GE T35 UR Series

GE Multilin T35 Transformer Protection System 9-5
9 COMMISSIONING 9.2 DIFFERENTIAL CHARACTERISTIC TEST EXAMPLES
9
b) TEST FOR ZERO DIFFERENTIAL CURRENT
1. Inject the following currents into the relay:
2. These are determined as follows:
(EQ 9.7)
From the Current Distribution diagram above, there is a secondary current for HV
phases B and C, and a secondary current for LV phases b and c.
3. The relay should display the following differential and restraint currents and the element should not operate:
c) MINIMUM PICKUP TEST
Reduce the restraint current I
r
to a value lower than 0.67 pu (the restraint corresponding to the intersection of Slope 1 and
the pickup). This is obtained from , where 0.1 is the differential setting of minimum pickup, and
0.15 is the setting of Slope 1. Note that
(EQ 9.8)
where I
ri
is an intersection of Minimum PKP and Slope 1 calculated as PKP/Slope 1 value.
4. Change the current magnitude as follows:
5. The following differential and restraint current should be read from the T35 actual values menu:
The relay will not operate since I
d
is still lower that the 0.1 pu MINIMUM PICKUP setting.
6. Increase I
1
to 0.2 A. The differential current increases to and .
7. Verify that the Percent Differential element operates and the following are displayed in the actual values menu:
WINDING 1 WINDING 2
PHASE SINGLE CURRENT (I
1
) PHASE SINGLE CURRENT (I
2
)
A0 A ∠0° A 0 A ∠0°
B 0.434 A ∠0° B 0.8 A ∠–180°
C 0.434 A ∠–180° C 0.8 A ∠0°
PHASE DIFFERENTIAL CURRENT (I
d
) PHASE RESTRAINT CURRENT (I
r
)
A0 ∠0° A 0 ∠0°
B0 ∠0° B 0.801 pu ∠–180°
C0 ∠0° C 0.801 pu ∠0°
WINDING 1 WINDING 2
PHASE SINGLE CURRENT (I
1
) PHASE SINGLE CURRENT (I
2
)
A0 A ∠0° A 0 A ∠0°
B 0.15 A ∠0° B 0.23 A ∠–180°
C 0.15 A ∠–180° C 0.23 A ∠0°
PHASE DIFFERENTIAL CURRENT (I
d
) PHASE RESTRAINT CURRENT (I
r
)
A0 ∠0° A 0 ∠0°
B 0.044 pu ∠0° B 0.275 pu ∠–180°
C 0.044 pu ∠0° C 0.275 pu ∠0°
PHASE DIFFERENTIAL CURRENT (I
d
) PHASE RESTRAINT CURRENT (I
r
)
A0 ∠0° A 0 ∠0°
B 0.136 ∠0° B 0.367 pu ∠–180°
C 0.136 ∠0° C 0.367 pu ∠0°
I
n
w
1
()
20 10
6
× VA
311510
3
× V×
---------------------------------------------
100.4 A, I
n
w
2
()
20 10
6
× VA
3 12.47 10
3
× V×
--------------------------------------------------
925.98 A== = =
0.866 pu 100.4 A 200⁄× 0.434 A=
0.866 pu 925.98 A 1000⁄× 0.8 A=
I
r
0.1 0.15⁄ 0.67 pu==
0 I
r
I
ri
<<
I
d
0.136 pu Min PKP>=
I
r
0.67 pu<