Split Wye Unbalance Detection Discussion
(Article # GES19233)
Section 1 | Section 2 | Library Index

Split wye is actually ungrounded wye bank, split into two wye sections, with the neutrals of each section connected through a current transformer (for current detection) or potential Transformer (for voltage detection.) Sections generally are made of equal, or nearly equal in size. The current transformer should have voltage rating and class of the system.

1. Not sensitive to system unbalance.

2. Not affected by triple harmonic currents since ungrounded wye capacitor banks does not provide any path for these currents.

3. Standard overcurrent relays can be used for detection.

4. On larger banks, having more than one series group, splitting the wye sections into two separate sections decreases the discharge current during unit failure.

For our calculations we will use the ungrounded wye examples from section 3 of 5. This was with two 100 KVAR, 7200 volts capacitors per phase for one wye and three of the same size capacitors per phase for the other wye configuration.

For current detection we need to find the currents in each leg, phase A will have a larger decrease than phases B and C. The reduced current in phases B and C will be equal.

To find the current in the affected phase (A) we will use the following:

From Section 3 of 5 we found the following:

The actual current in this leg will be

Example from figure 5, with one unit removed in phase A

Then

This confirms the results of formula (4) in Section 3 of 5.

Another way to look at this is using Table 2 from Section 3. In this table we find when one unit is removed with two units in parallel the neutral voltage shifts 120%. We can view this as the delta of V_N.

As stated

This gives us

From (1)

A detail analysis of this approach is shown in Figure 6. With this information we need to see what the voltage will be on the other two phases. From figure 6 we see the following ratio of this neutral shift:

The voltage shift as stated is

The voltage from "N" to the mid point as shown in figure 6 is 0.5 or 3600 Volts.

With the voltage shift of 120% the voltage from "N’" to the mid point is or

Or

and

Or this confirms the 16.6667 Amps in phase A

The base triangle from A to the mid point is or

This gives us two sides of the new triangle created with the shift in the neutral.

If the table is not available the voltage from N’ to the mid point can be arrived at with

This will be

With either method we can now calculate the new voltage on phase B & C.

Again, from figure 6, and as developed from the information in table 2 we have B’ to C as 2160 Volts and A’ to C as 6235 Volts. The base side (A’ to C) will always be .

Using the Pythagorean theorem

This can be obtained with the following

This will yield

To determine the current in phases B & C we need to determine the KVAR drop in these legs.

Then

To determine the unbalance current in the neutral we need to take the difference of the currents in the phases (I_A’-I_B’)

The above calculations looked at two different WYE sections one with two units in parallel and the other with three. We developed methods to look at unbalance conditions. Now what will the results be if we connect the neutral of both WYE sections together and place a CT to detect the unbalance.