Voltage drop workout

[Deleted member 72666]

Out of interest what is this 'some form of machinery of 2200 watts' you speak of?

 

[PiousJim]

I just picked a random machine from screw fix.

http://m.screwfix.com/p/record-power-pt107-265mm-planer-thicknesser-240v/48940

 

[PiousJim]

I've created a essay/document to test my knowledge and that I could refer back to.

I was designing the circuit. This thread is in relation to powering a shed.

 

[Richard Burns]

I know this is veering slightly from the plan.
Load of 27A on a buried SWA cable run for 25m from the origin of the installation.

Load is 27A so a 32A breaker required, cable must be able to take 32A.
Current carrying capacity of SWA method D no other rating factors in table 4D4A:
2.5mm² has 29A not enough for 32A breaker.
4.0mm² has 37A OK so the minimum cable size so far is 4.0mm².

So for a 4.0mm² cable the volt drop is 11 mV/A/m.

Using the values we now have in VD = (mV/A/m x L x A)/1000
Volt drop = (11 mV/A/m * 25 m * 27 A)/1000
Volt drop = 7.425 V

maximum volt drop for power is 11.5 V (5% of 230 V) so this is OK for power.
maximum volt drop for lighting is 6.9 V (3% of 230V) so this is not OK for lighting

OK so move up to 6.0 mm² cable with a volt drop of 7.3 mV/A/m as it is a bigger cable the current carrying capacity is going to be OK.

Using the values we now have in VD = (mV/A/m x L x A)/1000
Volt drop = (7.3 mV/A/m * 25 m * 27 A)/1000
Volt drop = 4.93 V

maximum volt drop for power is 11.5 V (5% of 230 V) so this is OK for power.
maximum volt drop for lighting is 6.9 V (3% of 230V) so this is OK for lighting.

Therefore for this run of cable, assuming the cable comes from the origin and does not have a significant circuit length to supply the sockets and lights and machinery in the shed, and for this installation method carrying a design curent of 27A, the minimum size cable that can be used to supply lighting is 6 mm².

 

[PiousJim]

Thanks Richard I was struggling with the calculation it self.

I wasn't happy with myself for getting the answer but not understanding the calculation.


Thank you for clearing this up for me. I think I owe you a few by now.
Thanks again Richard

 

[Gavin A]

Also, watch for the surge current on start up of that machinery. Put a big lathe or something in there and the start up current could well be 3 times the nominal rated current, so you could end up with a 15% VD on start up.

I remember one fault report with an inverter on one of the forums that turned out to be caused by the inverter tripping out due to the massive voltage swing when a big bit of machinery started up in the shed it was connected in to.

Also when doing these calcs, if you could double check them to be sure to meet the 1% VD limit for a 16A solar PV supply that'd be much appreciated. Nowt worse* than having to dig the driveway up to install solar on the garage because the existing supply to the garage doesn't quite cut it for our purposes.


*ok there are actually quite a few things worse than that.