Added(A)/Deleted(D) following books

 A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter1.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter10.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter11.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter2.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter3.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter4.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter5.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter6.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter7.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter8.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/Chapter9.ipynb
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/screenshots/Chapter1.png
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/screenshots/Chapter3.png
A  Energy_Management_by_W._R._Murphy_and_G._A._Mckay/screenshots/Chapter9.png
A  Introduction_to_flight_by_J_D_Anderson/11._Hypersonic_vehicles.ipynb
A  Introduction_to_flight_by_J_D_Anderson/2._Fundamental_Thoughts.ipynb
A  Introduction_to_flight_by_J_D_Anderson/3._The_Standard_Atmosphere.ipynb
A  Introduction_to_flight_by_J_D_Anderson/4._Aerodynamics.ipynb
A  Introduction_to_flight_by_J_D_Anderson/5._Airfoils,_Wings_and_Other_Aerodynamic_shapes.ipynb
A  Introduction_to_flight_by_J_D_Anderson/6._Elements_of_Airplane_Performance.ipynb
A  Introduction_to_flight_by_J_D_Anderson/7._Principles_of_Stability_and_Control.ipynb
A  Introduction_to_flight_by_J_D_Anderson/8._Space_Flight_(Astronautics).ipynb
A  Introduction_to_flight_by_J_D_Anderson/9._Propulsion.ipynb
A  Introduction_to_flight_by_J_D_Anderson/Appendix_A.ipynb
A  Introduction_to_flight_by_J_D_Anderson/Appendix_B.ipynb
A  Introduction_to_flight_by_J_D_Anderson/Appendix_C.ipynb
A  Introduction_to_flight_by_J_D_Anderson/Appendix_D.ipynb
A  Introduction_to_flight_by_J_D_Anderson/Appendix_E.ipynb
A  Introduction_to_flight_by_J_D_Anderson/README.txt
A  Introduction_to_flight_by_J_D_Anderson/screenshots/1.png
A  Introduction_to_flight_by_J_D_Anderson/screenshots/2.png
A  Introduction_to_flight_by_J_D_Anderson/screenshots/3.png
A  Modern_Compressible_Flow_with_historical_perspective_by_John_D_Anderson/1._Compressible_Flow-Some_History_and_Introductory_Thoughts.ipynb
A  Modern_Compressible_Flow_with_historical_perspective_by_John_D_Anderson/3._One_Dimentional_Flow.ipynb
A  Modern_Compressible_Flow_with_historical_perspective_by_John_D_Anderson/4._Oblique_Shock_and_Expansion_Waves.ipynb
A  Modern_Compressible_Flow_with_historical_perspective_by_John_D_Anderson/5._Quasi-One-Dimensional_Flow.ipynb
A  Modern_Compressible_Flow_with_historical_perspective_by_John_D_Anderson/README.txt
A  Modern_Compressible_Flow_with_historical_perspective_by_John_D_Anderson/screenshots/1.png
A  Modern_Compressible_Flow_with_historical_perspective_by_John_D_Anderson/screenshots/2.png
A  Modern_Compressible_Flow_with_historical_perspective_by_John_D_Anderson/screenshots/3.png

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diff --git a/Introduction_to_flight_by_J_D_Anderson/11._Hypersonic_vehicles.ipynb b/Introduction_to_flight_by_J_D_Anderson/11._Hypersonic_vehicles.ipynb
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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 11: Hypersonic vehicles"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Percentage efficiency:  30.3 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 11.1\n",
+    "'''Determine the efficiency of a Rankine cycle using steam as the working fluid in which the\n",
+    "condenser pressure is 10 kPa. The boiler pressure is 2 MPa. The steam leaves the boiler\n",
+    "as saturated vapor.\n",
+    "In solving Rankine-cycle problems, we let w p denote the work into the pump per\n",
+    "kilogram of fluid flowing and q L denote the heat rejected from the working fluid per\n",
+    "kilogram of fluid flowing.\n",
+    "To solve this problem we consider, in succession, a control surface around the pump,\n",
+    "the boiler, the turbine, and the condenser. For each, the thermodynamic model is the steam\n",
+    "tables, and the process is steady state with negligible changes in kinetic and potential\n",
+    "energies. First, consider the pump:\n",
+    "Control volume:\n",
+    "Inlet state:\n",
+    "Exit state:\n",
+    "Pump.\n",
+    "P 1 known, saturated liquid; state fixed.\n",
+    "P 2 known.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P2 = 2000\t\t\t#Exit pressure in kPa\n",
+    "P1 = 10\t\t\t\t#Inlet pressure in kPa\n",
+    "v = 0.00101\t\t\t#specific weight of water in m**3/kg\n",
+    "h1 = 191.8\t\t\t#Enthalpy in kJ/kg from table\n",
+    "h3 = 2799.5\t\t\t#Enthalpy in kJ/kg\n",
+    "s4 = 6.3409\t\t\t#kJ/kg\n",
+    "sf = 0.6493\t\t\t#Entropy at liquid state in kJ/kg\n",
+    "sfg = 7.5009\t\t#Entropy difference for vapor and liquid state in kJ/kg\n",
+    "hfg = 2392.8\t\t#Enthalpy difference in kJ/kg for turbine\n",
+    "\n",
+    "#Calculations:\n",
+    "wp = v*(P2-P1)\t\t#work done in pipe in kJ/kg\n",
+    "h2 = h1+wp\t\t\t#enthalpy in kJ/kg\n",
+    "x4 = (s4-sf)/sfg\t#x-factor\n",
+    "h4 = h1+x4*hfg\t\t#Enthalpy in kJ/kg\n",
+    "nth = ((h3-h2)-(h4-h1))/(h3-h2)\n",
+    "\n",
+    "#Results:\n",
+    "print 'Percentage efficiency: ',round(nth*100,1),\"%\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Percentage efficiency:  35.3 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 11.2\n",
+    "''' In a Rankine cycle, steam leaves the boiler and enters the turbine at 4 MPa and 400\n",
+    "◦\n",
+    "C.\n",
+    "The condenser pressure is 10 kPa. Determine the cycle efficiency.\n",
+    "To determine the cycle efficiency, we must calculate the turbine work, the pump\n",
+    "work, and the heat transfer to the steam in the boiler. We do this by considering a control\n",
+    "surface around each of these components in turn. In each case the thermodynamic model\n",
+    "is the steam tables, and the process is steady state with negligible changes in kinetic and\n",
+    "potential energies.\n",
+    "Control volume:\n",
+    "Inlet state:\n",
+    "Exit state:\n",
+    "Pump.\n",
+    "P 1 known, saturated liquid; state fixed.\n",
+    "P 2 known.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P2 = 4000\t\t\t#Exit pressure in kPa\n",
+    "P1 = 10\t\t\t\t#Inlet pressure in kPa\n",
+    "v = 0.00101\t\t\t#specific weight of water in m**3/kg\n",
+    "h1 = 191.8\t\t\t#Enthalpy in kJ/kg from table\n",
+    "h3 = 3213.6\t\t\t#Enthalpy in kJ/kg from table\n",
+    "s4 = 6.7690\t\t\t#Entropy in kJ/kg from table\n",
+    "sf = 0.6493\t\t\t#Entropy at liquid state in kJ/kg from table\n",
+    "sfg = 7.5009\t\t#Entropy difference for vapor and liquid state in kJ/kg from table\n",
+    "hfg = 2392.8\t\t#Enthalpy difference in kJ/kg for turbine\n",
+    "\n",
+    "#Calculations:\n",
+    "wp = v*(P2-P1)\t\t#work done in pipe in kJ/kg\n",
+    "h2 = h1+wp\t\t\t#Enthalpy in kJ/kg\n",
+    "x4 = (s4-sf)/sfg\t#x-factor\n",
+    "h4 = h1+x4*hfg\t\t#Enthalpy in kJ/kg\n",
+    "nth = ((h3-h2)-(h4-h1))/(h3-h2)\n",
+    "\n",
+    "#Results:\n",
+    "print 'Percentage efficiency: ',round(nth*100,1),'%'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Percentage efficiency:  35.9 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 11.3\n",
+    "''' Consider a reheat cycle utilizing steam. Steam leaves the boiler and enters the turbine at\n",
+    "4 MPa, 400\n",
+    "◦\n",
+    "C. After expansion in the turbine to 400 kPa, the steam is reheated to 400\n",
+    "◦\n",
+    "C\n",
+    "and then expanded in the low-pressure turbine to 10 kPa. Determine the cycle efficiency.\n",
+    "For each control volume analyzed, the thermodynamic model is the steam tables,\n",
+    "the process is steady state, and changes in kinetic and potential energies are negligible.\n",
+    "For the high-pressure turbine,\n",
+    "Control volume:\n",
+    "Inlet state:\n",
+    "Exit state:\n",
+    "High-pressure turbine.\n",
+    "P 3 , T 3 known; state fixed.\n",
+    "P 4 known.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P2 = 4000\t\t\t#Exit pressure in kPa\n",
+    "P1 = 10\t\t\t\t#Inlet pressure in kPa\n",
+    "v = 0.00101\t\t\t#specific weight of water in m**3/kg\n",
+    "h1 = 191.8\t\t\t#Enthalpy in kJ/kg from table\n",
+    "h3 = 3213.6\t\t\t#Enthalpy in kJ/kg from table\n",
+    "s4 = 6.7690\t\t\t#Entropy in kJ/kg from table\n",
+    "sf1 = 1.7766\t\t#Entropy at liquid state in kJ/kg from table\n",
+    "sfg1 = 5.1193\t\t#Entropy difference for vapor and liquid state in kJ/kg from table\n",
+    "hf = 604.7\t\t\t#Enthalpy of liquid state in kJ/kg\n",
+    "hfg1 = 2133.8\t\t#Enthalpy difference in kJ/kg for turbine\n",
+    "sf = 0.6493\t\t\t#Entropy in liquid state in kJ/kg for turbine\n",
+    "h5 = 3273.4\t\t\t#Enthalpy in kJ/kg \n",
+    "s5 = 7.8985\t\t\t#Entropy in kJ/kg\n",
+    "sfg = 7.5009\t\t#Entropy diff in kJ/kg \n",
+    "hfg = 2392.8\t\t#Enthalpy difference for low pressure turbine in kj/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "wp = round(v*(P2-P1))#Work done in pipe in kJ/kg\n",
+    "h2 = h1+wp\t\t\t#Enthalpy in kJ/kg\n",
+    "x4 = (s4-sf1)/sfg1\t#x-factor\n",
+    "h4 = hf+x4*hfg1\t\t#Enthalpy in kJ/kg\n",
+    "x6 = (s5-sf)/sfg\t#x-factor\n",
+    "h6 = h1+x6*hfg\t\t#Entropy in kg/kg\n",
+    "wt = (h3-h4)+(h5-h6)#Work output in kJ/kg\n",
+    "qh = (h3-h2)+(h5-h4)\n",
+    "nth = (wt-wp)/qh\n",
+    "\n",
+    "#Results:\n",
+    "print 'Percentage efficiency: ',round(nth*100,1),'%'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Refrigerator Efficiency:  37.5 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 11.4\n",
+    "''' Consider a regenerative cycle using steam as the working fluid. Steam leaves the boiler\n",
+    "and enters the turbine at 4 MPa, 400\n",
+    "◦\n",
+    "C. After expansion to 400 kPa, some of the steam is\n",
+    "extracted from the turbine to heat the feedwater in an open feedwater heater. The pressure\n",
+    "in the feedwater heater is 400 kPa, and the water leaving it is saturated liquid at 400 kPa.\n",
+    "The steam not extracted expands to 10 kPa. Determine the cycle efficiency.\n",
+    "The line diagram and T–s diagram for this cycle are shown in Fig. 11.12.\n",
+    "As in previous examples, the model for each control volume is the steam tables, the\n",
+    "process is steady state, and kinetic and potential energy changes are negligible.\n",
+    "From Examples 11.2 and 11.3 we have the following properties:\n",
+    "h 5 = 3213.6 h 6 = 2685.6\n",
+    "h 7 = 2144.1 h 1 = 191.8\n",
+    "For the low-pressure pump,\n",
+    "Control volume:\n",
+    "Inlet state:\n",
+    "Exit state:\n",
+    "Low-pressure pump.\n",
+    "P 1 known, saturated liquid; state fixed.\n",
+    "P 2 known.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "h1 = 191.8\t\t\t#kJ/kg\n",
+    "h5 = 3213.6\t\t\t#kg/kg\n",
+    "h6 = 2685.7\t\t\t#kJ/kg\n",
+    "h7 = 2144.1\t\t\t#kJ/kg\n",
+    "h3 = 604.7\t\t\t#kJ/kg\n",
+    "P2 = 400\t\t\t#Exit pressure in kPa\n",
+    "P1 = 10  \t\t\t#Inlet pressure in kPa\n",
+    "v = 0.00101\t\t\t#Specific weight of water in m**3/kg\n",
+    "h1 = 191.8\t\t\t#Enthalpy in kJ/kg from table\n",
+    "P3 = 400\t\t\t#KPa\n",
+    "P4 = 4000\t\t\t#KPa\n",
+    "v = 0.001084\t\t#Specific heat for 3-4 process in m**3/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "wp1 = v*(P2-P1)\t\t#work done for low pressure pump in kJ/kg\n",
+    "h2 = h1+wp1\t\t\t#Enthalpy in kJ/kg\n",
+    "y = (h3-h2)/(h6-h2)\t#Extraction fraction\n",
+    "wt = (h5-h6)+(1-y)*(h6-h7)#Turbine work in kJ/kg\n",
+    "wp2 = v*(P4-P3)\t\t#Work done for high pressure pump\n",
+    "h4 = h3+wp2\t\t\t#Enthalpy in kJ/kg\n",
+    "wnet = wt-(1-y)*wp1-wp2\n",
+    "qh = h5-h4\t\t\t#Heat output in kJ/kg\n",
+    "nth = wnet/qh\n",
+    "\n",
+    "#Results:\n",
+    "print 'Refrigerator Efficiency: ',round(nth*100,1),'%'\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Cycle Efficiency:  29.2 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 11.5\n",
+    "''' A steam power plant operates on a cycle with pressures and temperatures as designated in\n",
+    "Fig. 11.17. The efficiency of the turbine is 86%, and the efficiency of the pump is 80%.\n",
+    "Determine the thermal efficiency of this cycle.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "h5 = 3169.1\t\t\t\t#kJ/kg\n",
+    "s5 = 6.7235\t\t\t\t#kJ/kg\n",
+    "sf = 0.6493\t\t\t\t#Entropy for liquid state in kJ/kg\n",
+    "sfg = 7.5009\t\t\t#Entropy difference in kJ/kg\n",
+    "hf = 191.8\t\t\t\t#kJ/kg\n",
+    "hfg = 2392.8\t\t\t#Enthalpy difference in kJ/kg\n",
+    "nt = 0.86\t\t\t\t#turbine efficiency given\n",
+    "np = 0.80\t\t\t\t#pump efficiency given\n",
+    "v = 0.001009\t\t\t#specific heat in m**3/kg\n",
+    "P2 = 5000\t\t\t\t#kPa\n",
+    "P1 = 10\t\t\t\t\t#kPa\n",
+    "h3 = 171.8\t\t#in kJ/kg from table\n",
+    "h4 = 3213.6\t\t#kJ/kg from table\n",
+    "\n",
+    "#Calculations:\n",
+    "s6s = s5\n",
+    "x6s = (s6s-sf)/sfg\t\t#x-factor\n",
+    "h6s = hf+x6s*hfg\t\t#kJ/Kg at state 6s\n",
+    "wt = nt*(h5-h6s)\n",
+    "wp = v*(P2-P1)/np\t\t#Work done in pump in kJ/kg\n",
+    "wnet = wt-wp\t\t\t#net work in kJ/kg\n",
+    "qh = h4-h3\n",
+    "nth = wnet/qh\n",
+    "\n",
+    "#Results:\n",
+    "print 'Cycle Efficiency: ',round(nth*100,1),'%'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Refrigeration Capacity:  3.89 KW\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 11.6\n",
+    "'''  Consider an ideal refrigeration cycle that uses R-134a as the working fluid. The temperature\n",
+    "of the refrigerant in the evaporator is −20\n",
+    "◦\n",
+    "C, and in the condenser it is 40\n",
+    "◦\n",
+    "C. The refrigerant\n",
+    "is circulated at the rate of 0.03 kg/s. Determine the COP and the capacity of the plant in\n",
+    "rate of refrigeration.\n",
+    "The diagram for this example is shown in Fig. 11.21. For each control volume\n",
+    "analyzed, the thermodynamic model is as exhibited in the R-134a tables. Each process is\n",
+    "steady state, with no changes in kinetic or potential energy.\n",
+    "Control volume:\n",
+    "Inlet state:\n",
+    "Exit state:\n",
+    "Compressor.\n",
+    "T 1 known, saturated vapor; state fixed.\n",
+    "P 2 known (saturation pressure at T 3 ). \n",
+    "'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P2 = 1017\t\t\t#KJ/Kg\n",
+    "h1 = 386.1 \t\t\t#KJ/Kg\n",
+    "s1 = 1.7395  \t\t#KJ/Kg\n",
+    "T2 = 47.7 \t\t\t#°C\n",
+    "h2 = 428.4 \t\t\t#KJ/Kg\n",
+    "h3 = 256.5 \t\t\t#KJ/Kg\t\n",
+    "\n",
+    "#Calculations:\n",
+    "s2 = s1\n",
+    "wc = h2-h1\t\t\t\t#Work done by compressor in KJ/Kg\n",
+    "h4 = h3\n",
+    "qL = h1-h4 \t\t\t\t#From Energy equation in KJ/Kg\n",
+    "B = round(qL/wc)\n",
+    "RC = qL*B/100\t\t\t#Refrigeration capacity in KW\n",
+    "\n",
+    "#Results:\n",
+    "print 'Refrigeration Capacity: ', round(RC,2), 'KW'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 11.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "COP of refrigerator :  1.45\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 11.7\n",
+    "'''A refrigeration cycle utilizes R-134a as the working fluid. The following are the properties\n",
+    "\tat various points of the cycle designated in Fig. 11.22:\n",
+    "\tP 1 = 125 kPa,\n",
+    "\tP 2 = 1.2 MPa,\n",
+    "\tP 3 = 1.19 MPa,\n",
+    "\tP 4 = 1.16 MPa,\n",
+    "\tP 5 = 1.15 MPa,\n",
+    "\tP 6 = P 7 = 140 kPa,\n",
+    "\tP 8 = 130 kPa,\n",
+    "\tT 1 = −10 ◦C\n",
+    "\tT 2 = 100◦C\n",
+    "\tT 3 = 80◦C\n",
+    "\tT 4 = 45◦C\n",
+    "\tT 5 = 40◦C\n",
+    "\tx 6 = x 7\n",
+    "\tT 8 = −20◦C\n",
+    "\tThe heat transfer from R-134a during the compression process is 4 kJ/kg. Determine\n",
+    "\tthe COP of this cycle.\n",
+    "\tFor each control volume, the R-134a tables are the model. Each process is steady\n",
+    "\tstate, with no changes in kinetic or potential energy.\n",
+    "\tAs before, we break the process down into stages, treating the compressor, the\n",
+    "\tthrottling value and line, and the evaporator in turn.\n",
+    "\tControl volume:\n",
+    "\tInlet state:\n",
+    "\tExit state:\n",
+    "\tCompressor.\n",
+    "\tP 1 , T 1 known; state fixed.\n",
+    "\tP 2 , T 2 known; state fixed.\n",
+    "'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "h1 = 394.9 \t\t\t#KJ/Kg\n",
+    "h2 = 480.9 \t\t\t#KJ/Kg\n",
+    "q = -4 \t\t\t\t#KJ/Kg\n",
+    "h5 = 256.4 \t\t\t#KJ/Kg\t\n",
+    "h8 = 386.6\t\t\t#KJ/Kg\n",
+    "\n",
+    "#Calculations:\n",
+    "wc = h2-h1-q\t\t#Work done by compressor in KJ/Kg\n",
+    "h6 = h5\n",
+    "h7 = h6\n",
+    "qL = h8-h7\n",
+    "B = qL/wc\t\t\t#Coefficient of performance\n",
+    "\n",
+    "#Results:\n",
+    "print 'COP of refrigerator : ', round(B,2)"
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/2._Fundamental_Thoughts.ipynb b/Introduction_to_flight_by_J_D_Anderson/2._Fundamental_Thoughts.ipynb
new file mode 100644
index 00000000..328a5430
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/2._Fundamental_Thoughts.ipynb
@@ -0,0 +1,365 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 2: Fundamental Thoughts"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Weight of person is: 9.75 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 2.1\n",
+    "'''What is the weight of a 1 kg mass at an altitude where the local acceleration of gravity is\n",
+    "9.75 m/s2?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "m = 1 \t\t#Mass in kg\n",
+    "g = 9.75 \t\t#Acc.due to gravity in m/s**2\n",
+    "\n",
+    "#Calculation:\n",
+    "F = m*g \t\t#Weight of 1 kg mass in N\n",
+    "\n",
+    "#Result\n",
+    "print \"Weight of person is:\",round(F,2),\"N\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Average specific volume is:  0.001325 m**3/kg\n",
+      "Overall density is: 755.0 kg/m**3\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 2.2\n",
+    "'''A1m3 container, shown in Fig. 2.9, is filled with 0.12 m3 of granite, 0.15 m3 of sand, and\n",
+    "0.2 m3 of liquid 25◦C water; the rest of the volume, 0.53 m3, is air with a density of 1.15\n",
+    "kg/m3. Find the overall (average) specific volume and density.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Vliq = 0.2 \t\t      #volume of liquid in m**3\n",
+    "dliq = 997 \t\t      #density of liquid in kg/m**3\n",
+    "Vstone = 0.12 \t\t#volume of stone in m**3\n",
+    "Vsand = 0.15 \t\t#volume of sand in m**3\n",
+    "Vair = 0.53 \t\t#voume of air in m**3\n",
+    "dstone = 2750 \t\t#density of stone in kg/m**3\n",
+    "dsand = 1500 \t\t#density of sand in kg/m**3\n",
+    "Vtot = 1 \t\t      #total volume in m**3\n",
+    "dair = 1.1 \t\t      #density of air in kg/m**3\n",
+    "\n",
+    "#Calculation:\n",
+    "mliq = Vliq*dliq \t\t#mass of liquid in kg\n",
+    "mstone = Vstone*dstone #volume of stone in m**3\n",
+    "msand = Vsand*dsand \t#volume of sand in m**3\n",
+    "mair = Vair*dair \t\t#mass of air\n",
+    "mtot = mair+msand+mliq+mstone \t\t#total mass in kg\n",
+    "v = Vtot/mtot \t\t#specific volume in  m**3/kg\n",
+    "d = 1/v \t\t      #overall density in kg/m**3\n",
+    "\n",
+    "#Results:\n",
+    "print \"Average specific volume is: \",round(v ,6),\"m**3/kg\"\n",
+    "print \"Overall density is:\",round(d),\"kg/m**3\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Force that rod can push within the upward direction is: 932.9 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 2.3\n",
+    "'''The hydraulic piston/cylinder system shown in Fig. 2.11 has a cylinder diameter of D =\n",
+    "0.1 m with a piston and rod mass of 25 kg. The rod has a diameter of 0.01 m with an\n",
+    "outside atmospheric pressure of 101 kPa. The inside hydraulic fluid pressure is 250 kPa.\n",
+    "How large a force can the rod push within the upward direction?'''\n",
+    "from math import pi\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Dcyl = 0.1 \t\t#Cylinder diameter in m\n",
+    "Drod = 0.01 \t#Rod diameter in m\n",
+    "Pcyl = 250000 \t#Inside hydaulic pressure in Pa\n",
+    "Po = 101000 \t#Outside atmospheric pressure in kPa\n",
+    "g = 9.81 \t\t#Acc. due to gravity in m/s**2\n",
+    "mp = 25 \t\t#Mass of (rod+piston) in kg\n",
+    "\n",
+    "#Calculation:\n",
+    "Acyl = pi*Dcyl**2/4 \t\t#Cross sectional area of cylinder in m**2\n",
+    "Arod = pi*Drod**2/4 \t\t#Cross sectional area of rod in m**2\n",
+    "F = Pcyl*Acyl-Po*(Acyl-Arod)-mp*g \t#Force that rod can push within the upward direction in N\n",
+    "\n",
+    "#Result:\n",
+    "print \"Force that rod can push within the upward direction is:\",round(F,1),\"N\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Atmospheric pressure is: 99.54 KPa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 2.4\n",
+    "'''A mercury barometer located in a room at 25◦C has a height of 750 mm. What is the\n",
+    "atmospheric pressure in kPa?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "dm = 13534 \t\t       #Density of mercury in kg/m**3\n",
+    "H = 0.750 \t\t       #Height difference between two columns in metres\n",
+    "g = 9.80665 \t\t #Acc. due to gravity  in m/s**2\n",
+    "\n",
+    "#Calculation:\n",
+    "Patm = dm*H*g/1000 \t #Atmospheric pressure in kPa\n",
+    "\n",
+    "#Result:\n",
+    "print \"Atmospheric pressure is:\",round(Patm,2),\"KPa\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Absolute pressure inside vessel is: 1.3 atm\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 2.5\n",
+    "'''A mercury (Hg) manometer is used to measure the pressure in a vessel as shown in\n",
+    "Fig. 2.13. The mercury has a density of 13 590 kg/m3, and the height difference between the\n",
+    "two columns is measured to be 24 cm.We want to determine the pressure inside the vessel.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "dm = 13590 \t\t      #Density of mercury in kg/m**3\n",
+    "H = 0.24 \t\t      #Height difference between two columns in metres\n",
+    "g = 9.80665 \t\t#Acc. due to gravity  in m/s**2\n",
+    "Patm = 13590*0.750*9.80665 #Atmospheric Pressure in Pa\n",
+    "\n",
+    "#Calculation:\n",
+    "dP = dm*H*g \t\t#Pressure difference in Pa\n",
+    "Pvessel = dP+Patm \t#Absolute Pressure inside vessel in Pa\n",
+    "\n",
+    "#Result:\n",
+    "print \"Absolute pressure inside vessel is:\",round(Pvessel/101325,2),\"atm\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Pressure at the bottom of storage tank if liquid is Gasoline: 156.2 Kpa\n",
+      "Pressure at the bottom of storage tank if liquid is R-134a: 1089.0 Kpa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 2.6\n",
+    "'''What is the pressure at the bottom of the 7.5-m-tall storage tank of fluid at 25◦C shown\n",
+    "in Fig. 2.15? Assume that the fluid is gasoline with atmospheric pressure 101 kPa on the\n",
+    "top surface. Repeat the question for the liquid refrigerant R-134a when the top surface\n",
+    "pressure is 1 MPa.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "dg = 750 \t\t#Density of gaasoline in kg/m**3\n",
+    "dR = 1206 \t\t#Density of R-134a in kg/m**3\n",
+    "H = 7.5 \t\t#Height of storage tank in metres\n",
+    "g = 9.807 \t\t#Acc. due to gravity in m/s**2\n",
+    "Ptop1 = 101 \t#Atmospheric pressure in kPa\n",
+    "Ptop2 = 1000 \t\t#top surface pressure in kPa\n",
+    "\n",
+    "#Calculation:\n",
+    "dP1 = dg*g*H/1000\t#Pressure difference in kPa\n",
+    "P1 = dP1+Ptop1\n",
+    "dP2 = dR*g*H/1000#Pressure difference in kPa\n",
+    "P2 = dP2+Ptop2\n",
+    "\n",
+    "#Result:\n",
+    "print \"Pressure at the bottom of storage tank if liquid is Gasoline:\",round(P1,1),\"Kpa\"\n",
+    "print \"Pressure at the bottom of storage tank if liquid is R-134a:\",round(P2),\"Kpa\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Balancing force on second larger piston is: 122.3513 N\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 2.7\n",
+    "'''A piston/cylinder with a cross-sectional area of 0.01 m2 is connected with a hydraulic\n",
+    "line to another piston/cylinder with a cross-sectional area of 0.05 m2. Assume that both\n",
+    "chambers and the line are filled with hydraulic fluid of density 900 kg/m3 and the larger\n",
+    "second piston/cylinder is 6mhigher up in elevation. The telescope armand the buckets have\n",
+    "hydraulic piston/cylinders moving them, as seen in Fig. 2.16.With an outside atmospheric\n",
+    "pressure of 100 kPa and a net force of 25 kN on the smallest piston, what is the balancing\n",
+    "force on the second larger piston?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Po = 100\t\t#Outside atmospheric pressure in kPa\n",
+    "F1 = 25 \t\t#Net force on the smallest piston in kN\n",
+    "A1 = 0.01 \t\t#Cross sectional area of lower piston in m**2\n",
+    "d = 900 \t\t#Density of fluid in kg/m**3\n",
+    "g = 9.81 \t\t#Acc. due to gravity in m/s**2\n",
+    "H = 6 \t\t#Height of second piston in comparison to first one in m\n",
+    "A2 = 0.05 \t\t#Cross sectional area of higher piston in m**3\n",
+    "\n",
+    "#Calculation:\n",
+    "P1 = Po+F1/A1 \t#Fluid pressure in kPa\n",
+    "P2 = P1-d*g*H/1000 #Pressure at higher elevation on piston 2 in kPa\n",
+    "F2 = (P2-Po)*A2 \t#Balancing force on second piston in kN\n",
+    "\n",
+    "#Result:\n",
+    "print \"Balancing force on second larger piston is:\",round(F2,4),\"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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/3._The_Standard_Atmosphere.ipynb b/Introduction_to_flight_by_J_D_Anderson/3._The_Standard_Atmosphere.ipynb
new file mode 100644
index 00000000..368f5fbe
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/3._The_Standard_Atmosphere.ipynb
@@ -0,0 +1,546 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 3: The Standard Atmosphere"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Theoretical question\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.1\n",
+    "''' Determine the phase for each of the following water states using the tables in Appendix\n",
+    "B and indicate the relative position in the P–v, T–v, and P–T diagrams. '''\n",
+    "\n",
+    "print 'Theoretical question';"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Theoretical question\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.2\n",
+    "'''Determine the phase for each of the following states using the tables in Appendix B and\n",
+    "indicate the relative position in the P–v, T–v, and P–T diagrams, as in Figs. 3.11 and\n",
+    "3.12.'''\n",
+    "\n",
+    "print 'Theoretical question';"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Theoretical question\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.3\n",
+    "'''Determine the temperature and quality (if defined) for water at a pressure of 300 kPa and\n",
+    "at each of these specific volumes:\n",
+    "a. 0.5 m3/kg\n",
+    "b. 1.0 m3/kg'''\n",
+    "\n",
+    "print 'Theoretical question';"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Vapor on mass basis: 22.12 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.4\n",
+    "'''A closed vessel contains 0.1 m3 of saturated liquid and 0.9 m3 of saturated vapor R-134a\n",
+    "in equilibrium at 30◦C. Determine the percent vapor on a mass basis.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "vliq = 0.1 \t\t#Volume of saturated liquid in m**3\n",
+    "vf = 0.000843 \t#in m**3/kg\n",
+    "vvap = 0.9 \t\t#Volume of saturated vapor R-134a in equilbrium\n",
+    "vg = 0.02671 \t#in m**3/kg\n",
+    "\n",
+    "#Calculation:\n",
+    "mliq = vliq/vf \t#Mass of liquid in kg \n",
+    "mvap = vvap/vg \t#Mass of vapor in kg\n",
+    "m = mliq+mvap \t#Total mass in kg\n",
+    "x = mvap/m \t\t#Percentage of vapor on mass basis\n",
+    "\n",
+    "#Result:\n",
+    "print 'Vapor on mass basis:',round(x*100,2),'%'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Final pressure is(by interpolation): 945 KPa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.5\n",
+    "'''A rigid vessel contains saturated ammonia vapor at 20◦C. Heat is transferred to the system\n",
+    "until the temperature reaches 40◦C. What is the final pressure?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "v1 = 0.14922 \t#specific volume of sautrated ammonia in m**3/kg\n",
+    "P2 = 945 \t\t#final pressure in kPa\n",
+    "\n",
+    "#Calculation:\n",
+    "v2 = v1 \t\t#in m**3/kg\n",
+    "\n",
+    "print 'Final pressure is(by interpolation):',P2,'KPa'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Pressure is:  779.2 Kpa\n",
+      "Quality is:  0.2201\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.6\n",
+    "'''Determine the missing property of P–v–T and x if applicable for the following states.\n",
+    "a. Nitrogen: −53.2◦C, 600 kPa\n",
+    "b. Nitrogen: 100 K, 0.008 m3/kg'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "T1 = 273-53.2 \t#Given temperature in K\n",
+    "P1 = 600 \t\t#Given pressure in kPa\n",
+    "T2 = 100 \t\t#Given temp. in K\n",
+    "v2 = 0.008 \t\t#Given specific volume in m**3/kg\n",
+    "vf = 0.001452 \t#in m**3/kg\n",
+    "vg = 0.0312 \t#in m**3/kg\n",
+    "Psat = 779.2 \t#Saturation pressure in kPa\n",
+    "\n",
+    "#Calculation:\n",
+    "vfg = vg-vf \t#in m**3/kg\n",
+    "x = (v2-vf)/vfg \t#Quality\n",
+    "\n",
+    "#Result:\n",
+    "print \"Pressure is: \",round(Psat,2),\"Kpa\"\n",
+    "print \"Quality is: \",round(x,4)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Pressure of water: 534.2 KPa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.7\n",
+    "'''Determine the pressure for water at 200◦C with v = 0.4 m3/kg.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "vg = 0.12736 \t#Specific volume in m**3/kg for water at 200C\n",
+    "v = 0.4 \t\t#Specific volume in m**3/kg\n",
+    "P1 = 500 \t\t#in kPa\n",
+    "v1 = 0.42492 \t#Specific volume at P1 in m**3/kg\n",
+    "P2 = 600 \t\t#in kPa\n",
+    "v2 = 0.35202 \t#Specific volume at P2 in m**3/kg\n",
+    "\n",
+    "#Calculation:\n",
+    "P = P1+(P2-P1)*(v-v1)/(v2-v1) #Calculating pressure by interplotation\n",
+    "\n",
+    "#Result:\n",
+    "print 'Pressure of water:',round(P,1),'KPa'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Mass of air contained in room is: 280.43 Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.8\n",
+    "'''What is the mass of air contained in a room 6 m × 10 m × 4 m if the pressure is 100 kPa\n",
+    "and the temperature is 25◦C?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P = 100 \t\t#Pressure in kPa\n",
+    "V = 6*10*4 \t\t#Volume of room in m**3\n",
+    "R = 0.287 \t\t#in kN-m/kg-K\n",
+    "T = 25 \t\t#Temperature in Celsius\n",
+    "\n",
+    "#Calculation:\n",
+    "m = P*V/(R*(T+273.2)) \t#Mass of air contained in room\n",
+    "\n",
+    "#Result:\n",
+    "print \"Mass of air contained in room is:\",round(m,2),\"Kg\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Pressure inside tank is:  2066.0 KPa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.9\n",
+    "'''A tank has a volume of 0.5 m3 and contains 10 kg of an ideal gas having a molecular mass\n",
+    "of 24. The temperature is 25◦C. What is the pressure?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "V = 0.5 \t\t#Volumr of tank in m**3\n",
+    "m = 10 \t\t#Mass of ideal gas in kg\n",
+    "T = 25 \t\t#Temperature of tank in Celsius\n",
+    "M = 24 \t\t#Molecular mass of gas in kg/kmol\n",
+    "Ru = 8.3145 \t#Vniversal gas constant in kN-m/kmol-K\n",
+    "\n",
+    "#Calculation:\n",
+    "R = Ru/M \t\t#Gas constant for given ideal gas in kN-m/kg-K\n",
+    "P = m*R*(T+273.2)/V#Pressure inside tank\n",
+    "\n",
+    "#Result:\n",
+    "print \"Pressure inside tank is: \",round(P),\"KPa\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Volume flow rate is:  0.004054 m**3/s\n",
+      "Mass flow rate is:  0.00766 Kg/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.10\n",
+    "'''A gas bell is submerged in liquid water, with its mass counterbalanced with rope and\n",
+    "pulleys, as shown in Fig. 3.20. The pressure inside is measured carefully to be 105 kPa,\n",
+    "and the temperature is 21◦C. A volume increase is measured to be 0.75 m3 over a period\n",
+    "of 185 s. What is the volume flow rate and the mass flow rate of the flow into the bell,\n",
+    "assuming it is carbon dioxide gas?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "dt = 185 \t\t#Time period in seconds over which there is incrrease in volume \n",
+    "dV = 0.75 \t\t#Increase in volume in 0.75 in m**3\n",
+    "P = 105 \t\t#Pressure inside gas bell kPa\n",
+    "T = 21 \t\t#Temperature in celsius\n",
+    "R = 0.1889 \t\t#Ideal gas constant in kJ/kg-K\n",
+    "\n",
+    "#Calculation:\n",
+    "V = dV/dt \t\t#Volume flow rate in m**3/s\n",
+    "m = P*V/(R*(T+273.15))#Mass flow rate of the flow in kg/s\n",
+    "\n",
+    "#Result:\n",
+    "print \"Volume flow rate is: \",round(V,6),\"m**3/s\"\n",
+    "print \"Mass flow rate is: \",round(m,5),\"Kg/s\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Specific volume for R-134a using R-134a tables is:  0.0065 m**3/kg\n",
+      "Specific volume for R-134a using R-134a the ideal gas laws is: 0.01011 m**3/kg\n",
+      "Specific volume for R-134a using the generalized chart is:  0.00677 m**3/kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.12\n",
+    "'''Determine the specific volume for R-134a at 100◦C, 3.0 MPa for the following models:'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "T = 100 \t\t#Given temp.in 100 celsius\n",
+    "P = 3 \t\t#Given pressure in MPa\n",
+    "v1 = 0.0065 \t#Specific volume in m**3/kg using table\n",
+    "M = 102.3 \t\t#Molecular mass in kg\n",
+    "R = 8.3145 \t\t#in kJ/K\n",
+    "Tr = 373.2/374.2 \t#Reduced temperature using generalized chart\n",
+    "Pr = 3/4.06 \t#Reduced pressure using generalized chart\n",
+    "Z = 0.67 \t\t#Compressibility factor \n",
+    "\n",
+    "#Calculation:\n",
+    "Ru = R/M \t\t#in kJ/K-kg\n",
+    "v2 = Ru*(T+273)/(P*1000)#Specific volume assuming R-134a to be ideal gas in m**3/kg\n",
+    "v3 = Z*v2 \t\t#Specific volume using generalized chart in m**3/kg\n",
+    "\n",
+    "#Result:\n",
+    "print \"Specific volume for R-134a using R-134a tables is: \",round(v1,5),\"m**3/kg\"\n",
+    "print \"Specific volume for R-134a using R-134a the ideal gas laws is:\",round(v2,5),\"m**3/kg\"\n",
+    "print \"Specific volume for R-134a using the generalized chart is: \",round(v3,5),\"m**3/kg\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3.13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Pressure is P: 850.0 KPa\n",
+      "Total propane mass is:  13.66 Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 3.13\n",
+    "'''Propane in a steel bottle of volume 0.1 m3 has a quality of 10% at a temperature of 15◦C.\n",
+    "Use the generalized compressibility chart to estimate the total propane mass and to find\n",
+    "the pressure.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Pc = 4250 \t\t#Critical pressure of propane in kPa\n",
+    "Tc = 369.8 \t\t#Critical temperature in K\n",
+    "T = 15 \t\t#Temperature of propane in celsius\n",
+    "Prsat = 0.2 \t#Reduced pressure \n",
+    "x = 0.1 \t\t#Given quality\n",
+    "Zf = 0.035 \t\t#from graph\n",
+    "Zg = 0.83 \t\t#from graph\n",
+    "V = 0.1 \t\t#Volume of steel bottle in m**3\n",
+    "R = 0.1887 \t\t#in kPa-m**3/kg-K\n",
+    "\n",
+    "#Calculation:\n",
+    "Tr = T/Tc \t\t#Reduced temperature\n",
+    "P = Prsat*Pc \t#Pressure in kPa\n",
+    "Z = (1-x)*Zf+x*Zg #Overall compressibility factor\n",
+    "m = P*V/(Z*R*(T+273))#Total propane mass in kg\n",
+    "\n",
+    "#Result:\n",
+    "print \"Pressure is P:\",round(P,2),\"KPa\"\n",
+    "print \"Total propane mass is: \",round(m,2),\"Kg\"\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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/4._Aerodynamics.ipynb b/Introduction_to_flight_by_J_D_Anderson/4._Aerodynamics.ipynb
new file mode 100644
index 00000000..d2461f95
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/4._Aerodynamics.ipynb
@@ -0,0 +1,229 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 4: Aerodynamics"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work done during the isobaric process is:  12.0 KJ\n",
+      "Work done in isothermal process is:  7.33 KJ\n",
+      "Work done during the described process is:  6.41 KJ\n",
+      "Work done in the isovolumic process is:  0.0 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 4.1\n",
+    "'''Consider as a system the gas in the cylinder shown in Fig. 4.7; the cylinder is fitted with\n",
+    "a piston on which a number of small weights are placed. The initial pressure is 200 kPa,\n",
+    "and the initial volume of the gas is 0.04 m3.'''\n",
+    "from math import log\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P1 = 200 \t\t#Initial pressure inside cylinder in kPa\n",
+    "V2 = 0.1 \t\t#in m**3\n",
+    "V1 = 0.04 \t\t#Initial volume of gas in m**3\n",
+    "W4 = 0 \t\t#Work done in isovolumic process\n",
+    "\n",
+    "#Calculation:\n",
+    "W1 = P1*(V2-V1) \t#Work done in isobaric process in kJ\n",
+    "W2 = P1*V1*log(V2/V1) #Work done in isothermal process in kJ\n",
+    "P2 = P1*(V1/V2)**(1.3)\t#Final pressure according to the given process\n",
+    "W3 = (P2*V2-P1*V1)/(1-1.3)\n",
+    "\n",
+    "#Result:\n",
+    "print \"Work done during the isobaric process is: \",round(W1,2),\"KJ\"\n",
+    "print \"Work done in isothermal process is: \",round(W2,2),\"KJ\"\n",
+    "print \"Work done during the described process is: \",round(W3,2),\"KJ\"\n",
+    "print \"Work done in the isovolumic process is: \",round(W4,2),\"KJ\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work produced by ammonia is:  12.71 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 4.3\n",
+    "'''The cylinder/piston setup of Example 4.2 contains 0.5 kg of ammonia at −20◦C with\n",
+    "a quality of 25%. The ammonia is now heated to +20◦C, at which state the volume\n",
+    "is observed to be 1.41 times larger. Find the final pressure and the work the ammonia\n",
+    "produced.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Psat = 190.2 \t\t#in kPa\n",
+    "vf = 0.001504 \t\t#in m**3/kg\n",
+    "vfg = 0.62184 \t\t#in m**3/kg\n",
+    "x1 = 0.25 \t\t      #Quality\n",
+    "P2 = 600 \t\t      #Pressure in state 2 in kPa\n",
+    "m = 0.5 \t\t      #Mass of ammonia in kg\n",
+    "\n",
+    "#Calculation:\n",
+    "P1 = Psat \t\t      #Saturation pressure in state 1\n",
+    "v1 = vf+x1*vfg \t\t#Specific volume at state 1 in m**3/kg\n",
+    "v2 = 1.41*v1 \t\t#Specific volume at state 2 in m**3/kg\n",
+    "W = m*(P1+P2)*(v2-v1)/2#Work produced by ammonia in kJ\n",
+    "\n",
+    "#Result:\n",
+    "print \"Work produced by ammonia is: \",round(W,2),\"KJ\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work in the overall process is:  -17.71 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 4.4\n",
+    "'''The piston/cylinder setup shown in Fig. 4.12 contains 0.1 kg of water at 1000 kPa, 500◦C.\n",
+    "The water is now cooled with a constant force on the piston until it reaches half the initial\n",
+    "volume. After this it cools to 25◦C while the piston is against the stops. Find the final water\n",
+    "pressure and the work in the overall process, and show the process in a P–v diagram.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "v1 = 0.35411 \t#Specific volume at state 1 in m**3/kg\n",
+    "m = 0.1 \t\t#Mass of water in kg\n",
+    "P1 = 1000 \t\t#Pressure inside cylinder in kPa\n",
+    "\n",
+    "#Calculation:\n",
+    "v2 = v1/2 \n",
+    "W = m*P1*(v2-v1) \t#in kJ\n",
+    "\n",
+    "#Result:\n",
+    "print \"Work in the overall process is: \",round(W,2),\"KJ\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Rate of heat transfer in the glass and convective layer is:  1105.0 KW\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 4.7\n",
+    "'''Consider the constant transfer of energy from a warm room at 20◦C inside a house to the\n",
+    "colder ambient temperature of −10◦C through a single-pane window, as shown in Fig.\n",
+    "4.19. The temperature variation with distance from the outside glass surface is shown\n",
+    "by an outside convection heat transfer layer, but no such layer is inside the room (as a\n",
+    "simplification). The glass pane has a thickness of 5 mm (0.005 m) with a conductivity of\n",
+    "1.4W/mKand a total surface area of 0.5m2. The outside wind is blowing, so the convective\n",
+    "heat transfer coefficient is 100 W/m2 K.With an outer glass surface temperature of 12.1◦C,\n",
+    "we would like to know the rate of heat transfer in the glass and the convective layer'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "k = 1.4 \t\t#Conductivity of glass pane in W/m-K\n",
+    "A = 0.5 \t\t#Total surface area of glass pane\n",
+    "dx = 0.005 \t\t#Thickness of glasspane in m\n",
+    "dT1 = 20-12.1 \t#Temperature difference between room air and outer glass surface temperature in celsius\n",
+    "h = 100 \t\t#Convective heat transfer coefficient in W/m**2-K \n",
+    "dT = 12.1-(-10)  \t#Temperature difference between warm room and colder ambient in celsius\n",
+    "\n",
+    "#Calculation:\n",
+    "Q = -k*A*dT1/dx \t#Conduction through glass slab in W\n",
+    "Q2 = h*A*dT \t#Heat transfer in convective layer in W\n",
+    "\n",
+    "#Result:\n",
+    "print \"Rate of heat transfer in the glass and convective layer is: \",round(Q2,2),\"KW\""
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/5._Airfoils,_Wings_and_Other_Aerodynamic_shapes.ipynb b/Introduction_to_flight_by_J_D_Anderson/5._Airfoils,_Wings_and_Other_Aerodynamic_shapes.ipynb
new file mode 100644
index 00000000..23c8b51f
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/5._Airfoils,_Wings_and_Other_Aerodynamic_shapes.ipynb
@@ -0,0 +1,533 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 5: Airfoils, Wings and Other Aerodynamic shapes"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Required height of car:  37.1 m\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.1\n",
+    "''' A car of mass 1100 kg drives with a velocity such that it has a kinetic energy of 400 kJ (see\n",
+    "Fig. 5.4). Find the velocity. If the car is raised with a crane, how high should it be lifted in\n",
+    "the standard gravitational field to have a potential energy that equals the kinetic energy? '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "m = 1100 \t\t#Mass of car in kg\n",
+    "ke = 400 \t\t#Kinetic energy of car in kJ\n",
+    "g = 9.807 \t\t#Acc. due to gravity in m/s**2\n",
+    "\n",
+    "#Calculations:\n",
+    "V = (2*ke*1000/m)**0.5 #Velocity of car in m/s\n",
+    "H = ke*1000/(m*g) \t#Height to which the car should be lifted so that its potential energy equals its kinetic energy\n",
+    "\n",
+    "#Results:\n",
+    "print \"Required height of car: \", round(H,1),\"m\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Change in internal energy is:  3590.0 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.2\n",
+    "'''A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel\n",
+    "is 5090 kJ. The heat transfer from the tank is 1500 kJ. Consider the tank and the fluid\n",
+    "inside a control surface and determine the change in internal energy of this control mass.  '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "W = -5090 \t\t#Work input to paddle wheel in kJ\n",
+    "Q = -1500 \t\t#Heat transfer from tank in kJ\n",
+    "\n",
+    "#Calculations:\n",
+    "dU = Q-W \t\t#Change in internal energy in kJ\n",
+    "\n",
+    "#Results:\n",
+    "print \"Change in internal energy is: \",round(dU),\"KJ\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "When stone is about to enter state 2, dKE:  1.0 KJ and dPE : -1.0 KJ\n",
+      "When stone has just come to rest in the bucket i.e. state 3, W = 0, dPE = 0, dKE1: -1.0 J and dU : 1.0 J\n",
+      "When stone has entered state 4, dPE = 0, W3 = 0,dKE = 0, dU:  -1.0 J Q3:  -1.0 J\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.3\n",
+    "''' Consider a stone having a mass of 10 kg and a bucket containing 100 kg of liquid water.\n",
+    "Initially the stone is 10.2 m above the water, and the stone and the water are at the same\n",
+    "temperature, state 1. The stone then falls into the water.\n",
+    "Determine \u0003U, \u0003KE, \u0003PE, Q, and W for the following changes of state, assuming\n",
+    "standard gravitational acceleration of 9.806 65 m/s2.\n",
+    "a. The stone is about to enter the water, state 2.\n",
+    "b. The stone has just come to rest in the bucket, state 3.\n",
+    "c. Heat has been transferred to the surroundings in such an amount that the stone and\n",
+    "water are at the same temperature, T1, state 4. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "g = 9.806 \t\t#Acceleration due to gravity in m/s**2\n",
+    "m = 10 \t\t#Mass of stone in kg\n",
+    "H1 = 10.2 \t\t#Initial height of stone above water in metres\n",
+    "H2 = 0 \t\t#Final height in metres\n",
+    "\n",
+    "#Calculations:\n",
+    "dKE1 = -m*g*(H2-H1) #Change in kinetic energy when stone enters state 2 in J\n",
+    "dPE1 = -1 \t\t#Change in potential energy when stone enters state 2 in J\n",
+    "dPE2 = 0 \t\t#Change in potential energy when stone enters state 3 in JQ2 = 0 \t\t#no heat transfer when stone enters state 3 in J\n",
+    "W2 = 0 \t\t#no work done when stone enters state 3 in J\n",
+    "dKE2 = -1 \t\t#Change in kinetic energy when stone enters state 3\n",
+    "dU2 = -dKE2 \t#Change in internal energy when stone enters state 3 in J\n",
+    "dKE3 = 0 \t\t#Change in kinetic energy when stone enters state 4\n",
+    "dPE = 0 \t\t#Change in potential energy when stone enters state 4 in J\n",
+    "W3 = 0 \t\t#No work done when stone enters state 4 in J\n",
+    "dU3 = -1 \t\t#Change in internal energy when stone enters state 4 in J\n",
+    "Q3 = dU3 \t\t#Heat transfer when stone enters state 4 in J\n",
+    "\n",
+    "#Results:\n",
+    "print \"When stone is about to enter state 2, dKE: \",round(dKE1/1000,2),\"KJ and dPE :\",round(dPE1,2),\"KJ\"\n",
+    "print \"When stone has just come to rest in the bucket i.e. state 3, W = 0, dPE = 0, dKE1:\",round(dKE2,2),\"J and dU :\",round(dU2,2),\"J\"\n",
+    "print \"When stone has entered state 4, dPE = 0, W3 = 0,dKE = 0, dU: \",round(dU3,2),\"J Q3: \",round(Q3,2),\"J\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Specific volume for water is:  0.06474 m**3/Kg\n",
+      "Therefore ,this state is in the two phase region with quality:  0.6456\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.4\n",
+    "'''Determine the missing property (P, T, or x) and v forwater at each of the following states:\n",
+    "a. T = 300◦C, u = 2780 kJ/kg\n",
+    "b. P = 2000 kPa, u = 2000 kJ/kg'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "T1 = 300 \t\t#Given temp. in Celsius\n",
+    "u1 = 2780 \t\t#Given specific internal enrgy in kJ/kg\n",
+    "P2 = 2000 \t\t#Given pressure in kPa\n",
+    "u2 = 2000 \t\t#Given specific intrernal energy in kJ/kg\n",
+    "uf = 906.4 \t\t#in kJ/kg\n",
+    "ug = 2600.3 \t#in kJ/kg \n",
+    "vf = 0.001177 \t#in m**3/kg \n",
+    "vg = 0.099627 \t#in m**3/kg\n",
+    "\n",
+    "#Calculation:\n",
+    "x2 = (u2-906.4)/(ug-uf) \n",
+    "v2 = vf+x2*(vg-vf)#Specific volume for water in m**3/kg\n",
+    "\n",
+    "#Result:\n",
+    "print \"Specific volume for water is: \",round(v2,5),\"m**3/Kg\"\n",
+    "print \"Therefore ,this state is in the two phase region with quality: \",round(x2,4)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "By interplotation we find that for steam, if vg = 0.09831 m**3/kg then pressure is 2.03 MPa\n",
+      "Heat transfer for the process is:  104932.6 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.5\n",
+    "'''A vessel having a volume of 5 m3 contains 0.05 m3 of saturated liquid water and 4.95 m3\n",
+    "of saturated water vapor at 0.1 MPa. Heat is transferred until the vessel is filled with\n",
+    "saturated vapor. Determine the heat transfer for this process.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Vliq = 0.05 \t\t#Volume of saturated liquid in m**3\n",
+    "vf = 0.001043 \t\t#in m**3/kg\n",
+    "Vvap = 4.95 \t\t#Volume of saturated water vapour in m**3\n",
+    "vg = 1.6940 \t\t#in m**3/kg\n",
+    "u1liq = 417.36 \t\t#Specific internal energy of liquid in kJ/kg\n",
+    "u1vap = 2506.1 \t\t#Specific internal energy of vapors in kJ/kg\n",
+    "V = 5 \t\t      #Total volume in m**3\n",
+    "u2 = 2600.5 \t\t#Specific internal energy at final state in kJ/kg\n",
+    "\n",
+    "#Calculation:\n",
+    "m1liq = Vliq/vf \t\t#Mass of liquid in kg\n",
+    "m1vap = Vvap/vg \t\t#Mass of vapors in kg\n",
+    "U1 = m1liq*u1liq+m1vap*u1vap #Total internal energy in kJ\n",
+    "m = m1liq+m1vap \t\t#Total mass in kg\n",
+    "v2 = V/m \t\t      #Final specific volume in m**3/kg\n",
+    "U2 = m*u2 \t\t      #Internal energy at final state in kJ\n",
+    "Q = U2-U1 \t\t      #Heat transfer for the process in kJ\n",
+    "\n",
+    "print 'By interplotation we find that for steam, if vg = 0.09831 m**3/kg then pressure is 2.03 MPa'\n",
+    "print \"Heat transfer for the process is: \",round(Q,2),\"KJ\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work done for the process:  91.0 KJ\n",
+      "Heat transfer:  771.1 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.6\n",
+    "'''A cylinder fitted with a piston has a volume of 0.1 m3 and contains 0.5 kg of steam at 0.4\n",
+    "MPa. Heat is transferred to the steam until the temperature is 300◦C, while the pressure\n",
+    "remains constant.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "V1 = 0.1 \t\t      #Volume of cylinder in m**3\n",
+    "m = 0.5 \t\t      #Mass of steam in kg\n",
+    "vf = 0.001084 \t\t#m**3/kg\n",
+    "vfg = 0.4614 \t\t#m**3/kg\n",
+    "hf = 604.74 \t\t#kJ/kg\n",
+    "hfg = 2133.8\t\t#kJ/kg\n",
+    "h2 = 3066.8 \t\t#Final specific heat enthalpy in kJ/kg\n",
+    "P = 400 \t\t      #Pressure inside cylinder in kPa\n",
+    "v2 = 0.6548 \t\t#Specific enthalpy in m**3/kg\n",
+    "\n",
+    "#Calculation:\n",
+    "v1 = V1/m \t\t      #Specific volume of steam in m**3/kg\n",
+    "x1 = (v1-vf)/vfg \t\t#Quality\n",
+    "h1 = hf+x1*hfg \t\t#Initial specific enthalpy in kJ/kg\n",
+    "Q = m*(h2-h1) \t\t#Heat transfer for this process in kJ\n",
+    "W = m*P*(v2-v1) \t\t#Work done for the process in kJ\n",
+    "\n",
+    "#Result:\n",
+    "print \"Work done for the process: \",round(W), \"KJ\"\n",
+    "print \"Heat transfer: \",round(Q,1), \"KJ\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Change in specific heat enthalpy if ideal gas tables are used is:  1267.0 KJ/Kg\n",
+      "If empirical equations are used:  1241.64 KJ/Kg\n",
+      "If specific heat is assumed to be constant and using its value at T1:  1106.4 KJ/Kg\n",
+      "If specific heat is assumed to be constant at its value at (T1+T2)/2:  1292.0 KJ/Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.8\n",
+    "'''Calculate the change of enthalpy as 1 kg of oxygen is heated from 300 to 1500 K. Assume\n",
+    "ideal-gas behavior.'''\n",
+    "from sympy import symbols,integrate\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "h1 = 273.2 \t\t#Specific heat enthalpy for oxygen at 300 K\n",
+    "h2 = 1540.2 \t#Specific heat enthalpy for oxygen at 1500 K\n",
+    "T1 = 300 \t\t#Initial temperature in K\n",
+    "T2 = 1500 \t\t#Final temparature in K\n",
+    "x = symbols('x')\n",
+    "\n",
+    "#Calculation:\n",
+    "Cp = 0.88-0.00001*x+0.54*x**2-0.33*x**3 #Expression for constant pressure specific heat enthalpy for oxygen\n",
+    "dh1 = h2-h1 \t#This change in specific heat enthalpy is calculated using ideal gas tables \n",
+    "dh2 = 1000*integrate(Cp,(x,T1/1000,T2/1000)) #Using empirical equation\n",
+    "dh3 = 0.922*(T2-T1) #It is claculated if we assume specific heat enthalpy to be constant and uses its value at 300K\n",
+    "dh4 = 1.0767*(T2-T1)#It is claculated if we assume specific heat enthalpy to be constant and uses its value at 900K i.e mean of initial and final temperature\n",
+    "\n",
+    "#Result:\n",
+    "print \"Change in specific heat enthalpy if ideal gas tables are used is: \",round( dh1,2),\"KJ/Kg\"\n",
+    "print \"If empirical equations are used: \",round( dh2,2),\"KJ/Kg\"\n",
+    "print \"If specific heat is assumed to be constant and using its value at T1: \",round(dh3,2),\"KJ/Kg\"\n",
+    "print \"If specific heat is assumed to be constant at its value at (T1+T2)/2: \",round(dh4),\"KJ/Kg\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Heat transfer for the above process is:  -4.21 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.9\n",
+    "'''A cylinder fitted with a piston has an initial volume of 0.1 m3 and contains nitrogen at 150\n",
+    "kPa, 25◦C. The piston is moved, compressing the nitrogen until the pressure is 1 MPa and\n",
+    "the temperature is 150◦C. During this compression process heat is transferred from the\n",
+    "nitrogen, and the work done on the nitrogen is 20 kJ. Determine the amount of this heat\n",
+    "transfer.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P = 150 \t\t#Pressure of nitrogen in cylinder in kPa\n",
+    "V = 0.1 \t\t#Initial volume of cylinder in m**3\n",
+    "T1 = 25 \t\t#Initial temperature of nitrogen in celsius\n",
+    "T2 = 150 \t\t#Final tempareture of nitrogen in celsius\n",
+    "R = 0.2968 \t\t#In kJ/kg-K\n",
+    "Cv = 0.745 \t\t#Constant volume specific heat for nitrogen in kJ/kg-K\n",
+    "W = -20 \t\t#Work done on nitrogen gas in kJ\n",
+    "\n",
+    "#Calculation:\n",
+    "m = P*V/(R*(T1+273))#Mass of nitrogen in kg\n",
+    "Q = m*Cv*(T2-T1)+W #Heat transfer during the process in kJ\n",
+    "\n",
+    "#Result:\n",
+    "print \"Heat transfer for the above process is: \",round( Q,2),\"KJ\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Rate of increase of internal energy is:  246.0 J/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.10\n",
+    "'''During the charging of a storage battery, the current i is 20 A and the voltage e is 12.8\n",
+    "V. The rate of heat transfer from the battery is 10 W. At what rate is the internal energy\n",
+    "increasing?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "W = -12.8*20 \t#Power consumed in  J/s\n",
+    "Q = -10 \t\t#Heat transfer rate from battery in J/s\n",
+    "\n",
+    "#Calculation:\n",
+    "r = Q-W \t\t#Rate of increase of internal energy\n",
+    "\n",
+    "#Result:\n",
+    "print \"Rate of increase of internal energy is: \",round( r,2),\"J/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5.11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Rate of change of temperature is:  0.0828 K/s\n",
+      "Time taken to reach a temperature of T:  11.0 min\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 5.11\n",
+    "'''A 25-kg cast-iron wood-burning stove, shown in Fig. 5.12, contains 5 kg of soft pine wood\n",
+    "and 1 kg of air. All the masses are at room temperature, 20◦C, and pressure, 101 kPa. The\n",
+    "wood now burns and heats all the mass uniformly, releasing 1500W. Neglect any air flow\n",
+    "and changes in mass of wood and heat losses. Find the rate of change of the temperature\n",
+    "(dT/dt) and estimate the time it will take to reach a temperature of 75◦C.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Q = 1500 \t\t#Power produced by burning wood in J/s\n",
+    "mair = 1 \t\t#Mass of air in kg\n",
+    "mwood = 5 \t\t#Mass of soft pine wood in kg \n",
+    "miron = 25 \t\t#Mass of cast iron in kg\n",
+    "Cvair = 0.717 \t#Constant volume specific heat for air in kJ/kg\n",
+    "Cwood = 1.38 \t#Constant volume specific heat for wood in kJ/kg\n",
+    "Ciron = 0.42 \t#Constant volume specific heat for iron in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "dT = 75-20 \t\t#Increase in temperature in Celsius\n",
+    "T = (Q/1000)/(mair*Cvair+mwood*Cwood+miron*Ciron) \t#Rate of change of temperature in K/s\n",
+    "dt = (dT/T)/60 \t#in minutes\n",
+    "\n",
+    "#Results:\n",
+    "print \"Rate of change of temperature is: \",round(T,4),\"K/s\"\n",
+    "print \"Time taken to reach a temperature of T: \",round(dt),\"min\""
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
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+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/6._Elements_of_Airplane_Performance.ipynb b/Introduction_to_flight_by_J_D_Anderson/6._Elements_of_Airplane_Performance.ipynb
new file mode 100644
index 00000000..b3942e06
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/6._Elements_of_Airplane_Performance.ipynb
@@ -0,0 +1,677 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 6: Elements of Airplane Performance"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The mass flow rate is: 0.0055 kg/s.\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.01\n",
+    "''' Air is flowing in a 0.2-m-diameter pipe at a uniform velocity of 0.1 m/s. The temperature\n",
+    "is 25◦C and the pressure is 150 kPa. Determine the mass flow rate. '''\n",
+    "from math import pi\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "R = 0.287 \t\t#in kJ/kg-K\n",
+    "T = 25 \t\t#temperature in celsius\n",
+    "P = 150 \t\t#pressure in kPa\n",
+    "v = R*(T+273.2)/P#specific volume in m**3/kg\n",
+    "D = 0.2 \t\t#diameter of pipe in metre\n",
+    "V = 0.1 \t\t#velocity of air in m/s\n",
+    "\n",
+    "#Calculations:\n",
+    "A = pi*D**2/4 \t#cross sectional area in m**2\n",
+    "m = V*A/v \t\t#mass flow rate in kg/s\n",
+    "\n",
+    "#Results:\n",
+    "print \"The mass flow rate is:\",round(m,4),\"kg/s.\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The work involved in this process is: 0.6 kJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.02\n",
+    "''' Assume we are standing next to the local city’s main water line. The liquid water inside\n",
+    "flows at a pressure of 600 kPa (6 atm) with a temperature of about 10◦C. We want to add\n",
+    "a smaller amount, 1 kg, of liquid to the line through a side pipe and valve mounted on the\n",
+    "main line. How much work will be involved in this process? '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P = 600 \t\t#pressure in kPa\n",
+    "m = 1 \t\t#in kg\n",
+    "v = 0.001 \t\t#specific volume in m**3/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "W = P*m*v \t\t#necessary work in kJ for adding the fluid \n",
+    "\n",
+    "#Results:\n",
+    "print \"The work involved in this process is:\",round(W,2),\"kJ\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The rate at which cooling water flows thorugh the condenser is: 0.919 kg/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.03\n",
+    "''' Consider a water-cooled condenser in a large refrigeration system in which R-134a is the\n",
+    "refrigerant fluid. The refrigerant enters the condenser at 1.0 MPa and 60◦C, at the rate of\n",
+    "0.2 kg/s, and exits as a liquid at 0.95 MPa and 35◦C. Cooling water enters the condenser\n",
+    "at 10◦C and exits at 20◦C. Determine the rate at which cooling water flows through the\n",
+    "condenser.\n",
+    "Control volume: Condenser.\n",
+    "Sketch: Fig. 6.6\n",
+    "Inlet states: R-134a—fixed; water—fixed.\n",
+    "Exit states: R-134a—fixed; water—fixed.\n",
+    "Process: Steady-state.\n",
+    "Model: R-134a tables; steam tables. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "hir = 441.89 \t\t#in kJ/kg for refrigerant using steam table \n",
+    "her = 249.10 \t\t#in kJ/kg for refrigerant using steam table\n",
+    "hiw = 42 \t\t     #in kJ/kg for water using steam table\n",
+    "hew = 83.95 \t     #in kJ/kg for water using steam table\n",
+    "mr = 0.2 \t\t     #the rate at which refrigerant enters the condenser in kg/s\n",
+    "\n",
+    "#Calculations:\n",
+    "mw = mr*(hir-her)/(hew-hiw) \t\t#rate of flow of water in kg/s\n",
+    "\n",
+    "#Results:\n",
+    "print \"The rate at which cooling water flows thorugh the condenser is:\",round( mw,3),\"kg/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The quality is:  0.99\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.04\n",
+    "''' Steam at 0.6 MPa and 200◦C enters an insulated nozzle with a velocity of 50 m/s. It leaves\n",
+    "at a pressure of 0.15 MPa and a velocity of 600 m/s. Determine the final temperature if\n",
+    "the steam is superheated in the final state and the quality if it is saturated.\n",
+    "Control volume: Nozzle.\n",
+    "Inlet state: Fixed (see Fig. 6.7).\n",
+    "Exit state: Pe known.\n",
+    "Process: Steady-state.\n",
+    "Model: Steam tables. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "hi = 2850.1 \t#initial specific heat enthalpy for steam in kJ/kg\n",
+    "Vi = 50 \t\t#initial velocity of steam in m/s\n",
+    "Ve = 600 \t\t#final velocity of steam in m/s\n",
+    "hf = 467.1 \t\t#at final state in kJ/kg\n",
+    "hfg = 2226.5 \t#at final state in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "he = hi+Vi**2/(2*1000)-Ve**2/(2*1000) #final specific heat enthalpy for steam in kJ/kg\n",
+    "xe = (he-hf)/hfg \t#quality of steam in final state\n",
+    "\n",
+    "#Results:\n",
+    "print \"The quality is: \",round(xe,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Quality of the ammonia leaving the expansion valve is: 16.38 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.05\n",
+    "''' Consider the throttling process across the expansion valve or through the capillary tube\n",
+    "in a vapor-compression refrigeration cycle. In this process the pressure of the refrigerant\n",
+    "drops from the high pressure in the condenser to the low pressure in the evaporator,\n",
+    "and during this process some of the liquid flashes into vapor. If we consider this\n",
+    "process to be adiabatic, the quality of the refrigerant entering the evaporator can be\n",
+    "calculated.\n",
+    "Consider the following process, in which ammonia is the refrigerant. The ammonia\n",
+    "enters the expansion valve at a pressure of 1.50MPa and a temperature of 35◦C. Its pressure\n",
+    "on leaving the expansion valve is 291 kPa. Calculate the quality of the ammonia leaving\n",
+    "the expansion valve.\n",
+    "Control volume: Expansion valve or capillary tube.\n",
+    "Inlet state: Pi, Ti known; state fixed.\n",
+    "Exit state: Pe known.\n",
+    "Process: Steady-state.\n",
+    "Model: Ammonia tables. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "hi = 346.8 \t\t#specific heat enthalpy for ammonia at initial state in kJ/kg\n",
+    "hf = 134.4 \t\t#at final state in kJ/kg\n",
+    "hfg = 1296.4\t#at final state in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "he = hi \t\t#specific heat enthalpy for ammonia at final state will be equal that at initial state because it is a throttling process\n",
+    "xe = (he-hf)/hfg \t#quality at final state\n",
+    "\n",
+    "#Results:\n",
+    "print \"Quality of the ammonia leaving the expansion valve is:\",round(xe*100,2),\"%\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The power output of the turbine is: 678.2 kW\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example:\n",
+    "''' The mass rate of flow into a steam turbine is 1.5 kg/s, and the heat transfer from the\n",
+    "turbine is 8.5 kW. The following data are known for the steam entering and leaving the\n",
+    "turbine.\n",
+    "            Inlet Conditions                Exit Conditions\n",
+    "Pressure        2.0 MPa                         0.1 MPa\n",
+    "Temperature     350◦C\n",
+    "Quality                                         100%\n",
+    "Velocity        50 m/s                          100 m/s\n",
+    "Elevation above 6 m                                3 m\n",
+    "reference plane\n",
+    "\n",
+    "g = 9.8066 m/s2\n",
+    "\n",
+    "Determine the power output of the turbine.\n",
+    "Control volume:         Turbine (Fig. 6.9).\n",
+    "Inlet state:            Fixed (above).\n",
+    "Exit state:             Fixed (above).\n",
+    "Process:                Steady-state.\n",
+    "Model:                  Steam tables. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "hi = 3137 \t\t#initial specific heat of enthalpy in kJ/kg\n",
+    "he = 2675.5 \t#final specific heat of enthalpy in kJ/kg\n",
+    "Vi = 50 \t\t#initial velocity of steam in m/s\n",
+    "Ve = 100 \t\t#final velocity of steam in m/s\n",
+    "Zi = 6 \t\t#height of inlet conditions in metres\n",
+    "Ze = 3 \t\t#height of exit conditions in metres\n",
+    "m = 1.5 \t\t#mass flow rate of steam in kg/s\n",
+    "g = 9.8066 \t\t#acc. due to gravity in m/s**2\n",
+    "Qcv = -8.5 \t\t#heat transfer rate from turbine in kW\n",
+    "\n",
+    "#Calculations:\n",
+    "Wcv = Qcv+m*(hi+Vi**2/(2*1000)+g*Zi/1000)-m*(he+Ve**2/(2*1000)+g*Ze/1000) \t#power output of turbine in kW\n",
+    "\n",
+    "#Results:\n",
+    "print \"The power output of the turbine is:\",round(Wcv,1),\"kW\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Heat transfer rate in the aftercooler is: 35.2 kW\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.07\n",
+    "''' The compressor in a plant (see Fig. 6.10) receives carbon dioxide at 100 kPa, 280 K,\n",
+    "with a low velocity. At the compressor discharge, the carbon dioxide exits at 1100 kPa,\n",
+    "500 K, with velocity of 25 m/s and then flows into a constant-pressure aftercooler (heat\n",
+    "exchanger) where it is cooled down to 350 K. The power input to the compressor is\n",
+    "50 kW. Determine the heat transfer rate in the aftercooler. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "V1 = 0 \t\t#we assume initial velocity to be zero because its given that it enters with a low velocity\n",
+    "V2 = 25 \t\t#final velocity with which carbon dioxide exits in m/s\n",
+    "h2 = 401.52 \t#final specific enthalpy of heat when carbon dioxide exits in kJ/kg\n",
+    "h1 = 198 \t\t#initial specific enthalpy of heat in kJ/kg\n",
+    "Wc = -50 \t\t#power input to the compressor in kW\n",
+    "h3 = 257.9 \t\t#final specific enthalpy of heat when carbon dioxide flows into a constant pressure aftercooler\n",
+    "\n",
+    "#Calculations:\n",
+    "w = h1-h2-V2**2/(2*1000)#in kJ/kg\n",
+    "m = Wc/w \t\t#mass flow rate of carbon dioxide in kg/s\n",
+    "Qcool = -m*(h3-h2) \t\t#heat transfer rate in the aftercooler in kW\n",
+    "\n",
+    "#Results:\n",
+    "print \"Heat transfer rate in the aftercooler is:\",round(Qcool,1),\"kW\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The pump requires power input of 840.0 W\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.08\n",
+    "''' A small liquid water pump is located 15 m down in a well (see Fig. 6.11), taking water\n",
+    "in at 10◦C, 90 kPa at a rate of 1.5 kg/s. The exit line is a pipe of diameter 0.04 m that\n",
+    "goes up to a receiver tank maintaining a gauge pressure of 400 kPa. Assume the process\n",
+    "is adiabatic with the same inlet and exit velocities and the water stays at 10◦C. Find the\n",
+    "required pump work. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "m = 1.5 \t\t#mass flow rate of water in kg/s\n",
+    "g = 9.807 \t\t#acceleration due to gravity in m/s**2\n",
+    "Zin = -15 \t\t#depth of water pump in well in metres\n",
+    "Zex = 0 \t\t#in metres\n",
+    "v = 0.001001 \t#specific volume in m**3/kg\n",
+    "Pin = 90 \t\t#in kPa\n",
+    "\n",
+    "#Calculations:\n",
+    "Pex = 400+101.3 \t#exit pressure in kPa\n",
+    "W = m*(g*(Zin-Zex)*0.001-(Pex-Pin)*v) \t\t#power input in kW\n",
+    "\n",
+    "#Results:\n",
+    "print \"The pump requires power input of\",-round(W,2)*1000,\"W\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Heat transfer in line between boiler and turbine is: -21.0 kJ/kg\n",
+      "Turbine work is: 640.7 kJ/kg\n",
+      "Heat transfer in condenser is: -2173.3 kJ/kg\n",
+      "Heat transfer in boiler is: 2831.0 kJ/kg. \n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.09\n",
+    "''' Consider the simple steam power plant, as shown in Fig. 6.12. The following data are for\n",
+    "such a power plant.\n",
+    "\n",
+    "Location                            Pressure            Temperature or Quality\n",
+    "Leaving boiler                      2.0 MPa                     300◦C\n",
+    "Entering turbine                    1.9 MPa                     290◦C\n",
+    "Leaving turbine,entering condenser  15 kPa                      90%\n",
+    "Leaving condenser,entering pump     14 kPa                      45◦C\n",
+    "Pump work = 4 kJ/kg\n",
+    "\n",
+    "Determine the following quantities per kilogram flowing through the unit:\n",
+    "a. Heat transfer in the line between the boiler and turbine.\n",
+    "b. Turbine work.\n",
+    "c. Heat transfer in the condenser.\n",
+    "d. Heat transfer in the boiler. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "h1 = 3023.5 \t\t#specific heat of enthalpy of steam leaving boiler in kJ/kg\n",
+    "h2 = 3002.5 \t\t#specific heat of enthalpy of steam entering turbine in kJ/kg\n",
+    "x = 0.9 \t\t     #quality of steam entering condenser\n",
+    "hf = 226 \t\t     #in kJ/kg\n",
+    "hfg = 2373.1 \t\t#in kJ/kg\n",
+    "h4 = 188.5 \t\t     #specific heat of enthalpy of steam entering pump in kJ/kg\n",
+    "w45 = -4 \t\t     #pump work in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "h3 = hf+x*hfg \t\t#specific heat of enthalpy of steam entering condenser in kJ/kg\n",
+    "q12 = h2-h1 \t\t#heat transfer in line between boiler and turbine in kJ/kg\n",
+    "w23 = h2-h3 \t\t#turbine work in kJ/kg\n",
+    "q34 = h4-h3 \t\t#heat transfer in condenser\n",
+    "h5 = h4-w45 \t\t#in kJ/kg\n",
+    "q51 = h1-h5 \t\t#heat transfer in boiler in kJ/kg\n",
+    "\n",
+    "#Results:\n",
+    "print \"Heat transfer in line between boiler and turbine is:\",round(q12,2),\"kJ/kg\"\n",
+    "print \"Turbine work is:\",round(w23,1),\"kJ/kg\"\n",
+    "print \"Heat transfer in condenser is:\",round(q34,1),\"kJ/kg\"\n",
+    "print \"Heat transfer in boiler is:\",round(q51,2),\"kJ/kg. \""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "The quality at the evaporator inlet is: 0.345\n",
+      "The rate of heat transfer to the evaporator is: 14.54 kW. \n",
+      "Rate of heat transfer from the compressor is: -0.21 kW. \n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.10\n",
+    "''' The refrigerator shown in Fig. 6.13 uses R-134a as the working fluid. The mass flow rate\n",
+    "through each component is 0.1 kg/s, and the power input to the compressor is 5.0 kW. The\n",
+    "following state data are known, using the state notation of Fig. 6.13:\n",
+    "P1 = 100 kPa,       T1 = −20◦C\n",
+    "P2 = 800 kPa,       T2 = 50◦C\n",
+    "T3 = 30◦C,          x3 = 0.0\n",
+    "T4 = −25◦C\n",
+    "\n",
+    "Determine the following:\n",
+    "a. The quality at the evaporator inlet.\n",
+    "b. The rate of heat transfer to the evaporator.\n",
+    "c. The rate of heat transfer from the compressor. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "hf4 = 167.4 \t#in kJ/kg\n",
+    "hfg4 = 215.6 \t#in kJ/kg\n",
+    "h3 = 241.8 \t\t#specific heat of enthalpy of R-134a entering expansion valve\n",
+    "h4 = h3 \t\t#specific heat of enthalpy of R-134a leaving expansion valve\n",
+    "h1 = 387.2 \t\t#in kJ/kg\n",
+    "h2 = 435.1 \t\t#in kJ/kg\n",
+    "m = 0.1 \t\t#mass flow rate in kg/s\n",
+    "Wcomp = -5 \t\t#power input to compressor in kW\n",
+    "\n",
+    "#Calculations:\n",
+    "x4 = (h3-hf4)/hfg4 #quality of R-134a at evaporator inlet\n",
+    "Qevap = m*(h1-h4) #rate of heat transfer to the evaporator\n",
+    "Qcomp = m*(h2-h1)+Wcomp #rate of heat transfer from compressor\n",
+    "\n",
+    "#Results:\n",
+    "print \"The quality at the evaporator inlet is:\",round(x4,3)\n",
+    "print \"The rate of heat transfer to the evaporator is:\",round(Qevap,2),\"kW. \"\n",
+    "print \"Rate of heat transfer from the compressor is:\",round(Qcomp,2),\"kW. \""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Mass of the steam that flows into the tank is: 1.263 kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.12\n",
+    "''' Let the tank of the previous example have a volume of 0.4m3 and initially contain saturated\n",
+    "vapor at 350 kPa. The valve is then opened, and steam from the line at 1.4 MPa and 300◦C\n",
+    "flows into the tank until the pressure is 1.4 MPa.\n",
+    "Calculate the mass of steam that flows into the tank. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "V1 = 0.4 \t\t#initial volume fo tank in m**3\n",
+    "v1 = 0.5243 \t#initial specific volume in m**3/kg\n",
+    "h1 = 3040.4 \t#initial specific enthalpy in kJ/kg\n",
+    "u1 = 2548.9 \t#initial specific internal energy in kJ/kg\n",
+    "V2 = 0.4 \t\t#final volume in m**3\n",
+    "T2 = 342 \t\t#final temperature in Celsius by interpolation\n",
+    "v2 = 0.1974 \t#final specific volume in m**3/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "m1 = V1/v1 \t\t#initial mass of steam in tank in kg\n",
+    "m2 = V2/v2 \t\t#final mass of the steam in the tank in kg\n",
+    "m = m2-m1 \t\t#mass of steam that flowsinto the tank\n",
+    "\n",
+    "#Results:\n",
+    "print \"Mass of the steam that flows into the tank is:\",round(m,3),\"kg\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6.13"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 12,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Mass of ammonia withdrawn is: 72.7 kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 6.13\n",
+    "''' A tank of 2 m3 volume contains saturated ammonia at a temperature of 40◦C. Initially the\n",
+    "tank contains 50% liquid and 50% vapor by volume. Vapor is withdrawn from the top of\n",
+    "the tank until the temperature is 10◦C. Assuming that only vapor (i.e., no liquid) leaves\n",
+    "and that the process is adiabatic, calculate the mass of ammonia that is withdrawn. '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "vf1 = 0.001725 \t\t#in m**3/kg\n",
+    "vf2 = 0.0016 \t\t#in m**3/kg\n",
+    "uf1 = 368.7 \t\t#in kJ/kg\n",
+    "uf2 = 226 \t\t     #in kJ/kg\n",
+    "vg1 = 0.08313 \t\t#in m**3/kg\n",
+    "vfg2 = 0.20381\n",
+    "ug1 = 1341 \t\t      #in kJ/kg\n",
+    "ufg2 = 1099.7 \t\t#in kJ/kg\n",
+    "Vf = 1 \t\t      #initial volume of liquid in m**3\n",
+    "Vg = 1 \t\t      #initial volume of vapor in m**3\n",
+    "he = 1461.1 \t\t#in kJ/kg\n",
+    "V = 2 \t\t      #volume of tank in m**3\n",
+    "\n",
+    "#Calculations:\n",
+    "mf1 = Vf/vf1 \t\t#initial mass of liquid in kg\n",
+    "mg1 = Vg/vg1 \t\t#initial mass of vapor in kg\n",
+    "m1 = mf1+mg1 \t\t#initial mass of liquid in kg\n",
+    "x2 = ((2*1461.1)-(2*226)-(0.00160*634706))/((634706*0.20381)+(2*1099.7)) \t\t#quality of ammonia\n",
+    "v2 = 0.00160+(0.20381*x2) #final specific volume in m**3/kg\n",
+    "m2 = V/v2 \t\t    #final mass of ammonia in kg\n",
+    "m = m1-m2 \t\t    #mass of ammonia withdrawn\n",
+    "\n",
+    "#Results:\n",
+    "print \"Mass of ammonia withdrawn is:\",round(m,1),\"kg\""
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/7._Principles_of_Stability_and_Control.ipynb b/Introduction_to_flight_by_J_D_Anderson/7._Principles_of_Stability_and_Control.ipynb
new file mode 100644
index 00000000..2057ecdb
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/7._Principles_of_Stability_and_Control.ipynb
@@ -0,0 +1,278 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "# Chapter 7: Principles of Stability and Control"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## Example 7.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total rate of energy rejected is:  233.0 KW\n",
+      "Rate of fuel consumption is:  0.0095 Kg/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 7.1\n",
+    "'''An automobile engine produces 136 hp on the output shaft with a thermal efficiency of\n",
+    "30%. The fuel it burns gives 35 000 kJ/kg as energy release. Find the total rate of energy\n",
+    "rejected to the ambient and the rate of fuel consumption in kg/s.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "W = 136*0.7355 \t#Output of automobile engine in kW\n",
+    "neng = 0.3 \t\t#Thermal efficiency of automobile engine\n",
+    "qh = 35000 \t\t#Energy output of fuel in kJ/kg\n",
+    "\n",
+    "#Calculation:\n",
+    "Qh = W/neng \t#Energy output of fuel in kW\n",
+    "Ql = Qh-W \t\t#Total rate of energy rejected to the ambient\n",
+    "m = Qh/qh \t\t#Rate of fuel consumption in kg/s\n",
+    "\n",
+    "#Result:\n",
+    "print \"Total rate of energy rejected is: \",round(Ql),\"KW\" \n",
+    "print \"Rate of fuel consumption is: \",round(m,4),\"Kg/s\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## Example 7.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Rate of energy taken out of the cold space is:  250.0 W\n",
+      "Coefficient of performance of the refrigerator is:  1.67\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 7.2\n",
+    "'''The refrigerator in a kitchen shown in Fig. 7.7 receives electrical input power of 150 W\n",
+    "to drive the system, and it rejects 400 W to the kitchen air. Find the rate of energy taken\n",
+    "out of the cold space and the COP of the refrigerator.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Qh = 400 \t\t#Heat rejected to kitchen air in W\n",
+    "W = 150 \t\t#Electrical input power in W\n",
+    "\n",
+    "#Calculation:\n",
+    "Ql = Qh-W \t\t#Rate of energy taken out to cold space in W\n",
+    "B = Ql/W \t\t#Coefficicent of performnace of refrigerator\n",
+    "\n",
+    "#Result:\n",
+    "print \"Rate of energy taken out of the cold space is: \",round(Ql,2),\"W\"\n",
+    "print \"Coefficient of performance of the refrigerator is: \",round(B,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## Example 7.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Celcius temperature at which pressure go to zero:  -273.15 °C\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 7.3\n",
+    "'''In a certain constant-volume ideal-gas thermometer, the measured pressure at the ice point\n",
+    "(see Section 2.11) of water, 0◦C, is 110.9 kPa and at the steam point, 100◦C, is 151.5 kPa.\n",
+    "Extrapolating, at what Celsius temperature does the pressure go to zero (i.e., zero absolute\n",
+    "temperature)?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P1 = 110.9          #Pressure at 1 in Kpa\n",
+    "P2 = 151.5          #Pressure at 2 in Kpa\n",
+    "T1 = 0              #Temperature at 1 in °C\n",
+    "T2 = 100            #Temperature at 2 in °C\n",
+    "P = 0               #pressure at which temperature is to be found out(KPa)\n",
+    "\n",
+    "#Calculation:\n",
+    "C =P1              #y-intercept\n",
+    "m = (P2-P1)/(T2-T1)#Slope of P-T graph\n",
+    "T = (P-C)/m        #Interpolating straight line P=mT+C and finding T in °C\n",
+    "\n",
+    "#Result:\n",
+    "print 'Celcius temperature at which pressure go to zero: ',round(T,2),\"°C\" "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## Example 7.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Engine efficiency is:  0.635\n",
+      "Energy discarded to the ambient surroundings is:  365.0 KW\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 7.4\n",
+    "'''Let us consider the heat engine, shown schematically in Fig. 7.25, that receives a heattransfer\n",
+    "rate of 1 MW at a high temperature of 550◦C and rejects energy to the ambient\n",
+    "surroundings at 300 K. Work is produced at a rate of 450 kW. We would like to know\n",
+    "how much energy is discarded to the ambient surroundings and the engine efficiency and\n",
+    "compare both of these to a Carnot heat engine operating between the same two reservoirs.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Qh = 1000 \t\t#Rate of heat transfer to heat engine in kW\n",
+    "W = 450 \t\t#Rate of production of work in kW\n",
+    "Tl = 300 \t\t#Temperature of surroundings in K\n",
+    "Th = 550 \t\t#Temperature of heat source in Celsius\n",
+    "\n",
+    "#Calculation:\n",
+    "Ql = Qh-W \t\t#Rate of heat rejected by heat engine in kW\n",
+    "nthermal = W/Qh \t#Efficiency from the definition of efficiency\n",
+    "ncarnot = 1-Tl/(Th+273)#Efficiency if heat engine is considered to be ideal carnot heat engine\n",
+    "W2 = ncarnot*Qh \t#Rate of work production if heat engine is assumed to be ideal carnot heat engine in kW\n",
+    "Ql2 = Qh-W2 \t#Rate of heat rejected by heat engine in kW if heat engine is assumed to be ideal carnot heat engine\n",
+    "\n",
+    "#Result:\n",
+    "print \"Engine efficiency is: \",round(ncarnot,3)\n",
+    "print \"Energy discarded to the ambient surroundings is: \",round(Ql2),\"KW\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "## Example 7.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Magnitude of reqiured work is:  0.15 KW\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 7.5\n",
+    "'''As one mode of operation of an air conditioner is the cooling of a room on a hot day, it\n",
+    "works as a refrigerator, shown in Fig. 7.26.Atotal of 4kWshould be removed from a room\n",
+    "at 24◦C to the outside atmosphere at 35◦C.We would like to estimate the magnitude of the\n",
+    "required work. To do this we will not analyze the processes inside the refrigerator, which\n",
+    "is deferred to Chapter 11, but we can give a lower limit for the rate of work, assuming it\n",
+    "is a Carnot-cycle refrigerator.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Tl = 24+273 \t\t#Room temperature in Kelvins\n",
+    "Th = 35+273 \t\t#Atmospheric temperature in Kelvins\n",
+    "Ql = 4 \t\t      #Rate of heat rejection from room\n",
+    "\n",
+    "#Calculation:\n",
+    "B = Tl/(Th-Tl) \t\t#Coefficient of performance of air conditioner\n",
+    "W = Ql/B \t\t      #Required work in kW\n",
+    "\n",
+    "#Result:\n",
+    "print \"Magnitude of reqiured work is: \",round(W,2),\"KW\""
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/8._Space_Flight_(Astronautics).ipynb b/Introduction_to_flight_by_J_D_Anderson/8._Space_Flight_(Astronautics).ipynb
new file mode 100644
index 00000000..85beaf65
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/8._Space_Flight_(Astronautics).ipynb
@@ -0,0 +1,548 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 8: Space Flight (Astronautics)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Pressure after compression is P2:  1487.1 KPa\n",
+      "Coefficient of performance of refrigerator:  5.55\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 8.1\n",
+    "''' Consider a Carnot-cycle heat pump with R-134a as the working fluid. Heat is absorbed\n",
+    "into the R-134a at 0◦C, during which process it changes from a two-phase state to saturated\n",
+    "vapor. The heat is rejected from the R-134a at 60◦C and ends up as saturated liquid. Find\n",
+    "the pressure after compression, before the heat rejection process, and determine the COP\n",
+    "for the cycle.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Th = 60 \t\t\t#temperature at which heat is rejected from R-134a\n",
+    "Tl = 0 \t\t\t\t#temperature at which heat is absorbed into the R-134a \n",
+    "s1 = 1.7262 \t\t#specific entropy at 0 Celsius\n",
+    "s3 = 1.2857 \t\t#specific entropy at 60 celsius\n",
+    "s4 = s3 \t\t\t#process of state change from 3-4 is isentropic\n",
+    "\n",
+    "#Calculations:\n",
+    "s2 = s1 \t\t\t#process of state change from 1-2 is isentropic \n",
+    "P2 = 1400+(1600-1400)*(1.7262-1.736)/(1.7135-1.736) \t\t#pressure after compression in kPa\n",
+    "B = (Th+273)/(Th-Tl)#coefficient of performance of refrigerator\n",
+    "\n",
+    "#Results:\n",
+    "print \"Pressure after compression is P2: \",round(P2,1) ,'KPa'\n",
+    "print \"Coefficient of performance of refrigerator: \",round(B,2)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work required to accomplish the process:  73.7 KJ\n",
+      "Heat transfer:  277.9 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 8.2\n",
+    "''' A cylinder/piston setup contains 1 L of saturated liquid refrigerant R-410a at 20◦C.\n",
+    "The piston now slowly expands, maintaining constant temperature to a final pressure of\n",
+    "400 kPa in a reversible process. Calculate the work and heat transfer required to accom-\n",
+    "plish this process'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "u1 = 87.94 \t\t\t\t\t#specific internal energy of R-12 at state 1 in kJ/kg\n",
+    "u2 = 276.44 \t\t\t\t#specific internal energy of R-12 at state 2 in kJ/kg\n",
+    "s1 = 0.3357 \t\t\t\t#specific entropy at state 1 in kJ/kg-K\n",
+    "s2 = 1.2108 \t\t\t\t#specific entropy at state 2 in kJ/kg-K\n",
+    "V = 0.001 \t\t\t\t\t#volume of saturated liquid in m**3\n",
+    "v1 = 0.000923 \t\t\t\t#specific volume in m**3/kg\n",
+    "T = 20 \t\t\t\t\t\t#temperature of liquid in celsius\n",
+    "\n",
+    "#Calculations:\n",
+    "m = V/v1 \t\t\t\t\t#mass of saturated liquid in kg\n",
+    "Q12 = m*(T+273.15)*(s2-s1) \t#heat transfer in kJ to accomplish the process\n",
+    "W12 = m*(u1-u2)+Q12 \t\t#work required to accomplish the process\n",
+    "\n",
+    "#Results\n",
+    "print \"Work required to accomplish the process: \",round(W12,1),'KJ'\n",
+    "print \"Heat transfer: \",round(Q12,1),'KJ'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Change in entropy assuming constant specific heat:  0.8958 KJ/Kg-K\n",
+      "Using steam table:  0.8959 KJ/Kg-K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import log\n",
+    "#Example: 8.3\n",
+    "''' One kilogram of liquid water is heated from 20◦C to 90◦C. Calculate the entropy change,\n",
+    "assuming constant specific heat, and compare the result with that found when using the\n",
+    "steam tables.\n",
+    "Control mass: Water.\n",
+    "Initial and final states: Known.\n",
+    "Model: Constant specific heat, value at room temperature.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "C = 4.184 \t\t# specific heat of water in kJ/kg-K\n",
+    "T1 = 20 \t\t#initial temperature of water in celsius\n",
+    "T2 = 90 \t\t#final temperature of water in celsius\n",
+    "\n",
+    "#Calculations:\n",
+    "dS1 = C*log((T2+273.2)/(T1+273.2)) \t\t#change in entropy in kJ/kg-K\n",
+    "dS2 = 1.1925-0.2966 \t\t#in kJ/kg-K using steam tables\n",
+    "\n",
+    "#Results\n",
+    "print \"Change in entropy assuming constant specific heat: \",round(dS1,4),'KJ/Kg-K'\n",
+    "print \"Using steam table: \",round(dS2,4) ,'KJ/Kg-K'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Change in entropy using ideal gas tables:  1.7228 KJ/Kg-K\n",
+      "Change in entropy using empirical equation:  1.7058 KJ/Kg-K\n",
+      "Change in entropy using the value of specific heat at 300K:  1.5586 KJ/Kg-K\n",
+      "Change in entropy assuming specific heat is constant at its value at 900K:  1.8076 KJ/Kg-K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import log\n",
+    "from sympy import symbols, integrate\n",
+    "#Example: 8.4\n",
+    "''' Consider Example 5.7, in which oxygen is heated from 300 to 1500 K. Assume that during\n",
+    "this process the pressure dropped from 200 to 150 kPa. Calculate the change in entropy\n",
+    "per kilogram.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "T1 = 300 \t\t#initial temperature in kelvins\n",
+    "T2 = 1500 \t\t#final temperature in kelvins\n",
+    "P1 = 200 \t\t#initial pressure in kPa\n",
+    "P2 = 150 \t\t#final pressure in kPa\n",
+    "R = 0.2598 \t\t# in kJ/kg-K\n",
+    "Cp = 0.922 \t\t#specific heat in kJ/kg-K at constant pressure\n",
+    "dsT2 = 8.0649 \t#in kJ/kg-K\n",
+    "dsT1 = 6.4168 \t#in kJ/kg-K\n",
+    "x = symbols('x')#Symbolic representation of theta \n",
+    "\n",
+    "#Calculations:\n",
+    "dS1 = dsT2-dsT1-R*log(P2/P1) \t\t#entropy change calculated using ideal gas tables\n",
+    "dS2 = integrate(0.88/x-0.0001+0.54*x-0.33*x**2,(x,0.3,1.5))-R*log(P2/P1) \t\t#entropy change calculated using empirical equation \n",
+    "dS3 = Cp*log(T2/T1)-R*log(P2/P1) \t#entropy change assuming constant specific heat in kJ/kg-K\n",
+    "dS4 = 1.0767*log(T2/T1)+0.0747 \t\t#entropy change assuming specific heat is constant at its value at 990K\n",
+    "\n",
+    "#Results\n",
+    "print \"Change in entropy using ideal gas tables: \",round(dS1,4),'KJ/Kg-K'\n",
+    "print \"Change in entropy using empirical equation: \",round(dS2,4),'KJ/Kg-K'\n",
+    "print \"Change in entropy using the value of specific heat at 300K: \",round(dS3,4),'KJ/Kg-K'\n",
+    "print \"Change in entropy assuming specific heat is constant at its value at 900K: \",round(dS4,4),'KJ/Kg-K'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Entropy change assuming constant specific heat:  0.7785 KJ/Kg-K\n",
+      "Assuming variable specific heat:  0.7897 KJ/Kg-K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import log\n",
+    "#Example: 8.5\n",
+    "''' Calculate the change in entropy per kilogram as air is heated from 300 to 600 K while\n",
+    "pressure drops from 400 to 300 kPa. Assume:\n",
+    "1. Constant specific heat.\n",
+    "2. Variable specific heat.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Cp = 1.004 \t\t\t#specific heat at constant pressure in kJ/kg-K\n",
+    "R = 0.287 \t\t\t#gas constant in kJ/kg-K\n",
+    "P1 = 400 \t\t\t#initial pressure in kPa\n",
+    "P2 = 300 \t\t\t#final pressure in kPa\n",
+    "T1 = 300 \t\t\t#initial temperature in K\n",
+    "T2 = 600 \t\t\t#final temperature in K\n",
+    "s1 = 6.8693 \t\t#specific entropy at T1\n",
+    "s2 = 7.5764 \t\t#specific entropy at T2\n",
+    "\n",
+    "#Calculations:\n",
+    "dS1 = Cp*log(T2/T1)-R*log(P2/P1)#entropy change assuming constant specific heat\n",
+    "dS2 = s2-s1-R*log(P2/P1) \t\t#entropy change assuming variable specific heat\n",
+    "\n",
+    "#Results\n",
+    "print \"Entropy change assuming constant specific heat: \",round(dS1,4),'KJ/Kg-K'\n",
+    "print \"Assuming variable specific heat: \",round(dS2,4),'KJ/Kg-K'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work done by air:  106.96 KJ/Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import log\n",
+    "#Example: 8.6\n",
+    "''' One kilogram of air is contained in a cylinder fitted with a piston at a pressure of 400\n",
+    "kPa and a temperature of 600 K. The air is expanded to 150 kPa in a reversible adiabatic\n",
+    "process. Calculate the work done by the air.\n",
+    "Control mass: Air.\n",
+    "Initial state: P 1 , T 1 ; state 1 fixed.\n",
+    "Final state: P 2 .\n",
+    "Process: Reversible and adiabatic.\n",
+    "Model: Ideal gas and air tables, Table A.7.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "T1 = 600 \t\t\t\t#initial temperature of air in K\n",
+    "P1 = 400 \t\t\t\t#intial pressure of air in kPa\n",
+    "P2 = 150 \t\t\t\t#final pressure in kPa\n",
+    "u1 = 435.10 \t\t\t#specific internal energy at temperature T1 in kJ/kg\n",
+    "sT1 = 7.5764 \t\t\t#specific entropy at temperature T1 in kJ/kg-K\n",
+    "R = 0.287 \t\t\t\t#gas constant in kJ/kg-K\n",
+    "ds = 0\n",
+    "T2 = 457 \t\t\t\t#final temperature in K\n",
+    "u2 = 328.14 \t\t\t#specific internal energy at temperature T2 in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "sT2 = ds+sT1+R*log(P2/P1)#specific entropy at temperature T2 in kJ/kg-K\n",
+    "w = u1-u2 \t\t\t\t#work done by air in kJ/kg\n",
+    "\n",
+    "#Results\n",
+    "print \"Work done by air: \",round(w,2),'KJ/Kg'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work done is w12:  -130.5 KJ/Kg\n",
+      "Heat transfer are:  -32.2 KJ/Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 8.7\n",
+    "''' In a reversible process, nitrogen is compressed in a cylinder from 100 kPa and 20◦\n",
+    "C to 500 kPa. During this compression process, the relation between pressure and volume is\n",
+    "PV 1.3 = constant. Calculate the work and heat transfer per kilogram, and show this process\n",
+    "on P–v and T–s diagrams.\n",
+    "Control mass: Nitrogen.\n",
+    "Initial state: P 1 , T 1 ; state 1 known.\n",
+    "Final state: P 2 .\n",
+    "Process: Reversible, polytropic with exponent n < k.\n",
+    "Diagram: Fig. 8.14.\n",
+    "Model: Ideal gas, constant specific heat—value at 300 K.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P2 = 500  \t\t\t\t\t#final pressure in cylinder in kPa\n",
+    "P1 = 100 \t\t\t\t\t#initial pressure in cylinder in kPa\n",
+    "T1 = 20+273.2 \t\t\t\t#initial temperature inside cylinder in Kelvins\n",
+    "n = 1.3 \n",
+    "R = 0.2968 \t\t\t\t\t#gas constant in kJ/kg-K\n",
+    "Cvo = 0.745 \t\t\t\t#specific heat at constant volume in kJ/kg-K\n",
+    "\n",
+    "#Calculations:\n",
+    "T2 = (T1)*(P2/P1)**((n-1)/n)#final temperature inside cylinder in K\n",
+    "w12 = R*(T2-T1)/(1-n) \t\t#work in kJ/kg\n",
+    "q12 = Cvo*(T2-T1)+w12 \t\t#heat transfer in kJ/kg\n",
+    "\n",
+    "#Results\n",
+    "print \"Work done is w12: \",round(w12,1),\"KJ/Kg\"\n",
+    "print \"Heat transfer are: \",round(q12,1),\"KJ/Kg\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Increase in entropy of water plus surr: 1.522 KJ/K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 8.8\n",
+    "'''Consider an electric space heater that converts 1 kW of electric power into a heat flux of\n",
+    "1 kW delivered at 600 K from the hot wire surface. Let us look at the process of the energy\n",
+    "conversion from electricity to heat transfer and find the rate of total entropy generation.\n",
+    "Control mass: The electric heater wire.\n",
+    "State: Constant wire temperature 600 K.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "m = 1 \t\t\t\t\t#mass of saturated water vapour\n",
+    "sfg = 6.0480 \t\t\t#in kJ/K\n",
+    "T = 25 \t\t\t\t\t#temperature of surrounding air in celsius\n",
+    "hfg = 2257.0 \t\t\t#in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "dScm = -m*sfg \t\t\t#change in entropy of control mass in kJ/K\n",
+    "Qsurr = m*hfg \t\t\t#heat transferred to surr in kJ\n",
+    "dSsurr = Qsurr/(T+273.15)#in kJ/K\n",
+    "dSnet = dScm+dSsurr\t\t#net increase in entropy in kJ/K\n",
+    "\n",
+    "#Results:\n",
+    "print \"Increase in entropy of water plus surr:\",round(dSnet,4)  ,'KJ/K'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Entropy generation is:  0.00167 KW/K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 8.9\n",
+    "''' Consider an electric space heater that converts 1 kW of electric power into a heat flux of\n",
+    "1 kW delivered at 600 K from the hot wire surface. Let us look at the process of the energy\n",
+    "conversion from electricity to heat transfer and find the rate of total entropy generation.\n",
+    "Control mass: The electric heater wire.\n",
+    "State: Constant wire temperature 600 K.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Qout = 1 \t\t#value of heat flux generated by 1kW of electric power\n",
+    "T = 600 \t\t#temperature of hot wire surface in K\n",
+    "\n",
+    "#Calculations:\n",
+    "Sgen = Qout/T \t\t#entropy generation in kW/K\n",
+    "\n",
+    "#Results\n",
+    "print \"Entropy generation is: \",round(Sgen,5),'KW/K'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total entropy generated:  29.3 W/K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 8.10\n",
+    "''' Consider a modern air conditioner using R-410a working in heat pump mode, as shown in\n",
+    "Fig. 8.21. It has a COP of 4 with 10 kW of power input. The cold side is buried underground,\n",
+    "where it is 8◦C, and the hot side is a house kept at 21◦C. For simplicity, assume that the\n",
+    "cycle has a high temperature of 50◦C and a low temperature of −10◦C (recall Section\n",
+    "7.10). We would like to know where entropy is generated associated with the heat pump,\n",
+    "assuming steady-state operation.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "B = 4 \t\t\t\t\t#COP of air conditioner\n",
+    "W = 10 \t\t\t\t\t#power input of air conditioner in kW\n",
+    "Thigh = 323 \t\t\t#in Kelvin\n",
+    "Tlow = 263 \t\t\t\t#in Kelvin\n",
+    "Tl = 281 \t\t\t\t#in K\n",
+    "Th = 294 \t\t\t\t#in K\n",
+    "\n",
+    "#Calculations:\n",
+    "Qh = B*W \t\t\t\t#in kW\n",
+    "Ql = Qh-W \t\t\t\t#in kW\n",
+    "SgenHP = (Qh*1000/Thigh)-(Ql*1000/Tlow)\t#in W/K\n",
+    "SgenCV1 = Ql*1000/Tlow-Ql*1000/Tl \t\t#in W/K\n",
+    "SgenCV2 = Qh*1000/Th-Qh*1000/Thigh \t\t#in W/K\n",
+    "SgenTOT = SgenCV1+SgenCV2+SgenHP \t\t#in W/K\n",
+    "\n",
+    "#Results\n",
+    "print \"Total entropy generated: \",round(SgenTOT,1),'W/K'"
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/9._Propulsion.ipynb b/Introduction_to_flight_by_J_D_Anderson/9._Propulsion.ipynb
new file mode 100644
index 00000000..c24ed6eb
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/9._Propulsion.ipynb
@@ -0,0 +1,585 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 9: Propulsion"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work per kilogram of steam for this isentropic process:  377.5 KJ/Kg-K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 9.1\n",
+    "'''Steam enters a steam turbine at a pressure of 1 MPa, a temperature of 300\n",
+    "◦C, and avelocity of 50 m/s. The steam leaves the turbine at a pressure of 150 kPa and a velocity of\n",
+    "200 m/s. Determine the work per kilogram of steam flowing through the turbine, assuming\n",
+    "the process to be reversible and adiabatic.\n",
+    "Control volume: Turbine.\n",
+    "Sketch: Fig. 9.2.\n",
+    "Inlet state: Fixed (Fig. 9.2).\n",
+    "Exit state: P e , V e known.\n",
+    "Process: Steady state, reversible and adiabatic.\n",
+    "Model: Steam tables.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "hi = 3051.2 \t\t#initial specific heat of enthalpy of steam in kJ/kg\n",
+    "si = 7.1228 \t\t#initial specific entropy of steam in kJ/kg-K\n",
+    "Pe = 0.15 \t\t\t#final pressure in MPa\n",
+    "se = si \t\t\t#specific entropy in final state in kJ/kg-K\n",
+    "sf = 1.4335 \t\t#in kJ/kg-K\n",
+    "sfg = 5.7897 \t\t#in kJ/kg-K\n",
+    "vi = 50 \t\t\t#velocity with which steam enters turbine in m/s\n",
+    "ve = 200 \t\t\t#velocity with which steam leaves the turbine in m/s\n",
+    "hf = 467.1 \t\t\t#in kJ/kg\n",
+    "hfg = 2226.5 \t\t#in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "xe = (se-sf)/sfg \t#quality of steam in final state\n",
+    "he = hf+xe*hfg \t\t#final specific heat of enthalpy of steam in kJ/kg\n",
+    "w = hi+vi**2/(2*1000)-he-ve**2/(2*1000) \t\t#work of steam for isentropic process in kJ/kg\n",
+    "\n",
+    "#Results\n",
+    "print \"Work per kilogram of steam for this isentropic process: \",round(w,1),\"KJ/Kg-K\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Exit velocity of the steam from the nozzle: 737.0 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 9.2\n",
+    "''' Consider the reversible adiabatic flow of steam through a nozzle. Steam enters the nozzle\n",
+    "at 1 MPa and 300◦C, with a velocity of 30 m/s. The pressure of the steam at the nozzle\n",
+    "exit is 0.3 MPa. Determine the exit velocity of the steam from the nozzle, assuming a\n",
+    "reversible, adiabatic, steady-state process.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "hi = 3051.2 \t\t#initial specific heat of enthalpy in kJ/kg\n",
+    "si = 7.1228 \t\t#initial specific entropy in kJ/kg-K\n",
+    "Pe = 0.3 \t\t\t#final pressure in MPa\n",
+    "he = 2780.2 \t\t#final specific heat of enthalpy in kJ/kg-K\n",
+    "Te = 159.1 \t\t\t#final temperature in celsius\n",
+    "vi = 30 \t\t\t#velocity with which steam enters the nozzle in m/s\n",
+    "\n",
+    "#Calculations:\n",
+    "se = si \t\t\t#final specific entropy \n",
+    "ve = ((2*(hi-he)+(vi**2/1000))*1000)**0.5 \t\t#final velocity of steam with which it exits in m/s\n",
+    "\n",
+    "#Results\n",
+    "print \"Exit velocity of the steam from the nozzle:\",round(ve),'m/s'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Specific work required:  -271.0 KJ/Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 9.4\n",
+    "''' An air compressor in a gas station (see Fig. 9.4) takes in a flow of ambient air at 100 kPa,\n",
+    "290 K and compresses it to 1000 kPa in a reversible adiabatic process. We want to know\n",
+    "the specific work required and the exit air temperature.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Cp = 1.004 \t\t#specific heat of air at constant pressure in kJ/kg-K\n",
+    "Ti = 290 \t\t#initial temperature in kelvins\n",
+    "Pi = 100 \t\t#initial pressure in kPa\n",
+    "Pe = 1000 \t\t#final pressure in kPa\n",
+    "k = 1.4 \n",
+    "\n",
+    "#Calculations:\n",
+    "Te = Ti*(Pe/Pi)**((k-1)/k) #final temperature in kelvins\n",
+    "we = Cp*(Ti-Te) \t\t#required specific work in kJ/kg\n",
+    "\n",
+    "#Results:\n",
+    "print \"Specific work required: \",round(we),'KJ/Kg'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Entropy generated in this process:  0.072 KW/K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 9.5\n",
+    "''' A de-superheater works by injecting liquid water into a flow of superheated steam. With\n",
+    "2 kg/s at 300 kPa, 200◦C, steam flowing in, what mass flow rate of liquid water at 20◦C\n",
+    "should be added to generate saturated vapor at 300 kPa? We also want to know the rate of\n",
+    "entropy generation in the process.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "h1 = 2865.54 \t\t#specific heat of enthalpy at state 1 in kJ/kg\n",
+    "h2 = 83.94 \t\t\t#specific heat of enthalpy  at state 2 in kJ/kg\n",
+    "h3 = 2725.3 \t\t#specific heat of enthalpy at state 3 in kJ?kg\n",
+    "s1 = 7.3115 \t\t#specific entropy at state 1 in kJ/kg-K\n",
+    "s2 = 0.2966 \t\t#specific entropy at state 2 in kJ/kg-K\n",
+    "s3 = 6.9918 \t\t#specific entropy at state 3in kJ/kg-K\n",
+    "m1 = 2 \t\t\t\t#mass flow rate at state 1 in kg/s\n",
+    "\n",
+    "#Calculations:\n",
+    "m2 = m1*(h1-h3)/(h3-h2)\t#mass flow rate at state 2 in kg/s\n",
+    "m3 = m1+m2 \t\t\t\t#mass flow rate at state 3 in kg/s\n",
+    "Sgen = m3*s3-m1*s1-m2*s2#entropy generation in the process\n",
+    "\n",
+    "#Results:\n",
+    "print \"Entropy generated in this process: \",round(Sgen,3),'KW/K'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Total amount of work required to fill the tank:  -31.9 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import log\n",
+    "#Example: 9.6\n",
+    "'''Assume an air tank has 40 L of 100 kPa air at ambient temperature 17◦C. The adiabatic\n",
+    "and reversible compressor is started so that it charges the tank up to a pressure of 1000\n",
+    "kPa and then it shuts off. We want to know how hot the air in the tank gets and the total\n",
+    "amount of work required to fill the tank.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "T1 = 17+273 \t\t#initial temperature of tank in Kelvins\n",
+    "sT1 = 6.83521 \t\t#specific entropy in kJ/kg-K\n",
+    "R = 0.287 \t\t\t#gas constant in kJ/kg-K\n",
+    "P1 = 100 \t\t\t#initial pressure in kPa\n",
+    "P2 = 1000 \t\t\t#final pressure in kPa\n",
+    "T2 = 555.7 \t\t\t#from interplotation \n",
+    "V1 = 0.04 \t\t\t#volume of tank in m**3\n",
+    "V2 = V1 \t\t\t#final volume is equal to initial volume\n",
+    "u1 = 207.19 \t\t#initial specific heat of enthalpy in kJ/kg\n",
+    "u2 = 401.49 \t\t#final specific heat of enthalpy in kJ/kg\n",
+    "hin = 290.43 \t\t#in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "sT2 = sT1+R*log(P2/P1)\t#specific entropy at temperature T2 in kJ/kg-K\n",
+    "m1 = P1*V1/(R*T1) \t#initial mass of air in tank in kg\n",
+    "m2 = P2*V2/(R*T2) \t#final mass of air in tank in kg\n",
+    "Min = m2-m1 \t\t#in kg\n",
+    "W12 = Min*hin+m1*u1-m2*u2 \t\t#work required to fill the tank in kJ\n",
+    "\n",
+    "#Results:\n",
+    "print \"Total amount of work required to fill the tank: \",round(W12,1),'KJ'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Work required to pump water isentropically:  4.92 KJ/Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 9.7\n",
+    "''' Calculate the work per kilogram to pump water isentropically from 100 kPa, 30◦C to 5 MPa.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P1 = 100 \t\t\t#initial pressure in kPa\n",
+    "P2 = 5000 \t\t\t#final pressure in kPa\n",
+    "v = 0.001004 \t\t#specific volume in m**3/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "w = v*(P2-P1) \t\t#work required to pump water isentropically\n",
+    "\n",
+    "#Results:\n",
+    "print \"Work required to pump water isentropically: \",round(w,2),'KJ/Kg'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Ideal nozzle can generate upto Ve =  20.0 m/s in the exit flow.\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 9.8\n",
+    "'''Consider a nozzle used to spray liquid water. If the line pressure is 300 kPa and the water\n",
+    "temperature is 20◦C, how high a velocity can an ideal nozzle generate in the exit flow?'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "vf = 0.001002 \t#in m**3/kg\n",
+    "Pi = 300 \t\t#Line pressure in kPa\n",
+    "Po = 100 \t\t#in kPa\n",
+    "\n",
+    "#Calculations:\n",
+    "v = vf\n",
+    "Ve = (2*v*(Pi-Po)*1000)**0.5 \t\t#velocity in the exit flow\n",
+    "\n",
+    "#Results:\n",
+    "print \"Ideal nozzle can generate upto Ve = \",round(Ve),'m/s in the exit flow.'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 8,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Rate of entropy generation for this process:  0.00166 KW/K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 9.9\n",
+    "'''Saturated vapor R-410a enters the uninsulated compressor of a home central air-\n",
+    "conditioning system at 5◦C. The flow rate of refrigerant through the compressor is 0.08\n",
+    "kg/s, and the electrical power input is 3 kW. The exit state is 65◦C, 3000 kPa. Any heat transfer from the compressor is with the ambient at 30◦C. Determine the rate of entropy\n",
+    "generation for this process.\n",
+    "Control volume: Compressor out to ambient T 0 .\n",
+    "Inlet state: T i , x i known; state fixed.\n",
+    "Exit state: P e , T e known; state fixed.\n",
+    "Process: Steady-state, single fluid flow.\n",
+    "Model: R-410a tables, B.4.'''\n",
+    "#Variable Declaration: \n",
+    "hi = 280.6 \t\t\t\t#in kJ/kg\n",
+    "he = 307.8 \t\t\t\t#in kJ/kg\n",
+    "si = 1.0272 \t\t\t#in kJ/kg\n",
+    "se = 1.0140 \t\t\t#in kJ/kg\n",
+    "m = 0.08 \t\t\t\t#flow rate of refrigerant in kg/s\n",
+    "P = 3 \t\t\t\t\t#electrical power input in kW\n",
+    "To = 30 \t\t\t\t#in °C\n",
+    "\n",
+    "#Calculations:\n",
+    "Qcv = m*(he-hi)-P \t\t#in kW\n",
+    "Sgen = m*(se-si)-Qcv/(To+273.2)\t#Rate of entropy generation \n",
+    "\n",
+    "#Results:\n",
+    "print \"Rate of entropy generation for this process: \",round(Sgen,5),'KW/K'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.10"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 9,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Efficiency of the turbine:  80.9 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 9.10\n",
+    "''' A steam turbine receives steam at a pressure of 1 MPa and a temperature of 300\n",
+    "◦C. The steam leaves the turbine at a pressure of 15 kPa. The work output of the turbine is measured\n",
+    "and is found to be 600 kJ/kg of steam flowing through the turbine. Determine the efficiency\n",
+    "of the turbine.\n",
+    "Control volume: Turbine.\n",
+    "Inlet state: P i , T i known; state fixed.\n",
+    "Exit state: P e known.\n",
+    "Process: Steady state.\n",
+    "Model: Steam tables.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "hi = 3051.2 \t\t#initial specific heat of enthalpy in kJ/kg\n",
+    "si = 7.1228 \t\t#initial specific entropy in kJ/kg-K\n",
+    "sf = 0.7548 \t\t#in kJ/kg-K\n",
+    "sfg = 7.2536 \t\t#in kJ/kg-K\n",
+    "hf = 225.9 \t\t#in kJ/kg\n",
+    "hfg = 2373.1 \t\t#in kJ/kg\n",
+    "wa = 600 \t\t#actual work output of turbine in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "ses = si \t\t#final specific entropy is same as the initial\n",
+    "xes = (si-sf)/sfg \t\t#quality of steam when it leaves the turbine\n",
+    "hes = hf+xes*hfg \t\t#final specific heat of enthalpy in kJ/kg\n",
+    "ws = hi-hes \t\t#work output of turbine calculated ideally  in kJ/kg\n",
+    "nturbine = wa/ws \t\t#efiiciency of turbine \n",
+    "\n",
+    "#Results:\n",
+    "print \"Efficiency of the turbine: \",round(nturbine*100,1),'%'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.11"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Turbine inlet pressure:  2995.0 Kpa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import e\n",
+    "#Example: 9.11\n",
+    "''' Air enters a gas turbine at 1600 K and exits at 100 kPa, 830 K.The turbine efficiency is\n",
+    "estimated to be 85%. What is the turbine inlet pressure?\n",
+    "Control volume: Turbine.\n",
+    "Inlet state: T i known.\n",
+    "Exit state: P e , T e known; state fixed.\n",
+    "Process: Steady state.\n",
+    "Model: Air tables, Table A.7.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "hi = 1757.3 \t\t#initial specific heat of enthalpy of air in  kJ/kg\n",
+    "si = 8.6905 \t\t#initial specifc entropy of airin kJ/kg-K\n",
+    "he = 855.3 \t\t\t#final specific heat of enthalpy of air in kJ/kg\n",
+    "n = 0.85 \t\t\t#efficiency of turbine \n",
+    "Tes = 683.7 \t\t#final temperature in kelvins from air tables\n",
+    "ses = 7.7148 \t\t#in kJ/kg-K\n",
+    "R = 0.287 \t\t\t#gas constant in kJ/kg-K\n",
+    "\n",
+    "#Calculations:\n",
+    "w = hi-he \t\t\t#actual work done by turbine in kJ/kg\n",
+    "ws = w/n \t\t\t#ideal work done by turbine in kJ/kg\n",
+    "hes = hi-ws \t\t#from first law of isentropic process\n",
+    "Pi = 100/e**((si-ses)/-R) #turbine inlet pressure in kPa\n",
+    "\n",
+    "#Results:\n",
+    "print \"Turbine inlet pressure: \",round(Pi),'Kpa'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9.12"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 11,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Required work input:  -53.0 KJ/Kg\n",
+      "Exit temperature:  352.6 K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 9.12\n",
+    "'''Air enters an automotive supercharger at 100 kPa, 300 K and is compressed to 150 kPa.\n",
+    "The efficiency is 70%. What is the required work input per kilogram of air? What is the\n",
+    "exit temperature?\n",
+    "Control volume: Supercharger (compressor).\n",
+    "Inlet state: P i , T i known; state fixed.\n",
+    "Exit state: P e known.\n",
+    "Process: Steady state.\n",
+    "Model: Ideal gas, 300 K specific heat, Table A.5.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Pe = 150 \t\t#final pressure of air in kPa\n",
+    "Pi = 100 \t\t#initial presure of air in kPa\n",
+    "k = 1.4\n",
+    "Ti = 300 \t\t#initial temperature of air in kelvis\n",
+    "n = 0.7 \t\t#efficiency of automotive supercharger \n",
+    "\n",
+    "#Calculations:\n",
+    "Tes = Ti*(Pe/Pi)**((k-1)/k) \t\t#from second law\n",
+    "ws = 1.004*(Ti-Tes) \t\t#from first law of isentropic process\n",
+    "w = ws/n \t\t#real work input in kJ/kg\n",
+    "Te = Ti-w/1.004 \t\t#temperature at supercharger exit in K\n",
+    "\n",
+    "#Results:\n",
+    "print \"Required work input: \",round(w),'KJ/Kg'\n",
+    "print \"Exit temperature: \",round(Te,1),'K'"
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/Appendix_A.ipynb b/Introduction_to_flight_by_J_D_Anderson/Appendix_A.ipynb
new file mode 100644
index 00000000..352d2566
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/Appendix_A.ipynb
@@ -0,0 +1,517 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Appendix A"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "('Temperature T2  =  %.1f K', 556.4270854641527)\n",
+      "(' Compressor work  =  %.1f kJ/kg ', 269.2999938060093)\n",
+      "(' Temperature T3  =  %.1f K', 1373.2)\n",
+      "(' Temperature T4  =  %.1f K', 711.24546295203)\n",
+      "(' Turbine work  =  %.1f kJ/kg', 664.6023551961619)\n",
+      "(' Net work  =  %.1f kJ/kg', 395.30236139015256)\n",
+      "(' Thermal Efficiency of cycle  =  %.1f percent', 48.205253207687875)\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 12.1\n",
+    "'''  '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "\n",
+    "\n",
+    "\t\t#1-Inlet for compressor\n",
+    "\t\t#2-Exit for compressor\n",
+    "\t\t#T-Temperature at a state\n",
+    "\t\t#P-Pressure at a state\n",
+    "T1 = 288.2\t\t#K\n",
+    "P2 = 1000\t\t#kPa\n",
+    "P1 = 100\t\t#kPa\n",
+    "k = 1.4\n",
+    "T2 = T1*(P2/P1)**(1-1/k)\t\t#K\n",
+    "Cp = 1.004\t\t#Specific heat at constant pressure in kJ/kg\n",
+    "wc = Cp*(T2-T1)\t\t#compressor work in kJ/kg\n",
+    "print ('Temperature T2  =  %.1f K',T2)\n",
+    "print (' Compressor work  =  %.1f kJ/kg ',wc)\n",
+    "\t\t#3-Turbine Inlet\n",
+    "\t\t#4-Turbine Exit\n",
+    "P4 = P1\n",
+    "P3 = P2\n",
+    "T3 = 1373.2\t\t#K\n",
+    "T4 = T3*(P4/P3)**(1-1/k)\t\t#K\n",
+    "wt = Cp*(T3-T4)\n",
+    "wnet = wt-wc\n",
+    "print (' Temperature T3  =  %.1f K',T3)\n",
+    "print (' Temperature T4  =  %.1f K',T4)\n",
+    "print (' Turbine work  =  %.1f kJ/kg',wt)\n",
+    "print (' Net work  =  %.1f kJ/kg',wt-wc)\n",
+    "\t\t#2-Also high temperature heat exchanger Inlet\n",
+    "\t\t#3-(-do-) Exit\n",
+    "qh = Cp*(T3-T2)\t\t#Heat of source in kJ/kg\n",
+    "\t\t#4-high temp heat exchanger inlet\n",
+    "\t\t#1-(-do-) Exit\n",
+    "ql = Cp*(T4-T1)\t\t#Heat of sink in kJ/kg\n",
+    "nth = wnet/qh\n",
+    "print (' Thermal Efficiency of cycle  =  %.1f percent',nth*100)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 14,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "('Temperature T2  =  %.1f K', 623.4838568301909)\n",
+      "(' Compressor work  =  %.1f kJ/kg ', 336.6249922575117)\n",
+      "(' Temperature T3  =  %.1f K', 1373.2)\n",
+      "(' Temperature T4  =  %.1f K', 810.5386435092256)\n",
+      "(' Turbine work  =  %.1f kJ/kg', 564.9120019167375)\n",
+      "(' Net work  =  %.1f kJ/kg', 228.28700965922582)\n",
+      "(' Thermal Efficiency of cycle  =  %.1f percent', 30.32847854912508)\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 12.2\n",
+    "'''  '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "\t\t#Standard brayton cycle\n",
+    "\n",
+    "\n",
+    "\t\t#Calculation mistake in book\n",
+    "\t\t#1-Inlet for compressor\n",
+    "\t\t#2-Exit for compressor\n",
+    "\t\t#T-Temperature at a state\n",
+    "\t\t#P-Pressure at a state\n",
+    "T1 = 288.2\t\t#K\n",
+    "P2 = 1000\t\t#kPa\n",
+    "P1 = 100\t\t#kPa\n",
+    "k = 1.4\n",
+    "T2s = T1*(P2/P1)**(1-1/k)\t\t#K\n",
+    "nc = .80\t\t#Compressor Efficiency\n",
+    "T2 = T1+(T2s-T1)/0.80\n",
+    "Cp = 1.004\t\t#Specific heat at constant pressure in kJ/kg\n",
+    "wc = Cp*(T2-T1)\t\t#compressor work in kJ/kg\n",
+    "print ('Temperature T2  =  %.1f K',T2)\n",
+    "print (' Compressor work  =  %.1f kJ/kg ',wc)\n",
+    "\t\t#3-Turbine Inlet\n",
+    "\t\t#4-Turbine Exit\n",
+    "P4 = P1\n",
+    "P3 = P2\n",
+    "T3 = 1373.2\t\t#K\n",
+    "T4s = T3*(P4/P3)**(1-1/k)\t\t#K\n",
+    "nt = 0.85\t\t#turbine Efficiency\n",
+    "T4 = T3-(T3-T4s)*0.85\n",
+    "wt = Cp*(T3-T4)\n",
+    "wnet = wt-wc\n",
+    "print (' Temperature T3  =  %.1f K',T3)\n",
+    "print (' Temperature T4  =  %.1f K',T4)\n",
+    "print (' Turbine work  =  %.1f kJ/kg',wt)\n",
+    "print (' Net work  =  %.1f kJ/kg',wt-wc)\n",
+    "\t\t#2-Also high temperature heat exchanger Inlet\n",
+    "\t\t#3-(-do-) Exit\n",
+    "qh = Cp*(T3-T2)\t\t#Heat of source in kJ/kg\n",
+    "\t\t#4-high temp heat exchanger inlet\n",
+    "\t\t#1-(-do-) Exit\n",
+    "ql = Cp*(T4-T1)\t\t#Heat of sink in kJ/kg\n",
+    "nth = wnet/qh\n",
+    "print (' Thermal Efficiency of cycle  =  %.1f percent',nth*100)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "('Thermal efficiency  =  %.1f percent', 59.42413919202417)\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 12.3\n",
+    "'''  '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "\n",
+    "\n",
+    "wnet = 395.2\t\t#kJ/kg from example no 1\n",
+    "\t\t#Tx = T4\n",
+    "Tx = 710.8\t\t#K from example no 1\n",
+    "T3 = 1373.2\t\t#K from example no 1\n",
+    "Cp = 1.004\t\t#specific heat in kJ/kg \n",
+    "qh = Cp*(T3-Tx)\n",
+    "nth = wnet/qh\n",
+    "print ('Thermal efficiency  =  %.1f percent',nth*100) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 18,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "('Isothermal work in compressor  =  %.1f kJ/kg ', -190.4546418308537)\n",
+      "(' Isothermal work in turbine  =  %.1f kJ/kg', 907.4681268637344)\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 12.4\n",
+    "'''  '''\n",
+    "from math import log\n",
+    "#Variable Declaration: \n",
+    "\n",
+    "\n",
+    "R = 0.287\t\t#gas constant \n",
+    "T1 = 288.2\t\t#compressor temperature K\n",
+    "T2 = 1373.2\t\t#K turbine temperature K\n",
+    "\t\t#Pe/Pi = c = 10, Pi/Pe = 1/c from example 12.1\n",
+    "c = 10\n",
+    "wc = -R*T1*log(c)\n",
+    "print ('Isothermal work in compressor  =  %.1f kJ/kg ',wc)\n",
+    "wt = -R*T2*log(1/c)\n",
+    "print (' Isothermal work in turbine  =  %.1f kJ/kg',wt) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 20,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "('Velocity of air leaving the nozel  =  %.0f m/s', 889.4375751001304)\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 12.5\n",
+    "'''  '''\n",
+    "from math import sqrt\n",
+    "#Variable Declaration: \n",
+    "\n",
+    "\n",
+    "\t\t#1-compressor inlet\n",
+    "\t\t#2-Compressor exit\n",
+    "\t\t#P-Pressure at given point\n",
+    "\t\t#T-Temperature at given point\n",
+    "P1 = 100\t\t#kPa\n",
+    "P2 = 1000\t\t#kPa\n",
+    "T1 = 288.2\t\t#K\n",
+    "T2 = 556.8\t\t#K\n",
+    "wc = 269.5\t\t#from ex 12.1 work done in compressor in kJ/kg\n",
+    "\t\t#2-Burner inlet\n",
+    "\t\t#3-Burner exit\n",
+    "P3 = 1000\t\t#kPa\n",
+    "T3 = 1373.2\t\t#K\n",
+    "\t\t#wc = wt\n",
+    "Cp = 1.004\t\t#specific enthalpy of heat at constant pressure in kJ/kg\n",
+    "k = 1.4\n",
+    "T4 = T3-wc/Cp\n",
+    "P4 = P3*(T4/T3)**(1-1/k)\n",
+    "\t\t#from s4 = s5 and h4 = h5+v2/2 we get\n",
+    "T5 = 710.8\t\t#K, from second law\n",
+    "v = sqrt(2*Cp*1000*(T4-T5))\t\t#m/s\n",
+    "print ('Velocity of air leaving the nozel  =  %.0f m/s',v)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "('Coefficient of performance  =  %.3f ', 1.712857850240205)\n",
+      "(' Rate at which the air enter the compressor  =  %.3f kg/s ', 0.013985866348928795)\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 12.6\n",
+    "'''  '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "\n",
+    "\n",
+    "\t\t#1-compressor inlet\n",
+    "\t\t#2-compressor exit\n",
+    "P1 = 100\t\t#kPa\n",
+    "P2 = 500\t\t#kPa\n",
+    "k = 1.4\n",
+    "rp = P2/P1\n",
+    "cop = (rp**(1-1/k)-1)**-1\n",
+    "print ('Coefficient of performance  =  %.3f ',cop)\n",
+    "\t\t#3-Expander inlet\n",
+    "\t\t#4-Expander exit\n",
+    "P3 = P2\n",
+    "P4 = P1\n",
+    "T3 = 288.23\t\t#K, given and fixed\n",
+    "T4 = T3/(P3/P4)**(1-1/k)\n",
+    "T1 = 253.2\t\t#K, given\n",
+    "Cp = 1.004\t\t#Specific heat at cons pressure in kJ/kg\n",
+    "ql = Cp*(T1-T4)\t\t#heat released in kJ/kg\n",
+    "P = 1\t\t#power required in kW \n",
+    "ms = P/ql\t\t#kg/s\n",
+    "print (' Rate at which the air enter the compressor  =  %.3f kg/s ',ms) "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 23,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Temperature at compressor exit, T2  =  %.1f K  723.925669561\n",
+      " Pressure at compressor exit, P2  =  %.3f MPa   2.51188643151\n",
+      " Initial Temperature during heat additon process, T3  =  %.0f K  3234.38592061\n",
+      " Initial pressure during heat addition process, P3  =  %.3f MPa  11.222713118\n",
+      "  Final temperature during heat addition process, T4  =  %.1f K  1287.63222733\n",
+      " Final pressure during heat addition process, P4  =  %.4f MPa  0.446784256534\n",
+      " Thermal efficiency  =  %.1f percent  96.0189282945\n",
+      " Mean effective pressure  =  %.0f kPa  1455.36957602\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 12.7\n",
+    "'''  '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "\n",
+    "\n",
+    "\t\t#1-compressor inlet\n",
+    "\t\t#2-compressor exit\n",
+    "P1 = 100\t\t#kPa\n",
+    "T1 = 288.2\t\t#K\n",
+    "R = 0.287\t\t#gas constant\n",
+    "v1 = R*T1/P1\t\t#specific volume at inlet in m**3/kg\n",
+    "rv = 10\t\t#compression ratio given\n",
+    "k = 1.4\t\t#constant\n",
+    "T2 = T1*rv**(k-1)\t\t#K\n",
+    "print 'Temperature at compressor exit, T2  =  %.1f K ',T2\n",
+    "P2 = P1*rv**k\t\t#kPa\n",
+    "print ' Pressure at compressor exit, P2  =  %.3f MPa  ',P2/1000\n",
+    "v2 = v1/rv\t\t#specific heat at exit in m**3/kg\n",
+    "\t\t#23-heat addition process\n",
+    "\t\t#q23 = Cv*(T3-T2) = 1800 kJ/kg given\n",
+    "q23 = 1800\t\t#kJ/kg heat addition, given\n",
+    "Cv = 0.717\t\t#specific heat at constant volume in kJ/kg\n",
+    "T3 = T2+q23/Cv\t\t#K\n",
+    "print ' Initial Temperature during heat additon process, T3  =  %.0f K ',T3\n",
+    "P3 = P2*(T3/T2)\t\t#kPa\n",
+    "print  ' Initial pressure during heat addition process, P3  =  %.3f MPa ',P3/1000\n",
+    "r = 10\t\t#k = V4/V3 = P3/P4\n",
+    "T4 = T3*(1/r)**(k-1)\n",
+    "print '  Final temperature during heat addition process, T4  =  %.1f K ',T4\n",
+    "P4 = P3/r**k\t\t#kPa\n",
+    "print ' Final pressure during heat addition process, P4  =  %.4f MPa ',P4/1000\n",
+    "nth = 1-1/r**k\t\t#thermal efficiency\n",
+    "print ' Thermal efficiency  =  %.1f percent ',nth*100\n",
+    "q41 = Cv*(T1-T4)\t\t#/heat for process 4-1 in kJ/kg\n",
+    "wnet = q23+q41\n",
+    "mep = wnet/(v1-v2)\t\t#effective mean pressure n kPa\n",
+    "print ' Mean effective pressure  =  %.0f kPa ',mep"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 24,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Temperature at compressor exit, T2  =  %.1f K  955.225647797\n",
+      " Pressure at compressor exit, P2  =  %.3f MPa   6.62890803468\n",
+      " Initial Temperature during heat addition process, T3  =  %.0f K  2748.05433306\n",
+      " Final temperature during heat addition process, T4  =  %.0f K  1265.26371322\n",
+      " Thermal efficiency  =  %.1f percent  61.0802954233\n",
+      " Mean effective pressure  =  %.0f kPa  1399.18182622\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 12.8\n",
+    "'''  '''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "\n",
+    "\n",
+    "\t\t#1-compressor inlet\n",
+    "\t\t#2-compressor exit\n",
+    "P1 = 100\t\t#kPa\n",
+    "T1 = 288.2\t\t#K\n",
+    "R = 0.287\t\t#gas constant\n",
+    "v1 = R*T1/P1\t\t#specific volume at inlet in m**3/kg\n",
+    "rv = 20\t\t#compression ratio given\n",
+    "k = 1.4\t\t#constant\n",
+    "T2 = T1*rv**(k-1)\t\t#K\n",
+    "print 'Temperature at compressor exit, T2  =  %.1f K ',T2\n",
+    "P2 = P1*rv**k\t\t#kPa\n",
+    "print ' Pressure at compressor exit, P2  =  %.3f MPa  ',P2/1000\n",
+    "v2 = v1/rv\t\t#specific heat at exit in m**3/kg\n",
+    "\t\t#23-heat addition process\n",
+    "\t\t#q23 = Cv*(T3-T2) = 1800 kJ/kg given\n",
+    "q23 = 1800\t\t#kJ/kg heat addition, given\n",
+    "Cv = .717\n",
+    "Cp = 1.004\t\t#specific heat at constant pressure in kJ/kg\n",
+    "T3 = T2+q23/Cp\t\t#K\n",
+    "print ' Initial Temperature during heat addition process, T3  =  %.0f K ',T3\n",
+    "r = T3/T2\t\t#T3/T2 = V3/V2 = r\n",
+    "v3 = r*v2\n",
+    "T4 = T3/(v1/v3)**(k-1)\n",
+    "print ' Final temperature during heat addition process, T4  =  %.0f K ',T4\n",
+    "q41 = Cv*(T1-T4)\t\t#/heat for process 4-1 in kJ/kg\n",
+    "wnet = q23+q41\n",
+    "mep = wnet/(v1-v2)\t\t#effective mean pressure in kPa\n",
+    "qh = 1800\t\t#heat transfer in kJ/kg\n",
+    "nth = wnet/qh\t\t#thermal efficiency\n",
+    "\n",
+    "print ' Thermal efficiency  =  %.1f percent ',nth*100\n",
+    "print ' Mean effective pressure  =  %.0f kPa ',mep"
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/Appendix_B.ipynb b/Introduction_to_flight_by_J_D_Anderson/Appendix_B.ipynb
new file mode 100644
index 00000000..283ba0a6
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/Appendix_B.ipynb
@@ -0,0 +1,290 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Appendix B"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Mass of vapour:  2.77 Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 13.3\n",
+    "'''Consider 100 m 3 of an air–water vapor mixture at 0.1 MPa, 35◦C, and 70% relative\n",
+    "humidity. Calculate the humidity ratio, dew point, mass of air, and mass of vapor.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "r = 0.70\t\t\t#relative humidity\n",
+    "Pg = 5.628\t\t\t#saturation pressure in kPa\n",
+    "P = 100\t\t\t\t#net pressure kPa \n",
+    "V = 100\t\t\t\t#volume in m**3\n",
+    "Ra = 0.287\t\t\t#gas constant for water vapour\n",
+    "T = 308.2\t\t\t#Temperature in K\n",
+    "\n",
+    "#Calculations:\n",
+    "Pv = r*Pg\t\t\t#vapour pressure in kPa\n",
+    "Pa = P-Pv\t\t\t#Partial pressure of air\n",
+    "w = 0.622*Pv/Pa\t\t#humidity ratio formula\n",
+    "ma = Pa*V/(Ra*T)\t#mass in kg\n",
+    "mv = w*ma\t\t\t#mass of vapour\n",
+    "\n",
+    "#Results:\n",
+    "print 'Mass of vapour: ', round(mv,2),'Kg'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Mass of vapour condense 2.175 Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 13.4\n",
+    "'''Calculate the amount of water vapor condensed if the mixture of Example 13.3 is cooled\n",
+    "to 5◦C in a constant-pressure process.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "w1 = 0.0255\t\t#w1 = w, humidity ratio at initial temperature\n",
+    "ma = 108.6\t\t#mass of air in kg\n",
+    "P = 100\t\t\t#kPa net pressure\n",
+    "Pg2 = 0.8721\n",
+    "\n",
+    "#Calculations:\n",
+    "Pv2 = Pg2\n",
+    "w2 = 0.622*Pv2/(P-Pg2)\n",
+    "mc = ma*(w1-w2)\n",
+    "\n",
+    "#Results:\n",
+    "print 'Mass of vapour condense',round(mc,3) ,'Kg'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Heat transferred per unit mass:  -41.65  kJ/kg of dry air\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 13.5\n",
+    "''' An air-conditioning unit is shown in Fig. 13.5, with pressure, temperature, and relative\n",
+    "humidity data. Calculate the heat transfer per kilogram of dry air, assuming that changes\n",
+    "in kinetic energy are negligible.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "r1 = 0.80\t\t\t\t#realtive humidity at state 1\n",
+    "Pg1 = 4.246\t\t\t\t#saturation pressure of vapour in kPa\n",
+    "P1 = 105\t\t\t\t#net pressure at state 1 in kPa\n",
+    "P2 = 100\t\t\t\t#net pressure at state 2 in kPa\n",
+    "r2 = 0.95\t\t\t\t#relative humidity at state 2\n",
+    "Pg2 = 1.7051\t\t\t#saturation pressure of vapour in kPa\n",
+    "T1 = 30\t\t\t\t\t#C\n",
+    "T2 = 15\t\t\t\t\t#C\n",
+    "Cp = 1.004\t\t\t\t#specific heat of water vapour in kJ/kg\n",
+    "hv2 = 2528.9\t\t\t#enthalpy of vapourisation of vapour in kJ/kg\n",
+    "hv1 = 2556.3\t\t\t#enthalpy of vapourisation of vapour in kJ/kg\n",
+    "hl2 = 62.99\t\t\t\n",
+    "\n",
+    "#Calculations:\n",
+    "Pv1 = r1*Pg1\t\t\t#partial pressure of vapour in kPa\n",
+    "w1 = 0.622*Pv1/(P1-Pv1)\t#humidity ratio at state 1\n",
+    "Pv2 = r2*Pg2\t\t\t#partial pressure of vapour in kPa\n",
+    "w2 = 0.622*Pv2/(P2-Pv2)\t#humidity ratio at state 2\n",
+    "q = Cp*(T2-T1)+w2*hv2-w1*hv1+hl2*(w1-w2)\t\t#kJ/kg\n",
+    "\n",
+    "#Results:\n",
+    "print 'Heat transferred per unit mass: ',round(q,2),' kJ/kg of dry air'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Heat transferred -339.1 kJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 13.6\n",
+    "''' A tank has a volume of 0.5 m 3 and contains nitrogen and water vapor. The temperature\n",
+    "of the mixture is 50◦C, and the total pressure is 2 MPa. The partial pressure of the water\n",
+    "vapor is 5 kPa. Calculate the heat transfer when the contents of the tank are cooled to\n",
+    "10◦C.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Pn2 = 1995\t\t\t\t#Pressure of nitrogen in kPa\n",
+    "V = 0.5\t\t\t\t\t#Volume in m**3\n",
+    "Rn2 = 0.2968\t\t\t#Gas constant for nitrogen in kJ/kg.K\n",
+    "Rv = 0.4615\t\t\t\t#gas constant for vapour\n",
+    "T1 = 323.2\t\t\t\t#Temperature in K\n",
+    "T2 = 283.2\t\t\t\t#Temperature in K\n",
+    "Pv1 = 5\t\t\t\t\t#Pressure of water vapour in kPa at state 1\n",
+    "Pv2 = 1.2276\t\t\t#Pressure of water vapour in kPa at state 2\n",
+    "uv1 = 2443.1\t\t\t#specific internal energy of vapour in kJ/kg at state 1\n",
+    "uv2 = 2389.2\t\t\t#specific internal energy of vapour in kJ/kg at state 2\n",
+    "ul2 = 42.0\t\t\t\t#specific internal energy of liquid water in kJ/kg\n",
+    "Cv = 0.745\t\t\t\t#specific heat at constant volume in kJ/kg.K\n",
+    "\n",
+    "#Calculations:\n",
+    "mn2 = Pn2*V/(Rn2*T1)\t#mass of nitrogen\n",
+    "mv1 = Pv1*V/(Rv*T1)\t\t#mass of vapour in kg\n",
+    "mv2 = Pv2*V/(Rv*T2)\t\t#mass of vapour in kg\n",
+    "ml2 = mv1-mv2\t\t\t#mass of liquid condensed n kg\n",
+    "Q = mn2*Cv*(T2-T1)+mv2*uv2+ml2*ul2-mv1*uv1\n",
+    "\n",
+    "#Results:\n",
+    "print 'Heat transferred',round(Q,1),'kJ'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Relative humidity:  0.0107\n",
+      "Humidity ratio:  0.399\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 13.7\n",
+    "'''The pressure of the mixture entering and leaving the adiabatic saturator is 0.1 MPa, the\n",
+    "entering temperature is 30◦C, and the temperature leaving is 20◦C, which is the adiabatic\n",
+    "saturation temperature. Calculate the humidity ratio and relative humidity of the air–water\n",
+    "vapor mixture entering.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "P = 100\t\t\t\t#net pressure n kPa \n",
+    "Pg2 = 2.339\t\t\t#saturation pressure of vapour in kPa\n",
+    "Cpa = 1.004\t\t\t#specific heat n kJ/kg/K\n",
+    "T2 = 20\t\t\t\t#final temp in C\n",
+    "T1 = 30\t\t\t\t#initial temp in C\n",
+    "Hfg2 = 2454.1\t\t#specific heat difference at state 2 in kJ/kg\n",
+    "hv1 = 2556.3\t\t#enthalpy of water vapour at state 1 in kJ/kg\n",
+    "hl2 = 83.96\t\t\t#enthalpy of liquid water in kJ/kg\n",
+    "Pg1 = 4.246\t\t\t#saturation pressure at state 1 in kPa\n",
+    "\n",
+    "#Calculations:\n",
+    "Pv2 = Pg2\t\t\t#partial pressure of vapour\n",
+    "w2 = 0.622*Pv2/(P-Pg2)\n",
+    "w1 = (Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2)\n",
+    "Pv1 = 100*w1/(0.622+w1)\n",
+    "r = Pv1/Pg1\t\t\t#humidity ratio\n",
+    "\n",
+    "#Results:\n",
+    "print 'Relative humidity: ',round(w1,4)\n",
+    "print 'Humidity ratio: ',round(r,3)"
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/Appendix_C.ipynb b/Introduction_to_flight_by_J_D_Anderson/Appendix_C.ipynb
new file mode 100644
index 00000000..d2f0522f
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/Appendix_C.ipynb
@@ -0,0 +1,352 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Appendix C"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Sublimation Pressure:  0.00109 KPa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import exp\n",
+    "#Example: 14.1\n",
+    "'''Determine the sublimation pressure of water vapor at −60◦C using data available in the\n",
+    "steam tables.\n",
+    "Control mass: Water.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "#from table in appendix B.1.5\n",
+    "T1 = 213.2\t\t\t#K, Temperature at state 1\n",
+    "P2 = 0.0129\t\t\t#kPa, pressure at state 2\n",
+    "T2 = 233.2\t\t\t#K, Temperature at state 2\n",
+    "hig = 2838.9\t\t#kJ/kg, enthalpy of sublimation \n",
+    "R = .46152\t\t\t#Gas constant \n",
+    "\n",
+    "#Calculations:\n",
+    "P1 = P2*exp(-hig/R*(1/T1-1/T2)) #using relation log(P2/P1) = (hig/R)*(1/T1-1/T2) \n",
+    "\n",
+    "#Results:\n",
+    "print 'Sublimation Pressure: ',round(P1,5),'KPa'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 4"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Change in internal energy:  -159.4 J/Kg\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 14.4\n",
+    "'''The pressure on a block of copper having a mass of 1 kg is increased in a reversible process\n",
+    "from 0.1 to 100 MPa while the temperature is held constant at 15◦C. Determine the work\n",
+    "done on the copper during this process, the change in entropy per kilogram of copper, the\n",
+    "heat transfer, and the change of internal energy per kilogram.\n",
+    "Over the range of pressure and temperature in this problem, the following data can\n",
+    "be used:\n",
+    "Volume expansivity = αP = 5.0 × 10−5 K−1\n",
+    "Isothermal compressibility = βT = 8.6 × 10−12 m2/N\n",
+    "Specific volume = 0.000 114 m3/kg'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "#known data\n",
+    "ap = 5*10**-5\t\t\t\t#K**-1 Volume expansivity\n",
+    "bt = 8.6*10**-12\t\t\t#m**2/N, Isothermal compressibility\n",
+    "v = 0.000114\t\t\t\t#m**3/kg, specific volume\n",
+    "P2 = 100*10**6\t\t\t\t#pressure at state 2 in kPa\n",
+    "P1 = 100\t\t\t\t\t#pressure at state 1 in kPa\n",
+    "T = 288.2\t\t\t\t\t#Temperature in K\n",
+    "\n",
+    "#Calculations:\n",
+    "w = -v*bt*(P2**2-P1**2)/2\t#work done in J/kg\n",
+    "q = -T*v*ap*(P2-P1)\t\t\t#heat in J/kg\n",
+    "du = q-w\t\t\t\t\t#change in internal energy in J/kg\n",
+    "\n",
+    "#Results:\n",
+    "print 'Change in internal energy: ',round(du,1),'J/Kg'\n",
+    " "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Enthalpy change:  18.73 KJ/Kg\n",
+      "Temperature : 146.0 K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 14.5\n",
+    "'''Nitrogen is throttled from 20 MPa, −70◦C, to 2 MPa in an adiabatic, steady-state, steadyflow\n",
+    "process. Determine the final temperature of the nitrogen.\n",
+    "Control volume:\n",
+    "Inlet state:\n",
+    "Exit state:\n",
+    "Process:\n",
+    "Diagram:\n",
+    "Model:\n",
+    "Throttling valve.\n",
+    "P1, T1 known; state fixed.\n",
+    "P2 known.\n",
+    "Steady-state, throttling process.\n",
+    "Figure 14.8.\n",
+    "Generalized charts, Fig. D.2.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "#from table A.2\n",
+    "P1 = 20\t\t#pressure at state 1 in MPa\n",
+    "P2 = 2\t\t#pressure at state 2 in MPa\n",
+    "T1 = 203.2\t\t#Temperature at state 1 in K\n",
+    "Pr1 = P1/3.39\t\t#Reduced pressure at state 1\n",
+    "Pr2 = P2/3.39\t\t#Reduced pressure at state 2\n",
+    "Tr1 = T1/126.2\t\t#Reduced temperature\n",
+    "#from compressibility chart h1*-h1 = 2.1*R*Tc\n",
+    "#from zero pressure specific heat data h1*-h2* = Cp*(T1a-T2a)\n",
+    "#h2*-h2 = 0.5*R*Tc\n",
+    "#this gives dh = h1-h2 = -2.1*R*Tc+Cp*(T1a-T2a)+0.5*R*Tc\n",
+    "R = 0.2968\t\t#gas constant for given substance\n",
+    "Tc = 126.2\t\t#K, Constant temperature\n",
+    "Cp = 1.0416\t\t#heat enthalpy at constant pressure in kJ/kg\n",
+    "T2 = 146\t\t#temperature at state 2\n",
+    "\n",
+    "#Calculations:\n",
+    "dh = 0.5*R*Tc#\n",
+    "\n",
+    "#Results:\n",
+    "print 'Enthalpy change: ',round(dh,2),'KJ/Kg'\n",
+    "print 'Temperature :',round(T2,2),'K' "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Enthalpy change:  61.98 kJ/kg\n",
+      "Entropy Change:  1.0782 kJ/kg.K\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import log\n",
+    "\n",
+    "#Example: 14.6\n",
+    "'''Nitrogen at 8 MPa, 150 K, is throttled to 0.5 MPa. After the gas passes through a short\n",
+    "length of pipe, its temperature is measured and found to be 125 K. Determine the heat\n",
+    "transfer and the change of entropy using the generalized charts. Compare these results\n",
+    "with those obtained by using the nitrogen tables.\n",
+    "Control volume:\n",
+    "Inlet state:\n",
+    "Exit state:\n",
+    "Process:\n",
+    "Diagram:\n",
+    "Model:\n",
+    "Throttle and pipe.\n",
+    "P1, T1 known; state fixed.\n",
+    "P2, T2 known; state fixed.\n",
+    "Steady state.\n",
+    "Figure 14.10.\n",
+    "Generalized charts, results to be compared with those\n",
+    "obtained with nitrogen tables.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "#from table A.2\n",
+    "P1 = 8\t\t\t\t\t#pressure at state 1 in MPa\n",
+    "P2 = 0.5\t\t\t\t#pressure at state 2 in MPa\n",
+    "T1 = 150\t\t\t\t#Temperature at state 1 in K\n",
+    "Pr1 = P1/3.39\t\t\t#Reduced pressure at state 1\n",
+    "Pr2 = P2/3.39\t\t\t#Reduced pressure at state 2\n",
+    "Tr1 = T1/126.2\t\t\t#Reduced temperature\n",
+    "T2 = 125\t\t\t\t#Temperature at state 2\n",
+    "R = 0.2968\t\t\t\t#gas constant for given substance\n",
+    "Tc = 126.2\t\t\t\t#K, Constant temperature\n",
+    "Cp = 1.0416\t\t\t\t#Heat enthalpy at constant pressure in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "dh = (2.35)*R*Tc+Cp*(T2-T1)\t\t#\n",
+    "ds = 1.6*R-0.1*R+Cp*log(T2/T1)-R*log(P2/P1)\n",
+    "\n",
+    "#Results:\n",
+    "print 'Enthalpy change: ',round(dh,2),'kJ/kg'\n",
+    "print 'Entropy Change: ',round(ds,4),'kJ/kg.K'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Percentage deviation in specific volume using Kays rule:  4.8 %\n",
+      "Percentage deviation in specific volume using vander waals eqn:  6.4 %\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import sqrt\n",
+    "#Example: 14.7\n",
+    "'''A mixture of 80% CO2 and 20% CH4 (mass basis) is maintained at 310.94 K, 86.19 bar,\n",
+    "at which condition the specific volume has been measured as 0.006757 m3/kg. Calculate\n",
+    "the percent deviation if the specific volume had been calculated by (a) Kay’s rule and (b)\n",
+    "van der Waals’ equation of state.\n",
+    "Control mass:\n",
+    "State:\n",
+    "Model:\n",
+    "Gas mixture.\n",
+    "P, v, T known.\n",
+    "(a) Kay’s rule. (b) van der Waals’ equation.'''\n",
+    "#Keys:\n",
+    "#a-denotes C02\n",
+    "#b-denotes CH4\n",
+    "#c-denotes critical conditions\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "T = 310.94\t\t\t\t#Temperature of mixture K\n",
+    "P = 86.19\t\t\t\t#Pressure of mixture in MPa\n",
+    "Tca = 304.1\t\t\t\t#K\n",
+    "Tcb = 190.4\t\t\t\t#K\n",
+    "Pca = 7.38\t\t\t\t#MPa\n",
+    "Pcb = 4.60\t\t\t\t#MPa\n",
+    "Ra = 0.1889\t\t\t\t#gas constant for a in kJ/kg.K\n",
+    "Rb = 0.5183\t\t\t\t#gas constant for b in kJ/kg.K\n",
+    "xa = 0.8\t\t\t\t#fraction of CO2\n",
+    "xb = 0.2\t\t\t\t#fraction of CH4\n",
+    "Ma = 44.01\t\t\t\t#molecular mass of a\n",
+    "Mb = 16.043\t\t\t\t#molecular mass of b\n",
+    "Zm = 0.7\t\t\t\t#Compressiblity from generalised compressibility chart\n",
+    "ve = 0.0006757\t\t\t#experimental specific volume in m**3/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "Rm = xa*Ra+xb*Rb\t\t#mean gas constant in kJ/kg.K\n",
+    "ya = xa/Ma/(xa/Ma+xb/Mb)#mole fraction of a\n",
+    "yb = xb/Mb/(xa/Ma+xb/Mb)#mole fraction of b\n",
+    "Tcm = ya*Tca+yb*Tcb\t\t#mean critical temp in K\n",
+    "Pcm = ya*Pca+yb*Tcb\t\t#mean critical pressure n MPa\n",
+    "Trm = T/Tcm\n",
+    "Prm = P/Pcm\n",
+    "vc = Zm*Rm*T/P/1000\t\t#specific volume calculated in m**3/kg\n",
+    "pd1 = (ve-vc)/ve*100\t#percent deviation\n",
+    "Aa = 27*Ra**2*Tca**2/(64*Pca*1000)\n",
+    "Ba = Ra*Tca/(8*Pca*1000)\n",
+    "Ab = 27*Rb**2*Tcb**2/(64*Pcb*1000)\n",
+    "Bb = Rb*Tcb/(8*Pcb*1000)\n",
+    "Am = (xa*sqrt(Aa)+xb*sqrt(Ab))**2\n",
+    "Bm = (xa*Ba+xb*Bb)\n",
+    "vc = 0.0006326\t\t\t#calculated specific volume in m**3/kg\n",
+    "pd2 = (ve-vc)/ve*100\n",
+    "\n",
+    "#Results:\n",
+    "print 'Percentage deviation in specific volume using Kays rule: ',round(pd1,1),'%'\n",
+    "print 'Percentage deviation in specific volume using vander waals eqn: ',round(pd2,1),'%' "
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/Appendix_D.ipynb b/Introduction_to_flight_by_J_D_Anderson/Appendix_D.ipynb
new file mode 100644
index 00000000..91a87652
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/Appendix_D.ipynb
@@ -0,0 +1,125 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Appendix D"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 2"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Change in gibbs free energy at 298K : 457.179 KJ\n",
+      "Change in gibbs free energy at 298K : 271.04 KJ\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 16.2\n",
+    "'''Determine the value of \u0002G0 for the reaction 2H2O \u0003\u0004 2H2 + O2 at 25◦C and at 2000 K,\n",
+    "with the water in the gaseous phase.'''\n",
+    "\n",
+    "#Keys:\n",
+    "#1-H2\n",
+    "#2-O2\n",
+    "#3-H2O\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "def dG(T1,Hf1,Hf2,Hf3,Sf1,Sf2,sf3):\n",
+    "\tdH = 2*Hf1+Hf2-2*Hf3\t#Change in enthalpy in kJ\n",
+    "\tdS = 2*Sf1+Sf2-2*sf3 \t#Change in entropy in J/K\n",
+    "\tdG = dH-T1*dS/1000\t#change n gibbs free energy in kJ\n",
+    "\treturn dG\n",
+    "\n",
+    "#Results:\n",
+    "print 'Change in gibbs free energy at 298K :',round(dG(298,0, 0, -241.826, 130.678,205.148,188.834),3),\"KJ\"\n",
+    "print 'Change in gibbs free energy at 298K :',round(dG(2000,52.942, 59.176, -241.826+72.788,188.419,268.748,264.769),3),\"KJ\"\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Equilibrium constant at 298K:  -184.51\n",
+      "Equilibrium constant at 2000K:  -16.299\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 16.3\n",
+    "'''Determine the equilibrium constant K, expressed as ln K, for the reaction 2H2O <--->\n",
+    "2H2 + O2 at 25◦C and at 2000 K.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "dG1 = -457.166\t\t\t#change in gibbs free energy at temp 298 K from example2 in kJ\n",
+    "dG2 = -271.040\t\t\t#change in gibbs free energy at temp 2000 K from example2 n kJ\n",
+    "T1 = 298\t\t\t\t#K\n",
+    "T2 = 2000\t\t\t\t#K\n",
+    "R = 8.3145\t\t\t\t#gas constant\n",
+    "\n",
+    "#Calculations:\n",
+    "K1 = dG1*1000/(R*T1)\n",
+    "K2 = dG2*1000/(R*T2)\n",
+    "\n",
+    "#Results:\n",
+    "print 'Equilibrium constant at 298K: ',round(K1,2)\n",
+    "print 'Equilibrium constant at 2000K: ',round(K2,3)"
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/Appendix_E.ipynb b/Introduction_to_flight_by_J_D_Anderson/Appendix_E.ipynb
new file mode 100644
index 00000000..f6035999
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/Appendix_E.ipynb
@@ -0,0 +1,421 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Appendix E"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 1"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Stagnation Temperature:  319.9 K\n",
+      "Stagnation Pressure: 187.9 KPa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 17.1\n",
+    "'''Air flows in a duct at a pressure of 150 kPa with a velocity of 200 m/s. The temperature\n",
+    "of the air is 300 K. Determine the isentropic stagnation pressure and temperature.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "T = 300\t\t\t\t\t#Temperature of air in K\n",
+    "P = 150\t\t\t\t\t#Pressure of air in kPa\n",
+    "v = 200\t\t\t\t\t#velocity of air flow n m/s\n",
+    "Cp = 1.004\t\t\t\t#specific heat at constant pressure in kJ/kg\n",
+    "\n",
+    "#Calculations:\n",
+    "To = v**2/(2000*Cp)+T\t#stagnation temperature in K\n",
+    "k = 1.4\t\t       \t\t#constant\n",
+    "Po = P*(To/T)**(k/(k-1))#stagnation pressure in kPa\n",
+    "\n",
+    "#Results:\n",
+    "print 'Stagnation Temperature: ',round(To,1),'K'\n",
+    "print 'Stagnation Pressure:',round(Po,1),'KPa'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 3"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Thrust acting in x direction:  10.68 KN\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 17.3\n",
+    "'''A jet engine is being tested on a test stand (Fig. 17.5). The inlet area to the compressor is\n",
+    "0.2 m2, and air enters the compressor at 95 kPa, 100 m/s. The pressure of the atmosphere\n",
+    "is 100 kPa. The exit area of the engine is 0.1 m2, and the products of combustion leave the\n",
+    "exit plane at a pressure of 125 kPa and a velocity of 450 m/s. The air–fuel ratio is 50 kg\n",
+    "air/kg fuel, and the fuel enters with a low velocity. The rate of air flow entering the engine\n",
+    "is 20 kg/s. Determine the thrust, Rx, on the engine.'''\n",
+    "\n",
+    "#Keys\n",
+    "#i = inlet\n",
+    "#e = exit\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "#using momentum equation on control surface in x direction\n",
+    "me = 20.4\t\t#mass exiting in kg\n",
+    "mi = 20\t\t\t#mass entering in kg\n",
+    "ve = 450\t\t#exit velocity in m/s\n",
+    "vi = 100\t\t#exit velocity in m/s\n",
+    "Pi = 95\t\t\t#Pressure at inlet in kPa\n",
+    "Pe = 125\t\t#Pressure at exit in kPa\n",
+    "Po = 100\t\t#surrounding pressure in kPa\n",
+    "Ai = 0.2\t\t#inlet area in m**2\n",
+    "Ae = 0.1\t\t#exit area in m**2\n",
+    "\n",
+    "#Calculations:\n",
+    "Rx = (me*ve-mi*vi)/1000-(Pi-Po)*Ai+(Pe-Po)*Ae\t\t#thrust in x direction in kN\n",
+    "\n",
+    "#Results:\n",
+    "print 'Thrust acting in x direction: ',Rx,'KN' "
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 5"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Speed of sound at 300K:  347.2 m/s\n",
+      "Speed of sound at 1000K:  633.9 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import sqrt\n",
+    "#Example: 17.5\n",
+    "'''Determine the velocity of sound in air at 300 K and at 1000 K.'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "k = 1.4\t\t\t#constant\n",
+    "R = 0.287\t\t#gas constant\n",
+    "#At 300K\n",
+    "T1 = 300\t\t#K\n",
+    "T2 = 1000\t\t#K\n",
+    "\n",
+    "#Calculations:\n",
+    "c1 = sqrt(k*R*T1*1000)\n",
+    "c2 = sqrt(k*R*T2*1000)\n",
+    "\n",
+    "#Results:\n",
+    "print 'Speed of sound at 300K: ',round(c1,1),'m/s'\n",
+    "print 'Speed of sound at 1000K: ',round(c2,1),'m/s'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 6"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Mass flow rate at the throat section:  1.0646 Kg/s\n",
+      "Mass flow rate at the exit section:  0.8711 Kg/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import sqrt\n",
+    "#Example: 17.6\n",
+    "'''A convergent nozzle has an exit area of 500 mm2. Air enters the nozzle with a stagnation\n",
+    "pressure of 1000 kPa and a stagnation temperature of 360 K. Determine the mass rate of\n",
+    "flow for back pressures of 800 kPa, 528 kPa, and 300 kPa, assuming isentropic flow'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "k = 1.4\t\t\t\t#constant\n",
+    "R = 0.287\t\t\t#gas constant\n",
+    "To = 360\t\t\t#stagnation Temperature in K \n",
+    "P = 528\t\t\t\t#stagnation pressure in kPa\n",
+    "A = 500*10**-6\t\t#area in m**2\n",
+    "Me = 0.573\t\t\t#Mach number\n",
+    "Pe = 800\t\t\t#exit pressure in kPa\n",
+    "\n",
+    "#Calculations:\n",
+    "T = To*0.8333\t\t#Temperature of air in K, 0.8333 stagnation ratio from table\n",
+    "v = sqrt(k*R*T*1000)#velocity in m/s\n",
+    "d = P/(R*T)\t\t\t#stagnation density in kg/m**3\n",
+    "ms = d*A*v\t\t\t#mass flow rate in kg/s\n",
+    "Te = To*0.9381\t\t#exit temperature in K, ratio from table\n",
+    "ce = sqrt(k*R*Te*1000)#exit velocity of sound in m/s\n",
+    "ve = Me*ce\n",
+    "de = Pe/R/Te\n",
+    "mse = de*A*ve\n",
+    "\n",
+    "#Results:\n",
+    "print 'Mass flow rate at the throat section: ',round(ms,4),'Kg/s'\n",
+    "print 'Mass flow rate at the exit section: ',round(mse,4),'Kg/s'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 7"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "__When diverging section act as a nozzle__\n",
+      "Exit pressure:  93.9 Kpa\n",
+      "Exit Temperature:  183.2 K\n",
+      "Exit velocity:  596.1 m/s\n",
+      "__When diverging section act as a diffuser__\n",
+      "Exit pressure:  93.6 Kpa\n",
+      "Exit Temperature:  353.2 K\n",
+      "Exit velocity:  116.0 m/s\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "from math import sqrt\n",
+    "#Example: 17.7\n",
+    "'''A converging-diverging nozzle has an exit area to throat area ratio of 2. Air enters this\n",
+    "nozzle with a stagnation pressure of 1000 kPa and a stagnation temperature of 360 K. The\n",
+    "throat area is 500 mm2. Determine the mass rate of flow, exit pressure, exit temperature,\n",
+    "exit Mach number, and exit velocity for the following conditions:\n",
+    "a. Sonic velocity at the throat, diverging section acting as a nozzle.\n",
+    "(Corresponds to point d in Fig. 17.13.)\n",
+    "b. Sonic velocity at the throat, diverging section acting as a diffuser.\n",
+    "(Corresponds to point c in Fig. 17.13.)'''\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Po = 1000\t\t      \t#stagnation pressure in kPa\n",
+    "To = 360\t\t      \t#stagnation temperature in K\n",
+    "#when diverging section acting as nozzle\n",
+    "Pe1 = 0.0939*Po\t\t\t#exit pressure of air in kPa\n",
+    "Te1 = 0.5089*To\t\t\t#exit temperature in K\n",
+    "k = 1.4\t\t      \t\t#constant\n",
+    "R = 0.287\t\t      \t#gas constant for air\n",
+    "Me = 2.197\t\t      \t#mach number from table\n",
+    "#when diverging section act as diffuser\n",
+    "Me2 = 0.308\n",
+    "Pe2 = 0.0936*Po\t\t#exit pressure of air in kPa\n",
+    "Te2 = 0.9812*To\t\t#exit temperature in K\n",
+    "\n",
+    "#Calculations:\n",
+    "ce = sqrt(k*R*Te1*1000)\t#velocity of sound in exit section in m/s\n",
+    "ve1 = Me*ce\t\t\t\t#velocity of air at exit section in m/s\n",
+    "ce = sqrt(k*R*Te2*1000)\t\t#velocity of sound in exit section in m/s\n",
+    "ve2 = Me2*ce\n",
+    "\n",
+    "#Results:\n",
+    "print '__When diverging section act as a nozzle__'\n",
+    "print 'Exit pressure: ',round(Pe1,1),\"Kpa\"\n",
+    "print 'Exit Temperature: ',round(Te1,1),\"K\"\n",
+    "print 'Exit velocity: ',round(ve1,1),\"m/s\"\n",
+    "print '__When diverging section act as a diffuser__'\n",
+    "print 'Exit pressure: ',round(Pe2,1),\"Kpa\"\n",
+    "print 'Exit Temperature: ',round(Te2,1),\"K\"\n",
+    "print 'Exit velocity: ',round(ve2,1),\"m/s\"\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 8"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Static Pressure in downstream:  512.7 Kpa\n",
+      "Static Temperature in downstream:  339.7 K\n",
+      "Stagnation Pressure in downstream:  630.0 Kpa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 17.8\n",
+    "'''Consider the convergent-divergent nozzle of Example 17.7, in which the diverging section\n",
+    "acts as a supersonic nozzle (Fig. 17.16). Assume that a normal shock stands in the exit\n",
+    "plane of the nozzle. Determine the static pressure and temperature and the stagnation\n",
+    "pressure just downstream of the normal shock.'''\n",
+    "\n",
+    "#Variable Declaration:\n",
+    "Px = 93.9 \t\t\t#Static Pressure in Upstream(Kpa)\n",
+    "Tx = 183.2 \t\t\t#Static Temperature in Upstream(K)\n",
+    "Pox = 1000\t\t\t#Total Pressure in Upstream(Kpa)\n",
+    "Mx = 2.197\t\t\t#X-direction Mach No (Using table A.13)\n",
+    "My = 0.547\t\t\t#Y-direction Mach No (Using table A.13)\n",
+    "rP = 5.46\t\t\t#Py/Px (Using table A.13)\n",
+    "rT = 1.854\t\t\t#Ty/Tx (Using table A.13)\n",
+    "rPo = 0.63\t\t\t#Poy/Pox (Using table A.13)\n",
+    "\n",
+    "#Calculations:\n",
+    "Py = rP*Px\n",
+    "Ty = rT*Tx\n",
+    "Poy = rPo*Pox\n",
+    "\n",
+    "#Results:\n",
+    "print 'Static Pressure in downstream: ',round(Py,1),'Kpa'\n",
+    "print 'Static Temperature in downstream: ',round(Ty,1),'K'\n",
+    "print 'Stagnation Pressure in downstream: ',round(Poy,1),'Kpa'"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 9"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Exit pressure:  669.6 Kpa\n",
+      "Exit temperature:  327.8 K\n",
+      "Exit stagnation pressure:  929.8 Kpa\n"
+     ]
+    }
+   ],
+   "source": [
+    "# -*- coding: utf8 -*-\n",
+    "from __future__ import division\n",
+    "#Example: 17.9\n",
+    "'''Consider the convergent-divergent nozzle of Examples 17.7 and 17.8. Assume that there\n",
+    "is a normal shock wave standing at the point where M = 1.5. Determine the exit-plane\n",
+    "pressure, temperature, and Mach number. Assume isentropic flow except for the normal\n",
+    "shock (Fig. 17.18).'''\n",
+    "\n",
+    "#Key\n",
+    "#x = inlet\n",
+    "#y = exit\n",
+    "\n",
+    "#Variable Declaration: \n",
+    "Mx = 1.5\t\t\t\t#mach number for inlet\n",
+    "My = 0.7011\t\t\t\t#mach number for exit\n",
+    "Px = 272.4\t\t\t\t#inlet pressure in kPa\n",
+    "Tx = 248.3\t\t\t\t#inlet temperature in K\n",
+    "Pox = 1000\t\t\t\t#stagnation pressure for inlet\n",
+    "\n",
+    "#Calculations:\n",
+    "Py = 2.4583*Px\t\t\t#Exit Pressure in kPa\n",
+    "Ty = 1.320*Tx\t\t\t#Exit temperature in K\n",
+    "Poy = 0.9298*Pox\t\t#Exit pressure in kPa\n",
+    "\n",
+    "#Results:\n",
+    "print 'Exit pressure: ',round(Py,1),\"Kpa\"\n",
+    "print 'Exit temperature: ',round(Ty,1),\"K\"\n",
+    "print 'Exit stagnation pressure: ',round(Poy,1),\"Kpa\""
+   ]
+  }
+ ],
+ "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": "2.7.6"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
diff --git a/Introduction_to_flight_by_J_D_Anderson/README.txt b/Introduction_to_flight_by_J_D_Anderson/README.txt
new file mode 100644
index 00000000..990e4f2c
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/README.txt
@@ -0,0 +1,10 @@
+Contributed By: Jay Chakra
+Course: btech
+College/Institute/Organization: IITB
+Department/Designation: Aerospace
+Book Title: Introduction to flight
+Author: J D Anderson
+Publisher: McGraw-Hill; 6th edition (February 1, 2008)
+Year of publication: 2008
+Isbn: 978-0071263184
+Edition: 6
\ No newline at end of file
diff --git a/Introduction_to_flight_by_J_D_Anderson/screenshots/1.png b/Introduction_to_flight_by_J_D_Anderson/screenshots/1.png
new file mode 100644
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+++ b/Introduction_to_flight_by_J_D_Anderson/screenshots/1.png
diff --git a/Introduction_to_flight_by_J_D_Anderson/screenshots/2.png b/Introduction_to_flight_by_J_D_Anderson/screenshots/2.png
new file mode 100644
index 00000000..5ff8cc76
--- /dev/null
+++ b/Introduction_to_flight_by_J_D_Anderson/screenshots/2.png
diff --git a/Introduction_to_flight_by_J_D_Anderson/screenshots/3.png b/Introduction_to_flight_by_J_D_Anderson/screenshots/3.png
new file mode 100644
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+++ b/Introduction_to_flight_by_J_D_Anderson/screenshots/3.png