CT Saturation calculation

[pukar12345]

I have a CT with following data

Ratio : 300:5
Class : C100
Burden : 0.1 ohm

I don't have magnetisation curve for this CT. I want to calculate whether CT will saturate at fault current 3000A.

Total Burden (Including CT secondary, CT leads and relay) is 0.45 ohm

Excitation voltage (V) = (Fault current/CT ratio)* Total burden
V = (3000/60)*0.45
V = 22.5V

I don't have magnetion curve to compare this voltage.

Is there any other method so that I can be assured the CT won't be saturate.
V =

 

[scottf]

C100 means that the CT can develop 100V on secondary without exceeding 10% error (meaning excitation current below 10A).

This is based on 20 times rated secondary current of 5A into a 1 ohm burden (20 x 5A x 1 ohm = 100V)

Now, you use the term "saturate" and if you really mean staying in the linear region below the knee-point, then you would need to check the curve to be sure. However, from a practical perspective, you should be no where near the knee-point with 3000A primary current and 0.45 ohms total burden, which is 22.5 A, as you calculated.

A good rule of thumb is the actual knee-point voltage is approx. 70-75% of the accuracy limiting voltage (the "100" in C100). So the knee-point of this CT is in the area of 70-75V using the IEEE definition of knee-point (tangent at 45deg).

 

[pwrtran]

Kind of misleading answer.

The secondary winding resistance, remanence and X/R ratio of the system should be also evaluated.

 

[scottf]

pwrtran-

I read that the secondary resistance was included in the 0.45 ohms he listed.

The system x/r ratio and the remanence factor have no bearing on at what voltage the CT will saturate at.

They can play a part in determining what voltage might be present on the CT secondary during a fault, especially if considering a reclosing operation.

How is the answer misleading?

 

[pwrtran]

Yes, I missed 0.45 ohms that includes the secondary winding resistance, sorry about that.

When selecting a CT, the calculated saturation voltage should be > Is*Z.s*(1+ X/R). X/R ration also have effect on time-to-saturate

If you have some residual flux permanently left in a CT, say a C100 class CT, do you think you can drive the CT to the saturation point by build up to same level of voltage or in other words pass same level of the primary current?

 

[AnotherEskomite]

It was quite rightly stated that all the peripherals do not influence the voltage at which the CT will saturate, but I think it is important to note that under steady state fault conditions a CT might not saturate but during transient conditions it might. And it depends on X/R as well as % remanence in the core. This will affect time to saturation unless you have a very big core. Perhaps the question was not asked in that detail, but there are more to CT saturation than just knee point voltage.

 

[scottf]

I agree with the sentiment in both pwrtran and anothereskomite's posts.

A few points:
- I was giving a practical answer. Normal ranges of X/R ratio would not cause this CT to saturate at 3000A primary.
- % remanence is really only a factor for reclosing actions, i.e. breaker open and then close back into a fault after xxx ms. I assumed since he didn't mention anything complicated that it was a simple over-current protection question.

CTs that become magnetized (remenance) will tend to "work off" the remenance over time as a result of varying current levels.