GE T35 Instruction Manual page 219

8 COMMISSIONING
b) TEST FOR ZERO DIFFERENTIAL CURRENT
1. Inject the following currents into the relay:
WINDING 1
PHASE SINGLE CURRENT (I
0 A ∠0°
A
0.434 A ∠0°
B
0.434 A ∠–180°
C
2. These are determined as follows:
20
( )
I w -------------------------------------------- -
=
n 1
×
3 115 10
From the Current Distribution diagram above, there is a
phases B and C, and a 0.866 pu 925.98 A 1000
3. The relay should display the following differential and restraint currents and the element should not operate:
PHASE DIFFERENTIAL CURRENT (I
0 ∠0°
A
0 ∠0°
B
0 ∠0°
C
c) MINIMUM PICKUP TEST
Reduce the restraint current I
the pickup). This is obtained from
0.15 is the setting of Slope 1. Note that
4. Change the current magnitude as follows:
WINDING 1
PHASE SINGLE CURRENT (I
0 A ∠0°
A
0.15 A ∠0°
B
0.15 A ∠–180°
C
5. The following differential and restraint current should be read from the T35 actual values menu:
PHASE DIFFERENTIAL CURRENT (I
0 ∠0°
A
0.044 pu ∠0°
B
0.044 pu ∠0°
C
The relay will not operate since I
6. Increase I to 0.2 A. The differential current increases to
1
7. Verify that the Percent Differential element operates and the following are displayed in the actual values menu:
PHASE DIFFERENTIAL CURRENT (I
0 ∠0°
A
0.136 ∠0°
B
0.136 ∠0°
C
GE Multilin
WINDING 2
) PHASE SINGLE CURRENT (I
1
0 A ∠0°
A
0.8 A ∠–180°
B
0.8 A ∠0°
C
6
×
10 VA
100.4 A, I
=
3
×
V
× ⁄
=
) PHASE RESTRAINT CURRENT (I
d
0 ∠0°
A
0.801 pu ∠–180°
B
0.801 pu ∠0°
C
to a value lower than 0.67 pu (the restraint corresponding to the intersection of Slope 1 and
r
⁄
I 0.1 0.15 0.67 pu
= =
r
< < (
0 I I intersection of Minimum PKP and Slope 1
r r
WINDING 2
) PHASE SINGLE CURRENT (I
1
0 A ∠0°
A
0.23 A ∠–180°
B
0.23 A ∠0°
C
) PHASE RESTRAINT CURRENT (I
d
0 ∠0°
A
0.275 pu ∠–180°
B
0.275 pu ∠0°
C
is still lower that the 0.1 pu MINIMUM PICKUP setting.
d
) PHASE RESTRAINT CURRENT (I
d
0 ∠0°
A
0.367 pu ∠–180°
B
0.367 pu ∠0°
C
T35 Transformer Management Relay
8.2 DIFFERENTIAL CHARACTERISTIC TEST EXAMPLES
)
2
6
×
20 10 VA
( )
w ------------------------------------------------- -
=
n 2
3
× ×
3 12.47 10 V
× ⁄
0.866 pu 100.4 A 200
=
0.8 A secondary current for LV phases b and c.
)
r
, where 0.1 is the differential setting of minimum pickup, and
)
)
2
)
r
>
I 0.136 pu Min PKP
=
d
)
r
925.98 A
(EQ 8.7)
=
0.434 A secondary current for HV
(EQ 8.8)
<
and I 0.67 pu .
r
8
8-5