{ "metadata": { "name": "", "signature": "sha256:0bd1584504a5182c99f46d576a77cfaa07f83a047faf3b55eaafd0cea74518f2" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter17-Ideal solutions" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example1-pg367" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#calculate Total pressure and composition of vapour phases and composition of last drop liquids\n", "##initialisation of variables\n", "Pa= 40. ##kPa\n", "Pb= 50. ##kPa\n", "na= 2. ##moles\n", "nb= 6. ##moles\n", "##CALCULATIONS\n", "a= Pb/Pa\n", "xa= na/(na+nb)\n", "xb= 1.-xa\n", "p= xa*Pa+xb*Pb\n", "y= (xa*Pa)/p\n", "ya= 1.-y\n", "Xa= a*xa/(1+(a-1)*xa)\n", "Xb= 1.-Xa\n", "##RESULTS\n", "print'%s %.1f %s'%('Total pressure=',p,'kPa')\n", "print'%s %.4f %s'%('composition of vapour phase=',y,'')\n", "print'%s %.4f %s'%('composition of vapour phase=',ya,'')\n", "print'%s %.4f %s'%('composition of last drop of liquid=',Xa,'')\n", "print'%s %.4f %s'%('composition of last drop of liquid=',Xb,'')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Total pressure= 47.5 kPa\n", "composition of vapour phase= 0.2105 \n", "composition of vapour phase= 0.7895 \n", "composition of last drop of liquid= 0.2941 \n", "composition of last drop of liquid= 0.7059 \n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example2-pg371" ] }, { "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": [ "Example3-pg373" ] }, { "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": [ "Example4-pg376" ] }, { "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": [ "Example5-pg377" ] }, { "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)))\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= 2160.4 kW\n", "irreversibility= 2160.4 kW\n" ] } ], "prompt_number": 7 } ], "metadata": {} } ] }