3 phase volt drop calcs

[wade88]

Hi guys,

Looked over some previous posts and done some further reading, and attempted the calcs, just want to check that i am not miles off the ball park here with this. From what i have read harmonics and current down the neutral will cause massive fluctuations in results not to mention unbalanced phases but i would just like to get a broad figure so i have reference.

Cable run is 50 meters, 4 core SWA 25mm, clipped direct, 105 amps per phase.

Calculation i have been going on is:

mV/A/m x length x load / 1000 to find the approximate single phase volt drop in this circuit

and then multiply by 0.866 to take account for any current in the neutral

so im getting:

1.5 x 50 x 105/1000 = 7.875 x 0.866 = 6.81Volts.

Any one free to confirm they get similar results. Or have i missed an important part of the 3 phase calc process here.

Thanks guys

 

[wade88]

Or should it be 2.5 x 50 x 105/1000 x 0.866 = 11.36V

Cant decipher accurately what the mV is from the tables i have. Two tables are providing two different data sets but are titled the same? So not positive on which figure i should be using.

 

[Brightspark2]

In the AEI cable table (I'm a poet and I know it) that I have to hand I have 25mm XLPE SWA as 1.65 mV/A/m...

Just to confuse you more...where are you getting your data from?...Im guessing it's XLPE your using?...

 

[telectrix]

the figure i would use is from table 6E2, OSG. is 1.5mV/A/m.

 

[Jonny66]

Hi Wade88,
According to the IEE design guide the formula (mV/A/m)z x L x Ib /1000 should be used, ie, the z value of 1.5mV/A/m if the power factor of the load is not known. As above:smilewinkgrin:

 

[Guest55]

the reason different volt drop figures for 2 & 3/4 core cables are given in the regs tables
is that it makes the adjustment for you.
follow Tels figures , youre making it too complex.

 

[Jonny66]

Just read your post...it should be single phase (2 core) volt drop times by square route of 3 divided by 2 (0.866) will give the 3 or 4 core volt drop in mV/A/m....
So it would be in your case single phase volt drop of 1.75 x 0.866 = 1.5mV/A/m

 

[Knobhead]

What is the supply for?

The neutral current calculation is:
In=√((Ia²+Ib²+Ic²)-((Ia*Ib)+(Ia*Ic)+(Ib*Ic)))

 

[wade88]

lol oh thanks Tony....ill just jot that down on my lunch box for next time :ihih:

Yea 1.5 is what i had to begin with, so im happy with my results after that.

Thanks fellas.

I wish you all a great Christmas and a boozy new year.

Ben