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Pressure decay as a function of time

  1. Mar 13, 2009 #1
    Problem: Model a leaking pressurized cylinder volume filled with an ideal gas (density ρ and molecular weight w) charged to a pressure P at temperature T. The rigid cylinder of fixed volume V has a small leak (adiabatic – T doesn’t change significantly) of area a. The leak rate will diminish with time since as material escapes, the driving pressure decreases. What function describes the pressure decay as a function of time, t?


    Solution: The model in the attached pdf uses the ideal gas law and Bernoulli flow equation to find that

    dP/dt = -constant * SQRT(P)

    I'm not sure what the solution to this is but see that it will decay more slowly than the exponential solution I expected to find.

    Question: Is the above differential solution correct, and if so, what is P(t)? If it is incorrect, what is a correct approach?
     

    Attached Files:

  2. jcsd
  3. Mar 13, 2009 #2
    In equation (7) in the attachment, you have sqrt(density) * sqrt(pressure)

    Because density is proportional to pressure,
    this factor is proportional to pressure, not sqrt(pressure).

    I think this will now be a standard exponential decay.
     
  4. Mar 14, 2009 #3
    I don't think density is a function of pressure in context of this problem, but am open to understanding why you do.

    The reason I don't think so is that density appears in eq 2 where Bernoulli's flow equation is applied: the left side of the equation where Pressure appears represents the stored (akin to potential) energy prior to the leak; the right side represents the kinetic energy of the molecules after they have been released. (Both sides of the equation are actually energy densities.) Therefore the (mass) density in eq 7 is of the molecules outside the cylinder and not a function of the pre-discharge pressure (where the mass density is very much a function of pressure).
     
  5. Mar 15, 2009 #4
    If you have a rigid container of volume V with N molecules in it at t = 0, and you let L molecules out over a time t because of a slow leak, then there are only N-L molecules left in the container at time t. So the density of gas in the container has changed from N/V to (N-L)/V. If the density of the gas in the container is continuously dropping, shouldn't the density of the leaking gas also change? If the density of the leaking gas is always constant, at some time it will be higher than the density of gas in the container. This sounds illogical. So maybe the density in your attachment equation (2) should be a function of time.
     
  6. Mar 22, 2009 #5
    The statement "the density of the leaking gas is always constant, at some time it will be higher than the density of gas in the container" is not accurate; the leak will slow and stop before the density inside is less than it is outside. The leak is driven by the pressure differential and, except for diffusion, gas only leaks from the cylinder when the pressure in the cylinder is higher than the surrounding (and same for density). With time, the gauge pressure of the leaking cylinder, and thus the leak rate, approach zero. And if the volume around the leaking cylinder is much larger than the volume of the cylinder (let the outside volume go to infinity) then the leaked gas density never changes.

    When Bernoulli's flow equation is applied to a leaking cylinder with the limit that the leak area is insignificantly small compared to the cross sectional area of the leaking cylinder, then the differential equation I published originally in this thread is obtained:

    dP/dt = -constant*SQRT(P)

    where P is the gauge pressure of the leaking cylinder. I’m still hoping for a validation or correction of the analysis presented in the original post (ref. pdf attachment). And if the solution above is valid, what is P(t)?
     
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