Buried Delta tertiary- Zero Sequnce 3

[so0774]

I am an electrical engineer in training working for an engineering firm.
Currently I am working on some technical aspects of a substation of a Wind Farm Project.
The main transformer for this substation has a rating of 34.5/138 KV (wye/wye configuration) Rating with buried Delta in Tertiary.

I have checked Electrical Transmission and Distribution Reference Book from Westinghouse which shows that for Star/Star/Delta
Solidly Grounded Neutrals configurations, calculated impedance parameters for the zero-sequence are the same as for positive sequence.

Now I have been asked to determine whether the Buried delta in the Tertiary winding will have any contributions to the zero-sequence under
Fault conditions. Any ideas?

Thanks

so0774

 

[mpparent]

Zero sequence currents circulate in delta windings.

Mike

 

[Bung]

The tertiary acts as a shunt from the H and L terminals to the reference bus in the zero phase sequence circuit. How much it affects the primary and secondary windings depends on the impedance between the tertiary and each of the primary and secondary (as shown in the Westinghouse book).


Bung
Life is non-linear...

 

[cuky2000]

[blue] Now I have been asked to determine whether the Buried delta in the Tertiary winding will have any contributions to the zero-sequence under fault conditions.[/blue]


The delta tertiary winding do not have any contribution to the zero-sequence under fault conditions.

The delta-winding of the Yyd -transformers permits no zero-phase current flow in the incoming lines of the buried delta-winding. As a result, the delta-connection in the single-phase zero sequence circuit has the effect of an interior short-circuit connection and interruption of the connecting line as shown in the enclose figure from the Westinghouse/ABB T&D Reference Book.

 

[Bung]

The delta winding does remove the zero sequence component from the current into the transformer, but it still has a "contribution" in that it will affect the fault current flowing from the secondary to the fault (assuming a secondary side fault).The higher ZL0, the lower the earth fault current.


Bung
Life is non-linear...

 

[ijl]

The line to earth fault in a YYD-transformer is an interesting, complicated and strange case. Numerical calculations show that the fault current can be up to 1.5 times as high when there is a delta tertiary in the transformer compared with a transformer without the delta. I found the following lengthy explanation to this result:

First, consider a two winding grounded wye - delta transformer. Connect the wye to a zero sequence source: same magnitude and same phase in all windings. The current is then the same in all windings of the wye-side. These currents induce an equal voltage in all windings of the delta-side. Because of the delta-connection, these voltages are in series, and short circuited. The result is the often-mentioned circulating current. The important conclusion is that the delta-side essentially short-circuits the zero sequence current, it does not get any further.

Next, consider the three winding grounded wye - grounded wye - delta transformer. The primary wye is connected to a normal three phase voltage source with voltage U. There is a line to earth fault in the phase A of the secondary wye. The delta is not connected to anywhere. Assume, for simplicity, that the number of turns in all windings is the same, and that the transformer is ideal. The source must have some nonzero impedance Z, otherwise the currents would not be reasonable.

The fault current can be decomposed into positive, negative and zero sequence components. This can be interpreted so that there are a positive sequence, a negative sequence, and a zero sequence current source (or drain, if you like) at the location of the fault. These sources are fictitious sources, not physical sources, of course. Only the zero sequence is considered here. The zero sequence source causes a zero sequence current to flow in the windings of the secondary wye, but the current is shorted and stopped in the delta-connected tertiary windings, as explained above. Thus, there is no zero sequence current in the windings of the primary wye.

Because the transformer is ideal, and because there is a line to earth fault in the phase A of the secondary side, the voltage over the primary winding of the phase A is zero. The current in the phase A of the primary side is thus U/Z, where U is the voltage and Z the impedance of the voltage source. The current in the phase A of the primary winding induces the circulating current in the tertiary delta-winding. (Remember, the voltage source in the primary side is the only physical source. The "sources" at the fault were fictitious, and were needed only to show that there is no zero sequence current in the primary side.) The circulating current again induces currents in the windings of the phases B and C of the primary wye.

Because there is no zero sequence current in the primary wye (no current to the ground), the sum of the currents in the phases A, B and C must be zero. Because the currents in the phases B and C can be thought to be induced by the circulating current, they must be equal. The magnitude of these currents is half of that of the current in phase A, and the phase is opposite to the phase of current in phase A (so that the sum of the currents is zero).

There is some current in all windings of the phase A of the transformer. In the primary side the current is equal to U/Z. The magnitude of the current in the tertiary delta is half of that. (Because the number of turns was assumed to be the same in all windings, the magnitude of the circulating current is equal to that of the currents in the phases B or C. ) The fault current flows in the secondary winding. Again, because the transformer is ideal, and the number of turns are the same, the secondary current must be equal to the sum of the primary and tertiary currents. Thus, the fault current is equal to 1.5*U/Z. When there is no tertiary delta winding, the fault current is equal to the primary current = U/Z. Thus, the fault current with the delta is 1.5 times the fault current without the delta. (QED

 

[jghrist]

ijl,

I don't see why you say there is no zero-sequence current in the primary winding for a secondary wye ground fault on the wye-wye-delta transformer.

Use equivalent circuit C-3 from Westinghouse T&D (in cuky's post). Add a system impedance ahead of terminal a in each network. For a phase-ground fault on the secondary a', connect the pos-seq (with source a-n), neg-seq (a-n shorted), and zero-seq (a-n shorted) networks in series. Terminal a' of pos-seq connected to neutral of zero-seq, neutral of pos-seq connected to a' of neg-seq, and neutral of neg-seq connected to a' of zero-seq networks. there will be current flowing in all three windings of the zero-seq network.

The total impedance around the loop will be 2·Zs + 2·ZM1 + 2·ZH1 + ZH0 + [parallel combination of (Zs0 + ZM0) and ZL0].

 

[ijl]

I assumed an ideal transformer.

 

[davidbeach]

Ideal transformer has nothing to do with it, there will be zero sequence current in the primary grounded wye for a ground fault in the secondary grounded wye circuit. There will also be circulating current in the tertiary delta. Star for jghrist.

 

[ijl]

An ideal transformer has zero impedances. Consider the parallel connected components (Zs0 + ZM0) and ZL0 in jghrist's network. What happens to the current through Zs0, when ZL0 becomes zero?

And yes, there is a circulating current in the tertiary. That was the first thing I explained in my posting.

You can check the currents with a (good) short circuit program. Set the impedances of the transformer to a small value. (The program probably does not accept zero impedances.) The smaller the impedances, the smaller the zero sequence current in the primary.

 

[davidbeach]

If you have a transformer with zero impedance, the transformer will not provide any limitation to the fault current, it won't make it go away. If there are circulating currents in the delta tertiary, there have to be corresponding zero sequence currents in both the primary and secondary grounded wye connections. Current can not flow in one side of a transformer winding without appearing in the other.

 

[ijl]

davidbeach

The impedance of the source limits the currents. It is true that currents must flow on both sides of a transformer, but this does not mean that there should be a zero sequence current both in the primary and in the secondary in the case of a wye-wye-delta transformer.

I suggest that you try to find the answer to the question in my previous posting. You should also check the currents with a short circuit program, as I suggested.

Alternatively, you could solve the currents in the transformer directly by writing the equations for the transformer and the network around it in a straightforward way, without the symmetrical components.