I have a question re relay coordina

[redox]

I have a question re relay coordination on electromechanical IDMTL earthfault relays.

Take a system with 10P10 1000/1 CT's.
Relay is 2xOC and 1xEF.
Take a earth fault setting of 10%, meaning that the relay will pick-up at 100A. Most relays claim to be effective up to x20 plugsetting, in this case 2000A. But how can you coordinate when a worst cast earth fault current of say 8000A can flow? (assume a earth fault occur and the system is solidly earthed)

When you coordinate between electromechanical relays you have to take a time of 0,4sec. How do you coordinate a upstream E/F relay, do you take time at 2000A and add 0.4sec? (Tripping time at 2000A will be roughly the same as at 8000A) Or do you have to take a setting of 40%(highest setting) to be in the range of 8000A, take then time at 8000A and add 0.4sec?


Thanks for any help

 

[dpc]

If you have determined that the necessary coordination time interval is 0.4 s, this applies to the maximum fault current that the relays will see. This is often a problem with induction disk relays.

To increase the separation at higher fault currents, slower relays can be used - e.g., instead of very inverse, try astandard inverse relays.

Instantaneous elements on the downstream relays can also help with this problem. With digital relays, time delays can be added to the instantaneous elements to provide additional flexibility.

In the end, it may be necessary to reduce the interval from 0.4 s for high fault currents. For calibrated relays, an interval of 0.3 s may be satisfactory.

Hope that helps.

 

[jbartos]

Suggestion: It is always better to have the faults cleared in the shortest possible time to avoid damages by the fault current. The 0.3sec or shorter separation time and faster relay characteristics can be applied, if there are not other restrictions preventing it.

 

[redox]

dpc:

Thanks for the reply!

I have no problem with the time-interval between the relays.
My problem is on which current to add the 0,4sec.

In the example I've used when I started this thread (10% setting), the relay will be effective up to 2000A, but on higher currents it will not - I think the triptime will be almost the same on a 2000A fault as on a 8000A fault. The electromechanical relays is still much in use today and the earth setting is between 10% and 40% - if my plugsetting is low, I can't grade effectively, but if it is high, my pick-up setting will also be too high. I have no problem with the overcurrent coordination, but I can't see how to do the earth coordination.

 

[jghrist]

As dpc wrote, the current to use is the maximum seen by the relays. Actually you have to look at the entire range of possible fault levels that can be seen by both relays, but the critical value is normally the highest at the downstream relay.

The relay is effective at higher fault currents, but as you say, may have the same operating time. You will have to either increase the pickup or the time dial or both to achieve coordination. With 8000A available ground fault, you don't have to set the pickup at 100A. The lower you set it, the better chance you have to clear high impedance ground faults, but you might have to set it higher in order to achieve coordination. You can also increase the time dial, but at the expense of slower clearing for faults ahead of the downstream device. Same goes for using a less inverse curve, although with a less inverse curve, you could use a lower time dial and get faster clearing for lower current faults.

You could also use an instantaneous o/c in conjunction with reclosing. This way you could get fast tripping for close-in permanent faults.

Relay coordination is full of compromises.

 

[stevenal]

I may be understanding the question a little different. Seems to me the quaestion may be regarding the time/overcurrent graph, which ends at a point less than 80X pickup. In most cases the curve is pretty flat at this point, and won't be much faster if extended to 80X. It won't be slower, so using the right most point on the curve should be fine. The coordinating interval is a minimum value, so the small speed increase unaccounted for will only improve coordination.

 

[jghrist]

For the upstream relay, assuming the curve to be flat after 30 x pickup will not be conservative because the downstream curve has to fit under it. I'm not sure what the official IEC curve definition is, but in the SEL-351 manual, the IEC curves are shown flat after 30 x pickup. There is nothing in the time characteristic equation to indicate it being flat, however.

For example, an inverse curve with TD=0.2 would have a trip time of 0.398 sec at 30 x pickup. This might be enough to coordinate with an instantaneous relay. At 80 x pickup, however, the equation would give a trip time of 0.306 sec which would not provide a 0.3 sec margin over an instantaneous relay and a 3 cycle breaker.

 

[davidbeach]

The SEL-351 changes equations at 30 x pickup and goes to a flat "curve" (definite time) for greater values.

 

[redox]

Thanks again for all the replies

Maybe my original question is a bit confusing - so let me try again

10P10 1000/1 CT's
Earth fault current 8000A

E/F relay setting on 10% (pick-up 100A) but relay is effective up to x20 plugsetting (effective up to 2000A - earth fault current is 8000A (x80 of plugsetting)

How do I compare the effective x20 plugsetting with the aultcurrent of x80? Have I to choose a higher plugsetting? can I use the original plugsetting and assume the time will be the same on the x20 and x80 faultcurrent, and coordinate with that time?

with a Normal Inverse curve:

time (s) = 0.14*k/((Multiple of PS)^0.02-1)

k = 0.2

1. Multiple of PS = 20 time = 0.45 sec
2. Multiple of PS = 80 time = 0.31 sec

But practical the relay is effective up to x20 of PS, so I assume time on x80 fault will also be 0.453 sec. My relay higher have to trip then in 0.45 + 0.4 = 0.85 sec?

Thanks again for all valuable inputs, I hope I am now more understandable!

 

[jghrist]

Basically your are correct for coordinating with an upstream relay. As dpc and jbartos have suggested, 0.3 sec coordinating interval may be adequate, but you must add in breaker clearing time. Consider using an instantaneous relay if you don't have to coordinate with anything downstream.

If the relay manufacturer does not show times beyond 20x pickup, assuming the 20x pickup value would be appropriate for coordinating with upstream relays. The SEL-351 manual shows the curve flat after 30x pickup and accuracy is only defined up to 30x pickup.

davidbeach, I can't find anything in the SEL manual that says that the relay changes curves after 30x pickup although I guess you could imply that from the flattened curves.

 

[davidbeach]

Well, I don't find the 30x pickup change in the 351-5,-6,-7 manual on-line, but I believe I've seen it, though it may have been for one of their other relays, and since they use the same equations for inverse time for all of their relays that I have looked at, if it applies to one, it is fairly safe to say it applies to all. I'm away from all of my manuals for a week, but I'll try to remember to check them next week.

 

[davidbeach]

On page 3-11 of the SEL-501-2 Instruction Manual (Date Code 20020506), the Time-Overcurrent Elements listing includes:

Timing Accuracy: +/- 4% +/- 1.5 cycles for 2<=M<=30; Curves operate on definite time for multiples above 30 or currents above 16 times nominal current.

This relay uses exactly the same formuli for the Time-Overcurrent Elements and hase the same TCC curves as the SEL-351 family, the SEL-300G, and the SEL587 relay; and presummably all the other SEL relays that I don't have manuals for.