Dual Appliance Connection Unit

[hightower]

So I was doing my usual morning browse this morning, and I have no idea how but I ended up reading a load of threads from years ago about the Dual Appliance Connection Units - it left me with a question that I couldn't find an answer to within those threads, and not wanting to reopen a 3 year old thread, here we are.

So, these:



It looks to me that the supply is thicker than the two loads. Let's say the supply is 6mm on a 32A MCB, and the load is two 4mm cables. If the 4mm goes directly to the appliance (for arguments sake it doesn't have a plug end) how do they get away with using a smaller cable size without fusing it down.

Is it because:

a) There is little chance of overload on these appliances

or

b) A 4mm cable clipped direct will still carry 37A (sorry if slightly incorrect, using older regs) and therefore doesn't need fusing down

or

c) Something else altogether

 

[Andy78]

You are right in that 4mm T&E will take 37A installed in ref method C, I have in the past installed 32A 4mm circuits for cookers.

You need to be looking at 433.3 Omission of overload protection where the cables are not to be afforded adequate overload protection with the supply circuit OCPD though.
For instance, if your cooker came with a 1.5mm flex supplied and your hob with a 2.5mm flex would you drop the MCB to a 16A to protect the 1.5mm flex ?

 

[hightower]

Thanks Andy! I think that's what I was trying to say for point a), just not very well.

So essentially, provided the CCC of 2.5 is sufficient for the load of the cooker or hob, you could use that to connect up from the connection unit on the load side?

(Ignore this last comment - of course that's what it means, duh!)

 

[Andy78]

Yep pretty much.

433.3 (ii) is the bit applicable which relates to loads unlikely to draw an overload through the conductors. This can be applied to fixed loads such as purely resistive heating elements.

If the flex has been supplied with the appliance it can be assumed that it is capable of carrying any current draw associated with that appliance.
If no cable is supplied then you would either install one as per the instructions supplied or calculate your own cable size as normal.

 

[Risteard]

So essentially, provided the CCC of 2.5 is sufficient for the load of the cooker or hob, you could use that to connect up from the connection unit on the load side?

Once again, it is not only the nature of the load being fixed (and therefore cannot overload) which is important. You must also demonstrate that short circuit protection will be afforded by the arrangement.

 

[hightower]

Once again, it is not only the nature of the load being fixed (and therefore cannot overload) which is important. You must also demonstrate that short circuit protection will be afforded by the arrangement.

Which would be done getting the r1+rn to work out the resistance, and therefore the max current under short circuit conditions, and make sure this triggers the breaker in the time required?

 

[telectrix]

you'd normally use Zs. the fault current would be equal or higher, L-N ( depending if the cpc was smaller csa than the N).

 

[hightower]

Ah, of course. Cheers Tel

 

[Risteard]

Which would be done getting the r1+rn to work out the resistance, and therefore the max current under short circuit conditions, and make sure this triggers the breaker in the time required?

Well you would need to work out the operating time of the protective device and ensure that the cable can withstand the fault current until disconnection occurs.