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For example copper resistivity in +75 would be calculated as follows:
With this given value most common used copper wires resistance per meter in 75 are:
1.5 mm
2
2.5 mm
2
4 mm
2
6 mm
2
0.0135 Ω/m
0.00812 Ω/m
0.00508 Ω/m
0.00338 Ω/m
These values were calculated with formula:
Suggestion is that for calculating the CT burden the worst case scenario is used. For most cases these
75ºC values can be used. If in your application ambient temperature is higher than 75ºC , then the
resistance should be calculated for that temperature.
Important is also to know the wiring of the CTs in that point that is there common return wire used or
are the CTs both ends wired to the terminal connector. If the case is as usual that four wires come from
the CTs to terminal then length per phase is distance from the CT to relay added with the distance to
common coupling point. If from all CTs are both sides wires connected to relay or to the terminal then
the length of the wiring is two times the distance from the CTs to relay. In case if the connection is
mixture of these then the length can be estimated by increasing the distance by proportion of the six
wires / four wires connection. For example if 30% of the wiring is made from the CTs to terminal with six
wires and from the terminal wiring continues with four wires then the wire length estimate should be 1.3
times the distance between of relay and CTs.
Next loading factor is the resistance of the relay measuring input. In this IED type it is 0.0005 for
current input. This gives about 0.001VA with 1A current.
Now how to calculate the accuracy limit factor rst the CT nominal accuracy limit factor needs to be
known. As mentioned before that number in the CT rating (after the P) gives the current overload as a
factor of nominal rated value which can still be said that the CT output will be in its rated accuracy 5%
(5P) or 10% (10P) gives the accuracy limiting factor applicable at that overload of the CT.
Actual accuracy limit factor can be calculated as follows (this is common method):
In this formula the S values are in VA. Biggest problem in this equation is to know the internal
resistance of the CT secondary for calculation of the S
CTRN
.
AQ-T216
Instruction manual
Version: 2.00
139
© Arcteq Relays Ltd
Содержание AQ 200 Series
Страница 1: ...AQ T216 Transformer protection IED Instruction manual ...
Страница 243: ...Figure 7 1 162 AQ T216 variant with binary input modules AQ T216 Instruction manual Version 2 00 Arcteq Relays Ltd 242 ...
Страница 264: ...Figure 8 11 184 Panel cut out and spacing of the IED AQ T216 Instruction manual Version 2 00 263 Arcteq Relays Ltd ...
Страница 284: ...10 Ordering information AQ T216 Instruction manual Version 2 00 283 Arcteq Relays Ltd ...
Страница 285: ...AQ T216 Instruction manual Version 2 00 Arcteq Relays Ltd 284 ...