conductance calculation

General discussion
1

Dear all,

Is there chamber conductance calculation example by use of MolFLow ?

Thanks for the help in advance

2

Hello Suidong,

When you use the “quick pipe” from the menu, what you get is in fact a conductance example. You see a pipe with sticking=1 on both sides, and desorption from one side. In the bottom right corner, you can see the transmission probability (percentage of molecules traversing the tube) calculated.
Conductance is ENTRANCE_CONDUCTANCE * TRANSMISSION_PROBABLITY. You can see the entrance conductance by clicking on the entrance with sticking=1, then the pumping speed displayed will be exactly the conductance. Watch out: conductance depends on the gas used, so make sure you set the correct gas mass in Global Settings. Get back here if you need more help, Marton

3

Dear Marton,
Thank you for the quick reply and helpful hints. The conductance is supposed to be a function of gas pressure, how do i get a curve showing this effect?

regards

Suidong

4

Hello,
While it is true that across high pressure/viscous/molecular regimes the conductance changes, Molflow - as its name suggests - works only in molecular flow regime (when the particles don’t interact with each other anymore, typically pressure below 10^-3mbar). In this regime the conductance only depends on the geometry and the gas velocity (temperature and mass), but not on the pressure.
There is an analytical formula for calculating conductances for straight pipes (http://lss.fnal.gov/archive/other/ssc/ssc-gem-tn-93-382.pdf) and Molflow was tested against it and gives the correct results.

5

Dear Marton, Could I ask another silly question: in the quick pipe example, the calculated trans. proba is 0.28, does it mean that only 28% particles are pumped out, that 72% of them are escape from the desorption port ?

regards

Suidong

6

Exactly.
If you click on the “…” button near the calculated 0.28 value, you can see how the formula is calculated: the expression is “A2/SUMDES” which is “number of molecules absorbed on facet 2 divided by total number of desorbed molecules”.
It means that indeed, 28% makes it through the pipe and 72% comes back to where it came from.
The ratio can be also approximated by an analytical formula, the parameter will be the tube’s length/radius (L/R) ratio.
In our case, L/R=5, and the transmission probability is close to that of this paper: http://www-afs.secure-endpoints.com/afs/ipp-garching.mpg.de/u/rome/1.3153280.pdf (There is a small deviation because we have a pentagonal tube instead of a round one).