{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Test line-sinks with image wells" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "import matplotlib.pyplot as plt\n", "import numpy as np\n", "\n", "import timflow.transient as tft" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "Consider a well pumping in a phreatic aquifer with $S_y=0.1$. The hydraulic conductivity of the aquifer is 10 m/d and the saturated thickness may be approximated as constant and equal to 20 m. The well is located at $(x,y)=(0,0)$. The discharge of the well is 1000 m$^3$/d and the radius is 0.3 m. there is a very long river with a fixed head located along the line $x=50$ m. The head is computed at $(x,y)=(20,0)$ for the first 20 days after the well starts pumping. The solution for an image well is compared to the solution using a `RiverString` element of different lengths." ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "ml1 = tft.ModelMaq(kaq=10, z=[20, 0], Saq=[0.1], phreatictop=True, tmin=0.001, tmax=100)\n", "w1 = tft.Well(ml1, 0, 0, rw=0.3, tsandQ=[(0, 1000)])\n", "w2 = tft.Well(ml1, 100, 0, rw=0.3, tsandQ=[(0, -1000)])\n", "ml1.solve()\n", "t = np.linspace(0.1, 20, 100)\n", "h1 = ml1.head(20, 0, t)\n", "plt.plot(t, h1[0], label=\"river modeled with image well\")\n", "plt.xlabel(\"time (d)\")\n", "plt.ylabel(\"head (m)\");" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "plt.plot(t, h1[0], label=\"river modeled with image well\")\n", "for ystart in [-50, -100]:\n", " ml2 = tft.ModelMaq(\n", " kaq=10, z=[20, 0], Saq=[0.1], phreatictop=True, tmin=0.001, tmax=100\n", " )\n", " w = tft.Well(ml2, 0, 0, rw=0.3, tsandQ=[(0, 1000)])\n", " yls = np.arange(ystart, -ystart + 1, 20)\n", " xls = 50 * np.ones(len(yls))\n", " lss = tft.RiverString(ml2, xy=list(zip(xls, yls, strict=False)), tsandh=\"fixed\")\n", " ml2.solve()\n", " h2 = ml2.head(20, 0, t)\n", " plt.plot(t, h2[0], \"--\", label=f\"line-sink string from {ystart} to {-ystart}\")\n", "plt.title(\"head at (x,y)=(20,0)\")\n", "plt.xlabel(\"time (d)\")\n", "plt.ylabel(\"head (m)\")\n", "plt.legend();" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "The solution is repeated for the case where there is a long impermeable wall along $x=50$ m rather than a river." ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "ml1 = tft.ModelMaq(kaq=10, z=[20, 0], Saq=[0.1], phreatictop=True, tmin=0.001, tmax=100)\n", "w1 = tft.Well(ml1, 0, 0, rw=0.3, tsandQ=[(0, 1000)])\n", "w2 = tft.Well(ml1, 100, 0, rw=0.3, tsandQ=[(0, 1000)])\n", "ml1.solve()\n", "t = np.linspace(0.1, 20, 100)\n", "h1 = ml1.head(20, 0, t)\n", "plt.plot(t, h1[0], label=\"impermeable wall modeled with image well\")\n", "plt.xlabel(\"time (d)\")\n", "plt.ylabel(\"head (m)\");" ] }, { "cell_type": "code", "execution_count": null, "metadata": {}, "outputs": [], "source": [ "plt.plot(t, h1[0], label=\"river modeled with image well\")\n", "for ystart in [-100, -200, -400]:\n", " ml2 = tft.ModelMaq(\n", " kaq=10, z=[20, 0], Saq=[0.1], phreatictop=True, tmin=0.001, tmax=100\n", " )\n", " w = tft.Well(ml2, 0, 0, rw=0.3, tsandQ=[(0, 1000)])\n", " yls = np.arange(ystart, -ystart + 1, 20)\n", " xls = 50 * np.ones(len(yls))\n", " lss = tft.LeakyWallString(ml2, xy=list(zip(xls, yls, strict=False)), res=\"imp\")\n", " ml2.solve()\n", " h2 = ml2.head(20, 0, t)\n", " plt.plot(t, h2[0], \"--\", label=f\"line-doublet string from {ystart} to {-ystart}\")\n", "plt.title(\"head at (x,y)=(20,0)\")\n", "plt.xlabel(\"time (d)\")\n", "plt.ylabel(\"head (m)\")\n", "plt.legend();" ] } ], "metadata": { "language_info": { "name": "python" } }, "nbformat": 4, "nbformat_minor": 4 }