ABB 615 Series - Page 1144

Example 2
Assuming that the actions according to alternative two above are taken in order to improve
the actual accuracy limit factor:
F
IrCT
IrTR
F
a n
= ×
GUID-31A3C436-4E17-40AE-A4EA-D2BD6B72034E V2 EN (Equation 202)
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
n
30 (rated accuracy limit factor of the CT)
F
a
(IrCT / IrTR) * Fn (actual accuracy limit factor due to oversizing the CT) = (1500/1000) * 30 = 45
In 87T, it is important that the accuracy limit factors F
a
of the phase current transformers
at both sides correspond with each other, that is, the burdens of the current transformers
on both sides are to be as equal as possible. If high inrush or pickup 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
for the settings of the function block.
Section 12 1MAC059074-MB A
Requirements for measurement transformers
1138 615 series ANSI
Technical Manual