GE T35 Instruction Manual page 521

CHAPTER 8: COMMISSIONING
2. Read the following differential and restraint current values 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 does not operate even though I
large enough to make the I
3. Adjust the I current as follows (thereby increasing I
1
Winding 1
Phase Single current (I
A 0 A ∠0°
B 0.45 A ∠0°
C 0.45 A ∠–180°
4. Read the following differential and restraint current values in the T35 actual values menu.
Phase Differential Current (I
A 0 ∠0°
B 0.170 pu ∠0°
C 0.170 pu ∠0°
5. The actual I /I ratio is now 17%. Verify that the element operates correctly.
d r
8.2.2.5 Intermediate curve between Breakpoint 1 and Breakpoint 2
This procedure tests the intermediate section of the differential characteristic curve that lies between the Breakpoint 1 and
Breakpoint 2 points (points B
1
1. Inject currents so that the magnitude of I
Breakpoint 2; that is:
For this example, 2 pu < I
Winding 1
Phase Single current (I
0 A ∠0°
A
B 1.2 A ∠0°
C 1.2 A ∠–180°
2. Read the following differential and restraint current values from the T35 actual values menu.
Phase Differential current (I
0 ∠0°
A
1.287 pu ∠–180°
B
1.287 pu ∠0°
C
The I /I ratio is 36.77% and the Differential element does not operate because the actual I
d r
at I = 3.5 pu.
r
T35 TRANSFORMER PROTECTION SYSTEM – INSTRUCTION MANUAL
) Phase Restraint current (I
d
0 ∠0°
A
1 pu ∠–180°
B
1 pu ∠0°
C
/I ratio larger than the Slope 1 setting of 15%. The actual ratio is 11.3%.
d r
) and verify that the element operates.
d
Winding 2
) Phase Single current (I
1
A 0 A ∠0°
B 1 A ∠–180°
C 1 A ∠0°
) Phase Restraint current (I
d
A 0 ∠0°
B 1 pu ∠–180°
C 1 pu ∠0°
and B on the Differential Restraint Characteristic diagram).
2
is between the restraint magnitudes defined by Breakpoint 1 and
r
I (at Breakpoint 1) > I < I
r r
< 8 pu. Remember that the maximum current is the restraint current I
r
Winding 2
) Phase Single current (I
1
0 A ∠0°
A
B 3.5 A ∠–180°
C 3.5 A ∠0°
) Phase Restraint current (I
d
0 ∠0°
A
3.5 pu ∠–180°
B
3.5 pu ∠0°
C
DIFFERENTIAL CHARACTERISTIC TEST EXAMPLES
)
r
is larger than the Minimum Pickup, because I
d
)
2
)
r
(at Breakpoint 2)
r
)
2
)
r
is not
d
Eq. 8-10
= 3.5 pu.
r
= 1.287 pu is still too low
d
8-7
8