Sizing EGC for Voltage Drop
- Thread starter mjc1060
- Start date
- Status
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
#8.
NEC 250.122(B)
NEC 250.122(B)
kwired
Electron manager
- Location
- NE Nebraska
The fact the EGC needs increased proportionally to the increase of the ungrounded conductors makes this easy to figure out with 15, 20 and 30 amp circuits as the EGC for those is same size as minimum size ungrounded conductor (as a general rule anyway).
Where it gets more complicated is with larger then 30 amp circuits where the EGC isn't typically required to be same size as the ungrounded conductors.
Where it gets more complicated is with larger then 30 amp circuits where the EGC isn't typically required to be same size as the ungrounded conductors.
Carultch
Senior Member
- Location
- Massachusetts
Here's a simple example.
Consider a simple 40A circuit.
Default wire size is #8 Cu
Default ground size is #10 Cu
Suppose we need to upsize the ungrounded conductors to #6 Cu for voltage drop reasons. The ratio of cross sectional areas is 26.3 kcmil / 16.5 kcmil = 1.59. This ratio must also be the kcmil ratio of the EGC. #10 is approximately 10 kcmil. Multiply by 1.59 and you get 15.9 kcmil. The next standard wire size is #8 Cu, which has 16.5 kcmil of cross sectional area.
Conclusion:
#6 Cu wires with #8 Cu ground.
In general, if you increase the wire size by X number of sizes, you'll usually increase the ground size by X number of sizes.
kwired
Electron manager
- Location
- NE Nebraska
Now stir pot a little.
Take a 60 amp circuit -
default wire size 6 AWG
default EGC 10 AWG
Why is 10 AWG sufficient to protect circuit with 6 AWG conductor and 60 amp OCPD, yet if we leave everything else the same but change the OCPD to a 40 it is now too small?
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
Why does Mickey Mouse wear pants and no shirt but Donald Duck wears a shirt and no pants?
Answer to both: 'Cause that is the way it is.
- Location
- Illinois
- Occupation
- retired electrician
Because CMP 5 refuses to accept proposals that would tie the required size of the EGC to the size of the ungrounded conductors and not to the size of the OCPD. Such a change in 250.122 would eliminate this issue.
JFletcher
Senior Member
- Location
- Williamsburg, VA
:?
So, with same sized ungrounded conductor (6ga here), downsizing from 60A breaker to 40 requires EGC be upsized from #10 to, what, #6? What's the code section, and why does it have to be done?
Carultch
Senior Member
- Location
- Massachusetts
We're talking about an unintended loophole to this code rule, where the consequences of the rule as written can yield counterintuitive results. And there simply is no explanation for it, other than that is the way the NEC is written.
The underlying reason why this requirement exists, is that upsizing the conductors usually means that you've upsized them to curtail voltage drop due to significant length. And if the length is significant for the wires, it is also significant for the OCPD in its ability to provide an effective ground fault current path. So to improve the EGC's ability to fulfill its purpose when length is significant, it gets increased in size to improve the EGC conductance. What if the ungrounded conductors were increased in size for other reasons than voltage drop (using up leftover wire, terminal minimums that are larger than the NEC required size, etc)? Well, for the simplicity of the inspectors, the NEC says that it doesn't matter the reason.
Perhaps a way to make the rule to eliminate this blindspot, is if the ungrounded size exceeds the minimum size that has sufficient ampacity for the default EGC's maximum corresponding OCPD, then the EGC needs to be upsized proportionally for the ratio of the ungrounded conductor cross sectional area to that which would have sufficient ampacity for the EGC's maximum corresponding OCPD. I can't think of a way to clean up this language for a better understanding, so perhaps it needs an example.
***** Caution: this is not an NEC compliant example - it is a hypothetical example only **************
Given: a 40A circuit, which requires upsizing wires to #4 to curtail voltage drop.
Default wire=#8Cu
Default ground = #10 Cu
Maximum OCPD for a #10 Cu ground = 60A
Corresponding wire to a 60A OCPD = #6 Cu
Wire is upsized above what it could be for a 60A breaker (corresponding to the maximum breaker for #10 EGC) by a ratio of #4 to #6, which is a ratio of 41.7/26.3 = 1.59. Therefore, the EGC is upsized from #10 to #8.
This algorithm is very confusing, much more than the current rule. What it essentially does, is puts the OCPD standard sizes into "bins" based on the 250.122 table. And if you upsize above the minimum size for the largest breaker in the actual breaker's "bin", then you correspondingly upsize the ground.
Last edited:
Carultch
Senior Member
- Location
- Massachusetts
Here's another example to stir the pot. Consider parallel sets of conductors for very large circuits.
Given a 400A circuit, that requires a total of 800 kcmil to curtail voltage drop, yet the equipment terminals are limited to 750 kcmil. Also, good luck finding any 800 kcmil copper.
The default EGC is #3 Cu
For the ungrounded conductors, we have three options:
A: The default size for a single wire per phase is 600 kcmil Cu.
B: The default size for two wires per phase in the same conduit is #4/0 Cu
C: The default size for two wires per phase in separate conduits is #3/0 Cu
To achieve 800 kcmil, the solution is to use 2 parallel sets 400 kcmil.
So what then becomes our upsize ratio, that must be used for upsizing the EGC?
A: (400 + 400)/600 = 1.33333
B: (400 + 400)/(212 + 212) = 1.887
C: (400 + 400)/(168 + 168) = 2.38
A still meets the intent of the code rule, but the NEC wording doesn't clearly indicate that this is an acceptable way to do it. Certainly it would be the most economical solution, if allowed.
B would only make sense, if the intended installation has conductors routed in a shared conduit, just like the starting point for the calculation.
C would be the conservative solution, if the parallel sets are run in separate conduits
- Location
- Bremerton, Washington
Right, that's the way the physics of this would work, larger wire, more fault current.
I heard of an example for a 120V circuit that went 1/4 mile with 250 AL for a traffic signal flasher..
rlundsrud
Senior Member
- Location
- chicago, il, USA
I am confused, if you are running a distance of 500 feet, wouldn't you need at least #2 conductors? I can't imagine that #8 could handle that distance and maintain<3% vd. You might want to check your numbers.
jumper
Senior Member
- Location
- 3 Hr 2 Min from Winged Horses
OP never said VD had to be limited to <3%.
OP's load may be able to handle a greater VD.
Remember that except for a few cases, VD is not a NEC requirement.
rlundsrud
Senior Member
- Location
- chicago, il, USA
That would be a 16% VD, that seems implausible to me.
- Occupation
- Licensed Electrician
He never said what the load was.
Carultch
Senior Member
- Location
- Massachusetts
A larger wire doesn't necessarily cause more fault current. A larger wire just impedes the fault current less than what a smaller wire at the same length would do.
kwired
Electron manager
- Location
- NE Nebraska
Which will allow more fault current to flow
Source impedance may complicate this in some cases though.
Carultch
Senior Member
- Location
- Massachusetts
Which will allow more fault current to flow
Source impedance may complicate this in some cases though.
That's my point. Not counting motor contribution, the fault current is limited to what is available from the source.
- Location
- Massachusetts
Tom's point was simply that given the same source a larger wire results in more fault current.
I agree with him.
Carultch
Senior Member
- Location
- Massachusetts
True.
However, consider the entire reason the conductor is larger than it needs to be (for ampacity alone) in the first place. Considering identical sources, which do you think has greater fault current at the load end? A small conductor on a short length circuit? Or a large conductor on a long length circuit? Suppose they are both selected to have the same percentage voltage drop, and therefore would have the same total conductor resistance. Would you expect fault current to be less, greater, or equal on the larger/longer circuit?
- Status