ABB RELION 620 Series Technical Manual page 506

Section 4
Protection functions
500
T = 10 ms
m
K = 1/(1-0.4) = 1.6667
r
Equation 66 with these values gives the result:
F K Ik 0 9 .
> × ×
a r max
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If the actual burden of the current transformer S
cannot be reduced low enough to provide a sufficient value for F
alternatives to deal with the situation.
1. A current transformer with a higher rated burden S
means a higher rated accurate limit F
2. A current transformer with a higher nominal primary current I
rated burden) can be chosen.
Alternative 2 is more cost-effective and therefore often better, although the sensitivity
of the scheme is slightly reduced.
Example 2
Here the actions according to alternative 2 are taken to improve the actual accuracy
limit factor.
I CT
 
R
F F
=  ×

a n
I Motor
 R 
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I CT rated primary current of the CT, for example, 1500A
R
I Motor rated current of the motor under protection, for example, 1000A
R
F rated accuracy limit factor of the CT, for example, 30
n
F actual accuracy limit factor due to oversizing the CT, substituting the values in the equation, F
a
= 45
In differential protection it is important that the accuracy limit factors F
current transformers at both sides correspond with each other, that is, the burdens of
the current transformers on both sides are to be as close to each other as possible. If
high inrush or start currents with high DC components pass through the protected
object when it is connected to the network, special attention is required for the
performance and the burdens of the current transformers and the settings of the
function block.
T
−
m
T dc
( T ( 1 e ) 1
× × ω × − +
dc
in the accuracy limit factor equation
a
).
n
1MRS757644 F
) 24
≈
, there are two
a
can be chosen (which also
n
(but the same
1n
of the phase
a
Technical Manual
(Equation 67)
a
620 series