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I'm having trouble understanding an equation and solution method given in literature. The 1D flow equation for "high rate" linear gas flow through a porous medium is given as: $$\tag{1} p_1-p_2=\Delta p = \frac{\mu L}{k \beta} \left(\frac{w}{A}\right) + \frac{c_f L}{\sqrt{k} \beta} \left(\frac{w}{A}\right)^2$$ where $p_1-p_2=\Delta p$ is the difference in the upstream and downstream pressures over the length of flow $L$; $\mu$ is the gas viscosity, $k$ is the the porous medium permeability coefficient, $\beta$ is the gas isothermal compressibility, $w$ is the gas mass flow rate, $A$ is the cross-sectional area normal to the direction of flow, and $c_f$ is a dimensionless coefficient.

We assume that, for multiple values of $\Delta p$ (3 or more), all variables in (1) other than $k$ and $c_f$ are known.

Equation (1) is of the form $y = a_2x^2 + a_1x+c$. Its stated that to solve for $k$ and $c_f$, one plots $\bar p \times \Delta p$ as a function of $(w/A)$, determine the values of the coefficients $a_2$ and $a_1$ from the best fit 2nd-order polynomial to the data. From these $a_2$ and $a_1$ values $k$ and $c_f$ can be determined. For the the product $\bar p \times \Delta p$, $\bar p$ is the "average" pressure over the length of flow, taken as $=(p_1 + p_2)/2$.

I am having trouble seeing how (1) was derived, specifically using isothermal compressibility $\beta$ of the gas, and why the product $\bar p \times \Delta p$ is used in the plotting/solution methodology for solving for $k$ and $c_f$.

What I've tried: Starting with $$\tag{2} -\frac{dp}{dx}=\frac{\mu}{k}v+\frac{c_f}{\sqrt{k}}\rho v^2$$ where $v=q/A=w/(\rho A)$; $v$=superficial velocity, $q$=volumetric flow rate, $\rho$=fluid (gas) density. Therefore, (2) can be written in terms of mass rate and density: $$\tag{3} -\frac{d p}{d x}=\frac{\mu}{k}\frac{w}{\rho A}+\frac{c_f \rho}{\sqrt{k}}\left(\frac{w}{\rho A}\right)^2$$ Separating variables and integrating: $$\int_{p_1}^{p_2} d p=\int_0^L\frac{\mu}{k}\frac{w}{\rho A}d x+\int_0^L\frac{c_f \rho}{\sqrt{k}}\left(\frac{w}{\rho A}\right)^2 d x$$ Assume $k, A, c_f, w$ are constants, independent of pressure. Pull the presssure-dependent terms $\mu$ and $\rho$ from the integrals and evaluate them at the average pressure $\bar p$. We then have: $$\Delta p=\frac{\mu}{k}\frac{w}{\rho(\bar p) A}L+\frac{c_f}{\sqrt{k}\rho(\bar p)}\left(\frac{w}{A}\right)^2 L$$ At this point I'm not sure how we can go from here and swap the $1/\rho(\bar p)$ terms with isothermal compressibility $\beta=\frac{1}{\rho} \frac{\partial \rho}{\partial p}$ to obtain Eqn(1), nor do I see the reason why $\bar p\times\Delta p$ is plotted as a function of $w/A$ instead of $\Delta p$ as a function of $w/A$.

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  • $\begingroup$ $(3)$ and the fourth equation don't require partials ($\partial$). $\endgroup$
    – Gert
    Commented Jun 4, 2020 at 21:59

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You need to solve the differential equation properly, by approximating $\rho$ as a function of p: $$\rho=\rho(\bar{p})[1+\beta (p-\bar{p})]$$and moving that to the left side of the equation (with the dp/dx).

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  • $\begingroup$ would this be correct then? $-\left(\int_{p_1}^{p_2} \rho(\bar p)[1+\beta(p-\bar p)]\ \text{d}p \right)=\rho(\bar p)[p_1+\beta (\frac{p_1^2}{2} -p_1 \bar p)]-\rho(\bar p)[p_2+\beta (\frac{p_2^2}{2} - p_2\bar p)]$ This is messy. Still not sure how this can be simplified to (1) and used to show why we use the product $\bar p \times \Delta p$ for the plotting/solution method. Thoughts? $\endgroup$
    – Armadillo
    Commented Jun 5, 2020 at 17:47
  • $\begingroup$ The relation $\rho=\rho(\bar{p})[1+\beta (p-\bar{p})]$ is for "slightly" compressible fluids. Perhaps in this problem the authors assumed gases at low pressures, and thus the isothermal compressibility was approximated by $\beta=1/p$? If so, I'm still not seeing how I approximate $\rho$ as a function of $p$ using this different relation for $\beta$ prior to integration. Hints? $\endgroup$
    – Armadillo
    Commented Jun 5, 2020 at 19:14
  • $\begingroup$ The problem statement didn't say it was a gas. I naturally assumed it was a liquid. Otherwise, one would use the ideal gas law. $\endgroup$
    – Chet Miller
    Commented Jun 5, 2020 at 21:59
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    $\begingroup$ Even with the approximation, I still get the same final result as you (i.e., without the $\beta$s). The presence of the betas isn't even dimensionally correct. Otherwise, their procedure will be OK. $\endgroup$
    – Chet Miller
    Commented Jun 5, 2020 at 22:21
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    $\begingroup$ I saw that too (not dimensionally consistent). I realize that the literature to which I refer has made an error. Indeed, using the relation for ideal gas law $\rho=pM_w/(RT)$ will result in (after integrating, and neglecting the pressure dependence of $\mu$), $\bar p \Delta p \rho = \frac{\mu L w}{\beta k A}+\frac{c_f L w^2}{\beta \sqrt{k} A^2}$, where $\beta = 1/p$ $\endgroup$
    – Armadillo
    Commented Jun 5, 2020 at 23:22

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