Voltage drop workout

[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.

 

[Andy78]

Where the formula states mV/A/m, then you need to be putting in the mV/A/m value for the selected cable size and type from appendix 4 to work out your actual volt drop rather than using the max VD in volts.


EDIT: Having said that, it depends if you are trying to work out the actual volt drop of your selected cable, or trying to work out the minimum cable size for volt drop permitted.

 

[telectrix]

i calculated a VD of 7.425 with 2.5mm. OK for non-lighting, but just outside the 3% for lighting. as the actual supply voltage will be well above 230V, i'd go with 2.5mm. assuming the ccc of 2.5mm is OK depending on installation method.

edit: in the real world, however, i'd be inclined to use 6mm. this would allow for any future increase in load up to 40+A.

 

[PiousJim]

Thanks for the reply Andy78 and Telectrix.

Andy78 to be honest I was trying to workout which cable to use.

I should also keep a note of how to work out the voltage drop of a circuit.

Telectrix did you use a formula similar to my no 1?

 

[telectrix]

VD = (mV/A/m x L x A)/1000

 

[Andy78]

Thanks for the reply Andy78 and Telectrix.

Andy78 to be honest I was trying to workout which cable to use.

I should also keep a note of how to work out the voltage drop of a circuit.

Telectrix did you use a formula similar to my no 1?

If you are working out which cable size to be used for volt drop alone the the formula is
mV/A/m = (MaxVD x 1000) / (Ib x L)
If you are working out the volt drop on your selected cable the formula is
VD = (mV/A/m x Ib x L) / 1000

You have of course got to size your cable for CCC before any volt drop stage.

 

[PiousJim]

Ok so my no 1 is correct?
(11.5×1000)/ (20*25)
=11500 / 675
≠17.03
Page 339 table 4D4B would suggest 2.5mm

I'm sorry for being paranoid 2.5 feels to small. Could I use 4mm as a precaution?

 

[PiousJim]

I think Telectrix answered the question in the edit on post 3

 

[PiousJim]

To be honest I would still have went 6mm as well they told us at college to not go lower than 6 as a precaution.

 

[Andy78]

Well you are sizing your entire cable on the requirements for volt drop. You should be sizing your cable on requirements for CCC taking into account load and corection factors and then checking for compliance with VD, not the other way round.