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41-971.3M
sequence current entirely, and operate the relay on
negative-plus-zero sequence current. Tap A is avail-
able to operate in this manner. The relay picks up at
about tap value for all phase-to-phase faults, but is
unaffected by balanced load current or three-phase
faults.
For ground faults, separate taps G and H are avail-
able for adjustment of the ground fault sensitivity to
about 1/4 or 1/8 of the upper tap plate setting. For
example, if the upper tap plate is set at tap 4, the
relay pick-up current for ground faults can be either 1
or 1/2 ampere. It is possible to eliminate response to
zero sequence current. Tap F provides such an oper-
ation.
The response of the relay to various types of faults
are summarized in the following tables.
Table 1
Phase Faults
Sequence Components
in Sequence
Filter Output
Pos., Neg., Zero
Pos., Neg., Zero
Neg., Zero
Table 2
Ground Faults
Com
Lower Left
b.
Tap
1
C
2
B
3
A
The voltage, V
, impressed by the filter upon the sat-
F
urating transformer is:
V
=
C
I
+
C
I
+
F
1
A1
2
A 2
Table 3 shows the values of the C-constants in Eq. (1).
6
Pickup –
Multiples of T
AB
Taps
3Ø
CA
C
1
.86
B
2
.9
1.0
A
—
0
Ground Fault Pickup
Multiples of T
Tap
Tap G
Tap H
.25
.12
.20
.10
.20
.10
Equ. (1)
C
I
0
0
Type HCB-1 Pilot Wire Relay System
Constants For Equation (1)
Tap
A
B
C
F
G
H
For tap settings of C and H, the voltage, V
impressed by the filter upon the saturating trans-
former is:
V
=
–
.2I
F
4.1. SINGLE RELAY PICKUP (Pilot-wire Open) I
Single relay pickup, I
rent required to operate one relay with the pilot-wire
side of the insulating transformer open circuited (H1-
H4). The single relay pickup point in terms of filter
BC
voltage is:
.53
(
V
=
0.2T Tap C
.66
F
V
=
0.16T (Tap B)
1.0
F
0
where T is the saturating transformer tap value. Sin-
gle relay pickup is defined by equating (2) and (3):
0.2T
=
–
0.2I
Current I
varies with the type of fault. For example,
S
for a 3 phase fault, I
sequence current is present. Substituting I
Equ. (4) and rearranging, the 3 phase fault pickup is:
I
I
=
+
S
A1
For 4 Tap:
I
=
T
=
S
For a phase A to ground fault, if I
Table 3
C
C
1
2
0
0.26
-0.08
0.34
-0.20
0.46
—
—
—
—
—
—
+
.46I
+
4.9I
Volts
A1
A2
A0
, is defined as the phase cur-
S
)V
(
=
0.15T Tap A
F
+
.46I
+
4.9 I
A1
A2
A0
= I
, since only positive
S
A1
0.2T
T
---------- -
=
(3 phase fault)
.2
4 amp
(3 phase fault)
= I
A1
C
0
—
—
—
0
2.5
4.9
,
F
Equ. (2)
S
)
Equ. (3)
Equ. (4)
= I
in
S
A1
Equ. (5)
= I
A2
A0
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