EddyWirbelstrom
Electrical
If the calculated (prospective) d.c. component of symmetrical breaking current is less than 20%, then the breaker's required interrupt rating is defined by only its symmetrical breaking current. See IEC 62271-100
The definition of d.c. component is defined by the ratio d.c. current / ( sqrt(2)*symmetrical breaking current ).
If a breaker with a rated 30% d.c. component is required to interrupt a calculated symmetrical current of 27 kA with a d.c. component of 65% ,what is the breaker required rated symmetrical breaking current ?
The following formula equates the rated and calculated asymmetrical rms current:
Rated Asymmetrical = Calculated (prospective)Asymmetical
Isc ( 1 + 2 %DC^2 ) ^0.5 = Ib sym ( 1 + 2 % d.c.^2 ) ^0.5
Isc ( required rated symmetrical breaking current ) = 33.76 kA Use next highest standard IEC breaker rating of 31.5kA.
Assuming the actual breaker symmetrical breaking current rating was 33.76kA, then the max permissible d.c. component would be 0.30 * sqrt(2) * 33.76 = 14.32 kA.
A HV breaker can not interrupt d.c. current, however if the ratio of d.c. current / sqrt(2)* symmetrical breaking current is maintained, then sufficient current zeros occur to enable the breaker to extinguish the arc.
Can anyone out there in IEC land confirm the following ?
If the calculated (prospective) Ib sym is less than rated Isc, the above formula can not be used to calculate a higher permissible d.c. breaking current because te ratio d.c. current / sqrt(2)* symmetrical breaking current must not be less than the breaker d.c.component ratio.
See attached file for calculations.
The definition of d.c. component is defined by the ratio d.c. current / ( sqrt(2)*symmetrical breaking current ).
If a breaker with a rated 30% d.c. component is required to interrupt a calculated symmetrical current of 27 kA with a d.c. component of 65% ,what is the breaker required rated symmetrical breaking current ?
The following formula equates the rated and calculated asymmetrical rms current:
Rated Asymmetrical = Calculated (prospective)Asymmetical
Isc ( 1 + 2 %DC^2 ) ^0.5 = Ib sym ( 1 + 2 % d.c.^2 ) ^0.5
Isc ( required rated symmetrical breaking current ) = 33.76 kA Use next highest standard IEC breaker rating of 31.5kA.
Assuming the actual breaker symmetrical breaking current rating was 33.76kA, then the max permissible d.c. component would be 0.30 * sqrt(2) * 33.76 = 14.32 kA.
A HV breaker can not interrupt d.c. current, however if the ratio of d.c. current / sqrt(2)* symmetrical breaking current is maintained, then sufficient current zeros occur to enable the breaker to extinguish the arc.
Can anyone out there in IEC land confirm the following ?
If the calculated (prospective) Ib sym is less than rated Isc, the above formula can not be used to calculate a higher permissible d.c. breaking current because te ratio d.c. current / sqrt(2)* symmetrical breaking current must not be less than the breaker d.c.component ratio.
See attached file for calculations.