# Differential Relay CT Question



## Gman (Mar 22, 2021)

The question is about differential relay on a delta-wye transformer. I understand that the CTs need to be wye delta configuration to fix the phase shift. Therefore the CT secondary side current should be multiplied by sq root of 3 to get the relay current. On this PPI practice question the answer divided the secondary current by sq root of 3. Was my understanding wrong or this is an error on the answer key. Much appreciated!!!!


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## Zach Stone P.E. (Mar 22, 2021)

This looks like a mistake to me.

I'm not sure who drew this diagram but the secondary CT connections don't look correct. On the primary side of the transformer, the CTs are feeding a wye connection instead of being wye connected, and on the secondary side of the transformer, the CTs are again feeding a delta connection instead of just being delta connected. Kinda funny.

Edit - For example, here are wye and delta CT connections from our Live Class #9 - Protection:


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## RedRaider2020 (Mar 24, 2021)

I agree that it is drawn in a weird way but the answer is correct. You'll need to divide the secondary side by sq root of 3 to make the currents equal. You can do this by changing CT ratios or the relay taps. FYI newer microprocessor relays can handle this task for you without making any changes but those wont be on the test.


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## Zach Stone P.E. (Mar 24, 2021)

RedRaider2020 said:


> I agree that it is drawn in a weird way but the answer is correct. You'll need to divide the secondary side by sq root of 3 to make the currents equal. You can do this by changing CT ratios or the relay taps. FYI newer microprocessor relays can handle this task for you without making any changes but those wont be on the test.


The wording of the solution is pretty convoluted and hard to follow, but here is my take on it judging by the diagram and the math:

I_s = Power transformer secondary current
I_2 = Delta connected CT phase current on the secondary side of the power transformer
I_2,relay = Delta connected CT line current leaving the delta connection, entering the relay, on the secondary side of the power transformer

I_2,relay would be equal to √3·I_2, and not equal to I_2/√3 since it is the line current exiting the delta connected CTs:

I_2,relay = √3·I_2​
There would also be a lagging (negative) 30 phase shift in the phase angle of the I_2,relay current compared to I_2 assuming balanced and positive sequence conditions.

Another interesting observation is the solution setting I_1,relay equal to I_2,relay. This would only be true for a 1:1 ratio power transformer with equal primary and secondary CT ratios.

Just my 2 cents.


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## RedRaider2020 (Mar 25, 2021)

Zach Stone P.E. said:


> The wording of the solution is pretty convoluted and hard to follow, but here is my take on it judging by the diagram and the math:
> 
> I_s = Power transformer secondary current
> I_2 = Delta connected CT phase current on the secondary side of the power transformer
> ...


I agree with you and think that your calculation are correct. I'm not sure that you're understanding the problem correctly though. It would probably help to see the problem and not just the solution.
I believe that we are trying to make I1relay=I2relay so that there is no mismatch.
If there is mismatch you will inadvertently trip on a through fault.
In this problem from what I can tell we are not using the relay taps but instead trying to match the two sides with
the CT ratio. In practice this is usually not possibly so you have to use the relay taps also.

If you have a delta-wye transformer and you connect the CT's of the secondary wye side in a delta that is exactly the same as the way that the transformer primary delta is connected then the 30 deg phase shift and the Sqrt 3 from the delta should cancel each other out.
Since they cancel out we now just have to worry about getting the transformer ratio and the CT ratios to match up so that they cancel each other out.

Usually you're given line Voltages on transformers so using the rated voltages the transformer ratio would be
a=Vp/Vs/(Sqrt3)=Sqrt3 * Vp/Vs
Is=Ip*a
CTRatio= XXXXX/5 looks like 1200/5= 240/1 in this case.
I1relay=Ip/(CTRatio1)=I2relay=Is/(CTRatio2)=(Ip*Sqrt3*Vp/Vs)/(CTRatio2)
Ip/CTRatio1=Ip*(Sqrt3 * Vp/Vs)/CTRatio2
If Vp/Vs = 1
CTRatio1=CTRatio2/Sqrt3

So normally you would increase CTRatio2 by Sqrt3 * Vp/Vs to compensate for the transformer ratio but in this case you would only need to increase it by Sqrt3 if Vp/Vs=1.
So CTRatio2 will be Sqrt3 larger than CTRatio1 in this case, usually CTRatio2 is Sqrt3 *Vp/Vs larger than CTRatio1.
A most of the time you don't have the exact CT ratios that you need and will have to compensate for the amount of mismatch in the CT's with the relay taps.

Like I mentioned before with most of the new relays you'll wire the CT's all wye and possibly full CT ratio on all CT's then compensate for everything in the relay. The old way of wiring the CT's to compensate for the delta does make for an excellent test question though.


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## Kay155 (Oct 1, 2022)

Gman said:


> The question is about differential relay on a delta-wye transformer. I understand that the CTs need to be wye delta configuration to fix the phase shift. Therefore the CT secondary side current should be multiplied by sq root of 3 to get the relay current. On this PPI practice question the answer divided the secondary current by sq root of 3. Was my understanding wrong or this is an error on the answer key. Much appreciated!!!! View attachment 21493


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