ABB RED650 - Page 108

• Apparent source power at B side: Ss
B
= 1280 MVA
• Base current of differential current protection: I
Base
= 42 A
• Apparent power of transformer: S
n
= 10 MVA
• Short circuit impedance of transformer: e
k
= 10%
• Nominal voltage on transformer high voltage winding: U
n
= 138 kV
Fault current on the high voltage (HV) side of the tap transformer is calculated for
a three-phase fault on the low voltage (LV) side. 138 kV is chosen as calculation
voltage.
ZsA ZsB
Z
t
r
f
E
ZlA ZlB
IEC14000046-1-en.vsd
IEC14000046 V1 EN-US
Figure 47: Thevenin equivalent of the tap transformer
Converting the sources into impedances gives:
ZS
A
= =
138
1700
11 2
2
. Ω
EQUATION14000034 V1 EN-US (Equation 20)
ZS
B
= =
138
1200
15 9
2
. Ω
EQUATION14000035 V1 EN-US (Equation 21)
Calculating the short circuit impedance of the transformer gives:
Z
e U
S
trf
k n
n
= × = × =
100
0 1
138
10
190 4
2 2
. . Ω
EQUATION14000036 V1 EN-US (Equation 22)
Based on the Thevenin equivalent, it is possible to calculate the fault current on the
HV side of the transformer:
If
Z
res
138
138
3
=
×
EQUATION14000037 V1 EN-US
(Equation 23)
Section 6 1MRK 505 393-UEN B
Differential protection
102 Line differential protection RED650 2.2 IEC
Application manual