GE C70 Instruction Manual page 428
Capacitor bank protection and control system
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8.1 OVERVIEW
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 8.25 for the B and C-phase voltages are small numbers, while the factor for the V
close to 3. If so, equation 8.25 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 sum
0
of the phase voltages calculated by the relay), or directly measured neutral voltage component (open-corner-delta VT volt-
age).
c) SENSITIVITY
Now consider the consequences of a capacitor element failure in one leg, most conveniently leg C, making a small capaci-
tance change in leg-C capacitance. The effect on the operating signal can be calculated by taking the derivative of equation
8.25 with respect to C
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:
---------- - V
dC
Recall equation 8.29.
Differentiating this gives:
Substituting for V
from equation 8.34 gives:
X
This can be substituted into equation 8.33 to obtain:
8
For our purposes,
C
A
8-8
V
. In the general case, the derivative of the absolute value function is messy, but in our case where
C
d
1
d
-- -
---------- - V
1
k
=
+
OP
X
AB
3
dC
C
C
1
-- - 1
k
k
=
+
+
AB
AC
3
C
C
+
B
C
V
1
-------------------- -
3V
+
–
X
C
A
dV
C
C
+
X
B
C
---------- - 1
---------------------
V
+
+
X
dC
C
C
A
3V
d
0
---------- - V
-------------------------------------------------------------------------------------
=
X
dC
C
d
1
---------- - V
-- - 1
=
OP
dC
3
C
3V
1
=
-- -
-------------------------------------------------------------------------------------
3
C
C
, so
k
1
and
k
B
C
AB
d
---------- - V
OP
dC
C
C70 Capacitor Bank Protection and Control System
V
V
=
–
OP
X
0
3V
k
V
1 k
+
–
+
–
AC
0
B
AB
dV
X
---------- -
dC
C
C
C
B
C
V
1
------ -
V
1
-------
+
–
+
–
0
B
C
C
C
A
A
1
1
V
------ -
------ -
0
–
=
C
C
C
A
A
V
d
–
X
---------- - V
-------------------------------------------- -
=
–
X
dC
C
C
C
1
-------------------- -
+
A
V
d
–
X
---------- - V
----------------------------------------------- -
=
–
X
dC
C
1
k
+
C
A
V
1 k
V
2
k
–
–
–
+
B
AB
C
AB
2
C
1
k
k
+
+
A
AB
AC
dV
X
---------- -
k
k
+
+
AB
AC
dC
C
V
1 k
V
2
k
–
–
–
+
0
B
AB
C
AB
C
1
k
k
+
+
A
AB
AC
1
. This results in:
AC
V
V
1
–
-- -
C
0
------------------- -
3
C
C
1
1
-- -
-------
V
V
=
–
C
0
3
C
C
8 THEORY OF OPERATION
V
1 k
+
–
C
AC
0
=
V
C
C
+
B
C
C
A
V
C
k
+
AB
AC
voltage is
X
(EQ 8.32)
(EQ 8.33)
(EQ 8.34)
(EQ 8.35)
(EQ 8.36)
(EQ 8.37)
(EQ 8.38)
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