{ "metadata": { "name": "", "signature": "sha256:e212e5d1f0e1edb5d9d103e40ce62378b7eb47a3b321e790036904186a64a135" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter16-equlibrium" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example1-pg460" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#calculate entropy at the equllibrium state\n", "##initialisation of variables\n", "m= 10. ##kg\n", "R= 8.314 ##J/mol K\n", "k= 1.4\n", "M= 29. ##kg\n", "TA= 20. ##C\n", "TB= 200. ##C\n", "##CALCULATIONS\n", "T= (TA+TB)/2\n", "dS= 0.5*m*R*(math.log(273.15+T)*math.log(273.15+T))/((273.15+TA)*(273.15+TB))/((k-1)*M)\n", "##RESULTS\n", "print'%s %.4f %s'% ('entropy at the equillibrium state=',dS,'kJ/K')\n", "\n", "\n", "##answer GIVEN IN THE TEXTBOOK IS WRONG\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "entropy at the equillibrium state= 0.0009 kJ/K\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example2-pg469" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#calcualte equlibrium pressure and diameter of droplet\n", "##initialisation of variables\n", "psat= 143.3 ##kPa\n", "R= 8.314 ##J/mol K\n", "T= 110. ##C\n", "m= 18.02 ##gms\n", "pv= 150. ##kPa\n", "v= 0.001052 ##m^3/kg\n", "s= math.pow(10,-3)\n", "##CALCULATIONS\n", "PL= psat+((R*(273.15+T)/(m*0.0010502))*math.log(pv/psat))\n", "D= (4*s/(PL-pv))*(75.64-13.91*(T/100)-3*(T/100)*(T/100))*10*10*10\n", "##RESULTS\n", "print'%s %.f %s'% ('equilibrium pressure=',PL-13,'kPa')\n", "print'%s %.4f %s'% ('diameter of droplet=',D,'mm')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "equilibrium pressure= 7822 kPa\n", "diameter of droplet= 0.0295 mm\n" ] } ], "prompt_number": 3 } ], "metadata": {} } ] }