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#58191 11/03/05 08:54 PM
Joined: Oct 2003
Posts: 156
D
Member
redsy, I understand your confusion, but there is a purpose for the increase in size. The EGC has two serve two critical functions during a fault.

1. Is to be able to clear the fault quickly. Obviously even if the EGC were not increased in size, the standard EGC would be able to do that up to a point.

2. Limit the voltage at the point of the fault. This is where the size becomes important. If the supply conductor and EGC are the same size, then the voltage at the fault is roughly half the supply voltage. If the EGC is smaller, then the voltage at the fault is proportional higher.

#58192 11/03/05 09:17 PM
Joined: Mar 2001
Posts: 2,056
R
Redsy Offline OP
Member
Notice that Table 250.122 permits a #6 EGC for a 200-amp ckt.

A 3/0 conductor has an area of 0.173 sq.in.
A #6 has an area of 0.027 sq.in., which is about 15% the area of the 3/0.

So much for equal sizing. I still don't get it.

I think that increasing the EGC is a good idea, but some proportional value could be assigned for the conductors below a certain size.

Don, you got me thinking!

#58193 11/05/05 01:53 AM
Joined: Jul 2004
Posts: 625
S
Member
How about, call it a 40A circuit, and use #10 for the EGC? [Linked Image]

#58194 11/05/05 07:55 AM
Joined: Jan 2003
Posts: 4,391
I
Moderator
This code section can really be a costly mistake to those who run cables.

Use a larger cable assembly on a 15, 20 or 30 amp circuit and the factory EGC will be to small.

Run 8/2 UF for a 20 amp circuit to some far away load and the 10 AWG EGC will be a violation.


Bob Badger
Construction & Maintenance Electrician
Massachusetts
#58195 11/05/05 09:33 AM
Joined: Mar 2001
Posts: 2,056
R
Redsy Offline OP
Member
iwire,

That is why I wonder why it would truly be unsafe to run a "proportionally increased" EGC for smaller circuits that have been upsized for voltage drop.
Unless my math is off, further calculations reveal that:

60 amp ckt-- EGC = 40% circuit conductor size
100 amp ckt. 32%
200 amp ckt. 16%
300 amp ckt. 12%

I would suggest a conservative 40% ratio of increased EGC comared to increased circuit conductors, for smaller circuits, up to 60 amps.
Could this possibly be based on whatever engineering data was used to generate Table 250.122?

#58196 11/05/05 10:02 AM
Joined: Jan 2003
Posts: 4,391
I
Moderator
Redsy all I can say is put together a proposal for the 2011 NEC.

Personally I think making allowances for 15, 20 and 30 amp circuits, (the only circuits that are in issue) would only add confusion.

I do not think it is a hardship to run a full size ground on these size circuits.

I run into this requirement often with site lighting and we may run 10, 8, 6, 4 even 2 AWG for use with 20 or 30 amp circuits we know when we do that we will be running a full sized EGC.

I will also point again that a 10 AWG is only acceptable for a 60 amp breaker if you use 6 AWG CU for the ungrounded. If you bump up the ungrounded conductors to say 2 AWG for voltage drop the EGC will need to be 6 AWG on that 60 amp circuit. (a guess, I did not do the calacs).

My point is 250.122(B) is not about sizing the EGC to the OCP but about sizing the EGC for the VD.


Bob Badger
Construction & Maintenance Electrician
Massachusetts
#58197 11/05/05 10:17 AM
Joined: Mar 2001
Posts: 2,056
R
Redsy Offline OP
Member
Guys,

Thanks for the input.
In case you haven't figured it out, I did a project relevant to this issue. And I did run a properly increased EGC.
The topic is interesting, and took on a life of it's own.

#58198 11/05/05 10:32 AM
Joined: Mar 2001
Posts: 2,056
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Redsy Offline OP
Member
iwire,

Interesting point comes up when you say...

Use a larger cable assembly on a 15, 20 or 30 amp circuit and the factory EGC will be to small.

Run 8/2 UF for a 20 amp circuit to some far away load and the 10 AWG EGC will be a violation.

The point being that if you start with a short section of #14, 12, or 10 cable, (which have full size grounds) and then transition to your larger conductors, no violation would seem to occur because all the total line resistances would be equal, correct?

BTW,

How does one "clip a quote" to paste into a reply?

Thanks for everything guys!

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