Justification for use of higher GPR value
Justification for use of higher GPR value
(OP)
Due to high resistivity value I am not able to reduce the grid resistance. Although my touch and step potentials are within limits still GPR value is higher than specified value of 5kV.
I need to convince my client that GPR of 7.8 kV is a safe value.
I have tried arguing on the point that as communication is though fiber optic cable hence this is not a problem.
I would appreciate it if someone can give more arguments backed by literature which I can cite.
I need to convince my client that GPR of 7.8 kV is a safe value.
I have tried arguing on the point that as communication is though fiber optic cable hence this is not a problem.
I would appreciate it if someone can give more arguments backed by literature which I can cite.






RE: Justification for use of higher GPR value
“The ground potential rise of a substation MAY BE on the order of some 5000 V, which may be transferred out to a nonfault location by a ground conductor (or metal pipe, rail, etc.) leaving the station.”
In my opinion, in IEEE 367/2012- IEEE Recommended Practice for Determining the Electric Power Station Ground Potential Rise and Induced Voltage from a Power Fault – Redline it could be this limitation, but I have not this standard with me.
RE: Justification for use of higher GPR value
You need to work out the geographical Zone Of Influence (ZOI) and the impact of this GPR on all surrounding properties.
You can have unacceptably high GPR 'carried out' of the substation on neutrals that are externally bonded to 'remote earth' at a neighboring property's electrical service entrance, for example.
Distribution lines 'transiting' the ZOI need not be interconnected to your substation to be impacted. For example, at some distance from your substation, the worst-case GPR may be 3000V w.r.t. 'true earth' . Imagine a multi-grounded 4-wire distribution line is passing through there. The neutral will tend to rise up towards local (~3000V) earth potential, whereas a distant part of that circuit's neutral will be much closer to 0 V. Current will result.
Nice pictures in Link
Reference also at Link