The equation for the gas flow rate (in pressure*volume/time units) $$ Q = {kA\over d} (p_1-p_2) $$ where $A$ is the surface area, $d$ the wall thickness, $p_{1,2}$ the pressures on either side of the wall, and $k$ is the permeation conductivity.

How much pressure would I lose on one side of the wall? Because I get number (in pressure*volume/time units) and I need to know how much pressure I will lose over period of time.

thank you very much

  • $\begingroup$ if somebody could write an small example , i would be very thankful . $\endgroup$ – user118676 Jul 7 '16 at 15:58
  • $\begingroup$ what are 'pressure*volume/time' units? Usually gas flow rate is measured in 'volume/time' or perhaps 'mass/time' units. In any case if outside the container you have constant pressure (say atmospheric) and then you simply need a differential equation which relates the derivative of mass in the container with respect to time to the outflux of material $Q$ through the difference in pressure. $\endgroup$ – nluigi Jul 7 '16 at 18:29

Your question is short on details, so to answer your question I am assuming a particular configuration. Hope it helps with whatever your actual configuration is.

A closed porous container consists of a gas at partial pressure $p_1$ which is less than partial pressure $p_2$ of that gas in the ambient. We shall assume that $p_2$ is constant and to further simplify matters also assume that temperature of gas inside and outside the container is the same and constant with time, say $T$. Now if $p_1(t)$ is the pressure inside the container at $t\geq 0$ then its density is, $\rho_1(t)=f(T,p_1(t))$, where $f$ is some function that characterizes the gas (for e.g. you may assume ideal gas model, in which case $f(T,p_1(t))=\frac{p_1}{R_{gas} T}$). If container volume is $V$, then mass of gas inside that container at any particular time is, $m_1(t)=\rho_1(t)V$. Since pressure difference $p_1(t)-p_2$ is reducing with time so is the flow rate, $Q(t)$.

At a particular time instant, rate at which mass of gas is being lost from the container is, $\frac{dm_1}{dt}=-\rho_1(t)Q(t)=-f(T,p_1(t))\frac{kA}{d}(p_1-p_2)$. But since we also have $m_1(t)=\rho_1(t)V$ we get by differentiating, $\frac{dm_1}{dt}=\frac{d\rho_1}{dt}V=\frac{\partial f}{\partial p_1}\frac{dp_1}{dt}V$. Equating the two we get a differential equation for $p_1(t)$ which may be solved with the initial condition, $p_1(0)=p_0$.


The equation you wrote is for steady state flow through a porous wall of thickness d. Q is the volumetric flow rate. Q/A is the so-called superficial velocity. The equation inherently assumes that the pressures on both sides of the wall are constant, and not varying with time. The only way that you would get an increase in pressure difference with time is if the flowing fluid had suspended solids in it, so that you got a filter cake buildup.


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