Hi there everyone!
I'm currently doing a module titled '' Electrical Systems Protection '' and I'm just a little bit unsure of whether or not I'm correct with one question. I'll post the question and my attempt so far and maybe someone out there smarter than I could say yay or nay to to my approach. There is no diagram associated with the question.
Here's the question:
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.
Here's my attempt:
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 ( the original resistance value ) 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
I'm currently doing a module titled '' Electrical Systems Protection '' and I'm just a little bit unsure of whether or not I'm correct with one question. I'll post the question and my attempt so far and maybe someone out there smarter than I could say yay or nay to to my approach. There is no diagram associated with the question.
Here's the question:
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.
Here's my attempt:
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 ( the original resistance value ) 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