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Topic Title: Algorithm for IDMT when current is varying
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Created On: 03 February 2004 03:28 PM
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 03 February 2004 03:28 PM
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deleted_Ekij

Posts: 89
Joined: 03 February 2004

I know that the algorithm for an Inverse Definite Minimum Time Overcurrent is
TripTime = (TimeConstant * 0.15) / (((Current / StartingCurrentConstant) ^ 0.02) -1 )

The question is how should this be implemented if the current is changing?

Imagine a case where the 'relay' trips in 30 seconds at current A and at 10 seconds at current B.
If current A is applied for 15 seconds and then the current changes to B what happens.

Argument C:
At T=15 seconds, the TripTime is calculated at 10 seconds and the current trips as time (15) is beyond the trip time (10).
Argument D:
At T=15 the 'relay' has accumulated 50% of it's tripping time so it trips in another 5 seconds (t=20) as the relay needs 5 seconds at current B to accumulate the other 50%.

I know I prefer method 'D' but can anyone point me at any documentation that says how it should be implemented?

The problem is my supplier is trying to provide method 'C' and has put the onus on me to prove it's wrong.
 04 February 2004 07:08 AM
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rogerbryant

Posts: 867
Joined: 19 July 2002

The purpose of an overcurrent protection device is usually to limit the temperature rise of a conductor to a defined safe level.

The calculations are based on the conductor being at its maximum long term operating temperature.
In your case if there has been an overload of current A the conductor temperature will have risen above its maximum long term temperature. A further overload of current B for the full trip time will result in the conductor rising above its maximum safe short term temperature.
Your argument D is the appropriate one although things are not quite so linear.

A conventional fuse makes a good time*current^2 integrator and would follow D not C.

I am not sure of a good source of documentation, but I would suggest contacting a cable manufacturer and asking about overload protection for their cables.

Regards

Roger
 12 March 2004 12:36 PM
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deleted_RalphChristie

Posts: 42
Joined: 24 November 2003

Hi

Just something I noted, but....

there are different IDMT-curves (IEC), the most common types are the NI (normal inverse), VI (very inverse) and EI (extremely inverse) The formula for the mathematical trip time in seconds is:

t(sec) = a . k / (((Fault/measured Current / Set Current) ^ b ) - 1)

where:

    a (Alpha) :
      0.14 (Normal I)

      13.5 (Very I)

      80 (Extremly I)

    b (Beta) :
      0.02 (NI)

      1 (VI)

      2 (EI)

    k = time constant or time dial


Just note that for a NI-curve your formula is:

t (sec)= (TimeConstant * 0.14) / (((Current / StartingCurrentConstant) ^ 0.02) -1 )




To answer your question (in my opinion):

I'm not sure whether you mean that your tripping will be to fast or to slow in Argument C, but both (too fast or too slow) are not correct.
If I understand correctly, at Current A the relay will time for half of the time (15sec) and then with Current B it will time again (10sec), resulting in a total triptime of 25sec? (Too slow)
Or do you mean it will trip the moment the current change (15sec) because it is past the 10sec time of Current B? (Too fast)
For a too fast trip, you will struggle too coordinate with other feeders downstream, resulting in unnecessary downtime on a lot of feeders.
With a too slow trip you can damage the electrical equipment (Generators, Transformers, cables, switchgear, motors, etc.)

Argument D seems to be fine.

Sorry, I'm not aware of any documentation, I'm just thinking from a perspective out of the field



Regards
Ralph
 01 July 2004 07:28 AM
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deleted_judithkan

Posts: 1
Joined: 01 July 2004

Hi, I'm currently doing my thesis on the topic you posted. I was wondering if you found any further information regarding this topic. It would be greatly appreciated if you could give me some background information or any algorithm you figured out. My email is kan_judith@yahoo.com

Thank you so much for you help!
 05 December 2004 07:12 AM
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laxman1974

Posts: 4
Joined: 05 December 2004


Mr. Ralph,

The mathematical trip time formula you have mentioned is applicable to all IDMT over current fault relays of different make...such as westinghouse and schneider electric? could you please guide me.....

right now I am working on relay setting calculation for these above make relays...But I have only their standard char. curves....

regards

Laxman
 05 December 2004 07:17 AM
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laxman1974

Posts: 4
Joined: 05 December 2004


Mr. Ralph,

The mathematical trip time formula you have mentioned is applicable to all make of IDMT overcurrent relays such as westinghouse and schneider electric?

Could you please reply...right now I am working on these make relay's setting calculation..I have only thier standard char. curves information..your reply will help me very much...

Thanks®ards

Laxman

 19 February 2005 11:15 PM
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deleted_RalphChristie

Posts: 42
Joined: 24 November 2003

Hi

Sorry, I do not visit this site regulary.

But, regarding your question:

It depends on what kind of relay and with which brand of relay you are working. There are several (hundreds of) different trip curves been used on different relays throughout the years. The curves I were revering to are the IEC-trip curves. IEEE (In the USA) use different curves although their curve-name and the IEC curve-name can be the same, like:

IEC:
standard inverse
very inverse
extremely inverse
long-time inverse
short-time inverse

IEEE:
Moderately inverse
inverse
very inverse
extremely inverse
short-time inverse

Although they can have the same name, they do not have the same caracteristics.

The new microprocessor relays have a lot of different curves you can choose from. The SPAJ-series from ABB and the Argus-series from Reyrolle use IEC-curves. Some manufacturers like SEL use both IEC and IEEE-curves. With some of the SEL-relays you can even create your own curves. I am not familiar with Schneider Electric. I've checked their site quickly (www.schneider-electric.com) and saw they use Merlin Gerin and Square D products. You'll have to visit their website or find brochures/manuals on some of their relays to see what kind of curves they use.
The older elctromechanical relays are a bit different. You have to find their manuals to see their curve-details and then find an mathematical equation to draw it. I have seen approximate mathematical equations for the CO8 and CO9-series from Westinghouse.
If I know what relays you are using (like a CO8-electromechanical relay from Westinghouse or a SPAJ140C microprocessor relay from ABB) I will be able to help you easier. But, the best would be to find the relay-curve on a websites or in a brochure/manual.

Hope it helps

Regards
Ralph
 13 December 2006 10:25 AM
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deleted_smontom

Posts: 2
Joined: 13 December 2006

Originally posted by: Ekij

I know that the algorithm for an Inverse Definite Minimum Time Overcurrent is
TripTime = (TimeConstant * 0.15) / (((Current / StartingCurrentConstant) ^ 0.02) -1 )

The question is how should this be implemented if the current is changing?

Imagine a case where the 'relay' trips in 30 seconds at current A and at 10 seconds at current B.
If current A is applied for 15 seconds and then the current changes to B what happens.

Argument C:
At T=15 seconds, the TripTime is calculated at 10 seconds and the current trips as time (15) is beyond the trip time (10).
Argument D:
At T=15 the 'relay' has accumulated 50% of it's tripping time so it trips in another 5 seconds (t=20) as the relay needs 5 seconds at current B to accumulate the other 50%.

I know I prefer method 'D' but can anyone point me at any documentation that says how it should be implemented?

The problem is my supplier is trying to provide method 'C' and has put the onus on me to prove it's wrong. <img src="i/expressions/face-icon-small-sad.gif" border="0">
 13 December 2006 10:35 AM
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deleted_smontom

Posts: 2
Joined: 13 December 2006

When ever you are doing relay coordination and settings for a power system that is near to a generation plant, the fault current has a varing nature in time. This also highly dependent to the relay loctation, i.e. it is near or far from the generators. It also depends to the setting time of the relays.

Consider that you have calculated the fault current for 500ms in all over the system. For the relay that is close to the generator and must act in 1000ms for an especified settings, the fault current is not correct. On the other hand this fault calculation is not correct for the relay that trip in 200 msec since the current actually is higher in this condition.

Therefore, for exact setting you need a dynamic fault current.

Please refer to Link removed for a package that does the relay settings according to the dynmical fault current. It is also sujessted that do Transient stability calcultion in such situations.

Dr. S. Montaser Kouhsari

Edited: 13 December 2006 at 10:39 AM by deleted_smontom
 25 April 2007 07:53 AM
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deleted_db444

Posts: 14
Joined: 10 April 2007

Originally posted by: smontom


Please refer to Link removed for a package that does the relay settings according to the dynmical fault current. It is also sujessted that do Transient stability calcultion in such situations.



Dr. S. Montaser Kouhsari


and you dont happen to be an agent for this package? oops you are!

Al
 01 August 2007 07:22 PM
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pkumarbl

Posts: 1
Joined: 18 July 2003

I would like to know when to use between different curves of IEC and IEEE.during relay setting we need to choose some one.is there any guideline for this.
 30 August 2010 10:12 AM
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kiana

Posts: 1
Joined: 30 August 2010

Hello everybody
I'm newly working on Overcurrent Relays
please help me about the algorithm for IDMT curves.
and aso please help me about c code of them
tnx
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