * 8 / 13
 
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The result of 37.5 percent is close, but not because the problem was solved correctly. Looking at the percentage difference in resistance will not get you the right answer. This is a simple current divider. So dbuckley?s approach was correct. So was the answer of 37 amps (about 38.5 percent) and 23 amps (11.5%).
 
charlie b said:
The result of 37.5 percent is close, but not because the problem was solved correctly. Looking at the percentage difference in resistance will not get you the right answer. This is a simple current divider. So dbuckley?s approach was correct. So was the answer of 37 amps (about 38.5 percent) and 23 amps (11.5%).
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I don't see how you guy's got this. 38.5% and 11.5% is 50%. Wheres the other 50?
 
chris kennedy said:
I don't see how you guy's got this. 38.5% and 11.5% is 50%. Wheres the other 50?
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I don't know where those percentages came from, just use the ratios concept. For two resistances in parallel:
I1 = Itotal * R2/(R1 + R2)
I2 = Itotal * R1/(R1 + R2)

The fractions with the R's are the same ratio concept that dbuckley pointed out. Note that the subscript of the I and the top R are different. This makes intuitive sense, but it's the opposite of the voltage divider rule. Having the subscripts different means that the path with lower resistance has higher current, which makes sense.
 
charlie b said:
. . . the answer of 37 amps (about 38.5 percent) and 23 amps (11.5%).
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chris kennedy said:
I don't see how you guy's got this. 38.5% and 11.5% is 50%. Wheres the other 50?
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My bad. Tried to do mental math too quickly.

The answer is 37 amps (61.5%) and 23 amps (38.5%)
 
charlie b said:
My bad. Tried to do mental math too quickly.
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On Sunday Charlie? Take a break man! Clean the garage or clean the gutters.

Or shovel snow?

I still don't see why Ohm's Law won't solve this.
 
Talk to the Home owner

Talk to the Home owner

jason said:
Someone fed a 100 amp subpanel with a 6/3 and an 8/3 copper paralleled. These were protected (or not) by a 100 amp breaker. This is not part of the job I'm there to do, but it may be serious enough to fix regardless. I haven't mentioned it to the homeowner yet. This is a residence. However, it got me wondering. Lets say the load was 60 amps. How would the current be distributed on each wire? 30/30 40/20 60/0 ??? I have no idea and I know it's not right, just wondering.
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The fact is; it is a clear violation of the code to run any conductors in parallel if they are smaller than 1/0 (310.4) There are a few exceptions to 310.4, but none of them apply to this application. And, yes this is particularly dangerous due to the fact that the conductors are of different size with different resistances. A greater percentage of the total current will travel on the larger conductor. I would definitely talk to the home owner about this. If it were my home I would want to know. If they don't want to pay to have it fixed; have them sign off saying that they have been notified of the blatant code violation and opted not to fix it. This will clear you of any liability.
 
Shawny said:
The fact is; it is a clear violation of the code to run any conductors in parallel if they are smaller than 1/0 (310.4)
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I disagree - the way the code is worded, the (intended) prohibition from using conductors smaller than 1/0 is missing when you apply 90.5 to it. :)


Welcome to the forum. :D
 
jason said:
Someone fed a 100 amp subpanel with a 6/3 and an 8/3 copper paralleled. These were protected (or not) by a 100 amp breaker. This is not part of the job I'm there to do, but it may be serious enough to fix regardless. I haven't mentioned it to the homeowner yet. This is a residence. However, it got me wondering. Lets say the load was 60 amps. How would the current be distributed on each wire? 30/30 40/20 60/0 ??? I have no idea and I know it's not right, just wondering.
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As it was mentioned before it is a code violation, the parallel conductors must be sized identical besides of other issues. The current will split in inverse relationship to the conductors resistance, ignoring inductance. Roughly 60/40%.
 
georgestolz said:
I disagree - the way the code is worded, the (intended) prohibition from using conductors smaller than 1/0 is missing when you apply 90.5 to it. :)
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You lost me there. How do you figure 90.5 can be used to allow parallelling small conductors? This is a "shall be permitted" on the topic of parallelling larger conductors. The notion of "shall not be permitted" for parallelling smaller conductors just naturally follows, IMHO.
georgestolz said:
Welcome to the forum. :D
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OK. That part I agree with. :grin:
 
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