PX type CT Specs – Traditional differential protection
We are now ready for the specification for Class PX CTs. The following example is applicable for electromechanical differential relays. It is also applicable to electronic relays based on traditional high impedance protection scheme. Examples of such relays are Schneider Sepam 100LD and Siemens 7SJ8.
Example 1
A substation three-phase busbar has two incomers and four outgoing feeders with no bus section breaker. Other system data is as given below:
Maximum bus fault current (Isc) = 30 kA
Relay rated current (In) = 5 A
CT rated currents = 2000 / 5 A (CT ratio ‘n’ = 400)
CT Resistance (Rct) = 1.76 Ω (0.0044 Ω / turn)
CT Insulation Voltage = 3000 V
Total CT cabling Resistance (RL) = 2 x 15 x 0.02 / 2.5 = 0.24 Ω. (The farthest CT is 15m away and is connected using 2.5 sqmm copper cable)
Note that the CT ratio for busbar protection is not based on rated load current. Usually a higher value of CT ratio is used to reduce the knee point voltage requirement. This reduces the cost and size of current transformers.
Step 1 – CT specification calculations
Assume Kdc = 1 and calculate the knee point voltage using Equation 5.
Knee point voltage (Ek) ≥ 2 (Isc / CT Ratio) (Rct + RL)
≥ 2 (30000 / 400) (1.76 + 0.24)
≥ 2 (75) (2)
≥ 300 V
Allowing for a margin of 10%, let us choose Ek = 330 V.
The knee point voltage is dependent on the CT winding resistance. It is a good practice to use a conservative value of CT resistance for calculations. However, the manufacturer can be requested to quote an alternative (cheaper) CT, based on the actual CT resistance.
Class PX specification also includes the specification of magnetising current (Ik) at the specified knee point voltage (Ek). The value of ‘Ik’ depends on the CT design and is often left to the manufacturer. A value of 0.05 A is used here for illustration.
The complete specification of PX type CT is as given below:
0.05 PX 330 Rct 1.76 Ω
where
Ik = 0.05 A, Ek = 330 V & Rct = 1.76 Ω
Note: The following calculations are NOT required for Class PX CT specification. It has been included here to illustrate high impedance protection concepts.
Step 2 – Stabilising resistor calculations
Firstly, we need to establish the differential pickup current setting for the relay. The relay rated current (In) is 5 A. Let us allow 5% tolerance for each of the two incomer and four feeder CTs. There are six CTs connected in parallel per phase!
Relay pickup current setting (Id) = 6 x (5% of 5 A) = 1.5 A
The stabilising resistance is calculated using Equation 8.
Stabilising resistance (Rs) = (Ek / 2) / (Id) = (330 V / 2) / 1.5 A
= 110 Ω
Step 3 – Surge voltage calculations
The theoretical maximum surge voltage (peak to peak) can be calculated using the following equation – assuming that the relay resistance is negligible.
Vp(theoretical) = 2 x sqrt(2) x (Isc/n) x (Rs)
However, a practical value of surge voltage (peak to peak) considering the CT saturation is calculated using the following empirical equation:
Vp = 2 x sqrt(2) x sqrt( Vk x [(Isc/n) x (Rct+RL+Rs) – Vk] )
= 2 x sqrt(2) x sqrt( 330 * [(30000/400)*(1.76+0.24+110) – 330] )
= 2 x 1.4142 x sqrt( 330 * 8070 )
= 4,615 V
The surge voltage is 4,615 V. The CT insulation level is given to be 3,000 V. Allowing for a safety margin, the maximum allowable voltage is, say, 2500 V. Hence, it is necessary to install a surge limiter.
Note: The calculation steps in Example 1 are also applicable for traditional biased differential protection with operating and restraining coils. Note that appropriate ‘Kdc’ value should be used for calculations. The stabilising resistor and surge limiter calculations are not relevant for biased differential protection.
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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