Current Transformer Specification 1

[NickParker]

Why would someone specify PX class CTs for an exciting current at a fraction of the knee point voltage, that is at Ie<= 30mA @ Vk/2? Why 30mA and not other values? Why at Vk/2 and Why not at Vk? This was specified in a project at Middle East.

Does this reduce the cost or size?

 

[NickParker]

In this article Link, under section 10.4, the author says that,

Ideally for a internal fault and suppose the fault current supplied by nuetral CT and there is no current from the phase CTS . under such a fault condition the neutral CT,apart from its own maganetising current has to supply the parallel conncetd CTs ie another 4CTs . Hence the relay will not operate for 4 times the magnetising current plus the set current of the relay ( converted to primary ). In otherwords the relay is not sensitive up to the above current.

This is new to me as the neural CT supplying exciting currents to other CTs.

Can experts clarify this?. I thought this is related with the CT specification of exciting current < 30mA

Thank you!

 

[RRaghunath]

With regard to your first question:
The High impedance differential protection relays (specifically the old Electromehanical ones) needed the Vk to be twice the setting voltage Vs, as a minimum to achieve the relay to operate in <40ms. This is the background of excitation current requirement at Vk/2 (which in other words is Vs). The excitation current at the setting voltage Vs (read Vk/2) is required to estimate effective primary operating current for the protection as part of protection calculation.
It need not be 30mA and can be lower as well. I have seen 15mA specified by some. It is all a matter of what effective primary operating current is planned and what is feasible from manufacturing point of view.
With regard to your second question:
The subject is about Restricted Earthfault protection (ANSI 87N). The neutral CT is specified to have same parameters as the phase CTs. In other words, neutral CT shall be identical to phase CTs. In case of internal earth fault, the neutral CT will see the earth fault current but not the phase CT, if the protected equipment happens to be a generator or a transformer secondary winding.
In case of external fault, the corresponding phase CT as well as neutral CT will see the fault. In case the phase CT happens to saturate during the fault, the neutral CT will be supplying excitation current to all the phase CTs (in addition to its own excitation current) as the neutral CT secondary voltage is high when compared to that of the phase CTs.

 

[NickParker]

In case of external fault, the corresponding phase CT as well as neutral CT will see the fault. In case the phase CT happens to saturate during the fault, the neutral CT will be supplying excitation current to all the phase CTs (in addition to its own excitation current) as the neutral CT secondary voltage is high when compared to that of the phase CTs.

If the produced secondary voltage from any of the parallel CTs is higher than the other, it would supply the excitation currents for the remaining CTs in parallel - Both for Internal and External faults. Is it correct?

 

[rcw retired EE]

"If the produced secondary voltage from any of the parallel CTs is higher than the other, it would supply the excitation currents for the remaining CTs in parallel - Both for Internal and External faults. Is it correct? "

For an external fault, one CT will have current out of the zone and at least one will have current into the zone. Assume there is only one source and one faulted outgoing circuit. Each CT will create the necessary secondary voltage at its terminals to push the required current through its secondary circuit, minus its excitation losses. I like to think that one CT is pushing current and the other is pulling it so they are sharing the burden. The two internal voltages will be 180 degrees out of phase. If the CT's are similar, the combined secondary voltage will be about half of the voltage for a an internal fault. The other CT's in parallel will have the resulting voltage impressed on their secondary terminals. The non-linear impedance of each CT, as depicted on its Voltage vs. Current excitation curve, determines how much excitation current it draws from the secondary circuit. Any leftover current goes through the protection relay coil.

For an internal fault, only one CT is pushing current through the secondary circuit of the relay coil and the paralleled CT secondaries. Its excitation voltage is higher so the excitation currents drawn by the other CT's will also be higher. If any CT's terminal voltage goes above the knee of its V/I curve its excitation current is much larger due to the saturation. At some point, the leftover current through the relay coil is below the expected pickup.

For external faults, the driving voltage impressed on the secondary circuit is much less.