GE G30, Инструкция по эксплуатации

Страница: 465 / 698

GE Multilin G30 Generator Protection System 9-9
9 APPLICATION OF SETTINGS 9.1 SETTING EXAMPLE
9
Sequential tripping logic can be implemented in FlexLogic™ as follows:
Figure 9–4: SEQUENTIAL TRIPPING FLEXLOGIC™
9.1.8 SYSTEM BACKUP OVERCURRENT
System backup protection is implemented using a phase time overcurrent element with voltage restraint enabled. The
NEUTRL source will be chosen for this element. The pickup of this element should be set at a safe margin above the max-
imum load expected on the machine.
(EQ 9.17)
The selection of all standard curves (and FlexCurves™) is allowed for easy coordination with system relaying. For the
example system, an IEEE extremely inverse curve will be used and a setting will be chosen such that the operate time for a
three phase fault on the high side of the transformer is 0.75 seconds. For simplicity, the power system contribution is not
considered.
(EQ 9.18)
(EQ 9.19)
(EQ 9.20)
(EQ 9.21)
(EQ 9.22)
The equation for an IEEE extremely inverse curve is as follows:
(EQ 9.23)
where A = 28.2, B = 0.1217, and p = 2. Solving for TDM, we have:
(EQ 9.24)
56
57
54
55
AND(2)
= SEQ TRIP (VO9)
DIR POWER 1 STG2 OP
INLET VALVE OFF(H7C)
AND(2) = SEQ TRIP (VO9)
830731A2.CDR
PICKUP 1.5
Generator Nominal Current
CT Primary
------------------------------------------------------------------------ ×
1.5
6800
8000
------------- ×
1.275 pu
= = =
Transformer Impedance 0.10
162
153
---------- × 0.106 pu on machine base
= =
Impedance to Fault 0.216 0.106
+
0.322 pu
= =
Fault Current
V
X
----
1
0.322
---------------
6800
8000
------------- ×
2.64 pu
= = =
Fault Voltage Nominal Voltage
Transformer Impedance
Total Impedance
---------------------------------------------------------------- × 13.8 kV
10.6
32.2
----------- × 4543 V
= = =
Pickup Reduction
Fault Voltage
Generator Nominal Voltage
------------------------------------------------------------------------
4543 V
13.8 kV
--------------------
0.329
= = =
T TDM
A
I
I
pickup
Pickup Reduction
×
---------------------------------------------------------------------
 
 
p
1
–
---------------------------------------------------------------------------------------
B
+
×
=
TDM
0.75
28.2
2.64
1.275 0.329 ×
------------------------------------
 
 
2
1
–
------------------------------------------------------
0.1217
+
-------------------------------------------------------------------------------
0.88
= =