GE T35

CHAPTER 8: COMMISSIONING DIFFERENTIAL CHARACTERISTIC TEST EXAMPLES
T35 TRANSFORMER PROTECTION SYSTEM – INSTRUCTION MANUAL 8-11
8
8.2.5 Test example 4
8.2.5.1 D/D0° transformer with phase b to c fault on the secondary delta winding
Transformer data — D/D0°, 20 MVA, 115/12.47 kv, CT1 (200:1), CT2 (1000:1)
Test Phase Injected current Displayed current Status
W1 current W2 current Differential Restraint
Balanced
Condition
A0.25 ∠0° 0 ∠0° 0 ∠0° 0 ∠0° Not Applicable
B0.5 ∠–180° 0.8 ∠0° 0 ∠0° 0.8 ∠0°
C0.25 ∠0° 0.8 ∠–180° 0 ∠0° 0.8 ∠–180°
Min Pickup:
change the Min
PKP to 0.2 pu
A0.25 ∠0° 0 ∠0° 0 ∠0° 0 ∠0° Block
I
d
= 0.051 < Min PKP
B0.5 ∠–180° 0.95 ∠0° 0.154 ∠0° 0.948 ∠0°
C0.25 ∠0° 0.95 ∠–180° 0.155 ∠0° 0.950 ∠–180°
Minimum
Pickup
A0.25 ∠0° 0 ∠0° 0 ∠0° 0 ∠0° Operate
I
d
= 0.102 > Min PKP
B0.5 ∠–180° 1.05 ∠0° 0.253 ∠0° 1.049 ∠0°
C0.25 ∠0° 1.05 ∠–180° 0.255 ∠0° 1.050 ∠–180°
Slope 1:
return the Min
PKP to 0.1 pu
A0.25 ∠0° 0 ∠0° 0 ∠0° 0 ∠0° Block
I
d
/I
r
= 13.2%
B0.5 ∠–180° 0.92 ∠0° 0.123 ∠0° 0.919 ∠0°
C0.25 ∠0° 0.92 ∠–180° 0.123 ∠0° 0.919 ∠–180°
Slope 1 A 0.25 ∠0° 0 ∠0° 0 ∠0° 0 ∠0° Operate
I
d
/I
r
= 15.9%
B0.5 ∠–180° 0.95 ∠0° 0.153 ∠0° 0.948 ∠0°
C0.25 ∠0° 0.95 ∠–180° 0.153 ∠0° 0.948 ∠–180°
Intermediate
Slope 1 & 2
A2 ∠0° 0 ∠0° 0 ∠0° 0 ∠0° Block
I
d
/I
r
= 84.3%
< 86.6%
computed
B4 ∠–180° 1 ∠0° 5.37 ∠–180° 6.37 ∠0°
C2 ∠0° 1 ∠–180° 5.37 ∠0° 6.37 ∠–180°
Intermediate
Slope 1 & 2
A2 ∠0° 0 ∠0° 0 ∠0° 0 ∠0° Operate
I
d
/I
r
= 87.5%
> 86.6%
computed
B4 ∠–180° 0.8 ∠0° 5.57 ∠–180° 6.37 ∠0°
C2 ∠0° 0.8 ∠–180° 5.57 ∠0° 6.37 ∠–180°
Slope 2 A 4 ∠0° 0 ∠0° 0 ∠0° 0 ∠0° Block
I
d
/I
r
= 93.7%
< Slope 2 = 95%
B8 ∠–180° 0.8 ∠0° 11.93 ∠–180° 12.73 ∠0°
C4 ∠0° 0.8 ∠–180° 11.93 ∠0° 12.73 ∠–180°
Slope 2 A 4 ∠0° 0 ∠0° 0 ∠0° 0 ∠0° Operate
I
d
/I
r
= 95.7%
> Slope 2 = 95%
B8 ∠–180° 0.6 ∠0° 12.13 ∠–180° 12.73 ∠0°
C4 ∠0° 0.6 ∠–180° 12.13 ∠0° 12.73 ∠–180°