Discuss Does BS7671 allow fuse rating to exceed cable rating in the UK Electrical Forum area at ElectriciansForums.net
Yes providing a few conditions are met.Does BS7671 allow fuse rating to exceed cable rating where it is not required to offer overload protection ??
As in the case of a 100A fuse/16mm tails: supplying a CU with just a few final rcbo circuits ?
Could you give me any pointers ? Reg numbers ?Yes providing a few conditions are met.
Not specifics as the BYB is in the office but probably around the 433 mark in the BYB.Could you give me any pointers ? Reg numbers ?
Okay thanks great, !Not specifics as the BYB is in the office but probably around the 433 mark in the BYB.
The best practise guide (number 6 I think -consumer unit replacement) states in section 6.1 that 16mm meter tails with a 100A cutout could be deemed adequate if the maximum demand of the installation does not exceed the ccc of the tails AND also that the requirements of Regulation 434.5.2 for protection of the tails against fault current are met.Could you give me any pointers ? Reg numbers ?
Ps... can someone please tell me how to write a sqaured sign on an android device? Also an ohm! I used to be able to do it but the hash thing doesnt work any more!
Hi - the earth fault current for TT is very low. With typical Ze ranging 10 to 100 Ohms the earth fault current won't bother 16mm tails... if the maximum demand of the installation does not exceed the ccc of the tails AND also that the requirements of Regulation 434.5.2 ... Not sure about a TT but K sqared x S squared is never going to be > I squared x t so I guess its not acceptable for a TT.
Does anyone actually contact the supplier to confirm anything with them like ever??
Will their fuse protect our tails, can we extend our tails greater than 3meters, do we have any special considerations to think about when applying protective bonding in a pme installation other than using bs7671 etc?
Ω
Testing to see if this works..
K²S²>I²t
S = (√I²t) ÷ K
think that gave be a min x section of 2.2 mm. (using the virtual 230V and a disconnection time of 0.1 sec.Ω
Testing to see if this works..
K²S²>I²t
S = (√(I²t)) / K
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