Regarding the Relay Settings

Micom P 111 relay ia an AREVA make Relay.

CT rating :300/1A is the Current Transformer ratio

Settings of I> :

Current setting I> : 1A i.e set for 300 Amps i.e PSM (Plug Setting Multiplier) is 100%

Time setting(sec) : 0.2Sec this as per the relay is not the operating time but it is the Time multiplier Setting

Characteristics : IDMT NI, as per the relay, this implies that this is a equivalent of CDG (Areva make) 3 second relay

Settings of I>> :

Current setting I>> :4A ,this is the highset current i.e set to 1200 Amps

Time setting(Sec) : 0.3sec, this is the definite minimum time setting

Characteristics : DMT

Your query :If 3-ph fault would happen and its fault current is about 1000A(case-1) and 2000A(case-2)

Let us consider Case 1. i.e 1000 Amps fault current. In this case the I>> element does not pick up as it had been set to 1200 Amps. Only the I> shall protect and the relay operating time shall be approximately as per the relay manual

t =TX( (K/((I/Is)^{alpha _{- 1))+L)}}

^{_{K= 0.14 Alpha =0.02, L=0}}

^{_{or approximately}}

t= 3 X TMS /log( Fault current / Equivalent value of current setting in relay)

t= 3 X 0.2 /log(1000/300) = 1.147 seconds

Let us consider Case 2. i.e 2000 Amps fault current. In this case both the I> and the I>> element shall pick up as the fault current is greater than both the settings

I> operating time shall be approximately

t= 3 X TMS /log( Fault current / Equivalent value of current setting in relay)

t= 3 X 0.2 /log(2000/300) = 0.728 seconds

and I>> shall operate after definite minimum time of 0.3 Seconds.

So in Case 2 the I>> protection is faster and shall clear the fault

Note: Only for fault currents above 30000 Amps, the I> protection shall be faster than I>>.

So, if the fault current is limited to less than 30KA, your protection settings are OK for this feeder, you can consider reducing the DMT of I>> if the fault currents are still higher, however, proper coordination for the entire system shall be required and has to be carefully studied

Fault current has basically 3 stages, one is the sub transient stage,that lasts only1-2 cycles i.e 40 ms at the max, where the current will be very high

second is the transient stage where the fault currrent reduces slowly in 3-4 cycles i.e 60-80 ms

third is the steady stage fault current that remains till the fault is cleared.

The currents recorded by the relay pertain to which state is not indicated or given in the manual, however system study experts say that it records the 1st stage, i.e the sub transient stage, where the current is maximum.

From the details given above if the fault current of 6695 Amps persisted continously, then, I> can clear the fault only after 444 milliseconds.

Therefore, it is evident that the fault had been cleared by I>> only.

The time indicated 315 Milli seconds in the relay , is it the relay operating time or the total fault duration ?

If it is the total fault duration, it includes the breaker opening time also.

As I> setting the relay operating time is 444 milliseconds, which is higher than 300 milliseconds set for I>>, hence I>> had cleared the fault after 300 Milliseconds.

More important is that any relay cannot control the fault current, it can only sense it and provide protective relaying, rather in this case only after the fault currents have risen to such a high magnitude in the very first cycle of the fault, the relay had started to sense it as an abnormality.

At the time of fault the there will be heavy /severe voltage dip and this has a cascading effect as all the AC contactors for motors drop off at low voltages, hence tripping of motors at time of fault is common. However our efforts should be to reduce/eliminate the potential threats of fault which are in our control.

Also the relay has the disturbance recording facilty, if you retreive its record, the same will give you a better idea of the fault parameters and much better than the event record values.