To determine the time resolution of instrumentation such as BLMs and BPMs, it is useful to try to estimate how fast the machine protection system will need to react to sudden changes (e.g. beam loss) in the machine.
Q:How fast does the MPS system need to react, and hence what is the time resolution needed for instruments that feed signals into it?
A: One way of doing this is to calculate how much energy deposition is needed to damage (i.e. melt) material.
Assuming ESS runs at 5MW with 20Hz rep rate, this means each pulse is 250kJ. There are about 7.1 105 bunches is a pulse, so each bunch corresponds to 0.35J. This is at top energy (end of the linac).
Melting point of steel is about 1360C, the temperature coefficient is about 460J/kg/C and the density is 7800 kg/m3 (7.8 g/cm3). To raise the temperature of 1 cm3 of steel from ambient temperatures to the melting point (i.e. a delta T of 1340C) requires 460*7800*1340*10-6 = 4.8kJ, or the energy contained in 1.4 104 bunches.
In reality, the transverse beam size is less than 1 mm2, and the energy loss varies along the beam path in material.
A 2.5 GeV proton is essentially a MIP, and hence loses 1.5 MeV/g cm2 (or 11.7 MeV/cm in steel) initially. The energy deposited in the first cm is 0.5% of the total, and the corresponding volume is 0.25% of a cm3 (assuming the transverse beam size is 0.5 mm). Therefore, neglecting heat conductance and radiation, this small initial volume would be heated to the melting point if it were traversed by about 7000 bunches (or 1% of a full pulse).
Note that the thickness does not matter in the MIP approximation (for the first mm, the loss is 0.05%, and the volume 0.025%, so the outcome is the same).
On the other hand, for protons most energy is not deposited at the entrance but around the Bragg peak. The range for a 2.5GeV proton is about 1.3 m in iron. In this distance, however, the protons undergo multiple scattering and other effects that tend to spread out the beam distribution transversely, making it hard to evaluate the energy density analytically. This needs to be evaluated using a Monte Carlo code such as Fluka or MARS.
It is useful to note that the length of the ESS linac is about 500 m, which corresponds to about 1.5 s or 500 bunches. This sets a lower limit to how fast a signal can be sent back to the front end in order to turn off the beam.
Thus it seems that the response time of instruments feeding into the machine protection system should be of the order of a few s, or about 500-1000 bunches.
Note that the above argument assumes that the high energy end of the linac has the biggest problem. At lower energies, the beam size is larger and each particle carries less energy. However, due to the Bragg effect and the shorter range (and thus less multiple scattering), this may not necessarily translate into less energy density deposited. This needs to be studied by Monte Carlo simulations.