grounding grid in cellar below water table

[mosesnbklyn]

we have a cogen plant, GSU, and MV switchgear/GCB in the cellar of a new commercial/institutional building. we dont have the soil resistivity yet, but the equipment elevation is below the water table. how does this affect the ground grid design?

 

[7anoter4]

Since the grounding grid made of copper [or copper clad or else] is located somewhere at 0.5-0.7 m depth I don't think it will be so important. You may consult- at informative level-
IEEE Std 142-2007 Table 4-3—Effect of moisture content on soil resistivity and:
IEEE Std 80-2000 ch.12.4 Effect of moisture, temperature and chemical contains.
If it could be a danger of flooding see also NEC Art.555.

 

[mosesnbklyn]

I am familiar with outside plant grounding grid design which IEEE addresses. Inside plants get complex. I am assuming a reasonably low soil resistance (~25ohm-m) and high moisture content.

We use ETAP for grounding grid calculations, we have successfully modeled a particular design based on these assumptions at a depth of ~4ft below the slab, but this translates to ~30ft below grade, and most likely below the water table.

If at a particular season, the ground grid below the slab is completely submerged, how does this change the performance? does it become an equipotential mesh or is the depth now reduced by the high water table and higher hazards (step/touch potentials) at the walking surface?



M. Nissen,P.E.
Senior Electrical Engineer
Waldron Engineering & Construction

 

[7anoter4]

I don't know how is built the cellar floor. Usually it is a water sealing layer of polyethylene under the slab[vapor barrier] and this will insulate the floor-isolate it from the earth. I think it does not matter if the earth will be more moisturized. The building foundation grounding[Ufer] is formed by outside part of wall foundation and the column. However, this is not the "Effective Ground-Fault Current Path" as stated by NEC but only a grounding electrode. The Effective Ground-Fault Current Path has to be the grounding jumper connected to SSBJ –from utility provided along the supply cables.
Art.250.52 Grounding Electrodes. (3) Concrete-Encased Electrode. states how to provide a
Foundation grounding electrode.

 

[mosesnbklyn]

We have a 13.8kV switchgear located 11 stories above grade. When we run grounding conductors to down to the cellar ground grid in question, the "company standard" is to use bare CU cable in PVC. I am concerned that the GPR (even in rare instances) could damage the cable, conduit, or any other metallic structures due to high GPR (calculated at 22kV with varying assumptions).

Not only would I prefer insulated cable, I was even debating using 5/15kV jumper cable based on the BIL ratings. I did consider that PVC has a high melting point, and that it is a good insulator. The problem is the use of it in electrical rooms and plant areas that could be subject to mechanical damage by the various trades and operators over time.

M. Nissen,P.E.
Senior Electrical Engineer
Waldron Engineering & Construction

M. Nissen,P.E.
Senior Electrical Engineer
Waldron Engineering & Construction

 

[7anoter4]

If I well undestood the 13.8 kV switchgear is directly supplied from the Utility.
Usually 13.8 kV side of the supply transformer is delta connected so no single line- to- ground fault is possible.Only line-to-ground-to line it is possible.In this case if one phase conductor on Utility location will be grounded and another phase will be grounded in the switchgear then in the case of a line-to-ground-to-line fault , zero sequence fault current is:
Io=E*(R2+jX2)/(R1+jX1)/[(Ro+R2+3*Rf+j*(Xo+X2))]/[( R2+jX2)*(Ro+3*Rf+jXo)]
IEEE 80/2000 Formula (66) [Rf=Rg+Rgroundcable]
If R2=R1 X1=X2 [as “far from generator”] then Io=E/[(Ro+R2+3*Rf+j*(Xo+X2)+ Ro+3*Rf+jXo)]
If we shall put R2=R1=R0=0, X1=X2=Xo=0 then Io=E/[(3*Rf)+(3*Rf)]=E/(6*Rf).
Since R1,R2,Ro,X1,X2,Xo >0 obviously Io<E/(6*Rf) and GPR<E*Rg/(2*Rf)<E/2.
However, in the case of a single line-to-ground fault , zero sequence fault current is:
Io=E/(3*Rf+R1+R2+Ro+j*(X1+X2+Xo)) IEEE 80/2000 Formula (67)
It is obvious Io<E/3/Rf
GPR=3*Io*Rg<E/((Rg+Rgroundcable))*Rg<E
So, if the maximum voltage to ground will be 13.8/sqrt(3)=~8 kV GPR could not be more than 8 kV.
In order to protect the grounding cable you may use a pvc rigid conduit [even schedule 80] or a galvanized rigid steel conduit but in this case you have to bound it- with listed accessory-at both ends to the grounding conductor. The same proceeding with the insulation shield.

 

[mosesnbklyn]

7anoter4 - thanks for the manual calcs; I have to disagree, the GPR is dictated not by system voltage; it is almost purely fault current x impedance to ground. the R and X values are not zero, we calculated the Rgrid=1-2ohms.

thanks for confirming the use of PVC conduit for bare grounds, we could build a soffet around them..

I guess my original question stands; will the GPR be experienced on the riser cable to the upper level?



M. Nissen,P.E.
Senior Electrical Engineer
Waldron Engineering & Construction

 

[7anoter4]

First of all I agree with you the GPR=IG*Rg :
“GPR is equal to the maximum grid current times the grid resistance”. Definitions 3.14.
Second, what I said GPR could be not more then E [line-to-ground potential].
GPR actually cannot be measured and practical only step and touch potential are important and it may be feel.
As NEC requires an “Effective Ground-Fault Current Path” and the grounding electrode is not considered such a “path”, but a grounding conductor has to be connected between switchgear grounding bus and Utility grounding bus. A great part of the fault current will return to Utility through this conductor so GPR will be still less.
Since the fault current will flow through the grounding cable from switchgear grounding bus up to grounding grid [or grounding electrode] another potential drop will be added to GPR. If all metallic part which could be energized in case of a grounding fault is solid connected to the lower grounding grid the maximum touch potential will be this added potential drop.