Fault current ground path

[HamburgerHelper]

How come I never see the impedance of the earth in fault current calculations? If you have a fault, doesn't it have to makes its way in the ground back to your ground source or sources? I see stuff associated with creating a low effective fault paths in the NEC but nothing elsewhere for higher voltage stuff like in distribution and transmission. Is earth that great of a conductor for fault currents?

 

[davidbeach]

Earth resistance is accounted for the line Z0 impedance calculations.

 

[DistCoop]

It's an additional resistive component added to the impedance. Carson's paper derived a terrible series formula for approximating it based on frequency. Fortunately, at 60 Hz, you just add 0.0953 ohms/mile.

 

[HamburgerHelper]

In one of my text showing modeling a line with the return path, it shows the term, De = 658.368 *(rho/freq)^.5, inside of a logrithmic function for Zij. So, there would have to be a lot of variation in the impedance of the ground return path I imagine before anyone started caring about the soil resistivity enough to test it.

 

[DistCoop]

I believe you're correct... I've only ever assumed 100 ohm-meters in my world. As you get higher resistivity (thousands of ohm-meters), longer transmission circuits, you should give it a little more consideration. Unless you're in an area of known higher resistivity, I don't know that it's worth looking into.

 

[davidbeach]

100 ohm-meters may be much too high. I get a much better alignment between model results and actual faults with 10 ohm-meters. Too high a resistance number and you find your ground elements reach too far.

 

[cuky2000]

The IEEE Std 80 consider the split factor to determine the current path into the earth and the one to travel in the metallic return such as the shield wires of the line, neutral, etc. It is not unusual to consider a 50% fault current travelling into the ground.