ABB REG670 Applications Manual page 756

Section 24
Requirements
750
E 200
= =
E k
al
0.8 0.8
EQUATION2531 V2 EN
The rated current of the generator and the fault current for a three phase external short
circuit must be calculated.
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 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.
Check of CT1 and CT2:
The CTs must have a rated equivalent limiting secondary e.m.f. E
or equal to the maximum of the required rated equivalent limiting secondary e.m.f.
E and E below:
alreqRat alreqExt
I
³ = × ×
NG
E E 30 I
al alreqRat
I
pr
EQUATION2534 V2 EN
I
(
³ = × ×
tf
E E 2 I
al alreqExt sr
I
pr
EQUATION2535 V2 EN
In this application we can see that the CTs must have a rated equivalent secondary
e.m.f. E that is equal or larger than 190 V. As the existing CT1 has E
al
CT2 has E = 250 V we can conclude that the CTs fulfil the requirements for the
al
generator differential protection in REG670.
Calculation example 2
We are using the same example as before (Calculation example 1) but now the CT data
is not known and we shall specify the CTs and provide CT manufacturers with
necessary CT data.
The rated current of the generator and the fault current for a three phase external short
circuit is calculated.
Generator protection REG670 2.2 IEC and Injection equipment REX060, REX061, REX062
=
250 V
=
3.25 kA
( )
+ + =
R R R 30
sr ct w addbu
13000
)
+ + = ×
R R R 2
ct w addbu
4000
1MRK 502 071-UEN -
that is larger than
al
3250
( )
× × × + +
1 5 2.5 0.3
4000
( )
× × + + =
1 5 2.5 0.3 51 V
= 200 V and
al
Application manual
(Equation 286)
(Equation 287)
(Equation 288)
280 and
=
190 V
(Equation 289)
(Equation 290)