I will assume The total volume is 4000 l The circulation speed X lh Y lh 200 lh pump speed The volume is divided into N steps where NVX 20 with no diffusion between the steps ID: 781257
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Slide1
Fun games
with membranes and cold traps.
I will assume:The total volume is 4000 l.The circulation speed X l/h = Y l/h = 200 l/h pump speed.The volume is divided into N steps where N=V/X=20 with no diffusion between the steps.The total volume which is rejected at the exhaust is replaced by pure C4F10.The total volume, DVi, which is pumped out at stage i for one step, is transferred to stage i+1 as one volume and then treated at this membrane in 100 mini steps assuming that the partial pressures do not change significantly during a mini step. (This statement is perhaps overly optimistic, but I checked that there is no real difference between 100 and 500.)
1
(I
stopped
at n=4)
Slide2Neomechs
CO2 0.0762 l/h/mbar p1-p2 6000 mbar
Ar 0.0155 l/h/mbar p3-p4 1000 mbarO2 0.01485 l/h/mbar p5-p6 300 mbarN2 0.00641 l/h/mbar p7-p8 120 mbarCF4 0.00127 l/h/mbarC4F10 0.000532 l/h/mbar
5% air in the C
4
F
10
volume.
Numbers from an old logbook
2
Slide3Small losses of C
4
F10and reasonable purity. (1 % N2)but not fast.
3
Slide4Switch off Stage I and usep3-p4 6000 mbar
p5-p6 1000 mbarp7-p8 200 mbar
Still tolerable losses for the same purity.
4
Slide5Switch off Stage I and II and usep5-p6 6000 mbar
p7-p8 500 mbar
Not too surprisingly, the losses are now rather high.
5
Slide6CO2 is easy. For a 5% admixture usep1-p2 2000 mbarp3-p4 1000 mbar
p5-p6 500 mbarp7-p8 50 mbar
6
Slide7What about a steady state air leak?Try with 0.5 l/h. That is 0.5 l/h C4
F10 []and 0.5 l/h air []
usep1-p2 6000 mbarp3-p4 1000 mbarp5-p6 500 mbarp7-p8 120 mbarand start the membranes when there is 2% air in the volume.Not perhaps the most optimal device for such a state. Still, it will stabilise around 98.5% . The rest being N2.
7
Slide8The fiddling with the pressures is very depending on the actual (measured) throughput of the membranes. The pressures should probably also be adjusted as a function of the admixture in the gas.Look for another type in my old logbook: Generon
CO2 1.0152 l/h/mbar
Ar 0.0904 O2 0.174 N2 0.0394 CF4 0.00370 C4F10 0.00534 and usep1-p2 2000 mbarp3-p4 500 mbarp5-p6 100 mbarp7-p8 20 mbarStart out with 5 % air
8
Slide99
Will now have a look at gas scrubbing at low temperature. Have used:
Tables of Physical and Chemical ConstantsPhysical Properties and gas Solubilities, Journal of Chemical and Engineering Data Vol. 18 No. 4 1973 Solubility of gases in fluorocarbons, 3M pubM.F. Costa Gomes et al., Journal of Fluorine Chemistry 125(2004)1325Ostwald solubility coefficient.Volume of gas dissolved in unit volume at ambient temperature and pressure. In anesthetic practice, these are quoted in tables, assuming a body temperature of 37°C. Note that the volume of gas dissolved is only dependent on temperature, and not pressure (though the number of molecules and the activity of these is pressure-dependent). This differs from Bunsen's solubility coefficient (a) in that the amount of dissolved gas is expressed in terms of its volume at the temperature of the experiment, instead of STPD.
Friedrich Wilhelm Ostwald
Nobel Prize in Chemistry in 1909
The mole
Kingdom: Animalia
Phylum: Chordata
Class: Mammalia
Infraclass: Eutheria Order: Soricomorpha Family: TalpidaeWhich has taken me through a tour of mole units and Ostwald's coefficient. Just to get (cc gas) in (cc liquid) at a given temperature and partial pressure. It does indeed require some rather large extrapolations (and faith).
Slide1010
Water is well known.
Compare 3M numbers at one temperature with measurements at many different temperatures.Looks consistent.and then compare the solubility of the different gases in water and FC80. water FC80
a_N
2 1538 1304a_CO2 2625 2269
a
_O
2
1691 977
Known that the odd man out, O2 , has a relative high solubility in fluorocarbons.
As I know nothing about the temperature dependence of He in FC80, I will just use the
a
from water and fit to the 3M point.Solubility is nearly independent of the type of fluorocarbons. I will therefore assume that these curves holds for C4F10.
Slide1111
Use the same set-up as for the membranes.
That isTotal volume: 4000 lCirculation: 200 l/h → 20 stepsNo diffusion and no compressibility.Total volume of the scrubber: 2 l3 bar < scrubber pressure <3.5 barPressure stabilizing gas : He at 5 l/hEach step split in 500 mini steps.Checked that there is no real difference between 500 and 1000 steps (apart from time used on my PC).
Slide1212
Time (h)
C4F10 N2 O2 He0 95 4 1 020 98.69 1.02 0.22 0.0740 99.35 0.43 0.09 0.1360 99.61 0.2 0.04 0.1580 99.73 0.09 0.02 0.16100 99.78 0.04 0.01 0.17Composition (%)Start with 5 % airand the cold trap at - 60 degC
Slide1313
N
2 will also work fine, but will end up with some 2.5 %. This is less than what I would have expected from some measurements at COMPASS and could indicate that the solubility constants that I am using are too low by about a factor of two. C4F10 N2 O20 95 4 120 97.89 1.86 0.2540 97.8 2.08 0.1160 97.8 2.16 0.0580 97.79 2.19 0.02100 97.79 2.2 0.01
Slide1414
C4F10 N2 O2 He0 95 4 1 020 99.22 0.59 0.14 0.0540 99.67 0.18 0.04 0.1160 99.8 0.06 0.01 0.1280 99.85 0.02 0 0.13100 99.86 0.01 0 0.13
C
4F
10
N
2
O
2
He
0 95 4 1 0
20 99.56 0.32 0.09 0.0340 99.84 0.07 0.02 0.0860 99.89 0.01 0 0.0980 99.9 0 0 0.09100 99.91 0 0 0.09No real change in purity, but a rather dramatic increase in losses.CO2 will not work in a cold trap.
Slide1515
Some sort of a
Conclusion.Gas scrubbing, cold trap at (at least) -60 degC, is the best choice if the aim of the game is to get air content in the range of ppm.The triple membrane contraption will give oxygen content of about 0.1 % nitrogen of <1 %without any excessive loss of C4F10A steady state leak, would be better handled with a cold trap.The principle of Gas Scrubbing can easily be understood from one clear sketch found in freepatents.com