{
"metadata": {
"name": "",
"signature": "sha256:1ed89d68741f5e390e8fd1c09a55c0f77b1c405551af1981288ec865002f9662"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"CHAPTER04:INCOMPRESSIBLE FLOW OVER AIRFOILS"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E01 : Pg 126"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"import math\n",
"c = 0.64; # chord length of the airfoil\n",
"V_inf = 70.; # freestream velocity\n",
"L_dash = 1254.; # lift per unit span L'\n",
"rho_inf = 1.23; # density of air\n",
"mu_inf = 1.789*10.**-5.; # freestream coefficient of viscosity\n",
"q_inf = 1./2.*rho_inf*V_inf*V_inf; # freestream dynamic pressure\n",
"\n",
"# thus the lift coefficient can be calculated as\n",
"c_l = L_dash/q_inf/c;\n",
"\n",
"# for this value of C_l, from fig. 4.10\n",
"alpha = 4.;\n",
"\n",
"# the Reynold's number is given as\n",
"Re = rho_inf*V_inf*c/mu_inf;\n",
"\n",
"# for the above Re and alpha values, from fig. 4.11\n",
"c_d = 0.0068;\n",
"\n",
"# thus the drag per unit span can be calculated as\n",
"D_dash = q_inf*c*c_d;\n",
"\n",
"print\"c_l =\",c_l\n",
"print\"\\nfor this c_l value, from fig. 4.10we get alpha =\",alpha\n",
"print\"\\nRe =\",Re/1000000. \n",
"print\"\\nfor this value of Re, from fig. 4.11 we get c_d =\",c_d\n",
"print\"\\nD=\",D_dash,\"N/m\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"c_l = 0.650199104032\n",
"\n",
"for this c_l value, from fig. 4.10we get alpha = 4.0\n",
"\n",
"Re = 3.08015651202\n",
"\n",
"for this value of Re, from fig. 4.11 we get c_d = 0.0068\n",
"\n",
"D= 13.114752 N/m\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E02 : Pg 126"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"\n",
"c = 0.64; # chord length of the airfoil\n",
"V_inf = 70.; # freestream velocity\n",
"rho_inf = 1.23; # density of air\n",
"q_inf = 1./2.*rho_inf*V_inf*V_inf; # freestream dynamic pressure\n",
"c_m_ac = -0.05 # moment coefficient about the aerodynamic center as seen from fig. 4.11\n",
"\n",
"# thus moment per unit span about the aerodynamic center is given as\n",
"M_dash = q_inf*c*c*c_m_ac;\n",
"\n",
"print\"The Moment per unit span about the aerodynamic center is is M=\",M_dash,\"Nm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The Moment per unit span about the aerodynamic center is is M= -61.71648 Nm\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E04 : Pg 127"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"import math\n",
"from math import pi\n",
"alpha = 5.*pi/180.; # angle of attack in radians\n",
"\n",
"# from eq.(4.33)according to the thin plate theory, the lift coefficient is given by\n",
"c_l = 2.*pi*alpha;\n",
"\n",
"# from eq.(4.39) the coefficient of moment about the leading edge is given by\n",
"c_m_le = -c_l/4.;\n",
"\n",
"# from eq.(4.41)\n",
"c_m_qc = 0;\n",
"\n",
"# thus the coefficient of moment about the trailing can be calculated as\n",
"c_m_te = 3./4.*c_l;\n",
"\n",
"print\"(a)Cl =\", c_l\n",
"print\"(b)Cm_le =\",c_m_le\n",
"print\"(c)m_c/4 =\",c_m_qc\n",
"print\"(d)Cm_te =\",c_m_te"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)Cl = 0.548311355616\n",
"(b)Cm_le = -0.137077838904\n",
"(c)m_c/4 = 0\n",
"(d)Cm_te = 0.411233516712\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E06 : Pg 131"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"\n",
"alpha1 = 4.;\n",
"alpha2 = -1.1;\n",
"alpha3 = -4.;\n",
"cl_1 = 0.55; # cl at alpha1\n",
"cl_2 = 0; # cl at alpha2\n",
"c_m_qc1 = -0.005; # c_m_qc at alpha1\n",
"c_m_qc3 = -0.0125; # c_m_qc at alpha3\n",
"\n",
"# the lift slope is given by\n",
"a0 = (cl_1 - cl_2)/(alpha1-alpha2);\n",
"\n",
"# the slope of moment coefficient curve is given by\n",
"m0 = (c_m_qc1 - c_m_qc3)/(alpha1-alpha3);\n",
"\n",
"# from eq.4.71\n",
"x_ac = -m0/a0 + 0.25;\n",
"\n",
"print\"The location of the aerodynamic center is x_ac =\",x_ac"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The location of the aerodynamic center is x_ac = 0.241306818182\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E07 : Pg 139"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"import math \n",
"c = 1.5; # airfoil chord\n",
"Re_c = 3.1e6; # Reynolds number at trailing edge\n",
"\n",
"# from eq.(4.84), the laminar boundary layer thickness at trailing edge is given by\n",
"delta = 5*c/math.sqrt(Re_c);\n",
"\n",
"# from eq(4.86)\n",
"Cf = 1.328/math.sqrt(Re_c);\n",
"\n",
"# the net Cf for both surfaces is given by\n",
"Net_Cf = 2*Cf;\n",
"\n",
"print\"(a)delta =\",delta,\"m\"\n",
"print\"(b)Cf =\",Cf*10000 \n",
"print\"Net Cf =\",Net_Cf"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)delta = 0.00425971375685 m\n",
"(b)Cf = 7.5425331588\n",
"Net Cf = 0.00150850663176\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E08 : Pg 150"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"import math \n",
"c = 1.5; # airfoil chord\n",
"Re_c = 3.1e6; # Reynolds number at trailing edge\n",
"\n",
"# from eq.(4.87), the turbulent boundary layer thickness at trailing edge is given by\n",
"delta = 0.37*c/(Re_c**0.2);\n",
"\n",
"# from eq(4.86)\n",
"Cf = 0.074/(Re_c**0.2);\n",
"\n",
"# the net Cf for both surfaces is given by\n",
"Net_Cf = 2*Cf;\n",
"\n",
"print\"(a)delta =\",delta,\"m\"\n",
"print\"(b)Cf =\",Cf\n",
"print\"Net Cf =\",Net_Cf"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)delta = 0.0279267658904 m\n",
"(b)Cf = 0.00372356878539\n",
"Net Cf = 0.00744713757078\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example E09 : Pg 162"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# All the quantities are expressed in SI units\n",
"import math \n",
"from math import sqrt\n",
"c = 1.5; # airfoil chord length\n",
"Rex_cr = 5e5; # critical Reynold's number\n",
"Re_c = 3.1e6; # Reynold's number at the trailing edge\n",
"\n",
"# the point of transition is given by\n",
"x1 = Rex_cr/Re_c*c;\n",
"\n",
"# the various skin friction coefficients are given as\n",
"Cf1_laminar = 1.328/sqrt(Rex_cr);\n",
"Cfc_turbulent = 0.074/(Re_c**0.2);\n",
"Cf1_turbulent = 0.074/(Rex_cr**0.2);\n",
"\n",
"# thus the total skin friction coefficient is given by\n",
"Cf = x1/c*Cf1_laminar + Cfc_turbulent - x1/c*Cf1_turbulent;\n",
"\n",
"# taking both sides of plate into account\n",
"Net_Cf = 2*Cf;\n",
"\n",
"print\"The net skin friction coefficient is Net Cf=\",round(Net_Cf,5)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The net skin friction coefficient is Net Cf= 0.00632\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}