Grounding Electrode Question
- Thread starter C3PO
- Start date
- Status
- Location
- Bremerton, Washington
- Occupation
- Master Electrician
Sometimes with article 250 we think too hard. But knowing some of the theory helps.
cloudymacleod
Senior Member
well wouldn't the main disconnect be rated at 800a and the load centers for each unit be rated for 200a?Dennis Alwon said:
frizbeedog
Senior Member
- Location
- Oregon
But Why?
But Why?
In my area, a concrete encased electrode is required to be intalled for new construction. It is inspected and tagged prior to the concrete pour. Rebar is the method.
I can say that someone, somewhere figured this would be the minimum allowable electrode for new construction and better than just two ground rods.
So in this case the GEC is never required to be larger than #4 copper. 250.66 (B).
And the NEC is not going to require that a conductor be installed larger than is required for the particular effectivness of the available electrode.
And if your local inspectors don't know the what or the why of what they are enforcing, what is it exactly that they are enforcing? Let me guess. Their misunderstanding of purpose and intent.
But Why?
In my area, a concrete encased electrode is required to be intalled for new construction. It is inspected and tagged prior to the concrete pour. Rebar is the method.
I can say that someone, somewhere figured this would be the minimum allowable electrode for new construction and better than just two ground rods.
So in this case the GEC is never required to be larger than #4 copper. 250.66 (B).
And the NEC is not going to require that a conductor be installed larger than is required for the particular effectivness of the available electrode.
And if your local inspectors don't know the what or the why of what they are enforcing, what is it exactly that they are enforcing? Let me guess. Their misunderstanding of purpose and intent.
Cow
Senior Member
- Location
- Eastern Oregon
- Occupation
- Electrician
For the posters who are red tagged due to an inspectors "made-up" code violations, why are you not requiring code references on your inspection report? Pulling 3/0 to ground rods is ridiculous...:roll:
another call
another call
Even though four meters are in a pak, attention must be paid to how the water meter laterals are supplied to the 4 units. If the copper is individual to each water meter, then regardless of Ufer, each cu water electrode needs to be bonded back to each units water lines at the foundation entry.
I have seen rods used and bonded to the T&P valve vent copper in an attempt to jumper bond to the input Cu supply at the water heater. Some inspectors do not even see that violation, but count on most do. rbj
another call
Even though four meters are in a pak, attention must be paid to how the water meter laterals are supplied to the 4 units. If the copper is individual to each water meter, then regardless of Ufer, each cu water electrode needs to be bonded back to each units water lines at the foundation entry.
I have seen rods used and bonded to the T&P valve vent copper in an attempt to jumper bond to the input Cu supply at the water heater. Some inspectors do not even see that violation, but count on most do. rbj
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
080406-1036 EST USA
Suppose the resistance of a ground rod or equivalent was as low as 1 ohm. Very hard to achieve in most parts of the country. #6 copper is 0.4 ohms per 1000 ft or 0.004 ohms for 10 ft. Either of these is relatively insignificant to 1 ohm. At 10,000 A the 10 ft length will have a drop of 40 V.
The fusing current for #6 copper from the ITT handbook is 668 A. There are many variables that determine fusing current so this is only an approximation. Any very large current is going to be time limited. Lightning in particular is the most likely source of a current of 10,000 A in the type of application being discussed here. Thus, it is reasonable to expect that there would be no fusing problem here because the 668 figure is going to be based on a much longer time period.
Now consider a 200 KVA transformer with an internal impedance of 0.008 ohms. 200 KVA at 240 V is 834 A and the nominal full load transformer drop is 7.2 V. Somehow, and I do not know how this fault could naturally occur, but suppose that we apply 240 to 1.0 + 0.008 + 0.004 ohms, then the maximum current is 240/1.012 or 237 A well below the steady state fusing current of 668 A. This 237 is mostly a result from the high resistance of most earth and the interface between ground and the earth.
None of thiese assumptions would indicate any fallacy in the code specification of #6.
My above calculations are assuming all resistive components. In all cases there is inductance and for very fast waveforms, such as lightning, the impedance will be higher than the DC resistance.
In reality #6 is probably overkill going to a normal ground rod for anything other than lightning. Probably many ground rod installations are in the 10 ohms or higher range. However, using the same cross section of copper and making it broad and flat reduces the inductive effect and is useful in reducing voltage drop from lightning currents. At 240 V the current thru 10 ohms is 24 A.
How do you create a fault in wiring that will produce a large current into the ground rod. My home pole transformer has its center tap connected to a ground rod at the pole, and there is a ground at my main panel to my copper water supply, effectively a very good ground rod. At the time the house was built there was no requirement for a separate ground rod at the house. My house ground rod is essentially 150 ft of 1.25" copper pipe. If we open the neutral between the pole and house and short one of the hot wires to the water pipe with an assumption of 10 ohms ground resistance, then the fault current is 12 A. What else can I easily do to get a high current? I would have to get a much higher voltage somehow.
Here is an interesting calculation. Assume a lightning bolt is 10,000 A and the resistance to ground is 10 ohms. This means that at this instant the ground rod rises to a potential of 100,000 V relative to some place down in or across the earth. Also means everything connected to the building wiring rises this much relative to whatever that reference is.
.
Suppose the resistance of a ground rod or equivalent was as low as 1 ohm. Very hard to achieve in most parts of the country. #6 copper is 0.4 ohms per 1000 ft or 0.004 ohms for 10 ft. Either of these is relatively insignificant to 1 ohm. At 10,000 A the 10 ft length will have a drop of 40 V.
The fusing current for #6 copper from the ITT handbook is 668 A. There are many variables that determine fusing current so this is only an approximation. Any very large current is going to be time limited. Lightning in particular is the most likely source of a current of 10,000 A in the type of application being discussed here. Thus, it is reasonable to expect that there would be no fusing problem here because the 668 figure is going to be based on a much longer time period.
Now consider a 200 KVA transformer with an internal impedance of 0.008 ohms. 200 KVA at 240 V is 834 A and the nominal full load transformer drop is 7.2 V. Somehow, and I do not know how this fault could naturally occur, but suppose that we apply 240 to 1.0 + 0.008 + 0.004 ohms, then the maximum current is 240/1.012 or 237 A well below the steady state fusing current of 668 A. This 237 is mostly a result from the high resistance of most earth and the interface between ground and the earth.
None of thiese assumptions would indicate any fallacy in the code specification of #6.
My above calculations are assuming all resistive components. In all cases there is inductance and for very fast waveforms, such as lightning, the impedance will be higher than the DC resistance.
In reality #6 is probably overkill going to a normal ground rod for anything other than lightning. Probably many ground rod installations are in the 10 ohms or higher range. However, using the same cross section of copper and making it broad and flat reduces the inductive effect and is useful in reducing voltage drop from lightning currents. At 240 V the current thru 10 ohms is 24 A.
How do you create a fault in wiring that will produce a large current into the ground rod. My home pole transformer has its center tap connected to a ground rod at the pole, and there is a ground at my main panel to my copper water supply, effectively a very good ground rod. At the time the house was built there was no requirement for a separate ground rod at the house. My house ground rod is essentially 150 ft of 1.25" copper pipe. If we open the neutral between the pole and house and short one of the hot wires to the water pipe with an assumption of 10 ohms ground resistance, then the fault current is 12 A. What else can I easily do to get a high current? I would have to get a much higher voltage somehow.
Here is an interesting calculation. Assume a lightning bolt is 10,000 A and the resistance to ground is 10 ohms. This means that at this instant the ground rod rises to a potential of 100,000 V relative to some place down in or across the earth. Also means everything connected to the building wiring rises this much relative to whatever that reference is.
.
gar said:
Your system then will have a 'floating potential' if the Utility supply is plastic. Usually the street mains are metal and create a common path back to the zero reference AC source that will still operate the panel OCPD's. (backfed through a common transformer source in the neighborhood.)
The above does not pertain to a lightning stroke event that is an order of condition not well predicted. Sorry for the lack of a good answer. rbj
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Do not do this at home.
Do not do this at home.
Lifting the neutral and 'shorting' the metal water pipe to the AC supply source can be fatal should the experimenter touch a metal faucet and at the same time complete the 'circuit' to a 'ground' surface. rbj
Do not do this at home.
Lifting the neutral and 'shorting' the metal water pipe to the AC supply source can be fatal should the experimenter touch a metal faucet and at the same time complete the 'circuit' to a 'ground' surface. rbj
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
080408-2120 EST USA
gndrod:
I was presenting a hypothetical question to show that the maximum current would be arount 12 A. This is insignificant with respect to a #6 wire.
If you draw the circuit that my example illustrated there is a 10 ohm resistance from the transformer center tap to my water pipe. If I short one hot 120 V line to my water pipe, then there is 120 V across a 10 ohm resistance. Is there anything unsafe within my home? No. Is there a voltage gradient across my yard from the house to the pole? Yes.
There is a substantial voltage drop at the interface of each ground rod and the earth, but assume that is zero. Next assume the distance from the house to the pole is 30 ft, and the voltage is uniform from the house to the pole, then the gradient per foot is 120/30 = 4 volts per foot. A 3 foot pace would traverse a 12 V drop. Not particularly unsafe under typical conditions.
(edit) I need to clarify that comment. The nuetral in my home is tied to my ground rod, which is my water pipe. and by connecting a hot line to ground in the house the other phase voltage will become 240 V relative to nuetral and that will burnout many 120 V devices on that phase, and that would be unsafe. It does not make nuetral in the house a high voltage relative to ground in the house. Note: in my example I opened the nuetral between the pole and the house. This meant the transformer center tap was connected to the pole ground rod, and my house nuetral and house ground rod were still connected together.(end edit)
What happens to a neighbor with a good nuetral and their own ground rod? Their entire ground system from the pole to their house and internal to house will be approximately at the potential of the transformer center tap. Maybe there is a load current in the nuetral and thus a small voltage drop.
One item I have left out is the possibility of a low resistance path from my water pipe to the street, thru the main, thru the neighbor's water pipe to their nuetral and to the pole. Suppose this path was 1 ohm, I really have no idea what is realistic, and I do not plan to try to determine what it is. But the 1 ohm value with my shorting 120 to my water pipe produces approximately a 120 A load on that one side of the pole transformer. If this is a typical branch circuit the breaker will trip. Otherwise if directly from the main fuse probably not. The distribution of the voltage drops around that water pipe loop will be dependent upon the resistance in different parts of the loop. My guess is for big drops at joints in the pipe under the street.
My main point of presenting the example was to show that a #6 wire to a ground rod is probably a very realistic size. My street path circuit would not increase the the current in my ground rod if I had one. However, it does illustrate that if the 1 ohm guess was valid that I might see a large current thru the water pipe.
The major point is that aside from lightning a ground rod in many locations is unlikely to see a high fault current.
My pole transformer has no connection from the primary side to the pole ground rod. Thus, without a transformer short it is unlikely there is a voltage greater 240 that could cause a ground current. My pole transformer primary is from two legs of a 3 phase source.
.
gndrod:
I was presenting a hypothetical question to show that the maximum current would be arount 12 A. This is insignificant with respect to a #6 wire.
If you draw the circuit that my example illustrated there is a 10 ohm resistance from the transformer center tap to my water pipe. If I short one hot 120 V line to my water pipe, then there is 120 V across a 10 ohm resistance. Is there anything unsafe within my home? No. Is there a voltage gradient across my yard from the house to the pole? Yes.
There is a substantial voltage drop at the interface of each ground rod and the earth, but assume that is zero. Next assume the distance from the house to the pole is 30 ft, and the voltage is uniform from the house to the pole, then the gradient per foot is 120/30 = 4 volts per foot. A 3 foot pace would traverse a 12 V drop. Not particularly unsafe under typical conditions.
(edit) I need to clarify that comment. The nuetral in my home is tied to my ground rod, which is my water pipe. and by connecting a hot line to ground in the house the other phase voltage will become 240 V relative to nuetral and that will burnout many 120 V devices on that phase, and that would be unsafe. It does not make nuetral in the house a high voltage relative to ground in the house. Note: in my example I opened the nuetral between the pole and the house. This meant the transformer center tap was connected to the pole ground rod, and my house nuetral and house ground rod were still connected together.(end edit)
What happens to a neighbor with a good nuetral and their own ground rod? Their entire ground system from the pole to their house and internal to house will be approximately at the potential of the transformer center tap. Maybe there is a load current in the nuetral and thus a small voltage drop.
One item I have left out is the possibility of a low resistance path from my water pipe to the street, thru the main, thru the neighbor's water pipe to their nuetral and to the pole. Suppose this path was 1 ohm, I really have no idea what is realistic, and I do not plan to try to determine what it is. But the 1 ohm value with my shorting 120 to my water pipe produces approximately a 120 A load on that one side of the pole transformer. If this is a typical branch circuit the breaker will trip. Otherwise if directly from the main fuse probably not. The distribution of the voltage drops around that water pipe loop will be dependent upon the resistance in different parts of the loop. My guess is for big drops at joints in the pipe under the street.
My main point of presenting the example was to show that a #6 wire to a ground rod is probably a very realistic size. My street path circuit would not increase the the current in my ground rod if I had one. However, it does illustrate that if the 1 ohm guess was valid that I might see a large current thru the water pipe.
The major point is that aside from lightning a ground rod in many locations is unlikely to see a high fault current.
My pole transformer has no connection from the primary side to the pole ground rod. Thus, without a transformer short it is unlikely there is a voltage greater 240 that could cause a ground current. My pole transformer primary is from two legs of a 3 phase source.
.
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