CT Sizing 1

[mbk2k3]

I have a 600:5A CT, Rated C200. Total burden on CT circuit is 1.008Ohms.
C200 means, it holds 10% accuracy upto 20x secondary rated current.
If my fault level at this bus is 20kA, 20kA translates to 166A on the secondary, which is >20x secondary rated current.
Is this a clear-cut case that I need to increase the CT ratio, to a 1000:5A CT?

Am I right that it doesn't matter anymore that the total burden is 1.008Ohms, so the secondary voltage is 168V, still less than 200V, correct? This doesn't mean anything anymore since the accuracy of the CT is compromised?

 

[davidbeach]

Burden is generally not given in Amps.

If the CT performs reasonably well, how accurate do you need it to be? Is it enough to know for sure that the current is way above 2400A or is it necessary to distinguish 18,600 from 18,800A? For an overcurrent it’s probably fine to know orders of magnitude. The CT accuracy is also not a function of secondary current, it’s function of secondary voltage. If the burden is below the rated burden the 10% error point comes at a higher current.

 

[mbk2k3]

Sorry, 1.008 ohms. My typo.

This CT is for differential protection of a transformer with a long CT run (500ft). And I've heard that its important to get CTs for this type of setup sized properly so they behave predictably in fault conditions.

 

[bacon4life]

Some relay manufactures have whitepapers providing guidance on how to calculate acceptable CT class, ratio and burden.

 

[RRaghunath]

Standard burden for C200 as per IEEE C57.13 is B-2 that corresponds to 2 Ohm. Since connected burden is ~1 Ohm, the CT will not saturate up to say 40 times the rated current.
20kA fault level probably is the system fault level. The CT sizing for protection stability is done for through fault current, in case of transformer differential protection.
For CT adequacy check / sizing calculation, the transformer differential protection relay manuals include a formula which will take in to account system X/R levels as well. It is advisable to use the same.

 

[mbk2k3]

@ Rraghunath,

So if I understand this correctly:
its not simply the secondary current that determines whether the CT will saturate (i.e. >20x secondary doesn't mean it will saturate)

rather...

one needs to take the secondary current and multiply by the burden (V=IR), and only if V<IR will saturation take place?
i.e. if the voltage developed on the secondary for an internal or external/through fault is >200V (for a C200 relay), only then saturation will occur?

 

[cranky108]

Also, changing the CT ratio is not the only option. You can change to a higher than C200.

 

[RRaghunath]

That's right mbk2k3.

 

[stevenal]

For transformer differential protection, it's usually recommended to keep the voltage below 50% of the rating for a through fault.

 

[mbk2k3]

Thanks everyone. I think I've got it. Attached is my 2 pager calc I did.

Can someone share why is it that we make a distinction between through-faults (external) and differential-faults (internal)?

In my calc:
- For external faults (through-faults), the voltage on the secondary is FAR less than 200V (C200 CTs)
- For internal faults (in the differential zone), the voltage on the secondary creeps up to 166V, but still less than 200V (C200 CTs).

 

[davidbeach]

If it saturates for an external fault and causes a trip that’s not good. If it saturates for an internal fault but the diff still trips all is well.

 

[Parchie]

@mbk2k3,
There's something wrong with your lead resistance calcs! Remember that there are two lead wires connecting your device and the CT. So, your lead resistance should be twice of what you got there,1.31 ohms instead of 0.655 ohms. However, if you already took that into consideration,with 1.008 ohms, the required source voltage is = 168 volts (20,000A/120 x (1.008ohms) = 168 V).
Your setup looks fine because the computed knee voltage was 300 V (Vk = 20 x (50VA/5A + 1.008ohms x 5A = 300.8 V).

Try using 1.31 ohms on your lead resistance and find that Vs becomes 277 V while the Vk = 235 V! A bad design. Just choose a 1000/5 CT and be done with it.

 

[mbk2k3]

@Parchie, when you say two leads, you are referring to the return path / lead coming off both X1 and X2, right?

If so, yes I took the full length into account.

Thanks for looking at the calc.

 

[DM61850]

You can use high impedance bus differential if you want to get by using small CTs. There are trade-offs but out of zone faults become more secured with CT saturation. For inzone faults, you can't have CT saturation.

For other protection, it might not make sense to size CTs so that they don't saturate. Saturation due to X/R offset is only temporary and you might not care if the protection doesn't operate immediately, you sometimes would need a very large CT to avoid any saturation, microprocessor relays do a good job at filtering out just the 60 hz component and you need a decent amount of saturation to affect that the 60 hz component, and the magnitude of the DC offset depends on when the fault happens on the waveform so often usually not fully offset.


Here is a tool to see how much the fundamental component is impacted by saturation. It takes a lot of saturation to affect the fundamental component.



[URL unfurl="true"]https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwjB7-L_7MvkAhUHTawKHULIBdsQFjAAegQIBRAC&url=https%3A%2F%2Fwww.l-3.com%2Fwp-content%2Fuploads%2F2010%2F10%2FCT-SAT-Calculator-PSRC.xls&usg=AOvVaw36aPqxb3jPXBz04SFJ_Rh-[/url]


There is a IEEE paper series that was put together by a group of guys here in Houston that is very good. The series is "CT Saturation Calculations - Are they Applicable in the Modern World?". It is a 5 or 6 paper series. It is very good.


[URL unfurl="true"]https://ieeexplore.ieee.org/document/4132861/[/url]


Here are two good SEL and WECC papers. The SEL paper calculations do lead to a pretty beefy CT if you are trying to avoid all saturation.

[URL unfurl="true"]https://static.selinc.com/assets/Literature/Publications/Technical%20Papers/6576_SizingCurrent_DF_20130311_Web.pdf?v=20170324-134515[/url]

https://www.wecc.org/Reliability/White Paper on Relaying Current Transformer Application.pdf

 

[stevenal]

ABB/Westinghouse advocates applying 1.13 factor to the lead resistance to account for temperature rise during faults.

 

[scottf]

Have you considered using a larger gauge wire for the CT secondary wiring or running parallel wires?

 

[Parchie]

ABB/Westinghouse advocates applying 1.13 factor to the lead resistance to account for temperature rise during faults.

You are actually multiplying the required source voltage to prevent saturation by a factor 1+X/R of the system. In that case, the X/R=0.13 (7.4 degrees)!

 

[stevenal]

See page 8. The same explanation is used in Westinghouse Applied Protective Relaying and in ABB Protective Relaying Theory and Applications.

 

[Mario53]

A C200 (B2.0) CT will allow you to keep accuracy when the burden is up to 2 Ohms (200/(5*20) = 2 Ohm). The reason you still have a knee voltage below 200V is that your burden is only half of the maximum permitted. If the burden you calculated is correct, there is no reason to go for a 1000:5 CT. Remember that there is a small print when reading the "10% error @ 20x secondary rated current" and it is "[sub]for a maximum given burden[/sub]"

If your burden were 2 Ohms your knee voltage would be 333 V and in that case, you'd need to go for the 1000:5.

 

[DM61850]

Mario,

You have it backwards. A C200 can develop 200 V on the secondary without more than 10% error due to magnetizing current. That is just for AC saturation. If he had a 1 ohm burden and 200 A on his secondary he would have less than 10 % only if no DC offset was present. There are calculation for taking DC offset or the X/R ratio into account. If you don't want to take the DC offset into account, you might have to wait until the DC offset decays before your relay will operate.

The things that will hurt you on CT saturation:

Tapping down a CT
DC offset or high x/r ratio
High burden
High measured currents in comparison to the CT ratio.