volt drop?

[f1ngerz]

hi im struggling with volt drop , the circuit is 6mm 2core swa , run under ground , 220M long ,supplying a transformer charging batterys for a gate system rated at 5A.
Can anyone help me and give me the formula for the calculation? Thanks

 

[Smugley]

You need to know load or design current in Amps (Ib), Length in metres (L) and mV/A/m which for 6mm 2core armoured is 7.3.

Then use formula (7.3 x Ib x L) / 1000 = drop in Volts.

(7.3*5*220)/1000 = 8.03. So okay if 5A is your only load and no lights. Max allowable drop is 5% of 230V which is 11.5. Or 3% (6.9V) for lighting.

 

[Guest123]

Assuming the full 5A as the circuit design current (Ib) the calc is as follows......

VD = mV/A/m x Ib x L / 1000, which using your figures gives us......7.3 x 5 x 220 / 1000 = 8.03A. Surprisingly within the tolerance 5% of 230V which is 11.5V.

 

[Smugley]

** snap!

 

[f1ngerz]

Thanks all , where can I find this formula in the regs ? Can't bloody find it!

 

[Smugley]

OSG p124

 

[Deleted member 19807]

or Appendix 4 in the brb

 

[ian.settle1]

Can't remember if it is in the On-Site Guide or the Design book from the IET but the cable must be de-rated by 0.9 for being underground.

 

[Deleted member 19807]

Can't remember if it is in the On-Site Guide or the Design book from the IET but the cable must be de-rated by 0.9 for being underground.

sorry thats not correct, no derating for ground temp of 20c If ground is cooler then a positive rating see table 4B2 page 267 brb.