path: root/Introduction_To_Chemical_Engineering_Thermodynamics_by_G._Halder/Ch4.ipynb

{
 "cells": [
  {
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   "metadata": {},
   "source": [
    "# Chapter 4  - Heat effects"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.1 Page: 118"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.1 - Page: 118\n",
      "\n",
      "\n",
      "Value of Qv is -326.40 kcal\n",
      "\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "print \"Example: 4.1 - Page: 118\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "Qp = -327## [kcal]\n",
    "T = 27 + 273## [K]\n",
    "R = 2*10**(-3)## [kcal/K mol]\n",
    "#*************#\n",
    "\n",
    "# The reaction involved is:\n",
    "# C2H5OH(l) + 3O2(g) = 2CO2(g) + 3H2O(l)\n",
    "deltan = 2 - 3#\n",
    "Qv = Qp - deltan*R*T## [kcal]\n",
    "print \"Value of Qv is %.2f kcal\\n\"%(Qv)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.2 Page: 119"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.2 - Page: 119\n",
      "\n",
      "\n",
      "Heat produced in the reaction is -1019.9 kcal\n",
      "\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.2 - Page: 119\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "# Mg + (1/2)O2 = MgO ...............(1)\n",
    "deltaH1 = -610.01## [kcal]\n",
    "# 2Fe + (3/2)O2 = Fe2O3 ............(2)\n",
    "deltaH2 = -810.14## [kcal]\n",
    "#*************#\n",
    "\n",
    "# 3Mg + Fe2O3 = 3MgO + 2Fe .........(3)\n",
    "# Multiplying (1) by 3 and substracting from (2), we get (3):\n",
    "deltaH = 3*deltaH1 - deltaH2## [kcal]\n",
    "print \"Heat produced in the reaction is %.1f kcal\\n\"%(deltaH)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.3 Page: 121"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.3 - Page: 121\n",
      "\n",
      "\n",
      "The standard heat of formation of methane is -74.75 kJ/gmol\n",
      "\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.3 - Page: 121\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "# 2H2(g) + O2(g) ---------------> 2H2O .....................(1)\n",
    "deltaH1 = -241.8*2## [kJ/gmol H2]\n",
    "# C(graphite) + O2(g) =---------> CO2(g) ...................(2)\n",
    "deltaH2 = -393.51## [kJ/gmol C]\n",
    "# CH4(g) + 2O2(g) ---------------> CO2(g) + 2H2O(l) ........(3)\n",
    "deltaH3 = -802.36## [kJ/mol CH4]\n",
    "#*************#\n",
    "\n",
    "# For standard heat of formation of methane, (a) + (b) - (c)\n",
    "# C + 2H2 ------------------------> CH4\n",
    "deltaHf = deltaH1 + deltaH2 - deltaH3## [kJ/gmol]\n",
    "print \"The standard heat of formation of methane is %.2f kJ/gmol\\n\"%(deltaHf)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.4 Page: 122"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.4 - Page: 122\n",
      "\n",
      "\n",
      "Energy supplied by reaction A is -69.2 kJ\n",
      "\n",
      "Energy supplied by reaction B is -2802.8 kJ\n",
      "\n",
      "Reaction B supplies more energy to the organism\n",
      "\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.4 - Page: 122\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "deltaH_C6H12O6 = -1273## [kcal]\n",
    "deltaH_C2H5OH = -277.6## [kcal]\n",
    "deltaH_CO2 = -393.5## [kcal]\n",
    "deltaH_H2O = -285.8## [kcal]\n",
    "#**************#\n",
    "\n",
    "# C6H12O6(s) = 2C2H5OH(l) + 2CO2(g) ..........................(A)\n",
    "deltaH_A = 2*deltaH_C2H5OH + 2*deltaH_CO2 - deltaH_C6H12O6## [kJ]\n",
    "# C6H12O6(s) + 6O2(g) = 6CO2(g) + 6H2O(l) ...................(B)\n",
    "deltaH_B = 6*deltaH_CO2 + 6*deltaH_H2O - deltaH_C6H12O6## [kJ]\n",
    "print \"Energy supplied by reaction A is %.1f kJ\\n\"%(deltaH_A)#\n",
    "print \"Energy supplied by reaction B is %.1f kJ\\n\"%(deltaH_B)#\n",
    "if deltaH_A < deltaH_B:\n",
    "    print \"Reaction A supplies more energy to the organism\\n\"\n",
    "else:\n",
    "     print \"Reaction B supplies more energy to the organism\\n\"\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.5 Page: 122"
   ]
  },
  {
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   "execution_count": 6,
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     "text": [
      "Example: 4.5 - Page: 122\n",
      "\n",
      "\n",
      "Heat of formation of ZnSO4 is -233.48 kcal/kmol\n",
      "\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.5 - Page: 122\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "# Zn + S = ZnS ....................................................(A)\n",
    "deltaH_A = -44## [kcal/kmol]\n",
    "# ZnS + 3O2 = 2ZnO + 2SO2 .........................................(B)\n",
    "deltaH_B = -221.88## [kcal/kmol]\n",
    "# 2SO2 + O2 = 2SO3 ................................................(C)\n",
    "deltaH_C = -46.88## [kcal/kmol]\n",
    "# ZnO + SO3 = ZnSO4 ...............................................(D)\n",
    "deltaH_D = -55.10## [kcal/kmol]\n",
    "#***************#\n",
    "\n",
    "# Multiplying (A) by 2 & (D) by (2) and adding (A), (B), (C) & (D)\n",
    "# Zn + S + 2O2 = ZnSO4\n",
    "deltaH = 2*deltaH_A + deltaH_B + deltaH_C + 2*deltaH_D## [kcal/kmol for 2 kmol of ZnSO4]\n",
    "print \"Heat of formation of ZnSO4 is %.2f kcal/kmol\\n\"%(deltaH/2)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.6 Page: 124"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.6 - Page: 124\n",
      "\n",
      "\n",
      "Standard Heat of formation of NH3 is -11.8 kcal\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.6 - Page: 124\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "# HC : Heat of Combustion\n",
    "HC_NH3 = -90.6## [kcal]\n",
    "HC_H2 = -68.3## [kcal]\n",
    "#*************#\n",
    "\n",
    "# Heat of combustion of NH3:\n",
    "# 2NH3 + 3O = N2 + 3H2O ............................ (A)\n",
    "# Heat of combustion of H2:\n",
    "# H2 + O = H2O ..................................... (B)\n",
    "# Multiplying (B) by 3 & substracting from (A), we get:\n",
    "# 2NH3 = N2 + 3H2 .................................. (C)\n",
    "# Hf : Heat of Formation\n",
    "Hf_NH3 = -(2*HC_NH3 - 3*HC_H2)/2## [kcal]\n",
    "print \"Standard Heat of formation of NH3 is %.1f kcal\"%(Hf_NH3)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.7 Page: 125"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.7 - Page: 125\n",
      "\n",
      "\n",
      "The maximum attainable temperature is 2566.2 K\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.7 - Page: 125\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "# HC : Heat of Combustion\n",
    "HC_C2H2 = -310600# # [cal]\n",
    "#**************#\n",
    "\n",
    "# C2H2 + (5/2)O2 = 2CO2 + H2O\n",
    "Q = -HC_C2H2## [cal]\n",
    "# The gases present in the flame zone after combustion are carbon dioxide, water vapor and the unreacted nitrogen of the air.\n",
    "# Since (5/2) mole of oxygen were required for combustion, nitrogen required would be 10 mol.\n",
    "# Hence the composition of the resultant gas would be 2 mol CO2, 1  ol H2 & 10 mol N2.\n",
    "# Q = integrate('Cp(T)','T',T,298)#\n",
    "# On integrating we get:\n",
    "# Q = 84.52*(T - 298) + 18.3*10**(-3)*(T**2 - 298**2)\n",
    "#deff('[y] = f(T)','y = Q - 84.52*(T - 298) - 18.3*10**(-3)*(T**2 - 298**2)')#\n",
    "def f(T):\n",
    "    y = Q - 84.52*(T - 298) - 18.3*10**(-3)*(T**2 - 298**2)\n",
    "    return y\n",
    "from scipy.optimize import fsolve\n",
    "T = fsolve(f,7)## [K]\n",
    "print \"The maximum attainable temperature is %.1f K\"%(T)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.8 Page: 126"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.8 - Page: 126\n",
      "\n",
      "\n",
      "The theoretical temperature of combustion is 1906 degree Celsius\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.8 - Page: 126\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "Cp_CO2 = 54.56## [kJ/mol K]\n",
    "Cp_O2 = 35.20## [kJ/mol K]\n",
    "Cp_steam = 43.38## [kJ/mol K]\n",
    "Cp_N2 = 33.32## [kJ/mol K]\n",
    "# 2C2H6(g) + 7O2(g) = 4CO2(g) + 6H2O(g)\n",
    "deltaH_273 = -1560000## [kJ/kmol]\n",
    "#************#\n",
    "\n",
    "# Since the air is 25% in excess of the amount required,the combustion may be written as:\n",
    "# C2H6(g) + (7/2)O2(g) = 2CO2(g) + 3H2O(g)\n",
    "# 25% excess air is supplied.\n",
    "# Since the air contains N2 = 79% and O2 = 21%\n",
    "# C2H6(g) + 3.5O2(g) + 0.25*3.5O2(g) + (4.375*(79/21))N2 = 2CO2 + 3H2O + 0.875O2 + 16.46N2 .................. (A)\n",
    "# Considering the reaction (A),\n",
    "# Amount of O2:\n",
    "O2 = 3.5 + 3.5*0.25## [mol]\n",
    "# Amount of N2 required:\n",
    "N2 = 4.375*(79/21)## [mol]\n",
    "# Let the initial temperature of ethane and air be 0 OC and the temperature of products of combustion be T OC\n",
    "# Since heat librated by combustion = heat accumulated by combustion products\n",
    "Q = -deltaH_273## [kJ/mol K]\n",
    "T = Q/(2*Cp_CO2 + 3*Cp_steam + 0.875*Cp_O2 + N2*Cp_N2)## [OC]\n",
    "print \"The theoretical temperature of combustion is %d degree Celsius\"%(T)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.9 Page: 129"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.9 - Page: 129\n",
      "\n",
      "\n",
      "Laten heat of ice at -20 OC is 1266 cal/mol\n",
      "\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.9 - Page: 129\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "T1 = 273## [K]\n",
    "T2 = 253## [K]\n",
    "deltaH_273 = 1440## [cal/mol]\n",
    "Cp = 8.7## [cal/mol]\n",
    "#**************#\n",
    "\n",
    "deltaH_253 = deltaH_273 + Cp*(T2 - T1)## [cal/mol]\n",
    "print \"Laten heat of ice at -20 OC is %d cal/mol\\n\"%(deltaH_253)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.10 Page: 129"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.10 - Page: 129\n",
      "\n",
      "\n",
      "Heat of formation at 1273 K is -11172 cal\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.10 - Page: 129\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "T2 = 1273## [K]\n",
    "T1 = 300## [K]\n",
    "deltaH_300 = -11030## [cal/mol]\n",
    "#*************#\n",
    "\n",
    "# The chemical reaction involved is:\n",
    "# N2 + 3H2 = 2NH3\n",
    "# (1/2)N2 + (3/2)H2 = NH3\n",
    "# deltaH_1273 = deltaH_300 + integrate('Cp_NH3(T) - (1/2)*Cp_N2(T) - (1/2)*Cp_H2(T)','T',1273,300)#\n",
    "from sympy.mpmath import quad\n",
    "deltaH_1273 = deltaH_300 + quad(lambda T:(6.2 + 7.8*10**(-3)*T - 7.2*10**(-6)*T**2) - (1/2)*(6.45 + 1.4*10**(-3)*T) - (1/2)*(6.94 - 0.2*10**(-3)*T),[1273,300])## [cal]\n",
    "print \"Heat of formation at 1273 K is %d cal\"%(deltaH_1273)#"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example: 4.11 Page: 130"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Example: 4.11 - Page: 130\n",
      "\n",
      "\n",
      "Percent of excess air supplied is 39.9 %\n"
     ]
    }
   ],
   "source": [
    "print \"Example: 4.11 - Page: 130\\n\\n\"\n",
    "\n",
    "# Solution\n",
    "\n",
    "#*****Data*****#\n",
    "CO2 = 13.4## [percent by volume]\n",
    "N2 = 80.5## [percent by volume]\n",
    "O2 = 6.1## [percent by volume]\n",
    "#*************#\n",
    "\n",
    "# Basis : 100 cubic m of flue gas.\n",
    "Vol_N2_flue = N2## [Volume of Nitrogen in flue gas, cubic m]\n",
    "Vol_O2_flue = O2## [Volume of O2 in flue gas, cubic m]\n",
    "Vol_Air = N2/0.79## [Volume of air supplied, cubic m]\n",
    "Vol_O2 = Vol_Air*0.21## [Volume of O2 in air supply, cubic m]\n",
    "Vol_O2_cumbustion = Vol_O2 - Vol_O2_flue## [Volume of O2 used up in cumbustion of the fuel, cubic m]\n",
    "Excess_Air = Vol_O2_flue/Vol_O2_cumbustion * 100## [percent of excess air supplied]\n",
    "print \"Percent of excess air supplied is %.1f %%\"%(Excess_Air)#"
   ]
  }
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