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diff --git a/backup/Thermodynamics_for_Engineers_version_backup/Chapter_5.ipynb b/backup/Thermodynamics_for_Engineers_version_backup/Chapter_5.ipynb new file mode 100755 index 00000000..60d3de94 --- /dev/null +++ b/backup/Thermodynamics_for_Engineers_version_backup/Chapter_5.ipynb @@ -0,0 +1,199 @@ +{
+ "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": {}
+ }
+ ]
+}
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