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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 5:Absorption and Stripping"
     ]
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.1,Page number:287"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "\n",
      "\t# Component 'A' is to be absorbed #\n",
      "y_N1 = 0.018\t\t\t\t\t\t# [mole fraction 'A' of in entering gas]\n",
      "y_1 = 0.001\t\t\t\t\t\t# [mole fractio of 'A'in leaving gas]\n",
      "x_0 = 0.0001\t\t\t\t\t\t# [mole fraction of 'A' in entering \t\t\t\t\t\t\tliquid]\n",
      "m = 1.41\t\t\t\t\t\t# [m = yi/xi]\n",
      "n_1 = 2.115\t\t\t\t\t\t# [molar liquid to gas ratio at bottom, \t\t\t\t\t\t\tL/V]\t\n",
      "n_2 = 2.326\t\t\t\t\t\t# [molar liquid to gas ratio at top, L/V]\n",
      "E_MGE = 0.65 \n",
      "\n",
      "\n",
      "print\"Answer 5.1 (a)\"\n",
      "\t# Solution (a)\n",
      "\n",
      "\n",
      "A_1 = n_1/m\t\t\t\t\t\t# [absorption factor at bottom]\n",
      "A_2 = n_2/m\t\t\t\t\t\t# [absorption factor at top]\n",
      "import math\n",
      "from scipy.optimize import fsolve\n",
      "A = math.sqrt(A_1*A_2) \n",
      "\t# Using equation 5.3 to calculate number of ideal stages\n",
      "N = (math.log(((y_N1-m*x_0)/(y_1-m*x_0))*(1-1/A) + 1/A))/math.log(A)\t# [number of ideal \t\t\t\t\t\t\t\t\tstages]\n",
      "print\"Number of ideal trays is\",round(N,3)\n",
      "\t# Using equation 5.5\n",
      "E_o = math.log(1+E_MGE*(1/A-1))/math.log(1/A) \n",
      "\t# Therefore number of real trays will be\n",
      "n = N/E_o\n",
      "\n",
      "#Result\n",
      " \n",
      "print\"Number of real trays is\",round(n,2)\n",
      "\n",
      "print\"Since it is not possible to specify a fractional number of trays, therefore number of real trays is\",round(n)\n",
      "\n",
      "\n",
      "print\"\\nAnswer5.1 (b)\"\n",
      "\n",
      "\t# Solution (b)\n",
      "\n",
      "\t# Back checking the answer\n",
      "print\"Back checking the answer\"\n",
      "\n",
      "#Calculation\n",
      "\n",
      "N_o = E_o*n \n",
      "\t# Putting N_o in equation 5.3 to calculate y_1\n",
      "def f16(Z):\n",
      "    return(N_o-(math.log(((y_N1-m*x_0)/(Z-m*x_0))*(1-1/A) + 1/A))/math.log(A)) \n",
      "Z = fsolve(f16,0.001) \n",
      "print\"Mole fraction of A in leaving gas is\",Z[0],\"percent which satisfies the requirement that the gas exit concentration should not exceed 0.1 percent\"\n",
      "\n",
      "\t# For a tower diameter of 1.5 m, Table 4.3 recommends a plate spacing of 0.6 m\n",
      "Z = n*0.6\t\t\t\t\t\t# [Tower height, m]\n",
      "\n",
      "#Result\n",
      "print\"The tower height will be\",round(Z,1),\"m\""
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Answer 5.1 (a)\n",
        "Number of ideal trays is"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " 4.647\n",
        "Number of real trays is 7.79\n",
        "Since it is not possible to specify a fractional number of trays, therefore number of real trays is 8.0\n",
        "\n",
        "Answer5.1 (b)\n",
        "Back checking the answer\n",
        "Mole fraction of A in leaving gas is 0.001 percent which satisfies the requirement that the gas exit concentration should not exceed 0.1 percent\n",
        "The tower height will be 4.7 m\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Exmaple 5.3,Page number:295"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "\t# For tower diameter, packed tower design program of Appendix D is run using # the data \t\tfrom Example 5.2 and packing parameters from Chapter 4.\n",
      "\n",
      "\t# For a pressure drop of 300 Pa/m, the program converges to a tower diameter \n",
      "Db = 0.641  \t\t\t\t\t\t# [m]\n",
      "\t# Results at the bottom of tower\n",
      "fb= 0.733  \t\t\t\t\t\t# [flooding]\n",
      "ahb = 73.52 \t\t\t\t\t\t# [m**-1]\n",
      "Gmyb = 126  \t\t\t\t\t\t# [mol/square m.s]\n",
      "kyb = 3.417  \t\t\t\t\t\t# [mol/square m.s]\n",
      "klb = 9.74*10**-5  \t\t\t\t\t# [m/s]\n",
      "\n",
      "\t# From equation 2.6 and 2.11\n",
      "\t# Fg = ky*(1-y),   Fl = kx*(1-x)\n",
      "\t# Assume 1-y = 1-y1    1-x = 1-x1\n",
      "\t# let t = 1-y1  u = 1-x1\n",
      "\t# Therefore\n",
      "t = 0.926 \n",
      "u = 0.676 \n",
      "Fgb = kyb*t  \t\t\t\t\t\t# [mol/square m.s]\n",
      "rowlb = 780  \t\t\t\t\t\t# [kg/cubic m]\n",
      "Mlb = 159.12  \t\t\t\t\t\t# [gram/mole]\n",
      "c = rowlb/Mlb  \t\t\t\t\t\t# [kmle/cubic m]\n",
      "Flb = klb*c*u  \t\t\t\t\t\t# [mol/square m.s]\n",
      "\t# From equ 5.19\n",
      "Htgb = Gmyb/(Fgb*ahb)  \t\t\t\t\t# [m]\n",
      "import math\n",
      "from scipy.optimize import fsolve\n",
      "from numpy import *\n",
      "from pylab import *\n",
      "\t# Now, we consider the conditions at the top of the absorber\n",
      "\t# For a pressure drop of 228 Pa/m, the program converges to a tower        # diameter\n",
      "Dt = 0.641  \t\t\t\t\t\t# [m]\n",
      "\t# Results at the top of tower\n",
      "ft = 0.668  \t\t\t\t\t\t# [flooding]\n",
      "aht = 63.31  \t\t\t\t\t\t# [m**-1]\n",
      "Gmyt = 118  \t\t\t\t\t\t# [mol/square m.s]\n",
      "kyt = 3.204  \t\t\t\t\t\t# [mol/square m.s]\n",
      "klt = 8.72*10**-5  \t\t\t\t\t# [m/s]\n",
      "\n",
      "rowlt = 765  \t\t\t\t\t\t# [kg/cubic m]\n",
      "Mlt = 192.7  \t\t\t\t\t\t# [gram/mole]\n",
      "cl = rowlt/Mlt  \t\t\t\t\t# [kmole/cubic m]\n",
      "Fgt = kyt*0.99  \t\t\t\t\t# [mole/square m.s]\n",
      "Flt = klb*cl*0.953  \t\t\t\t\t# [mole/square m.s]\n",
      "\t# From equ 5.19\n",
      "Htgt = Gmyt/(Fgt*aht)  \t\t\t\t\t# [m]\n",
      "Htg_avg = (Htgb+Htgt)/2  \t\t\t\t# [m]\n",
      "Fg_avg = (Fgt+Fgb)/2  \t\t\t\t\t# [mole/square m.s]\n",
      "Fl_avg = (Flb+Flt)*1000/2  \t\t\t\t# [mole/square m.s]\n",
      "\n",
      "\t# The operating curve equation for this system in terms of mole fractions\n",
      "\t# y = \n",
      "\n",
      "\t# From Mathcad program figure 5.3\n",
      "x1 = 0.324 \n",
      "x2 = 0.0476 \n",
      "n = 50 \n",
      "dx = (x1-x2)/n \n",
      "me = 0.136 \n",
      "T = zeros((50,2)) \n",
      "y=zeros((50))\n",
      "x=zeros((50))\n",
      "yint=zeros((50))\n",
      "fd=zeros((50))\n",
      "for j in range(1,51):\n",
      "    x[j-1] = x2+j*dx \n",
      "    y[j-1] = (0.004+0.154*x[j-1])/(1.004-0.846*x[j-1]) \n",
      "    \n",
      "    def f12(yint):\n",
      "        return((1-yint)/(1-y[j-1]) - ((1-x[j-1])/(1-yint/me))**(Fl_avg/Fg_avg)) \n",
      "    yint[j-1] = fsolve(f12,0.03) \n",
      "    fd[j-1] = 1/(y[j-1]-yint[j-1]) \n",
      "    T[j-1][0] = y[j-1] \n",
      "\n",
      "    T[j-1][1] = fd[j-1] \n",
      "\n",
      "\n",
      "#Result\n",
      "\n",
      "a1=plot(T[:,0],T[:,1]) \n",
      "\n",
      "xlabel(\"y\") \n",
      "ylabel(\"f = 1/(y-yint)\") \n",
      "\n",
      "yo = y[0] \n",
      "yn = y[49] \n",
      "\t# From graph between f vs y\n",
      "Ntg = 10.612 \n",
      "\t# Therefore\n",
      "Z = Htg_avg*Ntg\t\t  \t\t\t\t\t# [m]\n",
      "\t\n",
      "print\"The total packed height is\",round(Z),\"m.\"\t\n",
      "deltaPg = 300*Z \t \t\t\t\t\t# [Pa]\n",
      "Em = 0.60 \t\t\t\t\t\t\t # [mechanical efficiency]\n",
      "Qg = 1.0 \n",
      "Wg = (Qg*deltaPg)/Em  \t\t\t\t\t\t# [Power required to force the \t\t\t\t\t\t\t\tgas through the tower, W]\n",
      "L2 = 1.214  \t\t\t\t\t\t\t# [kg/s]\n",
      "g = 9.8  \t\t\t\t\t\t\t# [m/square s]\n",
      "Wl = L2*g*Z/Em  \t\t\t\t\t\t# [Power required to pump the \t\t\t\t\t\t\t\tliquid to the top of the \t\t\t\t\t\t\t\tabsorber, W]\n",
      "\n",
      "\n",
      "print\"The power required to force the gas through the tower is \",round(Wg),\"W.\\n\\n\"\n",
      "print\"The power required to pump the liquid to the top of the absorber is \",round(Wl),\"W.\\n\\n\"\n",
      "\n",
      "show(a1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The total packed height is 6.0 m.\n",
        "The power required to force the gas through the tower is  2996.0 W.\n",
        "\n",
        "\n",
        "The power required to pump the liquid to the top of the absorber is  119.0 W.\n",
        "\n",
        "\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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G6GiYPRuOHLE6mYh/0BiGyGl89hlMnmwOjD/2GAwZYnZjidRHGvQW8TLDgGXL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\n2poAAAAASUVORK5CYII=\n",
       "text": [
        "<matplotlib.figure.Figure at 0x79c25f8>"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.4,Page number:299"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "m = 0.57 \n",
      "D = 0.738  \t\t\t\t# [tower diameter, m]\n",
      "G = 180.0  \t\t\t\t# [rate of gas entering the tower, kmole/h]\n",
      "L = 151.5  \t\t\t\t# [rate of liquid leaving the tower, kmole/h]\n",
      "\t# Amount of ethanol absorbed \n",
      "M = G*0.02*0.97  \t\t\t# [kmole/h]\n",
      "\n",
      "#Calculation\n",
      "\n",
      "import math\n",
      "\t# Inlet gas molar velocity\n",
      "Gmy1 = G*4/(3600*math.pi*D**2)  \t# [kmole/square m.s]\n",
      "\t# Outlet gas velocity\n",
      "Gmy2 = (G-M)*4/(3600*math.pi*D**2)  \t# [kmole/square m.s]\n",
      "\t# Average molar gas velocity\n",
      "Gmy = (Gmy1+Gmy2)/2  \t\t\t# [kmole/square m.s]\n",
      "\n",
      "\t# Inlet liquid molar velocity\n",
      "Gmx2 = L*4/(3600*math.pi*D**2)  \t# [kmole/square m.s]\n",
      "\t# Outlet liquid molar velocity\n",
      "Gmx1 = (L+M)*4/(3600*math.pi*D**2)  \t# [kmole/square m.s]\n",
      "\n",
      "\t# Absorption factor at both ends of the column:\n",
      "A1 = Gmx1/(m*Gmy1) \n",
      "A2 = Gmx2/(m*Gmy2) \n",
      "\t# Geometric average\n",
      "A = math.sqrt(A1*A2) \n",
      "\n",
      "y1 = 0.02 \n",
      "\t# For 97% removal of the ethanol\n",
      "y2 = 0.03*0.02 \n",
      "\t# Since pure water is used \n",
      "x2 = 0 \n",
      "\t# From equation 5.24\n",
      "Ntog = math.log((y1-m*x2)/(y2-m*x2)*(1-1/A)+1/A)/(1-1/A) \n",
      "\n",
      "\t# From example 4.4\n",
      "\t# ky*ah = 0.191 kmole/cubic m.s\n",
      "\t# kl*ah = 0.00733 s**-1\n",
      "kyah = 0.191  \t\t\t\t# [kmole/cubic m.s]\n",
      "klah = 0.00733  \t\t\t# [s**-1]\n",
      "rowl = 986  \t\t\t\t# [kg/cubic m]\n",
      "Ml = 18.0  \t\t\t\t# [gram/mole]\n",
      "c = rowl/Ml  \t\t\t\t# [kmole/cubic m]\n",
      "kxah = klah*c  \t\t\t\t# [kmole/cubic m.s]\n",
      "\n",
      "\t# Overall volumetric mass transfer coefficient\n",
      "Kyah = (kyah**-1 + m/kxah)**-1  \t# [kmole/cubic m.s]\n",
      "\n",
      "\t# From equation 5.22\n",
      "Htog = Gmy/Kyah  \t\t\t\t# [m]\n",
      "\t# The packed height is given by equation 5.21,\n",
      "Z = Htog*Ntog  \t\t\t\t\t# [m]\n",
      "\n",
      "#Result\n",
      "print\"The packed height of an ethanol absorber is\",round(Z,2),\"m.\\n\\n\" "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The packed height of an ethanol absorber is 5.66 m.\n",
        "\n",
        "\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 3,
     "metadata": {},
     "source": [
      "Example 5.5,Page number:302"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Variable declaration\n",
      "# a = CH4 b = C5H12\n",
      "Tempg = 27 \t\t\t\t\t\t\t# [OC]\n",
      "Tempo = 0 \t\t\t\t\t\t\t# [base temp,OC]\n",
      "Templ = 35 \t\t\t\t\t\t\t# [OC]\n",
      "xa = 0.75 \t\t\t\t\t\t\t# [mole fraction of CH4 in gas]\n",
      "xb = 0.25 \t\t\t\t\t\t\t# [mole fraction of C5H12 in gas]\n",
      "M_Paraffin = 200 \t\t\t\t\t\t# [kg/kmol]\n",
      "hb = 1.884 \t\t\t\t\t\t\t# [kJ/kg K]\n",
      "\n",
      "\n",
      "Ha = 35.59 \t\t\t\t\t\t\t# [kJ/kmol K]\n",
      "Hbv = 119.75 \t\t\t\t\t\t\t# [kJ/kmol K]\n",
      "Hbl = 117.53 \t\t\t\t\t\t\t# [kJ/kmol K]\n",
      "Lb = 27820 \t\t\t\t\t\t\t# [kJ/kmol]\n",
      "\t\t# M = [Temp (OC) m]\n",
      "import math\n",
      "from scipy.optimize import fsolve\n",
      "from numpy import *\n",
      "from pylab import *\n",
      "\n",
      "#Calculation\n",
      "\n",
      "M = matrix([[20,0.575],[25,0.69],[30,0.81],[35,0.95],[40,1.10],[43,1.25]])\n",
      "\t# Basis: Unit time\n",
      "GNpPlus1 = 1.0 \t\t\t\t\t\t\t# [kmol]\n",
      "yNpPlus1 = 0.25 \t\t\t\t\t\t# [kmol]\n",
      "HgNpPlus1 = ((1-yNpPlus1)*Ha*(Tempg-Tempo))+(yNpPlus1*(Hbv*(Tempg-Tempo)+Lb)) # [kJ/kmol]\n",
      "L0 = 2.0 \t\t\t\t\t\t\t# [kmol]\n",
      "x0 = 0 \t\t\t\t\t\t\t\t# [kmol]\n",
      "HL0 = ((1-x0)*hb*M_Paraffin*(Templ-Tempo))+(x0*hb*(Templ-Tempo))\t # [kJ/kmol]\n",
      "C5H12_absorbed = 0.98*xb \t\t\t\t\t# [kmol]\n",
      "C5H12_remained = xb-C5H12_absorbed \n",
      "G1 = xa+C5H12_remained \t\t\t\t\t\t# [kmol]\n",
      "y1 = C5H12_remained/G1 \t\t\t\t\t\t# [kmol]\n",
      "LNp = L0+C5H12_absorbed \t\t\t\t\t# [kmol]\n",
      "xNp = C5H12_absorbed/LNp \t\t\t\t\t# [kmol]\n",
      "\t# Assume:\n",
      "Temp1 = 35.6 \t\t\t\t\t\t\t# [OC]\n",
      "Hg1 = ((1-y1)*Ha*(Temp1-Tempo))+(y1*(Hbv*(Temp1-Tempo)+Lb)) \t# [kJ/kmol]\n",
      "Qt = 0 \n",
      "def f30(HlNp):\n",
      "    return(((L0*HL0)+(GNpPlus1*HgNpPlus1))-((LNp*HlNp)+(G1*Hg1)+Qt)) \n",
      "HlNp = fsolve(f30,2) \n",
      "\n",
      "def f31(TempNp):\n",
      "    return(HlNp-(((1-x0)*hb*M_Paraffin*(TempNp-Tempo))+(x0*hb*(TempNp-Tempo)))) \n",
      "TempNp = fsolve(f31,35.6) \n",
      "\t# At Temp = TempNp:\n",
      "mNp = 1.21 \n",
      "yNp = mNp*xNp \t\t\t\t\t\t\t# [kmol]\n",
      "GNp = G1/(1-yNp) \t\t\t\t\t\t# [kmol]\n",
      "HgNp = ((1-yNp)*Ha*(TempNp-Tempo))+(yNp*(Hbv*(TempNp-Tempo)+Lb)) # [kJ/kmol]\n",
      "\t# From equation 5.28 with n = Np-1\n",
      "def f32(LNpMinus1):\n",
      "    return(LNpMinus1+GNpPlus1-(LNp+GNp)) \n",
      "LNpMinus1 = fsolve(f32,2) \t\t\t\t\t# [kmol]\n",
      "\n",
      "\t# From equation 5.29 with n = Np-1\n",
      "def f33(xNpMinus1):\n",
      "    return(((LNpMinus1*xNpMinus1)+(GNpPlus1*yNpPlus1))-((LNp*xNp)+(GNp*yNp))) \n",
      "xNpMinus1 = fsolve(f33,0) # [kmol]\n",
      "\n",
      "\t# From equation 5.30 with n = Np-1\n",
      "def f34(HlNpMinus1):\n",
      "    return(((LNpMinus1*HlNpMinus1)+(GNpPlus1*HgNpPlus1))-((LNp*HlNp)+(GNp*HgNp))) \n",
      "HlNpMinus1 = fsolve(f34,0) # [kJ/kmol]\n",
      "def f35(TempNpMinus1):\n",
      "    return(HlNpMinus1-(((1-xNpMinus1)*hb*M_Paraffin*(TempNpMinus1-Tempo))+(xNpMinus1*hb*(TempNpMinus1-Tempo)))) \n",
      "TempNpMinus1 = fsolve(f35,42) # [OC]\n",
      "\n",
      "\t# The computation are continued upward through the tower in this manner until the gas \tcomposition falls atleast to 0.00662.\n",
      "\t# Results = [Tray No.(n) Tn(OC) xn yn]\n",
      "Results = matrix([[4.0,42.3,0.1091,0.1320],[3,39.0,0.0521,0.0568],[2,36.8,0.0184,0.01875],[1,35.5,0.00463, 0.00450]]) \n",
      "figure(1)\n",
      "a1=plot(Results[:,0],Results[:,3]) \n",
      "xlabel(\"Tray Number\") \n",
      "ylabel(\"mole fraction of C5H12 in gas\") \n",
      "show(a1)\n",
      "figure(2)\n",
      "a2=plot(Results[:,0],Results[:,1]) \n",
      "\n",
      "xlabel(\"Tray Number\") \n",
      "ylabel(\"Temparature( degree C)\") \n",
      "\n",
      "show(a2)\n",
      "# For the cquired y1\n",
      "Np = 3.75\n",
      " \n",
      "\n",
      "#result\n",
      "print\"The No. of trays will be\",round(Np)\n",
      " "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The No. of trays will be 4.0\n"
       ]
      }
     ],
     "prompt_number": 1
    }
   ],
   "metadata": {}
  }
 ]
}