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An often quoted figure is that the LHC magnets take a month to completely cool and a month to warm. There is never an explanation as to why that is. I can conjure any number of reasons (slow changes to prevent stress, very low temperature deltas, gremlins can hold their breath for 29 days, etc) but I can't find any facts.

I don't suspect the answer will be complex, but I'd just like to know why it takes a month and not, say, 14.6 hours.

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have a look at this espace.cern.ch/acc-tec-sector/Chamonix/Chamx2012/papers/… to get an idea of the complexity of the system, the effort to avoid quenches and beam dumps in the magnets. This is the acceleration and technology page at CERN espace.cern.ch/acc-tec-sector/default.aspx .you could search there for real answers. –  anna v Jan 5 at 6:44

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up vote 5 down vote accepted

This should be a comment as I'm not knowledgeable about the LHC refrigeration, but it's too long for a comment and I can hazard a pretty good guess.

Sheer heat capacity likely accounts for a great deal of this time. Assuming the total amount of kit that needs to be cooled is, say $2\times10^4$ tonnes, and if it has roughly a $1{\rm kJ K^{-1} kg^{-1}}$ heat capacity, that means we have to extract $20{\rm GJ}$ for every degree K that we cool.

Once we get down below $100{\rm K}$ were going to see some serious multipliers happenning. An ideal heat pump needs work input $W = Q_{LHC} \left(\frac{T_{out}}{T_{LHC}} - 1\right)$ to pump out heat $Q_{LHC}$ from the kit at temperature $T_{LHC}$ and dump it to the environment at $T_{out}$: this is the reversible heat pump.

So if we're drawing this heat out and dumping it at $300K$, say, the energy needed to get from $300K$ to $5K$ we get as a rough estimate (assuming heat capacities stay constant, which they won't, but there won't be any phase changes of most of the kit):

$$W_{total} = \sigma \int\limits_{T_{LHC}}^{T_{out}} \left(\frac{T_{out}}{T} - 1\right)\,{\rm d}T$$

where $\sigma$ is the $20{\rm GJ K^{-1}}$ total heat capacity I estimated above. Plugging in the numbers $T_{LHC} = 5K$ (not everything will need to be cooled all the way down to $1.9K$) and $T_{out} = 300K$ we get:

$$W_{total} = \sigma \left(T_{out}\left(\log\left(\frac{T_{out}}{T_{LHC}}\right)-1\right)+T_{LHC}\right) = 933 \sigma \approx 20{\rm TJ}$$

This is the total output of a $5{\rm GW}$ power station for over an hour, roughly the energy released by the first of the only two nuclear weapons brought to bear in combat. $5GW$ electricity generation is the electricity consumption of two million Australians, and we are extremely greedy electricity users by world standards, so I don't know how many normal people this would represent. The LHC Site quotes a peak power consumption of $180MW$ and about $30MW$ is used for cryogenics. $30{\rm TJ}$ at $30MW$ is about ten days.

Another factor is stresses in the kit induced by too swift cooling or warming. I am slightly familiar with the design of some of the magnetic beamsteering hardware, and much of this kit is toleranced to within tens of microns. One can't brook even the tiniest of any irreversible, plastic deformations of the kit and still have it work properly. So it's likely that heat transfer couldn't go much faster than this even if the refrigeration capacity were there.

There will also be economic considerations too. Even if one can cool faster than with $30MW$ refrigeration and still meet technical / engineering constraints, refrigeration capacity is expensive, particularly if you are only using this full capacity for cooldown during maintenance. The rest of the time the refrigeration needs are much less, so you have an economic tradeoff between capital spent on capacity that is unused most of the time and the cost of project delays arising from downtime. I'm absolutely sure exactly this calculation has been done, as it ones like it are done for all soundly managed engineering projects.

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I think there is a typo in the integral $$ W_{total} = \sigma \int\limits_{T_{LHC}}^{T_{out}} \left(\frac{T_{out}}{T} - 1\right)\,{\rm d}T $$ and it is equal to $$ W_{total} = \sigma \left(T_{out}\log\left(\frac{T_{out}}{T_{LHC}}\right) -(T_{out}-T_{LHC})\right) = \sigma \left(T_{out}\left(\log\left(\frac{T_{out}}{T_{LHC}}\right) - 1 \right) + T_{LHC})\right)$$ –  user31748 Jan 8 at 23:08
@user31748 You're right. The calculation still gives the same order of magnitude though. –  WetSavannaAnimal aka Rod Vance Jan 9 at 0:15

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