•
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
Ztr
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:
Z
S
A
=
=
138
1700
11 2
2
.
Ω
EQUATION14000034 V1 EN-US
(Equation 20)
Z
S
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
Summary of Contents for RED650
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