{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 16: Combustion"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.10: Air_fuel_ratio_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Air fuel ratio = %.1f lbm air/lbm fuel',AF)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.11: Air_fuel_ratio_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Air fuel ratio = %.1f lbm air/lbm fuel',AF)\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.12: Air_fuel_ratio_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Air fuel ratio = %.2f',AF)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.13: Equation_formulation.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Nitrogen = %.1f percent',N2)\n",
"printf('\n Equation 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": [
			"## Example 16.14: Higher_heating_value.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Higher heating value = %d Btu',dQ)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.15: Lower_heating_value.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"disp('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",
"printf('Lower heating value = %d Btu/lbm',Lhv)\n",
"disp('The answers are a bit different due to rounding off error in textbook.')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.16: Heat_of_reactio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"disp('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",
"printf('Heat of reaction = %d Btu',Q)\n",
"disp('The answers are a bit different due to rounding off error in textbook.')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.17: Temperature_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"n1=8 //Moles of CO2\n",
"n2=9 //Moles of H2O\n",
"n3=47 //Moles of N2\n",
"disp('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",
"printf('Upon interpolating, T2 = %d R',T2)\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.18: Equilibrium_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('In case 1, Equilibrium constant = %.1f ',Kp1)\n",
"printf('\n In case 2, Equilibrium constant = %.1f ',Kp2)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.19: Dissociation_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"kp=5 \n",
"//calculations\n",
"x=poly(0,'x')\n",
"vec=roots(24*x^3 + 3*x-2)\n",
"x=vec(3)\n",
"y=poly(0,'y')\n",
"vec2=roots(249*y^3 +3*y-2)\n",
"y=vec2(3)\n",
"//results\n",
"printf('percentage of dissociation = %.1f percent',x*100)\n",
"printf('\n If pressure =10 . degree of dissociation = %d percent',y*100)\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.1: Molecule_formulatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Molecule is C %d H %d',ad,bd+1)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.20: Extent_of_reaction.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//Initialization of variables\n",
"x=poly(0,'x')\n",
"vec=roots(24*x^3 +48*x^2 + 7*x -4)\n",
"x=vec(3) *100\n",
"//results\n",
"printf('Extent of reaction= %d percent',100-x)\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.2: Molecule_formulatio.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Oxygen = %.2f and Nitrogen = %.2f',O2,N2)\n",
"printf('\n Equation 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": [
			"## Example 16.3: Air_fuel_ratio_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Air fuel ratio = %.2f lbm air/lbm fuel',AF)\n",
"printf('\n In dry air, Air-fuel ratio = %.1f lbm air/lbm fuel as fired',AF2)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.4: Mass_and_energy_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Mass of dry flue gases = %.2f lbm dry flue gas/lbm fuel ash and moisture free',m/100)\n",
"printf('\n Mass of dry flue gases = %.2f lbm dry flue gas/lbm fuel as fired ',ratio)\n",
"printf('\n Energy carried away = %.1f btu/mol coal as fired which is same as  = %.1f Btu/lbm mol coal ',H, hrat)\n",
"disp('The answers are a bit different due to rounding off errors in textbook')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.6: Percentage_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Final orsat composition is %d CO2 + %.2f H20 + %.2f N2',n1, x, n2)\n",
"printf('\n Percentage of co2 on a wet basis = %.1f percent',y1*100)\n",
"printf('\n percentage of co2 on a dry basis = %.2f percent',y2*100)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.7: Air_fuel_ratio_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Air fuel ratio = %.1f lbm air/lbm fuel',AF)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.8: Air_fuel_ratio_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Air fuel ratio = %.1f lbm air/lbm fuel as fired',AF)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 16.9: Air_fuel_ratio_calculations.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clc\n",
"clear\n",
"//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",
"printf('Air fuel ratio = %.1f lbm air/lbm fuel',AF)\n",
""
   ]
   }
],
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		   "display_name": "Scilab",
		   "language": "scilab",
		   "name": "scilab"
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		  "language_info": {
		   "file_extension": ".sce",
		   "help_links": [
			{
			 "text": "MetaKernel Magics",
			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
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		   "mimetype": "text/x-octave",
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