Theoretical Analysis of One-Dimensional Pressure Diffusion from a Constant Upstream Pressure to a Constant Downstream Storage

Abstract

The one-dimensional diffusion equation was solved to understand the pressure and flow behaviors along a cylindrical rock specimen for experimental boundary conditions of constant upstream pressure and constant downstream storage. The solution consists of a time-constant asymptotic part and a transient part that is a negative exponential function of time. This means that the transient flow exponentially decays with time and is eventually followed by a steady-state condition. For a given rock sample, the transient stage is shortest when the downstream storage is minimized. For this boundary condition, a simple equation was derived from the analytic solution to determine the hydraulic permeability from the initial flow rate during the transient stage. The specific storage of a rock sample can be obtained simply from the total flow into the sample during the entire transient stage if there is no downstream storage. In theory, both of these hydraulic properties could be obtained simultaneously from transient-flow stage measurements without a complicated curve fitting or inversion process. Sensitivity analysis showed that the derived permeability is more reliable for lower-permeability rock samples. In conclusion, the constant head method with no downstream storage might be more applicable to extremely low-permeability rocks if the upstream flow rate is measured precisely upstream.

Appendix 1: Particular Solution of the 1-D Differential Equation

A procedure for solving the one-dimensional diffusion equation is briefly described here. Applying the Laplace transform to the partial differential Eq. (2) yields

$$\int\limits_{0}^{\infty } {e^{{ - {\text{st}}}} } \frac{{\partial^{2} p}}{{\partial x^{2} }}{\text{d}}t - \frac{1}{k}\int\limits_{0}^{\infty } {e^{{ - {\text{st}}}} } \frac{\partial p}{\partial t}{\text{d}}t = 0.$$

(25)

Equation (25) with the initial condition, Eq. (3), can be expressed in the form of an ordinary differential equation

$$\frac{{{\text{d}}^{2} \overline{p} }}{{{\text{d}}x^{2} }} - q^{2} \overline{p} = 0\quad{\text{with}}\quad q^{2} = \frac{s}{k},$$

(26)

where $\overline{p}$ is the Laplace transform of p. The general solution of Eq. (26) is

$$\overline{p} = C_{1} e^{qx} + C_{2} e^{ - qx}$$

(27)

Taking the Laplace transforms of the boundary conditions, Eqs. (4) and (5), we can determine C ₁ and C ₂ in Eq. (27), so that Eq. (27) becomes

$$\overline{p} = \frac{P}{s}\frac{{q\cosh (qx) + s\lambda_{d} \sinh (qx)}}{{q\cosh (qL) + s\lambda_{d} \sinh (qL)}}$$

(28)

The inverse Laplace transform of $\overline{p}$ is

$$p(x,t) = \frac{1}{2\pi i}\int\limits_{\gamma - i\infty }^{\gamma + i\infty } {e^{st} \overline{p}(s){\text{d}}s} ,$$

(29)

where γ has to be sufficiently large that all the singularities of $\overline{p} (s)$ lie to the left of the line (γ − i∞, γ + i∞). The contour integral (Eq. (29)) can then be evaluated by computing the contour to the left and summing residues. From the residue theorem, Eq. (29) is rewritten as

$$p(x,t) = \sum\limits_{m} {\text{Re} s} (s_{m} ),$$

(30)

where s _m are poles of the integrand, and Re s(s _m ) are the associated residues. In Eq. (29), the integrand, $e^{st} \overline{p} (s)$, has a simple pole at s = 0:

$$\text{Re} s(0) = \mathop {\lim }\limits_{s \to 0} e^{st} \overline{p}(s)$$

(31)

Taking the first two terms in the Taylor series of sinhx, coshx, and e ^st yield

$\sinh qx = qx + \frac{{q^{3} x^{3} }}{3!},\cosh qx = 1 + \frac{{q^{2} x^{2} }}{2!},$, and e ^st = 1 + st, respectively. Higher-order terms do not contribute to the pole at s = 0. Now Eq. (28) becomes

$${\overline{p} (s) = \frac{P}{s}\frac{{6k + 3kx^{2} s + 6k\lambda_{d} xs + \lambda_{d} x^{3} s^{2} }}{{6k + 3kL^{2} s + 6k\lambda_{d} Ls + \lambda_{d} L^{3} s^{2} }}}$$

(32)

The residue at a simple pole of order m > 1 at z = a of a function f (z) is given by Kreyszig (1983, p. 721),

$$\text{Re} s\text{ }\mathop {f\left( z \right)}\limits_{z = a} = \frac{1}{(m - 1)!}\mathop {\lim }\limits_{s \to 0} \left\{ {\frac{{{\text{d}}^{m - 1} }}{{{\text{d}}Z^{m - 1} }}\left[ {(z - a)^{m} f(z)} \right]} \right\}.$$

(33)

The residue of $e^{st} \overline{p} (s)$ at s = 0 for m = 2 is

$$\text{Re} s(0) = \mathop {\lim }\limits_{s \to 0} \frac{{P(6k + 3kx^{2} s + 6k\lambda_{d} xs + \lambda_{d} x^{3} s^{2} )(1 + st)}}{{{\text{d}}/{\text{d}}s(6ks + 3kL^{2} s^{2} + 6k\lambda_{d} Ls^{2} + \lambda_{d} L^{3} s^{3} )}} = P,$$

(34)

where $S_{d} \ne \infty$. If $S_{d} = \infty$ for the boundary condition, Eq. (32) will be

$${\overline{p} (s) = P\frac{{6kx + x^{3} s}}{{6kLs + L^{3} s^{2} }}}$$

(35)

From the same procedure,

$$\text{Re} s(0) = \mathop {\lim }\limits_{s \to 0} \frac{{P(6kx + x^{3} s)(1 + st)}}{{{\text{d}}/{\text{d}}s(6kLs + L^{3} s^{2} )}} = P\frac{x}{L}$$

(36)

In addition, if q is an imaginary the function, multiple poles occur when the denominator in Eq. (28) vanishes; that is,

$$q\cosh (qL) + s\lambda_{d} \sinh (qL) = 0$$

(37)

Because q is imaginary, we can write

$$q = i\phi \quad{\text{and}}\quad q^{2} = - \phi^{2} ,$$

(38)

where ϕ is a real number. Then, from Eq. (26) it is seen that s is always negative:

$$s = - k\phi^{2}$$

(38)

Equation (37) can be rewritten by substituting Eq. (38) into Eq. (37):

$$\tan \phi L = \frac{1}{\delta \phi L},$$

(39)

where $\delta = \frac{{S_{d} }}{{S_{s} AL}}$ ϕ _m are roots of Eq. (40). Substituting q = iϕ and s = −κϕ², the integrand can be rewritten as

$$e^{st} {\overline{p} (s) = \frac{{N(s_{m} )}}{{D(s_{m} )}} = \frac{ - P}{k}\frac{{\cos \phi_{m} x - \delta \phi_{m} L\sin \phi_{m} x}}{{\phi^{2}_{m} \cos \phi_{m} L - \delta \phi^{3}_{m} L\sin \phi_{m} L}}} \exp ( - k\phi^{2}_{m} t)$$

(40)

The multiple poles are then determined by the roots ϕ _m of Eq. (40). Now $e^{st} \overline{p} (s)$ is of the form N(s)/D’(s) and the residues are given by

$$\text{Re} s(S_{m} ) = \frac{{N(S_{m} )}}{{D'(S_{m} )}} = \frac{{ - 2P(\cos \phi_{m} x - \delta \phi_{m} L\sin \phi_{m} x)\exp ( - k\phi_{m}^{2} t)}}{{(\delta \phi_{m}^{2} L^{2} - 2)\cos \phi_{m} L + (3\delta \phi_{m} L + \phi_{m} L)\sin \phi_{m} L}}, ,$$

(41)

where $S_{d} \ne \infty$. If $S_{d} = \infty$, the residues are given by

$$\text{Re} s(S_{m} ) = \frac{{N(S_{m} )}}{{D'(S_{m} )}} = \frac{{ - 2P\sin \phi_{m} x}}{{(2\sin \phi_{m} L - \phi_{m} L\cos \phi_{m} L)}}\exp ( - k\phi_{m}^{2} t)$$

(42)

Because $\phi_{m}$ are roots of Eq. (40), $\phi_{m} L = m\pi$ where $m = 1,{ 2},{ 3}, \ldots$. Then Eq. (43) becomes

$$\text{Re} s(S_{m} ) = \frac{2P}{\pi }\sum\limits_{m = 1}^{\infty } {\frac{{( - 1)^{m} }}{m}} \sin \left( {m\pi \frac{x}{L}} \right)\exp \left( { - \frac{{km^{2} \pi^{2} t}}{{L^{2} }}} \right)$$

(43)

The complete equation for p is

$$p(x,t) = \text{Re} s(0) + \sum\limits_{m = 1}^{\infty } {\text{Re} s(S_{m} )} .$$

(44)

When $S_{d} \ne \infty$, the complete analytic solution of the diffusion equation is

$$p(x,t) = P - 2P\sum\limits_{m = 1}^{\infty } {\frac{{\cos \phi_{m} x - \delta \phi_{m} L\sin \phi_{m} x}}{{(\delta \phi^{2}_{m} L^{2} - 2)\cos \phi_{m} L + (3\delta \phi_{m} L + \phi_{m} L)\sin \phi_{m} L}}\exp ( - k\phi_{m}^{2} t)}$$

(45)

and when $S_{d} = \infty$,

$$p(x,t) = P - 2P\sum\limits_{m = 1}^{\infty } {\frac{{\cos \phi_{m} x - \delta \phi_{m} L\sin \phi_{m} x}}{{(\delta \phi^{2}_{m} L^{2} - 2)\cos \phi_{m} L + (3\delta \phi_{m} L + \phi_{m} L)\sin \phi_{m} L}}\exp ( - k\phi_{m}^{2} t)}$$

(46)