K
r
1
When the values are substituted in Equation 201, the result is:
F
a
> × × × × − + ≈
−
K Ik T e
r dc
T T
m dc
max
/
( ( ) )
ω
1 1 40
GUID-7F1019C5-C819-440B-871B-5CFD1AF88956 V1 EN
2. Re-energizing against a fault occurring further down in the network:
The protection must be stable also during re-energization against a fault on the line.
In this case, the existence of remanence is very probable. It is assumed to be 40
percent here.
On the other hand, the fault current is now smaller and since the ratio of the resistance
and reactance is greater in this location, having a full DC offset is not possible.
Furthermore, the DC time constant (T
dc
) of the fault current is now smaller, assumed
to be 50 ms here.
Assuming a maximum fault current being 30 percent lower than in the bus fault and
a DC offset 90 percent of the maximum.
Ik
max
0.7* 10 = 7 (I
r
)
T
dc
50 ms
ω 100π Hz
T
m
10 ms
K
r
1/(1-0.4) = 1.6667
When the values are substituted in the equation, the result is:
F
a
> × × × × × − + ≈
−
K Ik T e
r dc
T T
m dc
max
/
. ( ( ) )0 9 1 1 40
ω
GUID-9B859B2D-AC40-4278-8A99-3475442D7C67 V1 EN
If the actual burden of the current transformer (S
a
) in Equation 200 cannot be reduced
low enough to provide a sufficient value for F
a
, there are two alternatives to deal with
the situation:
• a CT with a higher rated burden S
n
can be chosen (which also means a higher
rated accuracy limit F
n
)
• a CT with a higher nominal primary current I
1n
(but the same rated burden) can
be chosen
1MAC059074-MB A Section 12
Requirements for measurement transformers
615 series ANSI 1137
Technical Manual
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