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How To Calculate Time To Fill Gas In Closed Vessel/storage

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#1 tnjtnj


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Posted 28 July 2009 - 09:54 PM

I want to derive equation to determine time to fill gas in closed vessel.
- Gas will feed to vessel with constant pressure at P1.
- Gas will feed to vessel until pressure vessel reach to feed pressure Pi-->P
- Time to fill gas in vessel until pressure vessel = feed pressur is required.

I can derive equation by using "law of conservation of mass" as below;

rho*v1*Apipe = [Vt*Mw/(ZRT)]*d(P-Pi)/dt

but I cannot find the function of v1 with differential pressure between P1 and P.

Anyone please kindly help me to clarify how to determine the function of feed gas velocity with pressure drop of (P1-P)

Thank you for your kindly help.

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#2 clarenceyue


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Posted 29 July 2009 - 12:15 PM

i certainly think you may assume that the feed rate is constant, so it should solve your problem in this case, i.e.:

rho*v1*Apipe = CONSTANT

And your mass balance is incorrect, why do you need d(P-Pi)/dt. Please check it.


#3 latexman


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Posted 29 July 2009 - 01:34 PM

QUOTE (clarenceyue @ Jul 29 2009, 12:15 PM) <{POST_SNAPBACK}>
i certainly think you may assume that the feed rate is constant, so it should solve your problem in this case, i.e.:

rho*v1*Apipe = CONSTANT

No, the feed rate will vary. At time = 0+, the flow will be a maximum. It may even be choked. As the pressure in the vessel increases, the driving force for flow (delta P) decreases, and flow decreases accordingly.

#4 Art Montemayor

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Posted 29 July 2009 - 03:07 PM

This is a STUDENT FORUM. It is one of my favorite forums on this website. It is one of my favorites because I have seen so many countless students confronted with reality and practical experience for the first time in their academic lives. This, I have always believed, is of priceless values to future Chemical Engineers who represent the promise of technical and managerial leadership in the world's industry.

I have taken time out from my present travel plans to fly to Bilbao tomorrow morning at 8 AM because I have found a valuable contribution to all students in Latexman's valued comments. PLEASE, all students, read carefully what Latexman has written. These are pearls of wisdom hidden in simply phrases. If you value your future career and your ability to succeed in your enginering goals, then read, study, and reflect on what he has written. This is a lesson is common, simple horse sense. When you have a batch process (as is obvious, by what the OP described as a "closed vessel"), then you have the classical chemical engineering effect of "accumulation". This is a complex and differentially expressed process that requires a unique and differential calculus approach to the problem. You certainly cannot, by any practical assumption, accept that there is "constant flow rate" into a closed vessel. As Latexman has stated, the feed rate will vary. In fact, with assumed permission from Latexman, I will add that the flow rate will DIFFERENTIALLY BECOME LESS AND LESS as the filling of the vessel approaches an end. Think carefully about it. You have to assume that the initial gas pressure is constant (and greater than that of the vessel). Therefore, the delta pressure (the driving force) will be decreasing as the filling increases. That is conceptually what is happening the moment you have a batch, dead-ended process where you have basic accumulation of matter.

I consider it very important for all students to study and understnad the basic chemical engineering facts that Latexman has put forth as the means to come to a real solution of this problem.

#5 clarenceyue


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Posted 30 July 2009 - 10:11 AM

Dear Mr. Art,

Thank you for pointing out my ignorance, and for bringing my ignorance awareness to a higher level.

However, I'm wondering that suppose a gear type of rotary compressor is used to supply the feed gas into the vessel - the flow from the compressor is constant, (as far as I am aware and I stand corrected, if I am wrong about), then the flow in this case would not be dictated by a pressure differential? I'm a little confused about this. I do seek your enlightenment and from all the more experienced folks/members.

I look forward to your esteemed replies.

#6 iceman


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Posted 07 September 2009 - 10:56 AM

Yes I think it will always be on pressure differential but with the a continuously higher delivery pressure by compressor to counter pressure rise in vessel if flow is constant.
A more realistic process would be of constant delivery pressure but with falling flow rate as poressure rises in vessel.

#7 black friday

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Posted 16 September 2009 - 09:53 PM

Wouldn't v1 be a function of the pressure drop here? The LHS of the equation should be taken to as much detail as the RHS. Use the equation for compressible gas flow in a pipe for the LHS? You can't assume that the mass influx is constant by any stretch. At least not with a centrifugal or axial compressor. Maybe with a positive displacement pump you could but that isn't so common in gas transport as far as I know.
If Q is mass flow rate, M is mass in tank, then Q(P)=dM(P)/dt=MdP/dt, therefore t=integral(M/Q)dP or something along those lines, between P1 and pump supply pressure.
Note: in your equation, it should be d(Pi-P)/dt. Driving force decreases with time as stated above.
Going with an ideal gas, because I don't want to complicate things with z, M=P.V.MW/R/T. Simplify to M=k1.P
Q=((pi^2.D^5)/(8.rho.f.L))^.5(Psupply-P)^.5 using Bernoulli's. Lets simplify to Q=k2.(Psupply-P)^.5
Therefore I have t=integral(k1.P)/(k2.(Psupply-P)^.5)dt=[k1/k2.(P-Psupply.ln(Psupply-p)](Integrate between the P that indicates a full tank and P at t=0)
This is all rather quick and dirty, so if I've misremembered the gas flow through a pipe equation or made an error in my calcs feel free to weigh in but this is my take on things.
You could always do a numerical solution to get a feel for the answer too if the calculus is a bit much. I admit I had to google the integral of P/sqrt(Psupply-P). Matlab or whatever it is you use.
Hopefully I'm vaguely on track here. I'm really curious to see what the correct answer is actually.

Edit: I think I've used the calc for liquid flow in a pipe, but the idea should be roughly the same.

Edited by black friday, 16 September 2009 - 10:53 PM.

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