Addition to Earlier Post 4

[SilverArc]

Hi Everybody,
I had sneaked in an earlier post"Grounding in Rocky areas" and Jghrist was kind enough to answer my question there about the potential variation shown by dotted lines in fig. 1 found in the link below:

http://www.era.co.uk/Docs/ECS/Ground_ROD_Tutorial.pdf#search="touch voltage step potential + grid"

Hi Jghrist,
Thanks for these valuable comments. We dont have any body at our work, who has even a figment of imagination about grounding. So, I try to work out concepts on my own or some times people in this forum are kind enough to advise me.
After reading through explanation,
"The potential shown by the dashed lines is not the potential beneath the earth. This is the surface potential as it varies along a line on the surface and it is shown below the surface for convenience"
we have a uniform copper mesh below the substation, where an equal amount of current will flow, So The potential should be same also ?
I thought through it and came up with explanation. The variation in potential on the grid its self is depending on the fault happens. Is this the reason ?

Q# Why do we say the "Mesh potential"(the potential in the centre of the grid) is always maximum.

Q# Eventually to give more practical shape to my dilemna, I did a simulation in ETAP and came across the plot below for touch profile.

http://img245.imageshack.us/my.php?image=gpr1do8.jpg

Could you please explain me, what are these red circles at different places. What do they highlight. I though a lot about it but could not get it.

Thanks.

 

[dgbpeakpwr]

SilverArc
What you have done is the infinite bus method, it is only a good first estimate. In most cases it will give you an upper limit on the non-source side of the transformer. I use it to start my grid design while waiting for information. The maximum fault current depends on the strength of the system and not necessarily the let through of the transformer.

 

[apowerengr]

If your 10 MVA 7% Z transformer is connected to a source with a maximum three phase fault of 400MVA, the resultant secondary three-phase fault current is reduced to approximately 4400 amps vs. your 5981 amp example above. That's why you need to know the source impedance. In the case of ground grid design, greater source impedance helps with lowering the grid current however with typical inverse time-overcurrent characteristics on the protective devices lower fault current can result in greater clearing times which can result in higher grid potentials and higher arc-flash energies (if you are performing arc-flash calculations too). Somewhere there's a best case tradeoff between fault current levels and clearing times which results in the minimum grid potentials.