Cable size calculation ..

[L Plate]

Evening all

I need help please for some assignment .. can anyone please show me how to work it out and i will try my very best to do the rest of my homework . I have tried to worked this out using the red book but I cant figure it out .

this requires to carry out cable calculation from the information given using the tables in BS7671 , we need to justify our choice of cable size and the appropriate rating of the protective device . we need to show the voltage drop for the circuit ..

>>>> the costumer has asked for a double socket-outlet and a light in her shed, which is at the bottom of the garden, 39 m from the CCU . it is used for sorting small gardening tools and plant seeds. the power point will be used to run her flymo type mower ..

I tried to work out this way ,.. since she asked for a double socket outlet , i would say that the design current is 26A ?? and I tried to work out base from table 4E4B, page 291. since voltage drop = (mV/A/m) x lbxL divide by 1000 .. so i tried this way

26Ax10x39 divide by 1000

= 0.1014 Vd ? so my calculation for sca is 4mm??

I am not sure with my calculation because it says that the nominal Voltage is 230V. I hope I am not confusing everyone .

A 10KW shower has been installed by a plumber in the bathroom. Only the water supply to the shower has been installed. the shower unit is 25m from CCU . the bathroom has does not have a window and the loft space is insulated with 150mm of insulation.

since the bathroom has no window , i will surely need to put a extractor fan, again where and how to find what specific fan needed .

thank you all

All your understanding and help is most appreciated ...

Evening all .

Lplate

 

[dempstar]

I take it your load current is 10 as is the 2nd number in your aquation as had a question on my 302 exam a 3kw 230v imersion heater is supplied by 15m of multicore 70 pvc cable clipped direct to the surface Protection is bsen 60898 type c ambient temp is 30 determine load current so 3000 divide 230 is 13.04 then maz allowed volt drop is 230 x 5% is 11.5v, found the minimum cable rating in osg table 6e1. Actual volt drop was worked out at 29x13.04x15 divide by 1000 equaled 5.67v, If that helps you