BS88 Fuses

Well, yes it is. But it gets worse, not better ... the 160 A fuse will pass an infinite amount of energy at 160 A ... it would never blow, so t = ∞, therefore I²t = ∞.

So, I'm not taking the micky by stating the above... I'm trying to illustrate that let-through energy takes on a different meaning with long-duration faults, over short-duration faults, or indeed when there is no fault at all. Similarly, the adiabatic criterion k²S²≥I²t is valid only for short-duration faults, and for longer duration faults, other methods (non-adiabatic) would be needed to take into account that the cable loses heat to the outside world (air or other material around the cable) as well as gaining it from the energy transferred by the passage of electrical current.

So, to illustrate what's going on, a cable rated for 160 A can get rid of the heat generated within it by passing 160 A quickly enough through its outer insulation or sheath, so it never gets above a certain conductor operating temperature (usually 70 deg C, occasionally higher if the cable is terminated in products that can cope with a higher temperature).

Can I help further, by asking why you are thinking about at a 70,000 second fault?

Do not worry - this is one that trips a few folk when they first meet the idea of adiabatic measurements

Well it will need to be pretty well lagged for  I^2*t to apply over 70,000 seconds. The whole idea of constant let-through energy really only works for things that happen really fast, so the energy is all in one place    - only in the melting fusewire, (or are in some super-lagged vacuum flask so no heat can escape.)
As a homely analogy consider that a mars bar and a small stick of dynamite both pack about 1megajoule, but the chemical decomposition of the mars bar does you less damage, as the energy is allowed to spread around the body and  dissipate slowly as a small temperature rise over hours. For things like fuses that hold in the hand, I2t is only really constant at times of a few seconds or less.

The cables too will cool over a few tens of seconds to minutes depending on size and installation method.

So look at the fuse curve again - your cable needs to carry 160A for ever, and maybe twice that for an hour or two. 35mm2 might do, maybe 50mm2. That is the near steady-state part of the operation.
Now slide you fingers along to the 'fires in seconds' part of the curve. Now we have thousands of amps, but not for very long at all,  and so we do not need a cable that can take those kilo-amps all day, just for the duration of the blowing time. That is where the approximation of a near constant I^2*t comes in.

Come back if there is no 'ahah' moment after you read this and I and the others can try again..

All the best
Mike

70,000 seconds is a very long time in the thermal world - the usual adiabatic equations are only good to around 5s - beyond that heat starts to be lost from the cable and you actually need a much smaller conductor than k²S² would suggest. If you applied the adiabatic to 160A and infinite time it would suggest you need an infinite c.s.a. cable - which clearly isn't correct or helpful.

Generally relatively small overcurrents like this are regarded as overloads rather than fault currents - and in most cases simply ensuring that the cable's Iz is at least In is sufficient to ensure it's protected - there being suitable margins in the tables and construction of the cables to ensure reasonable safety (thermally at least). Small corrections might be needed for BS 3036 fuses and buried cables, where the margins are otherwise a bit thin.

   - Andy.

Thank you for your reply.

Didnt know adiabatic equation was valid only for short duration faults - presumably in the range of seconds only ?

Not thinking about a 70,000 second fault, just trying to understand sizing of cables and protection etc.

Hi Andy, thanks for getting back.

It was the small overcurrent scenario I was thinking off.

I didnt realise it only applied to short duration faults and not the small overcurrent scenario, which take can take hours to blow the fuse.

Some good analogies above, for although the energy transferred for a small overcurrent may be large, it it the time overwhich it occurs keeps temperature rise down.

Thanks, Derek