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Topic Title: Earth fault Current - Current Transformer
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Created On: 06 June 2013 03:10 PM
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 06 June 2013 03:10 PM
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jackdaniel

Posts: 69
Joined: 29 November 2012

Hi, Im confused how the relay calculate the unbalanced current, for the 3CT configuration for each phase, the earth fault relay will operate if the unbalance current exist, let say we have 3 phase unbalance current, phase A is 5A, phase B=0 and phase C=0, because only phase A is connected to the load, the unbalanced current exist, but there is no fault current, will the earth fault relay operate?
 06 June 2013 08:38 PM
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Zuiko

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Joined: 14 September 2010

Say you have unit protection, then CTs (or VTs or both) at each end of the circuit are compared; and if there is a difference the relay will operate. (this is Kirchoffs law, if there is a difference, the current must have gone somewhere, and that somewhere must be to earth).



Zero sequence discrimination would involve a CT between the transformer star point and earth. This will detect any earth fault current and trip the circuit.

Edited: 09 June 2013 at 10:14 AM by Zuiko
 06 June 2013 09:03 PM
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dlane

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Joined: 28 September 2007

Remember you also have to take into consideration the ratio of the CT to find the current going to the relay.

For example where I have circuits with 3CTs connected together to form the earth protection, they have a ratio of 1200/5A. If the relay is set to operate at 5A then I would actually need 1200A flowing in one of the phases which is an extremely large imbalance to have.

Kind regards

Donald Lane
 07 June 2013 11:10 AM
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jackdaniel

Posts: 69
Joined: 29 November 2012

Originally posted by: Zuiko

An earth fault relay will not operated if there is unbalanced current as you describe.



Say you have unit protection, then CTs (or VTs or both) at each end of the circuit are compared; and if there is a difference the relay will operate. (this is Kirchoffs law, if there is a difference, the current must have gone somewhere, and that somewhere must be to earth).







Zero sequence discrimination would involve a CT between the transformer star point and earth. This will detect any earth fault current and trip the circuit.



Zuico,

I guess u were explaining differential relay, which is 2 relays are install in 1 phase, i was asking about ground fault relay,..each phase only has a CT, its operate as OC and earth fault protection..hope u can help me to explain...
 07 June 2013 11:48 AM
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Zuiko

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Joined: 14 September 2010

you might be thinking of an IDMT relay;

here the relay trips a breaker on overcurrent; the time it takes to trip is dependent on the magnitude of the current; there are characteristic curves associated with different types of circuit such as inverse, and extremely inverse.

An earth fault will result in an overcurrent for a period of time. The greater the fault the quicker the trip.

Some long distance over head lines will have sensitive earth fault which is programmed to trip for smaller overcurrents that last a long time - due to the high impedence of the fault.

cheers
 09 June 2013 01:51 AM
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simongallagher

Posts: 148
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Hi,

I think I know what you are asking. If there are only three CTs on the phase conductors, how can a relay detect current flowing to earth?

If the situation you describe existed (5 amp on the red, 0 on the yellow and blue)' then there has to be an earth fault, becaus those 5 amps have to be flowing back somewhere.

On a three phase system operating normally, the current in the phase conductors will sum to zero, no matter how unbalanced. If some is flowing back through earth, then the current will no longer sum to zero.

Simon
 09 June 2013 07:36 AM
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jackdaniel

Posts: 69
Joined: 29 November 2012

Originally posted by: simongallagher

Hi,



I think I know what you are asking. If there are only three CTs on the phase conductors, how can a relay detect current flowing to earth?



If the situation you describe existed (5 amp on the red, 0 on the yellow and blue)' then there has to be an earth fault, becaus those 5 amps have to be flowing back somewhere.



On a three phase system operating normally, the current in the phase conductors will sum to zero, no matter how unbalanced. If some is flowing back through earth, then the current will no longer sum to zero.



Simon



Hi Simon,

Understood about that, but how the relay calculate in order to detect unbalance?, there must be some sort of calculation?
 09 June 2013 10:13 AM
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Zuiko

Posts: 521
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Consider three CTs, one around each phase conductor.

The CTs are connected to the primary winding of a summation transformer; the secondary winding giving a single phase output.

The primary is wound as such:

The CT output from L1 (lets say red) is connected across the the full winding. If we give the winding 5 turns then the output from the primary is proportional to 5 X Ired (i.e 5 times the output from the red CT)

L2 CT outputs (yellow phase) are connected across 4 turns of the primary winding.

L3 CT outputs (blue phase) are connected across 3 turns of the primary winding.

The output of the secondary winding will be dependent on the current flowing through the primary and from this we can see what sort of fault is occuring.

for example:

5I is a red-earth fault

4I is a yellow-earth fault

3I is a blue-earth fault

2I is a red-blue fault

I is a red-yellow or a yellow-blue fault

sqrt3I is a three-phase or three-phase to earth fault.


cheers

W

Edited: 09 June 2013 at 12:28 PM by Zuiko
 10 June 2013 02:44 PM
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jackdaniel

Posts: 69
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Originally posted by: Zuiko

Consider three CTs, one around each phase conductor.



The CTs are connected to the primary winding of a summation transformer; the secondary winding giving a single phase output.



The primary is wound as such:



The CT output from L1 (lets say red) is connected across the the full winding. If we give the winding 5 turns then the output from the primary is proportional to 5 X Ired (i.e 5 times the output from the red CT)



L2 CT outputs (yellow phase) are connected across 4 turns of the primary winding.



L3 CT outputs (blue phase) are connected across 3 turns of the primary winding.



The output of the secondary winding will be dependent on the current flowing through the primary and from this we can see what sort of fault is occuring.



for example:



5I is a red-earth fault



4I is a yellow-earth fault



3I is a blue-earth fault



2I is a red-blue fault



I is a red-yellow or a yellow-blue fault



sqrt3I is a three-phase or three-phase to earth fault.





cheers



W



Still doesnt explain, in reality CT ratio is not like that...
 10 June 2013 02:56 PM
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Zuiko

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The CT ratios are not mentioned here because it is not important for the sake of the explanation of how the summation transformer is wound.

It is the the primary winding of the summation transformer, and the ratio of how much of that winding each CT is connected to that is important.

The three CTs are connected to the transformer primary, and there is a single, summated output on the secondary. This summation will tell you what the state of the circuit is; as described above.

The CTs outputs are connected as such
L1 - all of the winding (5 /5ths)
L2 - 4/5ths of the winding
L3 - 3/5ths of the winding

The secondary output is therefore the vector sum

5L1 + 4L2 + 3L3



Try drawing out what is explained and it will make more sense; even better, download a piece of free software called QUCS - you can quickly make this circuit and see how it works.

cheers
W

Edited: 10 June 2013 at 03:05 PM by Zuiko
 10 June 2013 03:03 PM
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jackdaniel

Posts: 69
Joined: 29 November 2012

Originally posted by: Zuiko

The CT ratios are not mentioned here because it is not important for the sake of the explanation of how the summation transformer is wound.



It is the the primary winding of the summation transformer, and the ratio of how much of that winding each end of each CT is connected to that is important. The three CTs are connected to the transformer primary, and there is a single, summated output on the secondary. This summation will tell you what the state of the circuit is; as described above.



Try drawing out what is explained and it will make more sense; even better, download a piece of free software called QUCS - you can quickly make this circuit and see how it works.



cheers

W



thanks for the info bro
 11 June 2013 10:47 PM
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MickeyB

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Have a look at this

http://www.schneider-electric....5en/pdfs/page_080.pdf

The 3 CT's are wired in a star/ wye formation with an artificial NE reference where the current from your single phase load would be 'balanced' in the neutral leg and therefore not presented as a fault.....
 12 June 2013 02:10 AM
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jackdaniel

Posts: 69
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Originally posted by: MickeyB

Have a look at this



http://www.schneider-electric....fs/page_080.pdf



The 3 CT's are wired in a star/ wye formation with an artificial NE reference where the current from your single phase load would be 'balanced' in the neutral leg and therefore not presented as a fault.....



Hi Bro,


Now im clear, CT must be connected to earth in order for them to sense the unbalance condition...i was thinking CT without earth connection..thanks again..
 12 June 2013 10:48 AM
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Zuiko

Posts: 521
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Jack,
All the CTs are doing is transforming current to a suitable magnitude for the summation transformer and relay to work with - after all you don't want thousands of amps fault current in your sensitive electronic relays.

The principle is based on the vector sum of the three currents, as explained above. In a balanced, healthy circuit the vector sum of all phase currents is zero. If the vector sum is non zero then current is leaking out of the circuit.

Edited: 12 June 2013 at 11:54 AM by Zuiko
 12 June 2013 11:33 AM
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Zuiko

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Jack,
Please see the diagram I have drawn
The top one shows a healthy three phase circuit. The feeder side is star wound, and the load is an unbalanced delta load.

Below the top diagram is a table showing the currents flowing through each phase. They are different in each phase, because the load is unbalanced. But the crucial thing is that they sum to zero - which I have set as the variable "y" in the equation. (okay, the software has summed it to 3.E - 14 which is zero to all intents and purposes. It is non-zero because the software is using calculus to work out the current, so as you increase the accuracy of the answer (ie more and more significant figures) the answer will tend towards zero) So. L1+L2+L3 = 0A


Below that, in the second diagram, I have introduced a resistive earth fault of 50 ohms onto the circuit. Note that I have called the phases L4, L5, L6 so as not to confuse with the first circuit.

Here there are two equations that are of interest. The total current flowing, t, is the sum of L4, L5, L6. Note now that it is not zero, It is 20A. So a summation transformer will see 20A flowing in the winding.

Also note that this 20A corresponds to the reading of ammeter PR1 which I have put on the earth fault.

The imbalance in the phases of 20A is flowing to earth.

Now look at Equation 3 and the variable f
f is the sum of the phase currents PLUS the earth current and the sum is zero, as you would expect. The sum of all currents in a circuit must be zero.

[IMG][/IMG]diagram

Edited: 12 June 2013 at 01:49 PM by Zuiko
 12 June 2013 02:58 PM
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Zuiko

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The 3 CT's are wired in a star/ wye formation with an artificial NE reference where the current from your single phase load would be 'balanced' in the neutral leg and therefore not presented as a fault.....


Mickey,
The single phase load would need a circuit return path, which in a healthy circuit would show as current in one of the phases (or a neutral if installed).

If you have three conductors and one CT per phase; current on L1 and zero current on L2 and L3; that means L1 has faulted to earth.

If there is a neutral conductor then there must be a CT around that as well, and the current in the neutral must be the vector sum of the current in the phases.


There are some installations where the earth is used as a return: traction or SWER circuits; but these require special consideration and a different appoach.
 13 June 2013 09:32 AM
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jackdaniel

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Originally posted by: Zuiko

Jack,

All the CTs are doing is transforming current to a suitable magnitude for the summation transformer and relay to work with - after all you don't want thousands of amps fault current in your sensitive electronic relays.



The principle is based on the vector sum of the three currents, as explained above. In a balanced, healthy circuit the vector sum of all phase currents is zero. If the vector sum is non zero then current is leaking out of the circuit.


Im very clear about CT functionality and relay...

The vector sum only zero if the 3 phase is in the balance condition.But in reality its impossible to achieve 3 phase balanced condition, if you sum the vector R,Y,B, the current will be produced or its consider as return current even in non fault condition.

So how the relay calculate to determine ground fault on the circuit?

Can we use 3 CT for for 3 phase, 4 wire circuit?, no CT to the neutral? - If yes, how relay going to perform the calculation as per above question.

For the 4 CT configuration, Im 100% Understand..
 13 June 2013 09:40 AM
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Zuiko

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Have another look at the diagram I provided, and I would suggest you look up, or remind yourself of, Kirchoffs Laws.

In the diagram I have showed a 3-phase circuit with an unbalanced load without a fault and with an earth fault.

The vector sum of RYB currents in a circuit in a non-fault condition, with no neutral, is zero.


The vector sum of RYB currents in a faulted circuit is non-zero, but the sum of RYB - E currents is zero.

This is encapsulated in Kirchoffs Law, which is paraphrashed, "what goes out, must come in!"

Think about it: the current at any point must be zero - this is what kirchoff explained (and is common sense).

Think of a point in a circuit: say all the currents flowing into the point are labelled +A, and all the currents flowing out, are - A

Extend this logically - you can think of the circuit as an almost infinite number of points, all with current flowing in and current flowing out, and at each point the sum is zero. If you add up all these infinite points you are just adding up an infinity of zeroes. Which of course, is zero.


If the sum is not zero, then where has the current gone? It must have leaked out of the (measured) circuit and gone to earth (the earth is now part of the circuit).

I have given you examples earlier of how to work this out when you have three CTs connected to a summation transformer.

I have also given an example where you can fit a single CT between the star point of the supply and earth to measure any current that has left the circuit.

I think it would be a good idea if you did a few calculations yourself to convince yourself that this is the case.

Edited: 13 June 2013 at 09:51 AM by Zuiko
 13 June 2013 09:56 AM
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jackdaniel

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Originally posted by: Zuiko

Have another look at the diagram I provided, and I would suggest you look up Kirchoffs Laws.



In the diagram I have showed an unbalanced circuit without a fault and withe a fault.



The vector sum of RYB circuit in a non-fault condition, with no neutral, is zero. It does not matter if the load is balanced equally across three phases! (see my diagram)

This is encapsulated in Kirchoffs Law, which is paraphrashed, "what goes out, must come in!"





If the vector sum is non-zero, then current must be flowing outside the circuit that is being measured - i.e. the circuit now contains an earth path.



I have given you examples earlier of how to work this out when you have three CTs connected to a summation transformer.





I think it would be a good idea if you did a few calculations yourself to convince yourself that this is the case.



I did my manual calculation, for unbalanced load, there will be the return current, thats the purpose of neutral wire...

From your simulation , the current is not exactly 0 for unbalanced load calculation, if you do the calculation based on balanced load, the sum of the vector will be zero.
 13 June 2013 10:10 AM
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Zuiko

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The return current cannot flow in the neutral wire when there is not one, so think about why we use neutral conductors?

Where do you normally see neutral conductors?

It find it easier to work the currents out using j notation, and I get (for the unbalanced circuit above)

L1A = - 375 - j130
L2A = 300 +j289
L3A = 75 -j159

which is exactly zero.

The simulator is using some much more complicated maths routines, meaning that the current is exactly zero when the measured precision is infinite....don't worry about 3 e-14, that number is so small in reality it is zero (it would be unmeasurable in any circuit outside of CERN!).


The vector sum of currents in ANY circuit is ZERO.
This is Kirchoffs Law, and it is fundamental to the understanding of circuits.
It is an example of the fundmental law of science - the conservation of momentum and energy.
Once you get your head around that, you will understand the way the three CTs and summation TX work.

Edited: 13 June 2013 at 10:19 AM by Zuiko
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