Vehicle chraging point - Section 722

[Brightspark2]

I've been looking at a job installing a vehicle charging point on a commercial premises. Earthing facility is PME, installation is 3 phase.

Now looking at 722.411.4.1, it says a PME earthing facility shall not be used.........except where i, ii or iii applies.

i states - A charging point forms part of a 3 phase installation that also supplies loads other than for electric charging and because of the characteristics of the load of the installation, the maximum voltage between the main earthing terminal of the installation and earth in the event of an open circuit fault in the PEN conductor is not likely to exceed 70v.

Now it gives a formula in Annex A722 to calculate first the maximum neutral current of the installation, which you can then use to either calculate the load balance to prove the above exception OR to use to calculate the maximum Ra value should you have to use an earth rod. Now I/L1 etc are all straight forward, but it has no multiply, plus or minus sign between 0.5(I/L2 + I/L3), same again after 0.866, so does that mean I just put the sum in the brackets to the end of 0.5 or do I multiply the sum from the brackets to the 0.5 or add the sum from the brackets to 0.5 etc?.

The formula A.722.1 is √[(I/L1 - 0.5(I/L2 + I/L3)][SUP]2 + [/SUP][0.866(I/L2 - I/L30)][SUP]2


[/SUP]Now I/L1 etc are all straight forward, but it has no multiply, plus or minus sign between 0.5(I/L2 + I/L3), same again after 0.866, so does that mean I just put the sum in the brackets to the end of 0.5 or do I multiply the sum from the brackets to the 0.5 or add the sum from the brackets to 0.5 etc?.

The earth rod would be a bit of a nightmare as the the socket is going onto the rear of a shop in a shopping center, with a tarmac car park behind.

Any help much appreciated...:crazy:

 

[yellowvanman]

In the example

0.5(I/L2+ I/L3)

You always calculate whats within the brackets first and then multiply that by what is next to the brackets.

In the bigger formula with the square root, again you calculate whats within the () first, then calculate whats within the [], square each of the [] products and then add them together and then square root!

Clear as mud! :smiley2:

Taking it step by step:

√[(I/L1 - 0.5(I/L2 + I/L3)][SUP]2 + [/SUP][0.866(I/L2 - I/L30)]2

√[(I/L1 - 0.5(A)][SUP]2 + [/SUP][0.866(B)]2

√[(I/L1 - C][SUP]2 + [/SUP][D]2

√[(E][SUP]2 + [/SUP][D]2

√F

 

[yellowvanman]

Did a one day course on this and if I remember correctly the Ra needs to be quite low to comply with the reg! Never did put the course into practise!

 

[DuaneMHunt1976]

I have no clue on this but i am very maths Based

Im = is the rms maximum neutral current o a 3-phase installation (in Amperes), including that of the electric vehicle charging load and any other loads, determined in accordance with Regulation 311.1

Im = SQR ( { [IL1 - 0.5(IL2+IL3)]^2 } + { [0.866(IL2-IL3)]^2 } )

if you want to break it down then
A = 0.5(IL2+IL3)
B = (IL1 - A)^2

X = 0.866(IL2-IL3)
Y = X^2

Im = SQR (B + Y)

IL1 IL2 IL3 = are the RMS values of current (in amperes) in lines 1, 2 and 3, respectively, that were used when determining the value of Im



I dont know were these I/L1 .. I/L3 comes from