I have a question about conductance calculation for a very short tube (h = 2mm). The gas flows into two concentric cylinders and it has a diameter restriction, while the outer diameter is constant.
I calculated the conductance with this method
sticking inlet face = 1
sticking outlet face = 1
Q inlet face = 1 mbar l/s
C=Cinlet*transm probability
The transmission probability, as expected, is very similar to the ratio between the outlet and the inlet Area (Aoutlet/Ainlet).
The conductance calculated is about 980 l/s. It seems a reasonable value to me, I expected a high conductance.
Then, I calculated the conductance with the other method:
sticking inlet face = 0
sticking outlet face = 0.1
Q inlet face = 1 mbar l/s
C=Q/(p inlet-p outlet)
The conductance calculated is about 91 l/s in this case.
I see that these two methods give different results when the tube is very short, while the value is similar when the tube is long. I tried to calculate C in the first simulation as QCONST*trasm.prob./dp but the results are still different
Please note that I don’t want to take into account the entrance conductance of the tube (this restriction is connected upstream and downstream with components that have the same diameter).
What are your thoughts on this? Can you explain why the two methods give different results and if the second approach is wrong?
Thanks for your help! I upload the two simulations and a picture of the gas domain (the wider Area is the inlet)
The first method is correct - you calculate the transmission probability with Molflow, then use C=C_in*transmission.
In the second simulation…
measuring the pressure on the inlet facet makes no sense as it varies with position
measuring the pressure on the outlet doesn’t make sense either as pressure can’t be calculated on sticking facet (it is underestimated as sticking particles don’t rebound, so they only deposit their impulse then stop). Also, the pressure varies radially here as well.
I see maarton, thanks for your answer! I didn’t thought to check the pressure distribution. So, when the pressure distribution varies a lot with position that formula aren’t going to work as I intended
And, just a clarification, does the formula C=C_in*transmission take into account the inlet conductance of the inlet diameter or not?
I mean, usually when I have attached the component to a wide chamber I use the formula
1/C_tot = 1/C_inlet + 1/C_component, where C_inlet is 1/4 * Area * mean thermal velocity
to calculate the total conductance, while when the component in part of, for example, a tube with the same upstream and downstream diameter I am intrested only in C_component and I don’t want to take into account for C_inlet
What am I obtaining from this simulation? I think 1/C_tot but I’m not sure. I hope that I explained it well! Thanks
“does the formula C=C_in*transmission take into account the inlet conductance of the inlet diameter or not?” - Yes, C_in is the inlet conductance.
You can calculate it by Molflow by typing “1” in the sticking factor field, then Molflow will show what pumping speed (=orifice conductance) would that correspond to. For example it shows 1473l/s for your example inlet.
From the first simulation, you obtain the transmission probability. If you multiply transmission probability with C_inlet, then you get C_component.
I believe the 1/C_tot = 1/C_inlet + 1/C_component equation would be wrong, since C_component already includes the inlet conductance, therefore you would account for C_inlet twice (that wrong equation would represent a physical system where two chambers are connected directly through a hole the size of the inlet, then the component is connected to the chamber)
Thanks you for the clarification, it is very useful.
Let’s change the nomenclature I used for a second and label the conductance calculated with molfow using the transm. prob. as C_molflow
The explanation leads me to conclude that, when I have a component with no area restriction with upstream and downstream tubes, such as in the figure, it will be more appropriate consider the conductance C to calculate the pressure difference between inlet and outlet from:
1/C = 1/C_molflow - 1/C_inlet
That’s because I don’t want to consider also C_inlet (no connection to a big chamber/inlet area restrictions). Do you agree?
Sorry, I don’t understand the logic. With two very large chamber up and downstream, C=C_molflow, which already includes the inlet’s conductance. You can’t decouple the inlet from a tube that physically contains the inlet.
I recommend to play around with Molflow - with the built in geometry editor, you can create upstream and downstream buffers, and test various setups, and the vacuum equations, by directly testing the measured pressures. I believe it is a better way than trying to predict theoretical setups analytically.
