
Software Technical Lead
Premium job
 Hanslope Park, Milton Keynes
 £40,722 to £53,575 + £5k Welcome Package
From the smallest details to leading the biggest decisions, your impact will be out of this world.
 Recruiter: HMGCC

Lead Planning Investment Engineer
 Aberdeen, Aberdeenshire
 up to £62,500
Lead Planning and Investment Engineer (514108) Location: Aberdeen, Perth, Inverness (flexible) Salary: up to £62,500 depending on skills and experi...
 Recruiter: SSE

Electronics Technical Engineer (Design and Operation)
 Switzerland
 £5106.81  £7292.30 per month
Founded in 1954 in Geneva, CERN ? the European Organization for Nuclear Research is the world's most advanced fundamental research institute for particle physics with currently 22 Member States.
 Recruiter: CERN
IET 
search :
help :
home


Latest News:


Topic Title: The need for a C type CB? Topic Summary: Created On: 28 November 2012 09:13 PM Status: Read Only 
Linear : Threading : Single : Branch 
Search Topic

Topic Tools

28 November 2012 09:13 PM


Could someone enlighten me as to an easy rule of thumb for the need to upgrade from a B to a C type circuit breaker?
This is specifically for commercial flourescent lighting. The current lighting is 6 x 80w 6' tubes, to be replaced by 6 x double 70w fittings. It is currently protected by a 6A type B. So, I am, roughly, doubling the current usage, so using 1.8 x the rated current, this gives 1512 watts/ ~ 6.6A, so the CB needs to be uprated, which isnt a problem, as it is run in 1.5mm T+E. The lights I am fitting give no indication of any 'inrush' current, so what factor is used to determine if a C type CB is required? Thanks Alan. 



28 November 2012 09:40 PM


Time/ current graphs




28 November 2012 09:41 PM


Well Alan  as a conservative estimate in the absense of any firm data aim for Ib to be less than 0.4 x In
So in your case with a design current of 6.6A you'll want something like a 16A MCB If you reduce your 1.8 to a more reasonable factor though, closer to actual running current you will have Ib of about 4A  so a 10A MCB. You'll recognise this as a typical light commercial solution when used with 1.5mm2 cable  with a type C of course Ideally you want inrush data for those ballasts  you can then determine what your peak current is and select a suitable breaker from there (allowing a crest factor). Didn't you recently do 2396  take a look in the electricians guide to calculations  Mr Cook explains it all in one of the later chapters Regards OMS  Let the wind blow you, across a big floor. 



28 November 2012 09:44 PM


Or if it trips a "B", fit a "C"! Not a very scientific approach but practical,
Dave. 



28 November 2012 09:51 PM


Thanks all, yes, I did do 2396, and have got various books, but, of course, it is far easier to ask a question than to actually do some reading!
Yes, I'll have a look through the books tonight to understand the theory, I was going to fit a C type anyway, but wasnt quite sure on the need for it, and how to calculate it. Thanks. 



28 November 2012 10:02 PM


There is a section towards the end of the book Alan  it's a short chapter.
Basically you need the inrush current of the load. from there you need to take the leading edge of the MCB characteristic so 3 x In for a Type B and 5 x In for a Type c and multiply that by a crest factor of say 1.404  so for a 10A type C you have 10 x 5 x 1.404 = 70A For a Type B it would be 10 x 3 x 1.404 = 32A If the total inrush is less than 70A (or 32A) the MCB will hold against the inrush  if it's more the MCB will probably trip but remember it could actually be operating towards the trailing edge so not gauranteed that it will Does that help ? Regards OMS  Let the wind blow you, across a big floor. 



29 November 2012 09:19 AM


Needless to say you'll need to ensure that the lower Zs limit can be complied with (presuming no RCD)  and check that the energy letthrough of a C type won't exceed what the 1.0mm² c.p.c. will withstand (especially for faults close to the DB).
 Andy. 



29 November 2012 10:32 AM


multiply that by a crest factor of say 1.404 Really? Regards 



29 November 2012 11:35 AM


multiply that by a crest factor of say 1.404 Really? Regards Ahem  I think I'll need to check my typing  that should read 1.414  Regards OMS  Let the wind blow you, across a big floor. 



29 November 2012 02:32 PM


Thanks all, yes the cable is pretty short so can cope with any VD/Zs/fault current etc. I found the relevant section in the IEE/IET calculations book,  all pretty simple, now I'm just about to ring up the manufacturer to get the inrush figures  clearly they dont think a 'typical' fitter will want to know, as the instructions, (if you can call them that) are quite sparse in their information.
Thanks Alan. 



04 December 2012 10:01 PM


My reply said time/current graphs. Isn't that correct? You need to know the load current,the starting current, and the inrush current. Each of the current demands must be within the lowest tripping current of the overload device.
I must admit I had never heard of crest factor until I read this post but I have now read the electricians guide that oms refered to. 



04 December 2012 10:39 PM


Well, not the ones in BS 7671 anyway  as they only give the trailing edge (worst case for guaranteed disconnection)  whereas to avoid nuisance tripping you need to look at the leading edge. Manufacturer's data usually give both. Even then the graphs aren't that helpful when you're considering the magnetic (instantaneous) trip of MCBs  at best they're just the vertical line bit. You can get just as much information from the multiplier for each type of MCB  e.g. 3x to 5x for a B type (again for avoiding nuisance tripping you'd want to keep the total inrush below the lower figure (with any adjustments for crest factor etc). In terms of cables withstanding fault currents where MCBs are involved, faults close to the DB are the worst case!  Andy. 


IET
» Wiring and the regulations
»
The need for a C type CB?

Topic Tools

New here?
See Also:
FuseTalk Standard Edition v3.2  © 19992017 FuseTalk Inc. All rights reserved.