GE C70 Instruction Manual page 475
Capacitor bank protection and control system
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CHAPTER 9: THEORY OF OPERATION
Also, convert from phase coordinates into sequence components as follows, and equation 9.60 becomes:
I
DIF
Substituting this into equation 9.54, we get:
I
=
I
OP
DIF
=
I
(
k
1
=
I
(
k
1
=
I
(
k
1
The capacitor bank positive sequence inherent unbalance factor setting k
This allows the first and second terms in the I
virtually vanishes as well. Under normal system conditions (non-fault), the zero-sequence voltage and thus the zero-
sequence current is small. The k
matched, and so are also small. In addition, it is very likely that they are of varying signs, and thus their sum is doubly small.
Thus the initial operating signal under normal operating conditions is seen to be virtually zero.
9.1.6.3 Sensitivity
Now consider the consequences of an element failure in a typical string, say string A1, making a small capacitance change
in the C
capacitance. The effect on the operating signal can be calculated by taking the derivative of equation 9.63 with
A1
respect to C
.
A1
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. We assume here that the
currents remain constant, which investigation has shown results in negligible error. The derivative is thus:
d
d
---------- -I
=
---------- - I
OP
dC
dC
A1
dk
=
---------- -I
dC
dk
=
---------- -I
dC
(
C
=
---------------------------------------------------------- - I
2C
=
---------------------------- -
(
C
A1
I
A
---
≅
×
2
The final step assumes
C
≅
A1
C70 CAPACITOR BANK PROTECTION AND CONTROL SYSTEM – INSTRUCTION MANUAL
2
a ˆ
a ˆ I
=
k
(
I
+
I
+
I
) k
+
(
I
+
A
1
2
0
B
1
2
a ˆ
a ˆ k
=
I
(
k
+
k
+
) I
+
(
k
1
A
B
C
2
A
*
(
k
I
k
I
)
–
–
1
1
1
2
2
a ˆ
a ˆ k
a ˆ k
+
k
+
) I
+
(
k
+
+
A
B
C
2
A
B
2
a ˆ
a ˆ k
+
k
+
k
) I
+
(
k
+
–
A
B
C
1
2
A
2
a ˆ
a ˆ k
+
k
+
k
) I
+
(
k
+
–
A
B
C
1
2
A
2
a ˆ
a ˆ k
k
=
k
+
k
+
1
A
B
C
C
C
C
–
2
--------------------- - a ˆ
A1
A2
--------------------- - a ˆ
B1
=
+
C
+
C
C
A1
A2
B1
equation to vanish. The last term, being the product of two small numbers,
OP
, k
, and k
value involve the difference between two capacitances that are factory
A
B
C
2
a ˆ
a ˆ k
(
(
k
+
k
+
k
) I
+
–
1
A
B
C
1
2
A1
dk
dk
A
A
A
+
---------- -I
+
---------- -I
1
2
0
dC
dC
A1
A1
A1
A
A
A1
)
(
)
+
C
C
C
–
–
A1
A2
A1
A2
A
2
(
C
+
C
)
A1
A2
I
A2
A
2
+
C
)
A2
1
------- -
C
A1
and replaces the phase current vector with its magnitude. This can be written as:
C
A2
I
A
---
dI
=
OP
2
2
a ˆ I
a ˆ
+
I
) k
+
(
+
I
+
I
)
2
0
C
1
2
0
2
a ˆ k
a ˆ
+
+
k
) I
+
(
k
+
k
+
k
B
C
0
A
B
C
2
a ˆ
k
) I
+
(
k
+
k
+
k
) k
I
–
C
0
A
B
C
1
1
2
*
a ˆ k
a ˆ
+
k
k
) I
+
(
k
+
k
+
–
B
C
1
0
A
B
*
2
a ˆ
a ˆ k
k
+
k
)
+
I
(
k
+
k
+
–
B
C
1
0
A
B
is chosen to be:
1
C
C
C
–
–
B2
C1
C2
+
--------------------- -
+
C
C
+
C
B2
C1
C2
*
2
a ˆ
a ˆ k
(
k
+
k
+
k
)
+
I
(
k
–
A
B
C
1
0
A
dC
A1
×
---------- -
C
A1
OVERVIEW
Eq. 9-62
)
*
k
I
–
1
2
Eq. 9-63
k
)
C
k
)
C
Eq. 9-64
+
k
+
k
)
)
B
C
Eq. 9-65
Eq. 9-66
9
9-15
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