GE C70 Instruction Manual page 468
Capacitor bank protection and control system
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OVERVIEW
9.1.4.2 Balanced case
To understand neutral voltage unbalance protection, we can start with Kirchhoff's current law for the neutral node of the
bank:
This expression can be rearranged as follows.
Multiplying both sides by –1 / C
As the left hand side of this equation is equal to zero, we can subtract it from the expression inside the absolute value
brackets in equation 9.25 to get:
1
V
=
-- V
(
1 k
OP
X
3
1
=
-- V
k
X
AB
3
The unbalance ratio k-values that reflect the initial or inherent bank unbalance are chosen as:
Therefore, as seen from the previous two equations, the initial operating signal is zero.
Note that the ratios of the capacitances between phase A and the two other phases are close to unity, and therefore the
correcting factors in equation 9.25 for the B and C-phase voltages are small numbers, while the factor for the V
close to 3. If so, equation 9.25 takes a familiar simplified form:
The V
term in the operate equation can be either the neutral component in the bus voltages (one-third of the vectorial
0
sum of the phase voltages calculated by the relay), or directly measured neutral voltage component (open-corner-delta VT
voltage).
9.1.4.3 Sensitivity
Now consider the consequences of a capacitor element failure in one leg, most conveniently leg C, making a small
capacitance change in leg-C capacitance. The effect on the operating signal can be calculated by taking the derivative of
equation 9.25 with respect to C
where the initial value is zero, the derivative of the absolute function is simply the absolute value of the derivative of its
argument. As such, the derivative is:
9
Recall equation 9.29:
9-8
(
I
+
I
+
I
=
jωC
V
–
A
B
C
A
A
V
(
C
+
C
–
X
A
B
V
(
C
+
C
+
C
)
+
V
C
–
X
A
B
C
A
A
V
(
C
+
C
+
C
)
+
C
–
X
A
B
C
A
and substituting the sum of the phase voltages with 3V
A
C
+
C
B
C
V
1
+
---------------- -
–
X
C
A
+
+
k
) 3V
+
V
(
1 k
–
–
AB
AC
0
B
AB
C
C
C
B
C
B
---- -
---- -
+
k
+
V
---- - k
–
–
–
AC
B
C
C
C
A
A
A
k
. In the general case, the derivative of the absolute value function is messy, but in our case
C
d
1
d
-------- - V
=
--
-------- - V
(
(
1 k
+
OP
X
dC
3
dC
C
C
1
=
-- 1 k
(
+
+
k
AB
AC
3
C
+
C
B
C
V
1
+
---------------- -
–
X
C
A
) jωC
(
) jωC
(
V
+
V
V
+
V
–
X
B
B
X
C
+
C
)
+
V
C
+
V
C
+
V
C
=
C
A
A
B
B
C
C
+
V
C
+
V
C
+
V
C
V
C
–
B
A
C
A
B
B
B
A
(
V
+
V
+
V
) V
+
(
C
C
) V
+
–
A
B
C
B
B
A
C
C
B
C
---- -
---- -
3V
+
V
1
+
V
1
–
–
0
B
C
C
C
A
A
C
+
C
B
C
) V
+
(
1 k
) V
1
+
---------------- -
–
–
C
AC
X
C
A
C
C
+
V
---- - k
–
AB
C
AC
C
A
C
C
B
C
=
---- -, k
=
---- -
AB
AC
C
C
A
A
V
=
V
V
–
OP
X
0
+
k
) 3V
+
V
(
1 k
) V
–
–
AB
AC
0
B
AB
dV
X
)
-------- -
dC
C
C
C
B
C
---- -
---- -
3V
+
V
1
+
V
1
–
–
0
B
C
C
C
A
A
C70 CAPACITOR BANK PROTECTION AND CONTROL SYSTEM – INSTRUCTION MANUAL
CHAPTER 9: THEORY OF OPERATION
)
V
=
0
–
C
X
0
+
V
C
V
C
=
0
–
C
C
C
A
(
C
C
)
=
0
–
C
C
A
yields:
0
=
0
C
C
V
B
+
3V
V
1
---- -
1
-----
–
–
–
–
0
B
C
C
C
A
+
(
1 k
)
)
–
C
AC
=
0
Eq. 9-27
Eq. 9-28
Eq. 9-29
C
A
Eq. 9-30
Eq. 9-31
voltage is
X
Eq. 9-32
Eq. 9-33
Eq. 9-34
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