7SR242 Siemens Relay Knee Point Voltage Cal

Questions of this nature are not properly asked anonymously.

The time to ask for advice - is before your do the purchasing. And you study and design the protection scheme also before you do the purchasing.

Now that you have it all in hand, . . . connect it and commission it.

You will likely find that the protection malfunctions (trips) on large 'trough-faults', situations when it should not trip.

The concept of knee point is to confirm that the CT is notget saturaed for a teriminal fualt in the transformer , which may draw maximum fault current . The formulae given in the manual is only a guide line for worst condition.

If you know the system fault level at the location of the transformer, you can very well test and Know what will be the reqirement for CT out put voltage .Liberely assume two times the present fault level to accomadate future growth.

If the voltage requirement is less than the Knee point of the Cts you have, then you can successfully commission the relays.

Even if get maginallylesser value, you have ways and means to achieve a solution.

Adopt a higher CT ratio and select a sutable ICT ratio so as circulate rated current in the relay, you can mange to reduce the Knee point requirement.

Best wishes

Using TWO CTs instead of ONE is a silly way of arranging reliable relaying protection. The ICT will be adding considerable burden to the main CT - thus reducing its performance. Also, the ICT will need to be properly sized so as not to saturate on through-faults prior to the main CT saturating.

You need not have seperateICT. The duobias relay 7SR242 have internal ratio correctors which can be used similar to providing ICT. please go through the relay manual ,where such such suggessions are given by the relay manufacturer himself

thank you for your information, can u pls. tell whts the concept of recommendation for diffntl and ref. since it is 4xVS for differential and 2xVS for ref. why it differs. i think you are familiar with 7SR242 realy.

thanks

Hi,

If I am not mistaken the differential protection will use CTs on the phase conductors of primary and secondary sides of the transformer, and not the neutrals.

If this is correct, it would seem therefore that Vk - 2 x Vs for the neutral CTs (for REF protection) would be ok, provided the phase CTs are Vk = 4 x Vs.

When calculating the max through current use I full load / Z% for the transformer, ignoring upstream impedance. this will give the maximum through current forever. Then there is no need to make allowances for future growth.

Fault level increase does not related to the transformer impedance, but due to the system growth in future (i.e)Due to growth in generating sources as well as long distance inter connection of power systems.

Hi,

The formula Ifl/Z provides the upper limit of through current for a transformer. Even if additional generation is added upstream of the transformer, the transfomer cannot pass more current than Ifl/Z for a secondary side fault.