CT choice for transformer feeder 1

[RalphChristie]

What are the rules of thumb to choose the CT-ratio of protection CT (O/C and E/F) on a feeder in a substation? I've seen CT's with a low ratio, but the faultlevel on that point is very high.

The faultlevel of our 20MVA, 66/11 kV Trf is approximatly 12kA. The 11kV Incomer have a ratio of 1200/5, 10P10. That seems to be OK. But one of the feeder's CT-ratio's (For a downstream 2MVA 11kV/380V trf) is 120/5, 10P20. Is this ratio correct? I'm a bit confused, because I thought the main aim of protection is to protect against high faultlevels, but I think this feeder's CT's will saturate under a big fault on the primary side of the trf. It seems as if this ratio is been chosen on the faultlevel of a through-going fault on the secondary side of the trf. (Dyn-11 trf, faultlevel 40kA on secondary side, will give 2.4kA on 11kV side) I've seen it on a lot of our substations and it is confusing.

1. On witch faultlevel do you choose the CT-ratio of feeders? ( Faultlevel at that point, or through-going faultlevel of downstream trf)
2.Setting of relay's? ( Also on faultlevel of that point, or on through-going faultlevel of downstream trf)

Any help or sugestions will be appreciated

Regards
Ralph

 

[tgott]

CT ratios are selected by the current they are expected to see at the point of installation; this is regardless of where a fault is located. Using your numbers: I = 20MVA/ ((sqrt 3) * 11kV) = 1049.7 A. So a CTR = 1200/5 is reasonable for the 11 kV side of the transformer. CTs installed on the 66 kV side of the transformer should have a CTR = 200/5; 20MVA/((sqrt 3)* 66kV)= 175 A.

For the downstream CT: I = 2MVA/((sqrt 3)* 11kV)= 105 A. So a CTR of 120/5 is a good choice.

CTs are made to withstand 20 times their rated current without having an error of more than 10%. So for 1200/5 CTs, 20*1200 = 24kA and for 120/5 CTs, 20*120 = 2.4 kA. In both cases the CTs should not have an error of more than 10% at the respective currents.

 

[RalphChristie]

tgott:

Thanks for the answer.
My problem is that if there is a big fault on the 11kV feeder (faultlevel is about 12kA) the 120/5 10P20 CT's will saturate. Like you've said, the CT's will withstand a current of 2.4kA without an error of more than 10%. But what will happen on 12kA? If I am correct the incomer will trip for faults on the 11kV feeder network, and that is not what we want. Am I correct?

Regards
Ralph

 

[scottf]

If the fault current is 12 kA and you need the CT not to saturate up to that level, then you need a ratio of 12,000/20 = 600:5A assuming a 10P20 rating.

One thing to consider however, there is a burden associated with the accuracy rating. The lower the burden applied to the CT secondary, the higher the saturation point. For example, if a CT is rated 10P20 - 30 VA, but the burden on the CT is only 15 VA, then you can reasonably expect the CT not to saturate before 40 times over current (30/15 * 20).

I question whether the CT needs to stay accurate up to the full fault current for overcurrent protection. It really depends on what jind of sensitivity you need. For the 120:5A CT mentioned above, when the CT starts to saturate at 2.4 kA on the primary, there would be 100 A on the secondary. Is this not enough to trip your relay with the current settings?? If this relay was being used for another application, then perhaps saturation would be a larger concern.

 

[bigamp]

The 120/1 CT fitted to the 2MVA transformer feeder definitely should not be saturating at currents up to the through fault current, for a fault on the LV side of the transformer. This is to ensure that there will be correct discrimination with the downstream protective devices.

Presumably the overcurrent & earth fault relay fitted to the 2MVA transformer feeder has an instantaneous trip as well? The instantaneous trip might well be set to about 125% of the through fault current. It is desirable for the CT to not saturate up to the setting of the instantaneous unit. At higher currents, I do not really think it matters if the CT saturates as the relay will be tripping instantaneously.

Relay manufacturers give very good detailed information regarding CT requirements. It is essential that their recommendations be followed.

Regards

 

[coyney]

You should not depend upon the primary side protection relays for a t/f secondary fault. The t/f ratio turns a secondary line/earth fault into a primary line-line fault.

The secondary protection should cover for major secondary faults.

The t/f will normally have specific protection e.g. buchholtz gas etc. The t/f tails will have REF protection.

The primary protection o/c & e/f protection can therefor be sized and used to cover the t/f feeder as a bulk load.

I trust this helps.

Regards,

 

[sivarama69]

As far as I understand, the voltage which comes across the secondary wdg. of the CT should be less than the Vdesign.
Vdesign = ALF * VA * If/CTR
where,
ALF = Accuracy limit factor i.e, in case of a 10P10 CT ALF=10;
VA = Rated VA burden capacity of the CT;
If = Fault current;
CTR = CT Ratio
Now this Vdesign ensures the CT does not saturate under fault conditions if the Vactual is less than this value. You can apply this to all your CTs and check on a case by case basis if this condition holds good in your applications.
Hope I have been able to address to your problem. However, if you have anything else to share, please feel free to write to me.