{ "metadata": { "name": "", "signature": "sha256:33e389243f3edbfa798e02801dff08e24da71a0ad10959698b7564f561b867d4" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter17-Ideal solutions" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1-pg 480" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#intilization variables\n", "x=0.25\n", "pa=40\n", "pb=50\n", "ya=0.25\n", "alpha=1.25\n", "#calculation\n", "P=x *pa+(1-x)*pb\n", "y=x*pa/P\n", "yb=(1-y)\n", "xa=alpha*y/(1+(alpha-1)*y)\n", "xb=(1-x)\n", "#results\n", "print'%s %.2f %s'%('total pressure of an ideal solution',P,'kpa')\n", "print'%s %.2f %s %.2f %s '%('composition of the gaseous phase',y,'' and ' ',yb,'')\n", "print'%s %.2f %s %.2f %s '%('the composition of last drop',xa,' ' and ' ',xb,'') " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "total pressure of an ideal solution 47.50 kpa\n", "composition of the gaseous phase 0.21 0.79 \n", "the composition of last drop 0.25 0.75 \n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example2-pg484" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "##initialisation of variables\n", "T= 290 ##K\n", "xa= 0.4\n", "xb= 0.6\n", "P= 600 ##kPa\n", "V= 60 ##L\n", "R= 8.314 ##J/mol K\n", "Mp= 44 ##kg/kmol\n", "Mb= 58.12 ##kg/kmol\n", "vp= 0.00171 ##m**3/kg\n", "vb= 0.00166 ##m**3/kg\n", "na= 0.1 ##kmol\n", "nb= 0.15 ##kmol\n", "V1= 0.04000 ##m**3\n", "xa= 0.4 \n", "np= 2\n", "Vc= 0.1 ##m**3\n", "##CALCULATIONS\n", "Pasat= math.e**(14.435-(2255/T))\n", "Pbsat= math.e**(14.795-(2770/T))\n", "P1= xa*Pasat+xb*Pbsat\n", "Na1= P*V/(100*R*T)\n", "Vp= vp*Mp\n", "Vb= vb*Mb\n", "V= na*Vp+nb*Vb\n", "Vv= V1-V\n", "nv= P1*Vv/(R*T)\n", "ya= xa*Pasat/P\n", "yb=1-ya\n", "Na= na+ya*nv\n", "Nb= nb+yb*nv\n", "##RESULTS\n", "print'%s %.2f %s'% (' initial pressure= ',P1,' kPa')\n", "print'%s %.2f %s'% (' moles of propane= ',Na1,' kmol')\n", "print'%s %.2f %s'% (' initial mole of propane= ',Na,' kmol')\n", "print'%s %.2f %s'% (' initial mole of butane= ',Nb,' kmol')\n", "print'%s %.2f %s'% (' numbar of phases= ',np,'')\n", "print'%s %.2f %s'% (' volume in final state=',Vc,' m^3')" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " initial pressure= 874.89 kPa\n", " moles of propane= 0.15 kmol\n", " initial mole of propane= 0.11 kmol\n", " initial mole of butane= 0.15 kmol\n", " numbar of phases= 2.00 \n", " volume in final state= 0.10 m^3\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example3-pg489" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\t\n", "#calculate pressure of the phase of pure A\n", "##initialisation of variables\n", "p0= 10. ##Mpa\n", "R= 8.314 ##J/mol K\n", "T= 30. ##C\n", "va= 0.02 ##m^3/kmol\n", "xa= 0.98\n", "##CALCULATIONS\n", "p= p0+(R*(273.15+T)*math.log(xa)/(va*1000.))\n", "##RESULTS\n", "print'%s %.2f %s'%('Pressure of the phase of pure A=',p,'Mpa')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Pressure of the phase of pure A= 7.45 Mpa\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example4-pg491" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#calculate the boiling point elevation\n", "##initialisation of variables\n", "hfg= 2257.0 ##kJ/kg\n", "Tb= 100 ##C\n", "R= 8.314 ##J/mol K\n", "m2= 10 ##gms\n", "M2= 58.5 ##gms\n", "m1= 90. ##gms\n", "M1= 18. ##gms\n", "##CALCULATIONS\n", "x2= (m2/M2)/((m2/M2)+(m1/M1))\n", "dT= R*math.pow(273.15+Tb,2)*x2/(M1*hfg)\n", "##RESULTS\n", "print'%s %.3f %s'%(' Boiling point elevation=',dT,'C')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Boiling point elevation= 0.942 C\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example5-pg494" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#calculate Osomatic pressures\n", "##initialisation of variables\n", "M1= 18.02 ##gms\n", "m1= 0.965 ##gms\n", "m2= 0.035 ##gms\n", "M2= 58.5 ##gms\n", "R= 8.314 ##J/mol K\n", "M= 18.02 ##kg\n", "T= 20. ##C\n", "vf= 0.001002 ##m^3\n", "x21= 0.021856 ##m^3\n", "##CALCULATIONS\n", "n1= m1/M1\n", "n2= m2/M2\n", "x1= n1/(n1+n2)\n", "x2= n2/(n2+n1)\n", "P= R*(273.15+T)*x2/(M*vf)\n", "P1= R*(273.15+T)*x21/(M*vf)\n", "##RESULTS\n", "print'%s %.1f %s'%('Osmotic pressure=',P,'kpa')\n", "print'%s %.1f %s'%('Osmotic pressure=',P1,'kpa')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Osmotic pressure= 1491.4 kpa\n", "Osmotic pressure= 2950.2 kpa\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example6-pg495" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#what is useful work in the process and heat interaction and maximum work and irreversibility\n", "##initialisation of variables\n", "W= 0.\n", "Q= 0.\n", "R= 8.314 ##J/mol K\n", "T0= 300. ##K\n", "x= 5./13.\n", "n1= 0.5 ##kmol/s\n", "n2= 0.8 ##kmol/s\n", "##CALCULATIONS\n", "W1= (n1+n2)*R*T0*(x*math.log(1/x)+(1-x)*math.log(1./(1.-x)))+470\n", "I= W1\n", "##RESULTS\n", "print'%s %.f %s'%('useful work of the process=',W,'kW') \n", "print'%s %.f %s'%('heat interaction=',Q,'kW') \n", "print'%s %.1f %s'%('maximum work=',W1,'kW') \n", "print'%s %.1f %s'%('irreversibility=',I,'kW')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "useful work of the process= 0 kW\n", "heat interaction= 0 kW\n", "maximum work= 2630.4 kW\n", "irreversibility= 2630.4 kW\n" ] } ], "prompt_number": 8 } ], "metadata": {} } ] }