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Thread: Fluid Dynamics Question- Gas Diffusion into a Container

  1. #1 Fluid Dynamics Question- Gas Diffusion into a Container 
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    I am trying to prove that the pressure/temperature profiles in the second compartment of a partitioned container follow a typical growth/logistics function (i.e. are sigmoidal) with respect to time. Essentially, the partitioned container is a single container with a punctured wall in the center. The first compartment is kept at steady pressure/temp., the second is a vacuum.

    I tried solving this by setting dp/dt = dp/dt(in) + dp/dt(out), where dp/dt (in)= constant (constant volume, temp. and pressure in 1st container) and dp/dt(out)=kT / V * dN/t, where dN/dt= PANa/ (2piMRT)1/2 (effusion equation) and then solving the subsequent differential equation. (k=boltzman's const. N=number molecules, A=area of hole, Na=avogadro's constant, M=molar mass)

    I imagine I went wrong by setting Temperature in dp/dt(in) constant.
    How can I fix this? Alternatively, are there any fluid dynamics equations that I can use? Any help would be appreciated.


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  3. #2  
    Universal Mind John Galt's Avatar
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    Michael, I have deleted your duplicate post in the chemistry section. Please do not make multiple threads of the same item.
    I could have answered your question forty years ago. In fact I think I did, in an exam. I've long since forgotten how. I know there are some on here who will be able to come to your aid.
    Good luck.


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  4. #3  
    Forum Isotope Bunbury's Avatar
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    The first compartment is kept at steady pressure/temp.,
    The only way to do this is to reduce the volume of the container (e.g. by having its end wall be a moving piston). If the container size is fixed and the container is adiabatic the pressure and temperature will do down simultaneously as the mass of gas is reduced by leakage through the hole.

    typical growth/logistics function (i.e. are sigmoidal) with respect to time.
    This might have meant something to me years ago, but unfortunately not today.
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    [quote]

    The only way to do this is to reduce the volume of the container (e.g. by having its end wall be a moving piston). If the container size is fixed and the container is adiabatic the pressure and temperature will do down simultaneously as the mass of gas is reduced by leakage through the hole.


    I am not implying that this is an isolated system. There is no piston, only a punctured separating wall. (The pressure/temp. can be kept constant by introducing more gas into the system (e.g. valves + sensors)- this is not an engineering question, however.)

    Theoretically, the temperature/pressure profile of the second compartment will rise quickly and then decay as a steady state is approached. I am trying to solve for this function, which I believe will be sigmoidal.
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  6. #5  
    Forum Radioactive Isotope MagiMaster's Avatar
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    I would think the relevant relationship would be something like , which would mean the pressure would be something more like exponential decay, , rather than a sigmoid function.
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    I would think the relevant relationship would be something like , which would mean the pressure would be something more like exponential decay, , rather than a sigmoid function.
    Agreed. After a second consideration I came to that conclusion as well.

    My current attempt at solving it looks something like this:

    (1) dP/dt = dP/dt(in) - dP/dt(out)

    (2) dP/dt(in) = kT/V1 * dN/dt (in) where dN/dt= ZA = P1*A*Na/sqrt(2piMRT)

    (Z=effusion rate, V1 and P1 are the volume and pressure of container 1, A=area of hole, Na=Avogadro's const., M=molar mass, k=Boltzman const.)

    (3) dP/dt(in) = B =constant, because P,V, and T are all constant

    (4) dP/dt(out) = kT/V2 *dN/dt(out) where dN/dt(out) =P(t)ANa/sqrt(2piMRT)

    (V2= volume container 2, P(t)=Pressure container 2)

    Thus,

    (5) dP/dt = B - CP(t) where C= kT/V2 * ANa/sqrt(2piMRT)

    This is a standard 1st order linear diff. eq. whose soln. is:

    (6) P(t) =1/exp(Ct) * BCexp(Ct) + D, D is a new constant generated from integration.

    Solving for D (P(0)=0) yields D=-BC, so

    (7) P(t)=BC *[exp(Ct)-1]/exp(Ct)


    The only problem that I have is my assumption T=constant in eqn.4 which is probably false (T1=constant, T2 probably is not) I don't know how to resolve this.
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  8. #7  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Physics isn't a strong point of mine, so someone else will have to help with those kind of details.
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