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diff --git a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER09.ipynb b/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER09.ipynb deleted file mode 100755 index f8dc65c6..00000000 --- a/Fundamentals_Of_Aerodynamics_by_J._D._Anderson_Jr./CHAPTER09.ipynb +++ /dev/null @@ -1,573 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:ef96c69329d4d090ccb3995280a3d98afa4f31033dd69fab00ee6053f89763a8"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "CHAPTER09:OBLIGUE SHOCK AND EXPANSION WAVES"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E01 : Pg 302"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math\n",
- "M = 2.; # mach number\n",
- "h = 16000.; # altitude of the plane\n",
- "\n",
- "# the mach angle can be calculated from eq.(9.1) as\n",
- "mue = math.asin(1./M); # mach angle\n",
- "\n",
- "d = h/math.tan(mue);\n",
- "\n",
- "print\"The plane is ahead of the bystander by a distance of:\\nd =\",d/1000.,\"km\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The plane is ahead of the bystander by a distance of:\n",
- "d = 27.7128129211 km\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E02 : Pg 302"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "from math import pi, sin,cos\n",
- "M1 = 2.; # mach number\n",
- "p1 = 1.; # ambient pressure\n",
- "T1 = 288.; # ambient temperature\n",
- "theta = 20.*pi/180.; # flow deflection\n",
- "\n",
- "# from figure 9.9, for M = 2, theta = 20\n",
- "b = 53.4*pi/180.; # beta\n",
- "Mn_1 = M1*sin(b); # upstream mach number normal to shock\n",
- "\n",
- "# for this value of Mn,1 = 1.60, from Appendix B we have\n",
- "Mn_2 = 0.6684; # downstream mach number normal to shock\n",
- "M2 = Mn_2/sin(b-theta); # mach number downstream of shock\n",
- "p2 = 2.82*p1;\n",
- "T2 = 1.388*T1;\n",
- "\n",
- "# for M = 2, from appendix A we have\n",
- "p0_2 = 0.8952*7.824*p1;\n",
- "T0_1 = 1.8*T1;\n",
- "T0_2 = T0_1;\n",
- "\n",
- "print\"M2 =\",M2,\"\\np2 =\",p2,\"atm\",\"\\nT2 =\",T2,\"K\",\"\\np0,2 =\",p0_2,\"atm\",\"\\nT0,2 =\",T0_2,\"K\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "M2 = 1.214211418 \n",
- "p2 = 2.82 atm \n",
- "T2 = 399.744 K \n",
- "p0,2 = 7.0040448 atm \n",
- "T0,2 = 518.4 K\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E03 : Pg 305"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "from math import pi,sin\n",
- "b = 30.*pi/180.; # oblique shock wave angle\n",
- "M1 = 2.4; # upstream mach number\n",
- "\n",
- "# from figure 9.9, for these value of M and beta, we have\n",
- "theta = 6.5*pi/180.;\n",
- "\n",
- "Mn_1 = M1*sin(b); # upstream mach number normal to shock\n",
- "\n",
- "# from Appendix B\n",
- "pressure_ratio = 1.513;\n",
- "temperature_ratio = 1.128;\n",
- "Mn_2 = 0.8422;\n",
- "\n",
- "M2 = Mn_2/sin(b-theta);\n",
- "\n",
- "print\"theta =\",theta*180./pi,\"degrees\",\"\\np2/p1 =\",pressure_ratio,\"\\nT2/T1 =\",temperature_ratio,\"\\nM2 =\",M2"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "theta = 6.5 degrees \n",
- "p2/p1 = 1.513 \n",
- "T2/T1 = 1.128 \n",
- "M2 = 2.11210524521\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E04 : Pg 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "from math import pi,sin\n",
- "b = 35.*pi/180.; # oblique shock wave angle\n",
- "pressure_ratio = 3.; # upstream and downstream pressure ratio\n",
- "\n",
- "# from appendix B\n",
- "Mn_1 = 1.64;\n",
- "M1 = Mn_1/sin(b);\n",
- "\n",
- "print\"The upstream mach number is:\\nM =\",M1"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The upstream mach number is:\n",
- "M = 2.85925274482\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E05 : Pg 309"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "from math import pi,sin\n",
- "M1 = 3.;\n",
- "b = 40.*pi/180.;\n",
- "\n",
- "# for case 1, for M = 3, from Appendix B, we have\n",
- "p0_ratio_case1 = 0.3283;\n",
- "\n",
- "# for case 2\n",
- "Mn_1 = M1*sin(b);\n",
- "\n",
- "# from Appendix B\n",
- "p0_ratio1 = 0.7535;\n",
- "Mn_2 = 0.588;\n",
- "\n",
- "# from fig. 9.9, for M1 = 3 and beta = 40, we have\n",
- "theta = 22.*pi/180.;\n",
- "M2 = Mn_2/sin(b-theta);\n",
- "\n",
- "# from appendix B for M = 1.9; we have\n",
- "p0_ratio2 = 0.7674;\n",
- "p0_ratio_case2 = p0_ratio1*p0_ratio2;\n",
- "\n",
- "ratio = p0_ratio_case2/p0_ratio_case1;\n",
- "\n",
- "print\"Ans =\",ratio"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Ans = 1.76130338105\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E06 : Pg 310"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "from math import pi,sin,tan\n",
- "M1 = 5.;\n",
- "theta = 15.*pi/180.;\n",
- "gam = 1.4;\n",
- "\n",
- "# for these values of M and theta, from fig. 9.9\n",
- "b = 24.2*pi/180;\n",
- "Mn_1 = M1*sin(b);\n",
- "\n",
- "# from Appendix B, for Mn,1 = 2.05, we have\n",
- "p_ratio = 4.736;\n",
- "\n",
- "# hence\n",
- "c_d = 4.*tan(theta)/gam/(M1**2.)*(p_ratio-1.);\n",
- "\n",
- "print\"The drag coefficient is given by:\\ncd =\",c_d"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The drag coefficient is given by:\n",
- "cd = 0.114406649477\n"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E07 : Pg 311"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "from math import pi,sin,tan\n",
- "M1 = 3.5;\n",
- "theta1 = 10.*pi/180.;\n",
- "gam = 1.4;\n",
- "p1 = 101300.;\n",
- "T1 = 288.;\n",
- "b=1.;#\n",
- "# for these values of M and theta, from fig. 9.9\n",
- "b1 = 24.*pi/180.;\n",
- "Mn_1 = M1*sin(b);\n",
- "\n",
- "# from Appendix B, for Mn,1 = 2.05, we have\n",
- "Mn_2 = 0.7157;\n",
- "p_ratio1 = 2.32;\n",
- "T_ratio1 = 1.294;\n",
- "M2 = Mn_2/sin(b1-theta1);\n",
- "\n",
- "# now\n",
- "theta2 = 10.*pi/180.;\n",
- "\n",
- "# from fig. 9.9\n",
- "b2 = 27.3*pi/180.;\n",
- "phi = b2 - theta2;\n",
- "\n",
- "# from Appendix B\n",
- "p_ratio2 = 1.991;\n",
- "T_ratio2 = 1.229;\n",
- "Mn_3 = 0.7572;\n",
- "M3 = Mn_3/sin(b2-theta2);\n",
- "\n",
- "# thus\n",
- "p3 = p_ratio1*p_ratio2*p1;\n",
- "T3 = T_ratio1*T_ratio2*T1;\n",
- "\n",
- "print\"p3 =\",p3/1e5,\"x 10**5 N/m2\",\"\\nT3 =\",T3,\"K\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "p3 = 4.67916856 x 10**5 N/m2 \n",
- "T3 = 458.013888 K\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E08 : Pg 312"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "from math import pi,sin,tan\n",
- "M1 = 1.5; # upstream mach number\n",
- "theta = 15.*pi/180.; # deflection angle\n",
- "p1 = 1.; # ambient pressure in atm\n",
- "T1 = 288.; # ambient temperature\n",
- "\n",
- "# from appendix C, for M1 = 1.5 we have\n",
- "v1 = 11.91*pi/180.;\n",
- "\n",
- "# from eq.(9.43)\n",
- "v2 = v1 + theta;\n",
- "\n",
- "# for this value of v2, from appendix C\n",
- "M2 = 2.;\n",
- "\n",
- "# from Appendix A for M1 = 1.5 and M2 = 2.0, we have\n",
- "p2 = 1./7.824*1.*3.671*p1;\n",
- "T2 = 1./1.8*1.*1.45*T1;\n",
- "p0_1 = 3.671*p1;\n",
- "p0_2 = p0_1;\n",
- "T0_1 = 1.45*T1;\n",
- "T0_2 = T0_1;\n",
- "\n",
- "# from fig. 9.25, we have\n",
- "fml = 41.81; # Angle of forward Mach line\n",
- "rml = 30. - 15.; # Angle of rear Mach line\n",
- "\n",
- "print\"p2 =\",p2,\"atm\",\"\\nT2 =\",T2,\"K\",\"\\np0,2 =\",p0_2,\"atm\",\"\\nT0,2 =\",T0_2,\"K\",\"\\nAngle of forward Mach line =\",fml,\"degrees\",\"\\nAngle of rear Mach line =\",rml,\"degrees\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "p2 = 0.469197341513 atm \n",
- "T2 = 232.0 K \n",
- "p0,2 = 3.671 atm \n",
- "T0,2 = 417.6 K \n",
- "Angle of forward Mach line = 41.81 degrees \n",
- "Angle of rear Mach line = 15.0 degrees\n"
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E09 : Pg 312"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math\n",
- "from math import pi\n",
- "M1 = 10.; # upstream mach number\n",
- "theta = 15.*pi/180.; # deflection angle\n",
- "p1 = 1.; # ambient pressure in atm\n",
- "# from appendix C, for M1 = 10 we have\n",
- "v1 = 102.3*pi/180.;\n",
- "# in region 2\n",
- "v2 = v1 - theta;\n",
- "# for this value of v2, from appendix C\n",
- "M2 = 6.4;\n",
- "# from Appendix A for M1 = 10 and M2 = 6.4, we have\n",
- "p2 = 1./(2355.)*1.*42440.*p1;\n",
- "print\"M2 =\",M2,\"\\np2 =\",p2,\"atm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "M2 = 6.4 \n",
- "p2 = 18.0212314225 atm\n"
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E10 : Pg 315"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "M1 = 10; # upstream mach number\n",
- "theta = 15*math.pi/180; # deflection angle\n",
- "p1 = 1; # ambient pressure in atm\n",
- "\n",
- "# from fig 9.9, for M1 = 10 and theta = 15 we have\n",
- "b = 20*math.pi/180;\n",
- "Mn_1 = M1*math.sin(b);\n",
- "\n",
- "# from Appendix B, for Mn,1 = 3.42\n",
- "Mn_2 = 0.4552;\n",
- "M2 = Mn_2/math.sin(b-theta);\n",
- "p2 = 13.32*p1;\n",
- "\n",
- "# from Appendix A, for M1 = 10\n",
- "p0_2 = 0.2322*42440*p1;\n",
- "\n",
- "print\"M2 =\",M2,\"\\np2 =\",p2,\"atm\",\"\\np0,2 =\",p0_2/1e3,\"x 10**3 atm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "M2 = 5.22283426943 \n",
- "p2 = 13.32 atm \n",
- "p0,2 = 9.854568 x 10**3 atm\n"
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example E11 : Pg 316"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "# All the quantities are expressed in SI units\n",
- "import math \n",
- "M1 = 3.; # upstream mach number\n",
- "theta = 5.*math.pi/180.; # deflection angle\n",
- "alpha = theta; # angle of attack\n",
- "gam = 1.4;\n",
- "\n",
- "# from appendix C, for M1 = 3 we have\n",
- "v1 = 49.76*math.pi/180.;\n",
- "\n",
- "# from eq.(9.43)\n",
- "v2 = v1 + theta;\n",
- "\n",
- "# for this value of v2, from appendix C\n",
- "M2 = 3.27;\n",
- "\n",
- "# from Appendix A for M1 = 3 and M2 = 3.27, we have\n",
- "p_ratio1 = 36.73/55.;\n",
- "\n",
- "# from fig. 9.9, for M1 = 3 and theta = 5\n",
- "b = 23.1*math.pi/180.;\n",
- "Mn_1 = M1*math.sin(b);\n",
- "\n",
- "# from Appendix B\n",
- "p_ratio2 = 1.458;\n",
- "\n",
- "# thus\n",
- "c_l = 2./gam/(M1**2.)*(p_ratio2-p_ratio1)*math.cos(alpha);\n",
- "\n",
- "c_d = 2./gam/(M1**2.)*(p_ratio2-p_ratio1)*math.sin(alpha);\n",
- "\n",
- "print\"The lift and drag coefficients are given by:\\ncl =\",c_l,\"\\ncd =\",c_d"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "The lift and drag coefficients are given by:\n",
- "cl = 0.124948402826 \n",
- "cd = 0.0109315687729\n"
- ]
- }
- ],
- "prompt_number": 11
- }
- ],
- "metadata": {}
- }
- ]
-}
\ No newline at end of file |