Voltage Drop

[Simonslimline]

When selecting a cable and taking voltdrop in to acount is it 3% for lighting or 5% for other loads of 230v


Because the voltage obviously changes from anything from 216V up to 253V.

Thanks

 

[Rockingit]

You always calculate anything voltage related from 230V (single phase), as this is the nominal voltage.

As far as voltage drop is concerned, it all depends on what is connected to the end of the cable. For example, if you were calculating a cable to feed a machine or maybe power sockets then you would work to 5%, however if you were installing a cable to a distribution panel that would feed circuits for lighting then you'd work with 3%.

 

[Simonslimline]

Ok here is a question I was wondering then...

Say you were calculating a feed for some lights and due to the load and distance that you had to use 2.5 rather than 1.5 because your volt drop was more than 6.90Volts.
Say you started with 240volts as is very common in situations, then dropping the 6.90Volts would still leave you 233.10volts. surely if you were not dropping below 3% of 230V you could use the smaller cable ie the 1.5 mm2 .

I hope this makes sense

 

[Rockingit]

Whilst that is practically the likely scenario, you have to remember that we have no control of the voltage we receive - it might be 245V on Monday and 235V by Friday, hence why we have to calculate to one value - the nominal 230V.

 

[telectrix]

i see what you mean. but you can't have it both ways. voltage could be <220V , so the average ( or nominal ) 230V is used. sounds silly when you got 240V to play with. in a borderline situation, i'd use the measured voltage if adding to an existing installation, but 230V for anything new.

 

[Simonslimline]

Cool thanks for clearing that up for me

 

[Spearmint]

DNOs have a tolerance between -6% to +10% (216.2V to 253V) on 230V at any given time, that's why it's always better to go that little bit further IMO. Working from worst case is foolproof.

 

[SirKit Breaker]

Voltage drop is a product of circuit load and length. You can only calulate it one way.

(mV/A/m) x Ib x L / 1000.

You then re-calculate your answer as a % drop of Uo, and decide from this figure if the volts drop is acceptable or not.

(mV/A/M) is the value of a specific cable and comes from the tables in BS7671.
Ib is the rating of the protective device for the circuit.
L is length in metres of the circuit, and this can be worked out if the circuit is continuity tested and compared against the appopriate table in BS7671 or GN3.

Enjoy

Cheers..................Howard