Voltage drop help please!

[ta-electrical]

Hi, i am confused about how voltage drop is worked out and need an answer for the following please;

Work out cable size and cpc size you need when you have:
Cable length = 50 metres
Ib = 32Amps
to get a voltage drop that complies with regs i.e. below 5%(11.5 Volts).
It's wired in singles and enclosed in conduit trunking.

I tried to use the on-site guide and the tables it provides (page 166), i was using 'trial and error' on different cable sizes but, don't think i've been doing it right. Any help would be very appretiated and please if you get an answer refer to any sources you used such as the on-site guide.
I was using the formula that is in the on-site guide on page 124,
(mV/A/m)xIbxL / 1000.
On page 130, table 6D2 shows volt drops but, my answers don't seem to match what it says.
Thanks in advance.

 

[telectrix]

transpose the equation. VD = ( mV/A/m )/1000 x Ib x L. put in the values VD=11.5, Ib=32, L=50. then

11.5 = (mV/A/m)/1000 x32 x50

mV/A/m = (11.5 x1000) / ( 32x50)

= 11500/ 1600

= 7.18

now look in osg for the casble size with a mV/A/M of 7.18 or less

 

[ta-electrical]

So your saying to reverse the formula i.e. (mV/A/m)x32x50 /1000 = 11.5 into 11.5x1000 /50 /32 = (mV/A/m) ?

If i do that i get a value of 7.2, so would i have to choose a cable size with a resistance less than or equal to 7.2. e.g. 6mm by 6mm ?

Thanks for the help btw.

 

[telectrix]

yep, you got it. it's just a case of juggling the equation you know 3 values so the unknown 4th value can be determined.