GE G60 Instruction Manual page 410

5.6 GROUPED ELEMENTS
Measurement of R is accomplished by injecting a voltage, V
G
algorithm must be capable of discriminating between capacitive current due to C
current due to a fault. The exciter voltage, V
essentially resistive. The current I
Figure 5–134: EQUIVALENT CIRCUIT, SINGLE-POINT AND DOUBLE-POINT INJECTION
Referring to the single-point injection circuit, the magnitude of I
5
impact on the voltage drop across Z
written:
Where I is the current flowing due to V
G1
For the double-point injection circuit, R
The V voltage is therefore composed of a square wave to create two levels of injection. Once the value of R
INJ
can be substituted into the V
cuit) is known through measurement then the location of the fault is simply V
percentage of field winding from negative terminal in case of single point injection. If double point injection is used, fault
location cannot be determined. The relay displays an invalid fault location for approximately 10% for such conditions. The
fault location cannot be determined if the field voltage is zero (that is, when the generator is not running). The fault location
is displayed only when the measured field ground resistance is less than 500 KΩ. Refer to Field Ground Module Instruction
Manual for details of wiring and installation.
GROUND UNDERCURRENT:
A brush lift-off condition will prevent the field ground detector from operating. This is detected by calculating the RMS value
of the ground current. It will normally have a nonzero value due to the capacitance of the field winding. A drop in this signal
indicates an open circuit in the injection path and the field ground under current feature detects this condition.
5-250
is a DC value with a small ripple such that the impedance of the field is
FLD
is a DC current with a range of tens, hundreds, or thousands of amps.
FLD
. Therefore if two values (V
F2
V I R
= +
INJ1 G1 G
V I R
= +
INJ2 G2 G
and I is the current flowing due to V
INJ1 G2
( V V )
–
INJ1 INJ2
----------------------------------------------------------------------------------------- -
R
=
G
I
G1
is defined as:
CL
R
=
CL
equation above to determine V
INJ
G60 Generator Protection System
and measuring the resulting current, I
INJ
(which can be significant) and resistive
F
makes it evident that V
FLD
and V ) are injected the following equations can be
INJ1 INJ2
I R V
+
G1 CL
F2
I R V
+
G2 CL
F2
. Solving for R
INJ2
( I I ) R ×
– –
G1 G2 CL
I
–
G2
R
C
-------
2
. If the V voltage (refer to the single-point injection cir-
F2 FLD
/ V . This gives the location of the fault as
F2 FLD
5 SETTINGS
. The measurement
G
cannot have a significant
INJ
(EQ 5.43)
we get:
G
(EQ 5.44)
(EQ 5.45)
is known it
G
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