Question 2 of 7: Diffusivity equation and boundary conditions

Question 2: Diffusivity equation and boundary conditions (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Approach. Combine Darcy's law for radial flow, mass conservation, and a slightly-compressible-fluid equation of state to get the governing PDE, then state the initial condition and pair it with the two candidate outer conditions (infinite-acting vs. bounded) plus the shared constant-rate inner condition.

Governing equation. For single-phase radial flow of a slightly compressible fluid ($c$ small and constant) in a homogeneous cylindrical reservoir, mass conservation combined with Darcy's law and the fluid/rock compressibility relation reduces – after dropping the second-order $(\partial p/\partial r)^2$ term, which is legitimate precisely because the fluid is only slightly compressible – to the linear radial diffusivity equation: $$\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial p}{\partial r}\right)=\frac{\phi\mu c_t}{0.0002637\,k}\frac{\partial p}{\partial t}$$ (field units: $p$ psia, $r$ ft, $k$ mD, $\phi$ fraction, $\mu$ cp, $c_t$ psi$^{-1}$, $t$ hr; the constant $0.0002637$ folds in the field-unit conversions, and is the same group appearing as $\eta=0.0002637k/(\phi\mu c_t)$ and $t_D=\eta t/r^2$ on this paper's own formula sheet). Equivalently, expanding the left side, $\dfrac{\partial^2p}{\partial r^2}+\dfrac{1}{r}\dfrac{\partial p}{\partial r}=\dfrac{1}{\eta}\dfrac{\partial p}{\partial t}$. The initial condition is uniform pressure throughout the reservoir before production begins: $$p(r,0)=p_i\quad\text{for all }r_w\le r\le r_e$$

Inner boundary condition (constant-rate production, no skin, no wellbore storage). With no skin and no wellbore storage the sandface flow rate equals the constant surface rate $q$ for all $t>0$, so Darcy's law applied at $r=r_w$ (or, for the line-source idealisation, the limit as $r\to0$) fixes the pressure gradient there directly in terms of $q$: $$\left.r\frac{\partial p}{\partial r}\right|_{r=r_w}=-\frac{q\mu B_o}{0.00708\,kh}\qquad(\text{constant for all }t>0)$$ the same $0.00708kh$ group used in every $p_D$ conversion on the formula sheet; the boundary condition is simply "Darcy's law evaluated at the wellbore, with the flux held fixed at the constant surface rate."

Outer boundary condition – infinite-acting reservoir. If the reservoir is large enough that the transient never reaches $r_e$ within the time of interest, the pressure far from the well remains undisturbed at its initial value: $$\lim_{r\to\infty}p(r,t)=p_i\qquad\text{for all }t$$ this is the condition that admits the exponential-integral ($Ei$) line-source solution used in Q7, and is valid for as long as the radius of investigation (Q1-b) stays below $r_e$.

Outer boundary condition – bounded (closed) reservoir. If instead the reservoir is finite and sealed at its outer radius $r_e$ (no aquifer support, no fluid crossing the boundary), the flux – and hence the pressure gradient – must vanish there for all time: $$\left.\frac{\partial p}{\partial r}\right|_{r=r_e}=0\qquad\text{for all }t$$ this no-flow condition is exactly what forces the reservoir into pseudosteady state once the transient reaches $r_e$ (Q1-a), and underlies the closed-reservoir radius-of-investigation problem in Q3 and the reservoir-limit test in Q5.