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Grounded Wye Unbalance Detection Discussion
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In the late 1950’s Mr. Harold Stone of Line Material ( later Line Material was merged by McGraw-Edison Company with Pennsylvania Transformer to become McGraw-Edison Power Systems, and is now Cooper Power Systems) developed and publish several papers on unbalance detection for capacitor banks. |
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For the grounded wye connected bank (See Figure 1) the following formulas were developed:
For the neutral to ground current... |
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 Amps |
(1) |
%IN (of rated phase current)= |
(2) |
The voltage on remaining units in one series section with "F" units removed.
%Vr =  |
(3) |
Neutral current when one complete series section is shorted
%In (of rated phase current) =  |
(4) |
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LEGEND |
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| IN = | Neutral Current |
| IU = | Rated current of one units |
| VT = | Applied line–to-neutral volts |
| V = | Rated voltage of capacitor units |
| VR= | Voltage on remaining units in a group with "F" units removed. |
| S = | Number of series sections per phase |
| N = | Number of parallel units per series section |
| F = | Number of units removed from one section |
It is rare that more than one series group is used in metal enclosed applications. If we make that assumption we can simplify the above formulas.
Looking a balance load (See Figure 2) with two capacitors per phase, we have the following:
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Let XC = 1
Let 
= 1V
Then 
= 
= 0.002653590254 farads = 1var
For 2 units in parallel will equal C1+C2 =0.00530519 farads =2vars
Or ZT = 
=
This gives us
= 2 amps or, 
= 
=
= 2amps
Each leg in a balanced grounded wye capacitor bank as configured above will be 2 amps, and will be zero amps at the ground node point
Looking at an unbalance load (See Figure 3)using the same parameters as the balance load given above we have the following:
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Let XC = 1
Let = 1V
Then =
= 2 amps
= 1amp
This means the current neutral to ground will be 1 amp and this will flow as unbalance current in the neutral.
Looking back at the formula for neutral current formula (1) we had
IU = 1 amp
Let VT = V
N = 2
S =1
F = 1
This gives us 1 amp in the neutral, which confirms our numbers. However, we can really reduce this formula when S = 1.
This reduces to
and even further
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(5) |
or  |
(7) |
and if VT = V, this is really simplified to 
Also the voltage on the remaining units given by
%Vr =
with S = 1, and VT = V, %Vr will always be 100%
For the percentages and per unit values see Table 1 below.
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Table 1 - UNBALANCE VERIFICATION |
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GROUNDED WYE |
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Ph-n Volts |
Cap Volts |
KVAR |
Unit Current |
Total Bank Current |
N |
F |
%Vn |
%In Multipler |
In-Multiplier |
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1 |
1 |
1 |
1 |
1 |
1 |
1 |
100 |
100 |
1 |
|
1 |
1 |
1 |
1 |
1 |
2 |
1 |
100 |
50 |
1 |
|
1 |
1 |
1 |
1 |
2 |
2 |
2 |
100 |
100 |
2 |
|
1 |
1 |
1 |
1 |
1 |
3 |
1 |
100 |
33.33 |
1 |
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1 |
1 |
1 |
1 |
2 |
3 |
2 |
100 |
66.67 |
2 |
|
1 |
1 |
1 |
1 |
3 |
3 |
3 |
100 |
100 |
3 |
|
1 |
1 |
1 |
1 |
1 |
4 |
1 |
100 |
25 |
1 |
|
1 |
1 |
1 |
1 |
2 |
4 |
2 |
100 |
50 |
2 |
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1 |
1 |
1 |
1 |
3 |
4 |
3 |
100 |
75 |
3 |
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1 |
1 |
1 |
1 |
4 |
4 |
4 |
100 |
100 |
4 |
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1 |
1 |
1 |
1 |
1 |
5 |
1 |
100 |
20 |
1 |
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1 |
1 |
1 |
1 |
2 |
5 |
2 |
100 |
40 |
2 |
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1 |
1 |
1 |
1 |
3 |
5 |
3 |
100 |
60 |
3 |
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1 |
1 |
1 |
1 |
4 |
5 |
4 |
100 |
80 |
4 |
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1 |
1 |
1 |
1 |
5 |
5 |
5 |
100 |
100 |
5 |
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1 |
1 |
1 |
1 |
1 |
6 |
1 |
100 |
16.67 |
1 |
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1 |
1 |
1 |
1 |
2 |
6 |
2 |
100 |
33.33 |
2 |
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1 |
1 |
1 |
1 |
3 |
6 |
3 |
100 |
50 |
3 |
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1 |
1 |
1 |
1 |
4 |
6 |
4 |
100 |
66.67 |
4 |
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1 |
1 |
1 |
1 |
5 |
6 |
5 |
100 |
83.33 |
5 |
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1 |
1 |
1 |
1 |
6 |
6 |
6 |
100 |
100 |
6 |
| N = Number of units in parallel F = Number of failed units |
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