Derivation of Single-phase pressure diffusion @model

We start with the reservoir flow continuity equation:

(1)	$\begin{array}{l}\displaystyle \frac{\partial (\rho \phi)}{\partial t} + \nabla \, ( \rho \, {\bf u}) = \sum_k \dot m_k(t) \cdot \delta({\bf r}-{\bf r}_k)\end{array}$

percolation model:

(2)	$\begin{array}{l}\displaystyle {\bf u} = - M \cdot ( \nabla p - \rho \, {\bf g})\end{array}$

and the reservoir boundary flow condition:

(3)	$\begin{array}{l}\displaystyle {\rm F}_{\Gamma}(p, {\bf u}) = 0\end{array}$

where

$\begin{array}{l}\Sigma_k\end{array}$	well-reservoir contact of the $\begin{array}{l}k\end{array}$ -th well
$\begin{array}{l}d {\bf \Sigma}\end{array}$	normal vector of differential area on the well-reservoir contact, pointing inside wellbore
$\begin{array}{l}\dot m_k(t)\end{array}$	mass flowrate at $\begin{array}{l}k\end{array}$ -th well $\begin{array}{l}\dot m_k(t) = \rho(p) \cdot q_k(t)\end{array}$
$\begin{array}{l}q_k(t)\end{array}$	sandface flowrate at $\begin{array}{l}k\end{array}$ -th well
$\begin{array}{l}\rho(p)\end{array}$	fluid density as function of reservoir fluid pressure $\begin{array}{l}p\end{array}$

Then use the following equality:

(4)

$\begin{array}{l}\displaystyle d(\rho \, \phi) = \rho \, d \phi + \phi \, d\rho = \rho \, \phi \, \left( \frac{d \phi }{\phi} + \frac{d \rho }{\rho} \right) = \rho \, \phi \, \left( \frac{1}{\phi} \frac{d \phi}{dp} \, dp + \frac{1}{\rho} \frac{d \rho}{dp} \, dp \right) = \rho \, \phi \, (c_{\phi} \, dp + c \, dp) = \rho \, \phi \, c_t \, dp\end{array}$

where

$\begin{array}{l}c_t = с_\phi+ c\end{array}$

total compressibility

to arrive at:

(5)	$\begin{array}{l}\displaystyle \rho \, \phi \, c_t \cdot \frac{\partial p}{\partial t} + \nabla \, ( \rho \, {\bf u}) = \sum_k \rho \, q_k(t) \cdot \delta({\bf r}-{\bf r}_k)\end{array}$

(6)	$\begin{array}{l}\displaystyle {\rm F}_{\Gamma}(p, {\bf u}) = 0\end{array}$

The left-hand side of equation (5) can be transformed in the following way:

(7)	$\begin{array}{l}\displaystyle \nabla \, ( \rho \, {\bf u}) = \rho \, \nabla \, {\bf u} + (\nabla \rho, \, {\bf u}) = \rho \, \nabla \, {\bf u} + \frac{d\rho}{dp} \cdot (\nabla p, \, {\bf u}) = \rho \, \nabla \, {\bf u} + \rho \, c \cdot (\nabla p, \, {\bf u})\end{array}$

where $\begin{array}{l}\displaystyle c(p) = \frac{1}{\rho} \frac{d\rho}{dp}\end{array}$ is fluid compressibility.

By using the Dirac delta function property: $\begin{array}{l}f(x) \cdot \delta(x-x_0) = f(x_0) \cdot \delta(x-x_0)\end{array}$ the right-hand side of equation (5) can be transformed in the following way:

(8)

$\begin{array}{l}\displaystyle \sum_k \rho(p(t, {\bf r})) \cdot q_k(t) \cdot \delta({\bf r}-{\bf r}_k) = \sum_k \rho(p(t, {\bf r}_k)) \cdot q_k(t) \cdot \delta({\bf r}-{\bf r}_k) = \sum_k \rho(p(t, {\bf r})) \cdot q_k(t) \cdot \delta({\bf r}-{\bf r}_k) = \rho(p) \cdot \sum_k q_k(t) \cdot \delta({\bf r}-{\bf r}_k)\end{array}$

Substituting (7) and (8) into (5) and reducing the density $\begin{array}{l}\rho(p)\end{array}$ one arrives to:

(9)	$\begin{array}{l}\displaystyle \phi \, c_t \cdot \frac{\partial p}{\partial t} + \nabla {\bf u} + c \cdot ( {\bf u} \, \nabla p) = \sum_k q_k(t) \cdot \delta({\bf r}-{\bf r}_k)\end{array}$

(10)	$\begin{array}{l}\displaystyle {\rm F}_{\Gamma}(p, {\bf u}) = 0\end{array}$

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Derivation of Single-phase pressure diffusion @model