embeded generator on the utility feeder 2

[pajce]

I ran some models with embeded generation on utility feeder and I found that SC level seen by the TS relay protection remained constant.
However I came across some papers that say relay will see SC current decrease with new generation connected downstream on the feeder. This doesn't make sense to me.

Any comments?
Should I expect TS relay to see more or less curent for baulted 3phase fault with generation on?

 

[davidbeach]

TS relay? Please define your terms.

 

[dpc]

More sources generally means more fault current.

What is an "embedded" generator?

 

[pajce]

Thanks guys.
To clarify:
Embeded generator is generator that normally work in parallel with the grid.

TS relay is Transformer Station relay, this is first upstream protection of the feeder with embeded generator

 

[dpc]

If this is relay is in between the normal grid power source and the local generator, it may never see the combined fault current from both sources. If the fault is downstream, it will see the normal source contribution. If it is upstream, it sees only the contribution from the local generator, and this will indeed likely be very small.

For coordination purposes, you will need to look at both fault conditions.

 

[jghrist]

The local generation and the transformer station source are basically in parallel.

TS----Z1------+------Z3-------X
     ---->    |     ---->
      ITS     |      IF
              |   ^
              Z2  | IG
              |   |
              |
              G

Current in the fault X comes from TS and from G. Current from G will increase voltage drop in Z3 (compared to drop without G) and reduce the current coming from TS.

 

[slavag]

Jghrist, I'm not understand, why current from G reduce current from TS.
Could you please explain.
Regards.
Slava

 

[pajce]

Jghrist:
thanks a lot.
So it is not that impedance will be apparent for the relay in the TS, but also overcurrent protection will be measuring apparent current which is less then the current that the same relay measures w/o the generation. Great !

 

[pajce]

Jghrist - star for you!

 

[jghrist]

slavag,

Use the diagram in my previous post and take an oversimplified example. Put a 150V source at both G and TS, tying the negative side back to the fault location X. Make all impedances 10 ohms. Without G, the current ITS = IF = 150/(10+10) = 7.5A. With G, the G and TS sources will be in parallel, IF = 150/(10+10/2) = 10A, half (5A) will come from G and half from TS.

 

[slavag]

Jghrist.
Thanks a lot.
Haha, with time we forgot also ohms law and parallel connection of resistors .
Regards.
Slava