Hello Marton,
I have constructed a pipe with a radius of 0.25 cm and a length of 20 cm. The left end is set with a constant outgassing rate of 2.28e-8 mbarL/s, while the right end is a pump that pump speed determined by the sticking factor.
When system reaches steady, the throughput of pump facet should equal to outgassing.
When sticking factor is 0.1, I calculated P_sticking * pump_speed = 2.20e-8 mbarL/s ~=2.28 mbarL/s.
However, as sticking factor increases, the value of P_stickingS become smaller, and reaching minimum at sticking=1 with P_sticking * pump_speed = 1.28e-8mbarL/s.
You can’t use the regular Q=pS formula on a strongly pumping facet. Your facet has 100% sticking.
Pressure calculation on 100% sticking facets only shows the molecular force exerted by incident particles, not by rebounds. It will show ~half of what you’d expect.
Pressure is not isotropic near the end of the pipe, it will be obvious if you visualize the average direction vectors:
I’m not sure from the top of my head how to calculate the throughput in this specific, non-isotropic case, without “cheating”. My intuition would be to measure the directionless density (it is tested to be the same even in non-isothermal parts regardless of a transparent test facet’s orientation), then convert it to pressure through the ideal gas equation.
The “cheating” method would be the measure what part of the incident gas leaves the system on that orifice:
You can see that I first calculate the transmission probability through the pump in question, the multiply it with the outgassing. Here the result is obvious (pumping=outgassing), but if I set sticking=1 at the outgassing facet, the result is actually meaningful (19% of the outgassing in case of a L/R=10 pipe):
One final thought about “When system reaches steady, the throughput of pump facet should equal to outgassing.” - it is true in our cases, but only because the gas doesn’t change the temperature. (pV/T=const.)
Hello Marton,
Thank you very much for your detailed explanation. While I understood part of it, I still have some confusion and would greatly appreciate your clarification.
Firstly, regarding the statement “Pressure is not isotropic near the end of the pipe”: near a strongly pumping facet, since there are no rebound molecules, the incident molecules are not isotropic. Since the formula Q=pS is only valid in isotropic situations, and therefore, it cannot be applied in such a facet. Is my understanding correct?
However, I conducted a simulation varying the sticking coefficient from 0.01 to 1, and analyzed the relationship between sticking and throughput. The results indicate that Q=pS is only valid when the sticking coefficient is below 0.01. This suggests that even for a facet with a moderate or small sticking coefficient (e.g., sticking = 0.1), Q=pS is not applicable. How should this situation be explained?
Secondly, regarding the statement “It will show ~half of what you’d expect”: does this mean that the pressure calculated by Molflow for a sticking facet is not accurate? Specifically, when sticking = 1, does the pressure calculated in the formula editor only represent half of the actual pressure?
I appreciate your guidance on these questions. Thank you in advance!
Yes, I confirm that Q=pS assumes that p is isotropic, and therefore cannot be applied next to 100% sticking pumps.
As mentioned in the previous message, pressure shouldn’t be measured on pump facets, since these facets don’t exist in reality: the mechanical definition of pressure - force exerted on a wall - can’t be applied as there’s no wall.
Molflow still shows a pressure on these facets, assuming that sticking particles lose their momentum, or in other words, come to a standstill (and thus contribute half of the pressure compared to rebounding ones). Therefore I can’t answer “does this mean that the pressure calculated by Molflow for a sticking facet is not accurate? Specifically, when sticking = 1, does the pressure calculated in the formula editor only represent half of the actual pressure?” - there is no physical quantity “pressure” in a pump facet, as there is no wall, and the pressure is not isotropic, therefore pressure can only be defined as a vectorial - direction-dependent quantity.
I’ve checked your file, but didn’t find info about your calculation method in the Formula Editor, so I assume you still measure pressure on the end caps (the pumps) directly.
Textbooks usually calculate the pressure from the ideal gas equation (pV=NkT), and in the real worlds vacuum gauges also measure density, then they convert it to pressure electronically. You can do the same in Molflow, as Molflow’s density quantity is isotropic, even near strong pumps.
You can also read this relevant thread, which was folowed up by a Zoom call, and we concluded that when there’s a stream of particles, defining pressure isn’t straightforward: How do you calculate Pressure on transparent surfaces?
Based on your reply and the TOPIC, I have gained a deeper understanding of pressure and throughput calculation, as well as vacuum knowledge. Thank you very much!
Since I didn’t see the answer to the last question in the POST, I still have some remaining questions. Here is my understanding of it:
For a non-isotropic situation, such as a stream of particles, we can use momentum exchange to calculate the pressure on a real wall. For a transparent wall, using density and the ideal gas law is a useful way to calculate pressure, and the ‘double counting’ method is no longer applicable.
Does Molflow still use ‘double counting’ to calculate pressure for transparent surfaces? If so, should I manually convert density to pressure to obtain the correct pressure value in non-isotropic situations?
Thank you for reading the long discussion with Alan. There is no conclusion (answer) to the topic, as we had a Zoom call instead. We agreed that when there’s a stream of particles, probably the correct way is to distinguish static (thermal) pressure from dynamic (stream) pressure. This is well-researched in fluid dynamics and also at higher pressures (Bernoulli principle), but as far as we understand, not that much in free molecular flow. We therefore decided to “sit down and think”.
Specifically for the “double counting” issue on transparent surfaces: yes, for pressure calculation, Molflow will assume a “specular reflection” on transparent surfaces, i.e. count the incident momentum twice, as if the particle bounced.
For density calculation, Molflow will get the correct density for 2-sided transparent facets, regardless of their orientation. Using one-sided transparent facets to measure density are only an estimate, and depend on the orientation in non-isotropic situations.
I would not use pressure here at all. You can just take the Sticking Factor x Impingement Rate x Area to get the number of particles exiting and it should equal the outgassing on the left. This will work for any type of gas distribution. The 1E-4 is because Z is in #/m2s and A is in cm2.
