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I Designing cryostats

  1. Feb 15, 2017 #1
    I'm looking into how cryostats are designed. There's an absolute wealth of information on how they work; less about how design choices are made. So I'm giving it a go myself, but I'm stuck on pretty much the first hurdle. I'd like the cryostat to have a boil off rate of helium of ##50##ml/hr max, but this could be a very complicated calculation if I try and work out the heat from individual sources like radiation and residual gas conduction. Instead I'd like to work out the total heat input that leads to the ##50##ml/hr figure. But I'm not sure that's possible.

    Instead, my thoughts were that I can find the rate of heat flow into the cryostat from an insert using Fourier's law:

    ##\dot{Q} = \frac{A}{L}(T_2-T_1) \bar{K}##

    Then do similar calculations for what I consider to be the other important sources to try and get a total heat flow into the cryostat, .

    And then use this equation: for heat input ##\frac{dQ}{dT}## the evaporation rate is ##\frac{1}{L} \frac{dQ}{dT}##

    I found that second equation in some obscure textbook and I can't derive it. I don't want to use it if I can't derive it, so could anyone point me in the direction of some resources on the subject or show me how to derive this? Thanks for any help! :)
     
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  3. Feb 15, 2017 #2

    ZapperZ

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    Please don't take this the wrong way, but I had a good chuckle after I read your post. This sounded like a theorist designing a piece of hardware! :)

    When you talk about "designing cryostats", are you talking about building one, or are you just doing a mental exercise of some generic heat-loss problem? When I "design" something (and yes, I've worked with cryostats before, and one of my grad students right now is in the middle of designing a cryostat system for his tunneling experiments), the first thing I do is to consider the hardware, and then make an estimation to see if it will work. So if I'm designing a cryo system, the first thing I will do is (i) find a cryostat shaft out of a catalog (this tells you all the specs of the heat exchange rate) (ii) figure out the dewar or insulated vessel to hold the cryogen, and then (iii) design the liquid transfer. Only after that, can I make a reasonable estimate of the rate of heat loss or cryogen loss. Even then, this will only be an estimate. One still has to do diagnostics as the system is being built.

    You can't estimate "Q" rate in your equation without knowing the insulation of your system. Some are good, some are poor, and this will dictate how quickly the He will evaporate. You are putting the cart before the horse here.

    Every cryostat system is different, and that is why you can't find one single, universal "design" and diagnostics.

    Zz.
     
  4. Feb 15, 2017 #3

    f95toli

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    Also, 50 ml/hour is quite a difficult goal. Most well designed transports dewars (with helium volumes of 60-120l) will boil off around 1 l/day; but that is just a container of helium hanging in a vacuum with lots of superinsulation wrapped around it; as soon as you start adding experimental wiring etc. the boil-off rate goes up (sometimes dramatically); some of the small cryostats I've worked with boiled off several liters a day (note that the area/volume ratio tends to favor larger dewars)
     
  5. Feb 15, 2017 #4
    Ah, you got me! I was just curious about how anyone would actually go about putting together a cryostat. As I said there's so much out there about how they work and literally nothing about how they're designed. I'm not actually going to build it. So to change the premise a bit then, what about cryostat inserts? Say I'm only designing one of them. Doesn't especially matter what for, it's just to simplify the problem, which I assume this does; because then I'm only interested in solid conduction, and change the problem to only ##50##ml/hr due to solid conduction in the insert. Then could I use the equations above? Life isn't that neat I presume, but would it give you a ballpark figure for an allowable amount of heat leak?
     
  6. Feb 16, 2017 #5

    f95toli

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    What do you mean by an insert? A probe that can be dipped into a regular transport dewar?

    I've made a few of those and their design isn't very sophisticated. However, I have never bother trying to actually calculate the heat loss, there is a well-established way of doing this and as long as you follow a few -fairly simple- design guidelines it isn't very hard. The probe itself is not necessarily the biggest cause of the increase in boil-off (the main structure should be main from thin-walled stainless steel tubes); conduction via experimental wiring can be a bigger issue (especially if you need e.g. a few lines of microwave coax). Thermal radiation is also always an issue, it is sometimes difficult to fit and thermalize the baffles when the wiring is in place.

    If you are interested in this topic I would suggest getting a copy of Pobel's "Matters and methods at low temperatures"; it is the best textbook I've come across in this area (it is getting a bit old, but the basics remain the same)

    Edit: Also, note that inserts are frequently used with dewars with a fairly wide neck (e.g. 50mm), and this in itself means that the boil-off rate from the dewar can be significant (3-4l per day).
     
  7. Feb 16, 2017 #6
    What's a transport dewar? By cryostat insert I mean something that you'd dip into the liquid helium in the cryostat to do experiments with. I assume they usually have some kind of device for low temperature thermometry, and then whatever else you want to do. Maybe a resonator of some sort.
     
  8. Feb 16, 2017 #7

    f95toli

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    liquid-helium-dewars-cmsh.jpg

    One of these (They are named after James Dewar). They are not usually referred to as cryostats since they are mainly used to transport and store helium. You would typically transfer helium from one of these into a cryostat (which is the thing used for measurements).

    However, you can also use a dewar for measurements in which case you use a long probe that can be dipped into the helium. However, most regular transport dewars are optimized for minimal loss which means that the neck diameter is quite narrow (something like 1") which makes it difficult to fit probes into them. You can however buy dewars with a wider neck, but the losses are then higher.

    You can put all sorts of things at the bottom of an experimental insert, there is not standard. Also, a thermometer is not always needed, if the sample is immersed in helium (or in a sealed volume but with helium exchange gas) you can be pretty sure that it will be at 4.2K. It is only if you need to go above 4.2K (with the help of a heater) that you need a thermometer).
     
  9. Feb 16, 2017 #8

    ZapperZ

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    @Kara386 : See? This is why, without a clear design of the hardware FIRST, it makes it difficult to have an idea the factors that are involved here, especially in how this is going to be done.

    Most experiments dealing with LHe will require a transfer system of LHe into another setup. So already there's loss due to the transfer and the cooling down of the other setup. How the thermal isolation of this setup where the experiment is done needs to be described in detailed. Then there is the insert probe containing the sample being cooled, and, presumably, a temperature sensor.

    The cryostat probe that I deal with is a closed-loop cooling channel that cools the sample block. I do not dip it into the cryogen. I "dip" it into an ultra-high vacuum chamber.

    So without some sort of a clear design to start with, I still don't know how you are going to do be able to do any kind of heat loss/cryogen loss estimate.

    Zz.
     
  10. Feb 19, 2017 #9
    I probably won't. Thanks for your answers though, throws up some important points to consider! So I've definitely learned something. :)
     
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