ASEA Overcurrent Relay Current Rating

[ThePunisher]

Hi all, we have an ASEA vintage relay: RXIDF 2H, Very Inverse, RK 473 005-AA.

The relay current rating (In) can be set to either 0.5A, 1A, 2.5A and 5A (reconnectible). This relay is connected to a CT rated at 600/5A.

My question is:

Is the relay current rating (In)dependent to the rated CT secondary (in our case is 5A)?

Or is this a setting where I can calibrate In..say if I choose 2.5A, then the CT secondary current of 5A becomes proportional to 2.5A in the relay?

I will appreciate any experience on this relay since this model is way out of my generation.

Thank you in advance

 

[rasevskii]

The In is the nominal current rating of the relay. In other words the relay does not know what CT it is connected to.If you have a CT of 5A secondary you have to use the In=5A tapping on the relay. For the further settings of the curves you have to use the relay data sheet.

I commissioned dozens of these RXIDF relays years ago. I do not know if they are still being made.

You can have a look at the ABB library site, where a lot of the old RK data sheets of ASEA can be downloaded.

regards, rasevskii

 

[ThePunisher]

Thanks Rasevski.

Im curious, if say I set it at In=2.5A, will this going to change the calibration to a 5A CT secondary as 2.5A/5A.

Wherein a pickup setting of 1.2 multiples would be:

Is = 1.2 x (2.5/5)x 600A

Any comments? Hopefully there are ABB old timers here.

 

[rbulsara]

Before we get further, I am assuming you actually mean current rating and not taps for pick up settings. Normally current rating of a relay is a fixed quantity and needs to be ordered that way. I am not familiar with this particular relay, however.


Rafiq Bulsara

http://www.srengineersct.com

 

[rbulsara]

Just to correct my math error:

If it is indeed the relay current rating, at 2.5A rating, the 100% will be 300A, using a 600/5A CT.

So if the pick up is set at 80%, the actual primary pick up current would be 240A.

0.8 x 300=240A

or 0.8*2.5= 2A, which is 2A * 600/5= 240A.

Rafiq Bulsara

http://www.srengineersct.com

 

[rasevskii]

It is not a good idea to use an In of 5A if it is not a 5A CT that you have. Usually CTs are made for 5A or 1A secondary ratings. But if this is some out-in-the-bush application where odd things have to be done because the correct equipment is not available, then you are leaving a problem that will have to be sorted out eventually. Be sure to document the tests and settings properly so that someone coming along later can figure it out.

In a certain large city several years ago there was a major blackout, where, on a 400KV system, someone had connected a 5A CT to a relay set and calibrated for a 1A CT. When the load increased due a parallel incomer being out of service, it tripped its own (the remaining) 400KV incomer..

rasevskii

 

[ThePunisher]

Thanks for all your responses.

Presently, the equivalent primary current where the relay is et to pickup is 720A. This relay is protecting a 10/13.33MVA, 25-4.16kV transformer with an ONAF primary rated current of 308A ONLY. My aim is to set the relay pickup close to 308A but not beyond it. This is an industrial applications so I want to limit operational loading to ONAF rating only.

The relay publication RK 47-11 E states that:

"The relay can be reconnected for different current ratings. To change the current rating of the relay, a screw connection is moved between four alternative terminals visible on the front of the relay."

This is all what the publication discuss about this screw settings. I am looking at this like a tap range selection similar to the CO-9.

 

[rasevskii]

Getting back to your original question, if you use the In=2.5A screw connection, then your 600/5 CT effectively becomes 300/5 as seen by the relay, as Rafiq has said. To get the 308 A pick up setting you have to set the Is (starting current) on the potentiometer dial. I do not have that RK available here, but it would seem to be a too low setting on the relay to be practical. It can be that the minimum Is is only 1.2x In (about 360A). But not having the RK I cannot be sure what the minimum Is setting actually is.

It would be perhaps better to protect the transformer by oil or winding temperature rather than a low overcurrent setting. There can be problems with large motors (if any) starting, etc.

rasevskii

 

[rasevskii]

Another thing is, what about the magnetizing inrush on energization causing an unwanted trip, with such a low setting? Could it be that is why there is the 720A setting now? Is there a history behind this...

regards, rasevskii

 

[ThePunisher]

I am still trying to get history on this 720A setting. I created a TCC with the 4kV motor starting plus a current adder (pre-starting maximum demand base loading) and the curve is adequately to the left of the relay curve.

Furthermore, the transformer magnetizing inrush (approx. 12 times ONAN primary current rating) is well below and away the relay curve.

Having that 720A is still a question to me.

 

[AMBMI]

A trip threshold equal to 2xIn = 616 A could make sense and is quite common in the practice. Please consider that the transformer damage characteristic is starting at about 2In....
In any case you can protect the transformer against overload with the relay at the secondary side set at about 110-115% In.
720 A is for my feeling probably a bit too much

 

[rasevskii]

You might want to consider adding a modern digital transformer protection additional to the RXIDF, which can be left as a backup. That would give the possibility of proper overload protection and other functions that are available now, such as a pre-overload alarm, etc. You are risking now a false trip, with the attendant disruption possibly.

With this expensive transformer, an investment in a modern protection system is likely advisable.

regards, rasevskii

 

[rasevskii]

UPDATE: Now I have found the catalog sheet for the RXIDF in my archives... In fact you can set the Is at 1 x In which is the lowest setting on the dial. So you can get the setting you want. Do a secondary test to get the exact pickup value.

Still not the best way to do this...see all the comments above.

regards, rasevskii

 

[ThePunisher]

Another factor that concerns me is the primary cable which is an old 25kV, PILC, 500MCM cable with a de-rated ampacity of 339A.

The transformer may have 49 devices to protect it from abnormal temperatures due to overloading, but the cable only relies on the relay pickup for overload protection.

I will check the secondary (4.16kV) relay pickup equivalent amps and reflect that to the 25kV primary and ensure that the value is less than this de-rated primary cable ampacity...any comments?

 

[rbulsara]

At medium voltage, cable overcurrent protection is intended for short circuits only, overload is protected by design. It is different from low voltage applications in that sense.

Also when setting relay amp pick up, compare them to a comparable MV Fuse curve. For example, 200E MV fuse is in fact equivalent of a 400A relay setting. You also need to leave room for coordination with downstream devices.

As for transformers, applying temperature monitoring for thermal damage of transformers is a better option.

Rafiq Bulsara

http://www.srengineersct.com

 

[ThePunisher]

Again my thanks to all your diligence.

Rafiq, what do you mean when you say "At MV, cable overload protection is protected by DESIGN".

Regards...

 

[rbulsara]

Means that short term thermal overload for cables is seldom a concern, the short circuit protection is. Because it is usually a part of a system whose load is typically known, such as a transformer or a set of those. Maximum possible load is usually known.

I can't explain it over this forum (too much to type) but refer to some good reference books on MV distribution or even IEEE buff book.

One of the reasons, the MV fuse characteristics are what they are, that is they do not even open below 200% of their rated current or takes hours to open if ever.



Rafiq Bulsara

http://www.srengineersct.com

 

[Lin1968]

Hi, I haven't got it. Regarding the same relay, my case is the In is 0.1, and the CT ratio is 2000/5, the I> is 1.5 and I>> 6.0. so, what is my I> and I>> pickup, this relay is upstream of a MCC, and the motor rated current is about 800A (6 MOTORS).

please help.