HRG sizing

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dicklaxt

Senior Member
How do you size the ground conductor and resistor for a "High Resistance Ground" system on a 3 phase 13.8KV/480V-Wye transformer?

dick
 

dicklaxt

Senior Member
Yes that is the reference but it doesn't get me to the size other than saying it can't be smaller than #8[(250.35(b)].If I knew the size of the resistor I could pick a conductor ampacity that would be equal to or greater than but then it would also have to be good for the system fault current or would it?Thats all beyond me,,,,,,,,,,,,,I'm confused as usual

thanks anyway

dick
 
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jim dungar

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480V systems typically (maybe >99%) have resistors that limit the ground current to 10A, with 5A being probably the most common, so just about any size conductor will work.

System fault current does not need to enter the selection process, unless you feel the resistor will fail in a short circuit mode.
 

dicklaxt

Senior Member
Thanks Jim,kinda thought something like that may surface because of 250.36 being so loosely written

dick
 

Smart $

Esteemed Member
Location
Ohio
... just about any size conductor will work.

...
250.36(B), in part says...

The grounded system conductor shall have an ampacity
of not less than the maximum current rating of the grounding
impedance but in no case shall the grounded system
conductor be smaller than 8 AWG copper or 6 AWG aluminum
or copper-clad aluminum.
 

Smart $

Esteemed Member
Location
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As was pointed out in post #2, and agreed to by the OP in post #3.

My answer was to the question of sizing other than code minimum.
I wasn't implying you were incorrect... just posting the actual text. Saves other readers from opening their Code references... :D
 

nollij

Member
Location
Washington
but then it would also have to be good for the system fault current or would it?


The grounding resistor limits all current traveling through the ground path to the source windings. So, you do not need to size it per the maximum fault current regardless of 1LG or 3LG fault.

The only reason to go above #8AWG would be if someone was utilizing an LRG or a higher current HRG (at which point, might as well be LRG as it would behave similarily). I am not sure what exactly the NEC section 250.36(B) is defining as: "maximum current rating of the grounding impedance." These grounding resistors for higher fault currents typically do not have a continuous rating and this section does not define what current rating it is looking for. You can always model it to make sure your conductor is protected from damage by the protective device (and the backup protective devices). Section 250.36(B) could potentially also mean that the conductor must be capable of carrying the continuous load of the grounding resistor... but, that seems a bit absurd.
 

iceworm

Curmudgeon still using printed IEEE Color Books
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Occupation
EE (Field - as little design as possible)
How do you size the ground conductor and resistor for a "High Resistance Ground" system on a 3 phase 13.8KV/480V-Wye transformer? ...
The NEC does not contain any information on designing an HRG. Their only issue is the mechanical strength of the grounding electrode conductor.

The reference I use is an IEEE paper on "High-Resistance Grounding", by Baldwin Bridger, 1983. It gives good methods for estimating the capacitive charging current, and sizing the grounding resistor. It also contains a section on field measuring capacitive charging current. A google search will turn up a copy with out having to go to the IEEE.

ice
 

nollij

Member
Location
Washington
In terms of sizing the resistor:

The point of the HRG is to run continuously despite a 1LG fault. To do this, the fault path current must be limited to prevent the fault from being destructive and turning into a LLG or 3LG fault. This puts an upper limit on the resistor limit. A common number I have come across is limiting the 3I0 to <10A. During a 1LG fault, the charging current going through the faulted conductor (and ground) is 3Ic (where Ic is the normal charging current of one phase during normal operation). So, sqrt(3Ic^2+Ir^2) < 10A is the limit (where Ir is the HRG current = V/R).

At the same time, the idea is for the system to not behave like an ungrounded system and have high overvoltages from reclosing operations or intermittent faults. This will also result in the fault propagating to LLG or 3LG as the insulation system deteriorates (although, probably not right away). This then puts a bottom limit on the current traveling through the resistor. The common number for this is: Ir > 3Ic. This takes knowledge in knowing what the capactive charging current for each system is or... Just use the highest current readily available from a manufacturer without exceeding the above limitation for 480V systems (MV will take an analysis).
 

Phil Corso

Senior Member
Dicklaxt...

Responding to the "resistor" part of your question:

The resistor-value, despite previous advice given you, is not a number one picks out of the air! Its Ohm-value should be equal to (or slightly less than) the sum of the Capacitative-Reactance of the capacitance-to-ground values of any electrical equipment or apparatus connected to the system i.e., cable, transformer(s), generator(s), surge-capacitors, transient suppressors!

Its kW-withstand rating should be greater than the "expected" duration af any phase-to-ground event!

Regards, Phil Corso
 
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