Finding the value of a neutral earthing resistor

  • Thread starter Neil Hayes
  • Start date
  • #1
15
2

Homework Statement



An 11 kV motor is fed by cables from a transformer via switchgear,
having a phase impedance of 0.3 + j0.3 ohm. The earth return path to
the transformer neutral has a resistance of 0.42 ohm. Determine a
suitable value of neutral resistance if the voltage rise at the motor in the
event of an earth fault at the motor is not to exceed 430 V. Neglect
transformer winding resistance.


Homework Equations




The Attempt at a Solution



Hi there everyone. This is a question that I'm working on at the moment. My attempt so far is as follows. Maybe someone can tell me if I'm on the right path?
[/B]
Well using the information given the fault current would be 6350/.42 = 15121 amps

The 6350 here is the phase voltage.

To meet the requirement in the question the current would need to be reduced to
430/.42 = 1023.8 amps

Using an earth path impedance of 6350/1023.8 would give 6.20 ohms

What I'm thinking of doing next is R = the square root of Z² - X²

This gives me 6.19 ohms and we already have .3 ohms in the circuit so we would need to add a resistor of value 6.16 ohms.

Would this be correct?

Any help would be greatly appreciated!

Neil
 
Last edited:

Answers and Replies

  • #2
15
2
Hello again everyone!

I'm actually just looking at the question there again and think that I made a mistake. I thought that the .42 ohms was the impedance but I think that the impedance from the transformer to the motor is Z = 0.3 + j0.3 which is .42 but I think that the earth return path to the transformer neutral is also .42 ohms which would give me an earth fault loop impedance of Z = 0.72 + j0.3.

Does that make more sense I wonder?
 

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