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 V
A). Now we can assume the CT secondary winding resistance to be 60 %
of R
b
. R
ct
= 3 Ω.
W
e 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.
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
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