{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 16:Reactive Systems"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.2:pg-675"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 16.2\n",
      "\n",
      "\n",
      " K is  0.314529177004  atm\n",
      "\n",
      " Epsilon is  0.611607081035\n",
      "\n",
      " The heat of reaction is  60974.6120608  kJ/kg mol\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "eps_e = 0.27  # Constant\n",
    "P = 1.0 # Atmospheric pressure in bar\n",
    "K = (4*eps_e**2*P)/(1-eps_e**2) \n",
    "P1 = 100.0/760.0 # Pressure in Pa\n",
    "eps_e_1 = math.sqrt((K/P1)/(4.0+(K/P1)))\n",
    "T1 = 318.0 # Temperature in K\n",
    "T2 = 298.0# Temperature in K\n",
    "R = 8.3143 # Gas constant\n",
    "K1 = 0.664 # dissociation constant at 318K\n",
    "K2 = 0.141# dissociation constant at 298K\n",
    "dH = 2.30*R*((T1*T2)/(T1-T2))*(math.log10(K1/K2))\n",
    "print \"\\n Example 16.2\\n\"\n",
    "print \"\\n K is \",K ,\" atm\"\n",
    "print \"\\n Epsilon is \",eps_e_1\n",
    "print \"\\n The heat of reaction is \",dH ,\" kJ/kg mol\"\n",
    "#The answers vary due to round off error\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.3:pg-675"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 16.3\n",
      "\n",
      "\n",
      " Equilibrium constant is  1.61983471074\n",
      "\n",
      " Gibbs function change is  -4812.22485358 J/gmol\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "v1 = 1.0 # Assumed\n",
    "v2 = v1# Assumed \n",
    "v3 = v2 # Assumed\n",
    "v4 = v2# Assumed\n",
    "e = 0.56 # Degree of reaction\n",
    "P = 1.0 # Dummy\n",
    "T = 1200.0 # Reaction temperature in K\n",
    "R = 8.3143 # Gas constant\n",
    "x1 = (1-e)/2.0 # \n",
    "x2 = (1-e)/2.0\n",
    "x3 = e/2.0 \n",
    "x4 = e/2.0\n",
    "K = (((x3**v3)*(x4**v4))/((x1**v1)*(x2**v2)))*P**(v3+v4-v1-v2) # Equilibrium constant\n",
    "dG = -R*T*math.log(K) #Gibbs function change\n",
    "\n",
    "print \"\\n Example 16.3\\n\"\n",
    "print \"\\n Equilibrium constant is \",K\n",
    "print \"\\n Gibbs function change is \",dG ,\"J/gmol\"\n",
    "#The answers vary due to round off error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.5:pg-678"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 16.5\n",
      "\n",
      "\n",
      " The value of equillibrium constant is  0.755668681281  atm\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "Veo = 1.777 # Ve/Vo\n",
    "e = 1.0-Veo # Degree of dissociation\n",
    "P = 0.124 # in atm\n",
    "K = (4*e**2*P)/(1.0-e**2)\n",
    "\n",
    "print \"\\n Example 16.5\\n\"\n",
    "print \"\\n The value of equillibrium constant is \",K ,\" atm\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.6:pg-680"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 16.6\n",
      "\n",
      "\n",
      " Cp is  4.48364424966  J/g mol K\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "v1 = 1.0 # Assumed\n",
    "v2 = 0 # Assumed\n",
    "v3 = 1.0 # Assumed\n",
    "v4 = 1.0/2.0# Assumed\n",
    "dH = 250560.0 # Enthalpy change in j/gmol\n",
    "e = 3.2e-03 # Constant\n",
    "R = 8.3143 # Gas constant\n",
    "T = 1900.0 # Reaction temperature\n",
    "Cp = ((dH**2)*(1+e/2)*e*(1+e))/(R*T**2*(v1+v2)*(v3+v4))\n",
    "print \"\\n Example 16.6\\n\"\n",
    "print \"\\n Cp is \",Cp ,\" J/g mol K\"\n",
    "#The answers vary due to round off error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.7:pg-681"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 16.7\n",
      "\n",
      "\n",
      " The composition of fuel is  14.7645650439  percent Hydrogen and  85.2354349561  percent Carbon\n",
      "\n",
      " Air fuel ratio is  23.9829146049\n",
      "\n",
      " Percentage of excess air used is  67.2268907563  percent\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "a = 21.89 # stochiometric coefficient\n",
    "y = 18.5 # stochiometric coefficient\n",
    "x = 8.9 # stochiometric coefficient\n",
    "PC = 100*(x*12)/((x*12)+(y)) # Carbon percentage\n",
    "PH = 100-PC # Hydrogen percentage\n",
    "AFR = ((32*a)+(3.76*a*28))/((12*x)+y) #Air fuel ratio\n",
    "EAU = (8.8*32)/((21.89*32)-(8.8*32)) # Excess air used\n",
    "\n",
    "print \"\\n Example 16.7\\n\"\n",
    "print \"\\n The composition of fuel is \",PH ,\" percent Hydrogen and \",PC ,\" percent Carbon\"#The answer provided in the textbook is wrong\n",
    "print \"\\n Air fuel ratio is \",AFR\n",
    "print \"\\n Percentage of excess air used is \",EAU*100 ,\" percent\"\n",
    "#The answers vary due to round off error\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.8:pg-682"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 16.8\n",
      "\n",
      "\n",
      " Heat transfer per kg mol of fuel is  -965198.0  kJ\n",
      "\n",
      " Q_cv is  -890324.0  kJ\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "hf_co2 = -393522.0 # Enthalpy of reaction in kJ/kg mol\n",
    "hf_h20 = -285838.0# Enthalpy of reaction in kJ/kg mol\n",
    "hf_ch4 = -74874.0# Enthalpy of reaction in kJ/kg mol\n",
    "D = hf_co2 + (2*hf_h20) #Heat transfer \n",
    "QCV = D-hf_ch4 # Q_cv\n",
    "\n",
    "print \"\\n Example 16.8\\n\"\n",
    "print \"\\n Heat transfer per kg mol of fuel is \",D ,\" kJ\"\n",
    "print \"\\n Q_cv is \",QCV ,\" kJ\"\n",
    "#The answers vary due to round off error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.9:pg-683"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 16.9 \n",
      "\n",
      "\n",
      " Fuel consumption rate is  38.5131749981  kg/h\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Below values are taken from table\n",
    "Hr = -249952+(18.7*560)+(70*540)\n",
    "Hp = 8*(-393522+20288)+9*(-241827+16087)+6.25*14171+70*13491\n",
    "Wcv = 150.0 # Energy out put from engine in kW\n",
    "Qcv = -205.0 # Heat transfer from engine in kW\n",
    "n = (Wcv-Qcv)*3600/(Hr-Hp)\n",
    "print \"\\n Example 16.9 \\n\"\n",
    "print \"\\n Fuel consumption rate is \",n*114 ,\" kg/h\"\n",
    "#The answers vary due to round off error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.11:pg-684"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 16.11 \n",
      "\n",
      "\n",
      " Reversible work is  47139  kJ/kg\n",
      "\n",
      " Increase in entropy during combustion is  3699.6688  kJ/kg mol K\n",
      "\n",
      " Irreversibility of the process  25056.8559091  kJ/kg\n",
      "\n",
      " Availability of products of combustion is  22082.1440909  kJ/kg\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Refer table 16.4 for values\n",
    "T0 = 298.0 # Atmospheric temperature in K\n",
    "Wrev = -23316-3*(-394374)-4*(-228583) # Reversible work in kJ/kg mol\n",
    "Wrev_ = Wrev/44 # Reversible work in kJ/kg\n",
    "Hr = -103847 # Enthalpy of reactants in kJ/kg\n",
    "T = 980.0 # Through trial and error\n",
    "Sr = 270.019+20*205.142+75.2*191.611 # Entropy of reactants\n",
    "Sp = 3*268.194 + 4*231.849 + 15*242.855 + 75.2*227.485 # Entropy of products\n",
    "IE = Sp-Sr # Increase in entropy\n",
    "I = T0*3699.67/44 # Irreversibility\n",
    "Si = Wrev_ - I# Availability of products of combustion \n",
    "\n",
    "print \"\\n Example 16.11 \\n\"\n",
    "print \"\\n Reversible work is \",Wrev_ ,\" kJ/kg\"\n",
    "print \"\\n Increase in entropy during combustion is \",Sp-Sr ,\" kJ/kg mol K\"\n",
    "print \"\\n Irreversibility of the process \",I ,\" kJ/kg\"\n",
    "print \"\\n Availability of products of combustion is \",Si ,\" kJ/kg\"\n",
    "#The answers vary due to round off error\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.12:pg-685"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 6.12\n",
      "\n",
      "\n",
      " The chemical energy of carbon is   410541.588354  kJ/k mol\n",
      "\n",
      " The chemical energy of hydrogen is   235211.889921  kJ/k mol\n",
      "\n",
      " The chemical energy of methane is   821580.156423  kJ/k mol\n",
      "\n",
      " The chemical energy of Carbon monoxide is   275364.910207  kJ/k mol\n",
      "\n",
      " The chemical energy of liquid methanol is   716698.69005  kJ/k mol\n",
      "\n",
      " The chemical energy of nitrogen is   691.0909601  kJ/k mol\n",
      "\n",
      " The chemical energy of Oxygen is   3946.64370597  kJ/k mol\n",
      "\n",
      " The chemical energy of Carbon dioxide is   20108.2320604  kJ/k mol\n",
      "\n",
      " The chemical energy of Water is   5.21177422707  kJ/k mol\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "T0 = 298.15 # Environment temperature in K\n",
    "P0 = 1 # Atmospheric pressure in bar\n",
    "R = 8.3143# Gas constant\n",
    "xn2 = 0.7567  # mole fraction of nitrogen\n",
    "xo2 = 0.2035 # mole fraction of oxygen\n",
    "xh2o = 0.0312 # mole fraction of water\n",
    "xco2 = 0.0003# mole fraction of carbon dioxide\n",
    "# Part (a)\n",
    "g_o2 = 0 # Gibbs energy of oxygen\n",
    "g_c = 0 # Gibbs energy of carbon\n",
    "g_co2 = -394380  # Gibbs energy of carbon dioxide\n",
    "A = -g_co2 + R*T0*math.log(xo2/xco2) # Chemical energy\n",
    "\n",
    "# Part (b)\n",
    "g_h2 = 0 # Gibbs energy of hydrogen\n",
    "g_h2o_g = -228590# # Gibbs energy of water\n",
    "B = g_h2 + g_o2/2 - g_h2o_g + R*T0*math.log(xo2**0.5/xh2o)\n",
    "# Chemical energy\n",
    "# Part (c)\n",
    "g_ch4 = -50790 # Gibbs energy of methane\n",
    "C = g_ch4 + 2*g_o2 - g_co2 - 2*g_h2o_g + R*T0*math.log((xo2**2)/(xco2*xh2o))\n",
    "# Chemical energy\n",
    "# Part (d)\n",
    "g_co = -137150# # Gibbs energy of carbon mono oxide\n",
    "D =  g_co + g_o2/2 - g_co2 + R*T0*math.log((xo2**0.5)/xco2)\n",
    "# Chemcal energy\n",
    "# Part (e)\n",
    "g_ch3oh = -166240 # Gibbs energy of methanol\n",
    "E = g_ch3oh + 1.5*g_o2 - g_co2 - 2*g_h2o_g + R*T0*math.log((xo2**1.5)/(xco2*(xh2o**2)))\n",
    "# Chemical energy\n",
    "# Part (f)\n",
    "F = R*T0*math.log(1/xn2)\n",
    "# Chemical energy\n",
    "# Part (g)\n",
    "G = R*T0*math.log(1/xo2)\n",
    "# Chemical energy\n",
    "# Part (h)\n",
    "H = R*T0*math.log(1/xco2)\n",
    "# Chemical energy\n",
    "# Part (i)\n",
    "g_h2o_l = -237180 #  Gibbs energy of liquid water\n",
    "I = g_h2o_l - g_h2o_g + R*T0*math.log(1/xh2o)\n",
    "# Chemical energy\n",
    "print \"\\n Example 6.12\\n\"\n",
    "print \"\\n The chemical energy of carbon is  \",A ,\" kJ/k mol\"\n",
    "print \"\\n The chemical energy of hydrogen is  \",B ,\" kJ/k mol\"\n",
    "print \"\\n The chemical energy of methane is  \",C ,\" kJ/k mol\"\n",
    "print \"\\n The chemical energy of Carbon monoxide is  \",D ,\" kJ/k mol\"\n",
    "print \"\\n The chemical energy of liquid methanol is  \",E ,\" kJ/k mol\"\n",
    "print \"\\n The chemical energy of nitrogen is  \",F ,\" kJ/k mol\"\n",
    "print \"\\n The chemical energy of Oxygen is  \",G ,\" kJ/k mol\"\n",
    "print \"\\n The chemical energy of Carbon dioxide is  \",H ,\" kJ/k mol\"\n",
    "print \"\\n The chemical energy of Water is  \",I ,\" kJ/k mol\"\n",
    "#The answers vary due to round off error"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex16.13:pg-686"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      " Example 6.13\n",
      "\n",
      "\n",
      " The rate of heat transfer from the engine =  -4.33120060702  kW,\n",
      " The second law of efficiency of the engine =  13.3396896634  percent\n"
     ]
    }
   ],
   "source": [
    "import math\n",
    "# Environmet\n",
    "T0 = 298.15 # Environment temperature in K\n",
    "P0 = 1.0 # Atmospheric pressure in atm\n",
    "R = 8.3143# Gas constant\n",
    "xn2 = 0.7567  # mole fraction of nitrogen\n",
    "xo2 = 0.2035 # mole fraction of oxygen\n",
    "xh2o = 0.0312 # mole fraction of water\n",
    "xco2 = 0.0003# mole fraction of carbon dioxide\n",
    "xother = 0.0083 # Mole fraction of other gases\n",
    "# Liquid octane\n",
    "t1 = 25.0 # Temperature of liquid octane in degree centigrade\n",
    "m = 0.57 # Mass flow rate in kg/h\n",
    "T2 = 670 # Temperature of combustion product at exit in K\n",
    "x1 = 0.114 # Mole fraction of CO2\n",
    "x2 = .029 # Mole fraction of CO\n",
    "x3 = .016 # Mole fraction of O2\n",
    "x4 = .841 # Mole fraction of N2\n",
    "Wcv = 1 # Power developed by the engine in kW\n",
    "print \"\\n Example 6.13\\n\"\n",
    "# By carbon balance \n",
    "b = 55.9 \n",
    "# By hydrogen balace\n",
    "c=9\n",
    "# By oxygen balance\n",
    "a = 12.58\n",
    "Qcv = Wcv- 3845872*(.57/(3600*114.22))\n",
    "E = 5407843.0 # Chemical exergy of C8H18\n",
    "nII = Wcv/(E*.57/(3600*114.22))\n",
    "print \"\\n The rate of heat transfer from the engine = \",Qcv ,\" kW,\\n The second law of efficiency of the engine = \",nII*100 ,\" percent\""
   ]
  }
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