With the changes to the C&G EVSE courses there is a sudden resurgence in interest in doing full circuit design calculations as electricians are failing the domestic and commercial course exam, because they can’t do the full circuit design calculation.

I did it all properly twenty years ago when I did the old C&G 2400 Design, Erection and Verification course, but have rarely done a full calculation since, so I am out of practice and having to think about it.

Looking at your calculation it seems you have not multiplied the wattage of the lamps by 1.8 as recommended for discharge lighting in the IET guides, such as the IET On-Site Guide on page 136.

You have simply said they are 400 watt floodlights, not exactly what they are, multiplying the wattage by 1.8 may need to be considered, we would almost have certainly been doing it twenty years ago.

not multiplied the wattage of the lamps by 1.8

I thought that was to account for an unknown, possibly quote poor,  PF (plus control gear) - the OP seems to have used PF=0.9 (presumably accurate for those particular fittings).

   - Andy.

yes you are absolutely correct, our national regulation also using the same thing shown in your appendix, that 1.8 multiplier and 66% current demand (allowance for diversity) for domestic.

Let say I include that in my previous calculation,

Ib = [ Power / ( Voltage x  Power factor) ] x 1.8 x 0.66

Ib = [ 2x400W / (230V x 0.9) ] x 1.8 x 0.66

Ib = [ 800W / 207 ] x 1.8 x 0.66

Ib = 3.86 A x 1.8 x 0.66

Ib = 4.6A

so theoretically Ib 4.6A < In 10A < Iz 17.5A (for 1.5sqmm). With this result and my previous understanding, most likely I will still use the 1.5sqmm cable. But after further explanation from Mr. AJJewsbury, it seems like I need to check back and consider the inrush current as well.

Thank you for your input. 

yes you are absolutely correct, our national regulation also using the same thing shown in your appendix, that 1.8 multiplier and 66% current demand (allowance for diversity) for domestic.

Let say I include that in my previous calculation,

Ib = [ Power / ( Voltage x  Power factor) ] x 1.8 x 0.66

Ib = [ 2x400W / (230V x 0.9) ] x 1.8 x 0.66

Ib = [ 800W / 207 ] x 1.8 x 0.66

Ib = 3.86 A x 1.8 x 0.66

Ib = 4.6A

so theoretically Ib 4.6A < In 10A < Iz 17.5A (for 1.5sqmm). With this result and my previous understanding, most likely I will still use the 1.5sqmm cable. But after further explanation from Mr. AJJewsbury, it seems like I need to check back and consider the inrush current as well.

Thank you for your input.