Differential Transformer Protection

[sobeys81]

Hi..
I am trying to study transformer differential protection and I guess I have gone through IEEE std and some noted book but I did not get a single document that could give me an example of a traditional approach. Everywhere it is more conceptual.

Could you advise me a book or an IEEE transaction that could help.

Thanks

 

[davidbeach]

Check the protection section of faq238-1287.

 

[sobeys81]

Thanks for the input Davidbeach.
Surprisingly enough... I guess I have most of this information in my laptop but when ever I get to find an actual example it is never there. Let me see if you can put a word and get me out of my ddilemna for a moment.

In case of differential protection, why do we need...

1. Phase Overcurrent(I understand for Internal Fault and Extended external fault)

2. Neutral Overcurrent

3. Ground Overcurrent

What is the differen between 2 and 3. Could you advise.
It should be either one.

 

[magoo2]

Here's a simple single-phase example to help you understand how a transformer differential relay works. Take 2 CTs and wire up the outputs to give you the difference current across a relay coil.

As long as the fault is external to the CTs, the currents will be equal and you won't get any current through the relay coil. If the fault is internal, the currents won't be equal and you'll get current through the relay coil.

 

[TurbineGen]

Sobeys81,

Your question has to do with your zones of protection and types of protection. A differential relay protects against conditions inside the transformer only such as an internal fault. See magoo2's example for how the relay performs this operation. Your other listed devices protect the transformer from external sources and/or different type of conditions.

1. Phase overcurrent - This is for an external fault or extended overload conditions. These are outside of the differential's protection zone.

2. Neutral overcurrent - This is current imbalance. This is to protect the unit from a severely imbalanced load. Again, this is external to the transformer so it is outside of the zone of protection of the differential relay.

3. Ground overcurrent - Neutral and ground current are not always the same (consider an ungrounded WYE). A condition that could occur to make this device operate is a tank ground in the transformer, which the neutral overcurrent would not pick up.

------------------------------------------------------------------------
If it is broken, fix it. If it isn't broken, I'll soon fix that.

 

[sobeys81]

Thanks. COuld you advise me how to calulcate the slope for the curve. I just could not figure this out. I looked at some GE white papers but did not help.
Thanks

 

[slavag]

Hi Sobey81.
Please attached thread
thread238-204138.
Hope that help.
Im always prefer max. setting w/o any calculation.
Internal trafo fault isn't few amps.
Best Regards and Good Luck.
Slava

 

[sobeys81]

Hi..

It looks like I am catching up to it.

What does these error reflect. Could you please elaborate on this one.

CT error = +/- 10%
LTC variation = 10%
Excitation current at rated voltage = 1.4%
Margin = 5%

Relay Minimum Operating Current Setting:
(CT error)@HV side+(CT error)@LV side +LTC + Ie +Margin =
0.1+0.1 +0.1+0.02 + 0.05=0.37 pu

Thanks

 

[TurbineGen]

Sobeys,

Your first two numbers (LTC variation and CT variation) are simply stating that there is some error in the returned value of these devices. For example take a 100/5 CT at +/- 10%. With 100 Amps the CT could read as low as 4.5 or as high as 5.5 Amps. The same thing goes for the LTC and voltage.

Excitation current: It takes some extra energy to magnetize the transformwer's iron core. This current is seen only on the supply side and is absorbed by the transformer. In other words it is internal to the unit and therefore the differential protection will be sensitive to it. I believe this is nearly constant.

Margin - This is the minimum difeerential setting that you want the transformer to be tripped. For example, let's take a perfect isolation transformer with no losses (I know it doesn't exist, but it makes the description easier). If you have 5% or more current in than current out, the device will trip the unit.

In any case as Slava stated, transformer internal faults typically do not carry small currents. I think 0.37 pu would be a good setting for minimum operating current.

------------------------------------------------------------------------
If it is broken, fix it. If it isn't broken, I'll soon fix that.

 

[slavag]

Hi.
Agree with TurbineGen.
Maybe, add to this formula +0.05 of relay CT errors/accuracy.
In all cases 37% ( 0.37pu) is good setting. More or less we used 35-45% of tafo nominal current.
Best Regards.
Slava

 

[sobeys81]

perfect. I got it now. So from here how do I calculate by slope. I came this far but this gives me only first point at my slope. How do I calculate by first and second slope based on this IEEE std. clip.

http://files.engineering.com/getfile.aspx?folder=7bbb4e20-0f81-42b6-87df-0715d5947094&file=Slope.bmp

 

[sobeys81]

I guess one more question. This .37 PU is in reference to primary or secondary full load current. I also found this slope equation:

Relay Slope Setting:

(CT error)@HV side+(CT error)@LV side +LTC + Margin =
0.1+0.1 +0.1+0.05 = 35%

Why it does not have Ie like in minimum pick up equation.
thanks