ETAP: Star-Delta increases the phase-earth fault level upstream.

[JoshuaG]

Hi,

I wonder if any of you could explain the phenomenon behind what is happening with my phase-earth fault currents?

I’m modelling short circuit currents on ETAP, on a large 33/22/11kV network. The problem I am finding is that the phase to earth fault currents are higher when a star-delta transformer is connected downstream of a fault than when the transformer is delta-delta, star-star or no transformer is connected.

An example network with both cases is shown in the picture attached.

In both figures, Bus2 is faulted, with the only difference being the grounding of the transformer T1. The maximum P-E fault current from the supply is 20kA.

The star delta transformer seems to increase the P-E fault current beyond the ability of the supply. This extra current must be coming from the other two phases, but I don’t see the mechanism behind this.

Can anybody give me an idea as to what is causing the current from the other phases to flow into the earth? Or if this is not the case, another explanation?

Thanks in advance.

Joshua

 

[waross]

For a line to ground fault on A phase, consider the B and C phase transformers as an open delta. The A phase secondary is now back-fed by the open delta and feeds back into the primary line, increasing the fault current. You should see increased currents on B and C phases to support the back fed currents.
You may avoid the back-feed by floating the primary wye point but then you may encounter over voltage switching surges.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[wroggent]

The grounded wye connection provides for a lower zero sequence impedance; there are two paths for the ground fault current to flow through, (1) the upstream utility network and (2) the grounded wye. Draw the sequence component networks and you will see that the transformer and utility impedances are in parallel, thus giving a result that appears to be beyond the capability of the utility.

 

[JoshuaG]

Thanks both of you for your (astoundingly quick!) replies 

@waross: You are indeed correct that when I look at all phase currents, the difference is made up by the other two phases. Does this mean that the secondary (delta) winding of T1 is being excited by the B and C phases, which is causing circular currents to flow in the secondary, therefore inducing a voltage in the A phase of the primary? Do you know of a way to mathematically calculate that effect at all?

@wroggent: If the fault current is going to earth in order to get back to the supply, how is it driven to form in the loop that is created between the fault, earth and the star point of T1?

Thanks both again for your answers!

 

[wroggent]

The driving voltage is produced in the way you have asked waross (e.g. circulating current in delta). Perhaps I should have worded my answer a bit differently to say there are two paths to ground (the fault location).

 

[waross]

This is how I visualize the situation.
Remove the transformer on the faulted phase from the transformer bank and consider the other two transformers. They are now an open delta connection. Now describe these transformers vectorily:
Voltage, Voltage drop under load, Available fault current. Because of the phase displacement between the two transformers, the characteristics of the virtual transformer across the open delta are equal to either of the other two transformers. If there is no load on the other two transformers then the impedance voltage and the available fault current of the phantom transformer will be equal to either of the other transformers. This will be one limit of the back-fed current.
Now put the transformer on the faulted phase back in the circuit. It will also have an impedance voltage equal to either of the other transformers. This is the second limit on the back-fed current. Thus the impedance limiting the back-fed current is twice the impedance of one transformer. I am not sure of the two times impedance. Perhaps you could verify that with ETAP. Note: ETAP may be also using the impedance of the distribution circuit from the fault location to the transformer bank location to further limit the fault current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ijl]

I think the explanation is in the circulating current of the delta-windings, see the attached single phase diagram. The fault current returns from earth to the neutral point of the Y-windings of the transformer, and splits there in three equal components in the a-,b-,and c-windings. The a,b,c- components must be equal because of the circulating current in the Delta-windings. The b- and c-components return to the generator "a" through generators "b" and "c", respectively. The a-component of the current circulates back to the earth fault, increasing the earth fault current. The total fault current is 30kA, if it is not restricted.

 

[JoshuaG]

Thanks for the picture and explanation! This does certainly give me something to think about.