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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 T60 actual values menu:
PHASE
DIFFERENTIAL CURRENT (I
A
0 ∠0°
B
0.113 pu ∠0°
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
C
0.45 A ∠–180°
4.
The following differential and restraint current should appear in the T60 actual values menu:
PHASE
DIFFERENTIAL CURRENT (I
A
0 ∠0°
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
T60 Transformer Protection System
is larger than the restraint current of 0.67 pu, corresponding to the
r
)
<
I
(
actual
) I
r
SINGLE CURRENT (I
)
2
0 A ∠0°
1 A ∠–180°
1 A ∠0°
PHASE
RESTRAINT CURRENT (I
A
0 ∠0°
B
1 pu ∠–180°
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
A
0 ∠0°
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
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