Split WYE Unbalance Detection Detection Discussion (Page 2 of 2)
First I would like to mention that IEEE C37.99 covers a configuration similar to this. Their formulas are for the WYE sections to be equal.
These Formulas are as Follows (Unbalance Calculation Formula)
1. The amount of current flowing between neutrals of the two equal halves. Current transformer is assumed to have zero impedance.
I_{N} = 
(9) 
%I_{N} =
(PERCENT OF RATED TOTAL BANK CURRENT) 
(10) 
2. The Voltage remaining on units in one series section with F units removed.
%VR =

(11)

3. Permissible number of units that can be removed from one series section to determine the result in a given %VR on the remaining units in that section.
F =
If F is fractional use the next lower whole number.

(12)

LEGEND

Vt 
Applied line to neutral voltage. 
V 
Rated voltage of the capacitor unit 
V_{r} 
Voltage on remaining units in affected group with F units removed.

I_{N} 
Current between neutrals of two banks halves. 
I_{U} 
Rated current of one unit. 
S 
Number of series sections per phase. 
N 
Number of parallel units in one series section. 
F 
Number of units removed from one series section. 
TABLE 3
SPLIT WYE UNBALANCE WITH CT CONNECTED BETWEEN NEUTRALS

System Volts 
Capacitor Volts 
# Series Sections 
# of Units per Section
per Phase 
# of Failed Units 
Unit KVAR 
Unit Current 
Current Between Neutrals 
Percent Volts
on Remaining Units 
1000 
1000 
1 
1 
1 
1 
1.00 
0.600 
120.00 
1000 
1000 
1 
2 
1 
1 
1.00 
0.545 
109.09 
1000 
1000 
1 
2 
2 
1 
1.00 
1.200 
120.00 
1000 
1000 
1 
3 
1 
1 
1.00 
0.529 
105.88 
1000 
1000 
1 
3 
2 
1 
1.00 
1.125 
112.50 
1000 
1000 
1 
3 
3 
1 
1.00 
1.800 
120.00 
1000 
1000 
1 
4 
1 
1 
1.00 
0.522 
104.35 
1000 
1000 
1 
4 
2 
1 
1.00 
1.091 
109.09 
1000 
1000 
1 
4 
3 
1 
1.00 
1.714 
114.29 
1000 
1000 
1 
4 
4 
1 
1.00 
2.400 
120.00 
1000 
1000 
1 
5 
1 
1 
1.00 
0.517 
103.45 
1000 
1000 
1 
5 
2 
1 
1.00

1.071 
107.14 
1000 
1000 
1 
5 
3 
1 
1.00 
1.667 
111.11 
1000 
1000 
1 
5 
4 
1 
1.00 
2.308 
115.38 
1000 
1000 
1 
5 
5 
1 
1.00 
3.000 
120.00 
1000 
1000 
1 
6 
1 
1 
1.00 
0.514 
102.86 
1000 
1000 
1 
6 
2 
1 
1.00 
1.059 
105.88 
1000 
1000 
1 
6 
3 
1 
1.00 
1.636 
109.09 
1000 
1000 
1 
6 
4 
1 
1.00 
2.250 
112.50 
1000 
1000 
1 
6 
5 
1 
1.00 
2.903 
116.13 
1000 
1000 
1 
6 
6 
1 
1.00 
3.600 
120.00 
Unequal WYE Consideration
Figure 7
However, these formulas need modification in order to work when the WYE sections (halves) are not equal.
Actually this method with section 1 with 3 units and section 2 with 2 units will give us five (5) units in parallel. The voltage on the remaining units using (12) is for equal halves. Therefore, we have to force this to work for us by making N=2 ½.
This Gives us
N=2 ½
S=1
F=1
V_{T}=V=7200 Volts
However, this will not give us the exact current when units are lost in different WYE sections. We will now develop the neutral current by removing units in phase A from the different WYE section.
For our work we will again only consider one series group.
From Section 2 with one unit removed in phase "A" we have
Removing one unit from "A" phase in section 2 we have the following
From (4)
This gives "A" phase voltage (V_{A’N’}) of
The current in "A" phase (I_{A’N’}) is
To determine the voltage on phase B AND C we can use (8)
The effective KVAR for this phase will be
The current in V_{B’N’} is
Section 1 neutral current is 26.8AMPS – 23.84AMPS = 2.96AMPS.
From section 1 the neutral current will be
Phase A voltage is the same as section 2, at 7714.08 Volts.
The phase "A" current will be
For I_{B’N’} the V_{B’N’} will be the same as section 2 = 6795Volts, again
Section 1 neutral current is 44.6AMPS –40.26AMPS = 4.34AMPS.
The neutral current with one unit removed in section 2 is 4.34 + 3.7 = 8.04 AMPS.
From section 1 with one unit removed in phase "A" we have
Removing one unit from "A" phase in section 1 we have the following
From (4)
This gives "A" phase voltage (V_{A’N’}) of
The current in "A" phase (I_{A’N’}) is
To determine the voltage on phase B AND C we can use (8)
The effective KVAR for this phase will be
The current in V_{B’N’} is
Section 1 neutral current is 40.26AMPS – 35.714AMPS = 4.55AMPS.
From section 2 the neutral current will be as follows
Phase A voltage is the same, as section 1, at 7714.08 Volts and multipliers will be the same as above.
Therefore, phase "A" current will be
1.714 * 27.777 = 29.761 AMPS
Phase "B" current will be
For I_{B’N’} the V_{B’N’ }will be the same as section 2 = 6795Volts, again
Section 2 neutral current is 29.76AMPS –26.84AMPS = 2.92AMPS.
The neutral current between sections with one unit removed in section (1) is 4.55 + 2.92 = 7.47 AMPS
If two units are remove in one of the phases we can see the effect.
Again using formula (2) we have
This gives use 1.1538 times 7200 = 8308 Volts across the affected leg. We will continue with phase "A" in section (1).
The current in this leg we use formula (5)
We will not go through the rigors of obtaining all the numbers but will look at the pertinent values.
The voltage on phase B and C will be 6714.55 Volts and the current in these legs will be 38.86 AMPS. The neutral current in this section is
38.86 Amps28.846 Amps = 10.014 Amps.
In section (2) the phase "A" voltage 8303 Volts, the effective KVAR is 266 KVAR and current in this leg is 32.017 amps.
Phase "B" and "C" voltage will be same section (1), 6714 Volts and the effective KVAR is 173.94 KVAR. The current in phase "B" and "C" will be 25.9 amps.
The neutral current in section (2) is
32.017 amps – 25.9 amps = 6.11 amps
The neutral current between sections with two units removed in section (1) is 10.014 + 6.11 = 16.12 amps
If two units were removed in section (2) the neutral current between sections would be approximately 33amps.
This table is with uneven number of capacitor units in each wye section
System Voltage: 7200
Cap. Voltage:7200
TABLE 4  SPLIT WYE UNBALANCE WITH CT CONNECTED BETWEEN NEUTRALS

#UNITS OF PAR. IN SEC. 1 
#UNITS FAILED IN SECT.1 
UNIT KVAR SECT.1 
UNIT CURRENT SECT.1 
VOLTAGE ON PH A 
#UNITS OF PAR. IN SEC. 2 
#UNITS FAILED IN SECT.2 
UNIT KVAR SECT.2 
UNIT CURRENT 2 
VOLTAGE ON PH A 
% VOLTS REMAING UNITS 
Total Neutral current 
2 
0 
100 
13.89 
7200.00 
3 
0 
100 
13.89 
7200.00 
1.00 
0.00 
2 
1 
100 
13.89 
8307.69 
3 
1 
100 
13.89 
8307.69 
1.15 
16.69 
2 
2 
100 
13.89 
9818.18 
3 
2 
100 
13.89 
9818.18 
1.36 
41.94 
3 
0 
100 
13.89 
7200.00 
4 
0 
100 
13.89 
7200.00 
1.00 
0.00 
3 
1 
100 
13.89 
7957.89 
4 
1 
100 
13.89 
7957.89 
1.11 
15.78 
3 
2 
100 
13.89 
8894.12 
4 
2 
100 
13.89 
8894.12 
1.24 
36.57 
3 
3 
100 
13.89 
10080.00 
4 
3 
100 
13.89 
10080.00 
1.40 
65.31 
4 
0 
100 
13.89 
7200.00 
5 
0 
100 
13.89 
7200.00 
1.00 
0.00 
4 
1 
100 
13.89 
7776.00 
5 
1 
100 
13.89 
7776.00 
1.08 
15.31 
4 
2 
100 
13.89 
8452.17 
5 
2 
100 
13.89 
8452.17 
1.17 
34.15 
4 
3 
100 
13.89 
9257.14 
5 
3 
100 
13.89 
9257.14 
1.29 
57.95 
4 
4 
100 
13.89 
10231.58 
5 
4 
100 
13.89 
10231.58 
1.42 
88.97 
5 
0 
100 
13.89 
7200.00 
6 
0 
100 
13.89 
7200.00 
1.00 
0.00 
5 
1 
100 
13.89 
7664.52 
6 
1 
100 
13.89 
7664.52 
1.06 
15.03 
5 
2 
100 
13.89 
8193.10 
6 
2 
100 
13.89 
8193.10 
1.14 
32.78 
5 
3 
100 
13.89 
8800.00 
6 
3 
100 
13.89 
8800.00 
1.22 
54.07 
5 
4 
100 
13.89 
9504.00 
6 
4 
100 
13.89 
9504.00 
1.32 
80.14 
5 
5 
100 
13.89 
10330.43 
6 
5 
100 
13.89 
10330.43 
1.43 
112.76 
6 
0 
100 
13.89 
7200.00 
7 
0 
100 
13.89 
7200.00 
1.00 
0.00 
6 
1 
100 
13.89 
7589.19 
7 
1 
100 
13.89 
7589.19 
1.05 
14.84 
6 
2 
100 
13.89 
8022.86 
7 
2 
100 
13.89 
8022.86 
1.11 
31.89 
6 
3 
100 
13.89 
8509.09 
7 
3 
100 
13.89 
8509.09 
1.18 
51.69 
6 
4 
100 
13.89 
9058.06 
7 
4 
100 
13.89 
9058.06 
1.26 
74.99 
6 
5 
100 
13.89 
9682.76 
7 
5 
100 
13.89 
9682.76 
1.34 
102.82 
6 
6 
100 
13.89 
10400.00 
7 
6 
100 
13.89 
10400.00 
1.44 
136.63 
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