Difference between revisions of "Library of analytic element solutions (under construction)"
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== Transient Phreatic (Unconfined) groundwater == | == Transient Phreatic (Unconfined) groundwater == |
Revision as of 12:19, 5 December 2007
Contents
analytical solutions (Bruggeman,1999 classification)
Steady State Confined/Unconfined Flow
Governing Equation (1D or 2D):
<math>\nabla \cdot \left(k\phi \nabla h\right)=-N </math>
Where <math>\phi</math> is the saturated thickness of the aquifer [<math>L</math>], <math>N</math> is the vertical influx to the aquifer (recharge/leakage) [<math>LT^{-1}</math>], and <math>h</math> [<math>L</math>] is the head measured with the respect to a flat aquifer base.
A common transformation (the Girinskii potential) can be used to simplified the governing equation, and make it equally valid for both confined flow (<math>\phi=H</math>) and unconfined flow (<math>\phi=h</math>):
<math> \Phi=\frac{1}{2}k h^2 </math>
if <math>h<H</math>, and
<math> \Phi=kHh-\frac{1}{2}kH^2 </math>
if <math>h>H</math>. This results in the Laplace/Poisson equation:
<math>\nabla ^2\Phi=-N </math>
The integrated discharge vector may be obtained from the gradient of the potential.
One-dimensional horizontal flow
Governing Equation:
<math>\frac{\partial^2 \Phi}{\partial x^2}=-N</math>
Solution #1: Steady-state / Two Dirichlet Boundaries
Boundary conditions: <math>\Phi(0)=\Phi_1; \frac{}{} \Phi(L)=\Phi_2</math> Solution: <math>\Phi(x)=-\frac{1}{2}Nx^2+\left(\frac{\Phi_2-\Phi_1}{L}+\frac{1}{2}NL\right)x+\Phi_1</math>
Solution #2: Steady-state / One Dirichlet, One Neumann Boundary
Boundary conditions: <math>\Phi(0)=\Phi_1; \frac{}{} Q_x(L)=\beta</math> Solution: <math>\Phi(x)=-\frac{1}{2}Nx^2+\left(-\beta+NL\right)x+\Phi_1</math>
Two-dimensional radially-symmetric horizontal flow
Governing Equation:
<math>\frac{\partial}{\partial r}\left(r\frac{\partial \Phi}{\partial r}\right)=-N</math>
Solution #1 (Steady State Thiem Solution):
Boundary conditions: <math>\underset{r\rightarrow 0}{\lim}\int_0^{2\pi} rkh\frac{\partial h}{\partial r}d\theta=\underset{r\rightarrow 0}{\lim}\int_0^{2\pi} rQ_rd\theta=Q</math> (i.e., the total flow out at r=0 is equal to the total flow from the well, <math>Q</math>) <math>\Phi(R)=\Phi_0\frac{}{}</math> Solution: <math>\Phi(r)=-\frac{Q}{2\pi}\ln\left(\frac{r}{R}\right)+\Phi_0</math>
Two-dimensional general horizontal flow
Governing Equation:
<math>\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial y^2}=-N</math>
Transient Phreatic (Unconfined) groundwater
Governing Equation (1D or 2D):
<math>\nabla \cdot \left(kh \nabla h\right)=S_y\frac{\partial h}{\partial t}-N </math>
Where <math>S_y</math> is the specific yield [-], <math>N</math> is the vertical influx to the aquifer (recharge/leakage) [<math>LT^{-1}</math>], and <math>h</math> [<math>L</math>] is both the head and the saturated thickness (i.e., the head is measured with the respect to a flat aquifer base).
Exact solutions
General two-dimensional horizontal flow
Transient Confined groundwater
Governing Equation (1D or 2D):
<math>\nabla \cdot \left(kH \nabla h\right)=S_sH\frac{\partial h}{\partial t}-N </math>
Where <math>S_s</math> is the specific storage [-], <math>N</math> is the vertical influx to the aquifer (recharge/leakage) [<math>LT^{-1}</math>], <math>H</math> is the saturated thickness of the aquifer and <math>h</math> [<math>L</math>] is the head, measured with respect to an arbitrary datum.
For piecewise constant properties, we can substitute in the confined Girinskii potential (<math>\Phi=kHh</math>) to obtain the following governing equation:
<math>\nabla ^2\Phi=\frac{1}{\alpha}\frac{\partial \Phi}{\partial t}-N </math>
Where <math>\alpha=\frac{k}{S_s}</math> is the hydraulic diffusivity [<math>LT^{-1}</math>]of the aquifer.
One-dimensional flow
Two-dimensional flow
Two-dimensional radially-symmetric flow
Solution #1 (Transient Theis Solution):
Boundary/initial conditions: <math>\Phi(r,0)=\Phi_0\frac{}{}</math> <math>\underset{r\rightarrow 0}{\lim}\int_0^{2\pi} rkH\frac{\partial h}{\partial r}d\theta=\underset{r\rightarrow 0}{\lim}\int_0^{2\pi} rQ_rd\theta=QH(t)</math> (i.e., the total flow out at r=0 is equal to the total flow from the well, <math>Q</math>). Here, <math>H(t)</math> is the heaviside function, equal to 1 for <math>t>0</math>, zero otherwise. <math>\underset{r\rightarrow \infty}{\lim}\Phi(r)=\Phi_0\frac{}{}</math> Solution: <math>\Phi(r,t) = \Phi_0-\frac{Q}{4\pi}E_1\left(\frac{r^2 S_s}{4kt}\right)</math> Where <math>E_1()</math> is the exponential integral (a.k.a. the "well function")