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-{


- "metadata": {


-  "name": "",


-  "signature": "sha256:4d90868c7316d94fc2ffdc05c5691234f77f3a832d71da7eb2108eaff0a215bc"


- },


- "nbformat": 3,


- "nbformat_minor": 0,


- "worksheets": [


-  {


-   "cells": [


-    {


-     "cell_type": "heading",


-     "level": 1,


-     "metadata": {},


-     "source": [


-      "Chapter14:THERMODYNAMIC RELATIONS"


-     ]


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex14.1:Pg-567"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#ques1\n",


-      "#to determine the sublimation pressure of water\n",


-      "import math\n",


-      "#from table in appendix B.1.5\n",


-      "T1=213.2;#K, Temperature at state 1\n",


-      "P2=0.0129;#kPa, pressure at state 2\n",


-      "T2=233.2;#K, Temperature at state 2\n",


-      "hig=2838.9;#kJ/kg, enthalpy of sublimation \n",


-      "R=.46152;#Gas constant \n",


-      "#using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) \n",


-      "P1=P2*math.exp(-hig/R*(1/T1-1/T2));\n",


-      "print\" Sublimation Pressure \",round(P1,5),\"kPa\""


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        " Sublimation Pressure  0.00109 kPa\n"


-       ]


-      }


-     ],


-     "prompt_number": 25


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex14.1E:Pg-567\n"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#ques1\n",


-      "#to determine the sublimation pressure of water\n",


-      "import math\n",


-      "#from table in appendix B.1.5\n",


-      "T1=-70+460.7;# R, Temperature at state 1\n",


-      "P2=0.0019 # lbf/in^2 pressure at state 2\n",


-      "T2=-40+460.7;# R, Temperature at state 2\n",


-      "hig=1218.7;#Btu/lbm, enthalpy of sublimation \n",


-      "R=85.67;#Gas constant \n",


-      "#using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) \n",


-      "P1=P2*math.exp(-hig*778/R*(1/T1-1/T2));\n",


-      "print\" Sublimation Pressure \",round(P1,5),\"lbf/in^2\""


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        " Sublimation Pressure  0.00025 lbf/in^2\n"


-       ]


-      }


-     ],


-     "prompt_number": 3


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex14.4:Pg-579"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#ques4\n",


-      "#Volume expansivity, Isothermal and Adiabatic compressibility\n",


-      "\n",


-      "#known data\n",


-      "ap=5*10**-5;#K^-1 Volume expansivity\n",


-      "bt=8.6*10**-12;#m^2/N, Isothermal compressibility\n",


-      "v=0.000114;#m^3/kg, specific volume\n",


-      "P2=100*10**6;#pressure at state 2 in kPa\n",


-      "P1=100;#pressure at state 1 in kPa\n",


-      "w=-v*bt*(P2**2-P1**2)/2;#work done in J/kg\n",


-      "#q=T*ds and ds=-v*ap*(P2-P1)\n",


-      "#so q=-T*v*ap*(P2-P1)\n",


-      "T=288.2;#Temperature in K\n",


-      "q=-T*v*ap*(P2-P1);#heat in J/kg\n",


-      "du=q-w;#change in internal energy in J/kg\n",


-      "print\" Change in internal energy =\",round(du,3),\"J/kg\"\n",


-      "\n",


-      "#the answer is correct within given limts\n",


-      " "


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        " Change in internal energy = -159.372 J/kg\n"


-       ]


-      }


-     ],


-     "prompt_number": 22


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex14.5:Pg-586"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#ques5\n",


-      "#adiabatic steady state processes\n",


-      "\n",


-      "#from table A.2\n",


-      "P1=20;#pressure at state 1 in MPa\n",


-      "P2=2;#pressure at state 2 in MPa\n",


-      "T1=203.2;#Temperature at state 1 in K\n",


-      "Pr1=P1/3.39;#Reduced pressure at state 1\n",


-      "Pr2=P2/3.39;#Reduced pressure at state 2\n",


-      "Tr1=T1/126.2;#Reduced temperature\n",


-      "#from compressibility chart h1*-h1=2.1*R*Tc\n",


-      "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n",


-      "#h2*-h2=0.5*R*Tc\n",


-      "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.5*R*Tc\n",


-      "R=0.2968;#gas constant for given substance\n",


-      "Tc=126.2;#K, Constant temperature\n",


-      "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n",


-      "T2=146;#temperature at state 2\n",


-      "dh=-1.6*R*Tc+Cp*(T1-T2);#\n",


-      "print\" Enthalpy change =\",round(dh,3),\"kJ/kg \\n\"\n",


-      "print\" Since Enthalpy change is nearly \",-round(dh),\"kJ/kg so Temperature =\",round(T2,3),\"K\""


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        " Enthalpy change = -0.35 kJ/kg \n",


-        "\n",


-        " Since Enthalpy change is nearly  0.0 kJ/kg so Temperature = 146.0 K\n"


-       ]


-      }


-     ],


-     "prompt_number": 19


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex14.6:Pg-589"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#ques6\n",


-      "#isothermal steady state processes\n",


-      "import math\n",


-      "#from table A.2\n",


-      "P1=8;#pressure at state 1 in MPa\n",


-      "P2=0.5;#pressure at state 2 in MPa\n",


-      "T1=150.0;#Temperature at state 1 in K\n",


-      "Pr1=P1/3.39;#Reduced pressure at state 1\n",


-      "Pr2=P2/3.39;#Reduced pressure at state 2\n",


-      "Tr1=T1/126.2;#Reduced temperature\n",


-      "T2=125.0;#temperature at state 2\n",


-      "#from compressibility chart h1*-h1=2.1*R*Tc\n",


-      "#from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)\n",


-      "#h2*-h2=0.5*R*Tc\n",


-      "#this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.15*R*Tc\n",


-      "R=0.2968;#gas constant for given substance\n",


-      "Tc=126.2;#K, Constant temperature\n",


-      "Cp=1.0416;#heat enthalpy at constant pressure in kJ/kg\n",


-      "dh=(2.35)*R*Tc+Cp*(T2-T1);#\n",


-      "print\" Enthalpy change =\",round(dh),\"kJ/kg\"\n",


-      "#change in entropy \n",


-      "#ds= -(s2*-s2)+(s2*-s1*)+(s1*-s1)\n",


-      "#s1*-s1=1.6*R\n",


-      "#s2*-s2=0.1*R\n",


-      "#s2*-s1*=Cp*log(T2/T1)-R*log(P2/P1)\n",


-      "#so\n",


-      "ds=1.6*R-0.1*R+Cp*math.log(T2/T1)-R*math.log(P2/P1);\n",


-      "print\" Entropy Change =\",round(ds,3),\"kJ/kg.K \""


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        " Enthalpy change = 62.0 kJ/kg\n",


-        " Entropy Change = 1.078 kJ/kg.K \n"


-       ]


-      }


-     ],


-     "prompt_number": 15


-    },


-    {


-     "cell_type": "heading",


-     "level": 2,


-     "metadata": {},


-     "source": [


-      "Ex14.7:Pg-596"


-     ]


-    },


-    {


-     "cell_type": "code",


-     "collapsed": false,


-     "input": [


-      "#ques7\n",


-      "#percent deviation using specific volume calculated by kays rule and vander waals rule\n",


-      "import math\n",


-      "\n",


-      "#a-denotes C02\n",


-      "#b-denotes CH4\n",


-      "T=310.94;#Temperature of mixture K\n",


-      "P=86.19;#Pressure of mixture in MPa\n",


-      "#Tc- critical Temperature\n",


-      "#Pc-critical pressure\n",


-      "Tca=304.1;#K\n",


-      "Tcb=190.4;#K\n",


-      "Pca=7.38;#MPa\n",


-      "Pcb=4.60;#MPa\n",


-      "Ra=0.1889;#gas constant for a in kJ/kg.K\n",


-      "Rb=0.5183;#gas constant for b in kJ/kg.K\n",


-      "xa=0.8;#fraction of CO2\n",


-      "xb=0.2;#fraction of CH4\n",


-      "Rm=xa*Ra+xb*Rb;#mean gas constant in kJ/kg.K\n",


-      "Ma=44.01;#molecular mass of a\n",


-      "Mb=16.043;#molecular mass of b\n",


-      "#1.Kay's rule\n",


-      "ya=xa/Ma/(xa/Ma+xb/Mb);#mole fraction of a\n",


-      "yb=xb/Mb/(xa/Ma+xb/Mb);#mole fraction of b\n",


-      "Tcm=ya*Tca+yb*Tcb;#mean critical temp in K\n",


-      "Pcm=ya*Pca+yb*Tcb;#mean critical pressure n MPa\n",


-      "#therefore pseudo reduced property of mixture\n",


-      "Trm=T/Tcm;\n",


-      "Prm=P/Pcm;\n",


-      "Zm=0.7;#Compressiblity from generalised compressibility chart\n",


-      "vc=Zm*Rm*T/P/1000;#specific volume calculated in m^3/kg\n",


-      "ve=0.0006757;#experimental specific volume in m^3/kg\n",


-      "pd1=(ve-vc)/ve*100;#percent deviation\n",


-      "print\" Percentage deviation in specific volume using Kays rule =\",round(pd1,1),\"percent \\n\"\n",


-      "\n",


-      "#2. using vander waals equation\n",


-      "#values of vander waals constant\n",


-      "Aa=27*(Ra**2)*(Tca**2)/(64*Pca*1000);\n",


-      "Ba=Ra*Tca/(8*Pca*1000);\n",


-      "Ab=27*(Rb**2)*(Tcb**2)/(64*Pcb*1000);\n",


-      "Bb=Rb*Tcb/(8*Pcb*1000);\n",


-      "#mean vander waals constant\n",


-      "Am=(xa*math.sqrt(Aa)+xb*math.sqrt(Ab))**2;\n",


-      "Bm=(xa*Ba+xb*Bb);\n",


-      "#using vander waals equation we get cubic equation \n",


-      "#solving we get\n",


-      "vc=0.0006326;#calculated specific volume in m^3/kg\n",


-      "pd2=((ve-vc)/ve)*100;\n",


-      "print\" Percentage deviation in specific volume using vander waals eqn =\",round(pd2,1),\"percent\""


-     ],


-     "language": "python",


-     "metadata": {},


-     "outputs": [


-      {


-       "output_type": "stream",


-       "stream": "stdout",


-       "text": [


-        " Percentage deviation in specific volume using Kays rule = 4.8 percent \n",


-        "\n",


-        " Percentage deviation in specific volume using vander waals eqn = 6.4 percent\n"


-       ]


-      }


-     ],


-     "prompt_number": 9


-    }


-   ],


-   "metadata": {}


-  }


- ]


-}
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