Earthing Requirments

[ukgraduate]

Hello,

I have been getting myself extremely confused about earthing. We are doing a design for a mine which will be all cables. I have been reading up about step/touch/earth potential rise etc etc, worrying about how to achieve this on my project.

My confusion is this: The standards require a earth grid impedance of a substation of X ohms, via electrodes earth mat etc. My understanding is that this provides a good path for the ground fault back to the source. However If the substations are going to be interconnected via MULTIPLE cables (with the screen as the earth), why do I even need to install earthing conductors etc. The earth grid are my cables which will carry the fault back to the source by the screen of the cable.

I understand that there may be a bad joint or two. However with multiple paths this is mitigated. I was also thinking "well you want a good earth because a load like a motor will only be connected via the one cable. If this has a bad joint or is not connected it will short to ground (literally). So the current will go via the earth back to the source". This motor may be far away from any earth grid so I again question why I am putting earth conductors everywhere to reduce the impedance of the earth grid?

I am sure there is a good reason but I am lost..please help

 

[acog]

A minimum grid resistance does not mean your installation is safe or functional. You need to consider what your fault scenarios are, i.e. where the current will come from and how it will return to the source/star point.

As a part of your design, you will determine an earth grid resistance value (usually an estimate based on the substation grid size and native soil resistivity measurements).

This resistance value, will let you determine the percentage of fault current which will go down your cable screens and earth conductors to remote earth grids and also what percentage will go directly in to the ground.

When you know the fault current, fault clearing time and percentage of this fault current which will enter the substation soil, you can then determine the step and touch voltages on the soil and structures within your facility and determine whether they are at a safe level.

I suggest you read IEEE 80, IEEE 81 and whatever BS or IEC documents which are relevant to your installation.

Good luck!


To answer your original question, you need to have a good (i.e. low resistance) connection to each piece of equipment so that the fault current is large enough for protection to operate (greater than load current). You also need a good earth connection to offer a return path for unbalanced currents and harmonics in the situation that no neutral is provided.

 

[ukgraduate]

Thank you for answering this.

I am currently reading IEEE 80 to gain a better understanding. Can you also tell me if the EPR can exceed the source voltage? As EPR = Ig X Rg. I am aware of Kirchofs Laws but I cant see anywhere where this is confirmed. And my belief is the EPR cannot be greater than the source voltage but the fact I have looked to confirm this and cant find it anywhere is making me alittle confused.

IEEE states

"The maximum electrical potential that a substation grounding grid may
attain relative to a distant grounding point assumed to be at the potential of remote earth. This voltage, GPR,
is equal to the maximum grid current times the grid resistance"

I have searched all day to find some documentation where it confirms thats the EPR cannot be greater than the source voltage. I.e greater than the LV side of the 415V transformer. I have only found this in forums.

Any suggestions?

 

[waross]

Your background must be electronics or communications.
In the real world cable screens are may not be suitable to carry fault currents. Installation practices for screens and protective earths are not the same. Screens are often insulated and left floating at the load end to inhibit circulating current losses.
"EPR = Ig X Rg" Yes but you have forgotten that Ig max is dependent on I=E/R Your reality check is that E/R can not drive enough I to push EPR = Ig X Rg above E source.
Please forgive my abrupt manner. I'm not picking on you. I'm just a crusty old fart with not much diplomacy.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[acog]

Bill,

The cable screens I am referring to are HV cable screens. These are good for fault currents of up to 10kA for 120mm2 3c Aluminium XLPE cables with copper screens.


ukgraduate,

When calculating the EPR or current distribution via kirchoffs laws, it is also important to understand that fault currents have a magnitude AND an angle.

EPR can be many times the source voltage under some circumstances.

 

[pwrtran]

acog,

There was a discussion a year ago on whether or not the GPR can exceed the source voltage and some people think it will. I explained why it cannot be true.

GPR or EPR is impossible higher than the source voltage, otherwise it against the Kirchoff's Voltage Law. Draw a fault equivalent impedance circuit diagram you will find out.

 

[7anoter4]

You are right pwrtran.
If Io=VL-L/sqrt(3)/sqrt((sum(Ri)+3*rground)^2+sum(Xi)^2)
then Io<VL-L/sqrt(3)/(3*rground)
GPR=3*Io*rground<3*VL-L/sqrt(3)/3/rground*rground<VL-L/sqrt(3)

 

[7anoter4]

I think this article:
"CURRENT REDUCTION FACTOR OF A COMPOUND CABLE LINE"

http://www.emo.org.tr/ekler/8895f0da0edf4da_ek.pdf

may be used in order to calculate Sf[as per IEEE-80] for a shielded medium voltage cable.