CT saturation 2

[Emadshaaban1987]

Hi Folk,
Here we’ve conducted CT (100/1 2.5 VA)saturation test that used step down TR 20kv/0.4kv 2.5MVA
The reason we made this test ,the old TR was burned.
The result of saturation test is I knee =50mA
V knee 60v.
What this result means?
Above which Amp the CT will get saturated?

 

[electricpete]

This is probably better off in the power forum, but I'll give you my two cents:

This is a secondary test to detrmine inflection point where Iexcitation takes off as you increase Vsec
The excitation current itself is not very meaningful since it represents an error current.
You're interested in Vknee.
Compute the corresponding secondary load current as I2 = Vknee/Zburden where Zburden is the impedance that you have connected to the secondary.
Double check if I2 >> I knee...
...If that is the case than you can compute I1 = I2*(N2/N1) and it will be approximately the primary current required to bring the CT to the knee of the curve which represents the onset of saturation where the exciting/error current starts to take off.


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(2B)+(2B)' ?

 

[che12345]

Dear Mr. Emadshaaban1987 (Electrical)(OP)13 Feb 21 18:41
Q1. "... we’ve conducted CT (100/1 2.5 VA)saturation test .... The result of saturation test is I knee =50mA
V knee 60v. What this result means?
A1. In the IEC world, the (knee-point) of the excitation curve of a CT is defined as [that point at which a further increase of 10% of secondary e.m.f would required an increment of exciting current of 50%].
In this case Vk=60V and Ik=50mA. After this knee-point, if the excitation voltage is increased to 1.1 x 60V= 66V; the excitation current would increase to 1.5 x 50mA = 75mA.
A1.1. Any point below the knee-point voltage, the [increment of the excitation current is approximately equal to the increment of the excitation voltage].
Q2. "... Above which Amp the CT will get saturated?..."
A2. I take it that you wish to know up to what (current level) that this CT would [saturate i.e. the ratio 100/1 fails].
A2.1. The current error in % at rated current, see IEC 60044-1 for detail.
e.g. a) an (accuracy class 1) instrument CT, current error < 1.5% at 20% of rated current.... and <1.0% at 120%.
Note: permissible with (higher error) at the [lower current end] but [higher accuracy at upper end, including 20% over-load].
b) a 5P10 protection CT, with the accuracy class 5P and the accuracy limit factor (ALF)10; the [current error at rated primary current +- 1%]. The (composite error) of [5% at rated 10 ALF]
Che Kuan Yau (Singapore).

 

[electricpete]

I answered the question from a theoretical point of view.... how would we convert secondary excitation test knee point measurements to predict the saturation current on primary IF we wanted to do that.

che12345's answer reminds me that as a practical matter (even in the US) we don't normally use excitation tests for this purpose. We usually use excitation test knee point measurements to check health of the current transformer like you were doing (compare the excitation results to the factory excitation results if available, or at least check for reasonable volts per turn at saturation), not to predict saturation for purposes of verifying the CT will perform accurately to support reliable protective relaying. To predict CT accuracy we normally use the CT accuracy class. Like here Figure 3 tells us that a CT with accuracy class 2.5C100 would have 2.5% accuracy at 20X the CT rated current if the burden was set to keep the secondary voltage at/below 100V with that current (current 20x the CT rating).... so if CT secondary rating were 5A then burden below |Z| = 100V/(20*5A) = 1 ohm.



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(2B)+(2B)' ?

 

[edison123]

Nice explanation, Che.

Muthu

http://www.edison.co.in