{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 16 Combustion"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.1 Pg:738"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Molecule is C7 H17\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "per=85\n",
    "#calculations\n",
    "a=per/12\n",
    "b=100-per\n",
    "ad=1.13*a\n",
    "bd=1.13*b\n",
    "#results\n",
    "print \"Molecule is C%d H%d\"%(ad,bd+1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.2 Pg:738"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Oxygen = 8.74 and Nitrogen = 32.90\n",
      "\n",
      "Equation is C7.333 H6 + 8.74 O2 + 32.85 N2 = 7.333 CO2 + 3 H2O + 0.03120 SO2 + 32.90 N2\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "per=0.071 #mass fraction of nitrogen\n",
    "#calculations\n",
    "O2=8.74\n",
    "N2=per/2 + 3.76*O2\n",
    "Nin=32.85\n",
    "CO2=7.333\n",
    "H2o=3\n",
    "So2=0.0312\n",
    "#results\n",
    "print \"Oxygen = %.2f and Nitrogen = %.2f\"%(O2,N2)\n",
    "print \"\\nEquation is C%.3f H%d + %.2f O2 + %.2f N2 = %.3f CO2 + %d H2O + %.5f SO2 + %.2f N2\"%(CO2,2*H2o,O2,Nin,CO2,H2o,So2,N2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.3 Pg:739"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Air fuel ratio = 12.06 lbm air/lbm fuel\n",
      "\n",
      "In dry air, Air-fuel ratio = 9.9 lbm air/lbm fuel as fired\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "M=29\n",
    "m1=8.74\n",
    "m2=32.85\n",
    "fuel=100 #lbm\n",
    "#calculations\n",
    "mass=M*(m1+m2)\n",
    "AF=mass/fuel\n",
    "a2=9.75\n",
    "b2=12.19\n",
    "AF2=mass/(fuel+a2+b2)\n",
    "#results\n",
    "print \"Air fuel ratio = %.2f lbm air/lbm fuel\"%(AF)\n",
    "print \"\\nIn dry air, Air-fuel ratio = %.1f lbm air/lbm fuel as fired\"%(AF2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.4 Pg:740"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Mass of dry flue gases = 12.50 lbm dry flue gas/lbm fuel ash and moisture free\n",
      "\n",
      " Mass of dry flue gases = 10.25 lbm dry flue gas/lbm fuel as fired \n",
      "\n",
      " Energy carried away = 187079.8 btu/mol coal as fired which is same as  = 1534.2 Btu/lbm mol coal \n",
      "The answers are a bit different due to rounding off errors in textbook\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "m1=322.3 #Mass of Co2\n",
    "m2=2 #Mass of SO2\n",
    "m3=926 #Mass of N2\n",
    "basis=121.94 #Basis taken\n",
    "#calculations\n",
    "m=m1+m2+m3\n",
    "ratio=m/basis\n",
    "dh=5777 #Btu/mol\n",
    "h1=dh*7.364\n",
    "h2=14037\n",
    "h3=130501\n",
    "H=h1+h2+h3\n",
    "hrat=H/basis\n",
    "#results\n",
    "print \"Mass of dry flue gases = %.2f lbm dry flue gas/lbm fuel ash and moisture free\"%(m/100)\n",
    "print \"\\n Mass of dry flue gases = %.2f lbm dry flue gas/lbm fuel as fired \"%(ratio)\n",
    "print \"\\n Energy carried away = %.1f btu/mol coal as fired which is same as  = %.1f Btu/lbm mol coal \"%(H, hrat)\n",
    "print \"The answers are a bit different due to rounding off errors in textbook\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.6 Pg:741"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Final orsat composition is 1 CO2 + 0.22 H20 + 7.52 N2\n",
      "\n",
      " Percentage of co2 on a wet basis = 11.4 percent\n",
      "\n",
      " percentage of co2 on a dry basis = 11.74 percent\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "p=14.7 #psia\n",
    "ps=0.363 #psia\n",
    "n2=7.52 #moles\n",
    "n1=1 #moles\n",
    "#calculations\n",
    "x= (n1+n2)*ps/p /(1-ps/p)\n",
    "n=n1+n2+x\n",
    "y1=n1/n\n",
    "y2=n1/(n1+n2)\n",
    "#results\n",
    "print \"Final orsat composition is %d CO2 + %.2f H20 + %.2f N2\"%(n1, x, n2)\n",
    "print \"\\n Percentage of co2 on a wet basis = %.1f percent\"%(y1*100)\n",
    "print \"\\n percentage of co2 on a dry basis = %.2f percent\"%(y2*100)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.7 Pg:742"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Air fuel ratio = 11.3 lbm air/lbm fuel\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "N2=78.1\n",
    "M=29\n",
    "co2=8.7\n",
    "co=8.9\n",
    "x4=0.3\n",
    "x5=3.7\n",
    "x6=14.7\n",
    "#calculations\n",
    "O2=N2/3.76\n",
    "Z=(co2+co+x4)/8\n",
    "AF=(O2+N2)*M/(Z*113)\n",
    "#results\n",
    "print \"Air fuel ratio = %.1f lbm air/lbm fuel\"%(AF)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.8 Pg:743"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Air fuel ratio = 10.2 lbm air/lbm fuel as fired\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "basis=100 #lbm\n",
    "x1=0.6\n",
    "ash=12 #lbm\n",
    "N2=79.7\n",
    "M=29\n",
    "#calculations\n",
    "x=ash/x1\n",
    "C=(1-x1)*x\n",
    "O2=N2/3.76\n",
    "a= (14.6+0.2)/(5.83-0.66)\n",
    "AF=(O2+N2)*M/(a*100)\n",
    "#results\n",
    "print \"Air fuel ratio = %.1f lbm air/lbm fuel as fired\"%(AF)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.9 Pg:744"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Air fuel ratio = 11.4 lbm air/lbm fuel\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "N2=78.1 #Moles of Nitrogen\n",
    "M=29 #Molar mass of Air\n",
    "ba=2.12 #Basis\n",
    "x4=0.3 #Moles of Ch4\n",
    "x5=3.7 #Moles of H2\n",
    "x6=14.7 #moles of H2o\n",
    "#calculations\n",
    "O2=N2/3.76\n",
    "O2=N2/3.76\n",
    "Z=(x4*4+x5*2+x6*2)/17\n",
    "AF=(O2+N2)*M/(Z*113)\n",
    "#results\n",
    "print \"Air fuel ratio = %.1f lbm air/lbm fuel\"%(AF)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.10 Pg:745"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Air fuel ratio = 11.3 lbm air/lbm fuel\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "N2=78.1 #Moles of Nitrogen\n",
    "M=29 #Molar mass of Air\n",
    "ba=2.12 #Basis\n",
    "x4=0.3 #Moles of Ch4\n",
    "x5=3.7 #Moles of H2\n",
    "x6=14.7 #moles of H2o\n",
    "#calculations\n",
    "O2=N2/3.76\n",
    "c=14.7\n",
    "b= x4*4 + x5*2 + x6*2\n",
    "a=b/ba\n",
    "AF=(O2+N2)*M/(a*12 + b)\n",
    "#results\n",
    "print \"Air fuel ratio = %.1f lbm air/lbm fuel\"%(AF)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.11 Pg:746"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Air fuel ratio = 11.3 lbm air/lbm fuel\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "co2=8.7 #Moles of CO2\n",
    "co=8.9 #Moles of CO\n",
    "N2=78.1 #Moles of Nitrogen\n",
    "M=29 #Molar mass of Air\n",
    "ba=2.12 #Basis\n",
    "x4=0.3 #Moles of Ch4\n",
    "x5=3.7 #Moles of H2\n",
    "x6=14.7 #moles of H2o\n",
    "#calculations\n",
    "O2=N2/3.76\n",
    "c=14.7\n",
    "Z=2.238\n",
    "X=(Z*17-x4*4-x5*2)/2\n",
    "a=co2+co/2+x4+x6/2\n",
    "b=3.764*a\n",
    "AF=(O2+N2)*M/(Z*113)\n",
    "#results\n",
    "print \"Air fuel ratio = %.1f lbm air/lbm fuel\"%(AF)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.12 Pg:747"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Air fuel ratio = 11.37\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "x1=8.7 #Moles of Co2\n",
    "x2=8.9 #Moles of CO\n",
    "x3=0.3 #Moles of O2\n",
    "N=78.1 #Moles of N2\n",
    "z=113 #Af factor\n",
    "M=29 #Molar mass of air\n",
    "#calculations\n",
    "co2=(x1+x2+x3)*100/(N+x1+x2+x3)\n",
    "a=2.325\n",
    "AF=103*M/(a*z)\n",
    "#results\n",
    "print \"Air fuel ratio = %.2f\"%(AF)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.13 Pg:748"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Nitrogen = 87.1 percent\n",
      "\n",
      "Equation is a(96 CH4 + 3 H2+ 1 CO) + 87.1/3.76 O2 + 87.1 N2 = 10.8 CO2 + 1.2 CO + 0.6 H2 + 0.3 CH4 + 87.1 N2\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "co=1.2 #Moles of CO\n",
    "co2=10.8 #Moles of CO2\n",
    "#calculations\n",
    "H2=co/2\n",
    "ch4=0.3\n",
    "N2=88-H2-ch4\n",
    "#results\n",
    "print \"Nitrogen = %.1f percent\"%(N2)\n",
    "print \"\\nEquation is a(96 CH4 + 3 H2+ 1 CO) + %.1f/3.76 O2 + %.1f N2 = %.1f CO2 + %.1f CO + %.1f H2 + %.1f CH4 + %.1f N2\"%(N2,N2,co2,co,H2,ch4,N2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.14 Pg:748"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Higher heating value = -2363996 Btu\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "dH=-2369859 #Btu\n",
    "r=1.986 #Gas constant\n",
    "dn=5.5 #Change in number of moles\n",
    "T=536.7 #R\n",
    "#calculations\n",
    "dQ=dH+dn*r*T\n",
    "#results\n",
    "print \"Higher heating value = %d Btu\"%(dQ)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.15 Pg:749"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "from steam tables,\n",
      "Lower heating value = -2203398 Btu/lbm\n",
      "The answers are a bit different due to rounding off error in textbook.\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "M2=18 #Molar mass of water\n",
    "M=170 #Molar mass of octane\n",
    "p=0.4593 #Pressure of octane #psia\n",
    "print \"from steam tables,\"\n",
    "vfg=694.9 \n",
    "J=778.2\n",
    "m=9*18 #Mass of water\n",
    "u1=-2363996 #Btu\n",
    "#calculations\n",
    "hfg=1050.4 #Btu/lbm\n",
    "ufg= hfg- p*vfg*144/J\n",
    "dU=ufg*m \n",
    "Lhv=u1+dU\n",
    "#results\n",
    "print \"Lower heating value = %d Btu/lbm\"%(Lhv)\n",
    "print \"The answers are a bit different due to rounding off error in textbook.\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.16 Pg:750"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "From Table B-10,\n",
      "Heat of reaction = -2202154 Btu\n",
      "The answers are a bit different due to rounding off error in textbook.\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "n1=8 #Moles of CO2\n",
    "n2=9 #Moles of H2O\n",
    "n3=1 #Moles of Octane\n",
    "n4=12.5 #Moles of Oxygen\n",
    "print \"From Table B-10,\"\n",
    "U11=3852 #Internal energy at 1000 R of CO2\n",
    "U12=115 #Internal energy at 537 R of CO2\n",
    "U21=3009 #Internal energy at 1000 R of H2O\n",
    "U22=101 #Internal energy at  537 R of H2O\n",
    "U31=24773 #Internal energy at 1000 R of Octane\n",
    "U32=640 #Internal energy at 537 R of Octane\n",
    "U41=2539 #Internal energy at 1000 R of Oxygen\n",
    "U42=83 #Internal energy at 537 R of Oxygen\n",
    "H=-2203389 #heat Btu\n",
    "#calculations\n",
    "dU1=n1*(U11-U12)+n2*(U21-U22)\n",
    "dU2=n3*(U31-U32)+n4*(U41-U42)\n",
    "Q=H+dU1-dU2\n",
    "#results\n",
    "print \"Heat of reaction = %d Btu\"%(Q)\n",
    "print \"The answers are a bit different due to rounding off error in textbook.\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.17 Pg:751"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "from table B-10,\n",
      "Upon interpolating, T2 = 5271 R\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "n1=8 #Moles of CO2\n",
    "n2=9 #Moles of H2O\n",
    "n3=47 #Moles of N2\n",
    "print \"from table B-10,\"\n",
    "h1=118 #Enthalpy of CO2\n",
    "h2=104 #Enthalpy of H2O\n",
    "h3=82.5 #Enthalpy of N2\n",
    "Q=2203279 #Btu\n",
    "#calculations\n",
    "U11=n1*h1+n2*h2+n3*h3\n",
    "U12=U11+Q\n",
    "T2=5271 #R\n",
    "#results\n",
    "print \"Upon interpolating, T2 = %d R\"%(T2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.18 Pg:752"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "In case 1, Equilibrium constant = 38.5 \n",
      "\n",
      "In case 2, Equilibrium constant = 1480.1 \n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "n1=0.95\n",
    "n2=0.05\n",
    "n3=0.025\n",
    "P=147 #psia\n",
    "pa=14.7 #psia\n",
    "#calculations\n",
    "n=n1+n2+n3\n",
    "p1=n1/n *P/pa\n",
    "p2=n2/n *P/pa\n",
    "p3=n3/n *P/pa\n",
    "Kp1= p1/(p2*p3**0.5)\n",
    "Kp2= p1**2 /(p2**2 *p3)\n",
    "#results\n",
    "print \"In case 1, Equilibrium constant = %.1f \"%(Kp1)\n",
    "print \"\\nIn case 2, Equilibrium constant = %.1f \"%(Kp2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:16.19 Pg:753"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "percentage of dissociation = 34.3 percent\n",
      "\n",
      " If pressure =10 . Degree of dissociation = 18 percent\n"
     ]
    }
   ],
   "source": [
    "from sympy.abc import x,y\n",
    "from sympy import solve,N\n",
    "#Initialization of variables\n",
    "kp=5 \n",
    "#calculations\n",
    "vec=solve(24*x**3 + 3*x-2,x)\n",
    "x=N(vec[2],6)\n",
    "vec2=solve(249*y**3 +3*y-2,y)\n",
    "y=N(vec2[2],6)\n",
    "#results\n",
    "print \"percentage of dissociation = %.1f percent\"%(x*100)\n",
    "print \"\\n If pressure =10 . Degree of dissociation = %d percent\"%(y*100)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Ex:16.20 Pg:754"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 41,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Extent of reaction= 78 percent\n"
     ]
    }
   ],
   "source": [
    "from sympy.abc import x,y\n",
    "from sympy import solve,N\n",
    "\n",
    "#Initialization of variables\n",
    "vec=solve(24*x**3 +48*x**2 + 7*x -4,x)\n",
    "x=N(vec[1],6) *100\n",
    "#results\n",
    "print \"Extent of reaction= %d percent\"%(100-x)"
   ]
  }
 ],
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