Min size conductor and 250.122(B)

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hhsting

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Glen bunie, md, us
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I have full load amps of 40A for continuous load. However plans show going to have four current carrying conductors including the one fir the continuous load.

So 40x1.25 = 50A. Since I am going to have four current carrying conductors 50/.8 = 62.5A. So the minimum size conductor would be #6.

However I am provided with #3 awg due to voltage drop on 50A breaker. So would #6 gnd be big?

So in order to figure above I would proptional (#3 awg area/#6 awg area) or (#3 awg area/#8 awg)? #8 awg is for 50A mini conductor
 

petersonra

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Northern illinois
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engineer
You have a 50 A breaker so the minimum conductor size would be #8, with a #10 EGC.

You increased it to #3.

#3 is 52633 circ mils
#8 is 16509 circ mils

52633/16509 = 3.188

#10 is 10383 CM so it needs to be increased to at least 33102 CM. That would be #4.
 

Dennis Alwon

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That's right figure the cm for 6 and 3 get that ration and do that for the equipment grounding conductor.

Edit-- Bob got it for you ....
 

hhsting

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Ok lets forget about 50A breaker post#1 for a minute. My intended cable size = minimum size I need per code.

I have 208V single phase load draws 40A and is continuous. The cables are going to be in one conduit with 120V single phase 20A load.
So I am going to have more than 3 current carrying conductors in the conduit. Precisely 4 CC.


What is the minimum size cable I need? To solve that I would do following:

40*1.25 = 50A

Now I do 50/.8=62.5A

So would I be correct to have #6 as my minimum size in the above example or not?
 
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Dsg319

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West Virginia
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Wv Master “lectrician”
Ok lets forget about 50A breaker post#1 for a minute. My intended cable size = minimum size I need per code.

I have 208V single phase load draws 40A and is continuous. The cables are going to be in one conduit with 120V single phase 20A load.
So I am going to have more than 3 current carrying conductors in the conduit. Precisely 4 CC.


What is the minimum size cable I need? To solve that I would do following:

40*1.25 = 50A

Now I do 50/.8=62.5A

So would I be correct to have #6 as my minimum size in the above example or not?
Never checked calculations but if they are correct as long as your equipment and conductors are suitable for 75degree terminations than yes.
 

petersonra

Senior Member
Location
Northern illinois
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engineer
Ok lets forget about 50A breaker post#1 for a minute. My intended cable size = minimum size I need per code.

I have 208V single phase load draws 40A and is continuous. The cables are going to be in one conduit with 120V single phase 20A load.
So I am going to have more than 3 current carrying conductors in the conduit. Precisely 4 CC.


What is the minimum size cable I need? To solve that I would do following:

40*1.25 = 50A

Now I do 50/.8=62.5A

So would I be correct to have #6 as my minimum size in the above example or not?
The code does not care why you increased the size of the conductors. It can be because of voltage drop or number of current carrying conductors or just because you had a certain size wire in your truck and decided to use it, or some combination thereof.
 

hhsting

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Glen bunie, md, us
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Junior plan reviewer
Alright then 250.122(B) says minimum size that has sufficient ampacity for intended installation which in this case is #6 awg Not #8 awg

So then now if cables are increased to #3 awg due to voltage drop then I would do to determine the equipment grounding 250.122(B):

((#3 awg area)/(#6 awg area))*(10awg area)

Would I be Correct or incorrect above?
 

hhsting

Senior Member
Location
Glen bunie, md, us
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Junior plan reviewer
The code does not care why you increased the size of the conductors. It can be because of voltage drop or number of current carrying conductors or just because you had a certain size wire in your truck and decided to use it, or some combination thereof.

Cable size increase due to Number of current carrying conductor rating is required by code. Code does care about that.

However other items voltage drop or whats in your truck code does not care about. Someone can chime in
 

wwhitney

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Berkeley, CA
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You have a 50 A breaker so the minimum conductor size would be #8, with a #10 EGC.
But since there are 4 CCCs, #8 would be too small for the installation, so the minimum size for the ungrounded is #6, not #8. That means #6 is the baseline for 250.122(B).

Also, there's no reason to look at circular mils. AWG is a geometric series of areas, so the area ratio of any two AWG sizes that differ by, say, 3 is always the same. Any slight deviation from that using circular mils from a table is due to rounding errors.

Bottom line, for #3 ungrounded with 4 CCCS on a 50A breaker and no other adjustment/correction, the minimum size EGC is #7.

Cheers, Wayne
 

hhsting

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Glen bunie, md, us
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Junior plan reviewer
Also, there's no reason to look at circular mils. AWG is a geometric series of areas, so the area ratio of any two AWG sizes that differ by, say, 3 is always the same. Any slight deviation from that using circular mils from a table is due to rounding errors.



Cheers, Wayne

Wayne I don’t get it what you mean by above? Can you elaborate or give example?
 

wwhitney

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Geometric series is, e.g., 1, 2, 4, 8, 16, 32. The ratio of consecutive numbers is always the same (2 in this case).

Conductor cross-sectional area by AWG is a geometric series. So area #4 / area #5 = area #5 / area #6, or any two consecutive sizes. Likewise for any two pairs of sizes that are each 2 apart. Or each 3 apart. Etc.

The upshot for 250.122(B) is that if the ungrounded is bumped up 3 sizes (#6 to #3), the EGC needs to be bumped up 3 sizes (#10 to #7). As #7 is not common, #6 was called out on your plans.

Cheers, Wayne
 

hhsting

Senior Member
Location
Glen bunie, md, us
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Junior plan reviewer
Geometric series is, e.g., 1, 2, 4, 8, 16, 32. The ratio of consecutive numbers is always the same (2 in this case).

Conductor cross-sectional area by AWG is a geometric series. So area #4 / area #5 = area #5 / area #6, or any two consecutive sizes. Likewise for any two pairs of sizes that are each 2 apart. Or each 3 apart. Etc.

The upshot for 250.122(B) is that if the ungrounded is bumped up 3 sizes (#6 to #3), the EGC needs to be bumped up 3 sizes (#10 to #7). As #7 is not common, #6 was called out on your plans.

Cheers, Wayne

interesting but this only works with awg not kcmil or combination of both?
 

wwhitney

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Berkeley, CA
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Retired
interesting but this only works with awg not kcmil?
Yes. Once you get to sizes identified by kcmil, you have to take the ratio of kcmil, rather than the difference in AWG. But it works for 1/0, 2/0, 3/0, 4/0 if you treat them as 0, -1, -2, and -3, respectively.

Cheers, Wayne
 

hhsting

Senior Member
Location
Glen bunie, md, us
Occupation
Junior plan reviewer
Yes. Once you get to sizes identified by kcmil, you have to take the ratio of kcmil, rather than the difference in AWG. But it works for 1/0, 2/0, 3/0, 4/0 if you treat them as 0, -1, -2, and -3, respectively.

Cheers, Wayne

Ok. So if its continous load of 40A and 4 CC Does one need to do which if the following to size cable

40/0.8

Or

(40x1.25)/0.8

Which one?
 

don_resqcapt19

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Illinois
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retired electrician
The code does not care why you increased the size of the conductors. It can be because of voltage drop or number of current carrying conductors or just because you had a certain size wire in your truck and decided to use it, or some combination thereof.
The required ampacity adjustments and/or corrections do not result in an in crease of the conductor size for the purposes of 250.122(B). Those adjustments and corrections change the ampacity of the conductor and change the minimum permitted size of the conductor.
The language change in 250.122(B) makes this very clear in the 2020 code.
B) Increased in Size.
If ungrounded conductors are increased in size for any reason other than as required in 310.15(B) or 310.15(C), wire-type equipment grounding conductors, if installed, shall be increased in size proportionately to the increase in circular mil area of the ungrounded conductors.
This new language was intended to be a clarification and not a change in the requirements from the 2017 code.
In the 2017 code the words "has sufficient ampacity for the intended installation" were used to convey the idea that you apply the adjustments and/or corrections to arrive at the minimum size conductor.

Note that you also have to look at the rules in 210.19(A)(1) and 215.2(A) as part of the process for determining the minimum size conductor.
 
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wwhitney

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Berkeley, CA
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But since there are 4 CCCs, #8 would be too small for the installation [. . .] Bottom line, for #3 ungrounded with 4 CCCS on a 50A breaker and no other adjustment/correction, the minimum size EGC is #7.
Actually this may be wrong, sorry, it's fairly subtle, and the answer may be different for a 40A continuous load than a 50A non-continuous load. It goes like this:

40A continuous load, 4 CCCs. The termination check uses the termination temperature (if lower than insulation temperature) and the 125% continuous factor, but not ampacity correction or adjustment. The "in the conduit" check uses the insulation temperature and ampacity correction and adjustment, but not the 125% factor.

So if the terminals are at least 75C and the insulation temperature is at least 75C, then #8 is sufficient: 50A >= 40A * 125% for the termination check; and 50A * 0.8 >= 40A for the "in the conduit" check. [Note that the latter would not be true for a 50A non-continuous load, which my previous answer was based on.]

That means for this case, #8 is the smallest size allowed. Since #3 was used, the EGC needs to be upsized 5 sizes, from #10 to #5 minimum.

Now if the terminals are only 60C rated, then I believe #6 would be required, and my previous answer applies.

Cheers, Wayne
 

don_resqcapt19

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Location
Illinois
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retired electrician
Ok. So if its continous load of 40A and 4 CC Does one need to do which if the following to size cable

40/0.8

Or

(40x1.25)/0.8

Which one?
You have to look at 210.19(A)(1) to find the minimum conductor for this circuit.
210.19(A)(1)(a) requires a 50 amp conductor.
210.19(A)(1)(b) also requires a 50 amp conductor
The rule says you pick the larger.
 

Dsg319

Senior Member
Location
West Virginia
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Wv Master “lectrician”
I came up with a #4awg EGC based off increased size conductors from #8awg to #3awg figuring on 75degree equipment and wire.
 

Dsg319

Senior Member
Location
West Virginia
Occupation
Wv Master “lectrician”
Am I wrong in thinking this? Do voltage drop calculation first, find size needed for ungrounded conductors. Than proportionally increase EGC.

Than do ampacity adjustments for 4CCC on new increased conductor size conductors . And in his scenario he will be fine with his conductor ampacity far greater than the load to be served.

Feel free to correct.. it’s been a long day lol
 
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