The tails are fine as far as BS7671 is concerned (and as long as your DNO don't have some stupid policy...)
60A fuse in switch fuse will protect the cables from overload, the service fuse has to protect them against fault currents, you want the formula from 434-03-03 which is:
t=(K²S²)/I² which tells you how fast the fuse needs to disocnnect, if we plug the values in, S is the size, K is a constant for type of cable, in this case 115, I is fault current which is 2300kA
so t=(115²*16²)/2300² which gives 0.6 seconds, a look at the graph on page 194 will tell you that 1800A will open it in 0.1sec (see figures in top right), so 2300A will open it even faster, and as 0.1 is a lot faster than 0.6, the tails are protected.
I actually perfer to re-arrange it as S²=(I²t)/K² to give conductor size (it actually appears in in this form in another section of BS7671, work it through with 2300A and 0.1s again (its going to be faster, but it simpifies things to assume worse than it is) gives a size of 6.4mm² for minimum conductor size
Another rearrangement is to compare S²K² to I²t and if the former is bigger it passes (this is usful when obtaining more exact I²t values from datasheets rather than calc'ing from fault level and clearance time (sometimes there is current limiting involved)
Like most things, it doesn't pay to calculate things down to the nth significant figure, but you can clearly see your tails pass by many miles
60A fuse in switch fuse will protect the cables from overload, the service fuse has to protect them against fault currents, you want the formula from 434-03-03 which is:
t=(K²S²)/I² which tells you how fast the fuse needs to disocnnect, if we plug the values in, S is the size, K is a constant for type of cable, in this case 115, I is fault current which is 2300kA
so t=(115²*16²)/2300² which gives 0.6 seconds, a look at the graph on page 194 will tell you that 1800A will open it in 0.1sec (see figures in top right), so 2300A will open it even faster, and as 0.1 is a lot faster than 0.6, the tails are protected.
I actually perfer to re-arrange it as S²=(I²t)/K² to give conductor size (it actually appears in in this form in another section of BS7671, work it through with 2300A and 0.1s again (its going to be faster, but it simpifies things to assume worse than it is) gives a size of 6.4mm² for minimum conductor size
Another rearrangement is to compare S²K² to I²t and if the former is bigger it passes (this is usful when obtaining more exact I²t values from datasheets rather than calc'ing from fault level and clearance time (sometimes there is current limiting involved)
Like most things, it doesn't pay to calculate things down to the nth significant figure, but you can clearly see your tails pass by many miles
