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I.L. 40-385.1B
• Phase current (∆I
A
12.5% change
• Ground current (∆I
• Voltage ( ∆Van, ∆Vbn, or ∆Vcn ) >7V and 12.5%
change with a current change of ∆I>0.5 A
When one of the above is true, MDAR starts fault pro-
cessing. In order to perform the above, apply a certain
value of current suddenly. If MDAR does not trip, turn
the current off, readjust to a higher value, and then
suddenly reapply current.
The current required to trip can be calculated using the
following:
V
I
=
-------------------------------------------------------------------------------------------
(
Z
COS PANG X
1G
From Table 6-1:
= 4.5 Ω
• Z1G
• PANG
= 75°
• ZR
= 3.0
using an X = 75° (lagging)
The current required to trip = 4.00A ± 5 % for fault cur-
rent lagging fault voltage by 75°. This is the maximum
torque angle test. For other points on the MHO circle,
change X to a value between 0° and 150°, and calcu-
late the value of I.
See Table 4-3, for a description of the following dis-
played fault data for:
• Fault Type (FTYP)
• Targets (BK1, Z1G)
• Fault Voltages (V
A
(I
, I
, I
, 3I
)
A
B
C
0
With the external jumper connected between TB1-13
and TB1-14, the BFIA-1, BFIA-2 should be closed. The
GS contact will be ON for approximately 50 ms. when
the fault is applied. Alarm 2 relay will be picked-up,
which can be reset by the RESET button after the fault
is removed. Change TTYP to 1PR; the RI2-1 and RI2-
2 should be CLOSED. Change TTYP to 2PR (or 3PR).
Repeat test; RI2-1 and RI2-2 should be closed.
Change TTYP to "OFF". Repeat test. RI2-1, RI2-2
should not be picked-up.
Repeat AG fault and measure the trip time, which
should be < 2 cycles. Change the setting of T1 from No
6-2
, ∆I
, or ∆I
) >1.0A peak and
B
C
)>0.5 A peak
0
LN
(
)
Z
–
1
R
) 1
–
+
-------------------- -
3
, V
, V
, 3V
) and Currents
B
C
0
to Yes. Repeat the test; the trip time should extend
for an additional 2 cycles. Reset T1 to zero.
The following formula can be used when PANG
GANG:
Vxg = (I
+ K
I
) Zcg
0
0
X
K
I
0
x
or
Vxg =
Zcg
I
+
----------- -
x
3
V
xg
or
I
=
------------------------------ -
X
K
o
Z
1
+
------ -
cg
3
where
(
)
Z
–
Z
oL
1L
K
=
----------------------------- -
0
Z
1L
∠
(
=
Z
GANG PANG
R
I
=
--------------------------------------------------------------------------------------------------------------- -
X
jPANG
Z
e
cg
or
I
=
--------------------------------------------------------------------------------------------- -
X
2
jPANG
-- - Z
e
cg
3
Example:
Vag = 30
Z1g = 4.5
GANG = 40
Ia
=
-------------------------------------------------------------------- -
2
(
)e
j85
-- - 4.5
+
3
=
----------------------------------------------------------------------------------------------------------------------------- -
(
)+ j 3 sin (85) + 4.5 cos (40) + j4.5 sin (40)
cos
3
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 6-1, Configurations 2 and 3,
repeat (preceding) Step 10 for BG and CG faults.
Note Targets.
) -1
–
V
xg
(
)
j GANG PANG
–
Zr e
–
1
1
+
--------------------------------------------------------------- -
3
V
xg
1
jGANG
+
-- - Z
Z r e
cg
3
PANG = 85
Zr = 3
30
1
(
) 3 ( )e
j40
-- - 4.5
3
30
)
–
(5/92)
≠
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