Voltage drop question

[yarcyk]

Hi everybody.

I have a question (still learning, passed 17th edition and part P, but I'm not yet registered).
If you use formula from On-Site Guide p.140 what the design current you put in the formula, to calculate max length of cable for 11.5V.
If 32amp then the maximum length of ring final would be around 20metres (both legs parallel) or 16 amps and then around 40 metres in total length of 2.5mm cable.

If this is right how would you wire second floor socket circuit if CU is in the basement (40 metres not quite enough for 3 rooms on 2nd floor).

Hope someone can explain.

Thanks

 

[Sintra]

For volt drop on a ring final the formula is

mV/A/M x ib x L / 1000 / 4

And for maximum length it is

L = 4 x max vd x 1000 / ib x mV/A/M

Try a value of 20A for your design current and work it out.

 

[yarcyk]

That's much better now. around 128metres. Where is that formula from?

 

[Sintra]

It came from my brain where it was drummed into me nearly 30 years ago at college.

 

[yarcyk]

Ok. Just can't figure out where the "4" comes from.
I found table 7.1 in On-Site Guide which is probably what I was looking for.
It says this table has been designed to enable a radial or ring final circuit to be installed without calculations.

 

[telectrix]

the divide by 4 comes from the same place as with R1+R2 measurements for a RFC. because its a ring, you divide by 2 because the you need the value at the mid point, and then again as you have 2 cables in parallel, sharing the load.

 

[i=p/u]

It can be but think says 90m for rfc in osg where if you work it out you can make it longer..

The 4 comes from double the CSA (2legs) and have the distance to teturn

 

[matthunt]

Easy way of thinking about it is a follows: VD = mv/a/m x ib x length / 1000 so if its a ring just divide the ib in half and the length in half or (divide all of it by 4), try it, it works

 

[yarcyk]

So it's like making calculation for radial CSA 5mm and then multiplying length by 2?

 

[matthunt]

Try the numbers, all will become apparent!

 

[matthunt]

Volt Drop = 18mv/a/m x 32amps ib x 50 mts length /1000 / 4 = 7.2 volts or Volt Drop = 18mv/a/m x 16amps ib x 25 mts length /1000 = 7.2 volts, take your pick.

 

[yarcyk]

Sort of. As I can't get values for 5mm cable but work out estimated it's about right. Take voltage drop as 8.5 and Ib 30amps it is roughly 90m.

 

[yarcyk]

8.5 is not a voltage drop, is value of voltage drop in mV/A/m

 

[matthunt]

Yarcyk, I am trying to explain to you how the (4) number is derived, the worked example I gave you is a worked example to help you understand!

 

[yarcyk]

8.5 is not a VD, it's volt drop on mV/A/m

 

[Sintra]

Sort of. As I can't get values for 5mm cable but work out estimated it's about right. Take voltage drop as 8.5 and Ib 30amps it is roughly 90m.

Why mess about trying to make up formulae when you have been given the correct ones in post #2

 

[yarcyk]

So "4" is there because you calculate VD on one leg (length/2) and as one leg has half of the load ( Ib/2) ?

 

[yarcyk]

Thanks for your time guys.

 

[Sintra]

So "4" is there because you calculate VD on one leg (length/2) and as one leg has half of the load ( Ib/2) ?

You are starting to see the light

 

[telectrix]

light on a ring?