GE UR Series C70, Инструкция по эксплуатации

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GE Multilin C70 Capacitor Bank Protection and Control System 9-15
9 THEORY OF OPERATION 9.1 OVERVIEW
9
(EQ 9.63)
The capacitor bank positive sequence inherent unbalance factor setting k
1
is chosen to be:
(EQ 9.64)
This allows the first and second terms in the I
OP
equation above to vanish. The last term, being the product of two small
numbers, virtually vanishes as well. Under normal system conditions (non-fault), the zero-sequence voltage and thus the
zero-sequence current is small. The
k
A
, k
B
, and k
C
value involve the difference between two capacitances that are factory
matched, and so are also small. In addition, it is very likely that they will be of varying signs, and thus their sum will be dou-
bly small. Thus the initial operating signal under normal operating conditions is seen to be virtually zero.
c) SENSITIVITY
Now consider the consequences of an element failure in a typical string, say string A1, making a small capacitance change
in the
C
A 1
capacitance. The effect on the operating signal can be calculated by taking the derivative of equation 8.63 with
respect to C
A 1
.
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:
(EQ 9.65)
The final step assumes , and replaces the phase current vector with its magnitude. This can be written as:
(EQ 9.66)
Alternately, we can say:
(EQ 9.67)
where
Δ
C (pu) represents the capacitance change as a per-unit value of the string capacitance, and I
OP
(pu) represents the
operating signal resulting from the failure in per-unit of the nominal current of the differential source. I
A
represents the
phase A terminal current on the same base. When the system is normal (no fault), I
A
may be taken as the capacitor bank's
rated primary per-phase current I
rated
converted to the differential source base. For example, I
A
= I
rated
/ I
base
, where I
base
is again the differential CT primary current rating, or with system normal:
I
OP
I
DIF
k
1
I
1
k
1
*
I
2
–
( )
– =
I
1
k
A
a
ˆ
2
k
B
a
ˆ
k
C
+ +
( )
I
2
k
A
a
ˆ
k
B
a
ˆ
2
k
C
+ +
( )
I
0
k
A
k
B
k
C
+ +
( )
k
1
I
1
k
1
*
I
2
– – + + =
I
1
k
A
a
ˆ
2
k
B
a
ˆ
k
C
k
1
– + +
( )
I
2
k
A
a
ˆ
k
B
a
ˆ
2
k
C
k
1
*
– + +
( )
I
0
k
A
k
B
k
C
+ +
( )
+ + =
I
1
k
A
a
ˆ
2
k
B
a
ˆ
k
C
k
1
– + +
( )
I
2
k
A
a
ˆ
2
k
B
a
ˆ
k
C
k
1
– + +
( )
*
I
0
k
A
k
B
k
C
+ +
( )
+ + =
k
1
k
A
a
ˆ
2
k
B
a
ˆ
k
C
+ + =
C
A 1
C
A 2
–
C
A 1
C
A 2
+
--------------------------- a
ˆ
2
C
B 1
C
B 2
–
C
B 1
C
B 2
+
---------------------------
a
ˆ
C
C 1
C
C 2
–
C
C 1
C
C 2
+
---------------------------
+ + =
d
dC
A 1
--------------
I
OP
d
dC
A 1
--------------
I
1
k
A
a
ˆ
2
k
B
a
ˆ
k
C
k
1
– + +
( )
I
2
k
A
a
ˆ
2
k
B
a
ˆ
k
C
k
1
– + +
( )
*
I
0
k
A
k
B
k
C
+ +
( )
+ +
( )
=
dk
A
dC
A 1
--------------
I
1
dk
A
dC
A 1
--------------
I
2
dk
A
dC
A 1
--------------
I
0
+ + =
dk
A
dC
A 1
-------------- I
A
=
C
A 1
C
A 2
+
( )
C
A 1
C
A 2
–
( )
–
C
A 1
C
A 2
+
( )
2
--------------------------------------------------------------------- I
A
=
2 C
A 2
I
A
C
A 1
C
A 2
+
( )
2
---------------------------------- =
I
A
2
----
1
C
A 1
----------
× ≅
C
A 1
C
A 2
≅
dI
OP
I
A
2
----
dC
A 1
C
A 1
--------------
×
=
I
OP
pu
( )
I
A
2
----
C pu
( ) Δ ×
=