With reference to 433.1.204 and cable as installed has min capacity of 20A if protected by a 30A or 32A. If the protective device is reduced to 20A how is the new minimum as installed capacity calculated or arrived at ? I've been looking in the Electrical Installation Design Guide, but the answer is avoiding my eyes.
The assumpmtion is about bunching. The worse case is when all the load is near one end, say within the first 10% of one of the limbs.
By the miracle of resistance scaling with length of cable, and the voltage drop from origin to load being the same both the long and short way round, the current split is always in inverse proportion to the ratio of the lengths.
So if the load is equally far from each end, then the load is shared perfectly, and 16A goes each way.
But let us consider an example of the bad layout case..
It is possible to have say 3 sockets each pulling 10A at 1m, 2m and 5m from the origin and then a 20m return leg with no load.
Now the first 10 amps is spit in the ratio of 1/25 to 24/25, those being the tow path lengths to that socket.
The second 10A is split 2/25 to 23/25
the third 5/25 and 20/25
So as the common denominator is 25ths of 10A is 400mA 'units'
Going left we have (24+23+20)* 0.4A = 26.8 amps
going right we have (1+2+5)* 400mA = 3.2A
Quick sanity check total = 30A.
So that last metre of cable from CU to first socket takes nearly 27A, and would be fine in plaster or clipped direct, but may age rather faster than we would like in thick insulation. It is the only stressed cable on the ring, and making it longer, or moving one load to the return leg would restore order, as would making that first socket a spur off the MCB, so its current is not adding to the short leg current.
(This is often a quicker fix)
If you can avoid having more than one socket in the first 20% of the ring length on either end, the the problem vanishes even in quite thick insulation.
The assumpmtion is about bunching. The worse case is when all the load is near one end, say within the first 10% of one of the limbs.
By the miracle of resistance scaling with length of cable, and the voltage drop from origin to load being the same both the long and short way round, the current split is always in inverse proportion to the ratio of the lengths.
So if the load is equally far from each end, then the load is shared perfectly, and 16A goes each way.
But let us consider an example of the bad layout case..
It is possible to have say 3 sockets each pulling 10A at 1m, 2m and 5m from the origin and then a 20m return leg with no load.
Now the first 10 amps is spit in the ratio of 1/25 to 24/25, those being the tow path lengths to that socket.
The second 10A is split 2/25 to 23/25
the third 5/25 and 20/25
So as the common denominator is 25ths of 10A is 400mA 'units'
Going left we have (24+23+20)* 0.4A = 26.8 amps
going right we have (1+2+5)* 400mA = 3.2A
Quick sanity check total = 30A.
So that last metre of cable from CU to first socket takes nearly 27A, and would be fine in plaster or clipped direct, but may age rather faster than we would like in thick insulation. It is the only stressed cable on the ring, and making it longer, or moving one load to the return leg would restore order, as would making that first socket a spur off the MCB, so its current is not adding to the short leg current.
(This is often a quicker fix)
If you can avoid having more than one socket in the first 20% of the ring length on either end, the the problem vanishes even in quite thick insulation.
