An analytical solution for leakage rate and depletion of aquitard influenced by the delayed yield phenomenon

Abstract

Delayed yield phenomenon (DYP) occurs when the drawdown of an aquitard is no longer in equilibrium owing to previous changes in the level of water in adjacent aquifers. In this study, an analytical solution to determine the distribution of the drawdown at any time and at any point in an aquitard, in which the water level was not in equilibrium at the initial time, and the water level of the aquifers immediately above and below the aquitard was time dependent was developed. The analytical solution for the flow rate at any time and location was presented based on variations in drawdown in the aquitard. In addition, the effects of the DYP on the leakage rate and depletion of the aquitard were studied. Subsequent comparison of the proposed analytical and numerical solutions revealed that they were in good agreement. The results of this study will be useful to future studies of groundwater flow and environmental problems, such as water budgets, rising sea levels, solute transport, heat conduction and ground subsidence.

Appendix A: Derivation of the solution to (1)–(4)

Because boundary conditions (3) and (4) are non-homogeneous, it was necessary to simplify the boundary conditions as follows:

$$u\left( {z,t} \right)\, = \,v\left( {z,t} \right) + w\left( {z,t} \right)$$

(A1)

$$w\left( {z,t} \right)\, = \,u_{1} \left( t \right) + \frac{{u_{2} \left( t \right) - u_{1} \left( t \right)}}{l}z$$

(A2)

Substituting u (z, t) with v (z, t) of Eq. (1), the initial condition and boundary condition leads to:

$$c_{v} \frac{{\partial^{2} v}}{{\partial z^{2} }} - \left( {\left( {1 - \frac{z}{l}} \right)\frac{{\partial s_{1} \left( t \right)}}{\partial t} + \frac{{\partial s_{2} \left( t \right)}}{\partial t}\frac{z}{l}} \right)\, = \,\frac{\partial v}{\partial t}$$

$$v\left( {z,0} \right)\, = \,u\left( z \right) - \frac{{u_{2} - u_{1} }}{l}z - u_{1} \quad 0 < z < l$$

$$v\left( {0,t} \right)\, = \,0 \quad t > 0$$

$$v\left( {l,t} \right)\, = \,0 \quad t > 0$$

(A3)

where, u ₁ and u ₂ are the drawdown of the aquifers lying above and below the aquitard, respectively.

The solution of Eq. (A3) was derived using the characteristic function method. The corresponding eigenfunctions of control equation (A3) were$\left\{ {\sin \frac{m\pi z}{l}} \right\}$, so the form solution of Eq. (A3) was:

$$v\left( {z,t} \right)\, = \,\sum\limits_{m = 1}^{\infty } {T_{m} } \left( t \right)\,\sin \frac{m\pi z}{l}$$

(A4)

where T _m (t) (m = 1, 2, 3…) is the unknown function. Obviously, (A4) satisfies the boundary conditions v (0, t) = v (l, t) = 0. Substituting Eq. (A4) into Eq. (A3) leads to

$$\sum\limits_{m = 1}^{\infty } {\left( {T^{\prime }_{m} \left( t \right) + \frac{{m^{2} \pi^{2} c_{v} }}{{l^{2} }}T_{m} \left( t \right)} \right)} \sin \frac{m\pi z}{l} = - \left( {\left( {1 - \frac{z}{l}} \right)\frac{{\partial u_{1} \left( t \right)}}{\partial t} + \frac{{\partial u_{2} \left( t \right)}}{\partial t}\frac{z}{l}} \right)$$

(A5)

Multiplying both sides of the equation by $\sin \frac{n\pi z}{l}\left( {n = 1,2,3 \ldots } \right)$ gives:

$$\begin{aligned} & \sum\limits_{m = 1}^{\infty } {\left( {T^{{\prime }}_{m} \left( t \right) + \frac{{m^{2} \pi^{2} c_{v} }}{{l^{2} }}T_{m} \left( t \right)} \right)} \int\limits_{0}^{l} {\sin \frac{m\pi z}{l}\sin \frac{n\pi z}{l}} {\text{d}}z \\ \, & \quad = \,\int\limits_{0}^{l} { - \left( {\left( {1 - \frac{z}{l}} \right)\frac{{\partial u_{1} \left( t \right)}}{\partial t} + \frac{{\partial u_{2} \left( t \right)}}{\partial t}\frac{z}{l}} \right)\sin \frac{n\pi z}{l}} {\text{d}}z \\ \end{aligned}$$

(A6)

According to the orthogonality of trigonometric function system, equation Eq. (A6) can be simplified to:

$$T^{{\prime }}_{n} \left( t \right) + \frac{{n^{2} \pi^{2} c_{v} }}{{l^{2} }}T_{n} \left( t \right)\, = \, - \frac{2}{l}\int\limits_{0}^{l} {\left( {\left( {1 - \frac{z}{l}} \right)\frac{{\partial u_{1} \left( t \right)}}{\partial t} + \frac{z}{l}\frac{{\partial u_{2} \left( t \right)}}{\partial t}} \right)\sin \frac{n\pi z}{l}} {\text{d}}z$$

(A7)

Solving Eq. (A7) gives:

$$T_{n} \left( t \right)\, = \,T_{n} \left( 0 \right)e^{{ - \frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} + e^{{ - \frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} \int\limits_{0}^{t} {u_{n} \left( {z,t} \right)e^{{\frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} {\text{d}}t}$$

(A8)

where,

$$u_{n} \left( {z,t} \right)\, = \, - \frac{2}{l}\int\limits_{0}^{l} {\left( {\left( {1 - \frac{z}{l}} \right)\frac{{\partial s_{1} \left( t \right)}}{\partial t} + \frac{z}{l}\frac{{\partial s_{2} \left( t \right)}}{\partial t}} \right)\sin \frac{n\pi z}{l}} {\text{d}}z = \frac{2}{n\pi }\left[ {\frac{{\partial s_{2} \left( t \right)}}{\partial t}\left( { - 1} \right)^{n} - \frac{{\partial s_{1} \left( t \right)}}{\partial t}} \right]$$

(A9)

The initial condition can be expressed as:

$$v\left( {z,0} \right)\, = \,\sum\limits_{m = 1}^{\infty } {T_{n} \left( 0 \right)\sin \frac{m\pi z}{l}} = u\left( z \right) - \frac{{u_{2} - u_{1} }}{l}z - u_{1}$$

(A10)

Therefore, T _n (0) can be expressed by Fourier series as follows:

$$T_{n} \left( 0 \right)\, = \,\frac{2}{l}\int_{0}^{l} {\left[ {u\left( \xi \right) - \frac{{u_{2} - u_{1} }}{l}\xi - u_{1} } \right]\sin \frac{n\pi \xi }{l}} {\text{d}}z$$

(A11)

Substituting Eq. (A9) and Eq. (A11) into (A8) leads to:

$$\begin{aligned} T_{n} \left( t \right)\, = \, & \frac{2}{l}e^{{ - \frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} \int_{0}^{l} {\left[ {u\left( \xi \right) - \frac{{u_{2} - u_{1} }}{l}\xi - u_{1} } \right]} \sin \frac{n\pi \xi }{l}{\text{d}}\xi + \\ & \frac{2}{n\pi }e^{{ - \frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} \int\limits_{0}^{t} {\left[ {\frac{{\partial u_{2} \left( t \right)}}{\partial t}\left( { - 1} \right)^{n} - \frac{{\partial u_{1} \left( t \right)}}{\partial t}} \right]e^{{\frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} {\text{d}}t} \\ \end{aligned}$$

(A12)

Substituting Eq. (A12) into Eq. (A4) enables the form solution of question (A3) to be obtained as follows:

$$\begin{aligned} v\left( {z,t} \right)\, = \, & \frac{2}{l}\sum\limits_{n = 1}^{\infty } {e^{{ - \frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} } \int_{0}^{l} {\left[ {u\left( \xi \right) - \frac{{u_{2} - u_{1} }}{l}z - u_{1} } \right]\sin \frac{n\pi \xi }{l}{\text{d}}\xi } \sin \frac{n\pi z}{l} \\ & + \frac{2}{\pi }\sum\limits_{n = 1}^{\infty } {\frac{1}{n}e^{{ - \frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} \int\limits_{0}^{t} {\left[ {\frac{{\partial u_{2} \left( t \right)}}{\partial t}\left( { - 1} \right)^{n} - \frac{{\partial u_{1} \left( t \right)}}{\partial t}} \right]e^{{\frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} {\text{d}}t} \cdot \sin \frac{n\pi z}{l}} \\ \end{aligned}$$

(A13)

Finally, adding Equations (A2) and (A13) leads to:

$$\begin{aligned} u\left( {z,t} \right) & = v\left( {z,t} \right) + w\left( {z,t} \right) = \left( {1 - \frac{z}{l}} \right)u_{1} \left( t \right) + \frac{z}{l}u_{2} \left( t \right) \\ &\quad + \frac{2}{l}\sum\limits_{n = 1}^{\infty } {e^{{ - \frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} } \int_{0}^{l} {\left[ {u\left( \xi \right) - \frac{{u_{2} - u_{1} }}{l}\xi - u_{1} } \right]\sin \frac{n\pi \xi }{l}{\text{d}}\xi } \sin \frac{n\pi z}{l} \\ &\quad + \frac{2}{\pi }\sum\limits_{n = 1}^{\infty } {\frac{1}{n}e^{{ - \frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} \int\limits_{0}^{t} {\left[ {\frac{{\partial u_{2} \left( t \right)}}{\partial t}\left( { - 1} \right)^{n} - \frac{{\partial u_{1} \left( t \right)}}{\partial t}} \right]e^{{\frac{{n^{2} \pi^{2} c_{v} t}}{{l^{2} }}}} {\text{d}}t} \cdot \sin \frac{n\pi z}{l}} \\ \end{aligned}$$

(A14)