EXPLANATION OF CT PERFORMANCE IN HIGH IMPEDANCE DIFFERENTIAL RELAY

[appunni]

SIR,
IN A HIGH IMPEDANCE DIFFERENATIAL RELAY CIRCUIT, DURING A SEVERE INTERNAL FAULT, THE OUTPUT CURRENTS OF CTS CONNECTED THE RELAY WILL BE ABSORBED AS MAGNETISING CURRENTS AND VERY HIGH VOLTAGE WILL BE APPEARED ACROSS THE IMPEDANCE WHICH WILL OPERATE THE RELAY.
IS THE ABOVE STATEMENT CORRECT? IF YES, HOW? NO SECONDARY CURRENTS AND HIGH MAGNETISING CURRENT MEANS CT HAS SATURATED WHICH WILL REDUCE OUTPUT VOLTAGE.
THE ABOVE STATEMENT IS SIMILAR TO SOME ELSE I READ IN WARRINGTON'S BOOK 'PROTECTIVE RELAYS:THEORY AND APPLICATION VOL.1'.
IF THERE IS ANY AMENDMENTS TO MY STATEMENTS , PLEASE POINT OUT THE SAME WITH EXPLANATIONS. THANKS.

 

[dpc]

http://www.geindustrial.com/products/applications/ger3961.pdf

There is a good description in J. Lewis Blackburn's Protective Relaying book as well.

Schweitzer has a new digital high-impedance bus diff relay out as well.

 

[peebee]

CT's look like current sources. No current therefore implies high voltage as the CT tries it's darndest to maintain a secondary current flow.

That's the same reason you should never open-circuit a CT.

 

[DiscoP]

appuni
That statement is correct.
However the CT will not saturate if they are correctly specified for the application.
CT requirements are much more onerous in a high impedance scheme, compared to say a low impedance biased scheme for this reason.
If the correct CTs are specified then this protection will be effective.
(The same for all types of protection !)

 

[dpc]

I have to disagree a little with discoP's last message.

The CT in the faulted line will usually saturate - this is part of the design of the high-impedance system and not a problem.

The CT requirements for high-impedance bus diff are not difficult, except that all CTs should be matched for best performance and secondary lead length should be similar.

 

[busbar]

IEEE C37.97 §3 gives a reasonable overview of current-differential relaying, and also some “gotchas” related to its application. Product-specific information can be found in most instructional literature as a single-function device, but the concept is the same—well understood and documented.

 

[DiscoP]

I agree dpc !!

Just after I posted it, I re-read it and realised it was not quite correct, but couldn't find an edit option :-(

If that sentence is deleted, I'll stand by the rest of the post.

I wasn't suggesting the CT requirements are difficult to determine, but they are sized much higer than other types of protection. Onerous musn't be a universally used term.

 

[appunni]

SIR,
THANKS TO ALL FOR YOUR VALUABLE REMARKS. YOU HAVE STATED THAT THE CTS WILL SATURATE. THEN HOW THE CTS CAN PRODUCE HIGH VOLTAGE ENOUGH FOR OPERATION? PLEASE NOTE THAT I AM MENTIONING ABOUT THE CONDITION OF A SEVERE INTERNAL (IN ZONE) FAULT WHICH OF COURSE NECESSITATE A HIGH VOLTAGE FOR TRIPPING OF HIGH IMPEDANCE RELAY. WHAT WILL HAPPEN FOR GENERATOR DIFFERENTIAL RELAY WHERE NO. OF INPUTS ARE ONLY TWO AND FOR BUS DIFFERENTIAL RELAY WHERE NO. OF INPUTS AE NORMALLY MORE THAN TWO?
WITH REGARDS,
APPUNNI

 

[DanDel]

apunni, you may be misunderstanding the normal operation of a CT and relay. Typically, the CT will try to match the secondary current to the primary current that passes through its window at its ratio. It does this by producing just enough secondary voltage to cause that amount of current to flow through the impedance of its secondary circuit. When the secondary impedance is high, it will produce a higher voltage(which is limited by its burden rating) to try to produce the correct amount of secondary current.
An excessive amount of primary current will cause the CT to saturate, which means it won't be able to linearly produce the secondary current in the same ratio to the primary current. This also means that the voltage it can produce is limited, but will remain at this maximum for any currents above saturation.
This is why burden calculations should be performed for a protective relay circuit using the CT rating to make sure that the relay will operate in all conditions of overcurrent or fault.

 

[dpc]

I have a little different understanding of the impact of saturation. Since the secondary voltage is produced by the rate of change of the flux, once the core saturates, the voltage output drops off dramatically. The secondary waveform of a saturated CT is a series of sharp pulses that occur at the lower levels of primary current in each cycle, before the core saturates.

But in the high impedance bus differential scheme, it doesn't really matter. The high impedance of the relay actually forces the difference current through the other CTs instead of the relay. This tends to produce an extremely high secondary voltage. In fact, the problem with high impedance bus diff relays is that the voltage becomes too high for internal faults, requiring use of surge arrestors in the relay to avoid destroying it.

Hope that helps.

 

[appunni]

THANKS FOR ALL WHO GAVE VALUABLE INFORMATIONS.