If an electrical enclosure main supply is 400/3/50 and I create 230V inside the enclosure by using the 400V leading phase and the 400V neutral would there be any issue with doing this? Pros/cons. The enclosure I have doesn't have adequate room for a transformer to create single phase 230V.

2nd part - how is the main 400V supply requirements calculated from this. For example if my 400V load for the enclosure is 45A and the 230V supply that is created has 8.5 FLA then what would the final 400V supply required for the enclosure.

This project is for a european customer.

 

> If an electrical enclosure main supply is 400/3/50 and I create 230V inside the enclosure by using the 400V leading phase and the 400V neutral would there be any issue with doing this? Pros/cons. The enclosure I have doesn't have adequate room for a transformer to create single phase 230V. <

Common practise in electical design to take a phase & neutral to make 230V. However good electrical design should attempt to balance the load across phases

> 2nd part - how is the main 400V supply requirements calculated from this. For example if my 400V load for the enclosure is 45A and the 230V supply that is created has 8.5 FLA then what would the final 400V supply required for the enclosure. <

It will add around 4.9 Amps to the 400V supply ie 8.5 / root 3.

 

Mike,

Your three phase + N supply is protected with a 45A protection. This means that each phase is individually protected for 45A. Assume that the current load is 10A on R, 15A on S and 20A on T than te final load if you connect a single phase of 8,5A on R will be 18,5 on R, 15 on S and 20A on T.

If you want to ensure that the neutral doesn't become overloaded make sure that you spread the loads between the phases. Equally loaded phases don't burden the neutral.

For protection you need to know how the network configuration is. A combined overload/short circuit protection can be used and if required a differential protection (check local requirements).

 

Responding to Mark Heath's Fri, Sep 30, 1:33pm reply... dividing the single-phase load by root 3 is incorrect:

If the 45A (3-ph) and 8.5A (1-ph) loads connected to ph-A have the same power factor, then, the total ph-A current is the arithmetic sum of their magnitudes.

If the 3-ph and 1-ph loads have different power factors, then, the total ph-A current is found by adding them vectorially!

Regards,
Phil Corso, PE {Boca Raton, FL, USA}
[[email protected]] ([email protected])