>> So if you are doing microscopy though a gas flow-cryostat and it looks
>> like your subject is below the surface of a pond, it likely means you
[quoted text clipped - 4 lines]
> an intermediate chamber, evaporating it there before the gas
> goes into the optical cell ?
-cut-
> are you introducing liquid ?
> have you a needle valve ?
> expansion valve ?
> is the cell at atmospheric pressure or is it pumped down ?
Well, from what I can tell (I inherited this lab from a doctoral student
who ran off after giving me only a cursory explanation of how the lab
worked), we're working with liquid helium that should evaporate before
it hits the cell.
Liquid helium enters the cryostat via a coaxial transfer arm and needle
valve. We pump the return helium out via the same coaxial arm (this
time on the outside), back into a recycling system. Somewhere along the
line, I think the helium is supposed to evaporate---and before it hits
the sample. There is no intermediate chamber that I know of. Is this
possible? Or do I think I have a cryostat design that was never
invented? This is an old off-the-shelf Oxford system, so it shouldn't
be anything crazy.
The cell is at atmospheric pressure flooded with warm helium, but the
flowing cold helium should feel the under-pressure from the helium
recycling system pump downstream. Sometimes the pumping is strong
enough to overpower the evaporation pressure of the boiling helium, and
introduces a vacuum in the dewar.
So to your bullet points: yes, yes, not that I know of, and it varies.
> i think you will have a factor of 800 or so in volume expansion if you
> turn up the heat to boiling. what it your helium consummption ?
> with that and the latent heat of evaporation (20 J/g if i recall)
> you should be able to calculate a heat input ...
Without heat, I can put away roughly 1 cubic meter of gas helium per
hour. We measure this directly through the recycling system. So that's
about 180 grams, or 3600 J, right?
>> little and turn up the resistive heat slightly, to make sure you boil
>> off the incoming helium before it gets into your optical path. I tried
>
> sounds like you need less heat, and a pressure gauge in the cell
The first part is easy, second part not so much. I do at least have a
0.05 bar over-pressure safety valve, although it seems like that valve
is cut-off from wherever the pressure builds when the helium boils, hence...
>> this today and built up enough pressure to blow out the sample insert
>> out of its position (luckily, nothing was damaged).
>
> heehee
>> How do I avoid this? My guess is that I need to make sure to get the
>> helium flow down to a minimum before I go heating up the element.
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> input; when there is liquid it will hardly change at all and
> will sit at 4.2K (at atm. pressure)
There was definitely lag between changing the heater voltage and the
temperature readout. It wasn't at 4.2K though, more like 6K, which I
think just means the sensor isn't positioned as well as I would like it
to be.
>> Does anyone have any other cryo tips? There's precious little
>> literature on this subject, so I'm making an appeal to the great usenet.
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> the first reference is primarily for 1K or lower, but has
> all maner of data for crypgenics.
I've only read Ekin "experimental techniques for low-temperature
measurements", which seemed a little too general. I will see if I can
get a copy of Cornell or Lounaasama.
Thanks for the help. From what I can tell, I'm not that far off to try
reducing flow and being more careful with the heater...well, back into
the lab I go...
sidd@situ.com - 23 Feb 2008 13:00 GMT
>who ran off after giving me only a cursory explanation
let im get away, did you? find a fone # is my first step
>Liquid helium enters the cryostat via a coaxial transfer arm and needle
>valve. We pump the return helium out via the same coaxial arm (this
>time on the outside), back into a recycling system.
snip--
>is no intermediate chamber that I know of. Is this
>possible?
oh yes
>This is an old off-the-shelf Oxford system, so it shouldn't
>be anything crazy.
well...i might not go quite that far...
>Without heat, I can put away roughly 1 cubic meter of gas helium per
>hour. We measure this directly through the recycling system. So that's
>about 180 grams, or 3600 J, right?
mmm i think the 20J/g is for just latent heat to change he4 from
liquid to a (very dense) gas.. then you have the expansion to atmospheric
pressure and temperature which eats many joules so i think you have to put
that in as well
snip--
>There was definitely lag between changing the heater voltage and the
>temperature readout. It wasn't at 4.2K though, more like 6K, which I
>think just means the sensor isn't positioned as well as I would like it
>to be.
at 6K i tink you are to the right of the (normal) liquid gas
critical point at 5.sumpn in the P-T plane
sidd