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The rated current of the generator and the fault current for a three phase external short
circuit is calculated.
=
=
=
×
×
n
NG
n
S
90
I
3.25 kA
3 U
3 16
EQUATION2532 V2 EN-US
(Equation 284)
=
=
=
NG
tf
g
I
3.25
I
13.0 kA
X
0.25
EQUATION2533 V2 EN-US
(Equation 285)
We decide that CT1 and CT2 shall be equal (not necessary according to the
requirements). The CT ratio is decided to 4000/1 A and the burden is the same as in
Example 1. So R
w
= 2.5 Ω (single length) and the total additional burden R
addbu
= 0.3
Ω for both CTs. As we do not know the CT secondary winding resistance R
ct
we must
assume a realistic value. The value can vary much depending on the design of the CT
but a realistic range is between 20 to 80 % of the rated burden. Therefore we first must
decide the rated burden of the CT.
Maximum burden for the CTs are:
=
+
=
+
=
W
b max
w
addbu
R
R
R
2.5 0.3
2.8
EQUATION2536 V2 EN-US
(Equation 286)
It is often economical favorable to specify a low rated burden and a higher overcurrent
factor instead of vice versa. In our case it can be suitable to decide the rated burden to
R
b
= 5 Ω (5 VA). Now we can assume the CT secondary winding resistance to be 60 %
of R
b
. R
ct
= 3 Ω.
We can now calculate the required secondary e.m.f. according to equation
and
.
As the 16 kV system is high impedance earthed the burden only needs to consider the
single length of the secondary wire.
Dimensioning of CT1 and CT2:
The CTs must have a rated equivalent limiting secondary e.m.f. Eal that is larger than
or equal to the maximum of the required rated equivalent limiting secondary e.m.f.
E
alreqRat
and E
alreqExt
below:
(
)
(
)
³
=
×
×
+
+
=
×
× ×
+
+
=
NG
al
alreqRat
sr
ct
w
addbu
pr
I
3250
E
E
30
I R
R
R
30
1 3 2.5 0.3
142 V
I
4000
EQUATION2537 V2 EN-US
(Equation 287)
Section 24
1MRK 502 071-UUS A
Requirements
802
Generator protection REG670 2.2 ANSI and Injection equipment REX060, REX061, REX062
Application manual
Summary of Contents for RELION 670 SERIES REG670
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