5 ohms Ground Resistance in Commercial Buildings

[ThePunisher]

Hi all,

Ground Potential Rise (GPR) = Ig * Rg

For 500 kVA, 13800-480/277V, Dyn effectively grounded at Z=5%, Ig = 12kA taking an infinite primary source.

The contractors are providing a loop grounding grid buried around the building and aiming to obtain 5 ohms maximum.

Based on 5 ohms, GPR = 12,000 * 5 = 60 kV

This is quite very high.

IEEE 80 provides formula to calculate the tolerable touch and step potentials....however, the design GPR is very high. When asked, the contractor informed us that it has been their practice to get Rg = 5 ohms and call it a day.

The Code is even specifying for a higher one ( I think 25 ohms).

Is this a valid design application? Is the GPR neglected (regardless of Ig) in the building design?

 

[davidbeach]

What do you have that can drive a GPR of 60kV? You don't have a source that can produce 60kV. You can get, supposedly, 12kA for a bolted fault and then you add in 5 ohms resistance and still get 12kA? New laws of electricity?

 

[ThePunisher]

Thanks David for the prompt response.

From the rule of the thumb premise that the bolted line to ground fault at the transformer secondary terminals = bolted 3 phase fault, this is where I took 12 kA as a line-to-ground fault. The total ground resistance Rg = 5 ohms.

I assume that the 12 kA line-to-ground fault is injected into the ground grid loop having a total Rg = 5 ohms.

Are you implying that I should treat the ground grid resistance as an impedance that limits the line-to-ground fault current?

 

[X49]

You are assuming that the entire 12kA fault current will use the ground grid as a return path. This will not happen. The 12kA fault current will return to the transformer on the neutral conductor, which should be terminated to ground at the service entrance.

Very little fault current will flow into the ground grid. The current that does will be no more than 277V/5ohm = 55A.

 

[dpc]

Look at it this way: 12,000 A x 5 ohms = 60,000 volts. Obviously, that can't happen, since you only have 277 V driving voltage. Ground fault current must flow in a complete path from source transformer back to the transformer neutral. 12,000 A a maximum fault current level based on a bolted fault - zero fault impedance. If you assume 5 ohms fault impedance, then the actual fault current will be quite a bit less.

 

[ThePunisher]

Thank you X49, dpc for your responses.

The neutral of the transformer secondary is directly connected to the ground grid loop. Hence, if there is a local line to ground fault at the transformer secondary terminals (or near to it). The line-ground fault current will get injected first to the ground and up to the transformer neutral conductor.

If there is a line-to-ground fault at the transformer secondary terminals, then the only limiting impedance would be the transfomer winding impedance + neutral conductor impedance (which is relatively short). I agree that the fault current will not be as high as 12 kA owing to the positive + negative + zero sequence impedances.

But looking into IEEE 80, the allowable touch and step potentials are calculated based on 50 or 70 kg person with soil/surface resistivities included. Then the GPR is compared against them. In the GPR calculations, the line-ground fault Ig is calculated based on the three sequence impedances. Since our LV system is having a neutral which is effectivley grounded, the Rg = 5 to 25 ohms seems to be high with respect to what the GPR and the line-ground current would be. I apologize for insisting for now and hope I get clarified even further as I am following IEEE 80 at the moment and looking at the building for applicability.

Thank you for your patience.

 

[jghrist]

The 5 ohm requirement probably comes from the Motorola R56 manual which is commonly used to determine site grounding requirements to help limit voltage surges caused by lightning.

 

[ScottyUK]

The absolute worst-case GPR would, depending upon the source configuration, be limited to 13.8kV which is the largest voltage present in the system you describe. For any practical installation it will be considerably less than this.

The 12kA LV fault level isn't of huge concern for the GPR, you need to consider the available fault from the HV system. If you have an impedance-earthed system then the GPR is likely to be quite low. Determine your available fault current and apply Ohm's law to see what happens to the GPR. Remember that you can't have GPR greater than the source voltage. With a typical resistance-earthed system of, say, 400A your GPR would be limited to 2000V through a 5[Ω] resistance IF the whole fault current flowed to earth and none returned via the cable shield or armour.

 

[odlanor]

ThePunisher ,
Ground Potential Rise (GPR) = Ig * Rg <== OK
Your interpretation of Ig is wrong.
See examples attached and which case is applied in your system