ABB REG670 Applications Manual page 757

1MRK 502 071-UEN -
Generator protection REG670 2.2 IEC and Injection equipment REX060, REX061, REX062
Application manual
S 90
= =
I n
NG
× ×
3 U 3 16
n
EQUATION2532 V2 EN
I 3.25
= = =
NG
I 13.0 kA
tf
X 0.25
g
EQUATION2533 V2 EN
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 = 2.5 Ω (single length) and the total additional burden R
w
Ω for both CTs. As we do not know the CT secondary winding resistance R
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:
= +
R R R
b max w addbu
EQUATION2536 V2 EN
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 = 5 Ω (5 VA). Now we can assume the CT secondary winding resistance to be 60 %
b
of R . R = 3 Ω.
b ct
We can now calculate the required secondary e.m.f. according to equation
281. 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 and E below:
alreqRat alreqExt
I
³ = × ×
E E 30 NG I
al alreqRat
I
pr
EQUATION2537 V2 EN
I
(
³ = × ×
tf
E E 2 I
al alreqExt sr
I
pr
EQUATION2538 V2 EN
The conclusion is that we need a CT with E
the rated burden 5 VA and R
(
³ = × ×
E 142 ALF I R
al sr ct
EQUATION2539 V2 EN
=
3.25 kA
= + = W
2.5 0.3 2.8
( )
+ + =
R R R 30
sr ct w addbu
13000
)
+ + = ×
R R R 2
ct w addbu
4000
> 142 V. For example a CT class 5P with
al
< 3 Ω shall fulfil the following:
CT
) ( )
+ = × × +
R ALF 1 3 5
b
Section 24
Requirements
3250
( )
× × × + +
1 3 2.5 0.3
4000
( )
× × + + =
1 3 2.5 0.3 38V
(Equation 291)
(Equation 292)
= 0.3
addbu
we must
ct
(Equation 293)
280 and
=
142 V
(Equation 294)
(Equation 295)
(Equation 296)
751