More fallacious grounding

mbrooke

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Dirt is matter and has electrical properties. Our currents travel in a limited amount of the earth's total matter.
Why though? Does it not take all paths? I've heard people measuring 50 cycle power here in 60 over seas in earth driven electrodes.
 

mbrooke

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#1:
At the sub the N conductor is at the neutral point. A A-N fault drags A to the neutral point and the B-N and C-N arrestors see P/sqrt(3) volts (line-neutral).

Far away the N conductor has a large impedance that allows it to float from the neutral point. An A-N fault drags N to A and the arrestors see B-A and C-A voltage (line-line).

#2:
The earth path has impedance.
True.

Though much less so for an MGN, right?
 

mivey

Senior Member
The net resistance is indeed not zero, but most if not all of that resistance is the term associated with the earth electrode resistance rather than the distance to the uniground point. If the two electrodes are within each others' sphere of influence the overall resistance will be more complicated. That can lead to the appearance of the net resistance depending on the earth path length.


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Not quite. If we could parallel enough dirt at no cost then we would have a perfect earth conductor. The real circuit does not behave that way. Nature tends to the lowest energy state and simply will not reach the whole earth for the transmission line. The current tends to maximize nearer the transmission line rather than going through China. This limits the parallel dirt and the net dirt resistivity becomes a significant factor.
 

mivey

Senior Member
By how much though?
You have to run the numbers.

I ran a model for a 1 mile 115 kV transmission line with a 0.0001 ohm N-E connection to remote earth on the source end. Beyond that, I ran a 20 mile transmission line with a 0.0001 ohm N-E connection to remote earth at the source and load end. In the 20 mile section, I paralleled the earth with a giant conductor to short the earth resistance during a fault for case #1. For case #2 I then opened the parallel conductor to run fault current through earth only.

If the earth is a perfect conductor, then we would expect most of the current to be in the earth instead of in the parallel path. But that is not the reality of how currents run through the transmission line and earth.

The fault I created was a A-N fault at the end of the 21 miles. Clearly the earth path is resisting the current even though we have a negligible connection resistance to remote earth.



With parallel path:

In the 20 mile section:
Phase A: 590 amps
parallel path: 395 amps
earth: 196 amps

In the 1 mile section:
Phase A: 590 amps
earth: 590 amps



Without parallel path:

In the 20 mile section:
Phase A: 430 amps
earth: 430 amps

In the 1 mile section:
Phase A: 430 amps
earth: 430 amps
 

mivey

Senior Member
Not quite. If we could parallel enough dirt at no cost then we would have a perfect earth conductor. The real circuit does not behave that way. Nature tends to the lowest energy state and simply will not reach the whole earth for the transmission line. The current tends to maximize nearer the transmission line rather than going through China. This limits the parallel dirt and the net dirt resistivity becomes a significant factor.
Not to say nothing runs in the China dirt but it might not even be measurable. i.e., The closer dirt gets much more current than the far dirt so soil resistivity is a significant factor.

This is modeled using Carson's Equations.
 

mbrooke

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You have to run the numbers.

I ran a model for a 1 mile 115 kV transmission line with a 0.0001 ohm N-E connection to remote earth on the source end. Beyond that, I ran a 20 mile transmission line with a 0.0001 ohm N-E connection to remote earth at the source and load end. In the 20 mile section, I paralleled the earth with a giant conductor to short the earth resistance during a fault for case #1. For case #2 I then opened the parallel conductor to run fault current through earth only.

If the earth is a perfect conductor, then we would expect most of the current to be in the earth instead of in the parallel path. But that is not the reality of how currents run through the transmission line and earth.

The fault I created was a A-N fault at the end of the 21 miles. Clearly the earth path is resisting the current even though we have a negligible connection resistance to remote earth.



With parallel path:

In the 20 mile section:
Phase A: 590 amps
parallel path: 395 amps
earth: 196 amps

In the 1 mile section:
Phase A: 590 amps
earth: 590 amps



Without parallel path:

In the 20 mile section:
Phase A: 430 amps
earth: 430 amps

In the 1 mile section:
Phase A: 430 amps
earth: 430 amps
Thank you!

Really like your method and choice of numbers. :)


In what program did you model this? Just curious.
 
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