We are to comission a system with 1200/1 10p10 CTs. The fault level is 18kA at 11kV. At what point will the CTs saturate? They are connected to Siemens 7SJ protection relays.  

Kevin Bosch
Rainbow Technologies
 

Typically the CT curves should be consulted.
However if it was an ANSI (which I don't believe it is) I would say between 12 to 15 times the set ratio.

See Chapter 6 in http://www.areva-td.com/scripts/solutions/publigen/content/templates/show.asp?P=1228&L=US
 

Some commissioning  procedures follow these steps:
1> The saturation level is determined from the CT specs.
2> A test set is used to verify that the CT meets the spec.
 

Bill
--------------------
"Why not the best?"
Jimmy Carter

1200/1A CT 10P10 spec will mean the CT error will be below 10% up to a current magnitude of 10x1200A = 12kA. Beyond 12kA, the accuracy is not guaranteed and hence theoretically any point above 12kA can be the point of saturation for the CT. For more precise picture, CT factory test report could help.

As said previously, the 10P10 means it will maintain decent accuracy to 10 x rated current when loaded at rated burden. You didn't mention the burden rating. Then loaded below the rated burden, the point of saturation will increase proportionally.

For example, if the unit is rated 10P10 - 30 VA and you actually only have 15 VA, it will perform like a 10P20. Half the burden means twice the accuracy limiting point.
 

Thanks scottf!! This is what I want to hear:) I have heard before that burden is related to the saturating point. I dont know what the burden is but its a differential input to an electronic relay (siemens 7SJ). Electronic relays have low burdens in general? 3-5VA? I know that modern day electronic relays pose less of a problem regarding saturation. If the CT has to be sized at 1800/1 because the incoming fault level is 18kA, then the outgoing feeders would also have to be sized just as high (same kA just off the main bus). Thats rediculous! If you are feeding a 1000kVA trf off an 18kA bus you cant use a 1800/1 ct to protect is as you need to set the EF at 5% of trf rating (for example) and the relay cant be set that low. I would like to see the formula to calculate minimum CT ratios wrt burden, fault level etc.

Kevin Bosch
Rainbow Technologies
 

THD-

Remember that you need to include the round-trip lead length in the burden calculation.

 

Please note you have include the internal burden (Rct) for calculation of the saturation at another burden as specified. For Class 5P.. and 10P.. CTs the result will be lower as given above!

Proceed as follows:
- retrieve the sec. winding resistance Rct from testreport or manufacturers datasheet
- evaluate the saturation voltage: Usat=ALF*Isn*(Rct+Zb),
where ALF is saturation factor as specified, Isn=rated sec. current, Zb=specified burden in Ohms.
- the ALF at another burden Zx will be: ALF=Usat/Isn/(Rct+Zx)

Example: Isn=1A, class 5P10 (ALF=10), specified burden 30VA: Zb=30Ohms, Rct=10Ohm.
Usat = 10*1*(10+30) = 400V
ALF at Zx=15VA wiil be: 400V/1/(10+15)=16
So with burden 15VA the CT will saturate at 16x rated current (class 5P16)

douwe-

Your formula's are correct and provide the correct theoretical answer.

However, from a practical perspective, when working backwards from a ALF to calculate a different ALF with a different burden, internal resistance can often be ignored.

For example, in the case of a 1200:1A CT, the Rct value would typically be in the range of 3-4 ohms. So when I say at half the burden, the ALF will double, it's correct. And that's due to 2 factors...one the CT design is rarely designed to saturate exactly at the prescribed ALF. Second, the internal resistance is normally dominated by the external burden.

So using your formula, ALF as Zx=15VA = 400V/1/(4+15)=21

scottf-

in the case of Rct=4Ohm, Usat would be 340V resulting in
ALF = 340V/1/(4+15)=18 (5P18) for Zx=15VA

My point is, Rct has to be taken into account. Depending on its value compared to the external burden its impact might be large.
 

sorry douwe, but that's not correct.

The issue here is a CT already rated 10P10...and we assumed a burden rating of 30 VA.

10 x 1A x 30ohms = Vkp > 300V (well it really means Iexc can't exceed 1A at 10 time over-current, but will simplify and say it's the knee-point for discussion's sake).

If half the burden is placed on the unit 300V / 15 ohms = 20 times over-current.

Remember we're working backwards here from a known Vkp measured at the CTs secondary terminals during routine tests to confirm the rated accuracy class.

Scott,

I had a similar concerns about CT excitation curve and knee point. Would you please explain/correct my understanding.

Here the CT is C200, 1200:5A at 4.16kV.

Let's say that total CT burden (including wire, relay, winding resistance) is 1 OHM.

According explanation from one course I took:

At maximum fault of say 24,000A at that point the CT will "see" 100A. Secondary CT Voltage = 100A*1 =100V. Now, let's say that knee point C200 excitation curve is say 70V. Since CT secondary voltage is 100V we will not get accurate results and CT will saturate, hence relay may not operate properly.

On the other hand, (If you possess IEEE, 242-2001, chapter 3); in IEEE they are determining total burden (in OHMs) evaluated at 20times of rated CT (5A-rated, i.e. 5*20(times) =100A);
In the example, CT voltage is calculated based on CT current at primary fault (in example it is at 24,000A such that CT "sees" 200A). Then, the reference is made to excitation curve (Fig 3-4) from where I can see the calculated voltage goes a bit beyond knee point. From there excitation current is determined I=.0.06A. Since the excitation current is less than 10A (10% of evaluation current of 100A) the relay will operate as expected.

I am a bit concern about the knee point, its importance and proper reference to the excitation curves.

Knee-point voltage can be calculated in a number of different ways. Many people see a C200 rating and assume that the knee-point voltage is 200V. In reality, C200 simply means that the CT must be capable of developing 200V across a 2 ohm burden at 20 times over-current (20 x 5A x 2ohms = 200V) and not exceed a 10% error, which means Iexc < 10A.

Measuring the knee-point voltage as listed in IEEE C57.13 normally puts the knee-point voltage around 80% of the accuracy limiting voltage, i.e. Vkp for a CT rated C200 is approx. 160V.

 

Generally speaking, the whole thing about nn% error at ## x rated current out to be trashed from everyone's mind when calculating saturation point of a CT. You get a much clearer picture of the saturation level of a CT if you look at a CT as a voltage source. and if IxZ is more than the voltage source rating, you have a problem. More specifically, if the calculation of
        V.ideal = (I.fault.pri/CTR)*(CT internal resistance + external)
gives you a voltage where the Iexc is no longer negligble, then you have a saturated CT. Due to DC offset giving transient saturation that is not included in this equation, ideally you want a V.ideal that is on the order of half of the knee point. Oh sure, you can point out that knee point is a poorly defined number, but any half bright person can get the idea that if V.ideal is too far up the excitation curve, or higher than the class of the CT, you have a problem.

Timesabroom-

Isn't the whole idea behind calculating the saturation point to be able to ensure nn% error at ## x rated current?

Seems to me you've got your thinking backwards as to what's really important.

 

Geee, Scottf, those words got me going a bit and wanting my point to be clearer. I typed up the dissertation below. Can you comment if my thinking below is backward?
   Suppose I have a 5A C10 CT.  The typical statement is that it will withstand 100A with 0.1 ohm burden. Fine, but who has a standard ohm burden? No one. It is possibly true that the CT could admirably put out 200A with <1% error if burden was low enough, and that the CT might saturate horribly at 50A if burden was high enough. How can I tell if the CT will work in my application? An example of the calc I have in mind:
  A data sheet for C10 200:5 CT shows an internal resistance of 0.064 ohms. The excitation curve shows the CT excitation current stays low till about 18V secondary. Assume the secondary is shorted or so close to the relay that external resistance is negligible, to keep the calc simple for this example.
  Calc: 18V/0.064ohm = 281A, and 281*200:5 = 11250A.
  In this case, the CT can relatively accurately reproduce an 11,250A fault and put out ~281A into the secondary, without serious saturation (if I ignore transient saturation during the DC offset period), till something burned up due to the high secondary current. If my available current is >11250, I have to estimate how much current will flow in the secondary and if error will be bad enough to be a problem. If I have <11250A available, all is well in the world, unless I get to wondering if DC offset will cause transient CT saturation. If I want to reproduce DC offset to at least some extent, I hope I have something less than 11250A available. Also, note a C10 CT can put out more current than some realize.

  Saying the CT was built for burden of 100A into a 0.1 ohm external burden (i.e. 5A C10) does not tell me near as much as the calculation of
  Isec.max ~ Vknee/(Rct+Zexternal) and
  Ipri.sat ~ CTR*Isec.max

  Side note: C57.13 states that CT mfrs are supposed to be able to supply the 1 second withstand of CTs with the secondary shorted, (rating is given in terms of primary current) but it will be pulling teeth to get it out of them. The 1 second rating is typically >> than 100A * CTR per my experience.
 

Timesabroom-

Using your example, if you a C10 rated CT it means that you can drive it up to 100 times over-current into a MAX CONNECTED burden of 0.1 ohms.

I'm not sure I follow your comment of "who has a standard burden". All one has to do is confirm the connected burden is less than 0.1 ohms and then you know the minimum performance of the CT.

If you want to know what the CT can do above 100 times rated current with a burden below 0.1 ohms, then we discussed that above.

Let me ask you, do you do a full calculation using Rct (internal resistance) very often? If so, I suspect you may be in the minority from my experience. I work for a company that supplies CTs and VTs in all shapes and sizes. We do not typically provide an Rct value on routine test reports or in product literature. We only give it (by calculation) when someone asks, which is very rare.

On your side note, sounds like you're talking about the thermal current rating, which is the 1 sec withstand of the CT and always expressed in primary current. This value varies widely with type of CT and manufacturer. For instance for wound-type CTs, i.e. what is normally seen for MV and HV applications, a value of 100 x Inom is somewhat typical (can go a good bit higher on HV units). For 600V ring-type CTs (i.e. bushing CTs, etc...) the values are normally a lot higher than that, as the rating is only a function of the secondary winding. A lot of bushing CTs don't have the thermal current value listed, since it is almost always limited by the primary conductor (breaker or transformer bushing for instance) and most OEMs don't want the customer seeing a rating on the bushing CT and getting it confused with the rating of the entire device.

Note that the thermal current 1-second rating is not related at all to the saturation characteristics of the CT.



 

No, I said you could drive it up to 281 A if your burden is negligible. If you have 0.1 ohm burden, the calc becomes 18/(0.064+0.100), or ~110A secondary, which gives some agreement with the C class rating.

Yes, the statement that "if burden is less than X you will get at least Y current before saturation" is an easy way to say your application is OK, in some cases at least. But what if I have a 200:5 CT with 0.05 ohm external burden and 10kA available? Given the CT knee point and Rct, I can get at least some idea pretty quickly. Or going the reverse direction, it tells me what knee point and/or Rct I need to look for.

I do the full calculation on every job, but "full" in this case is about 1 minute longer than the less full. I have to determine the burden on the CT no matter what, so by the time I have added up the burden, the calc I showed takes an additional 60 seconds, if the CT excitation curve and Rct is available. I am not sure if I am in the minority, but it depends on what crowd you count. Among those that I work with, I think I am just a typical engineer doing typical calcs.
  The CT excitation curve, including Rct, comes in the approval package for every swgr, xfmr, bkr, etc that my company works on. It is a requirement of the specs I believe, but no bidder ever takes any exception that I am aware of.  Me and my fellow engineers take a look at that data. We do the calc I mentioned and need Rct and the CT excitation curve. You will not be selling to the crowd I run with unless you decide to readily supply this data. Take a look at the ITI/GE lineup; you freely download such data.
 
I found your comment on wound primary CTs having a 1sec rating ~100xInom to be informative. I do not deal with wound primary CTs very often. I had not realized that they are more limited on fault withstand than window CTs.

I mentioned the matter on the CT short time rating to say that 281A will not destroy the CT, and you can go above 100A secondary, as long as you clear the fault quickly. The high CT current is a reason you have to clear the fault quickly.

Your response told me why the mfrs resist putting the short time rating on the CT. I had asked for it a few times, and got it, and like you said, it was off the charts, and have not asked for it since.








 

jghrist (Electrical),
                     I really appreciate the useful link you have provided.

Look here from Stanley Zocholl the master!  

• http://files.engineering.com/getfile.aspx?folder=68cae9ca-8335-47b2-adab-53

Ceast-

That is a good paper, but this statement is horribly wrong on many levels:

"Higher ratio cts (3000:5) contribute a resistance of 0.0025 ohms per turn to the burden and lower
ratios (300:5) 0.005 ohms per turn. Consequently, applying a 600 turn (3000:5) ct contributes an
internal resistance of 1.5 ohms."

First, unless the CT is a bushing CT, you can't be sure how many turns are on the secondary. For example, a 230 kV CT with a 300:5A ratio likely has 4 primary turns and 240 turns on the secondary.

Second, there isn't really a good way to give a rule of thumb for ohms/turn based on the ratio. The secondary wire size used is typically a function of the rated secondary current and the rating factor. In other words, the secondary wire is typically the same for the 3000:5A and a 300:5A CT. If anything, lower ratio CTs can often times have larger cross-section cores, which means the lead distance per turn is higher. Also, the diameter of the CT plays a role in secondary resistance, since windings are often equally distributed about the core.

Note sure why such a good paper would be so sloppy with a fairly important parameter.

 

Just an additional reference, this might support on your query.
 

• http://files.engineering.com/getfile.aspx?folder=bdde606f-0cca-4af0-8abd-01

scottf, I'll see if I can't get clarification from M Zocholl.