% Differential / Slope Characteristic Selection for 87 Differential Relay

[YoungMeezy]

Hello,

I have a question regarding how to select / determine the % differential characteristic slope to use in an 87 differential current relay set up.

I see references to "Select 20% slope from table" or "Select 50% slope from table" when reviewing old (circa 1980's) design notes. But there aren't any other notes as to WHY that particular slope was chosen.

What I don't understand is, given a table of 5 to 10 different characteristic slopes, how do I know / analyse which characteristic slope will provide the right balance of sensitivity vs. selectivity.

Any and all comments are appreciated!

 

[topgone]

Hello,

I have a question regarding how to select / determine the % differential characteristic slope to use in an 87 differential current relay set up.

I see references to "Select 20% slope from table" or "Select 50% slope from table" when reviewing old (circa 1980's) design notes. But there aren't any other notes as to WHY that particular slope was chosen.

What I don't understand is, given a table of 5 to 10 different characteristic slopes, how do I know / analyse which characteristic slope will provide the right balance of sensitivity vs. selectivity.

Any and all comments are appreciated!

The slope is the bias differential. Simply, you compute for the CT secondary currents on both sides of the protection and compute for the difference ratio.

Ex. 6.5 MVA, 13.8/4.16 kV, Dy using 400/5 on the primary and 2000/5 on the secondary CTs. Primary CT current will be 271.9(5/400) = 3.4A while the secondary CT current = 902.1(5/2000)(1.732)= 3.91A. Then your differential bias = (3.91-3.4)/3.91 x 100% = 13%. You will have to choose a 20% slope! Hope that helps.

 

[YoungMeezy]

The slope is the bias differential. Simply, you compute for the CT secondary currents on both sides of the protection and compute for the difference ratio.

Ex. 6.5 MVA, 13.8/4.16 kV, Dy using 400/5 on the primary and 2000/5 on the secondary CTs. Primary CT current will be 271.9(5/400) = 3.4A while the secondary CT current = 902.1(5/2000)(1.732)= 3.91A. Then your differential bias = (3.91-3.4)/3.91 x 100% = 13%. You will have to choose a 20% slope! Hope that helps.

That is a huge help thanks! The missing piece of the puzzle was the differential bias calculation. Thanks a tonne!

 

[Ingenieur]

No expert, just trying to understand
it looks like bias is a unitized/compensated gain
~ (delta input-output)/input
line currents are measured on but sides and adjusted for ct ratio
if balanced input=output and gain = 0
correct?

my ???
if both ct's are measuring line (vs phase, for delta: line = phase x sqrt3 and wye: phase = line)
why is the sec i x sqrt3?

 

[YoungMeezy]

No expert, just trying to understand
it looks like bias is a unitized/compensated gain
~ (delta input-output)/input
line currents are measured on but sides and adjusted for ct ratio
if balanced input=output and gain = 0
correct?

my ???
if both ct's are measuring line (vs phase, for delta: line = phase x sqrt3 and wye: phase = line)
why is the sec i x sqrt3?

I am also curious to see the math / reasoning behind multiplying the CT secondary current by root 3 on the wye side. I have seen it in other design notes and calcs before but never really questioned it.

 

[Ingenieur]

I am also curious to see the math / reasoning behind multiplying the CT secondary current by root 3 on the wye side. I have seen it in other design notes and calcs before but never really questioned it.

Guessing related to the delta-wye 30 degree phase shift

 

[Phil Corso]

YoungMeezy & InjunEar,

The Y-side CT's are Delta-connected, so current out of Delta (to the Diff'l relay) is multiplied by Sqrt(3)!

Stated another way... if Ph-A CT's secondary current is the vector Ia, and Ph-C CT's secondary current is the vector Ic, then, Idiff to Relay is Ia-Ic!

Regards, Phil Corso

 

[Ingenieur]

YoungMeezy & InjunEar,

The Y-side CT's are Delta-connected, so current out of Delta (to the Diff'l relay) is multiplied by Sqrt(3)!

Stated another way... if Ph-A CT's secondary current is the vector Ia, and Ph-C CT's secondary current is the vector Ic, then, Idiff to Relay is Ia-Ic!

Regards, Phil Corso

thanks
makes sense
does not apply to primary since delta-delta

did not consider summing done by the ct arrangement
that each was brought to the relat and done thru it

 

[Bugman1400]

There are many other factors involved in picking a slope like CT error, secondary burden, short circuit availability, CT class, etc. Typical slope is 30-40% but, a lot depends if the relay is an IED or E-M relay and if the application is for a xfmr or bus diff. Most IED relays have the capability to use two slopes in cases where high fault current is present.

 

[Phil Corso]

InjunEar...

I may have misinterpreted what you meant by Delta-Delta! But, if the winding is delta-connected, then, the CT's are wye-connected!

Phil

 

[Ingenieur]

Sometimes a sketch clears the fog lol

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[YoungMeezy]

YoungMeezy & InjunEar,

The Y-side CT's are Delta-connected, so current out of Delta (to the Diff'l relay) is multiplied by Sqrt(3)!

Stated another way... if Ph-A CT's secondary current is the vector Ia, and Ph-C CT's secondary current is the vector Ic, then, Idiff to Relay is Ia-Ic!

Regards, Phil Corso

Thanks for the reply. That clears it up!

 

[Bugman1400]

What CT configuration would you use for a delta/delta xfmr or a g-wye/g-wye xfmr?

 

[Phil Corso]

Bugman,

To whom is your is question addressed?

Phil

 

[Bugman1400]

Bugman,

To whom is your is question addressed?

Phil

To anyone that can provide an intelligent answer.

 

[Ingenieur]

To anyone that can provide an intelligent answer.

Won't vouch for intelligent but here goes

T = transformer winding config
CT = ct winding config
CT OUT = ct output
i = line current
n = turns ratio
ctr = ct ratio

T....CT....CT OUT
d....y....n i / ctr
d....d....sqrt3 n i / ctr
y....d....n i / ctr
y....y....n i / ctr

so any combo where the CT OUT's differ only by n is ok with the following caveats
It may be good to avoid d-d with both ct's being d since The output is i x sqrt3, not i
if ratioed may not matter
if grounded y-y ct's should be avoided?

 

[topgone]

What CT configuration would you use for a delta/delta xfmr or a g-wye/g-wye xfmr?

The problem we always wanted to avoid is having DC components in the differential protection circuit. Any arrangement will do but knowing a wye arrangement will not filter DC in the circuit, hence it would be wise to implement delta CT arrangements. Else we needed to add zero-sequence trap circuit to the CT circuits.

 

[Sahib]

What for is the restraining coil in the differential relay?
To account for saturation effect, DC component etc., so that the sensitivity of the relay is maintained.