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Patrick Griffin

Hi everyone,

I am intending on working in the uk shortly, i am from ireland and dont have the uk regs presently.

I am trying to prepare for the exmas i will face and the cable calculations/volt drop calculations are something i cannot remember doing here in ireland in my college time so my understanding is rather weak. I have been googling away like mad today and i have 1 or 2 questions someone maybe able to answer for me, i have no doubt they are basic, for you, so here it goes..

The formula mVxAxM divided by 1000= VD

I understand that A=amps, M=metres but what does the 'mV' mean and where is its figure derived from? I am assuming its in the regs somewhere but what it stands for i have no idea.

Thanks in advance!!
 
millivolt per metre dropped on cable length subject to cable size and install method

Tables back of regs. 4D1B
 
Millivolts is the first thing i thought of too just didnt see how it was applicable here, however i do now, was a bit obvious infact.

In order to find the required mV in the recs what information do i need? Cable size and install method only?
 
you need to knowe the design current and the length. also the permitted VD. ( 11.5V except for lighting, which is 6.9V.) this will then give you a value for mV/A/m, which you then match to cable size from 4D1, as snowhead stated.
 
Hi Patrick..

Have a look in appendix 4 (BS7671) for the Volt Drop tables. They will give you mV figure you are looking for.

The actuall figure from the tables is mV p/Amp p/metre (i.e mV/A/m) so the figure is the amount of millivolts that will be dropped per amp per meter. You find the figure by using the correct table (based on whether it is single/multi core etc) and then it is based on the CSA of your cable.

Pop the figure into the equation you already have and that will give you your volt drop for that installation.

Maximum permissable VD for sockets is 11.5V and for light it is 6.9V.

Some worked examples for you:

http://www.tlc-direct.co.uk/Book/4.3.11.htm

Good luck.
 
Thanks lads, the penny is dropping.

Happyhippydad (love the username btw) thanks alot for the link, you simplified it beautifully!!

Cheers again appreciate it.
 
And remember to divide the multiplied figures by 1000. (the volt drop equation is in millivolts not volts)

Also remember to include the correction factor figures for your installation methods.
Sometimes the difference is quite a surprise to the final figures.
 
Last edited by a moderator:
With regard to correction factors, is it common practice to use these or would one just use an educated guess for the required cable size and use the volt drop formula to ensure that the volt drop isnt over the
maximum value?
 
With regard to correction factors, is it common practice to use these or would one just use an educated guess for the required cable size and use the volt drop formula to ensure that the volt drop isnt over the
maximum value?

You would follow the guidance and formulas laid out in appendix 4, section 5.

When burying a cable in insulation can halve it's CCC you can't rely on rule of thumb for accurate cable sizing.
 
How old is that? As I notice it is recommending a 4% (9.2V) volt drop when cables are fully loaded as opposed to the 3% (not lights) and 5% (lighting) currently required by 525.101 and Appendix 4 section 6.4.

Crikey, did not spot that...
Sorry for posting up misleading information, glad you are on the ball Richard ! I did think the Cr factor should have been Cf, but thought it was just a "napit anomaly" rather than outdated information.
 

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