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-{
- "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\""
-   ]
-  }
- ],
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