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 "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": {}
  }
 ]
}