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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 3: Processes and Process Variables"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1-1, page no. 44"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "mass=215.0 #kg\n",
      "\n",
      "#Calculations and printing :\n",
      "\n",
      "density=13.546*62.43\n",
      "print '%s %.2f' %(\"density of mercury (lbm/ft^3) = \",density)\n",
      "#the multiplication factor is to convert density from gm/cc to lbm/ft^3.\n",
      "volume=mass/(.454*density)     #ft^3\n",
      "#the division by 0.454 is to convert mass in kg to lbm.\n",
      "print '%s %.2f %s %.2f' %(\" \\n The volume of \",mass,\"kg of mercury is(ft^3) = \",volume)\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "density of mercury (lbm/ft^3) =  845.68\n",
        " \n",
        " The volume of  215.00 kg of mercury is(ft^3) =  0.56\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 1,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1-2, page no. 45"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "T1=20.0                 # \u02daC\n",
      "T2=100.0                # \u02daC\n",
      "Vat20=0.560             #ft^3\n",
      "D=0.0208333             #ft\n",
      "\n",
      "#Calculations and printing :\n",
      "\n",
      "print(\"we know that V(T)=Vo[1+0.math.pow(18182x10,()-3)xT +0.math.pow(0078x10,()-6)xTxT]\")\n",
      "Vat0=Vat20/(1+0.18182*math.pow(10,-3)*T1 +0.0078*math.pow(10,-6)*T1*T1)\n",
      "#the function is defined with the variable as temperature\n",
      "def volume(T):\n",
      "    volume=Vat0*(1+0.18182* math.pow(10,-3) *T +0.0078* math.pow(10,-6) *T*T)\n",
      "    return volume\n",
      "\n",
      "print '%s %.3f' %(\" vat20 (ft^3) = \",volume(T1))\n",
      "print '%s %.3f' %(\" \\n vat100 (ft^3) =\",volume(T2))\n",
      "change=((volume(T2))-(volume(T1)))*4/(math.pi*D*D)\n",
      "print '%s %.3f' %(\" \\n change in the height of mercury level (ft) = \",change)\n",
      "#the answer is a bit different due to rounding off of volume(T2) in textbook\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "we know that V(T)=Vo[1+0.math.pow(18182x10,()-3)xT +0.math.pow(0078x10,()-6)xTxT]\n",
        " vat20 (ft^3) =  0.560\n",
        " \n",
        " vat100 (ft^3) = 0.568\n",
        " \n",
        " change in the height of mercury level (ft) =  23.931\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 2,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3-1, page no. 48"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "mass=100.0 #g of CO2\n",
      "M=44.01  #molecular weight\n",
      "\n",
      "#Calculations and print '%s %.3f' %ing :\n",
      "\n",
      "moles=mass/M\n",
      "print '%s %.3f' %(\"\\n no.of moles=\",moles)\n",
      "lbmole=moles/453.6\n",
      "print '%s %.3f' %(\"\\n no.of lb moles=\",lbmole)\n",
      "Cmoles=moles\n",
      "print '%s %.3f' %(\"\\n no.of moles of carbon=\",Cmoles)\n",
      "Omoles=2*moles\n",
      "print '%s %.3f' %(\"\\n no.of moles of oxygen=\",Omoles)\n",
      "O2moles=moles\n",
      "print '%s %.3f' %(\"\\n no.of moles of dioxygen=\",O2moles)\n",
      "gramsO=Omoles*16\n",
      "print '%s %.3f' %(\"\\n no.of grams of oxygen=\",gramsO)\n",
      "gramsO2=O2moles*32\n",
      "print '%s %.3f' %(\"\\n no.of grams of oxygen=\",gramsO2)\n",
      "moleculesCO2=moles*6.02*math.pow(10,(23))\n",
      "print '%s %.3E' %(\"\\n no.of molecules of CO2 =\",moleculesCO2)\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        " no.of moles= 2.272\n",
        "\n",
        " no.of lb moles= 0.005\n",
        "\n",
        " no.of moles of carbon= 2.272\n",
        "\n",
        " no.of moles of oxygen= 4.544\n",
        "\n",
        " no.of moles of dioxygen= 2.272\n",
        "\n",
        " no.of grams of oxygen= 72.711\n",
        "\n",
        " no.of grams of oxygen= 72.711\n",
        "\n",
        " no.of molecules of CO2 = 1.368E+24\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 3,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3-2, page no. 50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "xA=0.15 #mass fraction\n",
      "yB=0.20 #mole fraction\n",
      "mass=175.0 #kg of solution\n",
      "flowrate1=53.0 #lbm/h\n",
      "flowrate2=1000.0 #mol/min\n",
      "massofA=300.0 #lbm\n",
      "molarB=28.0 #kmolB/s\n",
      "\n",
      "#Calculations and printing :\n",
      "\n",
      "massA=mass*xA\n",
      "print '%s %d %s %.3f'%(\" \\n Mass of A in\",mass,\" kg of solution (kg A) =\",massA)\n",
      "flowrateA=flowrate1*xA\n",
      "print '%s %d %s %.3f'%(\" \\n Mass flow rate of A in a stream flowing at\",flowrate1,\" lbm/h (lbm A/h)=\",flowrateA )\n",
      "flowrateB=flowrate2*yB\n",
      "print '%s %d %s %.3f'%(\" \\n Molar flowrate of B in a stream flowing at\",flowrate2,\" mol/min (molB/min)= \",flowrateB)\n",
      "Totalflowrate=molarB/yB\n",
      "print '%s %d %s %.3f'%(\" \\n Total flow rate of a solution with\",molarB,\"kmolB/s=\",Totalflowrate)\n",
      "MassSolution=massofA/xA\n",
      "print '%s %d %s %.3f'%(\" \\n Mass of solution that contains\",massofA,\" lbm of A =\",MassSolution)\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " \n",
        " Mass of A in 175  kg of solution (kg A) = 26.250\n",
        " \n",
        " Mass flow rate of A in a stream flowing at 53  lbm/h (lbm A/h)= 7.950\n",
        " \n",
        " Molar flowrate of B in a stream flowing at 1000  mol/min (molB/min)=  200.000\n",
        " \n",
        " Total flow rate of a solution with 28 kmolB/s= 140.000\n",
        " \n",
        " Mass of solution that contains 300  lbm of A = 2000.000\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 4,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3-3, page no. 50"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "#let the total mass be 100\n",
      "massO2=16.0\n",
      "massCO=4.0\n",
      "massCO2=17.0\n",
      "massN2=63.0\n",
      "MO2=32.0\n",
      "MCO=28.0\n",
      "MCO2=44.0\n",
      "MN2=28.0\n",
      "\n",
      "#Calculations and print '%s %.3f' %ing :\n",
      "\n",
      "molO2=massO2/MO2\n",
      "molCO=massCO/MCO\n",
      "molCO2=massCO2/MCO2\n",
      "molN2=massN2/MN2\n",
      "TotalMol=molO2+molCO+molCO2+molN2\n",
      "print '%s %.3f' %(\" \\n molefraction of O2=\",molO2/TotalMol)\n",
      "print '%s %.3f' %(\" \\n molefraction of CO=\",molCO/TotalMol)\n",
      "print '%s %.3f' %(\" \\n molefraction of CO2=\",molCO2/TotalMol)\n",
      "print '%s %.3f' %(\" \\n molefraction of N2=\",molN2/TotalMol)\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " \n",
        " molefraction of O2= 0.152\n",
        " \n",
        " molefraction of CO= 0.044\n",
        " \n",
        " molefraction of CO2= 0.118\n",
        " \n",
        " molefraction of N2= 0.686\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 5,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3-4, page no. 51"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "yN2=0.79\n",
      "MN2=28.0                            #g/mol\n",
      "MO2=32.0                            g/mol\n",
      "xN2=0.767\n",
      "\n",
      "#Calculations and printing :\n",
      "\n",
      "Mbar=yN2*MN2+(1-yN2)*MO2\n",
      "print '%s %.3f' %(\" \\n average molecular weight of air from molar composition (g/mol) = \",Mbar)\n",
      "InvMbar=xN2/28 + (1-xN2)/32\n",
      "print '%s %.3f' %(\" \\n average molecular weight of air from mass composition (g/mol) = \",1./InvMbar)\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        " \n",
        " average molecular weight of air from molar composition (g/mol) =  28.840\n",
        " \n",
        " average molecular weight of air from mass composition (g/mol) =  28.840\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 6,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3-5, page no. 52"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "conc=0.5 #molar\n",
      "rate=1.25 #m^3/min\n",
      "D=1.03 #no units\n",
      "M=98.0\n",
      "\n",
      "#Calculations and print '%s %.3f' %ing :\n",
      "\n",
      "mass_conc=conc*98.\n",
      "print '%s %.3f' %(\"mass concentration of sulfuric acid (kg/m^3) = \",mass_conc)\n",
      "mass_flowrate=rate*mass_conc/60.\n",
      "print '%s %.3f' %(\" \\n Mass flow rate of sulfuric acid (kg/s) = \",mass_flowrate)\n",
      "massfraction=1/(rate*D*1000/60)\n",
      "print '%s %.3f' %(\" \\n Mass fraction of sulfuric acid=\",massfraction)\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "mass concentration of sulfuric acid (kg/m^3) =  49.000\n",
        " \n",
        " Mass flow rate of sulfuric acid (kg/s) =  1.021\n",
        " \n",
        " Mass fraction of sulfuric acid= 0.047\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 8,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4-1, page no. 55"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "Pressure=2.00*100000. #Pa\n",
      "\n",
      "#Calculations and printing :\n",
      "\n",
      "Pressure=Pressure*1000./(13600.*9.807)\n",
      "print '%s %.3E' %(\"Pressure (mm of Hg) = \",Pressure)\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Pressure (mm of Hg) =  1.500E+03\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 9,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4-2, page no. 55"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "P0=10.4*1.013*100000. /10.33            #m H2O\n",
      "D=1000.0                                #kg/m^3\n",
      "g=9.807                                 #m/s^2\n",
      "h=30.0                                  #m\n",
      "\n",
      "#Calculations and printing :\n",
      "\n",
      "Ph=P0+D*g*h\n",
      "print '%s %.3E' %(\"Pressure at the bottom of the lake (N/m^2) = \",Ph)\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Pressure at the bottom of the lake (N/m^2) =  3.962E+05\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 12,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4-3, page no. 59"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initilization of variables\n",
      "import math\n",
      "spg=1.05\n",
      "h1=382. #mm of fluid\n",
      "h2=374. #mm of fluid\n",
      "reading= -2 #in of Hg\n",
      "Patm=30. #in of Hg\n",
      "#Calculations and printing:\n",
      "deltaP=(spg-1)*980.7*(h1-h2)/10.\n",
      "print '%s %.1f' %(\"Pressure drop between points 1 and 2 (dynes/cm^2) = \",deltaP)\n",
      "P1=Patm+reading\n",
      "print '%s %.1f' %(\"\\n Absolute Pressure (in Hg) = \",P1)\n",
      "raw_input(\"Press the Enter key to quit\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Pressure drop between points 1 and 2 (dynes/cm^2) =  39.2\n",
        "\n",
        " Absolute Pressure (in Hg) =  28.0\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Press the Enter key to quit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 13,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.5-2, page no. 62"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initialization of variables\n",
      "import math\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "\n",
      "T1=20.0 #F\n",
      "T2=80.0 #F\n",
      "\n",
      "#Calculations and printing :\n",
      "\n",
      "#In this code I used a function to achieve the conversion\n",
      "def conversion(fahrenheit):\n",
      "    conversion=(fahrenheit-32)/1.8\n",
      "    return conversion;\n",
      "\n",
      "difference=conversion(T2)-conversion(T1)\n",
      "print '%s %.2f %s %.2f' %(\"Equivalent temperature of\",T2-T1,\" temperature in C =\",difference)\n",
      "deltaTF=T2-T1\n",
      "deltaTC=deltaTF/1.8\n",
      "print '%s %.2f' %(\" \\n By second method, result=\",deltaTC)\n",
      "raw_input('press enter key to exit')"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Equivalent temperature of 60.00  temperature in C = 33.33\n",
        " \n",
        " By second method, result= 33.33\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "press enter key to exit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 14,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.5-3, page no. 63"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "\n",
      "#Initilization of variables\n",
      "import math\n",
      "a=0.487\n",
      "b=2.29*math.pow(10, -4)\n",
      "print '%s %.2f %s %.1E %s' %(\"Given Cp(Btu/lbm.F) = \",a,\"+\",b,\"T (F)\")\n",
      "print(\"\\n Step 1\")\n",
      "af1=a+b*32\n",
      "bf1=b*1.8\n",
      "print(\"\\n Step 2\")\n",
      "af2=af1*1.8/(454*9.486*math.pow(10, -4))\n",
      "bf2=bf1*1.8/(454*9.486*math.pow(10, -4))\n",
      "print '%s %.2f %s %.1E %s' %(\"Final Cp(J/g.C) = \",af2,\"+\",bf2,\"T (C)\")\n",
      "raw_input(\"Press the Enter key to quit\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Given Cp(Btu/lbm.F) =  0.49 + 2.3E-04 T (F)\n",
        "\n",
        " Step 1\n",
        "\n",
        " Step 2\n",
        "Final Cp(J/g.C) =  2.07 + 1.7E-03 T (C)\n"
       ]
      },
      {
       "name": "stdout",
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Press the Enter key to quit\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 15,
       "text": [
        "''"
       ]
      }
     ],
     "prompt_number": 15
    }
   ],
   "metadata": {}
  }
 ]
}