From 41f1f72e9502f5c3de6ca16b303803dfcf1df594 Mon Sep 17 00:00:00 2001
From: Thomas Stephen Lee
Date: Fri, 4 Sep 2015 22:04:10 +0530
Subject: add/remove/update books

---
 .../Chapter3.ipynb                                 | 756 ---------------------
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-{
- "metadata": {
-  "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
-  {
-   "cells": [
-    {
-     "cell_type": "heading",
-     "level": 1,
-     "metadata": {},
-     "source": [
-      "Chapter 3: Properties of Pure Substances"
-     ]
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-1 ,Page No.128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "m=50;#mass in kg\n",
-      "T=90;#temperature in C\n",
-      "\n",
-      "#Values from Table A-4\n",
-      "P=70.183;#in kPa\n",
-      "v=0.001036;#in m^3/kg\n",
-      "\n",
-      "#Calculation\n",
-      "V=m*v;#equating dimensions\n",
-      "print'pressure is %f kPa'%round(P,3);\n",
-      "print'total volumne of tank becomes %f m^3'%round(V,4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "pressure is 70.183000 kPa\n",
-        "total volumne of tank becomes 0.051800 m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 2
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-2 ,Page No.128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "V=2.0;#volumne of saturated water vapor in ft^3\n",
-      "P=50.0;#pressure in psia\n",
-      "\n",
-      "#Values from Table A-5E\n",
-      "T=280.99;#in F\n",
-      "v=8.5175;#in ft^3/lbm\n",
-      "\n",
-      "#caluclation\n",
-      "m=V/v;#dimension analysis\n",
-      "print'Temperature inside cylinder %f F'%round(T,2);\n",
-      "print'mass of vapour inside cylinder %f lbm'%round(m,3);\n",
-      "\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Temperature inside cylinder 280.990000 F\n",
-        "mass of vapour inside cylinder 0.235000 lbm\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-3 ,Page No.128"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Constants used\n",
-      "Hfg=2257.5;#enthalpy of vaporization in kJ/kg\n",
-      "\n",
-      "#Given values\n",
-      "m=200.0/1000;#mass converting in kg\n",
-      "P=100;#Pressure at which process takes place in kPa\n",
-      "\n",
-      "#Values from Table A-5\n",
-      "vg=1.6941;#specific vol of sat liq\n",
-      "vf=0.001043;#specific vol of vapor\n",
-      "\n",
-      "#Caluclation\n",
-      "vfg=vg-vf;\n",
-      "V=m*vfg;\n",
-      "print'the volume change %f m^3'%round(V,4);\n",
-      "E=m*Hfg;\n",
-      "print'amount of energy transferred to the water %f kJ'%round(E,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the volume change 0.338600 m^3\n",
-        "amount of energy transferred to the water 451.500000 kJ\n"
-       ]
-      }
-     ],
-     "prompt_number": 4
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-4 ,Page No.131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "mt=10#mass of water in kg\n",
-      "mf=8;#mass of water in liquid form in kg\n",
-      "T=90;#temperature in C\n",
-      "\n",
-      "#Values from Table A-4\n",
-      "P=70.183;#in kPa\n",
-      "vf=0.001036;#in m^3\n",
-      "vg=2.3593;#in m^3\n",
-      "\n",
-      "#Caluclation\n",
-      "mg=mt-mf;\n",
-      "V=mf*vf+mg*vg;# V= Vg + Vf\n",
-      "print'the volume of the tank %f m^3'%round(V,2);\n",
-      "print'the pressure in the tank %f kPa'%round(P,3)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the volume of the tank 4.730000 m^3\n",
-        "the pressure in the tank 70.183000 kPa\n"
-       ]
-      }
-     ],
-     "prompt_number": 5
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-5 ,Page No.131"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "m=4;#mass of refrigerant-134a in kg\n",
-      "V=80.0/1000;#volumne converting into m^3\n",
-      "P=160;#pressure in kPa\n",
-      "\n",
-      "#Values from Table A-12\n",
-      "vf=0.0007437;\n",
-      "vg=0.12348;\n",
-      "T=-15.60;\n",
-      "hf=31.21;\n",
-      "hfg=209.90;\n",
-      "\n",
-      "#Caluclations\n",
-      "v=V/m;\n",
-      "#vg>v>vf therefore it is a saturated mix\n",
-      "#hence temp will same as saturation temp\n",
-      "print'the temperature %f celcius'%round(T,2)\n",
-      "x=(v-vf)/(vg-vf);#x=vg/vfg i.e the dryness fraction\n",
-      "print'the quality is %f'%round(x,3);\n",
-      "h=hf+x*hfg;\n",
-      "print'the enthalpy of the refrigerant %f kJ/kg'%round(h,1);\n",
-      "mg=x*m;\n",
-      "Vg=mg*vg;\n",
-      "print'the volume occupied by the vapor phase %f m^3'%round(Vg,4)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the temperature -15.600000 celcius\n",
-        "the quality is 0.157000\n",
-        "the enthalpy of the refrigerant 64.100000 kJ/kg\n",
-        "the volume occupied by the vapor phase 0.077500 m^3\n"
-       ]
-      }
-     ],
-     "prompt_number": 1
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-7 ,Page No.133"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "P=0.5;#pressure in MPa\n",
-      "h=2890.0;#enthaply in kJ/kg\n",
-      "\n",
-      "#from Table A\u20136\n",
-      "#at P=0.5 MPa\n",
-      "T1=200.0;\n",
-      "h1=2855.8;\n",
-      "T2=250;\n",
-      "h2=2961.0;\n",
-      "# we need linear interpolation \n",
-      "\n",
-      "#calculatiom\n",
-      "#by interpolation we can say that\n",
-      "#h=h1+(T-T1)/(T2-T1)*(h2-h1)\n",
-      "#we have to find T\n",
-      "T=(h-h1)/(h2-h1)*(T2-T1)+T1;\n",
-      "print'temperature of water %f celcius'%round(T,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "temperature of water 216.300000 celcius\n"
-       ]
-      }
-     ],
-     "prompt_number": 6
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-8 ,Page No.134"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Given values\n",
-      "T=80;#temperature of compressed liquid water in C\n",
-      "P=5;#pressure in KPa\n",
-      "\n",
-      "#from Table A\u20137\n",
-      "#at compressed liq given conditions\n",
-      "u=333.82;\n",
-      "\n",
-      "#from Table A-4\n",
-      "#at saturation\n",
-      "usat=334.97;\n",
-      "\n",
-      "#calcualtion\n",
-      "e=(usat-u)/u*100;\n",
-      "print'internal energy of compressed liquid water using data from the compressed liquid table %f kJ/kg '%round(u,2);\n",
-      "print'internal energy of compressed liquid water using saturated liquid data %f kJ/kg '%round(usat,2);\n",
-      "print'the error involved %f the second case'%round(e,2)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "internal energy of compressed liquid water using data from the compressed liquid table 333.820000 kJ/kg \n",
-        "internal energy of compressed liquid water using saturated liquid data 334.970000 kJ/kg \n",
-        "the error involved 0.340000 the second case\n"
-       ]
-      }
-     ],
-     "prompt_number": 18
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-9 ,Page No.135"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#part a\n",
-      "print('Part a');\n",
-      "\n",
-      "#given values\n",
-      "P=200;#in KPa\n",
-      "x=0.6;\n",
-      "\n",
-      "#from Table A-5\n",
-      "T=120.21;\n",
-      "uf=504.50;\n",
-      "ufg=2024.6;\n",
-      "\n",
-      "#calcualtions\n",
-      "u=uf+(x*ufg);\n",
-      "print'temperature %f Celcius '%round(T,2);\n",
-      "print'internal energy %f kJ/kg'%round(u,2);\n",
-      "print('saturated liquid\u2013vapor mixture at a pressure of 200 kPa');\n",
-      "\n",
-      "#part b\n",
-      "print('Part b');\n",
-      "\n",
-      "#given values\n",
-      "T=125;#in C\n",
-      "u=1600;#in kJ/kg\n",
-      "\n",
-      "#from Table A\u20134\n",
-      "uf=524.83;\n",
-      "ug=2534.3;\n",
-      "#ug>u>ufg so its aturated liquid\u2013vapor mixture\n",
-      "P=232.23;\n",
-      "\n",
-      "#calculation\n",
-      "ufg=ug-uf;\n",
-      "x=(u-uf)/ufg;\n",
-      "print'Pressure %f kPa'%round(P,2);\n",
-      "print'x is %f'%round(x,3);\n",
-      "print('saturated liquid\u2013vapor mixture at a temp of 125 of celcius');\n",
-      "\n",
-      "#part c\n",
-      "print('Part c');\n",
-      "\n",
-      "#given values\n",
-      "P=1000;#in kPa\n",
-      "u=2950;#in kJ/kg\n",
-      "\n",
-      "#from Table A\u20136\n",
-      "uf=761.39;\n",
-      "ug=2582.8;\n",
-      "#u>ug so its superheated steam\n",
-      "T=395.2;\n",
-      "\n",
-      "#calculation\n",
-      "print'temperature %f Celcius'%round(T,1);\n",
-      "print('superheated vapor at 1MPa');\n",
-      "\n",
-      "#part d\n",
-      "print('Part d');\n",
-      "\n",
-      "#given values\n",
-      "T=75;#in C\n",
-      "P=100;#in kPa\n",
-      "\n",
-      "#from Table A\u20135\n",
-      "Tsat=151.83;\n",
-      "#T<Tsat so it is a compressed liquid\n",
-      "#the given pressure is much lower than the lowest pressure value in the compressed liquid table i.e 5 MPa\n",
-      "#assuming, the compressed liquid as saturated liquid at the given temperature\n",
-      "\n",
-      "#from Table A-4\n",
-      "u=313.99;\n",
-      "print'Internal energy %f kJ/kg'%round(u,2);\n",
-      "print('the compressed liquid condition');\n",
-      "\n",
-      "#Part e\n",
-      "print('Part e');\n",
-      "\n",
-      "#given values\n",
-      "P=850;#in kPa\n",
-      "x=0;\n",
-      "\n",
-      "#x=0 therefore it is a saturateed liquid condition\n",
-      "#from Table A-5\n",
-      "T=172.94;\n",
-      "u=731.00;\n",
-      "print'temperature %f Celcius'%round(T,2);\n",
-      "print'Internal energy %f kJ/kg'%round(u,2);\n",
-      "print('saturateed liquid condition')\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "Part a\n",
-        "temperature 120.210000 Celcius \n",
-        "internal energy 1719.260000 kJ/kg\n",
-        "saturated liquid\u2013vapor mixture at a pressure of 200 kPa\n",
-        "Part b\n",
-        "Pressure 232.230000 kPa\n",
-        "x is 0.535000\n",
-        "saturated liquid\u2013vapor mixture at a temp of 125 of celcius\n",
-        "Part c\n",
-        "temperature 395.200000 Celcius\n",
-        "superheated vapor at 1MPa\n",
-        "Part d\n",
-        "Internal energy 313.990000 kJ/kg\n",
-        "the compressed liquid condition\n",
-        "Part e\n",
-        "temperature 172.940000 Celcius\n",
-        "Internal energy 731.000000 kJ/kg\n",
-        "saturateed liquid condition\n"
-       ]
-      }
-     ],
-     "prompt_number": 7
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-10 ,Page No.139"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#constants used\n",
-      "R=0.287# in kPa m^3/kg K\n",
-      "\n",
-      "#given values\n",
-      "l=4;#dimensions of room in m\n",
-      "b=5;#dimensions of room in m\n",
-      "h=6;#dimensions of room in m\n",
-      "P=100.0;#pressure in kPa\n",
-      "T=25+273.0;#temperature in Kelvin\n",
-      "\n",
-      "#calculation\n",
-      "V=l*b*h;\n",
-      "m=P*V/R/T;\n",
-      "print'the mass of the air %f kg'%round(m,1)\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "the mass of the air 140.300000 kg\n"
-       ]
-      }
-     ],
-     "prompt_number": 23
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-11 ,Page No.142"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given values\n",
-      "P=1;#pressure in MPa\n",
-      "T=50+273;#tempearture converting into Kelvin\n",
-      "vgiv=0.021796;#specific vol. given in m^3\n",
-      "\n",
-      "#from Table A-1\n",
-      "R=0.0815;\n",
-      "Pcr=4.059;\n",
-      "Tcr=374.2;\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "#Part A\n",
-      "v1=R*T/(P*1000);\n",
-      "print'specific volume of refrigerant-134a under the ideal-gas assumption %fm^3/kg'%round(v1,6);\n",
-      "e=(v1-vgiv)/vgiv;\n",
-      "print'an error of %f'%round(e,3);\n",
-      "\n",
-      "#Part B\n",
-      "#determine Z from the compressibility chart, we will calculate the reduced pressure and temperature\n",
-      "Pr=P/Pcr;\n",
-      "Tr=T/Tcr;\n",
-      "#from chart\n",
-      "Z=0.84;\n",
-      "v=Z*v1;\n",
-      "print'specific volume of refrigerant-134a under the generalized compressibility chart %f m^3/kg'%round(v,6);\n",
-      "e=(v-vgiv)/vgiv;\n",
-      "print'an error of %f'%round(e,3);\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "specific volume of refrigerant-134a under the ideal-gas assumption 0.026325m^3/kg\n",
-        "an error of 0.208000\n",
-        "specific volume of refrigerant-134a under the generalized compressibility chart 0.022113 m^3/kg\n",
-        "an error of 0.015000\n"
-       ]
-      }
-     ],
-     "prompt_number": 8
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-12 ,Page No.143"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given values\n",
-      "v=0.51431;\n",
-      "T=600;\n",
-      "\n",
-      "#from Table A-1E\n",
-      "R=0.5956;\n",
-      "Pcr=3200;\n",
-      "Tcr=1164.8;\n",
-      "\n",
-      "#calculation\n",
-      "\n",
-      "#Part A\n",
-      "#from Table A-6E\n",
-      "Pa=1000.0;#in psia\n",
-      "print'from the steam tables %i psia'%Pa;\n",
-      "\n",
-      "#Part B\n",
-      "T=1060;#converted into R from F\n",
-      "Pb=R*T/v;\n",
-      "print'from the ideal-gas equation %i psia'%round(Pb,0);\n",
-      "e=(Pb-Pa)/Pa;\n",
-      "print'treating the steam as an ideal gas would result in an error of %f'%round(e,3)\n",
-      "\n",
-      "#Part C\n",
-      "#calculating the pseudo-reduced specific volume and the reduced temperature\n",
-      "Vr=v/(R*Tcr/Pcr);\n",
-      "Tr=T/Tcr;\n",
-      "#from the compressibility chart\n",
-      "Pr=0.33;\n",
-      "P=Pr*Pcr;\n",
-      "print'from the generalized compressibility chart %i psia'%P\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "from the steam tables 1000 psia\n",
-        "from the ideal-gas equation 1228 psia\n",
-        "treating the steam as an ideal gas would result in an error of 0.228000\n",
-        "from the generalized compressibility chart 1056 psia\n"
-       ]
-      }
-     ],
-     "prompt_number": 12
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-13 ,Page No.147"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#Answer of part c-d are having slight difference due to approximation in molar volumne in the textbook which here is caluculated to the approximation of 7 decimal digits\n",
-      "\n",
-      "from math import exp\n",
-      "\n",
-      "#given values\n",
-      "T=175.0;#temperature in C\n",
-      "v=0.00375;#sp. volumne in m^3/kg\n",
-      "Pex=10000;#experimentaion determination\n",
-      "\n",
-      "#from Table A-1\n",
-      "R=0.2968# in kPa m^3/kg  K\n",
-      "\n",
-      "#calculating\n",
-      "\n",
-      "#Part-a\n",
-      "P=R*T/v;\n",
-      "print'using the ideal-gas equation of state %i kPa'%(round(P))\n",
-      "e=(P-Pex)/Pex*100;\n",
-      "print'error is %f percent'%e;\n",
-      "\n",
-      "#Part-b\n",
-      "#van der Waals constants from Eq. 3-23\n",
-      "a=0.175;\n",
-      "b=0.00138;\n",
-      "#from van der waal eq.\n",
-      "P=R*T/(v-b)-a/v**2;\n",
-      "print'using the van der Waals equation of state  is %i kPa'%(round(P));\n",
-      "e=(P-Pex)/Pex*100;\n",
-      "print'error is %f percent'%e;\n",
-      "\n",
-      "#Part-c\n",
-      "#constants in the Beattie-Bridgeman equation from Table 3\u20134\n",
-      "A=102.29;\n",
-      "B=0.05378;\n",
-      "c=4.2*10**4;\n",
-      "Ru=8.314;#in kPa m^3/kmol K\n",
-      "M=28.013;#molecular weight in kg/mol\n",
-      "vb=M*v;#molar vol.\n",
-      "P=(Ru*T)/(vb**2)*(1-((c)/(vb*T**3)))*(vb+B)-(A/vb**2);\n",
-      "print'using the Beattie-Bridgeman equationis %i kPa'%(round(P));\n",
-      "e=(P-Pex)/Pex*100;\n",
-      "print'error is %f percent'%e;\n",
-      "\n",
-      "#Part-d\n",
-      "#constants of Benedict-Webb-Rubin equation from Table 3\u20134\n",
-      "a=2.54;\n",
-      "b=0.002328;\n",
-      "c=7.379*10**4;\n",
-      "alp=1.272*10**-4;\n",
-      "Ao=106.73;\n",
-      "Bo=0.040704;\n",
-      "Co=8.164*10**5;\n",
-      "gam=0.0053;\n",
-      "P=((Ru*T)/vb)+((Bo*Ru*T)-Ao-Co/T**2)/vb**2+(b*Ru*T-a)/vb**3+(a*alp/vb**6)+(c/((vb**3)*(T**2)))*(1 + (gam/vb**2))*exp(-gam/vb**2);\n",
-      "print'using Benedict-Webb-Rubin equation %i kPa'%(round(P));\n",
-      "e=(P-Pex)/Pex*100;\n",
-      "print'error is %f percent'%e;\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "using the ideal-gas equation of state 13851 kPa\n",
-        "error is 38.506667 percent\n",
-        "using the van der Waals equation of state  is 9471 kPa\n",
-        "error is -5.288326 percent\n",
-        "using the Beattie-Bridgeman equationis 10109 kPa\n",
-        "error is 1.092970 percent\n",
-        "using Benedict-Webb-Rubin equation 10004 kPa\n",
-        "error is 0.039256 percent\n"
-       ]
-      }
-     ],
-     "prompt_number": 14
-    },
-    {
-     "cell_type": "heading",
-     "level": 2,
-     "metadata": {},
-     "source": [
-      "Example 3-14 ,Page No.152"
-     ]
-    },
-    {
-     "cell_type": "code",
-     "collapsed": false,
-     "input": [
-      "#given value\n",
-      "T=25;#air temperature over a lake in C\n",
-      "\n",
-      "#from table 3-1\n",
-      "Psat=3.17;#in kPa\n",
-      "\n",
-      "#calculations\n",
-      "\n",
-      "#Relative Humidity 10%\n",
-      "Pv1=0.1*Psat\n",
-      "#Relative Humidity 80%\n",
-      "Pv2=0.8*Psat\n",
-      "#Relative Humidity 100%\n",
-      "Pv3=1*Psat\n",
-      "\n",
-      "# from table 3-1 Tsat at these Pressures are\n",
-      "T1=-8.0;\n",
-      "T2=21.2;\n",
-      "T3=25.0;\n",
-      "print'with relative humidity 10, temperature is %i in C'%T1\n",
-      "print'with relative humidity 80, temperature is %i in C'%T2\n",
-      "print'with relative humidity 100, temperature is %i in C'%T3\n"
-     ],
-     "language": "python",
-     "metadata": {},
-     "outputs": [
-      {
-       "output_type": "stream",
-       "stream": "stdout",
-       "text": [
-        "with relative humidity 10, temperature is -8 in C\n",
-        "with relative humidity 80, temperature is 21 in C\n",
-        "with relative humidity 100, temperature is 25 in C\n"
-       ]
-      }
-     ],
-     "prompt_number": 15
-    }
-   ],
-   "metadata": {}
-  }
- ]
-}
\ No newline at end of file
-- 
cgit