Recalling Darcy's equation for gas flow, assuming ($z\mu_g$) = Constant

$$\tag{1} q_{g,sc}=\frac{kA}{L}\left(\frac{T_{sc}}{Tp_{sc}}\right)\left(\frac{1}{2z\mu_g}\right)(p_1^2-p_2^2)$$

For laboratory flow experiments, at low pressures (less than ~10 atmospheres?), we can assume $T=T_{sc}$ and $z=1$

$$\tag{2} q_{g,sc}=\frac{k}{\mu_g}\left(\frac{A}{p_{sc}}\right)\frac{(p_1^2-p_2^2)}{2L}$$

For Darcy flow, plotting $(q_{g,sc}p_{sc})/A$ vs. $(p_1^2-p_2^2)/(2L)$ results in a straight line.

  • line passes through origin when $q_{g,sc}=0$, then $(p_1^2-p_2^2)=0$
  • slope $=\frac{k}{\mu_g}$
  • behavior departs from straight line under turbulent flow conditions (high flow velocity)

enter image description here

Forchheimer proposed a flow equation to account for the non-linear effect of turbulence by adding a second order term

$$\tag{3} \frac{-dp}{ds}=\frac{\mu_g}{k}\left(\frac{q_g}{A}\right)+\beta \rho_g \left(\frac{q_g}{A}\right)^2$$

Based on Forchheimer's Equation a plotting method was developed to determine absolute permeability even with Non-Darcy effects

  • $(1/k_{gas})$ vs. $q_{g,sc}$, where $k_{gas}$ is the apparent permeability determined from Darcy's Law (incorrectly assuming Darcy flow) and is a function of $q_{g,sc}$
  • intercept $=(1/k_{abs})$; absolute permeability
  • slope $=[(\beta \rho_{g,sc})/(\mu_g A)]$

enter image description here

What is the explanation/math/assumptions made, when using Eqns. 2 and 3, to justify the proposed plotting method to determine the absolute permeability from non-darcy data? I.e., how was the plotting method developed?


1 Answer 1


All credit goes to Dr. Thomas A. Blasingame. This derivation is from his Petroleum Engineering 620 Fluid Flow in Reservoirs Course Notes on Non-Laminar Flow in Porous Media -- 24 September, 1997.

The Forchheimer equation for non-laminar flow in porous media is given by

$$\tag{1} -\frac{dp}{dx}=\frac{\mu}{k}v+c\beta \rho v^2$$


$c=(\frac{1}{1.01325 \times 10^6 \text{ dyne cm}^{-2}})(12 \frac{\text{in}}{\text{ft}})(2.54 \frac{\text{cm}}{\text{in}})$

$\beta=\text{``inertial flow coefficient," ft}^{-1}$

By inspection we note that the first term on the right-hand-side (RHS) of Eq.1 is simply Darcy's law (for laminar flow). The $c\beta\rho v^2$ term is the "add-on" term used to account for non-laminar flow.

The average velocity, $v$, is given as

$$\tag{2} v=\frac{1}{A}q_{res}=\frac{1}{A}q_{sc}B$$


$v=\text{average velocity}$

$A=\text{cross-sectional area}$ $\color{red}{\text{normal to the direction of flow}}$

$q_{res}=\text{volumetric flowrate (reservoir volume)}$

$q_{sc}=\text{volumetric flowrate (``standard" volume)}$ $\color{red}{\text{*this a mass flowrate}}$

$B=\text{formation volume factor, res vol/std vol}$

The following notes are in addition to T.A. Blasingame's derivation:

Darcy's Law describes volumetric flow rate of gas flow at reservoir conditions (in-situ).

For steady-state flow conditions in the reservoir, as flow proceeds along the flow path:

  • Mass flow rate, $q_{g,sc}$, is constant (subscripts $g=$ gas and $sc=$ standard conditions)
  • Pressure decreases
  • Density decreases
  • volumetric flow rate, $q_g$, increases

Given a volumetric gas flow rate at reservoir conditions, $q_g$, we need to determine the mass flow rate, $q_{g,sc}$. We cannot assume gas flow in the reservoir is incompressible. Gas density is determined from the Real Gas Law,

$$\rho_g=\frac{p \gamma_g M_{air}}{zRT}$$

Gas gravity, $\gamma_g$, is the ratio $\rho_g/\rho_{air}$ at standard temperature and pressure. The molecular weight of air, $M_{air}=28.9625$ g/mol. The product, $(\gamma_gM_{air})=M_{gas}$.


$$\rho_g=\frac{p M_{gas}}{zRT}$$

since $\rho_g=\frac{m}{V}$ and $M_g=\frac{m}{mol}$ then

$$\rho_g=\frac{m}{V}=\frac{pm}{znRT}\rightarrow pV=znRT$$

Gas density is a function of pressure (for isothermal reservoir conditions). From the Real Gas Law, the universal gas constant shows:

$$R=\left(\frac{pV}{znT}\right)_{\text{rsvr}}=\left(\frac{pV}{znT}\right)_{\text{standard conditions}}$$

To determine the mass flow rate, $q_{g,sc}$, we employ the gas formation volume factor, $B_g$, with units of reservoir cubic feet per standard cubic feet (rcf/scf). In short, scf is a specified mass of gas, i.e. number of moles. In long, at any specified temperature and pressure, a specified volume of two different gasses contains the same number of moles, if the Ideal Gas Law is valid. The Ideal Gas Law is valid at standard temperature (e.g. 60 F, Texas) and standard pressure (e.g. 14.65 psia, Texas).

This means that $z_{sc}=1$. At standard conditions (sc), a standard cubic foot (abbreviated as scf) is a measure of quantity of gas, equal to a cubic foot of volume. A standard cubic foot is thus not a unit of volume but of quantity, and the conversion to quantity (scf to moles) is straight forward; a standard cubic foot represents 1.19804 moles (453.59237 moles per lb-mol).

The gas formation volume factor is defined as:

$$B_g=\frac{V_{res}}{V_{sc}}=\frac{(znRT/p)_{res}}{(znRT/p)_{sc}}=\frac{P_{sc}Tz}{pT_{sc}} \ \ ; \ \ z_{sc}=1$$

therefore, the volumetric flow rate and mass flow rate can be equated as


substituting Eq.2 into Eq.1, we obtain

$$\tag{3} -\frac{dp}{dx}=\frac{\mu B}{k A}q_{sc}+ \frac{c\beta \rho B^2}{A^2}q_{sc}^2$$

For a dry gas, we have

$$\tag{4} \rho_g=\frac{1}{B_g}\rho_{g,sc}$$

substituting Eq.4 into Eq.3, we obtain

$$-\frac{dp}{dx}=\frac{\mu_g B_g}{k A}q_{sc}+ \frac{c\beta \rho B_g}{A^2}\rho_{g,sc}q_{sc}^2$$

Dividing through by $\mu_g B_g$

$$\tag{5} -\frac{1}{\mu_g B_g} \frac{dp}{dx}=\frac{1}{k A}q_{sc}+ \frac{c\beta}{\mu_g A^2}\rho_{g,sc}q_{sc}^2$$

Multiplying through by $\mu_n B_n$ $(\mu_n=\mu_g(p_n);B_n=B_g(p_n))$, we obtain

$$ -\frac{\mu_n B_n}{\mu_g B_g} \frac{dp}{dx}=\frac{\mu_n B_n}{k A}q_{sc}+ \frac{c\beta}{\mu_g A^2}\rho_{g,sc} \mu_n B_n q_{sc}^2$$


$$ -\frac{\mu_n B_n}{\mu_g B_g} dp=\left[\frac{\mu_n B_n}{k A}q_{sc}+ \frac{c\beta}{\mu_g A^2}\rho_{g,sc} \mu_n B_n q_{sc}^2\right]dx$$


$$ -\mu_n B_n \int_{p_1}^{p_2}\frac{1}{\mu_g B_g} dp=\left[\frac{\mu_n B_n}{k A}q_{sc}+ \frac{c\beta}{\mu_g A^2}\rho_{g,sc} \mu_n B_n q_{sc}^2\right]\int_{0}^{L}dx$$

The gas formation volume factor, $B_g$, is defined as

$$\tag{7} B_g=\frac{p_{sc}}{p}\frac{T}{T_{sc}}\frac{z}{z_{sc}}$$

Substituting Eq.7 into the integral on the left-hand-side (LHS) of Eq.6, we have

$$I=-\mu_n B_n \int_{p_1}^{p_2}\frac{1}{\mu_g B_g}dp=-\frac{\mu_n z_n}{p_n}\int_{p_1}^{p_2}\frac{p}{\mu z}dp \ \ (\text{assume} \ T_n=T)$$

Using the initial reservoir pressure, $p_i$, as the "normalizing" pressure, $p_n$, we have

$$I=\frac{\mu_i z_i}{p_i} \int_{p_2}^{p_1}\frac{p}{\mu z}dp \ \ \text{(reversing limits)}$$

Expanding the integral

$$I=\frac{\mu_i z_i}{p_i} \int_{p_{\text{base}}}^{p_1}\frac{p}{\mu z}dp-\frac{\mu_i z_i}{p_i} \int_{p_{\text{base}}}^{p_2}\frac{p}{\mu z}dp $$


$$\tag{9} I=p_p(p_1)-p_p(p_2)$$

$\color{red}{\text{*I believe this term is referred to as the real gas pseudopressure or the "real gas flow potential." The integral is evaluated a priori (ahead of time)}}$


$$\tag{10} p_p(p)=\frac{\mu_i z_i}{p_i} \int_{p_{\text{base}}}^{p}\frac{p}{\mu z}dp$$

Substituting Eq.9 into Eq.6, we obtain

$$p_p(p_1)-p_p(p_2)=\left[\frac{\mu_n B_n}{k A}q_{sc}+ \frac{c\beta}{\mu_g A^2}\rho_{g,sc} \mu_n B_n q_{sc}^2\right]\int_{0}^{L}dx$$

Completing the integration and rearranging

$$\frac{p_p(p_1)-p_p(p_2)}{L}=\frac{\mu_i B_i}{kA}q_{sc}+\frac{c\beta}{\mu_g A^2}\rho_{g,sc}\mu_i B_i q_{sc}^2$$

Dividing through by $\frac{\mu_i B_i}{A}q_{sc}$

$$\tag{11} \frac{A}{\mu_i B_i}\frac{1}{q_{sc}}\frac{(p_p(p_1)-p_p(p_2))}{L}=\frac{1}{k}+\frac{c\beta \rho_{g,sc}}{\mu_g A}q_{sc}$$

From Darcy's Law (i.e., steady-state laminar flow) we have,

$$q_{sc}=\frac{k_{DL}A}{\mu_i B_i}\frac{(p_p(p_1)-p_p(p_2))}{L}$$


$$\tag{12} \frac{1}{k_{DL}}=\frac{A}{\mu_i B_i}\frac{1}{q_{sc}}\frac{(p_p(p_1)-p_p(p_2))}{L}$$

Substituting Eq.12 into Eq.11

$$\tag{13} \frac{1}{k_{DL}}=\frac{1}{k}+\frac{c\beta \rho_{g,sc}}{\mu_g A}q_{sc}$$

where Eq.13 is of the form,



$$y=\frac{1}{k_{DL}}; \ \ x=q_{sc}; \ \ m=\frac{c\beta \rho_{g,sc}}{\mu_g A}; \ \ \text{and} \ \ b=\frac{1}{k}$$


$$\mu_g= \text{is taken at} \ \ \bar{p}=\frac{1}{2}(p_1+p_2)$$

A plot of Eq.13 gives

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.