path: root/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/chapter6.ipynb

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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 6:Distillation "
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.1,Page number:324"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "T1 = 303  # [K]\n",
      "P = 1  # [bar]\n",
      "D = 0.6 \n",
      "W = 0.4 \n",
      "zf = 0.5 \n",
      " \n",
      "# Parameters for componenr 'A'\n",
      "Tc_a = 540.3  # [K]\n",
      "Pc_a = 27.4  # [bar]\n",
      "A_a = -7.675 \n",
      "B_a = 1.371 \n",
      "C_a =-3.536 \n",
      "D_a = -3.202 \n",
      "\n",
      "# Parameters for component 'B'\n",
      "Tc_b = 568.8  # [K]\n",
      "Pc_b = 24.9  # [bar]\n",
      "A_b = -7.912 \n",
      "B_b = 1.380 \n",
      "C_b = -3.804 \n",
      "D_b = -4.501 \n",
      "\n",
      "import math\n",
      "from numpy import *\n",
      "from scipy.optimize import fsolve\n",
      "# Using equation 6.5\n",
      "# x_a = 1-(T/Tc_a) \n",
      "# P_a = Pc_a*math.exp((A_a*x_a+B_a*x_a**1.5+C_a*x_a**3+D_a*x_a**6)/(1-x_a))  # [bar]\n",
      "\n",
      "# x_b = 1-(T/Tc_b) \n",
      "# P_b = Pc_b*math.exp((A_b*x_b+B_b*x_b**1.5+C_b*x_b**3+D_b*x_b**6)/(1-x_b))  # [bar]\n",
      "\n",
      "# m_a = P_a/P \n",
      "# m_b = P_b/P \n",
      "\n",
      "# Solution of simultaneous equation\n",
      "def F(e):\n",
      "    f1 = e[1] - (e[2]*Pc_a*math.exp(((A_a*(1-(e[0]/Tc_a))+B_a*(1-(e[0]/Tc_a))**1.5+C_a*(1-(e[0]/Tc_a))**3+D_a*(1-(e[0]/Tc_a))**6))/(1-(1-(e[0]/Tc_a)))))/P \n",
      "    f2 = 1-e[1] - ((1-e[2])*Pc_b*math.exp((A_b*(1-(e[0]/Tc_b))+B_b*(1-(e[0]/Tc_b))**1.5+C_b*(1-(e[0]/Tc_b))**3+D_b*(1-(e[0]/Tc_b))**6)/(1-(1-(e[0]/Tc_b)))))/P \n",
      "    f3 = (-W/D) - ((e[1]-zf)/(e[2]-zf)) \n",
      "    return(f1,f2,f3)\n",
      "\n",
      "\n",
      "# Initial guess\n",
      "e = [400,0.6,0.4] \n",
      "y = fsolve(F,e) \n",
      "T = y[0]  # [K] \n",
      "Yd = y[1] \n",
      "Xw = y[2] \n",
      "\n",
      "print\"The composition of the vapor and liquid and the temperature in the separator if it behaves as an ideal stage are\",round(Yd,3),round(Xw,3),\"and\",round(T,1),\"K respectively\\n\\n\"\n",
      "\n",
      "# For the capculation of the amount of heat to be added per mole of feed\n",
      "T0 = 298  # [K]\n",
      "lambdaA = 36.5  # [Latent heats of vaporization at To = 298 K ,kJ/mole]\n",
      "lambdaB = 41.4  # [Latent heats of vaporization at To = 298 K ,kJ/mole]\n",
      "CpA = 0.187  # [kJ/mole.K]\n",
      "CpB = 0.247  # [kJ/mole.K]\n",
      "CLA1 = 0.218  # [ 298-303 K,  kJ/mole.K]\n",
      "CLB1 = 0.253  # [ 298-303 K,  kJ/mole.K]\n",
      "CLA2 = 0.241  # [ 298-386 K,  kJ/mole.K]\n",
      "CLB2 = 0.268  # [ 298-386 K,  kJ/mole.K]\n",
      "# Bubble point calculated when 'D' approaches 0 and Dew point calculated when 'D' approaches 1\n",
      "Tbp = 382.2 # [Bubble point of the mixture, K]\n",
      "Tdp = 387.9 # [Dew point of mixture, K]\n",
      "\n",
      "HF = (T1-T0)*(Xw*CLA1+CLB1*(1-Xw))  # [kJ/mole]\n",
      "HW = (Tbp-T0)*(Xw*CLA2+CLB2*(1-Xw))  # [kJ/mole]\n",
      "HG = (Tdp-T0)*(Yd*CpA+(1-Yd)*CpB) + Yd*lambdaA +(1-Yd)*lambdaB  # [kJ/mole]\n",
      "\n",
      "f =1 # [feed]\n",
      "# Using equation 6.4\n",
      "def f14(Q):\n",
      "    return(W/D + (HG-(HF+Q/f))/(HW -(HF+Q/f)))\n",
      "Q = fsolve(f14,40) \n",
      "print\"The amount of heat to be added per mole of feed is\",round(Q[0],2),\"kJ/mole\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The composition of the vapor and liquid and the temperature in the separator if it behaves as an ideal stage are 0.576 0.385 and 385.4 K respectively\n",
        "\n",
        "\n",
        "The amount of heat to be added per mole of feed is 42.08 kJ/mole\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.2,Page number:326"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# a-benzene   b-toluene   c-orthoxylene\n",
      "T = 373  # [K]\n",
      "P = 101.3  # [kPa]\n",
      "Pa = 182.7  # [kPa]\n",
      "Pb = 73.3  # [kPa]\n",
      "Pc= 26.7  # [kPa]\n",
      "Zfa = 0.5 \n",
      "Zfb = 0.25 \n",
      "Zfc = 0.25 \n",
      "\n",
      "import math\n",
      "from numpy import *\n",
      "from scipy.optimize import fsolve\n",
      "# Therefore\n",
      "ma = Pa/P \n",
      "mb = Pb/P \n",
      "mc = Pc/P \n",
      "# Let Feed is 1 kmole\n",
      "# Therefore D+W = 1\n",
      "\n",
      "# Solution of simultaneous equation\n",
      "def F(e):\n",
      "    f1 = e[0]+e[1]-1 \n",
      "    f2 = e[1]/e[0] + (e[2]-Zfa)/(e[3]-Zfa) \n",
      "    f3 = e[2]-ma*e[3] \n",
      "    f4 = e[4]-mb*e[5] \n",
      "    f5 = 1-e[2]-e[4] -mc*(1-e[3]-e[5]) \n",
      "    f6 = e[1]/e[0] + (e[4]-Zfb)/(e[5]-Zfb) \n",
      "\n",
      "    return(f1,f2,f3,f4,f5,f6)\n",
      "\n",
      "# Initial guess\n",
      "e = [0.326,0.674,0.719,0.408,0.198,0.272] \n",
      "y = fsolve(F,e) \n",
      "D = y[0] \n",
      "W = y[1] \n",
      "Yad = y[2] \n",
      "Xaw = y[3] \n",
      "Ybd = y[4] \n",
      "Xbw = y[5] \n",
      "Ycd = 1-Yad-Ybd \n",
      "Xcw = 1-Xaw-Xbw \n",
      "\n",
      "print\"The amounts of liquid and vapor products are D=\",round(D,3),\"and W=\",round(W,3),\"respectively\"\n",
      "print\"The vapor compositions of components A(Yad), B(Ybd) and C(Ycd) are\",round(Yad,3),round(Ybd,3),round(Ycd,3),\"respectively\"\n",
      "print\"The liquid composition of components A(Xaw), B(Xbw) and C(Xcw) are\",round(Xaw,3),round(Xbw,3),round(Xcw,3),\"respectively\\n\\n\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The amounts of liquid and vapor products are D= 0.326 and W= 0.674 respectively\n",
        "The vapor compositions of components A(Yad), B(Ybd) and C(Ycd) are 0.714 0.199 0.087 respectively\n",
        "The liquid composition of components A(Xaw), B(Xbw) and C(Xcw) are 0.396 0.275 0.329 respectively\n",
        "\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.3,Page number:328"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "\t#   n-heptane - a    n-octane - b\n",
      "P = 1  \t\t\t\t\t\t# [bar]\n",
      "\n",
      "\t# Basis:\n",
      "F = 100  \t\t\t\t\t#  [mole]\n",
      "\t# Therefore\n",
      "D = 60  \t\t\t\t\t# [mole]\n",
      "W = 40  \t\t\t\t\t# [mole]\n",
      "xf = 0.5 \n",
      "\n",
      "#Calculation\n",
      "\n",
      "\n",
      "import math\n",
      "from numpy import *\n",
      "from pylab import *\n",
      "y_star = [0.5,0.55,0.60,0.65,0.686,0.70,0.75] \n",
      "x = [0.317,0.361,0.409,0.460,0.5,0.516,0.577] \n",
      "#for i in range(1,7):\n",
      "#    f(i-1) = 1/(y_star(i-1)-x(i-1)) \n",
      "area = matrix([[0.317,5.464],[0.361,5.291],[0.409,5.236],[0.460,5.263],[0.5,5.376],[0.516,5.435],[0.577,7.78]]) \n",
      "\t# LHS of equation 6.11\n",
      "a = math.log(F/W) \n",
      "\n",
      "\n",
      "a1=plot(area[:,0],area[:,1],label='$area under curve$') \n",
      "legend(loc='upper left')\n",
      "xlabel(\"x\") \n",
      "ylabel(\"1/(y_satr-x)\") \n",
      "\n",
      "# When the area becomes equal to 0.916, integration is stopped  this occurs at \n",
      "xw = 0.33  # [mole fraction of heptane in residue]\n",
      "yd =( F*xf-W*xw)/D  # [mole fraction of heptane]\n",
      "#Result\n",
      "\n",
      "print\"The composition of the composited distillate and the residue are \",round(yd,3),\"and\",round(xw,3),\"respectively\\n\\n\"\n",
      "show(a1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The composition of the composited distillate and the residue are  0.613 and 0.33 respectively\n",
        "\n",
        "\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
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       "text": [
        "<matplotlib.figure.Figure at 0x762d0b8>"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.4,Page number:342"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "T = 298  \t\t\t\t\t\t\t\t# [K]\n",
      "Fa = 200  \t\t\t\t\t\t\t\t# [feed, kmole/hr]\n",
      "zf = 0.6 \n",
      "yd = 0.95  \n",
      "xd = yd \n",
      "xw = 0.05 \n",
      "q = 0.5  \t\t\t\t\t\t\t\t# [Lf/F]\n",
      "\n",
      "\n",
      "print\"Illustration 6.4(a)\"\n",
      "\t# Solution (a)\n",
      "\n",
      "import math\n",
      "from scipy.optimize import fsolve\n",
      "from numpy import *\n",
      "\n",
      "\t# Solution of simultaneous equation \n",
      "def F(e):\n",
      "    f1 = Fa - e[0]-e[1] \n",
      "    f2 = zf*Fa - yd*e[0] - xw*e[1] \n",
      "    return(f1,f2)\n",
      "\n",
      "\n",
      "# Initial guess\n",
      "e = [120,70] \n",
      "y = fsolve(F,e) \n",
      "D = y[0] \n",
      "W = y[1] \n",
      "print\"Quantity of liquid and vapor products are \",round(D,1),\"kmole/h and\",round(W,1),\"kmole/h respectively\"\n",
      "\n",
      "\n",
      "print\"Illustration 6.4(b)\"\n",
      "# Solution(b)\n",
      "\t# VLE data is generated in the same manner as generated in Example 6.1 by applying \tRaoult's law\n",
      "\t# VLE_data = [T,x,y]\n",
      "VLE_data = matrix([[379.4,0.1,0.21],[375.5,0.2,0.37],[371.7,0.3,0.51],[368.4,0.4,0.64],[365.1,0.5,0.71],[362.6,0.6,0.79],[359.8,0.7,0.86],[357.7,0.8,0.91],[355.3,0.9,0.96]]) \n",
      "\t# From figure 6.14\n",
      "\t# The minimum number of equilibrium stages is stepped off between the equilibrium curve and the 45 degree Iine, starting from the top, giving\n",
      "Nmin = 6.7 \n",
      "\n",
      "#Result\n",
      "\n",
      "print\"The minimum number of theoretical stages is\",Nmin\n",
      "\n",
      "print\"Illustration 6.4(c)\"\n",
      "\t# Solution(c)\n",
      "\t# Slope of q-line = Lf/F/(1-(Lf/F))\n",
      "s = q/(1-q) \n",
      "\t# For minimum reflux ratio\n",
      "\t# From figure 6.12 y-intercept is\n",
      "i = 0.457 \n",
      "# Therefore Rmin is\n",
      "Rmin = xd/i -1 \n",
      "\n",
      "#result\n",
      "\n",
      "print\"The minimum reflux ratio is\",round(Rmin,3),\"mole reflux/mole distillate\"\n",
      "\n",
      "print\"Illustration 6.4(d)\"\n",
      "\n",
      "\t# Solution(d)\n",
      "R = 1.3*Rmin \n",
      "\t# The y-intercept of the rectifying-section operating line is\n",
      "ia = xd/(R+1) \n",
      "\t# The operating line for the stripping section is drawn to pass through the point x = y = \txw =0.05 on the 45\" line and the point of intersection of the q-line   # and the \t\t\trectifying-section operating line.\n",
      "\t# Therefore from figure 6.15 \n",
      "Nact = 13 \n",
      "\t# But it include boiler\n",
      "Nact1 = Nact-1 \n",
      "print\"The number of equilibrium stages for the reflux ratio specified is\",Nact1\n",
      "\t# For the optimal feed-stage location, the transition from one operating line to the \tother occurs \tat the first opportunity\n",
      "\t# after passing the operating-line intersection \n",
      "\t# Therefore from figure 6.15 shows that \n",
      "\n",
      "\n",
      "print\"The optimal location of the feed stage for the reflux ratio specified is sixth from the top\"\n",
      "\n",
      "print\"Illustration 6.4(e)\"\n",
      "# Solution(e)\n",
      "L = R*D  \t\t\t\t\t\t\t\t# [kmole/h]\n",
      "V = L+D  \t\t\t\t\t\t\t\t# [kmole/h]\n",
      "# From equation 6.27\n",
      "Lst = L+q*Fa  \t\t\t\t\t\t\t\t# [kmole/h]\n",
      "# From equation 6.28\n",
      "Vst = V+(q-1)*Fa  \t\t\t\t\t\t\t# [kmole/h]\n",
      "\n",
      "\t# For 50% vaporization of the feed ( zf = 0.60), from calculations similar to those \t\tillustrated in Example 6.1, the separator temperature and the equilibrium     # \tcompositions are\n",
      "Tf = 365.5  \t\t\t\t\t\t\t\t# [K]\n",
      "yf = 0.707 \n",
      "xf = 0.493 \n",
      "\n",
      "\t# Latent heat vaporisation data at temperature T = 298 K\n",
      "lambdaA = 33.9  \t\t\t\t\t\t\t# [kJ/mole]\n",
      "lambdaB = 38  \t\t\t\t\t\t\t\t# [kJ/mole]\n",
      "\t# Heat capacities of liquids (298-366 K)\n",
      "Cla = 0.147  \t\t\t\t\t\t\t\t# [kJ/mole.K]\n",
      "Clb = 0.174  \t\t\t\t\t\t\t\t# [kJ/mole.K]\n",
      "\t# Heat capacities of gases, average in the range 298 to 366 K\n",
      "Cpa = 0.094  \t\t\t\t\t\t\t\t# [kJ/mole.K]\n",
      "Cpb = 0.118  \t\t\t\t\t\t\t\t# [kJ/mole.K]\n",
      "\t# Substituting in equation 6.6 gives\n",
      "Hf = 0 \n",
      "Hlf = (Tf-T)*(xf*Cla+(1-xf)*Clb)  \t\t\t\t\t# [kJ/mole of liquid \t\t\t\t\t\t\t\t\tfeed]\n",
      "\t# From equation 6.7\n",
      "Hvf = (Tf-T)*(yf*Cpa+(1-yf)*Cpb) + yf*lambdaA + (1-yf)*lambdaB  \t# [kJ/mole of vapor feed]\n",
      "\n",
      "Lf = Fa*q  \t\t\t\t\t# [kmole/h]\n",
      "Vf = Fa*(1-q)  \t\t\t\t\t# [kmole/h]\n",
      "\t# From equation 6.3\n",
      "Qf = (Hvf*Vf +Hlf*Lf-Fa*Hf)*1000.0/3600.0  # [kW]\n",
      "\n",
      "\n",
      "Tlo = 354.3  \t\t\t\t\t# [Bubble point temperature, K]\n",
      "T1 = 355.8  \t\t\t\t\t# [Dew point temperature, K]\n",
      "y1 = 0.95  \t\t\t\t\t# [composition of saturated vapor at dew point]\n",
      "x0 = 0.95  \t\t\t\t\t# [composition of saturated liquid at bubble \t\t\t\t\t\tpoint]\n",
      "Hv1 = (T1-T)*(y1*Cpa+(1-y1)*Cpb) + y1*lambdaA + (1-y1)*lambdaB  # [kJ/mole of vapor feed]\n",
      "Hlo = (Tlo-T)*(x0*Cla+(1-x0)*Clb)  \t\t# [kJ/mole of liquid feed]\n",
      "\n",
      "\t# An energy balance around condenser\n",
      "Qc = V*(Hv1-Hlo)*1000/3600  \t\t\t# [kW]\n",
      "\n",
      "\t# A flash-vaporization calculation is done in which the fraction vaporized is known \t(53.8/75.4 = \t0.714) and the concentration\n",
      "\t# of the liquid residue is fixed at xw = 0.05\n",
      "\t# The calculations yield\n",
      "Tr = 381.6  \t\t\t\t\t# [K]\n",
      "x12 = 0.093 \n",
      "y13 = 0.111 \n",
      "T12 = 379.7  \t\t\t\t\t# [Bubble point of the liquid entering in the \t\t\t\t\t\treboiler, K]\n",
      "\n",
      "Hl12 = (T12-T)*(x12*Cla+(1-x12)*Clb)  \t\t# [kJ/mole of liquid feed]\n",
      "Hv13 = (Tr-T)*(y13*Cpa+(1-y13)*Cpb) + y13*lambdaA + (1-y13)*lambdaB  # [kJ/mole of vapor feed]\n",
      "\n",
      "Hlw = (Tr-T)*(xw*Cla+(1-xw)*Clb)  # [kJ/mole of liquid feed]\n",
      "\n",
      "\t# An energy balance around the reboiler \n",
      "Qr = (Vst*Hv13+W*Hlw-Lst*Hl12)*1000.0/3600.0  \t# [kW]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"The thermal load of the condenser, reboiler, and feed preheater are\",round(Qc),\"kW\",round(Qr),\"kW\",\"and\", round(Qf),\"kW respectively\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Illustration 6.4(a)\n",
        "Quantity of liquid and vapor products are  122.2 kmole/h and 77.8 kmole/h respectively\n",
        "Illustration 6.4(b)\n",
        "The minimum number of theoretical stages is 6.7\n",
        "Illustration 6.4(c)\n",
        "The minimum reflux ratio is 1.079 mole reflux/mole distillate\n",
        "Illustration 6.4(d)\n",
        "The number of equilibrium stages for the reflux ratio specified is 12\n",
        "The optimal location of the feed stage for the reflux ratio specified is sixth from the top\n",
        "Illustration 6.4(e)\n",
        "The thermal load of the condenser, reboiler, and feed preheater are 2549.0 kW 1794.0 kW and 1466.0 kW respectively\n"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stderr",
       "text": [
        "C:\\Anaconda\\lib\\site-packages\\scipy\\optimize\\minpack.py:227: RuntimeWarning: The iteration is not making good progress, as measured by the \n",
        "  improvement from the last ten iterations.\n",
        "  warnings.warn(msg, RuntimeWarning)\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.7,Page number:357"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "\t# a-benzene   b-toluene\n",
      "xa = 0.46 \n",
      "xb = 0.54 \n",
      "Tb = 395  \t\t\t\t# [bottom temp., K]\n",
      "Tt = 360  \t\t\t\t# [top temp., K]\n",
      "alphab = 2.26 \n",
      "alphat = 2.52 \n",
      "D = 1.53  \t\t\t\t# [diameter of column, m]\n",
      "f = 0.81  \t\t\t\t# [flooding]\n",
      "deltaP = 700  \t\t\t\t# [average gas-pressure drop, Pa/tray]\n",
      "import math\n",
      "\n",
      "Tavg = (Tb+Tt)/2  \t\t\t# [K]\n",
      "alpha_avg = (alphab+alphat)/2 \n",
      "\n",
      "print \"Solution6.7(a)\" \n",
      "\t# Solution(a)\n",
      "\t\n",
      "\t# Constants for components 'a' and 'b'\n",
      "Aa = 4.612 \n",
      "Ba = 148.9 \n",
      "Ca = -0.0254 \n",
      "Da = 2.222*10**-5 \n",
      "ua =math.exp(Aa+Ba/Tavg+Ca*Tavg+Da*Tavg**2)  \t\t\t# [cP]\n",
      "\n",
      "Ab = -5.878 \n",
      "Bb = 1287 \n",
      "Cb = 0.00458 \n",
      "Db = -0.450*10**-5 \n",
      "\n",
      "\n",
      "#Calculation\n",
      "\n",
      "ub = math.exp(Ab+Bb/Tavg+Cb*Tavg+Db*Tavg**2)  \t\t\t# [cP]\n",
      "\n",
      "\t# At the average column temperature \n",
      "ul = ua**xa*ub**xb  \t\t\t\t\t\t# [cP]\n",
      "K = alpha_avg*ul \n",
      "\t# From the O\u2019Connell correlation\n",
      "Eo = 0.52782-0.27511*math.log10(K) + 0.044923*(math.log10(K))**2 \n",
      "\n",
      "#Result\n",
      "\n",
      "print\"The overall tray efficiency using the O\u2019Connell correlation is \",round(Eo,1) \n",
      "\n",
      "print \"Example 6.7(b)\"\n",
      "\t# Solution(b)\n",
      "\n",
      "#Calculation\n",
      "\n",
      "Nideal = 20  \t\t\t\t\t\t\t# [number of ideal stages]\n",
      "Nreal = Nideal/(Eo)  \t\t\t\t\t\t# [nnumber of real stages]\n",
      "print Nreal \n",
      "\t# Since real stages cannot be fractional, therefore\n",
      "Nreal = 34  \n",
      "\t# From Table 4.3 tray spacing \n",
      "t = 0.6  \t\t\t\t\t\t\t# [m]\n",
      "\t# Adding 1 m over the top tray as an entrainment separator and 3 m beneath # the bottom \ttray for bottoms surge capacity, the total column height is\n",
      "Z = 4+Nreal*t  \t\t\t\t\t\t\t# [m]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"The number of real trays and the total tower height are\",Nreal,\"and\",Z,\" m respectively\"\n",
      "\n",
      "print \"Solution 6.7(c)\"\n",
      "\t# Solution(c)\n",
      "\n",
      "\t# Total gas pressure drop\n",
      "deltaPc = deltaP*Nreal/1000  \t\t\t\t\t# [kPa]\n",
      "\n",
      "#Result\n",
      "\n",
      "print\"The total gas-pressure drop through the column is\",deltaPc,\"kPa\"  "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Solution6.7(a)\n",
        "The overall tray efficiency using the O\u2019Connell correlation is  0.6\n",
        "Example 6.7(b)\n",
        "32.1801990089\n",
        "The number of real trays and the total tower height are 34 and 24.4  m respectively\n",
        "Solution 6.7(c)\n",
        "The total gas-pressure drop through the column is 23 kPa\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.10,Page number:371"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from numpy import *\n",
      "\n",
      "#Variable declaration\n",
      "\t# A-toluene   B-1,2,3-trimethyl benzene   C-benzene\n",
      "\t# Solution of above three are ideal \n",
      "\t# Feed\n",
      "za = 0.40 \n",
      "zb = 0.30 \n",
      "zc = 0.30 \n",
      "\t# Bottom \n",
      "FRAd = 0.95  \t\t\t\t# [recovery of toluene in distillate]\n",
      "FRBw = 0.95  \t\t\t\t# [recovery of 1,2,3-trimethyl benzene in the bottom]\n",
      "P = 1  \t\t\t\t\t# [atm]\n",
      "\n",
      "\t# First estimate of distillate composition \n",
      "xc = 40/70 \n",
      "xa = 30/70 \n",
      "xb = 0 \n",
      "\t# The bubble point temperature for this solution is \n",
      "Tb = 390  \t\t\t\t# [K]\n",
      "\t# The corresponding parameters for benzene, toluene and 1,2,3-trimethyl benzene\n",
      "\t# For toluene\n",
      "Tc_a = 568.8  \t\t\t\t\t# [K]\n",
      "Pc_a = 24.9  \t\t\t\t\t# [bar]\n",
      "A_a = -7.912 \n",
      "B_a = 1.380 \n",
      "C_a =-3.804 \n",
      "D_a = -4.501 \n",
      "\t# For 1,2,3-trimethyl benzene\n",
      "Tc_b = 664.5  \t\t\t\t\t# [K]\n",
      "Pc_b = 34.5 \t\t\t\t\t# [bar]\n",
      "A_b = -8.442 \n",
      "B_b = 2.922 \n",
      "C_b =-5.667 \n",
      "D_b = -2.281 \n",
      "\t# For benzene\n",
      "Tc_c = 540.3  \t\t\t\t\t# [K]\n",
      "Pc_c = 27.4  \t\t\t\t\t# [bar]\n",
      "A_c = -7.675 \n",
      "B_c = 1.371 \n",
      "C_c =-3.536 \n",
      "D_c = -3.202 \n",
      "\n",
      "\n",
      "\t# At the estimated reboiler temperature of 449.3 K\n",
      "Tr = 449.3  \t\t\t\t\t# [K]\n",
      "\t# P = [Toluene 1,2,3-trimethyl benzene Benzene]\n",
      "\n",
      "\t# P = [Tc Pc A B C D]\n",
      "P1=zeros((3,6))\n",
      "P1= matrix([[568.8,24.9,-7.912,1.380,-3.804,-4.501],[664.5,34.5,-8.442,2.922,-5.667,2.281],[540.3,27.4,-7.675,1.371,-3.536,-3.202]]) \n",
      "\n",
      "for i in range(0,3):\n",
      "    P1[i]= P1[i,1]*math.exp((P1[i,2]*(1-Tr/P1[i,0])+P1[i,3]*(1-Tr/P1[i,0])**1.5+P1[i,4]*(1-Tr/P1[i,0])**3+P1[i,5]*(1-Tr/P1[i,0])**6)/(1-(1-Tr/P1[i,0]))) \n",
      "\n",
      "PA1 = P1.item(0)  \t\t\t\t\t# [bar]\n",
      "PB1 = P1.item(6)  \t\t\t\t\t# [bar]\n",
      "PC1 = P1.item(12)\t\t\t\t# [bar]\n",
      "alphaAB1 = PA1/PB1 \n",
      "alphaCB1 = PC1/PB1 \n",
      "\n",
      "\t\t# At the estimated distillate temperature of 390 K\n",
      "Td = 390  \t\t\t\t\t# [K]\n",
      "\t# P = [Toluene 1,2,3-trimethyl benzene Benzene]\n",
      "\t# P = [Tc,Pc,A,B,C,D]\n",
      "P2 = zeros((3,6)) \n",
      "P2= matrix([[568.8,24.9,-7.912,1.380,-3.804,-4.501],[664.5,34.5,-8.442,2.922,-5.667,2.281],[540.3,27.4,-7.675,1.371,-3.536,-3.202]]) \n",
      "for i in range(0,3):\n",
      "    P2[i] = P2[i,1]*math.exp((P2[i,2]*(1-Td/P2[i,0])+P2[i,3]*(1-Td/P2[i,0])**1.5+P2[i,4]*(1-Td/P2[i,0])**3+P2[i,5]*(1-Td/P2[i,0])**6)/(1-(1-Td/P2[i,0])))\n",
      "    \n",
      "PA2 = P2.item(0)  # [bar]\n",
      "PB2 = P2.item(6)  # [bar]\n",
      "PC2 = P2.item(12)  # [bar]\n",
      "alphaAB2 = PA2/PB2 \n",
      "alphaCB2 = PC2/PB2 \n",
      "# The geometric-average relative volatilities are\n",
      "alphaAB_avg = math.sqrt(alphaAB1*alphaAB2) \n",
      "alphaCB_avg = math.sqrt(alphaCB1*alphaCB2) \n",
      "\n",
      "# From equation 6.66\n",
      "Nmin = math.log(FRAd*FRBw/((1-FRAd)*(1-FRBw)))/log(alphaAB_avg) \n",
      "\n",
      "# From equation 6.67\n",
      "FRCd = alphaCB_avg**Nmin/((FRBw/(1-FRBw))+alphaCB_avg**Nmin)  # [fractional recovery of benzene in the distillate]\n",
      "\n",
      "#Results\n",
      "\n",
      "\n",
      "print\"The number of equilibrium stages required at total reflux is\",round(Nmin,2)\n",
      "print\"The recovery fraction of benzene in the distillate is \",round(FRCd,3)\n",
      "print\"\\n\\nThus, the assumption that virtually all of the LNK will be recovered in the distillate is justified\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The number of equilibrium stages required at total reflux is 4.32\n",
        "The recovery fraction of benzene in the distillate is  0.997\n",
        "\n",
        "\n",
        "Thus, the assumption that virtually all of the LNK will be recovered in the distillate is justified\n"
       ]
      }
     ],
     "prompt_number": 41
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.11,Page number:376"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "\t# 1-toluene   2-1,2,3--trimethylbenzene   3-benzene\n",
      "\t# Basis: 100 kmol/h of feed\n",
      "F = 100  \t\t\t\t\t\t# [kmole/h]\n",
      "\t# Since feed is saturated, therefore\n",
      "q = 0 \n",
      "\t# From example 6.10\n",
      "x1d = 0.3 \n",
      "x2d = 0.3 \n",
      "x3d = 0.4 \n",
      "a12 = 3.91 \n",
      "a32 = 7.77 \n",
      "a22 = 1  \n",
      "\n",
      "#Calculation\n",
      "\n",
      "import math\n",
      "from scipy.optimize import fsolve\n",
      "\t# Equ 6.78 gives\n",
      "def f14(Q):\n",
      "    return(1- a12*x1d/(a12-Q)-a22*x2d/(a22-Q)-a32*x3d/(a32-Q)) \n",
      "Q = fsolve(f14,2) \n",
      "\n",
      "\t# From the problem statement\n",
      "\t# d1 = D*x1d    d2 = D*x2d\n",
      "d1 = F*x1d*0.95  \t\t\t\t\t# [kmol/h]\n",
      "d2 = F*x2d*0.05  \t\t\t\t\t# [kmol/h]\n",
      "d3 = F*x3d*0.997  \t\t\t\t\t# [kmol/h]\n",
      "\n",
      "\t\t# Summing the three distillate, d1,d2 and d3\n",
      "D = d1+d2+d3  \t\t\t\t\t\t# [kmole/h]\n",
      "\n",
      "Vmin = a12*d1/(a12-Q)+a22*d2/(a22-Q)+a32*d3/(a32-Q) \n",
      "\n",
      "\t\t# From the mass balance \n",
      "Lmin = Vmin-D  \t\t\t\t\t\t# [kmol/h]\n",
      "\t# Minimum reflux ratio\n",
      "Rmin = Lmin/D \n",
      "\n",
      "#Results\n",
      "print\"The minimum reflux ratio is \",round(Rmin[0],3) "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The minimum reflux ratio is  0.717\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.12,Page number:377"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# x-mole fraction  a-relative volatility\n",
      "xA = 0.25   \n",
      "aA = 4.08 \n",
      "xB = 0.11   \n",
      "aB = 1.00 \n",
      "xC = 0.26   \n",
      "aC = 39.47 \n",
      "xD = 0.09   \n",
      "aD = 10.00 \n",
      "xE = 0.17   \n",
      "aE = 2.11 \n",
      "xF = 0.12   \n",
      "aF = 0.50 \n",
      "\n",
      "FRlkd = 0.98 \n",
      "FRhkd = 0.01 \n",
      "import math\n",
      "from scipy.optimize import fsolve\n",
      "from numpy import *\n",
      "\t# For methane\n",
      "D_CR = (aC-1)/(aA-1)*FRlkd + (aA-aC)/(aA-1)*FRhkd \n",
      "\t# For ethane\n",
      "D_DR = (aD-1)/(aA-1)*FRlkd + (aA-aD)/(aA-1)*FRhkd \n",
      "\t# For butane\n",
      "D_ER = (aE-1)/(aA-1)*FRlkd + (aA-aE)/(aA-1)*FRhkd \n",
      "\t# For hexane\n",
      "D_FR = (aF-1)/(aA-1)*FRlkd + (aA-aF)/(aA-1)*FRhkd \n",
      "\t# Since the feed is 66% vaporized\n",
      "q = 1-0.66 \n",
      "\n",
      "\t# Now equation 6.82 is solved for two values of Q\n",
      "def f14(Q1):\n",
      "    return(0.66 - aA*xA/(aA-Q1)-aB*xB/(aB-Q1)-aC*xC/(aC-Q1)-aD*xD/(aD-Q1)-aE*xE/(aE-Q1)-aF*xF/(aF-Q1)) \n",
      "Q1 = fsolve(f14,1.2) \n",
      "\n",
      "def f15(Q2):\n",
      "    return(0.66 - aA*xA/(aA-Q2)-aB*xB/(aB-Q2)-aC*xC/(aC-Q2)-aD*xD/(aD-Q2)-aE*xE/(aE-Q2)-aF*xF/(aF-Q2)) \n",
      "Q2 = fsolve(f15,2.5) \n",
      "\n",
      "\t# Basis: 100 mole of feed\n",
      "F = 100  \t\t\t\t\t\t\t# [mole]\n",
      "\t# Let d1 = Dxad, d2 = Dxbd, d3 = Dxcd, and so on\n",
      "d1 = F*xA*FRlkd  \t\t\t\t\t\t# [moles of propane]\n",
      "d2 = F*xB*FRhkd  \t\t\t\t\t\t# [moles of pentane]\n",
      "d3 = F*xC  \t\t\t\t\t\t\t# [moles of methane]\n",
      "d4 = F*xD  \t\t\t\t\t\t\t# [moles of ethane]\n",
      "d6 = F*xF*0  \t\t\t\t\t\t\t# [moles of hexane]\n",
      "\t# And d5 is unknown\n",
      "\t# Applying equation 6,78 for each value of Q\n",
      "\n",
      "\t# Solution of simultaneous equation \n",
      "#Vmin=aA*d1/(aA-Q1)+aB*d2/(aB-Q1)+aC*d3/(aC-Q1)+aD*d4/(aD-Q1)+aE*d5/(aE-Q1)+aF*d6/(aF-Q1)\n",
      "#Vmin=aA*d1/(aA-Q2)+aB*d2/(aB-Q2)+aC*d3/(aC-Q2)+aD*d4/(aD-Q2)+aE*d5/(aE-Q2)+aF*d6/(aF-Q2)\n",
      "# we get\n",
      "d5=-(72.243-121.614)/(2.494+2.863)\n",
      "Vmin=72.243 + 2.494*D5\n",
      "\n",
      "\t# From equ 6.84\n",
      "D = d1+d2+d3+d4+d5+d6  \t\t\t\t\t\t# [mole]\n",
      "# From mass balance \n",
      "Lmin = Vmin-D  \t\t\t\t\t\t\t# [mole]\n",
      "# For minimum reflux ratio\n",
      "Rmin = Lmin/D \n",
      "print\"The minimum reflux ratio is\",round(Rmin,3)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The minimum reflux ratio is 0.384\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.13,Page number:380"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from scipy.optimize import fsolve\n",
      "from numpy import *\n",
      "\n",
      "#Variable declaration\n",
      "\t# A-benzene  B-toluene  C-1,2,3-trimethylbenzene\n",
      "\t# From example 6.10\n",
      "Nmin = 4.32  # [stages]\n",
      "\t# From example 6.11\n",
      "Rmin = 0.717  \t\t\t\t\t\t# [minimum reflux ratio]\n",
      "\t# For R = 1\n",
      "R = 1 \n",
      "X = (R-Rmin)/(R+1) \n",
      "\t# From equ 6.88\n",
      "Y = 1-math.exp((1+54.4*X)/(11+117.2*X)*(X-1)/math.sqrt(X)) \n",
      "\t# Fro equ 6.86\n",
      "N = (Y+Nmin)/(1-Y) \n",
      "\t# From example 6.10 99.7% of the LNK (benzene) is recovered in the distillate# , 95% of \t\tthe light key is in the distillate, and 95% of the heavy key is in# the bottoms\n",
      "\n",
      "\t# For a basis of 100 mol of feed, the material balances for three components # are\n",
      "\t# For distillate\n",
      "nAd = 39.88  \t\t\t\t\t\t# [LNK, moles of benzene]\n",
      "nBd = 28.5  \t\t\t\t\t\t# [LK, moles of toluene]\n",
      "nCd = 1.50  \t\t\t\t\t\t# [HK, moles of 1,2,3-trimethylbenzene]\n",
      "nTd = nAd+nBd+nCd  \t\t\t\t\t# [total number of moles]\n",
      "xAd = nAd/nTd \n",
      "xBd = nBd/(nTd) \n",
      "xCd = nCd/(nTd) \n",
      "\n",
      "\t# For bottoms\n",
      "nAb = 0.12 \n",
      "nBb = 1.50 \n",
      "nCb = 28.50 \n",
      "nTb = nAb+nBb+nCb \n",
      "xAb = nAb/nTb \n",
      "xBb = nBb/nTb \n",
      "xCb = nCb/nTb \n",
      "\n",
      "D = nTd \n",
      "W = nTb \n",
      "\t# From problem statement\n",
      "Zlk = 0.3 \n",
      "Zhk = Zlk \n",
      "\t# Substituting in equation 6.89\n",
      "\t# T = Nr/Ns\n",
      "T = (Zhk/Zlk*W/D*(xBb/xCd)**2)**0.206 \n",
      "\n",
      "\t# Solution of simultaneous equation \n",
      "def H(e):\n",
      "    f1 = e[0]-e[1]*T \n",
      "    f2 = e[0]+e[1]-N \n",
      "    return(f1,f2)        \n",
      "\n",
      "\n",
      "\t# Initial guess\n",
      "e = [5,4] \n",
      "y = fsolve(H,e) \n",
      "Nr = y[0]  \t\t\t\t\t# [number of stages in rectifying section]\n",
      "Ns = y[1]  \t\t\t\t\t# [number of stages in stripping section]\n",
      "\n",
      "#Result\n",
      "\n",
      "print \"Nr=\",round(Nr,2),\"Ns=\",round(Ns,2),\"\\n\"\n",
      "print\"Rounding the estimated equilibrium stage requirement leads to 1 stage as a partial reboiler, 4 stages below the feed, and 5 stages above the feed\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Nr= 5.39 Ns= 4.53 \n",
        "\n",
        "Rounding the estimated equilibrium stage requirement leads to 1 stage as a partial reboiler, 4 stages below the feed, and 5 stages above the feed\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 6.14,Page number:387"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "\t# a-acetone  b-methanol  c-water\n",
      "yna = 0.2971  \n",
      "yn1a = 0.17  \n",
      "ynIa = 0.3521  \n",
      "mnIa = 2.759  \n",
      "xna = 0.1459 \n",
      "ynb = 0.4631  \n",
      "yn1b = 0.429  \n",
      "ynIb = 0.4677  \n",
      "mnIb = 1.225  \n",
      "xnb = 0.3865 \n",
      "ync = 0.2398  \n",
      "yn1c = 0.4010  \n",
      "ynIc = 0.1802  \n",
      "mnIc = 0.3673  \n",
      "xnc = 0.4676 \n",
      "\n",
      "Fabv = 4.927  \t\t\t\t\t\t# [mol/square m.s]\n",
      "Facv = 6.066  \t\t\t\t\t\t# [mol/square m.s]\n",
      "Fbcv = 7.048  \t\t\t\t\t\t# [mol/square m.s]\n",
      "aI = 50  \t\t\t\t\t\t# [square m]\n",
      "Vn1 = 188  \t\t\t\t\t\t# [mol/s]\n",
      "Vn = 194.8  \t\t\t\t\t\t# [mol/s]\n",
      "\n",
      "print \"Solution6.14(a)\" \n",
      "from numpy import *\n",
      "import math\n",
      "\t# Solution(a)\n",
      "\n",
      "ya = (yna+ynIa)/2 \n",
      "yb = (ynb+ynIb)/2 \n",
      "yc = (ync+ynIc)/2 \n",
      "\n",
      "Rav = ya/Facv+yb/Fabv+yc/Facv \n",
      "Rbv = yb/Fbcv+ya/Fabv+yc/Fbcv \n",
      "\n",
      "Rabv = -ya*(1/Fabv-1/Facv) \n",
      "Rbav = -yb*(1/Fabv-1/Fbcv) \n",
      "\t# Thus in matrix form\n",
      "Rv =matrix([[Rav,Rabv],[Rbav,Rbv]]) \n",
      "kv = Rv.I  \t\t\t# [inverse of Rv]\n",
      "\t# From equ 6.99\n",
      "b =matrix([[yna-ynIa],[ynb-ynIb]]) \n",
      "J = kv*b \n",
      "\n",
      "\t# From equ 6.98\n",
      "Jc = -sum(J)  \t\t\t\t# [mol/square m.s]\n",
      "\n",
      "print\"The molar diffusional rates of acetone, methanol and water are\",round(J[0][0],4),\"mol/square m.s\",round(J[1][0],4),\"mol/square m.s and\",round(Jc,3) ,\"mol/square m.s respectively\"\n",
      "print \"Solution 6.14(b)\\n\" \n",
      "\t# Solution(b)\n",
      "Ntv = Vn1-Vn  \t\t\t\t\t\t# [mol/s]\n",
      "\n",
      "\t# From equation 6.94\n",
      "Nta = aI*J[0][0]+ya*Ntv \n",
      "Ntb = aI*J[1][0]+yb*Ntv \n",
      "Ntc = aI*Jc+yc*Ntv \n",
      "print\"The mass transfer rates of acetone, methanol and water are\",round(Nta,1),\"mol/s\" ,round(Ntb,1),\" mol/s and\", round(Ntc),\" mol/s respectively\"\n",
      "\n",
      "print \"Example6.14(c)\\n\"\n",
      "\t# Solution(c)\n",
      "\n",
      "\t# Approximate values of Murphree vapor tray efficiency are obtained from   # equation \t6.105\n",
      "\n",
      "EMG_a = (yna-yn1a)/(mnIa*xna-yn1a) \n",
      "EMG_b = (ynb-yn1b)/(mnIb*xnb-yn1b) \n",
      "EMG_c = (ync-yn1c)/(mnIc*xnc-yn1c) \n",
      "\n",
      "print\"The Murphree vapor tray efficiencies for acetone, methanol and water are\",round(EMG_a,3), round(EMG_b,3),\"and\",round(EMG_c,3), \"respectively\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Solution6.14(a)\n",
        "The molar diffusional rates of acetone, methanol and water are"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " -0.3068 mol/square m.s -0.0824 mol/square m.s and 0.389 mol/square m.s respectively\n",
        "Solution 6.14(b)\n",
        "\n",
        "The mass transfer rates of acetone, methanol and water are -17.5 mol/s -7.3  mol/s and 18.0  mol/s respectively\n",
        "Example6.14(c)\n",
        "\n",
        "The Murphree vapor tray efficiencies for acetone, methanol and water are 0.547 0.767 and 0.703 respectively\n"
       ]
      }
     ],
     "prompt_number": 10
    }
   ],
   "metadata": {}
  }
 ]
}