Added(A)/Deleted(D) following books

 A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter1.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter10.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter11.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter12.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter13.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter14.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter15.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter16.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter17.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter18.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter19.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter2.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter4.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter5.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter7.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter8.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter9.ipynb
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/screenshots/chap11.png
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/screenshots/chap14.png
A  Basic_Engineering_Thermodynamics_by_Rayner_Joel/screenshots/chap7.png
A  Principles_Of_Electric_Machines_And_Power_Electronics_by_P._C._Sen/README.txt
A  sample_notebooks/PrashantSahu/Chapter-2-Molecular_Diffusion_-_Principles_of_Mass_Transfer_and_Separation_Process_by_Binay_K_Dutta_1.ipynb
A  "sample_notebooks/S PRASHANTHS PRASHANTH/Chapter_1_5.ipynb"

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 5 - Gases and Single Phase systems"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1: pg 98"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.1\n",
+      " The new pressure exerted on the air (mmHg) =  900.0\n",
+      " The difference in the two mercury column level (mm) =  135.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 98\n",
+    "#calculate the new pressure and difference in two levels\n",
+    "print('Example 5.1');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  new pressure exerted on the air and the difference in two mercury column level\n",
+    "\n",
+    "#  Given values\n",
+    "P1 = 765.;#  atmospheric pressure, [mmHg]\n",
+    "V1 = 20000.;#  [mm^3]\n",
+    "V2 = 17000.;#  [mm^3]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  using boyle's law P*V=constant\n",
+    "#  hence\n",
+    "P2 = P1*V1/V2;#  [mmHg]\n",
+    "\n",
+    "del_h = P2-P1;#  difference in Height of mercury column level\n",
+    "#results\n",
+    "print ' The new pressure exerted on the air (mmHg) = ',P2\n",
+    "print ' The difference in the two mercury column level (mm) = ',del_h\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2: pg 99"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.2\n",
+      " The new volume after expansion (m^3) =  0.7\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 99\n",
+    "#calculate the new volume\n",
+    "print('Example 5.2');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  the new volume\n",
+    "\n",
+    "#  Given values\n",
+    "P1 = 300;#  original pressure,[kN/m^2]\n",
+    "V1 = .14;#  original volume,[m^3]\n",
+    "\n",
+    "P2 = 60.;#  new pressure after expansion,[kn/m^2]\n",
+    "\n",
+    "#  solution\n",
+    "#  since temperature is constant so using boyle's law P*V=constant\n",
+    "V2 = V1*P1/P2;#  [m^3]\n",
+    "\n",
+    "#results\n",
+    "print ' The new volume after expansion (m^3) = ',V2\n",
+    "\n",
+    "#  End\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3: pg 101"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.3\n",
+      " The new volume of the gas trapped in the apparatus (mm^3) =  12302.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 101\n",
+    "#calculate the new volume\n",
+    "print('Example 5.3');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  the new volume of the gas\n",
+    "\n",
+    "#  Given values\n",
+    "V1 = 10000;#  [mm^3]\n",
+    "T1 = 273.+18;#  [K]\n",
+    "T2 = 273.+85;#  [K]\n",
+    "\n",
+    "#  solution\n",
+    "#  since pressure exerted on the apparatus is constant so using charle's law V/T=constant\n",
+    "#  hence\n",
+    "V2 = V1*T2/T1;#  [mm^3]\n",
+    "\n",
+    "#results\n",
+    "print ' The new volume of the gas trapped in the apparatus (mm^3) = ',round(V2)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4: pg 102"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.4\n",
+      " The final temperature of the gas (C) =  15.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 102\n",
+    "#calculate the final temperature\n",
+    "print('Example 5.4');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  the final temperature\n",
+    "\n",
+    "#  Given values\n",
+    "V1 = .2;#  original volume,[m^3]\n",
+    "T1 = 273+303;# original temperature, [K]\n",
+    "V2 = .1;# final volume, [m^3]\n",
+    "\n",
+    "#  solution\n",
+    "#  since pressure is constant, so using charle's law V/T=constant\n",
+    "#  hence\n",
+    "T2 = T1*V2/V1;#  [K]\n",
+    "t2 = T2-273;#  [C]\n",
+    "#results\n",
+    "print ' The final temperature of the gas (C) = ',t2\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5: pg 106"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.5\n",
+      "The new volume of the gas (m^3) =  0.0223\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 106\n",
+    "#calculate the new volume\n",
+    "print('Example 5.5');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  the new volume of the gas\n",
+    "\n",
+    "#  Given values\n",
+    "\n",
+    "#  initial codition\n",
+    "P1 = 140;#  [kN/m^2]\n",
+    "V1 = .1;#   [m^3]\n",
+    "T1 = 273+25;# [K]\n",
+    "\n",
+    "#  final condition\n",
+    "P2 = 700.;#  [kN/m^2]\n",
+    "T2 = 273.+60;#  [K]\n",
+    "\n",
+    "#  by charasteristic equation, P1*V1/T1=P2*V2/T2\n",
+    "\n",
+    "V2=P1*V1*T2/(T1*P2);#  final volume, [m^3]\n",
+    "\n",
+    "#results\n",
+    "print 'The new volume of the gas (m^3) = ',round(V2,4)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6: pg 106"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.6\n",
+      " The mass of the gas present (kg) =  0.118\n",
+      " The new temperature of the gas (C) =  651\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 106\n",
+    "#calculate the new temperature and mass of gas\n",
+    "print('Example 5.6');\n",
+    "\n",
+    "#  aim : To determine\n",
+    "#  the mas of the gas and new temperature\n",
+    "\n",
+    "#  Given values\n",
+    "P1 = 350;#  [kN/m^2]\n",
+    "V1 = .03;#  [m^3]\n",
+    "T1 = 273+35;#  [K]\n",
+    "R = .29;#  Gas constant,[kJ/kg K]\n",
+    "\n",
+    "#  solution\n",
+    "#  using charasteristic equation, P*V=m*R*T\n",
+    "m = P1*V1/(R*T1);#  [Kg]\n",
+    "\n",
+    "#  Now the gas is compressed\n",
+    "P2 = 1050;#  [kN/m^2]\n",
+    "V2 = V1;\n",
+    "# since mass of the gas is constant so using, P*V/T=constant\n",
+    "#  hence\n",
+    "T2 = T1*P2/P1#  [K]\n",
+    "t2 = T2-273;#  [C]\n",
+    "\n",
+    "#results\n",
+    "print ' The mass of the gas present (kg) = ',round(m,3)\n",
+    "print ' The new temperature of the gas (C) = ',t2\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7: pg 111"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.7\n",
+      " The heat transferred to the gas (kJ) =  172.8\n",
+      " The final pressure of the gas (kN/m^2) =  338.06\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 111\n",
+    "#calculate the final pressure and heat transferred\n",
+    "print('Example 5.7');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  the heat transferred to the gas and its final pressure\n",
+    "\n",
+    "#  Given values\n",
+    "m = 2;#  masss of the gas, [kg]\n",
+    "V1 = .7;#  volume,[m^3]\n",
+    "T1 = 273+15;#  original temperature,[K]\n",
+    "T2 = 273+135;#  final temperature,[K]\n",
+    "cv = .72;#  specific heat capacity at constant volume,[kJ/kg K]\n",
+    "R = .29;#  gas law constant,[kJ/kg K]\n",
+    "\n",
+    "#  solution\n",
+    "Q = m*cv*(T2-T1);#  Heat transferred at constant volume,[kJ]\n",
+    "\n",
+    "#  Now,using P1*V1=m*R*T1\n",
+    "P1 = m*R*T1/V1;#  [kN/m^2]\n",
+    "\n",
+    "#  since volume of the system is constant, so P1/T1=P2/T2\n",
+    "#  hence\n",
+    "P2 = P1*T2/T1;#  final pressure,[kN/m^2]\n",
+    "#results\n",
+    "print ' The heat transferred to the gas (kJ) = ',Q\n",
+    "print ' The final pressure of the gas (kN/m^2) = ',round(P2,2)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8: pg 114"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.8\n",
+      " The heat transferred to the gas (kJ) =  -31.84\n",
+      " Work done on the gas during the process (kJ) =  -9.19\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 114\n",
+    "#calculate the heat transferred and work done\n",
+    "print('Example 5.8');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  the heat transferred from the gas and the work done on the gas\n",
+    "\n",
+    "#  Given values\n",
+    "P1 = 275;#  pressure, [kN/m^2]\n",
+    "V1 = .09;#  volume,[m^3]\n",
+    "T1 = 273+185;#  initial temperature,[K]\n",
+    "T2 = 273+15;#  final temperature,[K]\n",
+    "cp = 1.005;#  specific heat capacity at constant pressure,[kJ/kg K]\n",
+    "R = .29;#  gas law constant,[kJ/kg K]\n",
+    "\n",
+    "#  solution\n",
+    "#  using P1*V1=m*R*T1\n",
+    "m = P1*V1/(R*T1);#  mass of the gas\n",
+    "\n",
+    "#  calculation of heat transfer\n",
+    "Q = m*cp*(T2-T1);#  Heat transferred at constant pressure,[kJ]\n",
+    "\n",
+    "#  calculation of work done\n",
+    "#  Now,since pressure is constant so, V/T=constant\n",
+    "# hence\n",
+    "V2 = V1*T2/T1;#  [m^3]\n",
+    "\n",
+    "W = P1*(V2-V1);#  formula for work done at constant pressure,[kJ]\n",
+    "#results\n",
+    "print ' The heat transferred to the gas (kJ) = ',round(Q,2)\n",
+    "print ' Work done on the gas during the process (kJ) = ',round(W,2)\n",
+    "\n",
+    "#  End\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9: pg 117"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.9\n",
+      " The new pressure of the gas (kN/m^2) =  1299.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 117\n",
+    "#calculate the new pressure\n",
+    "print('Example 5.9');\n",
+    "\n",
+    "#  aim : To determine\n",
+    "#  the new pressure of the gas\n",
+    "\n",
+    "#  Given values\n",
+    "P1 = 300.;#  original pressure,[kN/m**2]\n",
+    "T1 = 273.+25;#  original temperature,[K]\n",
+    "T2 = 273.+180;#  final temperature,[K]\n",
+    "\n",
+    "#  solution\n",
+    "#  since gas compressing according to the law,P*V**1.4=constant\n",
+    "#  so,for polytropic process,T1/T2=(P1/P2)**((n-1)/n),here n=1.4\n",
+    "\n",
+    "#  hence\n",
+    "P2 = P1*(T2/T1)**((1.4)/(1.4-1));#  [kN/m**2]\n",
+    "\n",
+    "#results\n",
+    "print ' The new pressure of the gas (kN/m^2) = ',round(P2)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 10: pg 118"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.10\n",
+      " The new temperature of the gas (C) =  25.0\n",
+      " there is minor error in book answer due to rounding off error\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 118\n",
+    "#calculate the new temperature\n",
+    "print('Example 5.10');\n",
+    "\n",
+    "#  aim : To determine\n",
+    "#  the new temperature of the gas\n",
+    "\n",
+    "#  Given values\n",
+    "V1 = .015;#  original volume,[m**3]\n",
+    "T1 = 273.+285;#  original temperature,[K]\n",
+    "V2 = .09;#  final volume,[m**3]\n",
+    "\n",
+    "#  solution \n",
+    "#  Given gas is following the law,P*V**1.35=constant\n",
+    "#  so process is polytropic with\n",
+    "n = 1.35; # polytropic index\n",
+    "\n",
+    "# hence\n",
+    "T2 = T1*(V1/V2)**(n-1);#  final temperature, [K]\n",
+    "\n",
+    "t2 = T2-273;#  [C]\n",
+    "\n",
+    "#results\n",
+    "print ' The new temperature of the gas (C) = ',round(t2,1)\n",
+    "\n",
+    "print ' there is minor error in book answer due to rounding off error'\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11: pg 119"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.11\n",
+      " (a) The original volume of the gas (m^3) =  0.0765\n",
+      "      and The final volume of the gas (m^3) =  0.306\n",
+      " (b) The final pressure of the gas (kN/m^2) =  231.0\n",
+      " (c) The final temperature of the gas (C) =  92.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 119\n",
+    "#calculate the final pressure, temperature and volume of gas\n",
+    "print('Example 5.11');\n",
+    "\n",
+    "#  aim : To determine the\n",
+    "#  (a) original and final volume of the gas\n",
+    "#  (b) final pressure of the gas\n",
+    "#  (c) final temperature of the gas\n",
+    "\n",
+    "#  Given values\n",
+    "m = .675;#  mass of the gas,[kg]\n",
+    "P1 = 1.4;#  original pressure,[MN/m**2]\n",
+    "T1 = 273+280;#  original temperature,[K]\n",
+    "R = .287;#gas constant,[kJ/kg K]\n",
+    "\n",
+    "#  solution and results\n",
+    "\n",
+    "#  (a)\n",
+    "#  using characteristic equation, P1*V1=m*R*T1\n",
+    "V1 = m*R*T1*10**-3/P1;#  [m**3]\n",
+    "#  also Given \n",
+    "V2 =  4*V1;# [m**3]\n",
+    "print ' (a) The original volume of the gas (m^3) = ',round(V1,4)\n",
+    "print '      and The final volume of the gas (m^3) = ',round(V2,3)\n",
+    "\n",
+    "#  (b)\n",
+    "#  Given that gas is following the law P*V**1.3=constant\n",
+    "#  hence process is polytropic with \n",
+    "n = 1.3; #  polytropic index\n",
+    "P2 = P1*(V1/V2)**n;#  formula for polytropic process,[MN/m**2]\n",
+    "print ' (b) The final pressure of the gas (kN/m^2) = ',round(P2*10**3)\n",
+    "\n",
+    "#  (c)\n",
+    "#  since mass is constant so,using P*V/T=constant\n",
+    "#  hence\n",
+    "T2 = P2*V2*T1/(P1*V1);#  [K]\n",
+    "t2 = T2-273;#  [C]\n",
+    "print ' (c) The final temperature of the gas (C) = ',round(t2)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 12: pg 120"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.12\n",
+      " (a) The change in internal energy of the air is del_U (kJ) =  30.73\n",
+      "since del_U>0, so it is gain of internal energy to the air\n",
+      " (b) The work done is (kJ) =  -49.1\n",
+      "since W<0, so the work is done on the air\n",
+      " (c) The heat transfer is Q (kJ) =  -18.4\n",
+      "since Q<0, so the heat is rejected by the air\n",
+      "The answer is a bit different from textbook due to rounding off error\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 120\n",
+    "#calculate the change in internal energy, work done and heat transfer\n",
+    "print('Example 5.12');\n",
+    "\n",
+    "#  aim : T0 determine \n",
+    "#  (a) change in internal energy of the air\n",
+    "#  (b) work done\n",
+    "#  (c) heat transfer\n",
+    "\n",
+    "#  Given values\n",
+    "m = .25;#  mass, [kg]\n",
+    "P1 = 140;#  initial pressure, [kN/m**2]\n",
+    "V1 = .15;#  initial volume, [m**3]\n",
+    "P2 = 1400;#  final volume, [m**3]\n",
+    "cp = 1.005;#  [kJ/kg K]\n",
+    "cv = .718;#  [kJ/kg K]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  (a)\n",
+    "#  assuming ideal gas\n",
+    "R = cp-cv;#  [kJ/kg K]\n",
+    "#  also, P1*V1=m*R*T1,hence\n",
+    "T1 = P1*V1/(m*R);#  [K]\n",
+    "\n",
+    "#  given that process is polytropic with \n",
+    "n = 1.25; #  polytropic index\n",
+    "T2 = T1*(P2/P1)**((n-1)/n);#  [K]\n",
+    "\n",
+    "#  Hence, change in internal energy is,\n",
+    "del_U = m*cv*(T2-T1);#  [kJ]\n",
+    "print ' (a) The change in internal energy of the air is del_U (kJ) = ',round(del_U,2)\n",
+    "if(del_U>0):\n",
+    "    print('since del_U>0, so it is gain of internal energy to the air')\n",
+    "else:\n",
+    "    print('since del_U<0, so it is gain of internal energy to the surrounding')\n",
+    "#  (b)\n",
+    "W = m*R*(T1-T2)/(n-1);#  formula of work done for polytropic process,[kJ]\n",
+    "print ' (b) The work done is (kJ) = ',round(W,1)\n",
+    "if(W>0):\n",
+    "    print('since W>0, so the work is done by  the air')\n",
+    "else:\n",
+    "    print('since W<0, so the work is done on the air')\n",
+    "\n",
+    "\n",
+    "#  (c)\n",
+    "Q = del_U+W;#  using 1st law of thermodynamics,[kJ]\n",
+    "print ' (c) The heat transfer is Q (kJ) = ',round(Q,2)\n",
+    "if(Q>0):\n",
+    "    print('since Q>0, so the heat is received by  the air')\n",
+    "else:\n",
+    "    print('since Q<0, so the heat is rejected by the air')\n",
+    "\n",
+    "print 'The answer is a bit different from textbook due to rounding off error'\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 13: pg 123"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.13\n",
+      "\n",
+      " The final volume of the gas is V2 (m^3) =  0.048\n",
+      "\n",
+      " The work done by the gas is (kJ) =  9.77\n",
+      "\n",
+      " The change of internal energy is (kJ) =  -9.77\n",
+      "since del_U<0, so this is a loss of internal energy from the gas\n",
+      "The answer is a bit different from textbook due to rounding off error\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 123\n",
+    "#calculate the final volume, work done and the change in internal energy\n",
+    "print('Example 5.13');\n",
+    "\n",
+    "#  aim : To determine the\n",
+    "#  final volume, work done and the change in internal energy\n",
+    "\n",
+    "#  Given values\n",
+    "P1 = 700.;#  initial pressure,[kN/m^2]\n",
+    "V1 = .015;# initial volume, [m^3]\n",
+    "P2 = 140.;#  final pressure, [kN/m^2]\n",
+    "cp = 1.046;#  [kJ/kg K]\n",
+    "cv = .752; #  [kJ/kg K]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "Gamma = cp/cv;\n",
+    "#  for adiabatic expansion, P*V^gamma=constant, so\n",
+    "V2 = V1*(P1/P2)**(1/Gamma);#  final volume, [m^3]\n",
+    "print '\\n The final volume of the gas is V2 (m^3) = ',round(V2,3)\n",
+    "\n",
+    "#  work done\n",
+    "W = (P1*V1-P2*V2)/(Gamma-1);#  [kJ]\n",
+    "print '\\n The work done by the gas is (kJ) = ',round(W,2)\n",
+    "\n",
+    "#  for adiabatic process\n",
+    "del_U = -W;#  [kJ]\n",
+    "print '\\n The change of internal energy is (kJ) = ',round(del_U,2)\n",
+    "if(del_U>0):\n",
+    "    print 'since del_U>0, so the the gain in internal energy of the gas '\n",
+    "else:\n",
+    "    print 'since del_U<0, so this is a loss of internal energy from the gas'\n",
+    "\n",
+    "print 'The answer is a bit different from textbook due to rounding off error'\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 14: pg 125"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.14\n",
+      "\n",
+      " (a) The heat transferred during compression is Q (kJ) =  -60.0\n",
+      "\n",
+      " (b) The change of the internal energy during the expansion is,del_U (kJ) =  -45.2\n",
+      "\n",
+      " (c) The mass of the gas is,m (kg) =  0.478\n",
+      " There is calculation mistake in the book\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 125\n",
+    "#calculate the heat transfer, change of internal energy and mass of gas\n",
+    "print('Example 5.14');\n",
+    "import math\n",
+    "#  aim : To determine the\n",
+    "#  (a)heat transfer\n",
+    "#  (b)change of internal energy\n",
+    "#  (c)mass of gas\n",
+    "\n",
+    "#   Given values\n",
+    "V1 = .4;#  initial volume, [m^3]\n",
+    "P1 = 100.;#  initial pressure, [kN/m^2]\n",
+    "T1 = 273.+20;#  temperature, [K]\n",
+    "P2 = 450.;#  final pressure,[kN/m^2]\n",
+    "cp = 1.0;#  [kJ/kg K]\n",
+    "Gamma = 1.4; #  heat capacity ratio\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  (a)\n",
+    "#  for the isothermal compression,P*V=constant,so\n",
+    "V2 = V1*P1/P2;#  [m^3]\n",
+    "W = P1*V1*math.log(P1/P2);#  formula of workdone for isothermal process,[kJ]\n",
+    "\n",
+    "#  for isothermal process, del_U=0;so\n",
+    "Q = W;\n",
+    "print '\\n (a) The heat transferred during compression is Q (kJ) = ',round(Q)\n",
+    "\n",
+    "\n",
+    "#  (b)\n",
+    "V3 = V1;\n",
+    "#  for adiabatic expansion\n",
+    "#  also\n",
+    "\n",
+    "P3 = P2*(V2/V3)**Gamma;#  [kN/m^2]\n",
+    "W = -(P3*V3-P2*V2)/(Gamma-1);#  work done formula for adiabatic process,[kJ]\n",
+    "#  also, Q=0,so using Q=del_U+W\n",
+    "del_U = -W;#  [kJ]\n",
+    "print '\\n (b) The change of the internal energy during the expansion is,del_U (kJ) = ',round(del_U,1)\n",
+    "\n",
+    "#  (c)\n",
+    "#  for ideal gas\n",
+    "#  cp-cv=R, and cp/cv=gamma, hence\n",
+    "R = cp*(1-1/Gamma);#  [kj/kg K]\n",
+    "\n",
+    "#  now using ideal gas equation\n",
+    "m = P1*V1/(R*T1);#  mass of the gas,[kg]\n",
+    "print '\\n (c) The mass of the gas is,m (kg) = ',round(m,3)\n",
+    "\n",
+    "print' There is calculation mistake in the book'\n",
+    "\n",
+    "\n",
+    "#  End\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 15: pg 128"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.15\n",
+      " The heat received or rejected by the gas during this process is Q (kJ) =  1.03\n",
+      "since Q>0, so heat is received by the gas\n",
+      "\n",
+      " The polytropic specific heat capacity is cn (kJ/kg K)  =  0.239\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 128\n",
+    "#calculate the the heat transferred and polytropic specific heat capacity\n",
+    "print('Example 5.15');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  the heat transferred and polytropic specific heat capacity\n",
+    "\n",
+    "#  Given values\n",
+    "P1 = 1;#  initial pressure, [MN/m^2]\n",
+    "V1 = .003;#  initial volume, [m^3]\n",
+    "P2 = .1;#  final pressure,[MN/m^2]\n",
+    "cv = .718;#  [kJ/kg*K]\n",
+    "Gamma=1.4;#  heat capacity ratio\n",
+    "\n",
+    "#  solution\n",
+    "#  Given process is polytropic with\n",
+    "n = 1.3;#  polytropic index\n",
+    "#  hence\n",
+    "V2 = V1*(P1/P2)**(1/n);#  final volume,[m^3]\n",
+    "W = (P1*V1-P2*V2)*10**3/(n-1);#  work done,[kJ]\n",
+    "#  so\n",
+    "Q = (Gamma-n)*W/(Gamma-1);#  heat transferred,[kJ]\n",
+    "\n",
+    "print ' The heat received or rejected by the gas during this process is Q (kJ) = ',round(Q,2)\n",
+    "if(Q>0):\n",
+    "    print 'since Q>0, so heat is received by the gas'\n",
+    "else:\n",
+    "    print 'since Q<0, so heat is rejected by the gas'\n",
+    "\n",
+    "#  now\n",
+    "cn = cv*(Gamma-n)/(n-1);#  polytropic specific heat capacity,[kJ/kg K]\n",
+    "print '\\n The polytropic specific heat capacity is cn (kJ/kg K)  = ',round(cn,3)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 16: pg 129"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.16\n",
+      "\n",
+      " (a) The initial partial pressure of the steam is (kN/m^2) =  7\n",
+      "\n",
+      "      The initial partial pressure of the air is (kN/m^2) =  93.0\n",
+      " \n",
+      "(b) The final partial pressure of the steam is (kN/m^2) =  200.0\n",
+      "\n",
+      "     The final partial pressure of the air is (kN/m^2) =  117.2\n",
+      "\n",
+      " (c) The total pressure after heating is (kN/m^2) =  317.2\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 129\n",
+    "\n",
+    "print('Example 5.16');\n",
+    "\n",
+    "# To determine the \n",
+    "#  (a) initial partial pressure of the steam and air\n",
+    "#  (b) final partial pressure of the steam and air\n",
+    "#  (c) total pressure in the container after heating\n",
+    "\n",
+    "#  Given values\n",
+    "T1 = 273.+39;#  initial temperature,[K]\n",
+    "P1 = 100.;#  pressure, [MN/m^2]\n",
+    "T2 = 273.+120.2;#  final temperature,[K]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  (a)\n",
+    "#  from the steam tables, the pressure of wet steam at 39 C is\n",
+    "Pw1 = 7;#  partial pressure of wet steam,[kN/m^2]\n",
+    "#  and by Dalton's law\n",
+    "Pa1 = P1-Pw1;#  initial pressure of air, [kN/m^2]\n",
+    "\n",
+    "print '\\n (a) The initial partial pressure of the steam is (kN/m^2) = ',Pw1\n",
+    "print '\\n      The initial partial pressure of the air is (kN/m^2) = ',Pa1\n",
+    "\n",
+    "#  (b)\n",
+    "#  again from steam table, at 120.2 C the pressure of wet steam is\n",
+    "Pw2 = 200.;#  [kN/m^2]\n",
+    "\n",
+    "#  now since volume is constant so assuming air to be ideal gas so for air  P/T=contant, hence\n",
+    "Pa2 = Pa1*T2/T1 ;#  [kN/m^2]\n",
+    "\n",
+    "print ' \\n(b) The final partial pressure of the steam is (kN/m^2) = ',Pw2\n",
+    "print '\\n     The final partial pressure of the air is (kN/m^2) = ',round(Pa2,2)\n",
+    "\n",
+    "#  (c)\n",
+    "Pt = Pa2+Pw2;#  using dalton's law, total pressure,[kN/m^2]\n",
+    "print '\\n (c) The total pressure after heating is (kN/m^2) = ',round(Pt,2)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 17: pg 130"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.17\n",
+      "\n",
+      " The partial pressure of the air in the condenser is (kN/m^2) =  6.0\n",
+      "\n",
+      " The partial pressure of the steam in the condenser is (kN/m^2) =  8\n",
+      "\n",
+      " The mass of air which will associated with this steam is (kg) =  1430.5\n",
+      " There is misprint in book\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 130\n",
+    "print('Example 5.17');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  the partial pressure of the air and steam, and the mass of the air\n",
+    "\n",
+    "#  Given values\n",
+    "P1 = 660.;#  vaccum gauge pressure on condenser [mmHg]\n",
+    "P = 765.;#  atmospheric pressure, [mmHg]\n",
+    "x = .8;#  dryness fraction \n",
+    "T = 273.+41.5;#  temperature,[K]\n",
+    "ms_dot = 1500.;#  condense rate of steam,[kg/h]\n",
+    "R = .29;#  [kJ/kg]\n",
+    "\n",
+    "#  solution\n",
+    "Pa = (P-P1)*.1334;# absolute pressure,[kN/m^2]\n",
+    "# from steam table, at 41.5 C partial pressure of steam is\n",
+    "Ps = 8;#  [kN/m^2]\n",
+    "#  by dalton's law, partial pressure of air is\n",
+    "Pg = Pa-Ps;#  [kN/m^2]\n",
+    "\n",
+    "print '\\n The partial pressure of the air in the condenser is (kN/m^2) = ',round(Pg)\n",
+    "print '\\n The partial pressure of the steam in the condenser is (kN/m^2) = ',Ps\n",
+    "\n",
+    "#  also\n",
+    "vg = 18.1;#  [m^3/kg]\n",
+    "#  so\n",
+    "V = x*vg;#  [m^3/kg]\n",
+    "#  The air associated with 1 kg of the steam will occupiy this same volume\n",
+    "#  for air, Pg*V=m*R*T,so\n",
+    "m = Pg*V/(R*T);#  [kg/kg steam]\n",
+    "#  hence\n",
+    "ma = m*ms_dot;#  [kg/h]\n",
+    "\n",
+    "print '\\n The mass of air which will associated with this steam is (kg) = ',round(ma,1)\n",
+    "\n",
+    "print' There is misprint in book'\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 18: pg 130"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.18\n",
+      " (a) The final pressure in the cylinder is (kN/m^2) =  707.1\n",
+      " (b) The final dryness fraction of the steam is  =  0.83\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 130\n",
+    "print('Example 5.18');\n",
+    "\n",
+    "#  aim : To determine the\n",
+    "#  (a) final pressure\n",
+    "#   (b) final dryness fraction of the steam\n",
+    "\n",
+    "#  Given values\n",
+    "P1 = 130.;#  initial pressure, [kN/m^2]\n",
+    "T1 = 273.+75.9;#  initial temperature, [K]\n",
+    "x1 = .92;#  initial dryness fraction\n",
+    "T2 = 273.+120.2;#  final temperature, [K]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  (a)\n",
+    "#  from steam table, at 75.9 C\n",
+    "Pws = 40.;#  partial pressure of wet steam[kN/m^2]\n",
+    "Pa = P1-Pws;#  partial pressure of air, [kN/m^2]\n",
+    "vg = 3.99#  specific volume of the wet steam, [m^3/kg]\n",
+    "# hence\n",
+    "V1 = x1*vg;#  [m^3/kg]\n",
+    "V2 = V1/5;#  [m^3/kg]\n",
+    "#  for air, mass is constant so, Pa*V1/T1=P2*V2/T2,also given ,V1/V2=5,so\n",
+    "P2 = Pa*V1*T2/(V2*T1);#  final pressure,[kN/m^2]\n",
+    "\n",
+    "#  now for steam at 120.2 C\n",
+    "Ps = 200.;#  final partial pressure of steam,[kN/m^2]\n",
+    "#  so by dalton's law total pressure in cylindert is\n",
+    "Pt = P2+Ps;#  [kN/m^2]\n",
+    "print ' (a) The final pressure in the cylinder is (kN/m^2) = ',round(Pt,1)\n",
+    "\n",
+    "#  (b)\n",
+    "#  from steam table at 200 kN/m^2 \n",
+    "vg = .885;#  [m^3/kg]\n",
+    "#  hence\n",
+    "x2 = V2/vg;#  final dryness fraction of the steam\n",
+    "print ' (b) The final dryness fraction of the steam is  = ',round(x2,2)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 19: pg 131"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.19\n",
+      " (a) The value of adiabatic index Gamma is  =  1.426\n",
+      " (b) The change in internal energy during the adiabatic expansion is U2-U1  (This is loss of internal energy) (kJ) =  -55.97\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 131\n",
+    "print('Example 5.19')\n",
+    "\n",
+    "#  aim : To determine the \n",
+    "#  (a) Gamma,\n",
+    "#  (b) del_U\n",
+    "import math\n",
+    "#  Given Values\n",
+    "P1 = 1400.;#  [kN/m^2]\n",
+    "P2 = 100.;#  [kN/m^2]\n",
+    "P3 = 220.;#  [kN/m^2]\n",
+    "T1 = 273.+360;#  [K]\n",
+    "m = .23;# [kg]\n",
+    "cp = 1.005;#  [kJ/kg*K]\n",
+    "\n",
+    "#  Solution\n",
+    "T3 = T1;#  since process 1-3 is isothermal\n",
+    "\n",
+    "#  (a)\n",
+    "#  for process 1-3, P1*V1=P3*V3,so\n",
+    "V3_by_V1 = P1/P3;\n",
+    "#  also process 1-2 is adiabatic,so P1*V1^(Gamma)=P2*V2^(Gamma),hence\n",
+    "#  and process process 2-3 is iso-choric so,V3=V2 and\n",
+    "V2_by_V1 = V3_by_V1;\n",
+    "#  hence,\n",
+    "Gamma = math.log(P1/P2)/math.log(P1/P3); #  heat capacity ratio\n",
+    "\n",
+    "print ' (a) The value of adiabatic index Gamma is  = ',round(Gamma,3)\n",
+    "\n",
+    "#  (b)\n",
+    "cv = cp/Gamma;#  [kJ/kg K]\n",
+    "#  for process 2-3,P3/T3=P2/T2,so\n",
+    "T2 = P2*T3/P3;#  [K]\n",
+    "\n",
+    "#   now\n",
+    "del_U = m*cv*(T2-T1);#  [kJ]\n",
+    "print ' (b) The change in internal energy during the adiabatic expansion is U2-U1  (This is loss of internal energy) (kJ) = ',round(del_U,2)\n",
+    "#   End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 20: pg 133"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.20\n",
+      " The mass of oxygen used (kg) =  5.5\n",
+      " The amount of heat transferred through the cylinder wall is (kJ) =  13.28\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 133\n",
+    "print('Example 5.20');\n",
+    "\n",
+    "#  aim : To determine \n",
+    "#  the mass of oxygen and heat transferred\n",
+    "\n",
+    "#  Given values\n",
+    "V1 = 300.;#  [L]\n",
+    "P1 = 3.1;#  [MN/m^2]\n",
+    "T1 = 273.+18;#  [K]\n",
+    "P2 = 1.7;#  [MN/m^2]\n",
+    "T2 = 273.+15;#  [K]\n",
+    "Gamma = 1.4; #  heat capacity ratio\n",
+    "#  density condition\n",
+    "P = .101325;# [MN/m^2]\n",
+    "T = 273.;# [K]\n",
+    "V = 1.;#  [m^3]\n",
+    "m = 1.429;# [kg]\n",
+    "\n",
+    "#  hence\n",
+    "R = P*V*10**3/(m*T);#  [kJ/kg*K]\n",
+    "#  since volume is constant\n",
+    "V2 = V1;# [L]\n",
+    "#  for the initial conditions in the cylinder,P1*V1=m1*R*T1\n",
+    "m1 = P1*V1/(R*T1);#  [kg]\n",
+    "\n",
+    "# after some of the gas is used\n",
+    "m2 = P2*V2/(R*T2);#  [kg]\n",
+    "#  The mass of oxygen remaining in cylinder is m2 kg,so\n",
+    "#  Mass of oxygen used is\n",
+    "m_used = m1-m2;#  [kg]\n",
+    "print ' The mass of oxygen used (kg) = ',round(m_used,1)\n",
+    "\n",
+    "#  for non-flow process,Q=del_U+W\n",
+    "#  volume is constant so no external work is done so,Q=del_U\n",
+    "cv = R/(Gamma-1);#  [kJ/kg*K]\n",
+    "\n",
+    "#  heat transfer is\n",
+    "Q = m2*cv*(T1-T2);#  (kJ)\n",
+    "print ' The amount of heat transferred through the cylinder wall is (kJ) = ',round(Q,2)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 21: pg 134"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.21\n",
+      " (a) The work transferred during the compression is (kJ) =  -28.1\n",
+      " (b) The change in internal energy is (kJ) =  14.2\n",
+      " (c) The heat transferred during the compression is (kJ) =  -14.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 134\n",
+    "print('Example 5.21');\n",
+    "\n",
+    "#  aim : To determine the\n",
+    "#  (a) work transferred during the compression\n",
+    "#  (b) change in internal energy\n",
+    "#  (c) heat transferred during the compression\n",
+    "\n",
+    "#  Given values\n",
+    "V1 = .1;#  initial volume, [m^3]\n",
+    "P1 = 120.;#  initial pressure, [kN/m^2]\n",
+    "P2 = 1200.; # final pressure, [kN/m^2]\n",
+    "T1 = 273.+25;#  initial temperature, [K]\n",
+    "cv = .72;#  [kJ/kg*K]\n",
+    "R = .285;#  [kJ/kg*K]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  (a)\n",
+    "#  given process is polytropic with\n",
+    "n = 1.2;  # polytropic index\n",
+    "#  hence\n",
+    "V2 = V1*(P1/P2)**(1/n);#  [m^3]\n",
+    "W = (P1*V1-P2*V2)/(n-1);#  workdone formula, [kJ]\n",
+    "print ' (a) The work transferred during the compression is (kJ) = ',round(W,1)\n",
+    "\n",
+    "#  (b)\n",
+    "#  now mass is constant so,\n",
+    "T2 = P2*V2*T1/(P1*V1);#  [K]\n",
+    "#  using, P*V=m*R*T\n",
+    "m = P1*V1/(R*T1);#  [kg]\n",
+    "\n",
+    "#  change in internal energy is\n",
+    "del_U = m*cv*(T2-T1);#  [kJ]\n",
+    "print ' (b) The change in internal energy is (kJ) = ',round(del_U,1)\n",
+    "\n",
+    "#  (c)\n",
+    "Q = del_U+W;#  [kJ]\n",
+    "print ' (c) The heat transferred during the compression is (kJ) = ',round(Q,0)\n",
+    "\n",
+    "#  End\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 22: pg 135"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.22\n",
+      " (a) The new pressure of the air in the receiver is (kN/m^2) =  442.0\n",
+      " (b) The specific  enthalpy of the air at 15 C is (kJ/kg) =  15.075\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 135\n",
+    "print('Example 5.22');\n",
+    "\n",
+    "#  aim : To determine the\n",
+    "#  (a) new pressure of the air in the receiver\n",
+    "#  (b) specific enthalpy of air at 15 C\n",
+    "\n",
+    "#  Given values\n",
+    "V1 = .85;#  [m^3]\n",
+    "T1 = 15.+273;#  [K]\n",
+    "P1 = 275.;# pressure,[kN/m^2]\n",
+    "m = 1.7;#  [kg]\n",
+    "cp = 1.005;#  [kJ/kg*K]\n",
+    "cv = .715;#  [kJ/kg*K]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  (a)\n",
+    "\n",
+    "R = cp-cv;#  [kJ/kg*K]\n",
+    "#  assuming m1 is original mass of the air, using P*V=m*R*T\n",
+    "m1 = P1*V1/(R*T1);#  [kg]\n",
+    "m2 = m1+m;#  [kg]\n",
+    "#  again using P*V=m*R*T\n",
+    "#  P2/P1=(m2*R*T2/V2)/(m1*R*T1/V1); and T1=T2,V1=V2,so\n",
+    "P2 = P1*m2/m1;#  [kN/m^2]\n",
+    "print ' (a) The new pressure of the air in the receiver is (kN/m^2) = ',round(P2)\n",
+    "\n",
+    "#  (b)\n",
+    "#  for 1 kg of air, h2-h1=cp*(T1-T0)\n",
+    "#  and if 0 is chosen as the zero enthalpy, then\n",
+    "h = cp*(T1-273);#  [kJ/kg]\n",
+    "print ' (b) The specific  enthalpy of the air at 15 C is (kJ/kg) = ',h\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 23: pg 136"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.23\n",
+      " (a) The characteristic gas constant of the gas is R (kJ/kg K) =  0.185\n",
+      " (b) The specific heat capacity of the gas at constant pressure cp (kJ/kg K) =  0.827\n",
+      " (c) The specific heat capacity of the gas at constant volume cv  (kJ/kg K) =  0.642\n",
+      " (d) The change in internal energy is (kJ) =  136.0\n",
+      " (e) The work transfer is W (kJ) =  39.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 136\n",
+    "print('Example 5.23');\n",
+    "\n",
+    "#  aim : T determine the\n",
+    "#  (a) characteristic gas constant of the gas\n",
+    "#  (b) cp,\n",
+    "#  (c) cv,\n",
+    "#  (d) del_u \n",
+    "#  (e) work transfer\n",
+    "\n",
+    "#  Given values\n",
+    "P = 1.;#  [bar] \n",
+    "T1 = 273.+15;#  [K]\n",
+    "m = .9;#  [kg]\n",
+    "T2 = 273.+250;#  [K]\n",
+    "Q = 175.;#  heat transfer,[kJ]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  (a)\n",
+    "#  using, P*V=m*R*T, given,\n",
+    "m_by_V = 1.875;\n",
+    "#  hence\n",
+    "R = P*100/(T1*m_by_V);#  [kJ/kg*K]\n",
+    "print ' (a) The characteristic gas constant of the gas is R (kJ/kg K) = ',round(R,3)\n",
+    "\n",
+    "#  (b)\n",
+    "#  using, Q=m*cp*(T2-T1)\n",
+    "cp = Q/(m*(T2-T1));#  [kJ/kg K]\n",
+    "print ' (b) The specific heat capacity of the gas at constant pressure cp (kJ/kg K) = ',round(cp,3)\n",
+    "\n",
+    "# (c)\n",
+    "#  we have, cp-cv=R,so\n",
+    "cv = cp-R;# [kJ/kg*K]\n",
+    "print ' (c) The specific heat capacity of the gas at constant volume cv  (kJ/kg K) = ',round(cv,3)\n",
+    "\n",
+    "#  (d)\n",
+    "del_U = m*cv*(T2-T1);#  [kJ]\n",
+    "print ' (d) The change in internal energy is (kJ) = ',round(del_U)\n",
+    "\n",
+    "#  (e)\n",
+    "# using, Q=del_U+W\n",
+    "W = Q-del_U;#  [kJ]\n",
+    "print ' (e) The work transfer is W (kJ) = ',round(W)\n",
+    "\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 24: pg 136"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.24\n",
+      " (a) The work transfer is W (kJ) =  198.2\n",
+      " (b) The change of internal energy is del_U (kJ) =  -157.3\n",
+      " (c) The heat transfer Q (kJ) =  40.8\n",
+      "The answer is a bit different due to rounding off error in textbook\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 136\n",
+    "print('Example 5.24');\n",
+    "\n",
+    "#  aim : To determine the\n",
+    "#  (a) work transfer,\n",
+    "#  (b)del_U  and,\n",
+    "#  (c)heat transfer\n",
+    "\n",
+    "#  Given values\n",
+    "V1 = .15;#  [m^3]\n",
+    "P1 = 1200.;#  [kN/m^2]\n",
+    "T1 = 273.+120;# [K]\n",
+    "P2 = 200.;#  [kN/m^2]\n",
+    "cp = 1.006;#[kJ/kg K]\n",
+    "cv = .717;#  [kJ/kg K]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  (a)\n",
+    "#  Given, PV^1.32=constant, so it is polytropic process with\n",
+    "n = 1.32;#  polytropic index\n",
+    "# hence\n",
+    "V2 = V1*(P1/P2)**(1./n);#  [m^3]\n",
+    "#  now, W\n",
+    "W = (P1*V1-P2*V2)/(n-1);# [kJ]\n",
+    "print ' (a) The work transfer is W (kJ) = ',round(W,1)\n",
+    "\n",
+    "#  (b)\n",
+    "R = cp-cv;#  [kJ/kg K]\n",
+    "m = P1*V1/(R*T1);#  gas law,[kg]\n",
+    "#  also for polytropic process\n",
+    "T2 = T1*(P2/P1)**((n-1)/n);#  [K]\n",
+    "#  now for gas,\n",
+    "del_U = m*cv*(T2-T1);#  [kJ]\n",
+    "print ' (b) The change of internal energy is del_U (kJ) = ',round(del_U,1)\n",
+    "\n",
+    "#  (c)\n",
+    "Q = del_U+W;#  first law of thermodynamics,[kJ]\n",
+    "print ' (c) The heat transfer Q (kJ) = ',round(Q,1)\n",
+    "\n",
+    "print 'The answer is a bit different due to rounding off error in textbook'\n",
+    "#  End\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 26: pg 141"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 25,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Example 5.26\n",
+      "heat transfer from the gas (kJ) =  -248.2\n",
+      " The volume of gas before transfer is (m^3) =  0.145\n",
+      " The volume of pressure vessel is (m^3) =  0.27\n"
+     ]
+    }
+   ],
+   "source": [
+    "#pg 141\n",
+    "print('Example 5.26');\n",
+    "\n",
+    "#  aim : To determine\n",
+    "#  the volume of the pressure vessel and the volume of the gas before transfer\n",
+    "\n",
+    "#  Given values\n",
+    "\n",
+    "P1 = 1400.;#  initial pressure,[kN/m^2]\n",
+    "T1 = 273.+85;#  initial temperature,[K]\n",
+    "\n",
+    "P2 = 700.;#  final pressure,[kN/m^2]\n",
+    "T2 = 273.+60;#  final temperature,[K]\n",
+    "\n",
+    "m = 2.7;# mass of the gas passes,[kg]\n",
+    "cp = .88;#  [kJ/kg]\n",
+    "cv = .67;#  [kJ/kg]\n",
+    "\n",
+    "#  solution\n",
+    "\n",
+    "#  steady flow equation is, u1+P1*V1+C1^2/2+Q=u2+P2*V2+C2^2/2+W  [1], \n",
+    "#  given, there is no kinetic energy change and neglecting potential energy term\n",
+    "W = 0;# no external work done\n",
+    "#  so final equation is,u1+P1*v1+Q=u2   [2]\n",
+    "# also u2-u1=cv*(T2-T1)\n",
+    "# hence Q=cv*(T2-T1)-P1*v1    [3]\n",
+    "#  and for unit mass P1*v1=R*T1=(cp-cv)*T1  [4]\n",
+    "#  so finally\n",
+    "Q = cv*(T2-T1)-(cp-cv)*T1;#  [kJ/kg]\n",
+    "#  so total heat transferred is\n",
+    "Q2 = m*Q;#  [kJ] \n",
+    "\n",
+    "#  using eqn [4]\n",
+    "v1 = (cp-cv)*T1/P1;#  [m^3/kg]\n",
+    "#  Total volume is\n",
+    "V1 = m*v1;#  [m^3]\n",
+    "\n",
+    "#  using ideal gas equation P1*V1/T1=P2*V2/T2\n",
+    "V2 = P1*T2*V1/(P2*T1);#  final volume,[m^3]\n",
+    "\n",
+    "print 'heat transfer from the gas (kJ) = ',round(Q2,1)\n",
+    "print ' The volume of gas before transfer is (m^3) = ',round(V1,3)\n",
+    "print ' The volume of pressure vessel is (m^3) = ',round(V2,2)\n",
+    " \n",
+    "#  End\n"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.11"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}