= = =
k
al
E 200
E 250 V
0.8 0.8
EQUATION2531 V2 EN-US (Equation 279)
The rated current of the generator and the fault current for a three phase external short
circuit must be calculated.
= = =
× ×
n
NG
n
S
90
I 3.25 kA
3 U 3 16
EQUATION2532 V2 EN-US (Equation 280)
= = =
NG
tf
g
I
3.25
I 13.0 kA
X 0.25
EQUATION2533 V2 EN-US (Equation 281)
We can now calculate the required secondary e.m.f. according to equation 273 and 274.
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
al
that is lar
ger 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 5 2.5 0.3 190 V
I 4000
EQUATION2534 V2 EN-US (Equation 282)
( ) ( )
³ = × × + + = × × × + + =
tf
al alreqExt sr ct w addbu
pr
I
13000
E E 2 I R R R 2 1 5 2.5 0.3 51 V
I 4000
EQUATION2535 V2 EN-US (Equation 283)
In this application we can see that the CTs must have a rated equivalent secondary
e.m.f. E
al
that is equal or larger than 190 V
. As the existing CT1 has E
al
= 200 V and
CT2 has E
al
= 250 V we can conclude that the CTs fulfil the requirements for the
generator differential protection in REG670.
Calculation example 2
GUID-B8CD4497-3C92-4C92-8C51-D9CD5CA2CCAB v4
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.
1MRK 502 071-UUS A Section 24
Requirements
Generator protection REG670 2.2 ANSI and Injection equipment REX060, REX061, REX062 801
Application manual
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