Volt drops

[timboharticus]

Re. Volt drop.

 

4% is used in buildings for 230V systems = 9.2V see Tommo from "Advise please"

 

Where does the 0.5V for 12V systems come from? Is it a regulation or good practise?

 

1.5 mmsq has volt drop of 29mV/A/m

2.5 mmsq 18

4 11

6 7.3

 

These are for copper pvc cables and are for single conductor so need to double up there and back. Actual volt drop is likely to be less as these figures are at max current carrying capacity of cable.

 

So, for a typical lighting load say 40W, volt drop of 50m of cable (total)

 

Power (W) = I(Amps) x V

 

Amps = W/V = 40/12 = 3.3 Amps

 

Current carrying capacity of 1.5mm cable is about 12Amps so no problem here.

 

Volt drop will be 29mV x 3.3 Amps x 50m = 4785mV = 4.8V too high if 12V system ok if 230V system.

 

If we half the load to 20W and use 2.5mm cable get

 

Amps = 20/12= 1.67 Amps

 

Volt drop = 18mV x 1.67Amps x 50m = 1.5Volts still a bit high for 12V system.

 

Suggest wiring in 2.5mm in a ring circuit. This would give a volt drop of about 1V.

 

Perhaps my assumed cable length is a bit long.

[rog guiver]

Could I suggest that, rather than lots of large cable runs, use "starter cables" from the service batteries to three or four fuse/CB panels along the length of the boat. Services can then be run from adjacent fuse/CB panel is sensible cable diameters.