Is sizing an MCBs to a motor's FLC or LRC? 1

[Aftermathematics]

I am trying to size miniature circuit breakers for 1kW 3-phase motors but I'm a bit confused as to what the breaker size should be.

The motor characteristics are as follows.
Rated power=1kW
Rated Voltage=400V
Power factor=0.8
The full load current FLC was calculated using the formula; I=P/(?3*v*pf) =1.8A ? 2A
The locked rotor current(starting current) LRC= 6*flc= 12A

My question is; should the breaker be sized to the FLC of the motor or to the LRC? i.e. should the CB be 2A or 12A?

I intend to use type C circuit breakers to EN 60898 given in BS7671 but it doesn't characteristics for 2A breaker. However, ABB manufactures MCBs as low as 2A so at least I know they are available off the shelf. I understand there are MCBs specifically designed for motors. Does this apply in to the British market as well?

What is the significance of motor service factor? Does it play a role in the sizing of circuit breaker?

Please bear in mind that I'm asking from the UK hence please provide answers to British or European standards

 

[jraef]

The breaker must be sized based on the FLC of the motor, but the LRC must be taken into account when chosing the Instantaneous Trips of the breaker. In the case of most MCCBs the IT setting is adjustable, but in the case of MCBs it is not. For MCBs, they are sold with various "Trip Curves" which relate to the IT values. For motors, you want to select a "C" trip curve, which allows for motor magnetizing inrush without as much risk of nuisance tripping, as long as that current does not exceed 10X the MCB rated current (Ir). For many newer energy efficient motors this may not be sufficient because they have much higher magnetizing current, but you can use a trip curve "D" for those, which allows 10-20X Ir.


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