GE T60, Инструкция по эксплуатации

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GE Multilin T60 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 T60 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
)
A 0 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
)
A 0
∠
0° A 0
∠
0°
B 0 ∠ 0° B 0.801 pu ∠ –180°
C 0
∠
0° C 0.801 pu
∠
0°
WINDING 1 WINDING 2
PHASE SINGLE CURRENT (I
1
) PHASE SINGLE CURRENT (I
2
)
A 0 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
)
A 0 ∠ 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
)
A 0
∠
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
3 115 10
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 <