
#1
Apr412, 11:09 PM

P: 6

Hi there,
I have been asked by my boss to the following question... I have a vessel (500L) that is pressurised to 0.8 barg. That vessel is then held at that pressure for an hour (It is a pressure hold test on the vessel). Over the course of that hour the pressure drops by 2 kPa. He has asked me to find out what sized hole (theoretically) would be in the vessel in order for it to drop to that pressure over the course of an hour. Assumptions: 1. The vessel is in an area of normal atmospheric pressure 2. The temperature of the vessel does not change during the operation 3. We are using air to pressurise the vessel I have some small experience with Chemical Engineering but this is not my current area of expertise so I am having trouble remembering how to do this. I have calulated the density of the air in the pressurised vessel at 0.8 barg and at 0.78 barg ( 2 kPa). The drop of air in the vessel is 10 L over the hour. That gives me a flowrate of 10L/hr. I am unable to work out the velocity in order to get the cross sectional area, and in turn the diameter of the hole. Any help on this would be much appreciated! 



#2
Apr512, 03:06 AM

PF Gold
P: 946

You may want to research the concept of "molecular effusion".
I'm no expert on this, but Blundell [1] gives the molecular flux (that is, particles per area per time) for an ideal gas as [tex] \Phi = \frac{p}{\sqrt{2\pi m k_B T}} [/tex] where [itex]p[/itex] is pressure, [itex]m[/itex] is the molecule mass (molar mass divided by the Avogadro number), [itex]k_B[/itex] is Boltzmann constant, and T is temperature (in Kelvin). They later indicate that the pressure in a tank of volume [itex]V[/itex] with a small hole of area [itex]A[/itex] for an ideal gas kept at temperature [itex]T[/itex] is exponentially decaying as [tex] p(t) = p(0)e^{t/\tau} [/tex] where [tex] \tau = \frac{V}{A}\sqrt{\frac{2\pi m}{k_B T}} [/tex] From this it seems you should be able to solve for [itex]A[/itex] to get an estimate of the area of the hole. Note that the above is derived for a single species gas (e.g. N_{2}). For atmospheric gas I am guessing that a correct approach will be to add the partial pressures for each species, that is the drop in total pressure is the sum of drop in the partial pressures for each species. This of course complicates finding the hole area, so you may have to revert to numerical methods to find it. [Later:] By the way, the above assumes that the tank is effusing into vacuum. Since you have the same gas and temperature on both sides of the hole and the outside pressure is constant, it appears to me that you can just use the tank overpressure (i.e. [itex]p  p_{atm}[/itex]) instead of absolute pressure in the equations above, that is, it is the tank overpressure that decays exponentially towards zero. [1] Concepts in Thermal Physics, Blundell, Oxford, 2010. 



#3
Apr512, 11:34 AM

P: 13

Maybe assumption 2 is incorrect? You could check the pressure drop
for longer periods, see if it levels off. I'm basing this on the assumption that while you are pressurizing the vessel, the temperature of the air inside will end up higher than ambient, and the pressure will drop as it cools. Of course, since you were asked about the hole, then it needs to be answered whether or not that is the cause :). Still.. you could have 2 answers. 



#4
Apr612, 07:40 PM

P: 6

Can you help me with an engineering question?
Thank you for your responses.
I proposed solving the problem in the following manner... I broke it into 3 basic steps. 1. I firstly needed to find out how much air is lost from the vessel when the pressure drops 2kPa. I calculated the density of air at 1.8 bar and then at 1.78 bar. I then calculated the difference in density and from that determined the amount of air loss in the vessel over the hour. I got an answer of 5.5L. I therefore had a flowrate of 5.5L/hr. 2. I now had a flowrate. I next needed the velocity of air travelling through a differential pressure of 0.8bar. I used Bernoulli's equation to calculate the velocity from a drop in pressure. I found a website that suggested I use this equation to calculate the velocity of a leak on a tank. As sugested, I put Z1Z2=0 and P1 is pressure inside the vessel, while P2 is the atmospheric pressure. With this I was able to find the velocity of approx 300 m/s. 3. I now had the flowrate and the velocity. With that I was able to simply calculate the crosssectional area of the hole, in turn giving me the diameter. I received a figure of 8.4 nm. Obviously as you suggested the most likely reason for the drop in pressure is due to decay. However, we are looking for a worst case scenario in which a hole is present, theoretically, in the vessel. We would like to know if a bacteria is capable of fitting through this hole. Hence the many assumptions used to calculate this. Could someone please verify if this is a plausible method for calculating the size of the hole in the vessel? Thanks 



#5
Apr712, 06:42 AM

PF Gold
P: 946

You should probably compare your result with the mean free path of air, λ, which I believe is around 65 nm. For holes significant less than λ you are in the microscale regime where kinetic gas theory is normally used and in in this regime you should not expect accurate results from a macroscale theory like fluid mechanics. For instance, using the effusion equations above I calculate that it will take around 60000 years for the pressure to drop due to effusion through a single 8.4 nm diameter hole for the conditions given with N_{2} gas instead of air to keep the calculation simple. To drop 2 kPa (while kept at ambient temperature) in 1 hour due to effusion of N_{2} through one or more very small holes, I get that the total area of the holes should be around 0.03 mm^{2} (square millimeter).
In addition, as softport already has mentioned, the tank pressure is sensitive to change in tank temperature. A change in temperature of around 3 K will give a pressure change of around 2 kPa assuming room temperature, so unless you somehow have taken measures to control or compensate for the temperature during and after pressurizing the tank some of the pressure change you see could easily come from a change in temperature. If you are using a pump to pressurize the tank you should expect the air temperature in the tank to start out above ambient temperature. Likewise, if let air into the tank from another high pressure tank initially at ambient temperature you should expect the air temperature in the tank to start out below ambient temperature. In both processes you could compensate by either using cooler air to pump, or by having the high pressure tank at a slightly higher temperature, such that the initial gas temperature in the tank right after pressurization is at the ambient temperature. Another way to compensate would be to only measure pressure after the tank has settled to ambient temperature, which, depending on how well insolated your tank is, may take a while (and it may also require you to measure the temperature inside the tank unless you want to model the time it takes for the temperature to settle). Of course all of this is "armchair theory". If your result has to be used in any way for ensuring quality or safety I strongly recommend that you try to get access to established industrial procedures for satisfying whatever such requirements you may have (for instance, that bacteria cannot cross the tank wall in normal operation). If the tank has been designed specifically for such requirements it may be that the manufacturer of the tank has procedures that will allow you to test if the tank is fit of normal or safe use. 



#6
Apr712, 08:08 AM

P: 13

I found this in Wikipedia:
but it's not big enough to cause a problem: how can you be sure that it won't get bigger over time? 



#7
Apr912, 06:33 PM

P: 6

Filip and softport,
Thanks so much for your help on this. You have clarified a lot of things fomr me which is great. Filip, Can i ask you for one more favour? I am unable to get Blundell online. Would you be able to send me the breakdown of how exactly you calculated the area size of 0.03mm2? Its been a while since i did engineering calculations and am unfamiliar with what all the terms in equations 2 and 3 from your first post...hence me being unable to complete it. I also found this link online that i hoped woudl help me solve the issue in the way i origionally proposed... http://www.piug.org.uk/IUGNewsletter32.pdf Unfortunatelly though i'm not sure how they are calculating the flowrate as my volume does not change, only the density of the gas changes. I manipiluated the idea slightly by using PV=nRT to find the flowrate. I then used the velocity equation proposed in this paper to calculate the voelocity and then worked out the area. I still received a very small number for the diameter...0.219 microns This differs to my first attempt only by using the velocity equation proposed in this equation (which appears to be a simpler form of Bernoulli's Equation). Filip, i would like to see the breakdown of your calculation as i feel the moldular flux theory is the most relevant to what is going on. Softport, you are right in saying that the hole may get bigger over time. and also in stating theat assumption 2 may be incorrect. This task is purely out of interest for the director of company. Its completely theoretical and the assumption are major ones it has to be said! But its what he has asked for so i guess its best to provide him with the answer! 



#8
Apr912, 06:34 PM

P: 6

Apologies for the poor spelling mistakes. Should have proof read before I posted.




#9
Apr1012, 08:01 AM

PF Gold
P: 946

Now, from the second equation of post #2 you can solve for [itex]\tau[/itex] to get [tex] \tau = \frac{t}{\ln(p(t)/p(0))} = \frac{t}{\ln p(0)  \ln p(t)} [/tex] where the pressures p(0) and p(t) are overpressure, that is, the absolute tank pressure minus ambient pressure outside the tank. From [itex]\tau[/itex] it is now easy to get the area A using the third equation. In my calculation I assumed T is around room temperature (293 K) and arrived at A = 0.03 mm^{2} or an equivalent single hole diameter of 195 μm. 



#10
Apr1012, 07:28 PM

P: 6

Thanks Filip,
Having difficulty hitting the 0.03mm2 Just to ensure I am correct... p(0) is ambient pressure = 100000 Pa (kg m1 s1) p (t) is absolute tank pressure = 180000 Pa (kg m1 s1) t is time = 3600s For the third equation... V is tank volume = 0.5 cubic metres m is molecular mass (nitrogen molar mass / avogrado number) = 4.649 x 10(23) g Kb is Boltzmann constant = 1.3807 x 10(23) J/K (or kg m2 s2 K1) T is temperature = 293K Really appreciate your help with this 



#11
Apr1112, 03:28 AM

PF Gold
P: 946

No, p(0) is the initial overpressure, that is, 0.8 bar in your case (initial pressure minus ambient pressure), and p(t) is the over pressure after time t, that is, 0.78 bar in your case.
The rest of the numbers looks good, but remember to change to kg on m before you start to calculate. 


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