Ungrounded Wye Unbalance Detection Discussion
Section 1 | Section 2 | Section 3 | Library Index

Figure 3 shows the two line voltages V_AB and V_BC, using Mesh analysis for currents I₁ and I_2, as shown, the matrix form of the mesh current equations is

This will give us

Solving this we find and

I_{1 =}= 16.665amps

I_{2 =}= 25.45719047amps

The line currents with positive directions towards the load given in terms of I₁ and I₂ as

Amps

The voltages across these impedances are

Therefore, is the phase shift in phase A

But in Section 3 we stated that Z₂ = Z₃ therefore this would become

In our example E_BA= , and E_BC

I_A =

*21598.67 happens to equal three times the phase to neutral voltage. Now when rounding off the phase to neutral voltage we can let this become (using only the resultant)

=

Z₁ increases when capacitors are lost in that phase.

To determine the voltage shift, multiple this current by the impedance in phase A. As stated on page 1, a 100 KVAR capacitor at 7200 volts is 518.4 ohms

V_A’N’ = 16.665 amps *518.4ohms = 8640. volts

Which confirms the line voltage calculated above, after a little rounding off.

With three capacitors per phase using the 100KVAR, 7200-volt units as above the total Z_t or in our case X_C will become as follows (See Section 1).

For the total KVAR in a phase we have

As stated earlier 100 KVAR equals 518.4
And 200 KVAR will equal 259.2

Losing one capacitor of the three in phase A we have

Then

The voltage shift will be

When losing two capacitors in phase A, we have

Then

The voltage shift will be

With further review, we can use the following:

Where

N =Number of units in parallel
F =Number of units failed

Therefore, with the example given above we have with one failed unit

and with two units failed

To determine the voltage shift in phase A, we have to multiple the current I_A by the capacitance with the failed capacitors (Z₁)

But from (5) we see that Z₁ =

Substituting in the require I_A*Z₁, we can develop the following:

For a single series group only, we can factor this formula further, and it will reduce to

With multiple series groups the formula becomes

= V_A’N’

V_{AN =} phase to neutral, or unit voltage
N =Number of units in parallel
F =Number of units failed
V_A’N’ = Voltage shift with F units out
S = number of series sections per phase.

Example:

V_A’N’=

Let V_AN = 7200 Volts
   N = 3
   F = 1

Then V_A’N’ =

To determine the amount of shift due to the failure of F units in phase A we simply subtract V_A’N’ BY V_AN

V_NS = 8100 volts – 7200 volts = 900 Volts

V_NS = Neutral shift in Volts

To determine the % of voltage shift we can use

Example

From example above

CONCLUSION

For single series groups only...

The voltage on the remaining capacitors in the affected phase is

=V_A’N’

The voltage in the neutral is

V_A’N’ –V_AN = neutral voltage

The percentage of voltage shift in the affected leg

At the beginning of my "Grounded Wye Unbalance detection Discussion" I mention Mr. Harold Stone of Line Material He also develop a formulas for ungrounded WYE unbalance. His are a little different then mine. But they both work his formula’s are as follows:

The amount of neutral shift due to the removal of F units in one series group is

and

The voltage on the remaining units in one group with F units removed

Neutral shift when one complete series section is shorted

LEGEND

V_t =Applied line to neutral voltage.
V = Rated voltage of the capacitor unit
V_r ==Voltage on remaining units in group with F units removed
V_ns = Neutral shift in ungrounded
S = Number series section per phase
N = Number of units in parallel per series section
F = Number on units removed from one section.