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
I_{1} = = 16.665 amps
I_{2} = = 25.45719047 amps
The Line Currents with Positive Directions Towards the Load Given in Terms of I_{1} and I_{2} as
Amps
The Voltages Across These Impedances are
V_{AO} = I_{A}X_{1} = volts.
V_{BO} = I_{B}X_{2} = volts.
V_{CO} = I_{C}X_{3} = 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 Z_{2} = Z_{3} Therefore this Would Become

(3)

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)
I_{A} =

(4)

=
Z_{1} 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
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, 7200volt 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
Using I_{A} =

(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 I_{A} by the capacitance with the failed capacitors (Z_{1})
But from (5) we see that Z_{1} =
Substituting in the Require I_{A}*Z_{1}, We Can Develop the Following

(6)

For a Single Series Group Only, We Can Factor this Formula Further, and it Will Reduce to
= V_{A’N’}

(7)

With Multiple Series Groups the Formula Becomes
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
=
Neutral shift when one complete series section is shorted
TABLE 2
UNGROUNDED WYE WITH PT NEUTRAL TO GROUND

Phn 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 
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 
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