Differential Slope Calculation 9

[nightfox1925]

I am rusty on this. I remember calculating the % CT mismatch error by comparing primary and secondary currents using an old GE manual for differential relaying.

If I remember it right, is the % slope the sum of this %CT primary-secondary mismatch plus %CT error plus tap changer range (say for example 20% for a +/-10% regulation) plus some % margin?

I appreciate if anyone can provide me a recap...Thanks

 

[Marmite]

Chapter 16 of the Areva NPAG explains the bias setting and contains a worked example towards the end.
Regards
Marmite

 

[cranky108]

Why add 20% for a + or - 10% LTC? Why not just 10%?

Be sure to add some % for NLTC, where they never seem to talk to the protection engineer before changing.

 

[slavag]

Hi.
Please see attached thread on the topic
thread238-204138
But of course setting is depend on the relay and you need manual of relay
Best Regards.
Slava

 

[veritas]

Hi nightfox1925

The attached is from a protection course I've written. Hope it helps. Any comments are welcome.

Kind regards.

Veritas

 

[nightfox1925]

Thank you all for all the references.

Veritas, are these calculations generic and can be used as a general calculation for any relay model?

 

[veritas]

Hi nightfox1925

Yes, it is meant to be as generic as possible. Even the names for the settings are my own descriptions. However, I have chosen these so they can be as clear as possible to any reader.

Please do note that the translation from the relay setting to the actual primary current would have to factor in current transformer ratio correction factors, phase shift correction and possibly zero sequence eliminations factors as well.

Regards.

 

[Ohmly]

Hi veritas

Thanks for posting that document - I found it interesting and informative.

Not sure if you are concerned, but we use two relays that have other ways of determining the restraint current (you have 3 listed on page 5).

The SR745 uses the maximum of I(HV) & I(LV)

The BE1-CDS240 also uses either the maximum (same as SR745), or mean restraint.

For mean restraint equation 4.2.11 is true for a two winding relay, but if it is a three winding configuration the magntiudes are added and divided by 3, and divided by 4 for 4 winding arrangement (most other relays still divide by 2 for 3+ winding configs)

Also equation 4.2.14 is true when the slope starts at the origin, but some relays (eg Areva) don't do this.

The formula for a line (y=mx + b) is a bit more generic

y = diff current
m = slope setting (divide by 100% if needed)
x = restraint current
b = starting point on y axis

(we tailor the formula to suit the terms used by the relay to training documentation)

This is a generic formula that holds true if the bias characteristic is a straight line.

For most modern relays b = 0 then equation matches yours.

This might overly complicate your document, so please feel free to dis-regard, but I thought I would mention just in case you are interested.

 

[nightfox1925]

Thank you ohmly. I will read the reference and combine with your technical input as soon as I get out from work. This gives me the basic understanding of how things work nowadays with the advent of microprocessor relays.

 

[slavag]

Hi Veritas.
Good job!!!!!
Star to you.
Yes, possible add additional few ways, as Ohmly wrote.
But it more or less same way for slope calculation.
You can add other logic for restraint, base on the angle priciple
Ir=sqrt(I1*I2*cos(I1;I2))
It's other way, really other way.

Best Regards.
Slava

 

[nightfox1925]

Looking into your document. In calculating the full load currents references, are we going to consider the FLA at OA rating or at FA rating (which gives the maximum possible current at no-fault operating conditions)?

Veritas (or anyone), for example if I use SR 745 relays, the equation for the three winding transformer is Ir = Imax(I1, I2, I3)therefore, the Ir will chosen to be the maximum among I1, I2 and I3 inclusive of CTR correction. Do you agree with me?

 

[nightfox1925]

Veritas, do you have similar instruction for three winding transformers? I would appreciate if you can also share them (for free).

From your equations, for three winding transformers:

I(second. winding) = IFLA(second wndg)*(CTRcflv/CTRlv)

I(tertiary winding) = IFLA(tertiary wndg)*(CTRcfter/CTRter)

Note: 'ter' denotes tertiary

Equation 4.2.10 modified to take reference to each secondary and tertiary winding.

During a fault within the differential zone:

1. IHV is flowing opposite to the SUM (Ilv + Itertiary)
2. Currents in above 1. calculated to apply CT correction factors.

Is this a valid approach?

 

[nightfox1925]

Hi Veritas,

I made a trial calculation using my understanding on the reference you provided for a three winding transformer.

I am calculating a slope 1 of 80% at an Idiff =0.883A. I doubt this is quite high so I think I might have made some errors.

Here is a sample of the trial I made:

I am intending to use SR 745 with following relay algorithm:

Idiff = Abs[ I1+I2+I3]

Slope = Idiff/Irestraint where I restraint=Max[I1; I2; I3]

I am looking the I diff as +I1 + (-I2) + (-I3

I am hoping for comments and reactions. Thank you very much.

 

[veritas]

Hi nightfox1925

I will have a look at your questions within the next day or two. Just sorting out a few issues right now.

Veritas.

 

[nightfox1925]

Sure Veritas, I would appreciate any possible assistance I can get.

 

[rockman7892]

This was a very informative thread.

I was wondering if there was also a similar standard or method for minimum pickup and slope settings when setting up differential protection for a cable differential. I would think in this case you may have to factor in charging current for some of the minimum pickup and slope settings?

 

[nightfox1925]

Hoping we'll get response from veritas soon. I find this thread very informative and hope my doubts will be cleared soon enough.

 

[veritas]

Hello nightfox1925

Sorry for the delay. Things get rather busy around here. I am lloking at all your questions and will hopefully have an answer soon. One question though - what does OA, FA1, FA2 stand for? I presume it has something to do with trfr ratings? Are they IEEE/Ansi nomenclature?

I live in the IEC world.

Thanks.

 

[nightfox1925]

Thanks Veritas, I thought you have totally forgotten all about it (just kidding).

OA is natural air cooling
FA is forced air cooling

In our application, we have 2 stage cooling and so there are two forced air cooling MVAs.

Your calculation approach for the two winding is clear and straight forward. I have some difficulty applying the same concept with three winding transformers.

 

[veritas]

Hi everyone

Please find my responses in the attached doc. Easiest way to do it for me I guess.

Hope it helps.

Nightfox1925 - will get back to you one one question and will review your calcs soon.

Regards.

V