Simple question about vector group YnD11 ?

[rayas91]

Friends,

I have a question about vector group YnD11.

If we assume current flows from primary to secodary, angles will be Y (0°) - D (30°).


But if current flows secondary to primary , which angles will occur ? Y (30°) - D (0°)?

Thanks for your time.

 

[ScottyUK]

Are we talking about Dyn11 or YNd11?

 

[rayas91]

ScottyUK , we re talking about YnD11.

 

[ScottyUK]

OK, but that isn't a recognised transformer designation. If you insist on talking about YnD11 then good luck because it's pretty meaningless.

Upper case letters indicate HV connections, lower case letters indicate LV connections. I can guess that the 'n' is associated with the star winding but I don't know which winding is the HV winding because your designation is incorrect.

 

[rayas91]

I know YnD11 means.and designation is ok.Y is HV side. I know normal conditions about vektor groups and angle differences.But if we assume that we will use D as a primary-as a reference , what would Y angle be?

 

[ScottyUK]

I'm sure you won't mind if I stick to the standard winding designations in IEC 60076-1.

In any '11' configuration the LV winding leads the HV winding by 30°. In a YNd11 transformer the delta winding leads the star winding by 30°, so if you use the delta winding as the reference then the star winding has a 30° lag.

 

[rayas91]

ScottyUK ..I mind your data.and I have many standarts and designations..But pratically when we re injection voltage, values are not as same as your or mine idea.

When we get Delta as reference , Y leads 30 degree.

So if delta winding is 0 degree, Y winding is +30 degree.
This is the problem.

 

[ScottyUK]

I'm not sure where that leaves us. It isn't a YnD11 transformer?

The standard test usually links the A and a terminals then applies a suitable (safe) test voltage to the HV winding, and the resulting voltages between terminals prove the vector group. The relative magnitudes are more important than the actual values. The procedure and interpretation of results are well-documented.

 

[rayas91]

ScottyUK thanks for your.time...I think I will call the manufacturer

 

[ScottyUK]

Ok, let us know how things turn out. Good luck.

 

[rayas91]

Much thanks again.

 

[stevenal]

Link

See Figure 15 on page 15 (pdf 17).

 

[rayas91]

stevenal I have that document .And I ve checked already.But It is not my answer.If you have time,can you check the file which is on attach.
I m at powerplant now, Combined Cycle Power Plant.I drawed the single line and CT polarity to paint.

The reason that I asked you , I m configurating GE T60-U Block-Unit Differential Relay.

Thank you.

 

[jghrist]

On your diagram, which side of the YnD11 transformer is Y? You are confusing the relay configurations by including two transformers. There could be different current angles even if both transformers were YY0 because of differences in load angles.

 

[rayas91]

Yeah you re right.

Delta is Generator Side , 15750 V side
Y is 400 Kv Side , 400 kV Side

Dont mind the DYn1 transformer, just consider Gen+Transformer (YnD11)

 

[HamburgerHelper]

The apparent power into a transformer will be close to the apparent power out of a transformer. The voltages will lag like what you said and the power flow will still be in the same power quadrant. If the currents were leading the voltages on the generator side by 30 degrees, the currents will lead the voltages close to 30 degrees on the high side (minus vars loss exciting the transformer). You will see the same if power is flowing the other way. This for for balanced load. For unbalanced load or faults, remember the negative sequence phase shift through the transformer is the opposite of the positive sequence phase shift.

 

[jghrist]

Reference stevenal's Figure 15. Voltage on the wye side (1U) lags the voltage on the delta side (2U) by 30°.

If the current into the delta side of the Ynd11 transformer is at angle 0°, then the current out of the wye side will be -30°. Current into the wye side is 180° from current out of the wye side, so it is at an angle of +150°.

 

[rayas91]

HamburgerHelper and jghrist , thank you for you time and interest.I know It s too simple.

Please check values than you will understand my point.

I will attach here my protection relay set screen .
And the values which I injected as secondary are like below;

Gen Side : HV (Y Side)

I1: 0,083 0° I1: 0,083 -150°
I2: 0,083 -120° I2: 0,083 -270°
I3: 0,083 -240° I3: 0,083 -30°

At the values, Relay dont trip. There is balanced and no differential .

According to jghrist and all informations which i know, It should be -30° ,not -150°.

 

[HamburgerHelper]

The 30 degree difference will look like +/- 150 degree difference if the CTs are not facing the same direction. Draw out the vectors and you'll see why it looks the way it does. SEL relays have a polarity setting. GE relays might have this setting. If it does have this setting, you might have programmed it to work correctly but don't realize that the relay acts on the angle difference due to the CT polarization and transformer connection.

 

[rayas91]

Thanks ,I got you. But GE relays has no polarity setting.
We can understand the polarity just from connections.There s no setting.Because of this ,I need to consider all CT polarities which are related.