Theoretical Analysis of Pore Pressure Diffusion in Some Basic Rock Mechanics Experiments

Abstract

Non-homogeneity of the pore pressure field in a specimen is an issue for characterization of the thermo-poromechanical behaviour of low-permeability geomaterials, as in the case of the Callovo-Oxfordian claystone (k < 10⁻²⁰ m²), a possible host rock for deep radioactive waste disposal in France. In tests with drained boundary conditions, excess pore pressure can result in significant errors in the measurement of material parameters. Analytical solutions are presented for the change in time of the pore pressure field in a specimen submitted to various loading paths and different rates. The pore pressure field in mechanical and thermal undrained tests is simulated with a 1D finite difference model taking into account the dead volume of the drainage system of the triaxial cell connected to the specimen. These solutions provide a simple and efficient tool for the estimation of the conditions that must hold for reliable determination of material parameters and for optimization of various test conditions to minimize the experimental duration, while keeping the measurement errors at an acceptable level.

Appendix: Calculation of Time-Dependent Pore Pressure Distribution

Assuming a fluid flow governed by Darcy’s law, the pressure change in porous media can be described through Eq. (16). The pressure change is therefore dependent on the pressure gradient and a volumetric pressure source q:

$$q = B\frac{{{\text{d}}\sigma }}{{{\text{d}}\theta }} + \varLambda \frac{{{\text{d}}T}}{{{\text{d}}\theta }} = \tau_{v} \left( {Bv_{\sigma } + \varLambda v_{T} } \right)$$

(58)

Considering a domain of the length 2H_d, the pore pressure diffusion problem can hence be described as a function of dimensionless time θ and dimensionless length Z:

$$\frac{{{\text{d}}p_{\text{f}} }}{{{\text{d}}\theta }} = \frac{{\partial^{2} p_{\text{f}} }}{{\partial Z^{2} }} + q$$

(59)

$$p_{\text{f}} (0,\theta ) = p_{\text{f}} (2,\theta ) = v_{\text{p}} \tau_{v} \theta ,\quad {\text{for}}\,\theta > 0$$

(60)

$$p_{\text{f}} (Z,0) = 0,\quad {\text{for }}0 < Z < 2$$

(61)

The pore pressure is initially equal to zero over the specimen length, whereas it is controlled at the specimen ends is using a certain pore pressure loading rate v_p.

In the following, this problem is solved in the same way as it was described in detail for heat diffusion with sources in Hancock (2006), Guenther and Lee (1996) and Myint-U and Debnath (2007). We have to find an equilibrium solution u_E(Z) that satisfies the PDE and the BCs:

$$0 = \frac{{\partial^{2} u_{E} }}{{\partial Z^{2} }} + q$$

(62)

$$\begin{array}{*{20}c} {u_{E} (0) = 0,} & {u_{E} (2) = 0,} & {{\text{for}}\,\theta > 0} \\ \end{array}$$

(63)

We define u(Z,θ) as:

$$u(Z,\theta ) = p_{\text{f}} (Z,\theta ) - u_{\text{E}} (Z)$$

(64)

Substituting u_E(Z,θ) from the above relation, we obtain:

$$\frac{{{\text{d}}u}}{{{\text{d}}\theta }} = \frac{{\partial^{2} u}}{{\partial Z^{2} }}$$

(65)

$$u(0,\theta ) = u(2,\theta ) = v_{\text{p}} \tau_{v} \theta ,\quad {\text{for}}\,\theta > 0$$

(66)

$$\begin{array}{*{20}c} {u(Z,0) = - u_{E} (Z)} & {{\text{for}}\,0 < Z < 2} \\ \end{array}$$

(67)

To solve a diffusion equation with time-dependent boundary conditions, such as u(Z,θ), Carslaw and Jaeger (1959, pp. 103) propose the following analytical solution for u_C over the length 0 < x < L, the time t (with $\frac{{{\text{d}}u_{\text{C}} }}{{{\text{d}}t}} = \kappa \frac{{\partial^{2} u_{\text{C}} }}{{\partial x^{2} }}$, where κ is the diffusion coefficient, ϕ₁(t) and ϕ₂(t) are the boundary conditions for u_c(0, t) and u_c(L, t), respectively, and f(x) is the initial distribution of u_c(x, 0)):

$$u_{\text{C}} \left( {x,t} \right) = \frac{2}{L}\sum\limits_{n = 1}^{\infty } {e^{{ - \kappa n^{2} \pi^{2} t/L^{2} }} \sin \left( {\frac{n\pi x}{L}} \right)\left[ {\int\limits_{0}^{L} {f\left( {x^{{\prime }} } \right)\sin \left( {\frac{{n\pi x^{{\prime }} }}{L}} \right){\text{d}}x^{{\prime }} } + \frac{\kappa n\pi }{L}\int\limits_{0}^{t} {e^{{\kappa n^{2} \pi^{2} \lambda /L^{2} }} \left( {\phi_{1} \left( \lambda \right) - \left( { - 1} \right)^{n} \phi_{2} \left( \lambda \right)} \right){\text{d}}\lambda } } \right]}$$

(68)

To obtain the solution for u(Z, θ), we insert L = 2, κ = 1 into Eq. (68) and replace t with θ and x with Z. For the boundary conditions, we can insert $\phi_{1} \left( \lambda \right) = \phi_{2} \left( \lambda \right) = v_{p} \tau_{v} \lambda$ and $f(x^{{\prime }} ) = - u_{E} (x^{{\prime }} )$.

$$u\left( {Z,\theta } \right) = \sum\limits_{n = 1}^{\infty } {e^{{ - n^{2} \pi^{2} \theta /4}} } \sin \left( {\frac{n\pi Z}{2}} \right)\left[ {\int\limits_{0}^{2} { - u_{\text{E}} (x^{{\prime }} )\sin \left( {\frac{{n\pi x^{{\prime }} }}{2}} \right){\text{d}}x^{{\prime }} } + \frac{n\pi }{2}\int\limits_{0}^{\theta } {e^{{n^{2} \pi^{2} \lambda /4}} v_{\text{p}} \tau_{v} \lambda \left( {1 - \left( { - 1} \right)^{n} } \right){\text{d}}\lambda } } \right]$$

(69)

The first integral in Eq. (69) can be solved by partial integration:

$$\int\limits_{0}^{2} { - u_{\text{E}} (x^{{\prime }} )\sin \left( {\frac{{n\pi x^{{\prime }} }}{2}} \right){\text{d}}x^{{\prime }} = \left[ {\frac{2}{n\pi }u_{\text{E}} (x^{{\prime }} )\cos \left( {\frac{{n\pi x^{{\prime }} }}{2}} \right)} \right]_{0}^{2} - \int\limits_{0}^{2} {\frac{2}{n\pi }\left( {\frac{{\partial^{2} u_{\text{E}} (x')}}{{\partial x^{{{\prime 2}}} }}} \right)\cos \left( {\frac{{n\pi x^{{\prime }} }}{2}} \right){\text{d}}x^{{\prime }} } }$$

(70)

The first term in the right-hand side of Eq. (70) disappears because of the boundary conditions u_E(0) = u_E(2) = 0. Partial integration applied on the remaining second term gives:

$$\int\limits_{0}^{2} { - u_{\text{E}} (x^{{\prime }} )\sin \left( {\frac{{n\pi x^{{\prime }} }}{2}} \right){\text{d}}x^{{\prime }} } = - \left[ {\left( {\frac{2}{n\pi }} \right)^{2} \left( {\frac{{\partial u_{\text{E}} (x^{{\prime }} )}}{{\partial x^{{\prime }} }}} \right)\sin \left( {\frac{{n\pi x^{{\prime }} }}{2}} \right)} \right]_{0}^{2} + \left( {\frac{2}{n\pi }} \right)^{2} \int\limits_{0}^{2} {\left( {\frac{{\partial^{2} u_{\text{E}} (x^{{\prime }} )}}{{\partial x^{{{\prime 2}}} }}} \right)\sin \left( {\frac{{n\pi x^{{\prime }} }}{2}} \right){\text{d}}x^{{\prime }} }$$

(71)

Here the first term of Eq. (71) vanishes, as sin(nπ) = 0. From Eq. (62), we can insert ${{\partial^{2} u_{E} (x')} \mathord{\left/ {\vphantom {{\partial^{2} u_{E} (x')} {\partial x'^{2} }}} \right. \kern-0pt} {\partial x'^{2} }} = - q$ to obtain:

$$\int\limits_{0}^{2} { - u_{\text{E}} (x^{{\prime }} )\sin \left( {\frac{{n\pi x^{{\prime }} }}{2}} \right){\text{d}}x^{{\prime }} } = q\left( {\frac{2}{n\pi }} \right)^{3} \left[ {\cos \left( {n\pi } \right) - 1} \right]$$

(72)

It can be seen that Eq. (72) becomes nonzero only for odd values of n. Note also that the second integral in Eq. (69) is nonzero only for odd values of n. For this reason, the variable n is replaced by (2n − 1) and we introduce:

$$m = \frac{(2n - 1)}{2}\pi$$

(73)

Equation (69) can then be rewritten to:

$$u\left( {Z,\theta } \right) = \sum\limits_{n = 1}^{\infty } {e^{{ - m^{2} \theta }} } \sin \left( {mZ} \right)\left[ { - \frac{2q}{{m^{3} }} + 2v_{p} \tau_{v} m\int\limits_{0}^{\theta } {e^{{m^{2} \lambda }} \lambda {\text{d}}\lambda } } \right]$$

(74)

The remaining integral in Eq. (74) can be solved by partial integration. We also make use of the relationship $\sum\nolimits_{n = 1}^{\infty } {\frac{1}{m}} \sin \left( {mZ} \right) = \frac{1}{2}$, giving:

$$u\left( {Z,\theta } \right) = - 2\sum\limits_{n = 1}^{\infty } {e^{{ - m^{2} \theta }} } \sin \left( {mZ} \right)q\frac{1}{{m^{3} }} + 2\sum\limits_{n = 1}^{\infty } {\left( {e^{{ - m^{2} \theta }} - 1} \right)} \sin \left( {mZ} \right)v_{\text{p}} \tau_{v} \frac{1}{{m^{3} }} + v_{\text{p}} \tau_{v} \theta$$

(75)

From the initial condition at θ = 0, we know that:

$$u\left( {Z,0} \right) = - u_{\text{E}} (Z) = - 2\sum\limits_{n = 1}^{\infty } {\sin \left( {mZ} \right)\frac{q}{{m^{3} }}}$$

(76)

This allows us to write down the final analytical solution for p_f, using Eqs. (64), (75) and (76):

$$p_{\text{f}} (Z,\theta ) = 2\left( {Bv_{\sigma } + \varLambda v_{T} - v_{\text{p}} } \right)\tau_{v} \sum\limits_{n = 1}^{\infty } {\left[ {\left( {1 - e^{{ - m^{2} \theta }} } \right)\frac{1}{{m^{3} }}\sin \left( {mZ} \right)} \right]} + v_{\text{p}} \tau_{v} \theta$$

(77)