# Integrating the quadratic Forchheimer equation

Starting with the commonly cited Forchheimer equation:

$$\tag{1} -\frac{dp}{ds}=\frac{\mu q}{kA}+\frac{\beta \rho q^2}{A^2}$$

multiplying through by density

$$\tag{2} -\rho \frac{dp}{ds}=\frac{\mu \rho q}{kA}+\frac{\beta \rho^2 q^2}{A^2}$$

noting that $\rho q = \dot m$

Dividing (2) through by viscosity:

$$\tag{3} -\frac{\rho}{\mu} \frac{dp}{ds}=\frac{\dot m}{kA}+\frac{\beta \dot m^2}{\mu A^2}$$

Here I see I cannot completely separate the pressure dependent variables ($\rho$ and $\mu$) as I have a viscosity term on the far right-hand side of (3).

Multiplying through by the differential $ds$ and then including the integration notation:

$$\tag{4} -\int_{p_b}^p \frac{\rho}{\mu} dp=\frac{\dot m}{kA}\int_0^L ds+\frac{\beta \dot m^2}{A^2} \int_0^L \frac{1}{\mu} ds$$

So here is my question: $\mu$ is a function of pressure and pressure varies as a function of length (or distance). Is it possible to solve for the integral on the far right-hand side of (4) noting the aforementioned fact? I would also like to know what this type of problem in called in mathematics, if there is such a term for this type of problem.

Pressure as a function of length (or distance) may be written as:

$$\tag{5} p_p (p_s)=p_p(p_0)+\frac{\dot m \mu}{kA}s$$

where

$$\tag{6} p_p(p)=\frac{\mu_i z_i}{p_i}\int_{p_b}^p \frac{p}{\mu z} dp$$

where $\mu_i$, $z_i$, and $p_i$ are the normalizing viscosity, real-gas z-factor, and pressure values, respectively, which are constants, selected as the values at the inlet of the porous medium where $s=0$.

Is it possible to invoke (5) into the $\int_0^L \frac{1}{\mu} ds$ term of (4) and perform the integration?

Eqn. 3 is variable separable and can be rearranged to $$\frac{dp}{f(p)}=ds$$ The left side of the equation has to be integrated numerically.
• Thanks Chester, I understand how to perform the numerical integration of the $-\frac{\rho}{\mu} dp$ term on the LHS of (3). But how do I handle the viscosity term on the far RHS of (3)? Is there a way I can get that viscosity term to the LHS so that I have an equation as you describe ($\frac{dp}{f(p)}=ds$)? Commented Mar 21, 2017 at 14:54
• $\frac{1}{f(p)}=-\left[\frac{\mu \dot m}{\rho kA}+\frac{\beta \dot m^2}{\rho A^2}\right]$ Commented Mar 21, 2017 at 16:12
• so, starting from (3), multiply through by $-\frac{\mu}{\rho}$ to obtain $\frac{dp}{ds}=-\left[\frac{\mu \dot m}{\rho kA}+\frac{\beta \dot m^2}{\rho A^2}\right]$, then rearrange further? I think I am not following, as where I'm at now, I have all the variables as a function of pressure on the RHS, which would then need to be integrated as a function of distance. My apologies for not following, I will give more time and thought to let it sink in. Commented Mar 21, 2017 at 17:04
• Suppose you have f(x)dx=kdy. Could you integrate the left side? Could you integrate the right side? The solution is $\int{f(x)dx}=ky + C$ where C is the constant of integration. Commented Mar 21, 2017 at 20:07
• $$\frac{\rho dp}{\left[\frac{\mu}{k }+\frac{\beta \dot{m}}{ A}\right]}=-\frac{\dot{m}ds}{A}$$ Commented Mar 22, 2017 at 3:22