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-{
- "metadata": {
-  "name": "",
-  "signature": "sha256:d3b8a40a5268a38ad3fc311ebed2078460c9192ee65f2f5e3922c8f65a52bf49"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 2: Properties of Pure Substances"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.1:pg-22"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "# initialization of variables\n",
-      "\n",
-      "m=1;  #mass of saturated water in kg\n",
-      "\n",
-      "    # All the necessary values are taken from table C-2\n",
-      " \n",
-      "# part (a)\n",
-      "\n",
-      "P=0.001; # Pressure in MPa\n",
-      "vf=0.001; #specific volume of saturated liquid at 0.001 Mpa in Kg/m^3\n",
-      "vg=129.2; # specific volume of saturated vapour at 0.001 Mpa in Kg/m^3\n",
-      "deltaV=m*(vg-vf) # by properties of pure substance \n",
-      "# result\n",
-      "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV,1),\" m^3/kg \\n\"\n",
-      "\n",
-      "# part (b) \n",
-      "\n",
-      "P=0.10;  # Pressure in MPa\n",
-      "vf=0.001; # specific volume of saturated liquid at 0.26 MPa( it is same from at 0.2 and 0.3 MPa upto 4 decimals)\n",
-      "vg=1.694;  # specific volume of saturated vapour at 0.1 Mpa\n",
-      "deltaV=m*(vg-vf) # by properties of pure substance\n",
-      "# result\n",
-      "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV,3),\" m^3/kg \\n\"\n",
-      "\n",
-      "# part (c) \n",
-      "P=10;  # Pressure in MPa\n",
-      "vf=0.00145; # specific volume of saturated liquid at 10 MPa\n",
-      "vg=0.01803; # specific volume of saturated vapour at 10 MPa\n",
-      "deltaV=m*(vg-vf) # by properties of pure substance \n",
-      "# result\n",
-      "print \"The Volume change at pressure \",(P),\" MPa is\",round(deltaV,5),\" m^3/kg \\n\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Volume change at pressure  0.001  MPa is 129.2  m^3/kg \n",
-        "\n",
-        "The Volume change at pressure  0.1  MPa is 1.693  m^3/kg \n",
-        "\n",
-        "The Volume change at pressure  10  MPa is 0.01658  m^3/kg \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.2:pg-23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#    initialization of variables\n",
-      "m=4.0 # mass of water in kg\n",
-      "V=1.0 # volume in m^3\n",
-      "T=150 # temperature of water in degree centigrade\n",
-      "\n",
-      "# TABLE C-1 is used for values in wet region\n",
-      "# Part (a)\n",
-      "P=475.8 # pressure in KPa in wet region at temperature of 150 *C\n",
-      "print \" The pressure is\",round(P,1),\" kPa \\n\"\n",
-      "\n",
-      "# Part (b)\n",
-      "#first we determine the dryness fraction\n",
-      "v=V/m # specific volume of water\n",
-      "vg=0.3928 #  specific volume of saturated vapour @150 degree celsius\n",
-      "vf=0.00109 # specific volume of saturated liquid @150 degree celsius\n",
-      "x=(v-vf)/(vg-vf); # dryness fraction\n",
-      "mg=m*x; # mass of vapour\n",
-      "print \" The mass of vapour present is\",round(mg,3),\" kg \\n\"\n",
-      "\n",
-      "# Part(c) \n",
-      "Vg=vg*mg; # volume of vapour\n",
-      "print \" The volume of vapour is\",round(Vg,4),\" m^3\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " The pressure is 475.8  kPa \n",
-        "\n",
-        " The mass of vapour present is 2.542  kg \n",
-        "\n",
-        " The volume of vapour is 0.9984  m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.3:pg-23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#    initialization of variables\n",
-      "m=4 # mass of water in kg\n",
-      "P=220 # pressure in KPa\n",
-      "x=0.8 # quality of steam\n",
-      "\n",
-      "# Table C-2 is used for values\n",
-      "\n",
-      "vg=(P-200)*(0.6058-0.8857)/(300-200)+0.8857 # specific volume of saturated vapour @ given pressure by interpolating\n",
-      "vf=0.0011 # specific volume of saturated liquid at 220 KPa\n",
-      "v=vf+x*(vg-vf)# property of pure substance\n",
-      "V=m*v # total volume\n",
-      "#result\n",
-      "print \"The Total volume of the mixture is \",round(V,3),\" m^3\""
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Total volume of the mixture is  2.656  m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Ex2.4:pg-23"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#   initialization of variables\n",
-      "m=2 # mass of water in lb\n",
-      "P=540 # pressure in psi\n",
-      "T=700 # temperature in degree fahrenheit\n",
-      " # Table C-3E is used for values\n",
-      "v=1.3040+(P-500)*(1.0727-1.3030)/(600-500) # specific volue by interpolatin between 500 and 600 psi\n",
-      "V=m*v # final volume\n",
-      "print \"The Final Volume is\",round(V,3),\" ft^3\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Final Volume is 2.424  ft^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 9
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.5:pg-25"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#    initialization of variables\n",
-      "V=0.6 # volume of tyre in m^3\n",
-      "Pgauge=200 # gauge pressure in KPa\n",
-      "T=20+273 # temperature converted to kelvin\n",
-      "Patm=100 # atmospheric pressure in KPa\n",
-      "R=287 # gas constant in Nm/kg.K\n",
-      "Pabs=(Pgauge+Patm)*1000 # calculating absolute pressue in Pa \n",
-      "\n",
-      "m=Pabs*V/(R*T)# mass from ideal gas equation\n",
-      "print \"The Mass of air is\",round(m,2),\" Kg\"\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "The Mass of air is 2.14  Kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 10
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Ex2.6:pg-26"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "import math\n",
-      "#    initialization of variables\n",
-      "T=500+273 # temperature of steam in kelvin\n",
-      "rho=24.0 # density in Kg/m^3\n",
-      "R=0.462 # gas constant from Table B-2\n",
-      "v=1/rho # specific volume and density relation\n",
-      "# PART (a)\n",
-      "P=rho*R*T # from Ideal gas equation\n",
-      "print \" PART (a) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook\n",
-      "\n",
-      "# PART (b)\n",
-      "a=1.703 #  van der Waal's constant a value from Table B-8\n",
-      "b=0.00169 # van der Waal's constant b value from Table B-8\n",
-      "P=(R*T/(v-b))-(a/v**2) # Pressure from van der Waals equation\n",
-      "print \" PART (b) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook\n",
-      "\n",
-      "# PART (c)\n",
-      "a=43.9 # van der Waal's constant a value from Table B-8\n",
-      "b=0.00117 # van der Waal's constant b value from Table B-8\n",
-      "\n",
-      "P=(R*T/(v-b))-(a/(v*(v+b)*math.sqrt(T))) # Redlich-Kwong equation\n",
-      "print \" PART (c) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook\n",
-      "\n",
-      "# PART (d)\n",
-      "Tcr=947.4 # compressibilty temperature from table B-3\n",
-      "Pcr=22100 # compressibility pressure from table B-3\n",
-      "\n",
-      "TR=T/Tcr # reduced temperature\n",
-      "PR=P/Pcr # reduced pressure\n",
-      "Z=0.93 # from compressiblility chart\n",
-      "P=Z*R*T/v # Pressure in KPa\n",
-      "print \" PART (d) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook\n",
-      "\n",
-      "# PART (e)\n",
-      "P=8000 # pressure from steam table @ 500*c and v= 0.0417 m^3\n",
-      "print \" PART (e) The Pressure is \",int(P),\" KPa \\n\"\n",
-      "# answer is approximated in textbook"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        " PART (a) The Pressure is  8571  KPa \n",
-        "\n",
-        " PART (b) The Pressure is  7952  KPa \n",
-        "\n",
-        " PART (c) The Pressure is  7934  KPa \n",
-        "\n",
-        " PART (d) The Pressure is  7971  KPa \n",
-        "\n",
-        " PART (e) The Pressure is  8000  KPa \n",
-        "\n"
-       ]
-      }
-     ],
-     "prompt_number": 13
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
-\ No newline at end of file