transformers

[redface]

now i think i should know this but i don't know wether logic will apply

a 230 to 110V transformer
the 110V machines will take about 30A which is about 3.6kva
now the cable will run about 20m and i was going to run in 10mm 3core swa to a 40A switch then i was going to run a 4mm tw/e ring now this will run at 30A constant that is why i have used that size cable and it is very sensitive equipment so i don't want much volt drop at all do you think this is ok or well over the top ?

the other question is its 30A on 110V side so is that the same on 230V side or should it be protected by a 16A breaker?

am i right the lower the Voltage the higher the volt drop can any one show me a calculation for this to work it out please

redface

 

[Knobhead]

A 20A type C or D to cope with inrush current. To honest I'd go for a type D

 

[Knobhead]

Regarding volt drop it’s a % of the load, so the drop is a % of the voltage.

 

[Richard Burns]

now i think i should know this but i don't know wether logic will apply

a 230 to 110V transformer
the 110V machines will take about 30A which is about 3.6kva
now the cable will run about 20m and i was going to run in 10mm 3core swa to a 40A switch then i was going to run a 4mm tw/e ring now this will run at 30A constant that is why i have used that size cable and it is very sensitive equipment so i don't want much volt drop at all do you think this is ok or well over the top ? If you are aiming for low volt drop then the bigger the better, you have no problems with CCC here. However you are doing a lollipop ring which is regulations compliant but not standard.

the other question is its 30A on 110V side so is that the same on 230V side or should it be protected by a 16A breaker?
Assuming the current you mention is on 110V side: volts*amps is equal on both sides of a transformer so 110*30 = 230*amps so amps =110*30/230 =14.3A

am i right the lower the Voltage the higher the volt drop can any one show me a calculation for this to work it out please
Voltage dropped over a resistance is V=IR, the (theoretical) resistance of the cables is known (for 10mm and 10m = 0.018 ohm for 8mm(2*4mm) and 10m =0.023 ohm total 0.041 ohm) current =30A do V=30*0.041=1.23V, these values are normally lower than doing volt drop = (mV/A/m*I*L)/1000.
As Tony says 1.23V dropped on 110V side is 1.11% (if the current was 30A on the 230V side then the Vd would be 0.53%)
redface

Listen to Tony more than me, at least he knows what he is talking about.

 

[redface]

tony i think knows more than u richard but thank u great help

 

[Knobhead]

I notice Richard gets a thanks and I don't!

 

[Moses]

I notice Richard gets a thanks and I don't!

That's life!!

The good looking ones always get looked at, even if when they completely ignore you.

 

[Knobhead]

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