uses also \require{cancel}

Hagen_5.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Ü 5.1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Für die Reaktion $A \\rightarrow R$ ist k = 0,02 $min^{-1}$. Es sollen in 10 h 4752 mol R produziert werden, $U_A = 0,99$. Die Rüstzeit beträgt 1,16 h, $c_{A,0} = 8 mol/L$. Welches Reaktions-\n",
    "volumen wird für einen diskontinuierlich betriebenen Rührkessel benötigt?\n",
    "\n",
    "dnA/dt = (-1) r Vr\n",
    "\n",
    "dnR/dt = (+1) r Vr\n",
    "\n",
    "(nA0 - nA)/nA0 = UA = 1-nA/nA0\n",
    "\n",
    "nA = (1-UA)nA0\n",
    "\n",
    "\n",
    "-nA0 dUA/dt = (-1) r Vr\n",
    "\n",
    "\n",
    "dUA/dt = k (nA/Vr) Vr/nA0\n",
    "\n",
    "dUA/dt = k (1-UA)\n",
    "\n",
    "d(1-UA)/dt = -dUA/dt = -k (1-UA)\n",
    "\n",
    "1/(1-UA) d(1-UA) = -k dt\n",
    "\n",
    "ln(1-UA) + ln(1-0) = -k tau\n",
    "\n",
    "1-UA = exp(-k tau)\n",
    "\n",
    "UA = 1-exp(-k tau)\n",
    "\n",
    "\n",
    "tau = -ln(1-UA)/k\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$A \\rightarrow P$\n",
    "\n",
    "$V_R = (t_R - t_V)\\frac{\\dot{n_p} (-\\nu_a)}{U_A S_P C_{A0} \\nu_P}$\n",
    "\n",
    "$S_P = \\frac{\\dot{n_P} (-\\nu_A)}{(n_{A,0}-n_A)\\nu_P}$\n",
    "\n",
    "$U_A = \\frac{n_{A,0}-n_A}{n_{A,0}} = \\frac{n_{A,0}-n_{A,0}-\\nu_A\\xi_1}{n_{A,0}} $\n",
    "\n",
    "$A_P = \\frac{n_P/\\nu_P}{n_A/(-\\nu_A)} = U_A S_P$\n",
    "\n",
    "==>> $V_R = (t_R + t_V)\\frac{\\dot{n_P} (-\\nu_A)}{\\left(\\frac{n_P/\\nu_P}{n_A/(-\\nu_A)} \\right)} = (t_R + t_V)\\left( \\frac{\\dot{n_P}}{n_P} \\right) \\left( \\frac{\\dot{n_{A,0}}}{c_{A,0}} \\right)$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "tR = 3.83764h\n",
      "tV = 1.16h\n",
      "UA = 0.99\n",
      "Vr= 299.859L\n",
      "nbatches=2.00094\n"
     ]
    }
   ],
   "source": [
    "import numpy as np\n",
    "ua = 0.99\n",
    "tau = -np.log(1-ua)/0.02 # min\n",
    "tv = 1.16*60\n",
    "ndotp = 4752/(10*60.0)\n",
    "vr = (tau+tv)*ndotp/(8*0.99) # L\n",
    "nbatches = 10*60/(tau+tv)\n",
    "print ('tR = ' + '{0:g}'.format(tau / 60.0) + 'h')\n",
    "print ('tV = ' + '{0:g}'.format(1.16) + 'h')\n",
    "print ('UA = ' + '{0:g}'.format(ua))\n",
    "print ('Vr= ' + '{0:g}'.format(vr) + 'L')\n",
    "print ('nbatches=' + '{0:g}'.format(nbatches))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Ü 5.3\n",
    "\n",
    "$A(g) \\rightarrow 3 B(g) \\hspace{2cm} -r_A = (0,6min^{-1}) C_A$\n",
    "\n",
    "$\\dot{V_0} C_{A,0} = 9mol/min$, $C_{A,0}=0,5mol/L$ , $U_A = 0,667$ \n",
    "\n",
    "==>> $\\dot{V_0} = \\dot{V_0} C_{A,0}/C_{A,0} = (9/0,5) L/min$\n",
    "\n",
    "**Mit Volumenveränderung**\n",
    "\n",
    "$Z=\\frac{P V}{n R T}$\n",
    "\n",
    "$\\frac{\\dot{V}}{\\dot{V_0}}=\\left(\\frac{\\dot{n}}{\\dot{n_0}}\\right)\\left(\\frac{T}{T_0}\\right)\\left(\\frac{P_0}{P}\\right)\\left(\\frac{Z}{Z_0}\\right)$\n",
    "\n",
    "$\\frac{\\dot n}{\\dot n_0}=\\frac{n_0 + (\\sum_i{\\nu_i})\\xi_1}{n_0} = 1 + \\frac{(\\sum_i{\\nu_i})\\xi_1}{n_0}$\n",
    "\n",
    "$U_A=\\frac{n_{A,0} - n_{A,0} - \\nu_A \\xi_1}{n_{A,0}} \\rightarrow \\xi_1=U_A\\frac{n_{A,0}}{(-\\nu_A)}$ \n",
    "\n",
    "$\\frac{\\dot n}{\\dot n_0}= 1 + \\frac{(\\sum_i{\\nu_i})}{(-\\nu_A)}\\frac{n_{A,0}}{n_0}U_A= 1 + \\frac{(\\sum_i{\\nu_i})}{(-\\nu_A)}x_{A,0}U_A$\n",
    "\n",
    "$\\epsilon_A \\equiv \\frac{(\\sum_i{\\nu_i})}{(-\\nu_A)}$\n",
    "\n",
    "Vorausgesetzt, dass das Zulaufstrom und Auslassstrom konstant bleiben (somit auch das Volumen des Reaktors). Im zeitunabhängigen Zustand sollen beide Ströme folgende Werte vertreten: Zulaufstrom $\\dot{V_0}(1+\\epsilon_A U_A)$, Auslassstrom $\\dot{V_0}(1+\\epsilon_A U_A)$. Wie ist es dazu gekommen, dass das Zulaufstrom wächst, ohne dass die Temperatur ansteigt?. Vielleicht sind variabler Druck oder variable Temperatur im Zulaufstrom für diesen Ansatz notwendig. Sonst müsste man ein unterschiedliches Volumen im Zulaufstrom eingeben, als man im Auslassstrom hinauszieht: $\\dot{V_0}=18L/min$, $\\dot{V_0}(1+\\epsilon_A U_A)=42.012L/min$.\n",
    "\n",
    "$\\frac{d n_A}{d t} = 0 = \\dot{n_{A,0}} - \\dot{n_A} +  \\nu_A \\times r \\times V_R$\n",
    "\n",
    "$0 = \\dot{V} C_{A,0} - \\dot{V} C_A   +  \\nu_A \\times r \\times V_R = \\dot{V_0} (1 + \\epsilon_A x_{A,0} U_A) U_A C_{A,0} +  \\nu_A \\times r \\times V_R$\n",
    "\n",
    "$V_R = \\frac{\\dot{V_0} U_A C_{A,0} (1 + \\epsilon_A U_A)}{(-\\nu_A)k C_A} = \\frac{\\dot{V_0} U_A }{(-\\nu_A) k \\frac{1-U_A}{1 + \\epsilon_A x_{A,0} U_A}}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Vr= 140.25L\n",
      "tau= 7.79168min\n"
     ]
    }
   ],
   "source": [
    "vdot0 = 9/0.5 # L/min\n",
    "vr = 0.667/(1-0.667)*(1+(+3-1)/(-(-1))*0.667)*vdot0/(\n",
    "    -(-1.0)*0.6) # L\n",
    "tau = vr/vdot0\n",
    "print 'Vr= ' + '{0:g}'.format(vr) + 'L'\n",
    "print 'tau= ' + '{0:g}'.format(tau) + 'min'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Ü 5.5\n",
    "$4 PH_3(g) \\rightarrow P_4(g) + 6 H_2(g)$\n",
    "\n",
    "A: $PH_3(g)$\n",
    "\n",
    "$-r_{PH_3} = \\nu_{PH_3} r_1 = 10 h^{-1} C_{PH_3}$\n",
    "\n",
    "Hagen: $\\dot{n_A} = (\\dot{n_A} + d \\dot{n_A}) - r_A dV$\n",
    "\n",
    "Vorgeschlagen: $\\dot{n_A} = (\\dot{n_A} + d \\dot{n_A}) + r_A dV$ \n",
    "\n",
    "$\\hspace{2cm} 0 = (\\{\\text{Zulaufstrom}\\} - \\{\\text{Auslassstrom}\\}) + \\{\\text{Durch chemische Reaktion verwendete oder gebildete Stoffmengenstrom}\\}$\n",
    "\n",
    "$\\frac{d \\dot{n_A}}{dV} = \\nu_{A1} r_1$\n",
    "\n",
    "Sonst wäre Abbildung 5-8 gar nicht möglich, denn die Steigerung $\\frac{d U_A}{dV}$ wäre negativ.\n",
    "\n",
    "$\\frac{d \\dot{n_A}}{dV} = \\frac{d \\dot{n}_{A,0}}{dV}(1-U_A) = -\\dot{n}_{A,0}\\frac{d U_A}{dV} = - 10 h^{-1} c_A$\n",
    "\n",
    "$c_A = \\dot{n_A}/\\dot{V} = (\\dot{n}_{A,0}/\\dot{V}) \\times (1-U_A) = (\\dot{n}_{A,0}/\\dot{V_0}) \\times \\frac{1-U_A}{1+\\epsilon_A x_{A,0} U_A}$ \n",
    "\n",
    "$\\int_{0}^{V_R}dV = V_R = \\int_{0}^{U_A} \\frac{\\dot{n}_{A,0}}{k_1 c_{A,0}} \\times \\frac{1+\\epsilon_A x_{A,0} U_A}{1-U_A} dU_A $\n",
    "\n",
    "$= \\frac{\\dot{n}_{A,0}}{k_1 c_{A,0}} \\times(-(1+\\epsilon_A x_{A,0})ln(1-U_A)-\\epsilon_A x_{A,0}U_A)$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
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      "text/latex": [
       "$$\\frac{a x + c}{b x + d} = - \\frac{a d}{b \\left(b x + 1\\right)} + \\frac{b x \\frac{a}{b}}{b x + 1} + \\frac{c}{b x + d} + \\frac{a d}{b \\left(b x + 1\\right)} = \\frac{a}{b} + \\frac{- \\frac{a d}{b} + c}{b x + d}$$"
      ],
      "text/plain": [
       "                               a                                  a⋅d    \n",
       "                           b⋅x⋅─                                - ─── + c\n",
       "a⋅x + c         a⋅d            b       c          a⋅d       a      b     \n",
       "─────── = - ─────────── + ─────── + ─────── + ─────────── = ─ + ─────────\n",
       "b⋅x + d     b⋅(b⋅x + 1)   b⋅x + 1   b⋅x + d   b⋅(b⋅x + 1)   b    b⋅x + d "
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "simplify back\n"
     ]
    },
    {
     "data": {
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      "text/latex": [
       "$$\\frac{a}{b} + \\frac{- \\frac{a d}{b} + c}{b x + d} = \\frac{a x + c}{b x + d}$$"
      ],
      "text/plain": [
       "      a⋅d              \n",
       "    - ─── + c          \n",
       "a      b        a⋅x + c\n",
       "─ + ───────── = ───────\n",
       "b    b⋅x + d    b⋅x + d"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "data": {
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      "text/latex": [
       "$$\\frac{\\partial}{\\partial x}\\left(\\frac{a x}{b} + \\frac{1}{b} \\left(- \\frac{a d}{b} + c\\right) \\log{\\left (b x + d \\right )}\\right) = \\frac{a x + c}{b x + d}$$"
      ],
      "text/plain": [
       "  ⎛      ⎛  a⋅d    ⎞             ⎞          \n",
       "  ⎜      ⎜- ─── + c⎟⋅log(b⋅x + d)⎟          \n",
       "∂ ⎜a⋅x   ⎝   b     ⎠             ⎟   a⋅x + c\n",
       "──⎜─── + ────────────────────────⎟ = ───────\n",
       "∂x⎝ b               b            ⎠   b⋅x + d"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import sympy\n",
    "from IPython.display import display\n",
    "from sympy import UnevaluatedExpr\n",
    "from sympy import oo\n",
    "sympy.init_printing()\n",
    "x, a, b, c, d = sympy.symbols('x, a, b, c, d')\n",
    "display(sympy.Eq(sympy.Eq(\n",
    "    (c+a*x)/(d+b*x), c/(d+b*x) + UnevaluatedExpr(\n",
    "        a/b)*b*x/(1+b*x) + UnevaluatedExpr(\n",
    "        a/b*d/(1+b*x) \n",
    "        )-a/b*d/(1+b*x)), (c-a/b*d)/(d+b*x) + a/b\n",
    "))\n",
    "print('simplify back')\n",
    "display(sympy.Eq((c-a/b*d)/(d+b*x) + a/b, \n",
    "                 sympy.simplify((c-a/b*d)/(d+b*x) + a/b)))\n",
    "display(sympy.Eq(\n",
    "    UnevaluatedExpr(sympy.Derivative(\n",
    "        (c-a/b*d)*1/b*sympy.log(d+b*x)+a/b*x,x)),\n",
    "    sympy.simplify(sympy.diff(\n",
    "        (c-a/b*d)*1/b*sympy.log(d+b*x)+a/b*x,x))\n",
    "))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Vr= 7.39804L\n"
     ]
    }
   ],
   "source": [
    "ca0 = 4.6*1e5/101325.0/(0.08207*(650+273.15)) # mol(A)/L\n",
    "ua = 0.80\n",
    "nadot0 = 2.0 # mol(A)/h\n",
    "k1 = 10.0 # h^-1\n",
    "eps_a = (-4+1+6)/(-(-4.0))\n",
    "vr = nadot0/(k1*ca0)*(\n",
    "    -(1+eps_a*1.0)*np.log(1-ua)-eps_a*1.0*ua)\n",
    "print('Vr= ' + '{0:g}'.format(vr) + 'L')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Ü 5.6\n",
    "Gasphasencracking bei 800°C, 6 bar:\n",
    "\n",
    "$C_2 H_6 \\rightarrow C_2H_4 + H_2$\n",
    "\n",
    "A: $C_2H_6$\n",
    "\n",
    "B: $C_2H_4$\n",
    "\n",
    "Bedingungen: Reaktion 1. Ordnung, k = $3,07 s^{–1}$ (800 °C). Es sollen 300000 t Ethylen/Jahr (1 Jahr = 300 Tage) hergestellt werden bei einem Umsatz von Ethan= 0,80.\n",
    "Berechne das notwendige Volumen eines Strömungsrohrreaktors für die genannte Reaktion; es gilt die ideale Gasgleichung.\n",
    "\n",
    "**Lösung**\n",
    "\n",
    "$Z=\\frac{P V}{n R T}$\n",
    "\n",
    "$\\frac{\\dot{V}}{\\dot{V_0}}=\\left(\\frac{\\dot{n}}{\\dot{n_0}}\\right)\\left(\\frac{T}{T_0}\\right)\\left(\\frac{P_0}{P}\\right)\\left(\\frac{Z}{Z_0}\\right)$\n",
    "\n",
    "$\\frac{\\dot n}{\\dot n_0}=\\frac{n_0 + (\\sum_i{\\nu_i})\\xi_1}{n_0} = 1 + \\frac{(\\sum_i{\\nu_i})\\xi_1}{n_0}$\n",
    "\n",
    "$Z=1$, isothermisch, isobarisch\n",
    "\n",
    "$\\frac{\\dot{V}}{\\dot{V_0}}=\\left(\\frac{\\dot{n}}{\\dot{n_0}}\\right)=1+(\\sum_i{\\nu_i})\\left(\\frac{\\dot{n}_B - \\dot{n}_{B,0}}{\\nu_B \\dot n_0}\\right)$\n",
    "\n",
    "$\\frac{d \\dot{n_B}}{dV} = \\nu_{B} r_1 = \\nu_{B} k_1 c_A$\n",
    "\n",
    "$c_A$ als funktion des Stoffmengenstromes $\\dot n_B$\n",
    "\n",
    "$c_A = \\frac{\\dot n_A}{\\dot V} = \\frac{p_A}{R T}$, \n",
    "$c_B = \\frac{\\dot n_B}{\\dot V} = \\frac{p_B}{R T}$\n",
    "\n",
    "$c_A + c_B + c_C = \\frac{p_A}{R T} + \\frac{p_B}{R T} + \\frac{p_C}{R T} = \\frac{P}{R T}$\n",
    "\n",
    "$c_A = \\frac{P}{R T} - \\frac{\\dot n_B}{\\dot V} - \\frac{\\dot n_C}{\\dot V}=\\frac{P}{R T} - \\frac{(\\dot {n}_B+\\dot {n}_C)}{\\dot V_0}\\times \\frac{1}{1+(\\sum_i{\\nu_i})\\left(\\frac{\\dot{n}_B - \\dot{n}_{B,0}}{\\nu_B \\dot n_0}\\right)}$\n",
    "\n",
    "Nach Ersatz:\n",
    "\n",
    "$\\frac{d \\dot{n_B}}{dV} = \\nu_{B} k_1 \\frac{P}{R T} -  \\nu_{B} k_1\\frac{\\dot {n}_B+\\dot {n}_C}{\\dot V_0}\\times \\frac{1}{1+(\\sum_i{\\nu_i})\\left(\\frac{\\dot{n}_B - \\dot{n}_{B,0}}{\\nu_B \\dot n_0}\\right)}$\n",
    "\n",
    "$V_R = \\int_{0}^{V_R} 1 dV = \\int_{0}^{\\dot {n}_B} \\frac{1 - \\frac{\\sum_i \\nu_i}{\\nu_B}\\frac{\\dot {n}_{B,0}}{\\dot {n}_0}+\\frac{\\sum_i \\nu_i}{\\nu_B}\\frac{\\dot {n}_{B}}{\\dot {n}_0}}{\\nu_B k_1 \\frac{P}{RT} \\times \\left(1+\\frac{\\sum_i \\nu_i}{\\nu_B}\\frac{\\dot {n}_{B}}{\\dot {n}_0}-\\frac{\\sum_i \\nu_i}{\\nu_B}\\frac{\\dot {n}_{B,0}}{\\dot {n}_0} -\\frac{RT}{P V_0}\\dot{n}_B -\\frac{RT}{P V_0}\\dot{n}_C  \\right) }d \\dot {n}_{B}$\n",
    "\n",
    "$V_R = \\frac{R T}{P}\\frac{1}{\\nu_B k_1} \\int_{0}^{\\dot {n}_B} \\frac{\\left(1 - \\frac{\\sum_i \\nu_i}{\\nu_B}\\frac{\\dot {n}_{B,0}}{\\dot {n}_0}\\right)+\\frac{\\sum_i \\nu_i}{\\nu_B}\\frac{1}{\\dot {n}_0}\\dot {n}_{B}}{\\left(1-\\frac{\\sum_i \\nu_i}{\\nu_B}\\frac{\\dot {n}_{B,0}}{\\dot {n}_0}\\right)-\\left(1-\\frac{\\sum_i \\nu_i}{\\nu_B} \\right)\\frac{1}{\\dot {n}_0} \\dot{n}_B - \\frac{1}{\\dot {n}_0} \\dot{n}_C }d \\dot {n}_{B}$\n",
    "\n",
    "$\\frac{\\sum_i \\nu_i}{\\nu_B}=\\frac{-1+1+1}{+1} = 1$\n",
    "\n",
    "$\\dot {n}_{B,0}=0$\n",
    "\n",
    "$\\dot {n}_{C,0}=0$\n",
    "\n",
    "$\\dot {n}_0=\\dot {n}_{A,0}+\\dot {n}_{B,0}+\\dot {n}_{C,0}=\\dot {n}_{A0}$\n",
    "\n",
    "$U_A = \\frac{\\dot {n}_{A,0}-\\dot{n}_A}{\\dot{n}_{A,0}} =\\frac{\\dot {n}_{A,0}-\\dot{n}_{A,0} -\\nu_A \\xi_1 }{\\dot{n}_{A,0}} =  \\frac{(-\\nu_A)}{\\nu_B}\\frac{\\dot {n}_B - \\dot {n}_{B,0}}{\\dot{n}_{A,0}}$\n",
    "\n",
    "$\\Longrightarrow \\dot{n}_{A,0} = \\dot{n}_{0} = \\frac{(-\\nu_A)}{\\nu_B}\\frac{\\dot {n}_B -0}{U_A}$\n",
    "\n",
    "$\\dot {n}_C=\\dot {n}_{C,0} + \\nu_C \\xi_1$\n",
    "\n",
    "$\\dot {n}_B=\\dot {n}_{B,0} + \\nu_B \\xi_1$\n",
    "\n",
    "$\\Longrightarrow \\dot {n}_C=\\frac{\\nu_C}{\\nu_B} \\dot {n}_B$\n",
    "\n",
    "$V_R = \\frac{R T}{P}\\frac{1}{\\nu_B k_1} \\int_{0}^{\\dot {n}_B} \\frac{1+\\frac{\\sum_i \\nu_i}{\\nu_B}\\frac{1}{\\dot {n}_0}\\dot {n}_{B}}{1 - \\frac{\\nu_C}{\\nu_B}\\frac{1}{\\dot {n}_0} \\dot {n}_B }d \\dot {n}_{B} = \\frac{R T}{P}\\frac{1}{\\nu_B k_1} \\left( -\\frac{\\sum_i \\nu_i}{\\nu_C} \\dot {n}_B - \\dot {n}_0 \\frac{\\nu_B}{\\nu_C}(1+\\frac{\\sum_i \\nu_i}{\\nu_C} ) \\times ln(1-\\frac{\\nu_C}{\\nu_B}\\frac{1}{\\dot {n}_0} \\dot {n}_B) \\right)$\n",
    "\n",
    "$V_R = \\frac{R T}{P}\\frac{\\dot {n}_B}{\\nu_B k_1} \\left( -\\frac{\\sum_i \\nu_i}{\\nu_C}  - \\frac{(-\\nu_A)}{\\nu_C} \\frac{1}{U_A}(1+\\frac{\\sum_i \\nu_i}{\\nu_C} ) \\times ln(1-\\frac{\\nu_C}{(-\\nu_A)}U_A) \\right)$\n",
    "\n",
    "$V_R =\\frac{R T}{P}\\frac{\\dot {n}_B}{\\nu_B k_1} \\left( -1  - \\frac{1}{U_A}(1+1) \\times ln(1-U_A) \\right)$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Vr = 6053.86L\n",
      "ca0 = na/V = P/RT = 0.0672482mol/L\n",
      "na0 = nb/U_A nua/nub = 516.7mol/L\n"
     ]
    }
   ],
   "source": [
    "nb = 300000.0 * 1e6 /(12.0*2+1.0*4) / (300.*24.*60.*60.) # molB/s\n",
    "ua = 0.80\n",
    "p = 6.0 # bar\n",
    "r = 8.314/100.0 # L kPa/(mol K) * 1bar/(100kPa)\n",
    "t = 800 + 273.15 # K\n",
    "k1 = 3.07 # s^-1\n",
    "nub = +1.\n",
    "nua = -1.\n",
    "nu = +1.+1.-1.\n",
    "vr = nb * r * t / (p * k1 * nub) * (\n",
    "    -1 - 1/ua * (1+1) * np.log( 1 - ua)\n",
    ")\n",
    "print ('Vr = ' + '{0:g}'.format(vr) + 'L')\n",
    "print ('ca0 = na/V = P/RT = ' + '{0:g}'.format(p / (r * t)) + 'mol/L')\n",
    "print ('na0 = nb/U_A nua/nub = ' + '{0:g}'.format(nb/ua) + 'mol/L')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Ü 5.7 [5]\n",
    "\n",
    "Die irreversible Zersetzung von Di-tert.-Butylperoxid wird in einem isotherm betriebenen Strömungsrohr ohne Druckverlust betrieben; $A \\rightarrow B + 2 C$. Der Einsatzstrom besteht aus  reinem  Peroxid und inertem  Stickstoff.  Das Reaktorvolumen ist  200  L, der Eintrittsstrom beträgt 10  L/min.  Die  Reaktion verläuft bezüglich A nach  1.  Ordnung, $k = 0,08 min^{-1}$.\n",
    "\n",
    "a)  Ermittle  den  Umsatz  als  Funktion  des  Reaktorvolumens  für  einen  Zulaufstrom von reinem  A mit einer  Konzentration von  1,0 mol/L.  Vergleiche den \n",
    "Umsatzverlauf,  wenn  der  Zulaufstrom  nur  5 %  A  enthält,  der  Rest  besteht aus  Stickstoff.\n",
    "\n",
    "b)  Vergleiche  die  Ergebnisse  für  eine  Reaktion  $3A  \\rightarrow B$  mit  der  selben \n",
    "Geschwindigkeitskonstante, alle anderen Bedingungen bleiben gleich."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Lösung**\n",
    "\n",
    "$\\frac{d \\dot {n}_A}{d V} = -k_1 c_A$\n",
    "\n",
    "$\\dot {n}_A = \\dot {n}_{A, 0} (1-U_A) = n_{A,0} + \\nu_A \\xi_1$\n",
    "\n",
    "$\\dot {V} = \\frac{R T}{P} \\sum_i{n_i} =  \\frac{R T}{P} \\sum_i(n_{i,0} + \\nu_{i,1} \\xi_1) = \\frac{R T}{P} (n_0 + \\frac{\\sum_i(\\nu_{i,1})}{-\\nu_A} U_A n_{A,0} ) =\\dot {V}_0 (1 + \\frac{\\sum_i(\\nu_{i,1})}{-\\nu_A} x_{A,0} U_A )$\n",
    "\n",
    "$\\epsilon_A \\equiv \\frac{\\sum_i(\\nu_{i,1})}{-\\nu_A}$\n",
    "\n",
    "$-\\frac{d U_A}{d V} = -k_1 \\frac{\\dot {n}_A}{\\dot {V}}  = \\frac{\\dot {n}_{A,0}}{\\dot {V}_0}\\frac{1 - U_A}{1 + \\epsilon_A x_{A,0} U_A}$\n",
    "\n",
    "$V_R = \\frac{\\dot {V}_0}{k_1}(-\\epsilon_A x_{A,0} U_A - (1+\\epsilon_A x_{A,0})ln(1-U_A))$\n",
    "\n",
    "$0 = k_1\\frac{V_R}{\\dot {V}_0} +\\epsilon_A x_{A,0} U_A + (1+\\epsilon_A x_{A,0})ln(1-U_A)$ ... solve for UA\n",
    "\n",
    "$A \\rightarrow B + 2C$\n",
    "\n",
    "$\\epsilon_A = (-1+1+2)/(-(-1)) = +2$\n",
    "\n",
    "$3A \\rightarrow B$\n",
    "\n",
    "$\\epsilon_A = (-3+1)/(-(-3)) = -2/3$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "scrolled": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "100% A\n",
      "UA = 0.609151\n",
      "5% A\n",
      "UA = 0.782528\n"
     ]
    },
    {
     "data": {
      "image/png": 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2MvC/nx/jvT1nOLBsMtZWBkaFexLp40xsiDuxQe54ONl2ug0aTU+no26oNgXB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\nLIwxxtTKkoUxxphaWbIwxhhTq/8HYNoDxlVf0jMAAAAASUVORK5CYII=\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x28b6e2486a0>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import matplotlib.pyplot as plt\n",
    "from scipy.optimize import root\n",
    "import numpy as np\n",
    "%matplotlib inline\n",
    "xa0 = 1.\n",
    "k1 = 0.08 # min^-1\n",
    "v0 = 10. # L/min\n",
    "vr = 200 # L\n",
    "def ua(vr, epsa, xa0):\n",
    "    return root(lambda ua_var: \n",
    "        vr / v0 * k1 + \n",
    "        epsa * xa0 * ua_var + \n",
    "        (1 + epsa * xa0) * np.log(1 - ua_var),\n",
    "        0.99\n",
    "    ).x\n",
    "ua = np.vectorize(ua) # Vektorisieren\n",
    "print ('100% A')\n",
    "print ('UA = ' + '{0:g}'.format(\n",
    "    ua(vr, epsa=(-1. +2. + 1.)/(-(-1.)), xa0=1.).item()\n",
    "))\n",
    "\n",
    "xa0 = 0.05\n",
    "print ('5% A')\n",
    "print ('UA = ' + '{0:g}'.format(\n",
    "    ua(vr, epsa=-1. +2. + 1., xa0=0.05).item()\n",
    "))\n",
    "\n",
    "vr = np.array(range(0,200,1))\n",
    "plt.subplots(nrows=2, ncols=1)\n",
    "plt.subplot(2,1,1)\n",
    "plt.plot(\n",
    "    vr, ua(vr, epsa=(-1. + 2. + 1.)/(-(-1.)), xa0=1.), '-.',\n",
    "    vr, ua(vr, epsa=(-1. + 2. + 1.)/(-(-1.)), xa0=0.05), '-'\n",
    ")\n",
    "plt.xlabel('Vr, L')\n",
    "plt.ylabel('UA')\n",
    "plt.legend(['reines A', 'mit Stickstoff'])\n",
    "plt.title('$A \\\\rightarrow B + 2 C$')\n",
    "plt.subplot(2,1,2)\n",
    "plt.plot(\n",
    "    vr, ua(vr, epsa=(-3. + 1.)/(-(-3.)), xa0=1.), '-.',\n",
    "    vr, ua(vr, epsa=(-3. + 1.)/(-(-3.)), xa0=0.05), '-'\n",
    ")\n",
    "plt.xlabel('Vr, L')\n",
    "plt.ylabel('UA')\n",
    "plt.legend(['reines A', 'mit Stickstoff'])\n",
    "plt.text(0.5,0.5, '$3A \\\\rightarrow B$');"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Fazit**\n",
    "\n",
    "a) Bezüglich der Reaktion, wobei eine höhere Stoffmenge an Produkt hergestellt wird, als der Reagenzienkonsum ist es bevorzugt, ein inertes Komponent einzuführen. Denn somit steigt der Gesamtdruck, und somit die Konzentration und folglich die Reaktionsgeschwindigkeit.\n",
    "\n",
    "b) Bei der Reaktion, die eine niedrigere Produktstoffmege herstellt, als sie verbraucht ist es begünstigt, keine zusätzlichen Komponente einzuführen. Denn die Reaktion selbst erniedrigt den Gesamtdruck, und somit die Konzentratrion und infolgedessen die Reaktionsgeschwindigkeit."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Ü 5.8\n",
    "Die  Gasphasendehydrierung von  Benzol  B  zu  Diphenyl  D und weiter zu Triphenyl T wird  in  einem  isotherm  betriebenen  Strömungsrohr  durchgeführt.  Ziel  ist,  die \n",
    "Produktion  von  D  zu  maximieren  und  die  Weiterreaktion  zu  T zu  unterdrücken. \n",
    "Die Reaktion wird bei Normaldruck durchgeführt.\n",
    "\n",
    "$\\begin{array}{ccccc}\n",
    "    2 C_6H_6 & & \\overset{k_1}{\\longrightarrow} &  C_{12}H_{10}+& H_2\\\\\n",
    "    B & & & D & H\\\\\n",
    "    C_6H_6 +& C_{12}H_{10} & \\overset{k_2}{\\longrightarrow} & C_{18}H_{14} +& H_2\\\\\n",
    "    B & D & & T & H\\\\\n",
    "\\end{array}$\n",
    "\n",
    "Die Geschwindigkeitsgleichungen lauten wie folgt:\n",
    "\n",
    "$r_1  = 14,96 \\cdot 10^6 e^{-15200/T} \\left(p_B^2- \\frac{p_D p_H}{K_1} \\right)   kg \\cdot h^{-1} \\cdot L^{-1}$\n",
    "\n",
    "$r_2 = -8,67 \\cdot 10^6 e^{-15200/T}\\left(p_B p_D - \\frac{p_T p_H}{K_2}\\right)$\n",
    "\n",
    "$K_1 = 0,312$\n",
    "\n",
    "$K_2 = 0,480$\n",
    "\n",
    "1.  Studiere den Effekt der Raumzeit $\\tau$ auf die Umsätze $U_1$ und  $U_2$ und ermittle die Stoffströme im Reaktor.\n",
    "2.  Bestimme die Reaktionsgeschwindigkeiten $r_1$ und $r_2$ als Funktion der Raumzeit $\\tau$.\n",
    "\n",
    "**Lösung**\n",
    "\n",
    "Raumzeit (hydrodynamische Verweilzeit) $\\tau = \\frac{V_R}{\\dot {V}}$\n",
    "\n",
    "Billanz\n",
    "\n",
    "$\\frac{d \\dot{n}_B}{d V} = \\nu_{B,1}r_1+\\nu_{B,2}r_2 = -r_1 - r_2$\n",
    "\n",
    "Als Funktion der Raumzeit und des Partialdrucks $n_B = \\frac{p_B \\dot {V}}{R T}$:\n",
    "\n",
    "$\\frac{d \\dot{n}_B}{d V} = \\frac{1}{R T}\\frac{d p_B \\dot {V}}{d (V)} =\\frac{1}{R T}\\frac{d p_B }{d (V/\\dot {V})}= \\frac{1}{R T}\\frac{d p_B }{d \\tau}$\n",
    "\n",
    "${\\boldsymbol \\nu} = \\left( \n",
    "\\begin{array}{cccc}\n",
    "\\nu_{B,1} & \\nu_{D,1} & \\nu_{T,1} & \\nu_{H,1} \\\\\n",
    "\\nu_{B,2} & \\nu_{D,2} & \\nu_{T,2} & \\nu_{H,2}\\\\\n",
    "\\end{array}\n",
    "\\right) = \n",
    "\\left( \n",
    "\\begin{array}{cccc}\n",
    "-2 & +1 & 0 & +1 \\\\\n",
    "-1 & -1 & +1 & +1\\\\\n",
    "\\end{array}\n",
    "\\right)$\n",
    "\n",
    "$\\frac{1}{R T}\\frac{d {\\bf p}}{d \\tau} = {\\bf \\nu} {\\bf r}$\n",
    "\n",
    "$\\frac{1}{R T}\\frac{d p_B}{d \\tau} = -2 k_1 e^{-15200/T}\\left(p_B^2- \\frac{p_D p_H}{K_1} \\right) - k_2 e^{-15200/T}\\left(p_B p_D - \\frac{p_T p_H}{K_2}\\right)$\n",
    "\n",
    "$\\frac{1}{R T}\\frac{d p_D}{d \\tau} = + k_1 e^{-15200/T}\\left(p_B^2- \\frac{p_D p_H}{K_1} \\right) - k_2 e^{-15200/T}\\left(p_B p_D - \\frac{p_T p_H}{K_2}\\right)$\n",
    "\n",
    "$\\frac{1}{R T}\\frac{d p_T}{d \\tau} = + k_2 e^{-15200/T}\\left(p_B p_D - \\frac{p_T p_H}{K_2}\\right)$\n",
    "\n",
    "$\\frac{1}{R T}\\frac{d p_H}{d \\tau} = + k_1 e^{-15200/T}\\left(p_B^2- \\frac{p_D p_H}{K_1} \\right) + k_2 e^{-15200/T}\\left(p_B p_D - \\frac{p_T p_H}{K_2}\\right)$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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BcLRrR1KvXiT27Gn9Us/Pt6a8PJxpkbs91RhDoLKShu3b907+rVvxbyrBt2kT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\ngg==\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x28b6e8df828>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "from scipy.integrate import odeint\n",
    "t = 1033. # K\n",
    "r = 8.314/1000. # J / (mol K) [=] 8.314 kPa / (mol K)\n",
    "# Angeblich sollen die Einheiten der Geschwindigkeitskonstante \n",
    "# kg h^{-1} L^{-1}, was aber nicht Sinn ergibt, denn sind die kg\n",
    "# in der zweiten Konstante auch kg? Wenn so, sind es kg von B oder\n",
    "# D? Es wäre verständlicher, mit Einheiten kgmol h^{-1} L^{-1}\n",
    "k1 = 14.96E+06 * np.exp(15200./t) # kgmol h^{-1} L^{-1}\n",
    "k2 = -8.67E+06 * np.exp(15200./t) # kgmol h^{-1} L^{-1}\n",
    "kp1 = 0.312 # \n",
    "kp2 = 0.480 # \n",
    "def eq_set(y, t):\n",
    "    pb, pd, pt, ph = y\n",
    "    res = np.empty(4, dtype=np.float)\n",
    "    res[0] = r * t * (\n",
    "        -2 * k1 * (pb**2 - pd*ph/kp1) + \n",
    "        -1 * k2 * (pb*pd - pt*ph/kp2))\n",
    "    res[1] = r * t * (\n",
    "        +1 * k1 * (pb**2 - pd*ph/kp1) +\n",
    "        -1 * k2 * (pb*pd - pt*ph/kp2))\n",
    "    res[2] = r * t * (\n",
    "        +1 * k2 * (pb*pd - pt*ph/kp2))\n",
    "    res[3] = r * t * (\n",
    "        +1 * k1 * (pb**2 - pd*ph/kp1) + \n",
    "        +1 * k2 * (pb*pd - pt*ph/kp2))\n",
    "    return res\n",
    "t = np.arange(0, 0.00001, 0.00000001)\n",
    "y0 = np.array([1.0, 0, 0, 0])\n",
    "y = odeint(eq_set, y0, t)\n",
    "plt.plot(t, y)\n",
    "plt.legend(['B', 'D', 'T', 'H'])\n",
    "plt.xlabel('Raumzeit, $\\\\tau$' + \"'\")\n",
    "plt.ylabel('$p_i$');"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Ü 5.9\n",
    "Eine  Reaktion 1. Ordnung mit $\\epsilon_A=0$ wird in einem  Kreislaufreaktor durchgeführt, $k \\tau = 5$. Welche Umsätze resultieren für ein Kreislaufverhältnis von 0, 5, 10 und $\\infty$?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<img src=\"Rückvermischung.jpg\">"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**Lösung**\n",
    "\n",
    "Raumzeit (hydrodynamische Verweilzeit). Sie gibt definitionsgemäß das Verhältnis des Reaktionsvolumens zum volumetrischen **Zulauf**strom an.\n",
    "\n",
    "$\\tau = \\frac{V_R}{\\dot V_0}$\n",
    "\n",
    "Reaktion 1. Ordnung\n",
    "\n",
    "$A \\rightarrow Produkte...\\hspace{10mm} r_1 = k c_A$\n",
    "\n",
    "$\\nu_{A1}=-1$\n",
    "\n",
    "Stoffbilanz am Reaktor Ausgangsstrom:\n",
    "\n",
    "$\\frac{d \\dot{n}_{A,2}}{d V} = \\nu_{A1}r_1$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "*Allgemeiner Umsatz*\n",
    "\n",
    "$U_A = \\frac{\\dot n_{A,0} - \\dot n_A}{\\dot n_{A,0}}=1-\\frac{\\dot n_A}{\\dot n_{A,0}}=1-\\left(\\frac{c_A}{c_{A,0}}\\right)\\left(\\frac{\\dot V}{\\dot V_0}\\right)$\n",
    "\n",
    "$\\rightarrow \\dot n_A = \\dot n_{A0}(1-U_A)$\n",
    "\n",
    "*Allgemeine Stoffbilanzen*\n",
    "\n",
    "$\\begin{array}{ll}\n",
    "\\dot n_A &= \\dot n_{A,0} + \\nu_{A1}\\xi_1 \\hspace{10mm} \\rightarrow \\xi_1 = \\frac{\\dot n_A}{(-\\nu_{A1})}U_A\\\\\n",
    "\\dot n &= \\dot n_0 + (\\sum_j \\nu_{j1})\\xi_1\\\\\n",
    "&= \\dot n_0 + \\left(\\frac{\\sum_j \\nu_{j1}}{-\\nu_A}\\right)\\dot n_{A0} U_A\\\\\n",
    "&= \\dot n_0 \\left(1+\\left(\\frac{\\sum_j \\nu_{j1}}{-\\nu_A}\\right) x_{A0} U_A \\right)\\\\\n",
    "&= \\dot n_0 \\left(1+\\epsilon_A x_{A0} U_A \\right)\\\\\n",
    "\\end{array}$\n",
    "\n",
    "*Innerliche Stoffblianzen*\n",
    "\n",
    "$\\begin{array}{ll}\n",
    "\\dot n_{A2} &= \\dot n_A + \\dot n_{Ar}\\\\\n",
    "&= (R+1)\\dot n_A\\\\\n",
    "\\dot n_{A1} &= \\dot n_{A0} + \\dot n_{Ar}\\\\\n",
    "\\end{array}$\n",
    "\n",
    "*Volumenänderung*\n",
    "\n",
    "$\\frac{P \\dot V}{\\dot n R T}=Z$\n",
    "\n",
    "$\\frac{P \\dot V}{Z R T} = \\frac{P_0 \\dot V_0}{Z_0 R T_0}\\left(1+\\epsilon_A x_{A0} U_A \\right)$\n",
    "\n",
    "$\\require{cancel}$\n",
    "$\\dot V = \\dot V_0 \\left(1+\\epsilon_A x_{A0} U_A \\right)\\times \\cancelto{1}{ \\frac{P_0}{P}}\\cancelto{1}{\\frac{T}{T_0}}\\cancelto{1}{\\frac{Z}{Z_0}}$ (I. G.)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "data": {
      "image/png": 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3LagfwQAgwakEKrH9BttvyF0HpoOOAZUwj2FjoWMAkCAYACQIBgAJggFAgouPqMT2VkmK\niCdy14L6EQwAEpxKoBLbl9u+PHcdmA46BlTCPIaNhY4BQIJgAJAgGAAkCAYACS4+AkjQMQBIEAwA\nEgQDgATBACBBMABIEAwAEgQDgATBACBBMABIEAwAEgQDgATBACBBMABIEAwAEgQDgATBACBBMABI\nEAwAEgQDgATBACBBMABIEAwAEgQDgMT/AW34wvEqdFKdAAAAAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x28b6e8bfb38>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import matplotlib\n",
    "import matplotlib.pyplot as plt\n",
    "fig = plt.figure()\n",
    "ax = fig.add_subplot(111)\n",
    "patch1 = matplotlib.patches.Circle(\n",
    "    [0.5,0.5],0.05\n",
    ")\n",
    "patch2 = matplotlib.patches.Rectangle(\n",
    "    [0.3,0.3],0.4, 0.4, alpha=0.5, \n",
    "    fill=False, edgecolor='black',\n",
    "    linestyle = '--'\n",
    ")\n",
    "arrow1 = matplotlib.patches.Arrow(\n",
    "    0, 0.5,0.45,0, width=0.05,\n",
    "    color='black'\n",
    ")\n",
    "arrow2 = matplotlib.patches.Arrow(\n",
    "    0.55, 0.5,0.45,0, width=0.05,\n",
    "    color='black'\n",
    ")\n",
    "line1 = matplotlib.lines.Line2D(\n",
    "    [0.5,0.5], [0,0.45],\n",
    "    linestyle='--', color='black'\n",
    ")\n",
    "text1 = matplotlib.text.Text(\n",
    "    0, 0.45, '$n_{A0}$\\n$V_0$\\n$U_A=0$'\n",
    ")\n",
    "text2 = matplotlib.text.Text(\n",
    "    0.8, 0.45, '$n_{A1}$\\n$V_1$\\n$U_{A1}$'\n",
    ")\n",
    "for artist in [\n",
    "    patch1,patch2,arrow1,arrow2,\n",
    "    line1,text1,text2\n",
    "]:\n",
    "    ax.add_artist(artist)\n",
    "ax.set_frame_on(False)\n",
    "ax.set_axis_off()\n",
    "ax.set_aspect(1.0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Umsatz am Knotenpunkt # 1.\n",
    "\n",
    "In folgender Weise wird der Umsatz 1 definiert. Definitionsgemäß nimmt er nur den Speisestrom und den Reaktoreneintrittstrom in Acht, ohne irgendeiner Absicht auf den Rücklaufstrom.\n",
    "\n",
    "$U_{A1} = \\frac{\\dot n_{A0} - \\dot n_{A1}}{\\dot n_{A0}} = 1 - \\frac{\\dot n_{A1}}{\\dot n_{A0}} = 1- \\frac{c_{A1}}{c_{A0}} \\left(\\frac{\\dot V_1}{\\dot V_0} \\right)$\n",
    "\n",
    "$\\dot V_1 = \\dot V_0(1+\\epsilon_A x_{A0}U_{A1})$\n",
    "\n",
    "$U_{A1} = 1-\\frac{c_{A1}}{c_{A0}}(1+\\epsilon_A x_{A0}U_{A1})$ \n",
    "\n",
    "$\\begin{array}{lll}\\Rightarrow & U_{A1} & = \\frac{1-c_{A1}/c_{A0}}{1+\\epsilon_A x_{A0}c_{A1}/c_{A0}}\\\\ & c_{A1} & =\\left(\\frac{1-U_{A1}}{1+\\epsilon_A x_{A0}U_{A1}}\\right)c_{A0}\\end{array}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {
    "slideshow": {
     "slide_type": "-"
    }
   },
   "outputs": [
    {
     "data": {
      "image/png": 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22V4p6fKI+JSKjkKSNkv6TETMRcRTI64TLQgGVOlCHRsMG1V0DNdIWmP7JkkbbK+SdJGk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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x28b6ea4a048>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import matplotlib\n",
    "import matplotlib.pyplot as plt\n",
    "fig = plt.figure()\n",
    "ax = fig.add_subplot(111)\n",
    "patch1 = matplotlib.patches.Circle(\n",
    "    [0.5,0.5],0.05\n",
    ")\n",
    "patch2 = matplotlib.patches.Rectangle(\n",
    "    [0.3,0.3],0.4, 0.4, alpha=0.5, \n",
    "    fill=False, edgecolor='black',\n",
    "    linestyle = '--'\n",
    ")\n",
    "arrow1 = matplotlib.patches.Arrow(\n",
    "    0, 0.5,0.45,0, width=0.05,\n",
    "    color='black'\n",
    ")\n",
    "arrow2 = matplotlib.patches.Arrow(\n",
    "    0.55, 0.5,0.45,0, width=0.05,\n",
    "    color='black'\n",
    ")\n",
    "arrow3 = matplotlib.patches.Arrow(\n",
    "    0.5, 0.0, 0,0.45, width=0.05,\n",
    "    color='black'\n",
    ")\n",
    "text1 = matplotlib.text.Text(\n",
    "    0, 0.45, '$n_{A0}$\\n$V_0$\\n$U_A=0$'\n",
    ")\n",
    "text2 = matplotlib.text.Text(\n",
    "    0.8, 0.45, '$n_{A1}$\\n$V_1$\\n$U_{A1}$'\n",
    ")\n",
    "text3 = matplotlib.text.Text(\n",
    "    0.55, 0.1, '$n_{Ar}$\\n$V_r$'\n",
    ")\n",
    "for artist in [\n",
    "    patch1,patch2,arrow1,arrow2,\n",
    "    arrow3,text1,text2,text3\n",
    "]:\n",
    "    ax.add_artist(artist)\n",
    "ax.set_frame_on(False)\n",
    "ax.set_axis_off()\n",
    "ax.set_aspect(1.0)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Konzentration am Knotenpunkt # 1.\n",
    "\n",
    "$c_{A1} = \\frac{\\dot n_{A0} + \\dot n_{A1}}{\\dot V_0 + \\dot V_{ar}} = \\frac{\\dot n_{A0} + R \\dot n_{A0}(1-U_A)}{\\dot V_0  + R \\dot V_0 (1+\\epsilon_A x_{A0}U_A)}=c_{A0}\\left(\\frac{1 + R -U_A R}{1 + R+\\epsilon_A x_{A0}U_A R} \\right)$\n",
    "\n",
    "$\\left(\\frac{1-U_{A1}}{1+\\epsilon_A x_{A0}U_{A1}}\\right)c_{A0}=c_{A0}\\left(\\frac{1 + R -U_A R}{1 + R+\\epsilon_A x_{A0}U_A R} \\right)$\n",
    "\n",
    "$\\Rightarrow U_A \\frac{R}{R+1} = U_{A1}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "*Mathematisches Modell*\n",
    "\n",
    "$\\frac{d \\dot{n}_{A,2}}{d V} = \\nu_{A1}r_1$\n",
    "\n",
    "$\\begin{array}{ll}\n",
    "\\frac{d \\dot{n}_{A,2}}{d V} &= \\frac{d (R+1)c_A \\dot V}{d V} = \\frac{d }{d V} \\left[ (R+1)c_A \\dot V_0 (1 + \\epsilon_A x_{A0}U_A)\\right]\\\\\n",
    " &= (R+1)\\frac{d }{d (V/V_0)} \\left[c_{A0} \\left(\\frac{1-U_A}{1+\\epsilon_A x_{A0}U_A}\\right) (1 + \\epsilon_A x_{A0}U_A) \\right]\\\\\n",
    " &= (R+1)c_{A0}\\frac{d (1-U_A) }{d \\tau}=-(R+1)c_{A0}\\frac{d U_A }{d \\tau} \\\\\n",
    "\\Rightarrow \\int_0^\\tau \\tau &= \\tau = \\frac{-(R+1)c_{A0}}{\\nu_{A1}} \\int_{U_{A1}}^{U_{A}} \\frac{1}{r_1} d U_A = \\frac{-(R+1)c_{A0}}{\\nu_{A1}} \\int_{U_{A1}}^{U_{A}} \\frac{1}{k_1 c_{A0}\\left(\\frac{1-U_A}{1+\\epsilon_A x_{A0}U_A} \\right)} d U_A \\\\\n",
    "\\frac{(-\\nu_{A1})k \\tau}{R+1} &= \\int_{U_{A1}}^{U_{A}} \\frac{1+\\epsilon_A x_{A0}U_A}{1-U_A} d U_A = \\int_{U_{A1}}^{U_{A}} \\frac{\\epsilon_A x_{A0}}{-1}d U_A + \\int_{U_{A1}}^{U_{A}} \\frac{\\epsilon_A x_{A0}+1}{1-U_A} d U_A\\\\\n",
    " &= -\\epsilon_A x_{A0} U_A \\Big |_{\\frac{R}{R+1}U_A}^{U_A} - (\\epsilon_A x_{A0} +1) \\times ln (1-U_A)\\Big |_{\\frac{R}{R+1}U_A}^{U_A} \\\\\n",
    " &= -\\epsilon_A x_{A0} U_A \\left(1-\\frac{R}{R+1} \\right) + (\\epsilon_A x_{A0} +1) \\times ln \\left(\\frac{1-\\frac{R}{R+1}U_A}{1-U_A} \\right) \\\\\n",
    " \\\\\n",
    "\\bf{\\epsilon_A} &= 0 \\\\\n",
    "\\Rightarrow \\frac{(-\\nu_{A1})k \\tau}{R+1}  &= ln \\left(\\frac{1 - \\frac{R}{R+1}U_A}{1-U_A}\\right)\\\\\n",
    "U_A &= \\frac{1-exp((-\\nu_{A1})k\\tau/(R+1))}{\\frac{R}{R+1}-exp((-\\nu_{A1})k\\tau/(R+1))}\n",
    "\\end{array}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "UA = ['0.993262', '0.886439', '0.863575', '0.833334']\n"
     ]
    }
   ],
   "source": [
    "ua = [(1-np.exp((-1*-1)*5.0/(r+1)))/(r/(r+1)-np.exp((-1*-1)*5.0/(r+1))) for \n",
    "      r in [0., 5., 10., 1e6]]\n",
    "print ('UA = ' + str(list(map(lambda x: '{0:g}'.format(x), ua))))"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "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": "3.6.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}