Is it OK to drive a ground rod at a machine if switch gear & buss plug have no ground

Status
Not open for further replies.

Sahib

Senior Member
Location
India
Nothing in your example is real.

1. No where I have ever seen or witnessed a 400 amp 120 volt circuit 1000 feet long. Anything requiring that much power at that distance is going to be 3-phase operating at much higher voltage.
2. 1000 feet of #3 AWG is .2 Ohms not the .3 you mention.
3. If the supply is 1000 feet of 750 the total combined impedance is of the 750 and #3 is .217 Ohms impedance resulting in 120 V / .217 ohms = 552 amps. Even if you could drive a rod to get 10 ohm's impedance would do nothing.
4. At the point of contact the 750 MCM impedance is so much lower than 3 AWG the voltage will be 552 amps x .2 ohms = 552 amps x .2 ohms = 110 volts at the point of the fault.
5. A 1000 feet circuit using 750 MCM with 400 amp load current would have a voltage drop of 14 volts or 12 % voltage drop.
6. Not compliant to any known code or practice. Minimum EGC for this make believe circuit would require a 250 MCM which would create a 2400 amp fault current which would operate the OCPD instantly.

Nough said.
But the code does not seem to forbid it. Plug in the values of 400A, 120V, 750kCM single phase circuit in
http://www.electrician2.com/calculators/wireocpd_ver_1.html and see the results.
Regarding your points 3&4: You are correct that the resistance of #3 EGC wire is 0.2 ohm. But still the voltage drop due to PARTIAL insulation breakdown is assumed to be only 60V for which the ground fault current is 60/0.2=300A, which is not likely to operate the OCPD quickly.
 

Sahib

Senior Member
Location
India
I guess it is probably safe to say that in many cases the voltage to remote earth at the point of the ground fault would be a shock hazard until the OCPD clears the fault.
In such cases a short circuit ground fault analysis may be required to assess if the ground potential rise is within safe limit until the OCPD clears the fault.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
In such cases a short circuit ground fault analysis may be required to assess if the ground potential rise is within safe limit until the OCPD clears the fault.
I am not talking about ground potential rise. I am talking about the voltage on the metal parts of the equipment at the point of the ground fault. This voltage will be equal to the voltage drop on the fault clearing path and will be a shock hazard until the fault is cleared.
 

Sahib

Senior Member
Location
India
I am not talking about ground potential rise. I am talking about the voltage on the metal parts of the equipment at the point of the ground fault. This voltage will be equal to the voltage drop on the fault clearing path and will be a shock hazard until the fault is cleared.
The ground potential rise is the same as the voltage on the metal parts of the equipment at the point of the ground fault with respect to remote ground. The touch voltage at a distance of 1 meter or so from the fault has around 80% of the ground fault voltage and depends on the ground resistances. By reducing the ground resistance, the touch voltage may be brought within safe limits.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Only if you reduce the ground electrode resistance well below the wire resistance of the ungrounded conductors. That seems to me to be an outrageously difficult task.

Tapatalk!
 

Sahib

Senior Member
Location
India
The resistance of ungrounded conductor plays a role in reducing the voltage available at the ground fault point; that voltage is shared at least between two grounds ( one equipment ground and other the service entrance ground) in direct proportion to their resistances. So if equipment ground resistance is lowered, the touch voltage at the equipment may also be lowered.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
The ground potential rise is the same as the voltage on the metal parts of the equipment at the point of the ground fault with respect to remote ground.
The ground potential rise only exists near the grounding electrode.
The touch voltage at a distance of 1 meter or so from the fault has around 80% of the ground fault voltage and depends on the ground resistances.
The touch voltage from the equipment, at the point of the fault, to the earth or anything connected to the electrical grounding system is equal to the voltage drop on the fault clearing path.
By reducing the ground resistance, the touch voltage may be brought within safe limits.
The voltage drop on the fault clearing path will almost never be within safe limits. The only way to make it safe is to clear the fault. The ground resistance has very little to do with reducing the voltage drop on a fault clearing path.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
... So if equipment ground resistance is lowered, the touch voltage at the equipment may also be lowered.
How do you propose lowering the equipment ground resistance at the equipment? Even if the fault clearing path has an impedance that is equal to the supply conductor impedance the voltage drop will be one half of the circuit voltage. It would be a rare circuit where one half of the circuit voltage is a touch safe voltage.
 

Sahib

Senior Member
Location
India
How do you propose lowering the equipment ground resistance at the equipment? Even if the fault clearing path has an impedance that is equal to the supply conductor impedance the voltage drop will be one half of the circuit voltage. It would be a rare circuit where one half of the circuit voltage is a touch safe voltage.
Let the equipment ground resistance and service entrance ground each be 100 ohms. Let the voltage drop along the EGC be 60Volts. The EGC resistance is in parallel with the sum of the two ground resistances. So the 60V is shared equally between the two resistances. So 30V is the touch voltage at the equipment. Suppose the equipment ground resistance is reduced to 50 Ohms. Now the total ground resistance is 150 ohms only. Earlier it was 200 ohms for which at 60V voltage drop the ground fault current is 60/200=0.3A and the touch voltage at equipment is 0.3*100=30V. But after reduction to 150 ohms at the same 60V voltage drop the ground fault current is 60/150=0.4A and the touch voltage at equipment is 0.4*50=20V only.
 
Last edited:

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
That is so fundamentally wrong that I am at a loss about where to start.
If the voltage drop on the EGC is 60 volts, the current will be enormous since the EGC resistance will be well below one ohm. Where did you get this hypothetical EGC resistance of 100 ohms?
And putting a resistance of 100 ohms in parallel with that is going to have near zero effect on the voltage.
It is too late at night for me to continue, so I will pass the baton (properly grounded of course) to the next runner.

Tapatalk!
 

Sahib

Senior Member
Location
India
I did not specify any resistance for the EGC in my last post: just voltage drop of 60V across it. But if you to want it have specified, let it be 0.1 ohm. This 0.1 ohm of EGC is in parallel with two ground resistances in series:eek:ne equipment ground and other service entrance ground. Now apply the calculation of my last post. Is it now clear?
 

kwired

Electron manager
Location
NE Nebraska
I did not specify any resistance for the EGC in my last post: just voltage drop of 60V across it. But if you to want it have specified, let it be 0.1 ohm. This 0.1 ohm of EGC is in parallel with two ground resistances in series:eek:ne equipment ground and other service entrance ground. Now apply the calculation of my last post. Is it now clear?

Let the equipment ground resistance and service entrance ground each be 100 ohms. Let the voltage drop along the EGC be 60Volts. The EGC resistance is in parallel with the sum of the two ground resistances. So the 60V is shared equally between the two resistances. So 30V is the touch voltage at the equipment. Suppose the equipment ground resistance is reduced to 50 Ohms. Now the total ground resistance is 150 ohms only. Earlier it was 200 ohms for which at 60V voltage drop the ground fault current is 60/200=0.3A and the touch voltage at equipment is 0.3*100=30V. But after reduction to 150 ohms at the same 60V voltage drop the ground fault current is 60/150=0.4A and the touch voltage at equipment is 0.4*50=20V only.

Sure looks like you did:huh:
 

Sahib

Senior Member
Location
India
No. Do not equate ground resistance (say, the equipment ground resistance=100 ohms) with EGC resistance(=0.1 ohm)
 

Sahib

Senior Member
Location
India
Oh, sorry. By equipment ground resistance, I mean the individual equipment ground rod resistance.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
Let the equipment ground resistance and service entrance ground each be 100 ohms. Let the voltage drop along the EGC be 60Volts. The EGC resistance is in parallel with the sum of the two ground resistances. So the 60V is shared equally between the two resistances. So 30V is the touch voltage at the equipment. Suppose the equipment ground resistance is reduced to 50 Ohms. Now the total ground resistance is 150 ohms only. Earlier it was 200 ohms for which at 60V voltage drop the ground fault current is 60/200=0.3A and the touch voltage at equipment is 0.3*100=30V. But after reduction to 150 ohms at the same 60V voltage drop the ground fault current is 60/150=0.4A and the touch voltage at equipment is 0.4*50=20V only.
The impedance that we are concerned with is that of the fault clearing path between the point of the fault and the main or system bonding jumper. The EGC will be the conduit or a wire type EGC. The impedance of any additional grounding electrodes that are installed at the equipment will by much much higher than that of the EGC itself and will make almost no difference in the voltage drop on the fault return path. The touch voltage at the equipment, unless you are standing on or very near the grounding electrode, will remain ~ 1/2 of the system line to neutral voltage no matter how many additional grounding electrodes you install at the service or at the equipment. (unless you install so many grounding electrodes at the equipment that you create an equipotenial plane)
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
No. Do not equate ground resistance (say, the equipment ground resistance=100 ohms) with EGC resistance(=0.1 ohm)
The voltage drop is based on the resistance of the fault clearing path, the EGC...if the EGC has a resistance of 0.1 ohm and the resistance of the grounding electrode installed at the equipment is 100 ohms, the fault return path now has a resistance of 0.999 ohms. The reduction of 0.0001 ohms in the fault clearing path creates an unmeasurable change in the voltage drop of the fault return path.
 

Sahib

Senior Member
Location
India
The voltage drop is based on the resistance of the fault clearing path, the EGC...if the EGC has a resistance of 0.1 ohm and the resistance of the grounding electrode installed at the equipment is 100 ohms, the fault return path now has a resistance of 0.999 ohms. The reduction of 0.0001 ohms in the fault clearing path creates an unmeasurable change in the voltage drop of the fault return path.
There are at least three resistances involved: one, EGC wire resistance=0.1ohm second, equipment ground rod resistance=100 ohm and third, service entrance ground rod resistance=100 ohm. The resistance areas of the latter two ground rod resistances are assumed not to overlap. The EGC wire resistance is connected in parallel with the two ground rod resistance in series. So even though the 60V voltage drop across the parallel combination does not change appreciably with changes in the two ground rod resistances, the voltage across each ground rod resistance may be changed by changing its resistance while the sum of the voltage drops across the two remaining constant and equal to 60V.
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
There are at least three resistances involved: one, EGC wire resistance=0.1ohm second, equipment ground rod resistance=100 ohm and third, service entrance ground rod resistance=100 ohm. The resistance areas of the latter two ground rod resistances are assumed not to overlap. The EGC wire resistance is connected in parallel with the two ground rod resistance in series. So even though the 60V voltage drop across the parallel combination does not change appreciably with changes in the two ground rod resistances, the voltage across each ground rod resistance may be changed by changing its resistance while the sum of the voltage drops across the two remaining constant and equal to 60V.
That does nothing to change the touch potential (or the touch/step potential, that is touching the equipment and standing on a "grounded" surface) at the point of the fault. If you are touching the equipment at the point of the fault and something else that is connected to earth you will receive a shock with a voltage that is equal to the voltage drop of the fault return path. You will not reduce this shock voltage by adding grounding electrodes or reducing the resistance of the grounding electrodes. This shock hazard will exist until the fault is cleared.
 

Sahib

Senior Member
Location
India
That does nothing to change the touch potential (or the touch/step potential, that is touching the equipment and standing on a "grounded" surface) at the point of the fault. If you are touching the equipment at the point of the fault and something else that is connected to earth you will receive a shock with a voltage that is equal to the voltage drop of the fault return path. You will not reduce this shock voltage by adding grounding electrodes or reducing the resistance of the grounding electrodes. This shock hazard will exist until the fault is cleared.
Then what is wrong with the calculation below?
Let the equipment ground resistance and service entrance ground each be 100 ohms. Let the voltage drop along the EGC be 60Volts. The EGC resistance is in parallel with the sum of the two ground resistances. So the 60V is shared equally between the two resistances. So 30V is the touch voltage at the equipment. Suppose the equipment ground resistance is reduced to 50 Ohms. Now the total ground resistance is 150 ohms only. Earlier it was 200 ohms for which at 60V voltage drop the ground fault current is 60/200=0.3A and the touch voltage at equipment is 0.3*100=30V. But after reduction to 150 ohms at the same 60V voltage drop the ground fault current is 60/150=0.4A and the touch voltage at equipment is 0.4*50=20V only.
 
Status
Not open for further replies.
Top