ABB RED615 Technical Manual page 418

Section 4
Protection functions
412
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
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 0.7* 10 = 7 (I )
max r
T 50 ms
dc
ω 100π Hz
T 10 ms
m
K 1/(1-0.4) = 1.6667
r
When the values are substituted in the equation, the result is:
> × × 0 9 .
F K Ik
a r max
GUID-9B859B2D-AC40-4278-8A99-3475442D7C67 V1 EN
If the actual burden of the current transformer (S
reduced low enough to provide a sufficient value for F
alternatives to deal with the situation:
• a CT with a higher rated burden S
higher rated accuracy limit F
• a CT with a higher nominal primary current I
burden) can be chosen
Example 2
Assuming that the actions according to alternative two above are taken in order to
improve the actual accuracy limit factor:
IrCT
F * F
=
a n
IrTR
GUID-31A3C436-4E17-40AE-A4EA-D2BD6B72034E V1 EN
IrTR 1000 A (rated secondary side current of the power transformer)
IrCT 1500 A (rated primary current of the CT on the transformer secondary side)
F 30 (rated accuracy limit factor of the CT)
n
F (IrCT / IrTR) * Fn (actual accuracy limit factor due to oversizing the CT) = (1500/1000) * 30 = 45
a
−
T / T
× ( × × − ( 1 m dc ) +
T e
ω
dc
can be chosen (which also means a
n
)
n
1YHT530004D05 D
) of the fault current is
dc
1 ) ≈ 40
) in Equation 53 cannot be
a
, there are two
a
(but the same rated
1n
Technical Manual
(Equation 55)
615 series