How do we specify Class PX CT?
Knee point voltage (Ek)
The equation for specifying the knee point voltage is as given below:
Ek ≥ Kdc 2 (Isc /n) (Rct + RL) … (5)
where,
Ek – Knee point voltage (V)
Kdc – Factor to account for ‘DC offset’ of fault current
Isc – Maximum fault current through the protected device (A)
n – CT Ratio
Rct – CT secondary winding resistance (Ω)
RL – Total CT secondary cable wiring resistance (Ω)
RL is often written as 2RL. In such cases, RL represents resistance of one cable.
The fault current in a power system has a ‘DC offset’ before the current settles down to a sinusoidal steady state. The DC offset essentially increases the fault current magnitude for the first few cycles. The DC offset depends on the instant of fault and the X/R ratio of the faulted circuit. The most common values found in the literature are as given below:
Kdc = 1 for High impedance differential protection
Kdc = 2 for Biased differential protection
The reason for the difference in ‘Kdc’ values is that the high impedance differential relays are voltage operated relays, whereas the biased differential relays are current operated relays.
Equation 5 is a popular equation which is found in most protection literature, to calculate the knee point voltage. An additional margin of 10% to 50% is commonly used to specify the knee point voltage (Ek).
An understanding of the differential protection is necessary to derive Equation 5. Hence, a brief overview of differential protection and the derivation of Equation 5 are presented in the next section.
Magnetising current (Ik)
The specification of magnetising current is quite often left to the CT manufacturer. Only the required knee point voltage (Ek) is specified. A value of 0.05 A for HV switchgear and a value is 0.15 A for MV switchgear can be used as a guidance.
Secondary winding resistance (Rct)
We need a value for secondary winding resistance (Rct) to calculate the knee point voltage. Following values can be used for the estimation of knee point voltage (Ek).
– 0.003 Ω / turn for 5 A CTs
– 0.005 Ω / turn for 1 A CTs
The final values of ‘Rct’ and ‘Ek’ are determined by the CT manufacturer. Hence, a good communication with the supplier is essential to ensure that the CT meets the site requirements. Consequently, the value of ‘Ek’ is often expressed as a function of ‘Rct’ for the purposes of CT specification and quotation.
Important Note: Equation 5 is applicable only for ‘traditional’ differential protection schemes and not for modern microprocessor-based relays. The equation for microprocessor-based relays is presented in a later section in this blog.
Very informative and clearly laid out. Thank you.
Thanks Alpana.
Regards,
Sesha
Hi
For clarity
The FIRST and FOREMOST criteria for diff prot is to make sure the relay will NOT OPERATE for a through fault even with one CT saturated.
That means you must get the the CT kneepoint Ek correct FIRST!
Assume CT2 is saturated.
Ek of CT1 must be at least sufficient to drive the secondary max fault current Ifmax through the saturated CT2
Ohm’s Law then says
Ek ≥ Is1max x (Rct1 + Rl1 + Rl2 + Rct2)
which simplifies to approximately
Ek ≥ 2 x Is1max x (Rct + Rl)
(you can see that the same equation applies for Ek of CT2 if CT1 is saturated)
Now we can consider the relay setting for that through fault condition where the relay MUST NOT operate!
Again, Ohm’s Law says if CT2 is saturated then the voltage at the relay is
Vs = Is1max x (Rl2 + Rct2)
So the effective relay setting Vr must be greater than Vs by a “margin”
If the relay is a voltage setting relay e.g. MFAC14, then the relay setting Vr must be greater than the calculated Vs.
No stabilising resistor is required as the relay is inherently a high impedance (MFAC14 operating current at any setting is ~20 mA).
If the relay is a current setting relay e.g. MCAG14 , then the relay setting is Id and we need the stabilising resistance.
The total minimum impedance is therefore
Rtot > Vs/Id
Rs = Rtot – Rrelay
If we assume the relay is zero impedance we can simplify to the correct version of Eq 8 as:
Rs > Vs/Id
It is just a Ohm’s Law mathematical coincidence that the Vr_minimum is half of the Ek_minimum
But it is not necessarily true that Vr_actual is half of Ek_actual
Hence Eq 8 is not truly correct that Rs is based on half the kneepoint voltage
The kneepoint voltage could be “5 times the minimum” which would result in a much larger than necessary Rs
That would mean the CT waveform for an internal fault is driven much harder and faster into saturation than necessary making it harder for the relay to operate.
It is really
Rs ≥ (Ek_min/2) / (Id)
I discuss this in my own technical reference site:
https://ideology.atlassian.net/l/c/6v3TpqKv
Hello Rod,
Thanks for your comments. I have updated the blog and included a note regarding the CT resistance.
Regards,
Sesha
I have to strongly disagree!
It is not feasible, and should not be done, to “specify Class P CTs for differential protection”
I agree that modern numerical relays provide some great features and settings that make things a lot easier.
Whilst the Ohm’s Law consideration of the CTs and settings for a Merz-Price Circulating Current High Impedance Differential do not apply in the same way to modern “low impedance relays” (i.e. relays with individual CT inputs not paralleled to the other CTs), the underlying principle of preventing unwanted diff relay operation for a through fault must still be applied.
That cannot be achieved by simply specifying P class CTs .
P class defines what you can connect to the CT in terms of total burden to achieve a certain accuracy at the ALF and rated burden.
PX class defines the construction of the CT to achieve a certain dynamic performance of a certain magnetisation curve and internal Rct.
You could have two CTs with the same P class specification but with VERY different Ek, Ie and Rct.
Consequently for the same through fault current connected to a very low burden such as a modern diff relay, at a particular “required” output current of the two CTs, the terminal voltage could be VERY different.
That means the Excitation current will be very different.
That means the real output current of each CT will be different … and that is not what we want fro a through fault as that means a false differential current calculation.
Yes, the relay settings MAY provide sufficient bias to compensate for differences of P class CTs.
But you have no idea what that difference is going to be when you specify P class CTs
You will ONLY know what the difference is when you test the actual P class CTs to determine their actual Ek, Ie and Rct
The real point of specifying PX is that we really need to know Vk, Ie and Rct anyway for diff applications and so we can ensure the dynamic performance based on the slope Vk/Ie is the same so that the “false differential current” is minimised
Hello Rod,
Thanks for your comments.
The aim of these blogs is to provide an in-depth exposure to CT fundamentals, theory and CT design. Based on the information provided, the protection engineer can make judicious decisions as per site requirements.
The motivation for including the Class P specifications for differential protection in this blog are as below:
1. The Schneider SEPAM manuals recommend the use of 5P20 Class P CTs for their microprocessor based differential protection. I have included a link to SEPAM manual information in this blog.
2. I had the personal experience of tailoring the Class P CT for differential protection. The site had specified and installed a 22kV Class P CT erroneously for transformer differential protection. This was discovered at the time of commissioning the project. Replacing the CT was not an option. I had to research the CT design details to tailor the Class P CT on the site for differential protection. Please refer to my blog on ‘CT Specifications – Gold Report’ for details.
Regards,
Sesha
Very informative
Rodney Huges response adds the value
Very detailed information, like the way it has been presented.
Easy to follow through with calculations.
Give a good insight on the IEEE & IEC CT sizing. Thank you.
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