DOes anyone know how to do this

[lofty84]

A circuit is protected by a BS88 fuse and the prospective fault current is 171 A. How do I determine the csa of the CPC assuming 70 oC themoplastic cables with copper conductors.

Can anyone tell me where I need to be looking in the BS 7671 or onsite guide please ?

I think it might be page 128 in the onsite guide but Im not 100 %

thank you
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[pushrod]

reg 543.1.4 of BS7671 don't think it is in OSG

edit you need to know the size of the fuse as well as the type (if my brain is in gear)

 

[i=p/u]

not in osg

 

[Steve1412]

Selection from 543.1.4 or calc from 543.1.3. Need more info..Easiest selecting...what size is phase conductor???

 

[lofty84]

i think the answer is

S = sq root of three x 0.1 / 115 = 0.47mm sq

 

[lofty84]

anyone brave enough to attempt to see if this is right ?

A water heater with an output of 5kw is connected to a 230v supply via a bs 60898 type b protective device. the cable is 70 oC PVC insulated thermoplastic coper conductors and the length of cable is 18m. the ambient temperature is 35oC. The cable runs for 300 mm through thermal insualtion. it runs together with one other cable for most of its length and is clipped direct to the wall. Determine the following and show all calcs and refs to tables.

1. Design Current

Ib = 5000/230 = 21.73 A

No diversity according to the on site guide (page 97)

Minimum protective device rateing

In is greater then or equal to Ib = 25 A mcb BS 60898

Correction Factors

It ? = In/caxcgxcixcc

ca=table 6a OSG/ cg = table 6c OSG/ ci = i think table 6b OSG/ cc = 1 i think page 124 OSG ??

It = 25 / 0.94 x 0.80 x 0.51 x 1 = 65.18 A

not sure what do do or say for the minimum cable rating ??

Minimum cable size

= 10mm2

Maximum allowable voltage drop

= 4.4 according to table 6D on page 130 OSG

Actual Voltage Drop

= (mv/a/m) x ib x l / 1000

= 4.4 x 21.73 x 18 / 1000

= 1.72 A

is it 5 % of 230 (11.5) so yes that means it doesnt need a bigger cable ?