Reservoir Engineering: Derivation of The Diffusivity Equation (part 2)
We have derived the general form of continuity equation for fluid flow in porous media in the previous post.
\[\frac{\partial{ \left( \rho\phi \right) }}{\partial{t}} = \frac{-1}{r} \frac{\partial}{\partial r}(r \rho u_{r})\]Consider Darcy’s Law,
\[u_s = -\frac{k}{\mu} \left( \frac{\partial P}{\partial s} - \rho g \frac{\partial z}{\partial s} \right)\]Keep in mind that Darcy’s Law only applies to laminar (or Darcy) flow.
Translating that form in \(r\) direction gives,
\[u_r = \frac{k}{\mu} \left( \frac{\partial P}{\partial r} \right)\]Introduce Darcy’s equation into continuity equation,
\[\frac{\partial{ \left( \rho\phi \right) }}{\partial{t}} = \frac{-1}{r} \frac{\partial}{\partial r}\left( r \rho \frac{k}{\mu} \frac{\partial P}{\partial r} \right)\]This form is by far the most general form of diffusivity equation before we introduce further assumptions into our derivation. Until now, we have made the following assumptions:
- Flow only happens in \(r\) direction
- Laminar (or Darcy) flow
From this point on, we’re gonna introduce more assumptions in order to further derive this equation. Notice on the right-hand side of the equation, we have variables \(\rho\), \(k\), \(\mu\) that are obviously \(f(P)\) (a function of pressure). Pressure itself is \(f(r)\) (a function of space, which again in this case only happens in \(r\)). The
Flow of single-phase liquid (small, constant compressibility)
We need to assume that the porous medium (i.e. the rock) has constant permeability and fluid’s viscosity stays constant.
\[\frac{\mu}{k} \frac{\partial(\rho\phi)}{\partial{t}} = \frac{-1}{r} \frac{\partial}{\partial r} \left( r \rho \frac{\partial P}{\partial r} \right)\]Evaluate the left-hand side,
\[\begin{align} \frac{\partial (\rho\phi)}{\partial t} &= \frac{\partial (\rho\phi)}{\partial P} \frac{\partial P}{\partial t} \\ &= \left( \rho \frac{\partial (\phi)}{\partial P} + \phi \frac{\partial (\rho)}{\partial P} \right) \frac{\partial P}{\partial t} \end{align}\]Recall the definition of compressibility of both fluid and rock (yes, rock can be compressible!),
\[\begin{align} c_f &= \frac{1}{\rho} \frac{d\rho}{dP} & c_r &= \frac{1}{\phi} \frac{d\phi}{dP}\\ \rho c_f &= \frac{d\rho}{dP} & \phi c_r &= \frac{d\phi}{dP} \end{align}\]Substituting the compressibility expression into the left-hand,
\[\begin{align} \frac{\partial (\rho\phi)}{\partial t} &= \left( \rho \phi c_r + \phi \rho c_f \right) \frac{\partial P}{\partial t} \\ &= \rho\phi (c_r + c_f) \frac{\partial P}{\partial t} \\ &= \rho\phi c_T \frac{\partial P}{\partial t} \end{align}\]where \(c_T\) denotes the total compressibility (\(c_T = (c_r + c_f)\)).
Substituting this left-hand expression back into the derivation,
\[\frac{\mu\rho\phi c_T}{k} \frac{\partial P}{\partial t} = \frac{-1}{r} \frac{\partial}{\partial r} \left( r \rho \frac{\partial P}{\partial r} \right)\]Evaluate the right-hand side,
\[\begin{align} \frac{\mu\rho\phi c_T}{k} \frac{\partial P}{\partial t} &= \frac{-1}{r} \left[ \frac{\partial r}{\partial r} \rho \frac{\partial P}{\partial r} + r \frac{\partial \rho}{\partial P} \left( \frac{\partial P}{\partial r} \right)^{2} + r\rho \frac{\partial^2 P}{\partial r^2} \right] \\ &= \frac{-1}{r} \left[ \frac{\partial r}{\partial r} \rho \frac{\partial P}{\partial r} + r c_f \rho \left( \frac{\partial P}{\partial r} \right)^{2} + r\rho \frac{\partial^2 P}{\partial r^2} \right] \\ &= \frac{-\rho}{r} \left[ \frac{\partial r}{\partial r} \frac{\partial P}{\partial r} + r c_f \left( \frac{\partial P}{\partial r} \right)^{2} + r \frac{\partial^2 P}{\partial r^2} \right] \\ \end{align}\]We also assume that fluid has small, constant compressibility. Furthermore, pressure-gradient-squared times compressibility, \(c_f \left( \frac{\partial P}{\partial r} \right)^{2}\), can be neglected.
\[\begin{align} \frac{\mu\phi c_T}{k} \frac{\partial P}{\partial t} &= \frac{-1}{r} \left[ \frac{\partial r}{\partial r} \frac{\partial P}{\partial r} + r \frac{\partial^2 P}{\partial r^2} \right] \\ \frac{\mu\phi c_T}{k} \frac{\partial P}{\partial t} &= \frac{-1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial P}{\partial r} \right) \\ \end{align}\]