ground grid design with 3000 ohm-m soil resistivity 1

[JLuc]

Hi all,

i'm working on the preliminary design for an outside substation ground grid. (underground incoming cables + 4 feeders and a capacitor bank). Metal-enclosed switchgear on a concrete slab.

I received the soil resistivity from the civil contractor and it is 3000 ohm-m !!

so I started playing with the number using IEEE 80-2000.

With an area of 2500 pi^2, and , 500 pi of buried conductor, I get a grid resistance of about 100 ohm!, which is way above the recommended 5 ohm. But it is almost impossible to go lower than that with a 3000 ohm-m soil resistivity.

The fault current is 2675A and L-N voltage is 14.4 kV.

What I understand is that with a grid resistance that high, almost 100% of the fault current will go back to the source through the shield of the incoming underground cables.

So my question is , what current do I use to calculate the step and mesh voltages (or what is the value of the split factor)...

Also, what happens with GPR? I know it cannot go higher than L-N voltage, but it will probably be way above the recommended 5000 V.

Thanks in advance for your comments and answers about ground grid design with very high soil resistivity.

JL

 

[joralmova]

I Think that resistivity is not correct. The first that you must be is reduce the resistivity because is very great. I send you file in exel for calculate the step and mesh voltages.

 

[JLuc]

Yes, I will investigate the soil resistivity value that they gave me. I also think that it is very high.

Thank you.

 

[JLuc]

I don't know if this will be possible in practice but , by enlarging the grid to 10 000 pi^2 and 1080 pi of buried conductor, the grid resistance is reduced to about 50 ohms.

This helps for the mesh and step voltages, but i'm still confuse about the high GPR and how to choose the correct split factor.

JL

 

[jghrist]

Put a perimeter ground loop around the concrete slab with a couple ground rods. Is there any particular reason that you need a low resistance? There's no way that you will get a low resistance in that soil, and you probably don't need it. The way to get an accurate current split factor is to use software such as CDEGS that models the mutual impedances involved. You will find as you surmised that most of the current returns in the cable shields/neutrals.

 

[Dumbo2929]

Joralmova - would you happen to have your spreadsheet in English. I would like to use it to verify another calculation. thanks

 

[ELEP]

The soil resistivity seems more, perhaps they might have measured on crushed rock or it might be dry soil. I suggest you to use soil resistivity enhancing material, like bentonovite powder or equivalent available in the market but it is bit costly.

 

[joralmova]

Dumbo2929 - if you want know that have the spreadsheet, you have to read the IEEE Std 80-2000 because the formulas that spreadsheet have is there.

 

[7anoter4]

Hi JLuc
I agree with jghrist a few grounding rods will improve the grid resistance.
And as I think ,16 grounding rods of 3/4" dia and 50 ft long may reduce the grounding grid to 48 ohms .
Also I think the fault current of 2675 A is only the three phase short-circuit current.
So the reactance Xk3=14.4/2.675= 5.4 ohms
The actual grounding current will be: 3*14.4/sqrt((2*Xk3)^2+48^2)=0.89 KA
The GPR will be 45.5 Kv and Touch potential 460 % of admissible.
So, is not in your benefit to reduce the ground grid resistance.
The medium voltage cable shield [25 kv rated] has a maximum impedance of 1.2 ohm/km. For 10 km will be 12 ohm.
If the Utility agreed to ground both end of the incoming cables it is possible to convey
a minimum current of 89% of this grounding current back to utility substation.
Never-the- less the grounding resistance should be 95.3 ohm and GPR 5.09 KV the Touch potential and the step potential would be tolerable.
A grounding cable buried 3 feet outside of the fence is recommended.
For example see attached exl file.
Best regards

 

[7anoter4]

Hi JLuc
I am sory! The xls file was not the file you need.
So see attached new file.
The grounding resistance has to be combined[parallel] with the incoming cable shield and to be multiply by 3 in the grounding current formula.
regards

 

[7anoter4]

Hi JLuc
As I converted a VisualBasic 6 into xls there was a lot of symbols to change.
I have to apologize for the inconvenient .
I think this will be the last one.
I change also the fault clearing time from 0.5 to 0.15 sec.
Regards

 

[waross]

I saw an instance within the last year where the entire substation site was covered to a depth of 1 meter with soil of lower resistivity to improve grounding.
respectfully

 

[JLuc]

Hi all,

I did my own excel file to calculate mesh and step voltage for the grid.(based on IEEE 80-2000)

I used 5% current split factor. The geometric design is the one proposed by jghrist. As you can see, i'm still borderline with the mesh voltage compared to the max touch voltage.

Two things that are not helping: 3000 ohm-m soil resistivity AND 0.6 sec clearing time.

The clearing time I use is the utility clearing time (worst case if the main breaker fails to open). Am I right to do so?

Anyone have any ideas to bring the mesh voltage below the max touch voltage? I don't really want to add much copper in the ground. What would you do in this situation?

Thanks for your help!

JL

 

[jghrist]

If all equipment is in a metal enclosure on a concrete slab, and you put a ground wire around the perimeter of the concrete slab, what is in the center of the meshes? Your total area (17x27 meters) is much larger than the probable size of metal-enclosed switchgear. Is there anything grounded to touch when standing in the center of the mesh? A reinforced concrete slab will be almost an equipotential surface.

How did you calculate the split factor. If this was changed from 5% to 3.9%, your mesh voltage would match the allowable touch voltage.

 

[JLuc]

No, in fact, there is a 15X25 meter fence around the switchgear and a ground wire (1 meter away) tapped on the fence (17x27 ground grid). The only grounded metal parts are the fence and the switchgear itself. So this means that anytime you touch a metal part, your either on the concrete slab (in the switchgear) or directly over a ground wire.

Is that mean that my design is Ok?

I would like to have a quick explanation on how to interpret mesh voltage in relation with max touch voltage, and when it can become dangerous.

Thank you.

JL

 

[jghrist]

Practically speaking, if you have a perimeter wire both on the inside and outside of the fence and a perimeter wire around the switchgear, you will probably not have touch voltage problems. Unfortunately, there is no way to prove this using the simplified IEEE Std 80 formulas for touch voltage. You could with a good software package such as SES CEDG or WinIGS (Advanced Grounding Concepts).

During a fault, the ground grid rises in voltage with respect to remote earth. The soil around the ground grid rises also, but not as much as the grid. This leads to voltage differences between the soil surface and things connected to the grid. Touch voltage is the difference between the soil surface where you stand and the grid voltage. This assumes that you can actually touch something with your hand from where you are standing that is connected to the grid.

The mesh voltage is the grid voltage minus the lowest surface voltage within the mesh. This is the touch voltage if you stand at the point where this occurs and touch something connected to the ground grid. This location is generally near the middle of the mesh.

Because the difference between the voltage in the soil and the grid voltage gets higher as you move away from the grid, the touch voltage is lowest directly over a grid wire. This is the surface point that is closest to the grid wire. There is still some touch voltage directly over the grid wire, but probably not more than the tolerable touch voltage.

The simplified IEEE Std 80 equations calculate the mesh voltage, not the voltage above the grid wire.