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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 13 - Nonreactive and reactive gaseous mixtures"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1 - Pg 271"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the mass and partial volumes of  of oxygen, nitrogen\n",
      "#Initialization of variables\n",
      "P=70. #psia\n",
      "Pt=110. #psia\n",
      "V=20. #cu ft\n",
      "R0=1545. #Universal gas constant\n",
      "T=540. #R\n",
      "M=32. #Molecular weight of Oxygen\n",
      "M2=28. #Molecular weight of Nitrgoen\n",
      "#calculations\n",
      "N=P*V*144/(R0*T)\n",
      "mo=M*N\n",
      "Pn=Pt-P\n",
      "N2=Pn*V*144/(R0*T)\n",
      "mn=N2*M2\n",
      "Vo=N*R0*T/(144*Pt)\n",
      "Vn=N2*R0*T/(144*Pt)\n",
      "Vn2=V-Vo\n",
      "#results\n",
      "print '%s %.2f %s' %(\"Mass of oxygen =\",mo,\" lb\")\n",
      "print '%s %.2f %s' %(\"\\n Mass of nitrogen =\",mn,\" lb\")\n",
      "print '%s %.2f %s' %(\"\\n Partial volume of oxygen =\",Vo,\" cu ft\")\n",
      "print '%s %.2f %s' %(\"\\n Partial volume of nitrogen =\",Vn,\" cu ft\")\n",
      "print '%s %.2f %s' %(\"\\n In case 2, Partial volume of nitrogen =\",Vn2,\" cu ft\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Mass of oxygen = 7.73  lb\n",
        "\n",
        " Mass of nitrogen = 3.87  lb\n",
        "\n",
        " Partial volume of oxygen = 12.73  cu ft\n",
        "\n",
        " Partial volume of nitrogen = 7.27  cu ft\n",
        "\n",
        " In case 2, Partial volume of nitrogen = 7.27  cu ft\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2 - Pg 274"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the change in entropy\n",
      "#Initialization of variables\n",
      "import math\n",
      "P=50. #psia\n",
      "V=4. #cu ft\n",
      "dv=3. #cu ft\n",
      "J=778.\n",
      "T=560. #R\n",
      "#calculation\n",
      "ds= 144*P*V*math.log((V+dv)/V) /(J*T)\n",
      "#results\n",
      "print '%s %.3f %s' %(\"Change in entropy =\",ds,\"unit\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Change in entropy = 0.037 unit\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3 - Pg 275"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate the change in entropy\n",
      "#Initialization of variables\n",
      "import math\n",
      "p1=50. #psia\n",
      "t1=100+460. #R\n",
      "R1=48.3\n",
      "R2=55.2\n",
      "v1=4. #cu ft\n",
      "p2=100. #psia\n",
      "v2=3. #cu ft\n",
      "t2=200.+460 #R\n",
      "cv1=0.157\n",
      "cv2=0.177\n",
      "cpm=0.219\n",
      "J=778.\n",
      "#calculations\n",
      "m1=144*p1*v1/(R1*t1)\n",
      "m2=144*p2*v2/(R2*t2)\n",
      "tf=(m1*cv1*(t1-460) + m2*cv2*(t2-460))/(m1*cv1+m2*cv2)\n",
      "Po2=v1/(v1+v2) *(tf+460)/t1 *p1\n",
      "ds=cpm*math.log((tf+460)/t1) - R1/J *math.log(Po2/p1)\n",
      "dss=ds*m1\n",
      "#results\n",
      "print '%s %.4f %s' %(\"Change in entropy =\",dss,\"unit\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Change in entropy = 0.0528 unit\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4 - Pg 277"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Calculate the change in entropy\n",
      "#Initialization of variables\n",
      "import math\n",
      "p1=30 #psia\n",
      "t1=80.+460 #R\n",
      "R1=48.3\n",
      "R2=55.2\n",
      "m1=20. #lb/min\n",
      "p2=50. #psia\n",
      "m2=35. #lb/min\n",
      "t2=160.+460 #R\n",
      "cp1=0.219\n",
      "cp2=0.248\n",
      "J=778\n",
      "#calculations\n",
      "tf=(m1*cp1*(t1-460) + m2*cp2*(t2-460))/(m1*cp1+m2*cp2)\n",
      "Po2=m1/32/(m1/32+m2/28) *p1\n",
      "ds=cp1*math.log((tf+460)/t1) - R1/J *math.log(Po2/p1)\n",
      "dss=ds*m1\n",
      "#results\n",
      "print '%s %.4f %s' %(\"Change in entropy =\",dss,\"units/min\")\n",
      "print '%s' %(\"The answer is a bit different fromm textbook due to rounding off error in textbook\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Change in entropy = 1.7754 units/min\n",
        "The answer is a bit different fromm textbook due to rounding off error in textbook\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5 - Pg 278"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the weight per mole and also the gas constant\n",
      "#Initialization of variables\n",
      "import numpy\n",
      "x=([0.15, 0.08, 0.77])\n",
      "M=([44, 32, 28])\n",
      "#calculations\n",
      "b=len(x)\n",
      "y=numpy.zeros(b)\n",
      "yt=numpy.zeros(b)\n",
      "mt=numpy.zeros(b)\n",
      "per=numpy.zeros(b)\n",
      "wt=numpy.zeros(b)\n",
      "R=numpy.zeros(b)\n",
      "for i in range (0,b):\n",
      "\ty[i]=x[i]/M[i]\n",
      "\n",
      "yt=sum(y)\n",
      "for i in range (0,b):\n",
      "\tmt[i]=y[i]/yt\n",
      "\tper[i]=mt[i]*100\n",
      "\n",
      "wt=1. /yt\n",
      "R=1545. /wt\n",
      "\n",
      "#results\n",
      "print '%s' %(\"Volumetric analysis\")\n",
      "print '%s' %('percent by volume')\n",
      "print (per)\n",
      "print '%s %.1f %s' %(\"Weight per mole =\",wt,\" lb\")\n",
      "print '%s %.1f' %(\"\\n Gas constant =\",R)\n",
      "print '%s' %(\"The answer is a bit different fromm textbook due to rounding off error in textbook\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Volumetric analysis\n",
        "percent by volume\n",
        "[ 10.20408163   7.4829932   82.31292517]\n",
        "Weight per mole = 29.9  lb\n",
        "\n",
        " Gas constant = 51.6\n",
        "The answer is a bit different fromm textbook due to rounding off error in textbook\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6 - Pg 283"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the theoretical air and amount of nitrogen\n",
      "#Initialization of variables\n",
      "x1=0.885 #mole fraction of Ch4\n",
      "x2=0.115 #mole fraction of c2h6\n",
      "x3=0.4/100. #mole fraction of N2\n",
      "n1=2. #Moles of Ch4\n",
      "n2=3.5 #Moles of c2h6\n",
      "n3=1. #moles of ch4 in case 2\n",
      "n4=2. #moles of c2h6 in case 2\n",
      "#calculations\n",
      "y1=n1*x1\n",
      "y2=n2*x2\n",
      "y=y1+y2\n",
      "vec2=([y1, y2])\n",
      "air=y/0.21\n",
      "y3=n3*x1\n",
      "y4=n4*x2\n",
      "yy=y3+y4\n",
      "vec3=([y3 ,y4])\n",
      "air2=y/0.21 *0.79\n",
      "#results\n",
      "print '%s %.2f %s' %(\"Theoretical air =\",air,\" moles of air per mole of fuel\")\n",
      "print '%s' %(\"Oxygen analysis\")\n",
      "print(vec2)\n",
      "print '%s %.2f %s' %(\"\\n Amount of nitrogen =\",air2,\" moles of nitrogen per mole of fuel\")\n",
      "print '%s' %(\"Dry analysis\")\n",
      "print(vec3)\n",
      "print '%s %.3f %s' %(\"total =\",yy,\" moles\")\n",
      "print '%s' %(\"The answer is a bit different fromm textbook due to rounding off error in textbook\")"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Theoretical air = 10.35  moles of air per mole of fuel\n",
        "Oxygen analysis\n",
        "[1.77, 0.4025]\n",
        "\n",
        " Amount of nitrogen = 8.17  moles of nitrogen per mole of fuel\n",
        "Dry analysis\n",
        "[0.885, 0.23]\n",
        "total = 1.115  moles\n",
        "The answer is a bit different fromm textbook due to rounding off error in textbook\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7 - Pg 283"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the theoretical air fuel ratio\n",
      "#Initialization of variables\n",
      "import numpy\n",
      "x=([0.74, 0.06, 0.01]) #mole fraction of C, H and S respectively\n",
      "y=([8./3., 8, 1]) #Pounds O2 per pound substance of C,H and S respectively\n",
      "oxy=0.08 #Oxygen in coal\n",
      "z=0.232 #mass of coal\n",
      "#calculations\n",
      "b=len(x)\n",
      "pou = numpy.zeros(b)\n",
      "for i in range (0,b):\n",
      "\tpou[i]=x[i]*y[i]\n",
      "\n",
      "tot=sum(pou)\n",
      "oxy2=tot-oxy\n",
      "air=oxy2/z\n",
      "#results\n",
      "print '%s %.2f %s' %(\"Theoretical air fuel ratio =\",air,\"lb of air per pound of coal\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Theoretical air fuel ratio = 10.27 lb of air per pound of coal\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8 - Pg 284"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate air fuel ratio and also molal analysis\n",
      "#Initialization of variables\n",
      "o2=12.5 #moles of O2\n",
      "h20=9 #moles of H2O\n",
      "x=0.21 #Mole fraction of Oxygen in air\n",
      "M=28.97 #Molar mass of air\n",
      "M2=56. #molar mass of C4H8\n",
      "M1=8*12+18. #molecular mass of c8h18\n",
      "#calculations\n",
      "air=o2/x\n",
      "pound=air*M\n",
      "AR=pound/M1\n",
      "y1=h20/M2 *100.\n",
      "y2=o2*(79./21) /M2 *100.\n",
      "#results\n",
      "print '%s %.2f %s' %(\"Air fuel ratio =\",AR,\"lb of air per pound of fuel\")\n",
      "print '%s %.2f %s %.2f %s' %(\"\\n Molal or volumetric analysis is\",y1, \"percent of CO2 and\",y2, \"percent N2\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Air fuel ratio = 15.13 lb of air per pound of fuel\n",
        "\n",
        " Molal or volumetric analysis is 16.07 percent of CO2 and 83.97 percent N2\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9 - Pg 285"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the volumetric analysis and also volume of wet products\n",
      "#Initialization of variables\n",
      "x=18.5 #Moles of O2\n",
      "c=12. #Moles of CO2\n",
      "vap=13. #moles of H2O\n",
      "P=15. #psia\n",
      "R=1545. #Universal gas constant\n",
      "#calculations\n",
      "excess=x*0.5\n",
      "M=12*12+2*vap\n",
      "n2=(x+excess)*79/21.\n",
      "nt=n2+excess+c\n",
      "dry=([c, x/2, n2]) \n",
      "b= len(dry)\n",
      "for i in range(0,b):\n",
      "\tdry[i] = dry[i] /nt *100\n",
      "\n",
      "wet=nt+vap\n",
      "fue=100./(M)\n",
      "mol=wet*fue\n",
      "vol=mol*R*1460./(144*P)\n",
      "#results\n",
      "print '%s' %(\"Volumetric analysis in percentage\")\n",
      "print ('       CO2         O2          N2')\n",
      "print (dry)\n",
      "print '%s %d %s' %(\"Volume of wet products =\",vol,\"cfm\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Volumetric analysis in percentage\n",
        "       CO2         O2          N2\n",
        "[9.55088118249005, 7.362137578169414, 83.08698123934053]\n",
        "Volume of wet products = 85167 cfm\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 10 - Pg 286"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the volumetric analysis\n",
      "#Initialization of variables\n",
      "import numpy\n",
      "from numpy import linalg\n",
      "from numpy.linalg import inv\n",
      "A=numpy.array([[1, 1], [0.5, 1]])\n",
      "B=numpy.array([[1],[ 0.9]])\n",
      "x=0.9\n",
      "#calculations\n",
      "N2=x*79./21.\n",
      "C=numpy.linalg.solve(A,B)\n",
      "vec= ([ C[0], C[1], N2])\n",
      "su=sum(vec)\n",
      "b= len(vec)\n",
      "vec2=numpy.zeros(b)\n",
      "for i in range (0,b):\n",
      "\tvec2[i]=vec[i]/su *100.;\n",
      "\n",
      "#results\n",
      "print '%s' %(\"Volumetric analysis\")\n",
      "print '%s' %('CO      CO2      N2')\n",
      "print (vec2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Volumetric analysis\n",
        "CO      CO2      N2\n",
        "[  4.56026059  18.24104235  77.19869707]\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11 - Pg 288"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Moles of dry products per pound of coal\n",
      "#Initialization of variables\n",
      "c=0.74\n",
      "ref=0.02\n",
      "co2=0.12\n",
      "co=0.1/100.\n",
      "M=12.\n",
      "#calcualtions\n",
      "carbon=c-ref\n",
      "car2=co2+co\n",
      "wt=car2*M\n",
      "amount=carbon/wt\n",
      "#results\n",
      "print '%s %.3f %s' %(\"Moles of dry products per pound of coal =\",amount,\" mole\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Moles of dry products per pound of coal = 0.496  mole\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 12 - Pg 289"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Number of moles of dry products per pound of fuel\n",
      "#Initialization of variables\n",
      "x1=0.128\n",
      "x2=0.035\n",
      "x3=0.002\n",
      "M=12.\n",
      "N=26.\n",
      "#calculations\n",
      "c=x1+x3\n",
      "mole=12./c\n",
      "wt=M*M+N\n",
      "num=mole/wt\n",
      "#results\n",
      "print '%s %.3f %s' %(\"Number of moles of dry products per pound of fuel =\",num,\" mole\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Number of moles of dry products per pound of fuel = 0.543  mole\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13 - Pg 289"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the weight of air and excess air percentage\n",
      "#Initialization of variables\n",
      "c=0.74\n",
      "ref=0.02\n",
      "co2=0.12\n",
      "co=0.1/100.\n",
      "o2=0.065\n",
      "M=12\n",
      "x=0.79\n",
      "M=28.97\n",
      "#calcualtions\n",
      "n2=1-(co2+co+o2)\n",
      "mol=n2/x\n",
      "wt=mol*M\n",
      "wt2=0.496\n",
      "pou=wt2*wt\n",
      "ta=10.27\n",
      "EA=(pou-ta)/ta *100\n",
      "#results\n",
      "print '%s %.2f %s' %(\"Weight of air per pound of fuel =\",pou,\"lb\")\n",
      "print '%s %.1f %s' %(\"\\n Excess air percentage =\",EA,\"percent\")\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Weight of air per pound of fuel = 14.81 lb\n",
        "\n",
        " Excess air percentage = 44.2 percent\n"
       ]
      }
     ],
     "prompt_number": 14
    }
   ],
   "metadata": {}
  }
 ]
}