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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 +} |