Star/Star TX - Zero Seq circuit 3

[CDG16]

Could someone please explain the equivalent zero sequence circuit for a 2-winding Star/Star TX (HV ungrounded and LV grounded). The confusion is that PRAG shows that no zero sequence current will flow for ph-E faults. But, the J&P TX Book shows a link to the zero bus on the LV side. (I found several other books showing these circuits both ways).

The only explanation for this is that J&P might assume the tank acting as a delta and therefore the closed link to the zero bus. The problem is, if you assume the PRAG is correct, then for a ph-E fault on the LV, no current will flow. But the winding has basically a short across it - it does not make sense...

Could someone please explain this better.

Thanks in advance.

 

[jghrist]

No current will flow. Think of the 3Ø circuit of a 1:1 ratio ungrd Wye-grd Wye transformer. It will help to draw a diagram of the transformer windings. Assume a short from secondary ØC to ground but no connection to the other secondary phases. Since ground is connected to the secondary neutral, there is a completed circuit shorting the CØ winding.

The same current will flow in the primary CØ winding as in the secondary. But with no connection to ground or a direct path to the source, the only place the CØ primary winding current can flow is through the neutral connection to the AØ winding or the BØ winding. Since there is no current in the secondary AØ or BØ windings, there can be no current in the primary windings. With no current in the primary AØ or BØ windings, there can be no current in the neutral connection to the CØ winding and no current in the CØ winding.

 

[isquaredr]

Jghrist
Just to clarify, with a Wye-grd Wye (Yyn), if a phase -ground fault occurs you say no fault current will flow as it cannot be supported by the primary winding.
Im struggling a little with this theory, have I read your reply correctly or have I got hold of the wrong end of the stick.
thanks

 

[jghrist]

Basically, because there is nowhere for the current to flow in the primary winding, there can be no current in the secondary winding.

The zero-sequence equivalent circuit is

                  Z0
   o-------o  o--[URL unfurl="true"]www---o[/URL]

   o-------------------o

No connection to the neutral bus.

 

[tinfoil]

Umm...
Assuming this is not some sort of April fools...

If you short one phase to ground on the secondary of a Y-Ygnd xmfr, you DO get significant fault current.

We often refer to these as single-line-to-ground faults.

They are very very common.

Imagine a single-phase xmfr with the neutral (X2) grounded on the secondary side. If you were to then ground the 'hot' (X1) bushing, you are effectively shorting out the full secondary winding with a very small impedance. I think it is obvious that there will be significant current flow if the xmfr is energized.

A Y-y transformer is effectively three separate 1-phase xmfrs sharing a common tank and core for three-phase applications.

 

[LionelHutz]

I'd say no current.

Look at jghrist's example a different way. If you short the C phase secondary winding then the primary winding becomes very low impedance compared to the other two windings. Therefore, the equivalent circuit is almost the same as directly connecting the neutral point to the C winding.

So, you will increase the primary and secondary voltages of the A and B phases. The primary current will likely increase due to saturating the transformer core but not to fault current levels.

The connection seems useless though. There's no way to keep the voltages on the LV side balanced.

 

[tinfoil]

@CDG16

Have a look at the diagrams on p492 of

http://www.ece.uidaho.edu/ee/classes/ECE526S05/text/book_2-2005.pdf

 

[tinfoil]

@LionelHutz

I must respectively disagree. My answer was not a theoretical one, but based on years of direct empirical observation working at several power utilities.

Many POCOs use Y-Ygnd xmfrs for all of their distribution (120/208V or 277/480V secondary) transformers. I can assure you that if one phase faults on the secondary cables or in the customer premise, you do get a spectacular display of fault current.

 

[RRaghunath]

This is what I find in IEC 60076-8 Table-1.

The current will flow and the zero sequence impedance for Yyn connected transformer is magnetising impedance of relatively high value ~ 60%.

 

[davidbeach]

tinfoil, what you are referring to is a wye-wye transformer that has both H0 and X0 grounded. This is as shown in figure 11-30 of your reference and performs as you describe. The OP question refers to the case when H0 is not grounded, as shown in figure 11-31 of your reference - note the open in the zero-sequence diagram. With an open in the zero-sequence network there can be no zero sequence current. Current can not exist on the secondary of the transformer if it does not exist on the primary.

 

[tinfoil]

I was of the impression that the 'star-points' (brought out as X0 and 'H0') inside most distribution-class Wye-Wye transformers (as well as small station-class units) were internally connected (and case-bonded) at the factory, which is why
a) The 'H0' bushing is often just a nominal 'stud' welded to the side of the case
b) The utility only needs to connect to neutral / ground the X0 bushing to achieve both high and low side grounding.

I made the assumption that the OP of this thread was observing the single point of grounding made outside of their xmfr, and therefore referred to it as a Y - y(gnd) unit.

 

[tinfoil]

@davidBeach:

Even if the primary side is ungrounded, then you are going to have the situation shown in fig 11.31, only with HV and LV 'reversed'. There is still a sequence path.

Looking at it from my three 1-phase xmfrs on a common core point of view, you have the X2 (X0 on the real 3 ph xmfr) and X1 (X1,X2,X3 on the 3ph) bushings connected across one leg of the secondary winding, and then 'shorted' through the fault impedance. There WILL be a current flow (or else how do you pull single-phase load through a 3ph xmfr?). The secondary fault current can be thought of as a big unbalanced 'load' from the primary point of view, but its magnitude will be largely limited by the xmfr impedance.

 

[jghrist]

If X0 and H0 are internally connected, grounding X0 will ground H0 and you have the normal grd wye - grd wye connection. I have seen ungrd Wye - grd Wye transformers with a delta tertiary. This is an entirely different case because the tertiary makes it a ground source.

The diagram in my previous post is from Fig. 13.17 of Elements of Power System Analysis, William D. Stevenson, Jr., 2nd Edition, 1962, p. 297.

I with the short on the HV (grounded) side of Fig. 11.31 of Tinfoil's source. This indicates that the transformer is a ground source. Fig. 11.31 is the same as Fig. 11.33 (grd Wye - Delta) except for the phase shift.

Maybe the University of Idaho has a text more up-to-date than mine, but I'll trust Stevenson.

 

[tinfoil]

@jghrist:

I assume you accept that a three-phase ungrd Wye - grd Wye xmfr is capable of delivering unbalanced current, if one of the three secondary phases had more load than the other two.

Stated differently, there will be more current on one leg of the secondary if you connect a smaller net impedance (more load) to it w.r.t. the other two phases.

A SLG fault is a special case of 'more load/less impedance', where this impedance is at or near zero. The only choke on this current level is the internal impedance of the transformer in series with the primary network impedance, plus the whatever we assume for fault impedance.

Otherwise, I believe that you are asserting that you can short one secondary leg of the xmfr and see no (significant) current flow.

If I am misunderstanding what you are saying, please correct me.

 

[dpc]

It's possible to have unbalanced current without ground current.

Current flowing into the ground requires zero sequence current. For any transformer, for current to flow in the secondary it must also flow in the primary. This applies to sequence currents the same as "real" current and is required because the corresponding flux required in the core. So in most cases, there will be very little ground fault current when the primary wye is not grounded.

But... for a three-phase core-type transformer, the zero sequence flux in the ungrounded primary can complete its path outside of the core. This does give rise to what can appear to be a high-impedance "phantom" delta tertiary that will allow some zero sequence current to flow in the secondary. GE data I have indicates an equivalent zero sequence reactance of about 5 times the positive sequence. This would not be true for a bank of three single-phase transformers or a shell-type transformer.

 

[LionelHutz]

There WILL be a current flow (or else how do you pull single-phase load through a 3ph xmfr?).

That was my point, the connection under discussion doesn't seem useful because it doesn't allow single phase or phase-neutral loads and it doesn't keep the voltages on the transformers balanced.

Now, according to dpc some current can flow if it's a common core transformer but I still say none will flow if it's 3 single phase transformers in use. Well, in reality magnetizing currents and load currents on the other phases can cause some current to flow but not the same fault levels as you'd see with the primary neutral connected.

 

[tinfoil]

Let's briefly set aside the symmetrical component model, including the zero-sequence artifact that we use to ease our computations, and look at the real equipment that we are modelling:

We have any three phase wye-wye transformer. Assume that Ho and Xo are not connected, and that neither Xo nor Ho is grounded. Assume the transformer is energized by the primary, and that the X1, X2, and X3 points are not connected to anything. Therefore, there WILL be (nominal)voltage from X1 to Xo.

We then place a finite impedance across X1 to Xo(such as the very small impedance we assume during a fault). Regardless of the state of the Xo and Ho grounds, current WILL flow from X1 to Xo due to the voltage difference, with the primary side H1 leg furnishing the power. If Ho is ungrounded, then the other two primary legs will have to furnish this current.

Grounding the Xo bushing and then introducing an X1-to-earth fault is the same thing, except that it involves the earth path in determining the finite impedance.

 

[LionelHutz]

Yes, some current will flow but not fault current. The voltage from X1 to X0 will collapse due to the load you just connected which will limit the possible current flow.

The voltages measured from H1, H2 and H3 to H0 will not remain equal with an unbalanced secondary load.

 

[stevenal]

Try this exercise: When present, I0 is equal in magnitude and direction in the three phases. Kirchoff tells us the sum of the currents entering a node (such as the wye point)is zero. This is easily satisfied if the wye point is connected; to ground for example. The I0 leaving on each phase is balanced by 3I0 returning through the ground. If the wye point is floating, though, the path carrying the 3I0 is open circuited. If 3I0 is zero, so is the I0 in the phases. Of course the two windings do not exist in isolation. If I0 = 0 on one side, it must be zero on the other side as well even if the wye point is connected on that side.

You can also look at the primary neutral to ground open circuit as an impedance that is transformed by the square of the turns ratio to the secondary side where it is still an open circuit.

 

[jghrist]

I agree with Lionel. The voltage across both primary and secondary windings will collapse (neglecting the effects of the phantom tertiary in core type transformers, magnetizing current, stray capacitance, and saturation). That's why there is no current in the shorted winding.

Yes, you can serve unbalanced load and get a current, but the voltages will adjust so that there is no neutral current. You can serve unbalanced load from a wye-delta transformer, but there won't be any zero-sequence current in the primary.

By the way, the next to last paragraph of my previous post should start "I disagree with the short on the HV (grounded) side of Fig. 11.31..." Certainly, the zero-sequence circuit of a wye-wye transformer is not the same as that of a delta-wye transformer. There is no short to neutral on the grd wye - grd wye diagram Fig 11.30.