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-{

- "metadata": {

-  "name": "",

-  "signature": "sha256:62da0bbd390eac5357f5f997103c8800ba34f828bad9afe3f011ec1444ac1334"

- },

- "nbformat": 3,

- "nbformat_minor": 0,

- "worksheets": [

-  {

-   "cells": [

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Chapter19-Chemical reactions"

-     ]

-    },

-    {

-     "cell_type": "heading",

-     "level": 1,

-     "metadata": {},

-     "source": [

-      "Example1-pg 404"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#caluclate fuel ratio and excess air and emass air-fuel ratio\n",

-      "##initialisation of variables\n",

-      "pN2= 79. ##percent\n",

-      "VN2= 82.3 ##m^3\n",

-      "VCO2= 8. ##m^3\n",

-      "VCO= 0.9 ##m^3\n",

-      "M= 32. ##gms\n",

-      "M1= 28. ##gms\n",

-      "##CALCULATIONS\n",

-      "P= (pN2/(100-pN2))\n",

-      "z= VN2/P\n",

-      "x= VCO2+VCO\n",

-      "w= VCO2+(VCO/2)+(VCO2/10)\n",

-      "y= 2*w\n",

-      "r= y/x\n",

-      "TO= x+(y/4)\n",

-      "X= (z/TO)-1\n",

-      "AF= z*(M+P*M1)/(12*x+y)\n",

-      "##RESULTS\n",

-      "print'%s %.3f %s'%('fuel ratio=',r,'')\n",

-      "print'%s %.3f %s'%('excess air=',X,'')\n",

-      "print'%s %.2f %s'%('emass air-fuel ratio=',AF,'')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "fuel ratio= 2.079 \n",

-        "excess air= 0.618 \n",

-        "emass air-fuel ratio= 23.98 \n"

-       ]

-      }

-     ],

-     "prompt_number": 2

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example2-pg 410"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#calculate heat interaction\n",

-      "##initialisation of variables\n",

-      "m1= 24. ##kg\n",

-      "M1= 32. ##kg\n",

-      "m2= 28. ##kg\n",

-      "M2= 28. ##kg\n",

-      "e= 0.5\n",

-      "T3= 1800. ##C\n",

-      "T0= 25. ##C\n",

-      "T1= 25. ##C\n",

-      "T2= 100. ##C\n",

-      "R= 8.314 ##Jmol K\n",

-      "cp= 4.57 ##J/mol K\n",

-      "cp1= 3.5 ##J/mol K\n",

-      "cp2= 3.5 ##J/mol K\n",

-      "hCO2= -393522. ##J\n",

-      "hCO= -110529. ##J\n",

-      "##CALCULATIONS\n",

-      "n1= m1/M1\n",

-      "n2= m2/M2\n",

-      "N= n1-0.5*e\n",

-      "N1= n2-e\n",

-      "N2= e\n",

-      "N3= N+N1+N2\n",

-      "y1= N/N3\n",

-      "Q= ((N*cp+N1*cp1+N2*cp2)*R*(T3-T0)-(n1*cp*(T1-T0)+n2*cp2*(T2-T1))+N*(hCO2-hCO))/60.\n",

-      "##RESULTS\n",

-      "print'%s %.f %s'%(' Heat interaction=',Q,'kW ')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        " Heat interaction= -940 kW \n"

-       ]

-      }

-     ],

-     "prompt_number": 4

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example3-pg 412"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#calculate adiabatic flame temperature\n",

-      "##initialisation of variables\n",

-      "T0= 25. ##C\n",

-      "T1= 220. ##C\n",

-      "hCO2= -393520 ##kJ/kg\n",

-      "hH2O= -241830 ##kJ/kg\n",

-      "hC3H8= -103850 ##kJ/kg= 1.4\n",

-      "R= 8.314 ##Jmol K\n",

-      "k= 1.4\n",

-      "k1= 1.29\n",

-      "##CALCULATIONS\n",

-      "T= T0+((15*(R*(k/(k-1)))*4.762*(T1-T0)-(3*hCO2+4*hH2O-hC3H8))/(R*((3+4)*(k1/(k1-1))+(10+56.43)*(k/(k-1)))))\n",

-      "##RESULTS\n",

-      "print'%s %.1f %s'%('adiabatic flame temperature=',T,'C ')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "adiabatic flame temperature= 1142.4 C \n"

-       ]

-      }

-     ],

-     "prompt_number": 1

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example5-pg 415"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#calculate enthalpy formation\n",

-      "##initialisation of variables\n",

-      "T= 25. ##C\n",

-      "hfT= -241820 ##kJ/kmol\n",

-      "R= 8.314 ##J/mol K\n",

-      "k= 1.4\n",

-      "cpH2O= 4.45\n",

-      "cpO2= 3.5\n",

-      "T1= 1000. ##C\n",

-      "##CALCULATIONS\n",

-      "S= (cpH2O-k*cpO2)\n",

-      "hfT1= hfT+S*(T1-T)\n",

-      "##RESULTS\n",

-      "print'%s %.f %s'%('enthalpy formation=',hfT1,'kJ/kmol ')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "enthalpy formation= -242259 kJ/kmol \n"

-       ]

-      }

-     ],

-     "prompt_number": 8

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example6-pg 418"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#calculate equlibrium constant at K and KT1\n",

-      "##initialisation of variables\n",

-      "R= 8.314 ##J/mol K\n",

-      "T= 25. ##C\n",

-      "gf= 16590. ##kJ/kmol\n",

-      "T1= 500. ##C\n",

-      "Cp= 4.157 ##J/mol K\n",

-      "hf= -46190 ##kJ/kmol\n",

-      "e=0.5\n",

-      "##CALCULATIONS\n",

-      "K=math.pow(math.e,gf/(R*(273.15+T)))\n",

-      "r= (1-((273.15+T)/(273.15+T1)))*((hf/(R*(273.15+T)))+(R/Cp))-2*math.log((273.15+T1)/(273.15+T))+0.6\n",

-      "KT1= K*math.pow(math.e,r)\n",

-      "##RESULTS\n",

-      "print'%s %.1f %s'%('equilibrium constant=',K,'bar^-1 ')\n",

-      "print'%s %.5f %s'%('equilibrium constant=',KT1,'bar^-1 ')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "equilibrium constant= 806.5 bar^-1 \n",

-        "equilibrium constant= 0.00797 bar^-1 \n"

-       ]

-      }

-     ],

-     "prompt_number": 9

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example7-pg 419"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#what equilibrium constant at T1 and T2\n",

-      "##initialisation of variables\n",

-      "uCO2= -394374 ##J/mol\n",

-      "uCO= -137150 ##J/mol\n",

-      "uO2= 0.\n",

-      "R= 8.314 ##J/mol K\n",

-      "T= 25. ##C\n",

-      "cpCO2= 4.57 ##J/mol K\n",

-      "cpCO= 3.5 ##J/mol K\n",

-      "cpO2= 3.5 ##J/mol K\n",

-      "T1= 1500. ##C\n",

-      "hf= -393522 ##kJ/kmol\n",

-      "gf= -110529 ##kJ/kmol\n",

-      "T2= 2500. ##C\n",

-      "##CALCULATIONS\n",

-      "r= -(uCO2-uCO-0.5*uO2)/(R*(273.15+T))\n",

-      "s= (cpCO2-cpCO-0.5*cpO2)\n",

-      "r1= (1-((273.15+T)/(273.15+T1)))*((hf-gf)/(R*(273.15+T))-s)+s*math.log((273.15+T1)/(273.15+T))\n",

-      "KT1= math.pow(math.e,r+r1)\n",

-      "r2= (1-((273.15+T)/(273.15+T2)))*((hf-gf)/(R*(273.15+T))-s)+s*math.log((273.15+T2)/(273.15+T))\n",

-      "KT2= math.pow(math.e,r+r2)\n",

-      "##RESULTS\n",

-      "print'%s %.f %s'%('equilibrium constant at T1=',KT1,'C ')\n",

-      "print'%s %.3f %s'%('equilibrium constant at T2=',KT2,'C ')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "equilibrium constant at T1= 3477 C \n",

-        "equilibrium constant at T2= 2.635 C \n"

-       ]

-      }

-     ],

-     "prompt_number": 10

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example8-pg422"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#what is maximum work of given variable \n",

-      "##initialisation of variables\n",

-      "Wc= 12. ##kg\n",

-      "hf= -393520 ##kJ/kmol\n",

-      "gf= -394360 ##kJ/kmol\n",

-      "##CALCULATIONS\n",

-      "Wmax= -gf/Wc\n",

-      "##RESULTS\n",

-      "print'%s %.f %s'%('maximum work=',Wmax,'kJ/kg of carbon ')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "maximum work= 32863 kJ/kg of carbon \n"

-       ]

-      }

-     ],

-     "prompt_number": 11

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example9-pg423"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#calculate the outlet temperature and energy of formation and energy out let and energy of the products\n",

-      "##initialisation of variables\n",

-      "T= 25 ##C\n",

-      "R= 8.314 ##Jmol K\n",

-      "k= 1.27\n",

-      "k1= 1.34\n",

-      "hf= -393520 ##kJ/kmol\n",

-      "M= 28 ##gms\n",

-      "gf= -394360 ##kJ/kmol\n",

-      "M= 12 ##gms\n",

-      "##CALCULATIONS\n",

-      "T1= T+(-hf/((R)*((k/(k-1))+(0.2+4.5144)*(k1/(k1-1)))))\n",

-      "Bin= 0\n",

-      "dh= (k1*R/(k1-1))*(T1-T)\n",

-      "dh1= (k1*R/(k1-1))*math.log((273.15+T1)/(273.15+T))\n",

-      "H= dh-(273.15+T)*dh1\n",

-      "h= (k*R/(k-1))*(T1-T)+hf\n",

-      "h1= (k*R/(k-1))*math.log((273.15+T1)/(273.15+T))+((hf-gf)/(273.15+T))\n",

-      "h2= h-(273.15+T)*h1\n",

-      "Bout= (h2+(0.2+4.5144)*H)/M\n",

-      "##RESULTS\n",

-      "print'%s %.2f %s'%('outlet temperature=',T1,'C')\n",

-      "print'%s %.f %s'%('energy of formation=',Bin,'J')\n",

-      "print'%s %.f %s'%('energy at outlet=',H,'kJ/kmol')\n",

-      "print'%s %.f %s'%('energy of the products=',Bout,'k')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "outlet temperature= 2057.82 C\n",

-        "energy of formation= 0 J\n",

-        "energy at outlet= 46519 kJ/kmol\n",

-        "energy of the products= -9961 k\n"

-       ]

-      }

-     ],

-     "prompt_number": 12

-    },

-    {

-     "cell_type": "heading",

-     "level": 2,

-     "metadata": {},

-     "source": [

-      "Example10-pg427"

-     ]

-    },

-    {

-     "cell_type": "code",

-     "collapsed": false,

-     "input": [

-      "import math\n",

-      "#calculate the change in energy and amount of air and gas and netchange in energy and percent change in energy\n",

-      "##initialisation of variables\n",

-      "b= 1475.30 ##kJ/kg\n",

-      "b0= 144.44 ##kJ/kg\n",

-      "h2= 3448.6 ##kJkg\n",

-      "h1= 860.5 ##kJ/kg\n",

-      "k= 1.27 \n",

-      "k1= 1.34\n",

-      "R= 8.314 ##J/mol K\n",

-      "hf= -393520 ##kJ/kmol\n",

-      "hg= 72596 ##kJ/kmol\n",

-      "Mc= 12 ##kg\n",

-      "n= 1.2 ##moles\n",

-      "n1= 3.76 ##moles\n",

-      "M= 32. ##gms\n",

-      "M1= 28. ##gms\n",

-      "M2= 44. ##gms\n",

-      "n2= 0.2 ##moles\n",

-      "n3= 4.512 ##moles\n",

-      "B1= 25592. ##kJ/kmol C\n",

-      "B2= 394360. ##kJ/kmol C\n",

-      "e= 0.008065\n",

-      "##CALCULATIONS\n",

-      "B= b-b0\n",

-      "Q= h2-h1\n",

-      "CpCO2= k*R/(k-1)\n",

-      "CpO2= k1*R/(k1-1)\n",

-      "Qcoal= (hg+hf)/Mc\n",

-      "mcoal= Q/(-Qcoal)\n",

-      "ncoal= mcoal/Mc\n",

-      "r= (n*M+n1*M1)/Mc\n",

-      "r1= (M2+n2*M+n3*M1)/Mc\n",

-      "mair= r*mcoal\n",

-      "mgas= r1*mcoal\n",

-      "Bfuel= (B1-B2)*e\n",

-      "Bnet= Bfuel+B\n",

-      "p= B*100/(-Bfuel)\n",

-      "##RESULTS\n",

-      "print'%s %.2f %s'% ('change in energy=',B,'kJ/kg ')\n",

-      "print'%s %.3f %s'%('amount of air=',mair,'kg/kg ')\n",

-      "print'%s %.3f %s'%('amount of gas=',mgas,'kg/kg ')\n",

-      "print'%s %.3f %s'%('net change in energy=',Bnet,'kg/kg steam ')\n",

-      "print'%s %.2f %s'%('percent energy in original fuel=',p,'percent ')\n"

-     ],

-     "language": "python",

-     "metadata": {},

-     "outputs": [

-      {

-       "output_type": "stream",

-       "stream": "stdout",

-       "text": [

-        "change in energy= 1330.86 kJ/kg \n",

-        "amount of air= 1.159 kg/kg \n",

-        "amount of gas= 1.425 kg/kg \n",

-        "net change in energy= -1643.254 kg/kg steam \n",

-        "percent energy in original fuel= 44.75 percent \n"

-       ]

-      }

-     ],

-     "prompt_number": 13

-    }

-   ],

-   "metadata": {}

-  }

- ]

-}
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