{ "metadata": { "name": "", "signature": "sha256:86a66a56e6c068c1b5488c8e726c9aecdd48422f90b429aec4b5cc311da3bb32" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 5 - Actual Gases" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 1 - Pg 74" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the volume occupied by the gas and also the tabulated volume\n", "#Initialization of variables\n", "m=20. #lbm\n", "P=1000. #psia\n", "T=580. #R\n", "R=35.12\n", "#calculations\n", "print '%s' %(\"From table 5-2,\")\n", "z=0.667\n", "V=z*m*R*T/(P*144.)\n", "vt=0.0935\n", "vtt=vt*m\n", "#results\n", "print '%s %.3f %s' %(\"Volume occupied =\",V,\"cu ft\")\n", "print '%s %.2f %s' %(\"\\n Tablulated value for volume =\",vtt,\" cu ft\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From table 5-2,\n", "Volume occupied = 1.887 cu ft\n", "\n", " Tablulated value for volume = 1.87 cu ft\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2 - Pg 74" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the pressure obtained and theoretical pressure\n", "#Initialization of variables\n", "m=90 #lbm\n", "T=200+459.7 #R\n", "Tc=232.7+459.7 #R\n", "R0=1545.\n", "M=120.9\n", "V=30. #cu ft\n", "#calculations\n", "R=R0/M\n", "print '%s' %(\"From fig 5.5\")\n", "z=0.883\n", "P=z*R*m*T/V/144.\n", "vc=V/m\n", "P2=156.1 #psia\n", "#results\n", "print '%s %.2f %s' %(\"Pressure obtained =\",P,\"psia\")\n", "print '%s %.1f %s' %(\"\\n Theoretical pressure =\",P2,\"psia\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "From fig 5.5\n", "Pressure obtained = 155.08 psia\n", "\n", " Theoretical pressure = 156.1 psia\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3 - Pg 84" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the heat added in both cases\n", "#Initialization of variables\n", "import math\n", "T1=140+460. #R\n", "T2=240 +460. #R\n", "N=1.\n", "#calculations\n", "Q=N*(9.47*(T2-T1)-3.47*1000. *math.log(T2/T1) -1.16*math.pow(10,6) *(1/T2-1/T1))\n", "Tm=(T1+T2)/2\n", "Cp=9.47-3.47*1000. /Tm +1.16*math.pow(10,6) /Tm/Tm\n", "Q2=N*Cp*(T2-T1)\n", "#results\n", "print '%s %d %s' %(\"Heat added in case 1 =\",Q,\"Btu\")\n", "print '%s %.1f %s' %(\"\\n Heat added in case 2 =\",Q2,\"Btu\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat added in case 1 = 688 Btu\n", "\n", " Heat added in case 2 = 687.7 Btu\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 4 - Pg 85" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the heat added in both cases\n", "#Initialization of variables\n", "import math\n", "Rj=1.985\n", "N=1\n", "T1=540.+460 #R\n", "T2=3540.+460 #R\n", "#calculations\n", "Q=N*(14.215*(T2-T1)-6.53*1000. *math.log(T2/T1) -1.41*math.pow(10,6) *(1/T2-1/T1))\n", "Tm=(T1+T2)/2\n", "Cv=14.215-6.53*1000. /Tm +1.41*math.pow(10,6) /Tm/Tm\n", "Q2=N*Cv*(T2-T1)\n", "#results\n", "print '%s %.1f %s' %(\"Heat added in case 1 =\",Q,\"Btu\")\n", "print '%s %.1f %s' %(\"\\n Heat added in case 2 =\",Q2,\"Btu\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat added in case 1 = 34650.0 Btu\n", "\n", " Heat added in case 2 = 35485.8 Btu\n" ] } ], "prompt_number": 6 } ], "metadata": {} } ] }