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I am looking at the results for an earth grid study for a 50Hz substation.

  • Primary Voltage (Vp) = 11,000V
  • Secondary Voltage (Vs) = 415V
  • Secondary Fault Current (If)= 24kA
  • Secondary Grid Current (Ig)= 2.4kA
  • Ground Resistance (Rg) = 1.7 Ohm
  • Calculated Ground Potential Rise (GPR) = 4080V

I understand that GPR = Ig x Rg, but

What is the relationship between Vp/Vs and GPR? If 4000+V is available at the earth grid, wouldn't current flow back up the faulted conductor to the source?

Is it related to different reference points?

Appreciate it anyone could help me get my head around this one?

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This looks like the specifications for a 1 MVA transformer (2.4kA*415V) with a Zo of 10% so the Short Circuit current If = 2.4k/10%

Vp/Vs=25.1 and secondary , if shorted to earth ground with If =24kA max and a GRP rise of 4kV max until the fault is tripped. Vp would also be attenuated while current rises 10x on the primary as the surge goes from ground to grid source.

When tripped there will also be some expected transient voltage. (V=LdI/dt) But I'm guessing lightning arrestors may clamp this.

schematic

simulate this circuit – Schematic created using CircuitLab

I am thinking of a grid with only 2 sources with a local fault and remote feeders.

There is a better way to show this. But 1st cut above.

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  • \$\begingroup\$ Hi Sunnyskyguy, thanks for the response but are you able to address the questions I raised? How does current flow from the fault into the ground if the voltage in the ground is higher than the source of the fault? \$\endgroup\$ – Questioneer Jan 8 at 11:32

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