{ "cells": [ { "cell_type": "code", "execution_count": 2, "metadata": { "editable": true, "slideshow": { "slide_type": "" }, "tags": [ "remove-input" ] }, "outputs": [], "source": [ "import numpy as np\n", "import matplotlib.pyplot as plt\n", "from myst_nb import glue\n", "%config InlineBackend.rc = {'figure.dpi': 300}\n", "%config InlineBackend.figure_format = 'svg'\n", "import sys\n", "sys.path.append(\"../py_figures\")\n", "import diagrams" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "(content:chap1:flow_two_rivers)=\n", "# Flow between two rivers" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The first problem that is solved is for confined flow between two long and parallel rivers that fully penetrate the aquifer ({numref}`Figure %s `). The rivers are a distance $L$ apart, and there is no areal recharge ($N=0$). The rivers are in direct contact with the aquifer so that the head in the aquifer at the river is equal to the head in the river. The head at $x=0$ is $h_l$, and the head at $x=L$ is $h_r$. \n", "\n", "$$\n", "h\\vert_{x=0} = h_l\n", "$$(eq_bc1_1.1)\n", "\n", "$$\n", "h\\vert_{x=L} = h_r\n", "$$(eq_bc2_1.1)" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "tags": [ "remove-input", "remove-output" ] }, "outputs": [ { "data": { "application/papermill.record/image/svg+xml": "\n\n\n", "application/papermill.record/text/plain": "

" }, "metadata": { "scrapbook": { "mime_prefix": "application/papermill.record/", "name": "fig_solution" } }, "output_type": "display_data" }, { "data": { "image/svg+xml": [ "\n", "\n", "\n" ], "text/plain": [ "

" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "fig_solution = diagrams.solution1_1()\n", "glue(\"fig_solution\", fig_solution, display=False)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "```{glue:figure} fig_solution\n", "---\n", "name: fig_flow_two_rivers11\n", "---\n", "Flow between two rivers.\n", "```" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ " Application of the boundary conditions to the general solution {eq}`eq_hgen1.0` gives\n", "\n", "$$\n", "A=\\frac{h_r-h_l}{L}\n", "$$(eq_A_1.1)\n", "\n", "$$\n", "B=h_0\n", "$$(eq_B_1.1)\n", "\n", "so that the solution for the head becomes a straight line\n", "\n", "$$\n", "h=\\frac{(h_r-h_l)x}{L} + h_l\n", "$$(eq_hcanels1.1)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The discharge vector $Q_x$ is obtained by differentiating Eq. ({eq}`eq_hcanels1.1`) with respect to $x$ and inserting the result into Eq. ({eq}`eq_Qx1.0`)\n", "\n", "$$\n", "Q_x = -T\\frac{\\text{d}h}{\\text{d}x} = - T\\frac{h_r-h_l}{L}\n", "$$(eq_Qx1dconfined1.1)\n", "\n", "Note that $Q_x$ is constant throughout the flow field.\n", "
The solution is computed below and plotted in {numref}`Figure %s `. For the novice Python user, please be aware that the code blocks below only run when the `numpy` and `matplotlib.pyplot` packages are imported first as `np` and `plt`, respectively. This holds for virtually all code blocks in this book; see {ref}`Appendix A ` for more details. \n", "Note that the porosity $n$ and elevation $z_b$ of the bottom of the aquifer are defined but not used here. They will be used when visualizing the flow field and computing the travel time in the following." ] }, { "cell_type": "code", "execution_count": 8, "metadata": {}, "outputs": [], "source": [ "# parameters\n", "L = 1000 # aquifer length, m\n", "H = 10 # aquifer thickness, m\n", "zb = -6 # aquifer bottom, m\n", "k = 10 # hydraulic conductivity, m/d\n", "n = 0.3 # porosity, -\n", "T = k * H # transmissivity, m^2/d\n", "hl = 6 # specified head at the left boundary, m\n", "hr = 4 # specified head at the right boundary, m" ] }, { "cell_type": "code", "execution_count": 9, "metadata": {}, "outputs": [], "source": [ "# solution\n", "x = np.linspace(0, L, 100)\n", "h = (hr - hl) * x / L + hl\n", "Qx = - T * (hr - hl) / L * np.ones_like(x)" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "tags": [ "remove-output" ] }, "outputs": [ { "data": { "image/svg+xml": [ "\n", "\n", "\n" ], "text/plain": [ "

" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# basic plot\n", "plt.subplot(121)\n", "plt.plot(x, h)\n", "plt.subplot(122)\n", "plt.plot(x, Qx);" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "tags": [ "remove-input", "remove-output" ] }, "outputs": [ { "data": { "application/papermill.record/image/svg+xml": "\n\n\n", "application/papermill.record/text/plain": "

" }, "metadata": { "scrapbook": { "mime_prefix": "application/papermill.record/", "name": "fig_flow_two" } }, "output_type": "display_data" }, { "data": { "image/svg+xml": [ "\n", "\n", "\n" ], "text/plain": [ "

" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# plot\n", "fig = plt.figure(figsize=diagrams.figdef2)\n", "plt.subplot(121)\n", "plt.plot(x, h)\n", "plt.xlabel('$x$ (m)')\n", "plt.ylabel('head (m)')\n", "plt.title('(a)', loc='left')\n", "plt.yticks(np.arange(4, 6.5, 0.5))\n", "plt.grid(True)\n", "plt.subplot(122)\n", "plt.plot(x, Qx, 'C1')\n", "plt.ylim(0, 0.5)\n", "plt.xlabel('$x$ (m)')\n", "plt.ylabel('$Q_x$ (m$^2$/d)')\n", "plt.title('(b)', loc='left')\n", "plt.grid(True)\n", "plt.tight_layout()\n", "glue('fig_flow_two', fig, display=False);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "```{glue:figure} fig_flow_two\n", "---\n", "name: fig_flow_two_rivers_result11\n", "---\n", "Flow between two rivers with $N=0$; head (left) and $Q_x$ (right).\n", "```" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ " The flow field in a vertical cross-section may be visualized using the `streamplot` function of `matplotlib`, which takes as input a two-dimensional grid of points, and for each grid point, the horizontal and vertical components of the specific discharge vector. The horizontal component of the specific discharge vector $q_x$ is computed from the discharge vector $Q_x$ as\n", "\n", "$$\n", "q_x = \\frac{Q_x}{H} = -k\\frac{h_r - h_l}{L}\n", "$$(eq_q_x1.1)\n", "\n", "and does not vary in the vertical direction. The vertical component of the specific discharge vector $q_z$ equals zero. The `streamplot` function draws lines in the direction of the flow and adds arrows to indicate the direction of flow. For this case with constant flow, a small grid of 2 by 2 is sufficient. The vertical exaggeration is set with the `aspect` keyword of the `subplot` function. The vertical exaggeration is noted as “VE” in the label of the vertical axis ({numref}`Figure %s `)." ] }, { "cell_type": "code", "execution_count": 14, "metadata": {}, "outputs": [], "source": [ "# solution\n", "xg, zg = np.meshgrid(np.linspace(0, L, 2), np.linspace(zb, zb + H, 2))\n", "qx = -k * (hr - hl) / L * np.ones_like(xg)\n", "qz = np.zeros_like(xg)" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "tags": [ "remove-output" ] }, "outputs": [ { "data": { "image/svg+xml": [ "\n", "\n", "\n" ], "text/plain": [ "

" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# basic stream plot\n", "plt.subplot(111, aspect=25)\n", "plt.streamplot(xg, zg, qx, qz, color='C1', density=0.2);" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "tags": [ "remove-input", "remove-output" ] }, "outputs": [ { "data": { "application/papermill.record/image/svg+xml": "\n\n\n", "application/papermill.record/text/plain": "

" }, "metadata": { "scrapbook": { "mime_prefix": "application/papermill.record/", "name": "fig_stream" } }, "output_type": "display_data" }, { "data": { "image/svg+xml": [ "\n", "\n", "\n" ], "text/plain": [ "

" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "# plot\n", "fig = plt.figure(figsize=diagrams.figdef2)\n", "ax = plt.subplot(111)\n", "aspect = 25\n", "ax.set_aspect(aspect)\n", "plt.streamplot(xg, zg, qx, qz, color='C1', density=0.2)\n", "plt.xlabel('$x$ (m)')\n", "plt.ylabel(f'$z$ (m) - VE:{aspect}')\n", "plt.xlim(0, 1000)\n", "plt.ylim(zb, zb + H)\n", "plt.tight_layout()\n", "glue('fig_stream', fig, display=False);" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "```{glue:figure} fig_stream\n", "---\n", "name: fig_streamplot1.1\n", "---\n", "Streamplot in a vertical cross-section for flow between two rivers with $N=0$.\n", "```" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ " The (average) velocity in the horizontal direction $v_x$ is obtained through division of $q_x$ by the porosity $n$ (obviously, the velocity in the vertical direction $v_z$ is zero). For this simple problem,the (average) travel time from the left river to the right river can be computed as $L/v_x$. More formally, the travel time can be obtained from the definition of the horizontal velocity vector\n", "\n", "$$\n", "v_x = \\frac{q_x}{n} = \\frac{\\text{d}x}{\\text{d}t}\n", "$$(eq_v_x1.1)\n", "\n", "Integration gives the travel time $t_\\text{tr}$ \n", "\n", "$$\n", "\\int_0^{t_\\text{tr}} \\text{d}t = \\int_{0}^L \\frac{\\text{dx}}{v_x} = \\int_{0}^L -\\frac{Ln}{k(h_r-h_l)}\\text{d}x\n", "$$(eq_int_tr1.1)\n", "\n", "which gives\n", "\n", "$$\n", "t_\\text{tr} = \\frac{L^2n}{k(h_l-h_r)} = \\frac{L}{v_x}\n", "$$(eq_t_tr1.1)" ] }, { "cell_type": "code", "execution_count": 19, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "mean velocity: 0.0667 m/d\n", "travel time from left river to right river: 15000 days\n" ] } ], "source": [ "# travel time\n", "vx = -k * (hr - hl) / (n * L)\n", "print(f'mean velocity: {vx:.4f} m/d')\n", "print(f'travel time from left river to right river: {L / vx:.0f} days')" ] } ], "metadata": { "celltoolbar": "Tags", "kernelspec": { "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.13.7" }, "latex_envs": { "LaTeX_envs_menu_present": true, "autoclose": false, "autocomplete": true, "bibliofile": "biblio.bib", "cite_by": "apalike", "current_citInitial": 1, "eqLabelWithNumbers": true, "eqNumInitial": 1, "hotkeys": { "equation": "Ctrl-E", "itemize": "Ctrl-I" }, "labels_anchors": false, "latex_user_defs": false, "report_style_numbering": false, "user_envs_cfg": false }, "varInspector": { "cols": { "lenName": 16, "lenType": 16, "lenVar": 40 }, "kernels_config": { "python": { "delete_cmd_postfix": "", "delete_cmd_prefix": "del ", "library": "var_list.py", "varRefreshCmd": "print(var_dic_list())" }, "r": { "delete_cmd_postfix": ") ", "delete_cmd_prefix": "rm(", "library": "var_list.r", "varRefreshCmd": "cat(var_dic_list()) " } }, "types_to_exclude": [ "module", "function", "builtin_function_or_method", "instance", "_Feature" ], "window_display": false } }, "nbformat": 4, "nbformat_minor": 4 }