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Ungrounded Wye Unbalanced Detection Discussion (Page 3 of 3)  Figure 1 Figure 2

Using the same parameters as shown in Figure 1, only allowing one capacitor to be removed in phase A, we can see the neutral shift caused by this. Also, we can see the increase in voltage on phase A because of the change in impedance. This case (see Figure 3) so the math is similar and would look like this.

The new numbers are the same for phase B and C.

200 KVAR per phase and the impedance is 259.2 Ohms.

Phase A is 100 KVAR and the impedance is 518.4 Ohms.

Figure 3 shows the Two Line Voltages VAB and VBC, Using Mesh Analysis for Currents I1 and I2, as Shown, the Matrix Form of the Mesh Current Equations is (1)

This Will Give Us  Solving this We Find and I1 = 16.665 amps

I2 = 25.45719047 amps

The Line Currents with Positive Directions Towards the Load Given in Terms of I1 and I2 as Amps  The Voltages Across These Impedances are

 VAO = IAX1 = volts. VBO = IBX2 = volts. VCO = ICX3 = volts. (Line Voltages)

Therefore, is the phase shift in phase A Figure 4

Simplifying this we should only look at the phase (A) that has the failed capacitor and develop the current in that phase in order to determine the voltage shift (See Figure 4).

From Section 2 of Three We Found the Current in Phase A was Determined by (2)

But in Section 3 We Stated that Z2 = Z3 Therefore this Would Become  (3)

In our example EBA , and EBC   IA *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)

 IA = (4) Z1 increases when capacitors are lost in that phase.

To Determine the Voltage Shift, Multiply this Current by the Impedance in Phase A. As Stated on Page 1, a 100 KVAR Capacitor at 7200 Volts is 518.4 Ohms VA’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 Zt or in our case XC 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 Using IA = (4)

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 (5)

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 IA by the capacitance with the failed capacitors (Z1)

But from (5) we see that Z1 = Substituting in the Require IA*Z1, We Can Develop the Following (6)

For a Single Series Group Only, We Can Factor this Formula Further, and it Will Reduce to = VA’N’ (7)

With Multiple Series Groups the Formula Becomes VAN = phase to neutral, or unit voltage
N =Number of units in parallel
F =Number of units failed
VA’N’ = Voltage shift with F units out
S = number of series sections per phase.

Example
VA’N’ Let VAN = 7200 Volts
N = 3
F = 1

Then VA’N’ = To determine the amount of shift due to the failure of F units in phase A we simply subtract VA’N’ BY VAN

VNS = 8100 volts – 7200 volts = 900 Volts
VNS = 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 = VA’N’

The voltage in the neutral is

VA’N’ –VAN = 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 = Neutral shift when one complete series section is shorted TABLE 2 UNGROUNDED WYE WITH PT NEUTRAL TO GROUND Ph-n Kvolts Number series groups Cap KVolts Harm detuning Freq. Rated unit KVAR Effect. unit KVAR Number of units in parallel per series group(N) Number of units remove per series group(F) Neutral shift %Vn Neutral shift in volts Voltage remaining on units in group with F units removed Rated Primary Volts on neutral PT Volts on PT sec with 120 v sec 1 1 1 1 1 1 1 1 150.00 500.00 1500.00 500.00 120.00 7.2 1 7.2 1 100 100.00 2 1 120.00 1440.00 8640.00 7200.00 24.00 7.2 1 7.2 1 100 100.00 2 2 150.00 3600.00 10800.00 500.00 864.00 7.2 1 7.2 1 100 100.00 3 1 112.50 900.00 8100.00 500.00 216.00 7.2 1 7.2 1 100 100.00 3 2 128.57 2057.14 9257.14 500.00 493.71 1 1 1 1 1 1 3 3 150.00 3600.00 10800.00 500.00 864.00 1 1 1 1 1 1 4 1 109.09 90.91 1090.91 500.00 21.82 1 1 1 1 1 1 4 2 120.00 200.00 1200.00 500.00 48.00 1 1 1 1 1 1 4 3 133.33 333.33 1333.33 500.00 80.00 1 1 1 1 1 1 4 4 150.00 500.00 1500.00 500.00 120.00 1 1 1 1 1 1 5 1 107.14 71.43 1071.43 500.00 17.14 1 1 1 1 1 1 5 2 115.38 153.85 1153.85 500.00 36.92 1 1 1 1 1 1 5 3 125.00 250.00 1250.00 500.00 60.00 1 1 1 1 1 1 5 4 136.36 363.64 1363.64 500.00 87.27 1 1 1 1 1 1 5 5 150.00 500.00 1500.00 500.00 120.00 1 1 1 1 1 1 6 1 105.88 58.82 1058.82 500.00 14.12 1 1 1 1 1 1 6 2 112.50 125.00 1125.00 500.00 30.00 1 1 1 1 1 1 6 3 120.00 200.00 1200.00 500.00 48.00 1 1 1 1 1 1 6 4 128.57 285.71 1285.71 500.00 68.57 1 1 1 1 1 1 6 5 138.46 384.62 1384.62 500.00 92.31 1 1 1 1 1 1 6 6 150.00 500.00 1500.00 500.00 120.00 Column "Harm. Detuning Freq."  If the bank is straight power factor use a 1 in this column otherwize use the detuning frequency in the column  Column "Rated Primary volts on neutral PT" use volts not kilovolts"

 LEGEND Vt Applied line to neutral voltage. V Rated voltage of the capacitor unit Vr = Voltage on remaining units in group with F units removed Vns 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

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