Is there a way around High GPR

[tctctraining]

Claculation shows that GPR is 11000V , very high fault current which is around 28000A.
Step and touch potentials are fine and within limits. would this cause a serious damage to the equipments?

 

[jghrist]

The GPR can cause problems if you have any cables entering the facility that are grounded remotely, like telephone cables.

 

[tctctraining]

Is that the only concern?
What if we use fiber optics for communications?

 

[tctctraining]

I also think when distribution line neutral wire or transmission line sky wire are connected to grid, high GPR would increase the voltage on those lines as well and all insulating equipments on Neutral line will be exposed to this high voltage.

 

[jghrist]

Also, you can have high transferred potential to equipment that is connected to the facility ground. See IEEE std 80 and

http://www.groundperfect.com/transfer.htm

 

[dpc]

Maybe some additional info here:

http://www.positronpower.com/en/

The telco will be concerned about GPR as a safety issue.

Use of fiber for data and phone lines can certainly reduce some of the concerns.

David Castor

http://www.cvoes.com

 

[tctctraining]

I just looked up some info and inspection authorities are asking for " Zone of Influence" !!!
any idea how this can be calculated?

 

[jghrist]

This is the area where the earth surface voltage is greater than a particular value, usually 300 volts. Earth surface voltage decreases with distance away from the grounding system. I have calculated the zone of influence using grounding analysis software and making a contour plot of earth surface voltage, with 300 volts as one contour.

If you have a GPR of 11,000 volts, your zone of influence probably covers the whole county.;-)

Are you sure that you are accounting for ground fault return current flowing in neutrals or shield wires instead of through the earth? If your grid resistance is 0.4 ohms and half of the fault current returns to the source through a metallic neutral conductor, then the GPR is only (1-0.5)·28000·0.4 = 5,600 volts.

 

[tctctraining]

There are two distribution line coming to the sub and the software calculates the overall impedance. I assumed a typical value of 100 ohm for tower footing resistance-to-ground and this didn't affect the overall impedance too much.It just brought it down from 0.4 to 0.37.
"
Are you sure that you are accounting for ground fault return current flowing in neutrals or shield wires instead of through the earth? If your grid resistance is 0.4 ohms and half of the fault current returns to the source through a metallic neutral conductor, then the GPR is only (1-0.5)·28000·0.4 = 5,600 volts."

CYMGRD doesnt give me the option of choosing how the return current goes back to source!!

 

[WindTurbinesAreFine]

CYMGRD will not calculate the current split factor? You can use the tables in IEEE-80-2000 to estimate the split factor based on some assumptions.

This will lower you GPR as jghrist has described, but each site is different and should be calculated.

 

[jghrist]

If CYMGRD calculates the overall impedance to remote earth (two neutrals and earth path in parallel), then you would use the total ground fault current to calculate GPR. This isn't the way IEEE Std 80 calculates GPR. I'd be curious to find out what method CYMGRD uses to calculate the overall impedance. What is the resistance of the ground grid by itself?

According to the CYMGRD brochure, it computes Rg and

Current Split Factor (SF) estimated from substation configuration data as per IEEE Std. 80-2000.

 

[dpc]

See IEEE Std 367 for guidance on determination of ZOI.



David Castor

http://www.cvoes.com