{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Response function parameters\n",
    "This notebook demonstrates the use of response function parameter calculation functions for various aquifer parameters."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "from pastas_plugins.responses.parameters import (\n",
    "    exponential_parameters,\n",
    "    hantush_parameters,\n",
    "    kraijenhoff_parameters,\n",
    "    theis_parameters,\n",
    ")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 0. Aquifer parameters"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "S = 0.1  # Storativity\n",
    "K = 10.0  # Hydraulic conductivity [L/T]\n",
    "D = 50.0  # Aquifer thickness [L]\n",
    "x = 0.0  # Location in aquifer (center at 0.0)\n",
    "L = 100.0  # Aquifer length [L]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 1. Kraijenhoff van de Leur Parameters\n",
    "Homogeneous aquifer between two parallel canals"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Kraijenhoff Parameters: A=2.5000, a=0.2026, b=0.0000\n"
     ]
    }
   ],
   "source": [
    "A, a, b = kraijenhoff_parameters(S, K, D, x, L)\n",
    "print(f\"Kraijenhoff Parameters: A={A:.4f}, a={a:.4f}, b={b:.4f}\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 2. Exponential Parameters\n",
    "Linear reservoir system"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Exponential Parameters: A=200.0000, a=20.0000\n"
     ]
    }
   ],
   "source": [
    "c = 200.0  # Drainage resistance [T]\n",
    "\n",
    "A, a = exponential_parameters(S, c)\n",
    "print(f\"Exponential Parameters: A={A:.4f}, a={a:.4f}\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 3. Hantush Parameters\n",
    "Well in a confined aquifer"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Hantush Parameters: A=-0.000773, a=50.0000, b=0.0025\n"
     ]
    }
   ],
   "source": [
    "c = 500.0  # Aquitard resistance [T]\n",
    "r = 50.0  # Distance from well [L]\n",
    "\n",
    "A, a, b = hantush_parameters(S, K, D, c, r)\n",
    "print(f\"Hantush Parameters: A={A:.6f}, a={a:.4f}, b={b:.4f}\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 4. Theis Parameters\n",
    "Well between two parallel canals (y=0)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Theis Parameters: A=-0.000159, a=0.2026, b=0.0000\n"
     ]
    }
   ],
   "source": [
    "A, a, b = theis_parameters(S, K, D, x, L)\n",
    "print(f\"Theis Parameters: A={A:.6f}, a={a:.4f}, b={b:.4f}\")"
   ]
  }
 ],
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