SEL-387, 487, 587 restraint elements
SEL-387, 487, 587 restraint elements
(OP)
Does anyone have a procedure for testing the restraint differential elements using single phase test source while on the Y-D configuration ?
Relay manual suggest using identical winding cofiguration which in effect defeats the matrix phase adjustments between windings.
Thanks for any suggestion.
Relay manual suggest using identical winding cofiguration which in effect defeats the matrix phase adjustments between windings.
Thanks for any suggestion.






RE: SEL-387, 487, 587 restraint elements
The current injected on the Y side is divided by the TAP setting × SQRT(3)
The current on the D side is divided by TAP only.
Note that it operates two phases at the same time, but I don't think this is an issue with numerical relays.
RE: SEL-387, 487, 587 restraint elements
With a YDAC configuration, high side leads low side by 30, my best guess is injection Ia and Ic on the Y side with m=o where Ic=-Ia, 1pu, single phase input. Ia at O and IC at 180 degree.
If I inject 1.732 pu -Ia into the D side with m=1 it breaks down into 1pu -Ia and Ic. Ia at 180 and Ic at 0. Vector Sum of all inputs and outpus should be zero and I can start from there ?
I have never tried it yet, but if I start to reduce D side input, I should get a trip at some point, with restraint below 2 pu.
I had problems testing DYAC configuration in the past and have "always" resorted to changing to YY to make testing easier but I do not think that was a proper or correct test when the in-service configuration is YDAC.
So I need a plan for next time.
RE: SEL-387, 487, 587 restraint elements
Aph on the Y side is 0 deg, the A Ph on the DAC side is 180 deg (meaning Bph is also 0 deg)
Inject TAP1×1.732 on the Y side, inject TAP2 on DAC side - it will give you zero differential current, with restrain current = 1pu
I assume you have two sources available ?
If you inject A-C on the Y side, you need a 2-1-1 on the DAC to get no differential current.
RE: SEL-387, 487, 587 restraint elements
| 1 -1 0|
CTC(1)= 1/(root 3)| 0 1 -1|
|-1 0 1|
Multiply Ia<180 and Ib<0
result in pu are:
Ia= 2.0<180
Ib= 1.0<0
Ic= 2.0<0
Ic is generated from the Ib component