How to find a suitable value for a neutral resistance?

[Neil Hayes]

Homework Statement

.
[/B]
An 11kv motor is fed by cables from a transformer via switchgrear, having a phase impedence 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

Hi there folks!

I would be very grateful for some suggestions in relation to the question above. I have asked my tutor for information on what exactly is being asked. The course material is vague and this was my best attempt at solving the problem. When I sent the attached pdf to my tutor they just replied :

Neutral resistor = Resistance path - 0.72

To me this seems too easy and also doesn't tie in with the information given in the associated lessons etc.

Maybe I'm on the right path but I'm just a bit unsure.

I would really appreciate any ideas :)

Neil

 

[magoo]

I think you need to start with a diagram showing the impedances.

A transformer on the 11 kV side is normally a wye (or star) connection.

The neutral resistor is connected from the neutral of the transformer to ground (earth).

Write the equations for the fault current with the neutral resistor and then for the voltage rise at the motor.

I'm not sure what you 430/0.72 represents. Did you mean 0.78 instead?

 

[Neil Hayes]

I think you need to start with a diagram showing the impedances.

A transformer on the 11 kV side is normally a wye (or star) connection.

The neutral resistor is connected from the neutral of the transformer to ground (earth).

Write the equations for the fault current with the neutral resistor and then for the voltage rise at the motor.

I'm not sure what you 430/0.72 represents. Did you mean 0.78 instead?

Hi there Magoo,

Thanks a million for the reply to my post :)

There is an example in my lesson notes and if I was to follow the process it would go as follows ( using the values given in the question ).

Impedance of cables feeding the motor and the Earth return path to the transformer is :

Z = 0.72 + j0.3 which gives us 0.78 ohms

In the event of an Earth fault at the motor given this Impedance value we will have a Fault Current of 6350/.78 = 8141 amps

The cables that feed the motor from the transformer have an Impedance of .424 ohms

We need our fault current to be reduced to 430/.424 = 1014.15 amps

So ...

6350/1014.15 gives us a resistance or 6.26 ohms

What I'm thinking I need to do now is take our original 0.78 ohms which is the Impedance of the original system and take that from our newly found figure of 6.26 ohms to give us a value of 5.48 ohms.

A neutral earthing resistor with a value of 5.48 ohms needs to be added to the system.

This is what the information in the lesson would lead me to think anyway.

Does this make sense?

Neil

 

[magoo]

Where did you get the value of 0.424? You need to show your work.

Since you didn't sketch a one-line diagram, see if this agrees with what you had in mind:

 

[Neil Hayes]

Where did you get the value of 0.424? You need to show your work.

Since you didn't sketch a one-line diagram, see if this agrees with what you had in mind:

View attachment 134356

Hi again Magoo,

Apologies for not including a drawing. I will hand draw out one tomorrow and post it on the thread here. I need to look at my notes to see if your drawing is similar to what I have.

The 0.424 ohms from above was the impedance of the cables from the transformer to the motor. The question tells me that this section has an Impedance of :

Z = 0.3 + j0.3. The square root of 0.3^2 + 0.3^2 is equal to 0.424. If we have a current of 1014.15 then the voltage rise at the motor will be R x I ( 0.424 x 1014.15 = 430 volts )

 

[magoo]

Your 0.424 ohms represents the cable impedance only. Under a ground fault, the fault path includes the cable impedance plus the Earth return. It looks like you neglected that.

It also looks like you neglected the reactance when you came up with 6.26 ohms.

 

[Neil Hayes]

Your 0.424 ohms represents the cable impedance only. Under a ground fault, the fault path includes the cable impedance plus the Earth return. It looks like you neglected that.

It also looks like you neglected the reactance when you came up with 6.26 ohms.

Please see attached photo for the impedance diagram which shows the system prior to the addition of the neutral earthing resistor.

The figures that I've written out are relate to fault conditions that are set out in the question.

On the right hand side I have 3419 V which is the potential difference from the motor area and the transformer. In order to reduce this potential difference to the 430 V asked for in the question I calculated that we need to reduce the current to 1024 amps (6350/.42).

We have a phase voltage of 6350 which I'm dividing by the 1024 and this gives me 6.2 ohms.

This 6.2 ohms will be the Impedance of the system with the neutral earthing resistor included so then to find the resistance value I then go down this route.

Z^2 = R^2 + X^2

6.2^2 = R^2 + 0.3^2

R^2 = 38.35

R = 6.19 ohms

From our original system we had a resistance of 0.72 ohms so I believe the value for the neutral earthing resistor is the 6.19 - 0.72 = 5.47 ohms

The information in is both vague and limited at times in the notes I have for the distance learning course which I'm doing.

The process I've used above is the same as that which I have been shown in my notes. The problem with the course notes is that it does the problem and gives the the final answer as an earth path impedance. I have found that to be 6.2 ohms ( using the method in my notes ) but as the questions specifically asks for a '' suitable value of neutral resistance ''

My apologies for the delay in replying to your last reply. I'm having trouble with my Laptop :( The appreciate you taking the time to try and point me in the right direction.

Neil

 

[Neil Hayes]

Hi there Magoo,

It seems that my last post seems to be the correct way of going about the problem. I won't know for sure as I've to submit my work tomorrow but I'll let you know. Thanks again for your help to date!

Neil

 

[magoo]

w/o the neutral resistor

Zphase = 0.3 + j 0.3

Zearth return = 0.42

For a phase to Earth fault, Z = 0.72 + j 0.3 |Z| = 0.78 ohms
Isc = V / Z = 6350 / 0.78 = 8141 A
Vacross motor = 6350 V
With neutral resistor

Want Vacross motor = 430

Need to reduce the sc current to

430/6350 * 6350/0.78 = 551.3 A
To do so requires an impedance of 11.52 ohms
You already have 0.72 + j 0.3 or 0.78 ohms
Need to solve for R that results in an impedance of 11.52 ohms

I just took 0.72 + R + j 0.3 and tried a few values for R that results in a Z of 11.52
I get an R of around 11 ohmsYou could also just neglect the X term and it will be a bit easier.

 

[magoo]

Neil,

Good luck tomorrow and please post what you find out.

 

[Neil Hayes]

Neil,

Good luck tomorrow and please post what you find out.

Will do and thanks again for your time to date :)

 

[Neil Hayes]

Will do and thanks again for your time to date :)

Hi there Magoo,

I just got my results back from the assignment. I submitted my effort on post no 8 and got it correct. I got a distinction on the assignment so I'm well chuffed. Only another 100 to go lol ( well it fell like 100 anyway! )

Thanks for taking the time to work with me on the site here!

Neil

 

[magoo]

That is terrific! Your perseverance paid off.

 

[Neil Hayes]

That is terrific! Your perseverance paid off.

It sure did. Thanks again