# Finding the value of a neutral earthing resistor

#### Neil Hayes

Joined May 23, 2015
14
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

#### Neil Hayes

Joined May 23, 2015
14
Hi again!

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?

#### MrAl

Joined Jun 17, 2014
8,976
Hello,

It sounds like the motor impedance is Zm=0.3+0.3*j and the neutral line resistance Rn is 0.42 ohms.
Now that means that the total impedance without any change is Zm+Rn=0.72+0.3*j.
The total impedance means that the Zm part will have a certain voltage Vm across it with no changes.
The change comes in when the motor neutral is grounded, which removes the Rn so the total impedance is back to Zm and that has the total line voltage VL across it now.

So the total change in voltage would be Vc=VL-Vm. That voltage CHANGE will likely be too high so they want you to add a resistance right to the motor neutral that will limit that voltage change (Vc) to 430v if the other end of the resistor was grounded.
So you have to figure out how to get an increase (at the motor itself) of 430v by adding a resistor in series with the motor neutral. You have to calculate the value of the resistor.
This would involve finding the value of a resistor that when placed in parallel to the preexisting line resistance (that goes to ground also) causes a increase of 430v max.

It's always better to draw the schematic because visual inspection often leads to simple answers.

That is what it sounds like to me anyway. Does that help?

• Neil Hayes

#### Neil Hayes

Joined May 23, 2015
14
Hello,

It sounds like the motor impedance is Zm=0.3+0.3*j and the neutral line resistance Rn is 0.42 ohms.
Now that means that the total impedance without any change is Zm+Rn=0.72+0.3*j.
The total impedance means that the Zm part will have a certain voltage Vm across it with no changes.
The change comes in when the motor neutral is grounded, which removes the Rn so the total impedance is back to Zm and that has the total line voltage VL across it now.

So the total change in voltage would be Vc=VL-Vm. That voltage CHANGE will likely be too high so they want you to add a resistance right to the motor neutral that will limit that voltage change (Vc) to 430v if the other end of the resistor was grounded.
So you have to figure out how to get an increase (at the motor itself) of 430v by adding a resistor in series with the motor neutral. You have to calculate the value of the resistor.
This would involve finding the value of a resistor that when placed in parallel to the preexisting line resistance (that goes to ground also) causes a increase of 430v max.

It's always better to draw the schematic because visual inspection often leads to simple answers.

That is what it sounds like to me anyway. Does that help?

Hey there MrAI,

Thanks for taking the time to answer my question there. What you wrote makes sense alright. I'll have a look at my work there now and give the question a final bash and submit my assignment to my tutor.

Fingers crossed lol

Neil

#### MrAl

Joined Jun 17, 2014
8,976
Hi,

Yeah It surprises me that they dont allow more instructor and student interaction. In the real world if there is an important quiestion you can usually contact someone to get more information, such as the client.
Some correspondence courses are a little more flaky though 