GE C70, Инструкция по эксплуатации

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8-8 C70 Capacitor Bank Protection and Control System GE Multilin
8.1 OVERVIEW 8 THEORY OF OPERATION
8
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
X
voltage is
close to 3. If so, equation 8.25 takes a familiar simplified form:
(EQ 8.32)
The V
0
term in the operate equation can be either the neutral component in the bus voltages (one-third of the vectorial sum
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
C
. In the general case, the derivative of the absolute value function is messy, but in our case 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:
(EQ 8.33)
Recall equation 8.29.
(EQ 8.34)
Differentiating this gives:
(EQ 8.35)
Substituting for V
X
from equation 8.34 gives:
(EQ 8.36)
This can be substituted into equation 8.33 to obtain:
(EQ 8.37)
For our purposes, , so and . This results in:
(EQ 8.38)
V
OP
V
X
V
0
– =
d
d C
C
-----------
V
OP
1
3
---
d
dC
C
-----------
V
X
1 k
AB
k
AC
+ +
( )
3 V
0
–
V
B
1 k
AB
–
( )
V
C
1 k
AC
–
( )
+ +
( )
=
1
3
---
1 k
AB
k
AC
+ +
( )
dV
X
dC
C
-----------
=
V
X
1
C
B
C
C
+
C
A
---------------------
+
 
 
3 V
0
–
V
B
1
C
B
C
A
-------
–
 
 
V
C
1
C
C
C
A
-------
–
 
 
+ +
0
=
dV
X
dC
C
-----------
1
C
B
C
C
+
C
A
---------------------
+
 
 
V
X
1
C
A
-------
 
 
V
C
–
1
C
A
-------
 
 
+
0
=
d
dC
C
-----------
V
X

V
X
V
C
–
C
A
1
C
B
C
C
+
C
A
---------------------
+
 
 
---------------------------------------------
– =
d
dC
C
-----------
V
X

V
X
V
C
–
C
A
1 k
AB
k
AC
+ +
( )
------------------------------------------------
– =
d
dC
C
----------- V
X
3 V
0
V
B
1 k
AB
–
( )
–
V
C
2 k
AB
+
( )
–
C
A
1 k
AB
k
AC
+ +
( )
2
-------------------------------------------------------------------------------------
=
d
dC
C
-----------
V
OP
1
3
---
1 k
AB
k
AC
+ +
( )
dV
X
dC
C
-----------
=
1
3
---
3 V
0
V
B
1 k
AB
–
( )
–
V
C
2 k
AB
+
( )
–
C
A
1 k
AB
k
AC
+ +
( )
-------------------------------------------------------------------------------------
=
C
A
C
B
C
C
≅ ≅
k
AB
1
≅
k
AC
1
≅
d
dC
C
-----------
V
OP
1
3
---
V
C
V
0
–
C
C
--------------------
≅
1
3
---
1
C
C
-------
V
C
V
0
–
× ×
=