Capacitor Bank Grounded Wye Unbalanced Detection Discussion
In the late 1950’s Mr. Harold Stone of Line Material (later Line Material was merged by McGrawEdison Company with Pennsylvania Transformer to become McGrawEdison Power Systems, and is now (Cooper Power Systems), has developed and publish several papers on unbalance detection for capacitor banks.
For the grounded WYE connected bank the following formulas were developed:
For the neutral to ground current...
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) 
LEGEND 
I_{N} = 
Neutral Current

I_{U }= 
Rated current of one units 
V_{T} = 
Applied line–toneutral volts

V = 
Rated voltage of capacitor units 
V_{R}= 
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 
METAL ENCLOSED CAPACITOR GROUNDED WYE APPLICATIONS
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 with two capacitors per phase, we have the following:
Let X_{C} = 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 using the same parameters as the balance load given above we have the following:
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:
I_{U} = 1 amp
Let V_{T} = 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

(5)

or

(6) 
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.
Table 1  Unbalance Verification 
Grounded WYE

PhN Volts 
Cap Volts 
KVAR

Unit Current

Total Bank Current

N 
F

%Vn

% In Multiplier

InMultiplier

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 
1 
1 
100 
33.33 
1 
1 
1 
1 
1 
2 
2 
2 
100 
66.67 
2 
1 
1 
1 
1 
3 
3 
3 
100 
100 
3 
1 
1 
1 
1 
1 
1 
1 
100 
25 
1 
1 
1 
1 
1 
2 
2 
2 
100 
50 
2 
1 
1 
1 
1 
3 
3 
3 
100 
75 
3 
1 
1 
1 
1 
4 
4 
4 
100 
100 
4 
1 
1 
1 
1 
1 
1 
1 
100 
20 
1 
1 
1 
1 
1 
2 
2 
2 
100 
40 
2 
1 
1 
1 
1 
3 
3 
3 
100 
60 
3 
1 
1 
1 
1 
4 
4 
4 
100 
80 
4 
1 
1 
1 
1 
5 
5 
5 
100 
100 
5 
1 
1 
1 
1 
1 
1 
1 
100 
16.67 
1 
1 
1 
1 
1 
2 
2 
2 
100 
33.33 
2 
1 
1 
1 
1 
3 
3 
3 
100 
50 
3 
1 
1 
1 
1 
4 
4 
4 
100 
66.67 
4 
1 
1 
1 
1 
5 
5 
5 
100 
83.33 
5 
1 
1 
1 
1 
6 
6 
6 
100 
100 
6 
