# Porous media flow equation in terms of fluid compressibility and determination of coefficients?

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 $$w/A$$.

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

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).
• 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? Jun 5, 2020 at 17:47
• 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? Jun 5, 2020 at 19:14
• 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. Jun 5, 2020 at 22:21
• 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 $\beta = 1/p$ Jun 5, 2020 at 23:22