Test Example 3 - GE T60 Instruction Manual

Transformer protection system ur series
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8.2 DIFFERENTIAL CHARACTERISTIC TEST EXAMPLES
Yg/D30° TRANSFORMER WITH PHASE B TO C FAULT ON THE DELTA SIDE.
Transformer: Y/D30°, 20 MVA, 115/12.47 kv, CT1 (200:1), CT2 (1000:1)
A
B
C
Figure 8–4: CURRENT DISTRIBUTION ON A YG/D30° TRANSFORMER WITH AN a TO b FAULT ON THE LV SIDE
Three adjustable currents are required in this case. The Phase A and C Wye-side line currents, identical in magnitude but
displaced by 180°, can be simulated with one current source passed through these relay terminals in series. The second
current source simulates the Phase B primary current. The third source simulates the delta "b" and "c" phase currents, also
equal in magnitude but displaced by 180°.
TEST
PHASE
Balanced
A
Condition
B
C
Min Pickup
A
change the
B
Min PKP to
0.2 pu
C
Minimum
A
Pickup
B
C
Slope 1
A
return the
B
Min PKP to
0.1 pu
C
Slope 1
A
8
B
C
Intermediate
A
Slope 1 & 2
B
C
Intermediate
A
Slope 1 & 2
B
C
Slope 2
A
B
C
8-10
I (f) = 0.5 pu –270°
A
I (f) = 1 pu –90°
B
I (f) = 0.5 pu –270°
C
INJECTED CURRENT
W1 CURRENT
W2 CURRENT
0.25 0°
0 0°
0.5 –180°
0.8 0°
0.25 0°
0.8 –180°
0.25 0°
0 0°
0.5 –180°
0.95 0°
0.25 0°
0.95 –180°
0.25 0°
0 0°
0.5 –180°
1.05 0°
0.25 0°
1.05 –180°
0.25 0°
0 0°
0.5 –180°
0.92 0°
0.25 0°
0.92 –180°
0.25 0°
0 0°
0.5 –180°
0.95 0°
0.25 0°
0.95 –180°
2 0°
0 0°
4 –180°
1 0°
2 0°
1 –180°
2 0°
0 0°
4 –180°
0.8 0°
2 0°
0.8 –180°
4 0°
0 0°
8 –180°
0.8 0°
4 0°
0.8 –180°
T60 Transformer Protection System
Y/d30° Transformer
I (f) = 0.866 pu –90°
I (f) = 0.866 pu –270°
c
DISPLAYED CURRENT
DIFFERENTIAL
RESTRAINT
0 0°
0 0°
0.8 0°
0 0°
0.8 –180°
0 0°
0.154 0°
0.948 0°
0.155 0°
0.950 –180°
0 0°
0.253 0°
1.049 0°
0.255 0°
1.050 –180°
0 0°
0.123 0°
0.919 0°
0.123 0°
0.919 –180°
0 0°
0.153 0°
0.948 0°
0.153 0°
0.948 –180°
0 0°
5.37 –180°
6.37 0°
5.37 0°
6.37 –180°
0 0°
5.57 –180°
6.37 0°
5.57 0°
6.37 –180°
0 0°
11.93 –180°
12.73 0°
11.93 0°
12.73 –180°
8 COMMISSIONING

8.2.4 TEST EXAMPLE 3

I (f) = 0
a
A
b
B
F
C
828738A1.CDR
STATUS
0 0°
Not Applicable
0 0°
Block
I
= 0.051 < Min PKP
d
0 0°
Operate
I
= 0.102 > Min PKP
d
0 0°
Block
I
I
/
= 13.2%
d
r
0 0°
Operate
I
I
/
= 15.9%
d
r
0 0°
Block
I
I
/
= 84.3%
d
r
< 86.6%
computed
0 0°
Operate
I
I
/
= 87.5%
d
r
> 86.6%
computed
0 0°
Block
I
I
/
= 93.7%
d
r
< Slope 2 = 95%
GE Multilin

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