Basically if you have a given fault, and lets assume it is a dead short at the end of your circuit, then if you know Zs you can find the lowest PFC from the minimum supply voltage as:
PFC = Umin / Zs
From this level of fault current you then look up the current-time curve of the breaker or fuse and that gives you the time to disconnect at that level. Once you know the fault current and corresponding time you have the let-through I2t "energy" and that allows you to compute what it will do to the cable.
The usual assumption is the adiabatic case - then the input heat energy is over a short period of time so the out going heat energy can be neglected, which is reasonable for a short circuit type of fault - and so if you assume a closed thermal system, then a given temperature rise depends on the material constant and input energy.
This is normally simplified even more by assuming only a couple standard conductors (Cu, Al, and Fe) and a few common thermal limits (e.g. for thermoplastic and for thermosetting insulation) as a table of a few fudge-factors to use.
It might be easier to give an example. Say you have 63A fuse (common for domestic in UK) and you find your Ze value was 0.35 ohms (typical upper limit for TN-C-S) and you wanted to size a copper earth wire. Lowest fault current would be 95% of U / Ze
PFC = 0.95 * 230 / 0.35 = 624A
Here is a typical fuse curve:
If you look up 624A prospective current for the 63A curve (as far as practical) you see the corresponding time is around 0.4 seconds, so we can compute our let-through energy as:
I2t = 624^2 * 0.4 = 156k (A2s units)
Looking at the values for copper and say 30C initial and thermoplastic (70C cable) we have k = 143 so we can size our earth using:
S >= sqrt(I2t) / k = 395 / 143 = 2.76 mm^2 CSA minimum
If we had Ze = 0.7 ohms (double the fault impedance) then our current is half at 312A but then our disconnect time is 7s hence I2t = 312^2 * 7 = 681k (A2s units) which is about 4.4 times larger and now our minimum earth conductor is given by:
S >= sqrt(681E3) / 143 = 5.77 mm^2 CSA
For a fuse (at least our BS88 ones) they limit the I2t let-through and so worst-case is at lowest fault currents when disconnection takes a long time. Of course once you get in to the ten seconds or more the adiabatic assumption no longer holds so eventually you end up with a steady-state current carrying requirement. But for a fuse the least fault energy is at max PFC.
For a breaker it is more complicated, and you have a massive difference between faults that hit the "instantaneous" trip and those that don't. Also MCB and MCCB are not fault-limiting to the same degree as fuses, so as PFC increases from that trip point you see a moderate increase in let-through energy.
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Just to add - that is why you do the lowest PFC (minimum supply voltage) for the adiabatic check as it gives the longest disconnect time and hence usually the biggest let-through energy.