GE T35 Instruction Manual page 346

9.2 DIFFERENTIAL CHARACTERISTIC TEST EXAMPLES
d) SLOPE 1 TEST
Inject current in such a manner that the magnitude of I
intersection of the minimum PKP and Slope 1 and smaller than the Breakpoint 1 setting; that is,
1. Change the current magnitudes as follows:
WINDING 1
PHASE SINGLE CURRENT (I
0 A 0°
A
0.48 A 0°
B
0.48 A –180°
C
2. The following differential and restraint current should be read from the T35 actual values menu:
PHASE DIFFERENTIAL CURRENT (I
0 0°
A
0.113 pu 0°
B
0.113 pu 0°
C
The Percent Differential element will not operate even though I
is not large enough to make the
NOTE
3. Adjust the I
1
WINDING 1
PHASE SINGLE CURRENT (I
0 A 0°
A
0.45 A 0°
B
0.45 A –180°
C
4. The following differential and restraint current should appear in the T35 actual values menu:
PHASE DIFFERENTIAL CURRENT (I
0 0°
A
0.170 pu 0°
B
0.170 pu 0°
C
5. The actual I
d
9
9-6

I intersection of Min PKP and Slope 1
r
WINDING 2
) PHASE
1
A
B
C
)
d
I
d
current as shown below (thereby increasing I
WINDING 2
) PHASE
1
A
B
C
)
d

I ratio is now 17%. Verify that the element operates correctly.
r
T35 Transformer Protection System
is larger than the restraint current of 0.67 pu, corresponding to the
r
    I
I actual
r
SINGLE CURRENT (I )
2
0 A 0°
1 A –180°
1 A 0°
PHASE RESTRAINT CURRENT (I
0 0°
A
1 pu –180°
B
1 pu 0°
C
d

I ratio larger than the Slope 1 setting of 15%. The actual ratio is 11.3%.
r
) and verify that the element operates.
d
SINGLE CURRENT (I )
2
0 A 0°
1 A –180°
1 A 0°
PHASE RESTRAINT CURRENT (I
0 0°
A
1 pu –180°
B
1 pu 0°
C
9 COMMISSIONING
  
Break 1
r
)
r
is larger than the Minimum Pickup, because I
)
r
(EQ 9.9)
d
GE Multilin