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I
=
------------------------------------------------------------------
a
2
(
)e
j85
-- - 4.5
+
3
=
---------------------------------------------------------------------------------------------------------------------------- -
(
)+ j 3 sin (85) + 4.5 cos (40) + j4.5 sin (40)
3
cos
85
∠
– (
)
=
4.31
57.76
This is the trip current (4.3A) at the maximum torque
angle of -57.76° (current lags voltage by 57.76°).
The following equation should be used for the angle
of x on the MHO circle:
30
Iax =
---------------------------------------------- -
(
6.96
cos
57.76 x
Step 11. Using Figure A-1, Configurations 2 and 3,
repeat (preceding) Step 10 for BG and CG faults.
Note Targets.
1.1.4
Zone 1 Test/Three-Phase
Step 12. Using Figure A-2, connect current and
voltage circuits and apply:
• V AN = 30V ∠0°
• V BN = 30V ∠−120° IB = 6.67 ∠−195°
• V CN = 30V ∠120°
Using a value of x = 75° (lagging), the current re-
quired to trip is calculated as follows (Note Targets):
V
LN
I
=
------------------------------------------------- - = 6.67 A
(
)
Z
cos
PANG X
–
1P
Since V2.6x uses I A + I B + I C to calculate 3I 0 , a bal-
anced 3-phase current source is recommended to
be used for this test (3I 0 = 0). Figure A-2 gives a con-
cept of the test. If a Doble or Multi-Amp test set is
used, be sure to synchronize the 3-phase currents.
If possible, use a multi-trace storage scope to verify
the waveforms of I
, I
A
B
In order to plot the MHO circle for different input an-
gle (X), the setting of RT (and RU, if OSB option is
included) should be at maximum (15 ohms). Refer to
OSB test for detailed information. Set TTYP = 3PR.
Repeat test; RI2-1 and RI2-2 should be closed. For
TTYP = OFF, or 1PR, or 2PR repeat the test. RI2-1
and RI2-2 should be open.
1.1.5
Zone 1 Test/Phase-To-Phase
Step 13. Two methods can be used for this test.
30
1
(
) 3 ( )e
j40
-- - 4.5
3
30
)
–
IA = 6.67 ∠−75°
IC = 6.67 ∠+45°
±
5%
and I
C.
a. Using T-connection (with Doble or Multi-Amp
Test Unit; refer to Figure A-3 for external termi-
nal connection and configuration 1.
• V
= 1/2 V
@
0°
A
F
• V
= 1/2 V
@
180°
B
F
• V
= 3/2 (70) = 105V @
C
Using V
= V
-V
= 30V
F
A
B
• V
= 15V @ 0°
A
• V
= 15V @ 180°
B
The following table is for BC and CA fault tests when
the T-connection is used:
BC
= 105 ∠90°
V
AN
∠0°
V
= 15
BN
∠180°
V
= 15
CN
= 3.33 ∠-75°
I
F
) required to trip = 3.33A ± 5%,
NOTE: Current (I
A
with an angle of -75 degrees.
b. Using Y-connection
NOTE: This test is actually for φφG testing (see
Figure A-3, configuration 1).
1
2
∠
V AN =
V F
-- -
------ -
0°
2
3
1
2
∠
V BN =
V F
-- -
------ -
–
120°
2
3
3
×
V CN =
-- -
70
–
V
AN
2
or
∠
V
= 17.3
0°
AN
∠
V
= 17.3
–
120°
BN
∠
V
= 96.4
120°
CN
For either T or Y connection, using a value of x = 75 °
(lagging), current required to trip:
I.L. 40-385.5
90°
lead
CA
∠180°
V
= 15
AN
= 105 ∠90°
V
BN
∠0°
V
= 15
CN
= 3.33 ∠−75°
I
F
120°
2
1
∠
2
-- - V
–
F
2
A-3
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