Non-Darcian flow toward a larger-diameter partially penetrating well in a confined aquifer

Abstract

In this study, non-Darcian flow to a larger-diameter partially penetrating well in a confined aquifer was investigated. The flow in the horizontal direction was assumed to be non-Darcian and described by the Izbash equation, and the flow in the vertical direction was assumed to be Darcian. A linearization procedure was used to approximate the nonlinear governing equation. The Laplace transform associated with the finite cosine Fourier transform was used to solve such non-Darcian flow model. Both the drawdowns inside the well and in the aquifer were analyzed under different conditions. The results indicated that the drawdowns inside the well were generally the same at early times under different conditions, and the features of the drawdowns inside the well at late times were similar to those of the drawdowns in the aquifer. The drawdown in the aquifer for the non-Darcian flow case was larger at early times and smaller at late times than their counterparts of Darcian flow case. The drawdowns for a partially penetrating well were the same as those of a fully penetrating well at early times, and were larger than those for a fully penetrating well at late times. A longer well screen resulted in a smaller drawdown in the aquifer at late times. A larger power index n in the Izbash equation resulted in a larger drawdown in the aquifer at early times and led to a smaller drawdown in the aquifer at late times. A larger well radius led to a smaller drawdown at early times, but it had little impact on the drawdown at late times. The wellbore storage effect disappears earlier when n is larger.

Appendix: Derivation of the analytical solution in the Laplace domain

Applying Laplace transform to Eq. (12) with considering the initial condition Eq. (2), one can obtain

$$\frac{{\partial^{2} \bar{s}(r,z,p)}}{{\partial r^{2} }} + \frac{n}{r}\frac{{\partial \bar{s}(r,z,p)}}{\partial r} + \frac{{K_{\text{z}} n}}{{K_{\text{r}} }}\left( {\frac{Q}{2\pi rB}} \right)^{n - 1} \frac{{\partial^{2} \bar{s}(r,z,p)}}{{\partial z^{2} }} \approx \frac{n}{{K_{\text{r}} }}\left( {\frac{Q}{2\pi rB}} \right)^{n - 1} S_{\text{s}} p\bar{s}(r,z,p),$$

(17)

in which p is the Laplace variable, over bar means the variable in Laplace domain. After using the finite cosine Fourier transform to deal with the second-order derivative of z, it can be expressed as:

$$\begin{aligned} & F_{c} \left( {\frac{{\partial^{2} \bar{s}(r,z,p)}}{{\partial z^{2} }},z \to N} \right) = \int\limits_{0}^{B} {\frac{{\partial^{2} \bar{s}(r,z,p)}}{{\partial z^{2} }}\cos \frac{N\pi z}{B}} d_{\text{z}} \\ & \quad = - \frac{{N^{2} \pi^{2} }}{{B^{2} }}\overline{{\hat{s}}} (r,N,p) - \frac{{\partial \bar{s}(r,0,p)}}{\partial z} - ( - 1)^{N} \frac{{\partial \bar{s}(r,B,p)}}{\partial z}\quad (N = 0,1,2,3 \ldots ) \\ \end{aligned}$$

(18)

The over ^sign means the variable in Fourier domain. The vertical boundary equations Eqs. (4) and (5) can be rewritten after the Laplace transform as:

$$\frac{{\partial \bar{s}(r,0,p)}}{\partial z} = 0,$$

(19)

$$\frac{{\partial \bar{s}(r,B,p)}}{\partial z} = 0,$$

(20)

Substituting Eqs. (19) and (20) to (18), one has

$$F_{\text{c}} \left( {\frac{{\partial^{2} \bar{s}(r,z,p)}}{{\partial z^{2} }},z \to N} \right) = - \frac{{N^{2} \pi^{2} }}{{B^{2} }}\overline{{\hat{s}}} (r,N,p) ,\quad N = 0, \, 1, \, 2 \ldots$$

(21)

With the finite cosine Fourier transform, Eq. (17) can be changed to the following equation after some simplifications:

$$\frac{{{\text{d}}^{2} \bar{\hat{s}}(r,N,p)}}{{{\text{d}}r^{2} }} + \frac{n}{r}\frac{{{\text{d}}\bar{\hat{s}}(r,N,p)}}{{{\text{d}}r}} = \frac{{nK_{z} N^{2} \pi^{2} + B^{2} npS_{s} }}{{K_{\text{r}} B^{2} }}\left( {\frac{Q}{2\pi rB}} \right)^{n - 1} \bar{\hat{s}}(r,N,p).$$

(22)

Eq. (22) is a Bessel equation, whose general solution can be expressed as:

$$\bar{\hat{s}}(r,N,p) = r^{{\frac{1 - n}{2}}} \left[ {C_{1} I_{{\frac{1 - n}{3 - n}}} \left( {\frac{2}{3 - n}r^{{\frac{3 - n}{2}}} \sqrt \delta } \right) + C_{2} K_{{\frac{1 - n}{3 - n}}} \left( {\frac{2}{3 - n}r^{{\frac{3 - n}{2}}} \sqrt \delta } \right)} \right],$$

(23)

in which $\delta = \frac{{nK_{\text{z}} N^{2} \pi^{2} + B^{2} npS_{\text{s}} }}{{K_{\text{r}} B^{2} }}\left( {\frac{Q}{2\pi B}} \right)^{n - 1}$, I _v(x) and K _v(x) are the first and second kinds of modified Bessel functions with order v, respectively. C ₁ and C ₂ are constants, which depend on the boundary conditions. The radial boundary conditions Eqs. (3) and (6) can be changed to

$$\overline{{\hat{s}}} (\infty ,N,p) = 0,$$

(24)

and after using the Laplace transform and the finite cosine Fourier transform, one has

$$\frac{{2\pi r_{\text{w}} (l - d)K_{r} }}{{\left[ {\frac{Q}{{2\pi r_{\text{w}} (l - d)}}} \right]^{n - 1} }}\frac{{{\text{d}}\bar{\hat{s}}(r_{\text{w}} ,N,p)}}{{{\text{d}}r_{\text{w}} }} - \pi r_{\text{w}}^{2} p\bar{\hat{s}}(r_{\text{w}} ,N,p) = - \frac{Q}{p}\frac{{\sin \left( {\frac{N\pi l}{B}} \right) - \sin \left( {\frac{N\pi d}{B}} \right)}}{N\pi /B}.$$

(25)

It should be pointed out that the linearization procedure has also been used in Eq. (25). With the consideration of Eq. (24) and the properties of the Bessel functions, one has C ₁ = 0. Then, the solution of Eq. (23) can be rewritten as:

$$\bar{\hat{s}}(r,N,p) = C_{2} r^{{\frac{1 - n}{2}}} K_{{\frac{1 - n}{3 - n}}} \left( {\frac{2}{3 - n}r^{{\frac{3 - n}{2}}} \sqrt \delta } \right).$$

(26)

With Eqs. (25) and (26), one can obtain C ₂ as:

$$C_{2} = \frac{{QB\left[ {\sin \left( {\frac{N\pi l}{B}} \right) - \sin \left( {\frac{N\pi d}{B}} \right)} \right]}}{{pN\pi \left\{ {\frac{{[2\pi r_{\text{w}} (l - d)]^{n} }}{{Q^{n - 1} }}K_{\text{r}} r_{\text{w}}^{1 - n} \sqrt \delta K_{{\frac{2}{3 - n}}} \left( {\frac{2}{3 - n}r_{\text{w}}^{{\frac{3 - n}{2}}} \sqrt \delta } \right) + \pi r_{\text{w}}^{{\frac{5 - n}{2}}} pK_{{\frac{1 - n}{3 - n}}} \left( {\frac{2}{3 - n}r_{\text{w}}^{{\frac{3 - n}{2}}} \sqrt \delta } \right)} \right\}}}$$

(27)

It should be pointed out that the following properties of the second kind of modified Bessel functions have been used to obtain C ₂: xdK _v (x)/dx + vK _v (x) = −xK _v−1(x), K _v (x) = K _−v (x), $K_{v} (x) \approx \frac{\varGamma (v)}{2}\left( \frac{x}{2} \right)^{ - v} ,\quad v > 0,x \to 0$. Therefore, the solution in Laplace–Fourier domain can be expressed as:

$$\bar{\hat{s}}(r,N,p) = \frac{{QB\left[ {\sin \left( {\frac{N\pi l}{B}} \right) - \sin \left( {\frac{N\pi d}{B}} \right)} \right]r^{{\frac{1 - n}{2}}} K_{{\frac{1 - n}{3 - n}}} \left( {\frac{2}{3 - n}r^{{\frac{3 - n}{2}}} \sqrt \delta } \right)}}{{pN\pi \left\{ {\frac{{[2\pi r_{\text{w}} (l - d)]^{n} }}{{Q^{n - 1} }}K_{\text{r}} r_{\text{w}}^{1 - n} \sqrt \delta K_{{\frac{2}{3 - n}}} \left( {\frac{2}{3 - n}r_{\text{w}}^{{\frac{3 - n}{2}}} \sqrt \delta } \right) + \pi r_{\text{w}}^{{\frac{5 - n}{2}}} pK_{{\frac{1 - n}{3 - n}}} \left( {\frac{2}{3 - n}r_{\text{w}}^{{\frac{3 - n}{2}}} \sqrt \delta } \right)} \right\}}}$$

(28)

With inverse Fourier transform, one can obtain the solution in Laplace domain finally:

$$\overline{s} (r,z,p) = \frac{1}{B}\overline{{\hat{s}}} (r,0,p) + \frac{2}{B}\sum\limits_{N = 1}^{\infty } {\overline{{\hat{s}}} } (r,N,p)\cos \left( {\frac{N\pi z}{B}} \right)$$

(29)