3 phase current

[ericmark]

Assuming a pf of 0.8, this gives 10.8 A per phase, so it looks like 2.5mm swa 4 core (heat pump controller needs a neutral) with a 16A mcb and 40A rccb will be man enough, including installation factors and volt drop.

That is probably the running current once the motor is up to speed. You should calculate voltage drop at the peak current that occurs as the motor comes up to speed from stationary.

525.4 A greater voltage drop than stated in Table 12A (Appendix 12) may be accepted for a motor during starting periods and for other equipment with high inrush currents. provided that it is verified that the voltage variations are within the limits specified in the relevant product standard for the equipment or. in the absence of a product standard.. in accordance with the manufacturers recommendations.

So no you do not need to calculate voltage drop at the peak current! And with a soft start that is unlikely to be an issue anyway.

 

[bernardgreen]

may be accepted for a motor during starting periods and for other equipment with high inrush currents. provided that it is verified that the voltage variations are within the limits specified in the relevant product standard for the equipment

The OP needs to verify this is the case. Some motors will take longer to run up to speed if the supply voltage on the motor terminals is lower than it should be. This puts more wear on the motor during the high current higher force acceleration period. It also increases the time that over current is flowing and this may lead to tripping of thermal breakers.

I know of one case where a heavy motor was frequently tripping the thermal breakers on start-up. The factory's in-experienced electrician was amazed when at the suggestion of an older and very experienced electrician the problem was solved by replacing the cables to the motor with larger cross section. Thicker cable would suggest that more current would flow which it did. But it was for a shorter period of time as the motor came up to speed in a shorter time. And only when the motor was at normal operating speed was the current reduced to the motor's normal current due to the speed dependent back EMF of the motor being fully established.

 

[davelx]

a / b / c (evaluated left to right) is the same as a / (b*c).

In the same way that a - b - c (evaluated left to right) is the same as a - (b + c)

The rule has a proper mathsy name which I've forgotten.

Liam

It's called operator precedence and it does not always flow left to right: a+b*4 is not the same as (a+b) * 4