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Example 4
U
= 12 kV;
U
GN
HV
HV system ungrounded.
I
= 15 A
Emax
U
12000
=
=
U
GN
1
n
3
3
12000
³
R
e
min
´
3
15
£
R
e
max
´
6
314 3 10
Since from this calculation R
protection is stable at the chosen current I
the resistor R
can be determined in relation to R
e
I
= 200 A
e
15
=
U
6930
2
n
´
3
200
It then follows that
³ 0.60 ´ 10
R
emin
£ 1.81 ´ 10
R
emax
= 1.80 W
R
e
At I
= 200 A, the voltage drop across the resistor R
e
= 1.8 ´ 200 = 360 V
U
= R
I
e
e
R e
Neglecting load current, the maximum voltage across the broken
delta windings is:
= 3 ´ 173 » 520 V
U = 3 U
2n
Specification:
1 resistor 1.80 W ;
3 v.t's
1 interposing v.t.
3-132
REG 316*4 1MRB520049-Uen / Rev. F
= 110 kV; C
= 3 x 10
12
=
6930
V
2
æ
3
U
ö
2
n
÷ ´
=
ç
0 7 0 60 10
.
.
è
ø
6930
´
3
0 05 12
.
æ
ç
-
è
´ ´
9
´
110
6930
is greater than R
emax
=
173
V
-4
2
´ 173
= 1.80 W
-4
2
´ 173
= 5.42 W
200 A;
10 s
12000
173 V
3
10 VA;
50 Hz;
F; w = 314 1/s
-9
-
4
2
´
´
U
2
n
2
U
ö
-
2
n
÷ =
´
4
´
181 10
.
ø
emin
and the value of
Emax
.
emin
is
e
single-phase insulated
520 / 100 V
2
U
2
n
, the
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