P
PiousJim
Hi guys. Those who don't know me I have recently started on my own.
I am testing my understanding, re Volt drop. And trying to stick to one formula. But I'm getting mixed results.
My example scenario would be to feed a CU in an outhouse / shed.
The CU will feed sockets, lights and some form of machinery of 2200 watts. Cable buried.
Length 25m
Design Current 27 amps (6200)
1.Using OSG P124 formula
VD=(mV/A/m) x lb x L ÷ 1000
I got
11.5v (5% single phase) x 27a x 25m
= 7.6
According to BS7176 p339 should use 4mm
2.On another thread someone suggested the following
11.5x1000 /25/27 = 17.03
According to BS7176 p339 should use 2.5
To be honest I would rather follow no 1 as this is how we were taught.
But no 2 is the result I got from tlc calculator and a few others.
Have I got something wrong or missed something?
In your opinion which would you go for and why?
Any advice is helpful.
I am testing my understanding, re Volt drop. And trying to stick to one formula. But I'm getting mixed results.
My example scenario would be to feed a CU in an outhouse / shed.
The CU will feed sockets, lights and some form of machinery of 2200 watts. Cable buried.
Length 25m
Design Current 27 amps (6200)
1.Using OSG P124 formula
VD=(mV/A/m) x lb x L ÷ 1000
I got
11.5v (5% single phase) x 27a x 25m
= 7.6
According to BS7176 p339 should use 4mm
2.On another thread someone suggested the following
11.5x1000 /25/27 = 17.03
According to BS7176 p339 should use 2.5
To be honest I would rather follow no 1 as this is how we were taught.
But no 2 is the result I got from tlc calculator and a few others.
Have I got something wrong or missed something?
In your opinion which would you go for and why?
Any advice is helpful.