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transformer differential testing

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igorigor

Electrical
Jan 9, 2020
6
Hello,
I have one differential relay and I'am trying to test transformer differential protection.
Using current injection test set I inject currents only from side of transformer.
For example, pick up value for differential protection is 500mA (secondary value):
- I inject in all three phases 500mA and I have a differential protection pick up. This is OK.
- If I inject in only one phase 750mA I get differential protection pick up. This is also OK.
(we have ''correction factor'' between 3 phase and 1 phase injection of 1.5, (500*1.5=750), and ok, I can find mathematical
explanation for this)
- for injection in two phases, we need to inject 565mA (500*1.13=565)
My question is, how to calculate this ''correction factor'' of 1.13 between 3 phase and 2 phase injection.
For example setting is for YNd11.
I hope you understand the question.
Thank you.
 
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Hi again,
I hope somebody can help me with my problem which I describe earlier...
I will try to write it more detail and with easier example....
So i have differential protection relay and I would like to test pick up of differential protection, for example, for Yy0 transformer vector group.
If the pick up setting for differential protection is 0.2 I/Irated (where Irated is 200A primary, and CT is 200/1A)
Using secondary injection test set I simulate currents on Y side of transformer( same is if I inject on the y side) and
I get following pick up values for Diff protection:
- 3 ph injection: I inject 200mA in each phase (this one is easy to calculate 0.2*(200/200) )
- 2 ph injection: I inject 226ma in phase L1 and L2
- 1 ph injection: I inject 300mA in phase L1
How to calculate pick up values for 2 ph and 1 ph injection? I can not find logical solution.
Can you please help me or at least give me some pointers....
Thanks a lot...
 
Transformer differential protection (87) Siemens 7UTx relay
 
Look at manual for 7UT.
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Thanks a lot.
This is very helpful.
Just one more thing. How come I get 226mA for 2Ph injection and 200mA for 3ph injection for HV side.
According to attached table, correction factor should be 1, but I have 226/200=1.13?
This is confusing me....
 
HI....
I finally found this factors in the manual...
Thank you for pointing me...
Now it's clear.
Can't believe I didn't see that.
Thanks one more time.
 
I was happy to soon...
This tables are working excellent for 3ph and 1ph fault, but for 2ph is not matching to the real injected current.
 
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