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Convert V-I excitation curve to p.u
3

Convert V-I excitation curve to p.u

Convert V-I excitation curve to p.u

(OP)
Hi,

Can somebody tell me how to convert saturation of voltage-current excitation curve in RMS to per unit values.
The only information I have are:
1. current transformer ratio, 1200:5
2. Class- C200, and
3. Data point (V-I) in rms

regards

RE: Convert V-I excitation curve to p.u

You can set the base for conversion to pu as whatever you want, but by convention:
1. Set your current base as 5A because your rated current on the secondary is 5A.

2. Set your voltage base as 10V.  The C200 rating of the CT means that the CT can supply the 20 times the rated current (5A) to a standard burden (for C200, 2 Ohms) with an error less than 10%, which not coincidentally will give you 200V secondary voltage.  Divide 200V by 20 (or multiply 5A by 2 Ohms) and you get 10V.

RE: Convert V-I excitation curve to p.u

(OP)
Thank for your help. But I need further clarification on your second point(2).

If C200 means CT can supply 20 times rated current, then

1.C400 means CT can supply 40 times rated current?
or
2. C400 means CT can supply 20 times rated current with at standard burden of 4 Ohm (i.e. standard burden = 400/(5X20))?

regards

RE: Convert V-I excitation curve to p.u

Number 2.

RE: Convert V-I excitation curve to p.u

The number portion of the C class is called the accuracy limiting voltage and it means the CT can develop that voltage on the secondary.

C200 - means 200V capable on the secondary, which is 20 times 5A (100A) into 2 ohm burden = 200V

C200 also means 40 times 5A (200V) into a 1 ohm burden = 200V

 

RE: Convert V-I excitation curve to p.u

(OP)
This means any combination resulting 200V (C200) will cause CT to saturate
i.e.
1.(20)(5A)(2 Ohm) =200 (saturate)
2.(20)(2A)(5 Ohm) =200 (saturate)
3.(20)(2A)(4 Ohm) =160 (not saturate)

Am I correct? and thank for everyone
 

RE: Convert V-I excitation curve to p.u

PrtectiveBoy-

In short yes.

However, at 200V, the CT should NOT be saturated...at least not to the point where the accuracy is more than 10% off.

By the IEEE definition, most CTs knee-point voltage is actually less than the accuracy limiting voltage, so technically it would be saturated, but not practically.

Normally, the Vkp is approx. 80% of the Val. So a C200 would technically have a Vkp of about 160V.

 

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