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diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER10.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER10.ipynb deleted file mode 100755 index 5e2e846a..00000000 --- a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER10.ipynb +++ /dev/null @@ -1,329 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:50b271f93e8a365fc7786fed67e8c1b9727566cca589daf5682afce2ce0beb0c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER10:COMPRESSIBLE FLOW THROUGH NOZZLES DIFFUSERS AND WIND TUNNELS"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E01 : Pg 345"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in Si units\n",
- "area_ratio = 10.25; # exit to throat area ratio\n",
- "p0 = 5; # reservoir pressure in atm\n",
- "T0 = 333.3; # reservoir temperature\n",
- "\n",
- "# from appendix A, for an area ratio of 10.25\n",
- "Me = 3.95; # exit mach number\n",
- "pe = 0.007*p0; # exit pressure\n",
- "Te = 0.2427*T0; # exit temperature\n",
- "\n",
- "print\"Me =\",Me\n",
- "print\"pe =\",pe,\"atm\"\n",
- "print\"Te =\",Te,\"K\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Me = 3.95\n",
- "pe = 0.035 atm\n",
- "Te = 80.89191 K\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E02 : Pg 346"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in Si units\n",
- "\n",
- "area_ratio = 2.; # exit to throat area ratio\n",
- "p0 = 1.; # reservoir pressure in atm\n",
- "T0 = 288.; # reservoir temperature\n",
- "\n",
- "# (a)\n",
- "# since M = 1 at the throat\n",
- "Mt = 1.;\n",
- "pt = 0.528*p0; # pressure at throat\n",
- "Tt = 0.833*T0; # temperature at throat\n",
- "\n",
- "# from appendix A for supersonic flow, for an area ratio of 2\n",
- "Me = 2.2; # exit mach number\n",
- "pe = 1./10.69*p0; # exit pressure\n",
- "Te = 1./1.968*T0; # exit temperature\n",
- "\n",
- "print\"At throat: Mt =\",Mt\n",
- "print\"\\nAt throat: pt =\",pt,\"atm\"\n",
- "print\"\\nAt throat: Tt = \",Tt,\"K\"\n",
- "print\"\\nAt throat: For supersonic exit:\",Me\n",
- "print\"\\nAt throat: pe =\",pe,\"atm\"\n",
- "print\"\\nAt throat: Te = \",Te,\"K\"\n",
- "\n",
- "# (b)\n",
- "# from appendix A for subonic flow, for an area ratio of 2\n",
- "Me = 0.3; # exit mach number\n",
- "pe = 1/1.064*p0; # exit pressure\n",
- "Te = 1/1.018*T0; # exit temperature\n",
- "\n",
- "print\"\\nFor subrsonic exit:\",Me\n",
- "print\"\\npe=\",pe,\"atm\"\n",
- "print\"\\nTe=\",Te,\"K\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "At throat: Mt = 1.0\n",
- "\n",
- "At throat: pt = 0.528 atm\n",
- "\n",
- "At throat: Tt = 239.904 K\n",
- "\n",
- "At throat: For supersonic exit: 2.2\n",
- "\n",
- "At throat: pe = 0.0935453695042 atm\n",
- "\n",
- "At throat: Te = 146.341463415 K\n",
- "\n",
- "For subrsonic exit: 0.3\n",
- "\n",
- "pe= 0.93984962406 atm\n",
- "\n",
- "Te= 282.907662083 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E03 : Pg 346"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in Si units\n",
- "\n",
- "area_ratio = 2.; # exit to throat area ratio\n",
- "p0 = 1.; # reservoir pressure in atm\n",
- "T0 = 288.; # reservoir temperature\n",
- "pe = 0.973; # exit pressure in atm\n",
- "\n",
- "p_ratio = p0/pe; # ratio of reservoir to exit pressure\n",
- "\n",
- "# from appendix A for subsonic flow, for an pressure ratio of 1.028\n",
- "Me = 0.2; # exit mach number\n",
- "area_ratio_exit_to_star = 2.964; # A_exit/A_star\n",
- "\n",
- "# thus\n",
- "area_ratio_throat_to_star = area_ratio_exit_to_star/area_ratio; # A_exit/A_star\n",
- "\n",
- "# from appendix A for subsonic flow, for an area ratio of 1.482\n",
- "Mt = 0.44; # throat mach number\n",
- "\n",
- "print\"Me =\",Me\n",
- "print\"Mt =\",Mt"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Me = 0.2\n",
- "Mt = 0.44\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E04 : Pg 352"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "p0 = 30.*101000.; # reservoir pressure\n",
- "T0 = 3500.; # reservoir temperature\n",
- "R = 520.; # specific gas constant\n",
- "gam = 1.22; # ratio of specific heats\n",
- "A_star = 0.4; # rocket nozzle throat area\n",
- "pe = 5529.; # rocket nozzle exit pressure equal to ambient pressure at 20 km altitude\n",
- "\n",
- "# (a)\n",
- "# the density of air in the reservoir can be calculated as\n",
- "rho0 = p0/R/T0;\n",
- "\n",
- "# from eq.(8.46)\n",
- "rho_star = rho0*(2/(gam+1))**(1/(gam-1));\n",
- "\n",
- "# from eq.(8.44)\n",
- "T_star = T0*2/(gam+1);\n",
- "a_star = math.sqrt(gam*R*T_star);\n",
- "u_star = a_star;\n",
- "m_dot = rho_star*u_star*A_star;\n",
- "\n",
- "# rearranging eq.(8.42)\n",
- "Me = math.sqrt(2/(gam-1)*(((p0/pe)**((gam-1)/gam)) - 1));\n",
- "Te = T0/(1+(gam-1)/2*Me*Me);\n",
- "ae = math.sqrt(gam*R*Te);\n",
- "ue = Me*ae;\n",
- "\n",
- "# thus the thrust can be calculated as\n",
- "T = m_dot*ue;\n",
- "T_lb = T*0.2247;\n",
- "\n",
- "# (b)\n",
- "# rearranging eq.(10.32)\n",
- "Ae = A_star/Me*((2/(gam+1)*(1+(gam-1)/2*Me*Me))**((gam+1)/(gam-1)/2));\n",
- "\n",
- "print\"(a)The thrust of the rocket is:T =\" ,T/1e6, \"N\"\n",
- "print\"\\n(b)The nozzle exit area is\",T_lb \n",
- "\n",
- "print\"\\nAe =\",Ae, \"m2\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "(a)The thrust of the rocket is:T = 2.1702872295 N\n",
- "\n",
- "(b)The nozzle exit area is 487663.540469\n",
- "\n",
- "Ae = 16.7097500627 m2\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E05 : Pg 353"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "p0 = 30.*101000.; # reservoir pressure\n",
- "T0 = 3500.; # reservoir temperature\n",
- "R = 520.; # specific gas constant\n",
- "gam = 1.22; # ratio of specific heats\n",
- "A_star = 0.4; # rocket nozzle throat area\n",
- "\n",
- "# the mass flow rate using the closed form analytical expression\n",
- "# from problem 10.5 can be given as\n",
- "m_dot = p0*A_star*math.sqrt(gam/R/T0*((2/(gam+1))**((gam+1)/(gam-1))));\n",
- "\n",
- "print\"The mass flow rate is: m_dot =\",m_dot, \"kg/s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The mass flow rate is: m_dot = 586.100122081 kg/s\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E06 : Pg 356"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "M = 2.; # Mach number\n",
- "# for this value M, for a normal shock, from Appendix B\n",
- "p0_ratio = 0.7209;\n",
- "# thus\n",
- "area_ratio = 1./p0_ratio;\n",
- "print\"The diffuser throat to nozzle throat area ratio is: =\",area_ratio"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The diffuser throat to nozzle throat area ratio is: = 1.38715494521\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-}
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