{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 8 - Membrane Separation Processes"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Page 460 Example 8.2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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QF12UShSTJ6e2iAMOyDsyM2tmrVCVBICkTYBfA2dFxNLi97Jv/8GbAXohpam+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\nVuL/AKOrVmLW7TWCAAAAAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x7f2751936d50>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Jo = 0.099 m/h\n",
      "\n",
      " Jf = 0.000   m/h\n",
      "\n",
      " Jav = 0.025   m/h\n",
      "\n",
      " Area = 79.650  m**2\n"
     ]
    }
   ],
   "source": [
    "%matplotlib inline\n",
    "from matplotlib.pyplot import plot,title,xlabel,ylabel,show\n",
    "from math import log\n",
    "\n",
    "#From the gel polarisation model:\n",
    "         #J = (1/A)(dV/dt) = hD ln(Cg/Cf)\n",
    "         #Cf = Co(Vo/V)\n",
    "#where Co and Vo are the initial concentration and volume,respectively and Cf and V are the values at subsequent times\n",
    "#Combining these eq gives\n",
    "        #dV/dt = A(hDln(Cg/Co)-hDln(Vo/V))\n",
    "V = [10, 5 ,3 ,2, 1]#\n",
    "y = [9.90, 13.64, 18.92, 27.30, 112.40]\n",
    "plot(V,y)\n",
    "title(\"Area under the curve is 184.4\")\n",
    "xlabel(\"Volume(m**3)\")\n",
    "ylabel(\"(J - hDln(Vo/V))**(-1)\")\n",
    "show()\n",
    "\n",
    "#(b)\n",
    "Jo = 0.04*log(250/20)#\n",
    "print\"\\n Jo = %.3f m/h\"%(Jo)#\n",
    "Jf = 0.04*log(250/200)#\n",
    "print\"\\n Jf = %.3f   m/h\"%(Jf)#\n",
    "Jav = Jf + 0.27*(0.101-0.008)#\n",
    "print\"\\n Jav = %.3f   m/h\"%(Jav)#\n",
    "#For the removal of 9m**3 filtrate in 4 hours\n",
    "Area = (9/4)/Jav#\n",
    "print\"\\n Area = %.3f  m**2\"%(Area)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Page 462 Example 8.3"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Cl = 4 kg/m**3\n",
      "\n",
      " below this concentration the membrane flux is 0.04 m/h\n",
      "\n",
      " A = 300 m**2\n",
      "\n",
      " the no of required modules are 10 \n",
      "\n",
      " The procedure is to use trial and error to estimate the value of C1 that gives the optimum values of A1 and A2\n",
      "\n",
      " If C1 = 5kg/m**3 then A1 = 62 m**2 and A2 = 23 m**2\n",
      "\n",
      " an arrangement of 3 modules −1 module is required.\n",
      "\n",
      "\n",
      "\n",
      "  If C1 = 4 kg/m**3 then A1 = 54m**2 and A2 = 30m**2\n",
      "\n",
      " an arrangement of 2 modules −1 module is almost sufficient.\n",
      "\n",
      "\n",
      "\n",
      "   If C1 = 4.5 kg/m**3 then A1 = 58m**2 and A2 = 26 m**2\n",
      "\n",
      " an arrangement of 2 modules −1 module which meets the requirement\n",
      "\n",
      "\n",
      " This arrangement requires the minimum number of modules.\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import log\n",
    "#It is assumed that Q0 is the volumetric flowrate of feed\n",
    "# Q2 the volumetric flowrate of concentrate\n",
    "#C0 the solute concentration in the feed\n",
    "# C2 the solute concentration in the concentrate\n",
    "# F the volumetric flowrate of membrane permeate\n",
    "# A the required membrane area.\n",
    "# It is also assumed that there is no loss of solute through the membrane.\n",
    "Cl = 3#\n",
    "while 1:\n",
    "    Clnew = Cl -(0.04-0.02*log(30/Cl))/(Cl**(-1)/50)#\n",
    "    if Clnew == Cl:\n",
    "        break#\n",
    "    \n",
    "    Cl = Clnew#\n",
    "\n",
    "print\"\\n Cl = %d kg/m**3\"%(Cl)#\n",
    "print\"\\n below this concentration the membrane flux is 0.04 m/h\"\n",
    "\n",
    "#This does not pose a constraint for the single stage as the concentration of solute C2 will be that of the final concentrate, 20 kg/m3.\n",
    "#Conservation of solute gives:QoCo = Q2C2\n",
    "#A fluid balance gives : Qo = F + Q2\n",
    "#Combining these eq and substituting Known values:\n",
    "A = (2.438/0.02)/log(30/20)#\n",
    "print\"\\n A = %d m**2\"%(A)#\n",
    "#The tubular membranes to be used are available as 30 m**2modules.\n",
    "print\"\\n the no of required modules are %d \"%(A/30)#\n",
    "\n",
    "#Part(b)\n",
    "\n",
    "#Conservation of solute gives = QoCo = Q1C1 = Q2C2\n",
    "#A fluid balance on stage 2 gives Q1 = Q2 + F2\n",
    "#A fluid balance on stage 2 gives Q1 = Q2 +F2\n",
    "#Substituting given values in above eqns\n",
    "#2.5 = 1.25/C1 + 0.02A1ln(30/C1)\n",
    "def a(C1):\n",
    "    A1 = (2.5-1.25/C1)/(0.02*log(30/C1))#\n",
    "    return A1\n",
    "def b(C1):\n",
    "    A2 = (1.25/C1 - 0.0625)/0.00811\n",
    "    return A2\n",
    "\n",
    "print\"\\n The procedure is to use trial and error to estimate the value of C1 that gives the optimum values of A1 and A2\"\n",
    "print\"\\n If C1 = 5kg/m**3 then A1 = %d m**2 and A2 = %d m**2\"%(a(5),b(5))#\n",
    "print\"\\n an arrangement of 3 modules −1 module is required.\"\n",
    "print\"\\n\\n\\n  If C1 = 4 kg/m**3 then A1 = %dm**2 and A2 = %dm**2\"%(a(4),b(4))#\n",
    "print\"\\n an arrangement of 2 modules −1 module is almost sufficient.\"\n",
    "print\"\\n\\n\\n   If C1 = 4.5 kg/m**3 then A1 = %dm**2 and A2 = %d m**2\"%(a(4.5),b(4.5))#\n",
    "print\"\\n an arrangement of 2 modules −1 module which meets the requirement\"\n",
    "print\"\\n\\n This arrangement requires the minimum number of modules.\""
   ]
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