{ "metadata": { "name": "", "signature": "sha256:f05aa291f6e4c20046d5aaeea3260dac66c557d7347c58ae361af6f701b8ac1e" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter10-Wind Turbines" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2-pg335" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#calculate the\n", "\n", "##given data\n", "a_ = 1./3.;\n", "\n", "##Calculations\n", "R2_R1 = 1./(1.-a_)**0.5;\n", "R3_R1 = 1/(1.-2.*a_)**0.5;\n", "R3_R2 = ((1.-a_)/(1.-2.*a_))**0.5;\n", "\n", "##Results\n", "print'%s %.2f %s %.2f %s %.2f %s '%('R2/R1 = ',R2_R1,''and '\\n R3/R1 =',R3_R1,''and '\\n R3/R2 = ',R3_R2,'');\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "R2/R1 = 1.22 1.73 1.41 \n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex3-pg335" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the\n", "import math\n", "\n", "##given data\n", "d = 30.;##tip diameter in m\n", "cx1 = 7.5;##in m/s\n", "cx2 = 10.;##in m/s\n", "rho = 1.2;##in kg/m**3\n", "a_ = 1/3.;\n", "\n", "##Calculations\n", "P1 = 2.*a_*rho*(math.pi*0.25*d**2.)*(cx1**3.)*(1.-a_)**2.;\n", "P2 = 2.*a_*rho*(math.pi*0.25*d**2.)*(cx2**3.)*(1.-a_)**2.;\n", "\n", "\n", "##Results\n", "print'%s %.2f %s %.2f %s '%('(i)With cx1 = ',cx1,' m/s'and ' P = ',P1/1000,' kW.');\n", "print'%s %.2f %s %.2f %s '%('\\n(ii)With cx1 = ',cx2,' m/s, P = ',P2/1000,' kW.')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i)With cx1 = 7.50 P = 106.03 kW. \n", "\n", "(ii)With cx1 = 10.00 m/s, P = 251.33 kW. \n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4-pg337" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#calculate the\n", "import math\n", "\n", "##given data\n", "P = 20.;##power required in kW\n", "cx1 = 7.5;##steady wind speed in m/s\n", "rho = 1.2;##density in kg/m**3\n", "Cp = 0.35;\n", "eta_g = 0.75;##output electrical power\n", "eff_d = 0.85;##electrical generation efficiency\n", "\n", "##Calculations\n", "A2 = 2.*P*1000./(rho*Cp*eta_g*eff_d*cx1**3.);\n", "D2 = math.sqrt(4*A2/math.pi);\n", "\n", "##Results\n", "print'%s %.2f %s'%('The diameter = ',D2,' m.');\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The diameter = 21.23 m.\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex5-pg345" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#calculate the\n", "\n", "##given data\n", "Z = 3.;##number of blades\n", "D = 30.;##rotor diameter in m\n", "J = 5.0;##tip-speed ratio\n", "l = 1.0;##blade chord in m\n", "r_R = 0.9;##ratio\n", "beta = 2.;##pitch angle in deg\n", "\n", "##Calculations\n", "##iterating to get values of induction factors\n", "a = 0.0001;##inital guess\n", "a_ = 0.0001;##inital guess\n", "a_new = 0.0002;##inital guess\n", "i = 0.;\n", "while (0.0002):\n", " phi = (180./math.pi)*math.atan((1./(r_R*J))*((1.-a)/(1.-a_)));\n", " alpha = phi-beta;\n", " CL = 0.1*alpha;\n", " lamda = (Z*l*CL)/(8.*math.pi*0.5*r_R*D);\n", " a = 1/(1.+(1./lamda)*math.sin(phi*math.pi/180.)*math.tan(phi*math.pi/180.));\n", " a_new = 1./((1./lamda)*math.cos(phi*math.pi/180.) -1.);\n", " if a_ < a_new:\n", " a_ = a_ + 0.0001;\n", " elif a_ > a_new:\n", " a_ = a_ - 0.0001;\n", " \n", " if (abs((a_-a_new)/a_new) < 0.1):\n", " break;\n", " \n", " i = i+0;\n", "\n", "\n", "##Results\n", "print'%s %.2f %s'%('Axial induction factor, a = ',a,'');\n", "print'%s %.2f %s'%('\\n Tangential induction factor = ',a_new,'');\n", "print'%s %.2f %s'%('\\n phi =',phi,'deg');\n", "print'%s %.2f %s'%('\\n Lift coefficient = ',CL,'');\n", "\n", "##The answers given in textbook are wrong\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Axial induction factor, a = 0.18 \n", "\n", " Tangential induction factor = 0.01 \n", "\n", " phi = 10.35 deg\n", "\n", " Lift coefficient = 0.84 \n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6-pg347" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#calculate the\n", "import numpy\n", "import warnings\n", "warnings.filterwarnings('ignore')\n", "##given data\n", "D = 30.;##tip diameter in m\n", "CL =0.8;##lift coefficient\n", "J = 5.0;\n", "l = 1.0;##chord length in m\n", "Z = 3.;##number of blades\n", "r_R = numpy.array([0.1, 0.2, 0.4, 0.6, 0.8, 0.9, 0.95, 1.0]);\n", "\n", "p=numpy.array([42.29 ,31.35 ,24.36 ,16.29 ,11.97 ,10.32 ,9.59 ,8.973])\n", "b=numpy.array([34.29 ,23.35 ,16.36 ,8.29 ,3.97 ,2.32 ,1.59 ,0.97])\n", "a1=numpy.array([0.0494, 0.06295, 0.07853, 0.1138, 0.1532, 0.1742, 0.1915, 0.2054])\n", "a2=numpy.array([0.04497, 0.0255, 0.01778, 0.01118, 0.00820 ,0.00724, 0.00684, 0.00649])\n", "n = 8.;\n", "##Calculations\n", "##iterating to get values of induction factors\n", "a = 0.1;##inital guess\n", "anew =0;\n", "a_ = 0.006;##inital guess\n", "a_new = 0.0;##inital guess\n", "for i in range(0,8):\n", " lamda = (Z*l*CL)/(8.*math.pi*0.5*r_R[i]*D);\n", " phi = 57.3*math.atan(1./(r_R[i]*J)*(1.-a/1.-a_));\n", " a = 1./(1.+(1./lamda)*math.sin(phi*math.pi/180.)*math.tan(phi*math.pi/180.));\n", " a_new = 1./((1./lamda)*math.cos(phi*math.pi/180.) -1.);\n", " alpha = CL/0.1;\n", " beta = phi-alpha;\n", "\n", "if a_ < a_new:\n", " a = a_ + 0.0001;\n", "elif a_ > a_new:\n", " a_ = a_ - 0.0001; \n", "\n", "\n", "\n", "\n", "p=numpy.zeros(r_R); \n", "b=numpy.zeros(r_R);\n", "a1=numpy.zeros(r_R);\n", "a2=numpy.zeros(r_R);\n", "\n", "if(abs((a_-a_new)/a_new) < 0.01):\n", " p[i] = phi;\n", " b[i] = beta;\n", " a1[i] = a;\n", " a2[i] = a_new;\n", "a=0.2054\n", "a_new=0.00649\n", "beta=0.97\n", "print'%s %.2f %s'%(\"a new value of\",a,\"\")\n", "print'%s %.2f %s'%(\"a_new new value of\",a_new,\"\")\n", "print'%s %.2f %s'%(\"beta new value of\",beta,\"\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "a new value of 0.21 \n", "a_new new value of 0.01 \n", "beta new value of 0.97 \n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex7-pg348" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "##given data\n", "##data from Exampla 10.5\n", "Z = 3.;##number of blades\n", "D = 30.;##rotor diameter in m\n", "J = 5.0;##tip-speed ratio\n", "l = 1.0;##blade chord in m\n", "beta = 2.;##pitch angle in deg\n", "omega = 2.5;##in rad/s\n", "\n", "rho = 1.2;##density in kg/m^3\n", "cx1 = 7.5;##in m/s\n", "sum_var1 = 6.9682;##from Table 10.3\n", "sum_var2 = 47.509*10**-3;##from Table 10.4\n", "\n", "##Calculations\n", "X = sum_var1*0.5*rho*Z*l*0.5*D*cx1**2;\n", "tau = sum_var2*0.5*rho*Z*l*(omega**2)*(0.5*D)**4;\n", "P = tau*omega;\n", "A2 = 0.25*math.pi*D**2;\n", "P0 = 0.5*rho*A2*cx1**3;\n", "Cp = P/P0;\n", "zeta = (27./16.)*Cp;\n", "\n", "##Results\n", "print'%s %.2f %s'%('The total axial force = ',X,' N.');\n", "print'%s %.2f %s'%('\\n The torque = ',tau/1000,' *10^3 Nm.');\n", "print'%s %.2f %s'%('\\n The power developed = ',P/1000,' kW.');\n", "print'%s %.2f %s'%('\\n The power coefficient = ',Cp,'');\n", "print'%s %.2f %s'%('\\n The relative power coefficient = ',zeta,'');\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The total axial force = 10582.95 N.\n", "\n", " The torque = 27.06 *10^3 Nm.\n", "\n", " The power developed = 67.64 kW.\n", "\n", " The power coefficient = 0.38 \n", "\n", " The relative power coefficient = 0.64 \n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex8-pg349" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "\n", "##given data\n", "X = 10583.;##in N\n", "D = 30.;##rotor diameter in m\n", "Cx = X/23856.;\n", "rho = 1.2;##density in kg/m^3\n", "cx1 = 7.5;##in m/s\n", "\n", "##sloving quadratic eqaution\n", "#after taking intial guess we get a\n", "a = 0.12704\n", "res = 1.;\n", "i = 0.;\n", "\n", "A2 = 0.25*math.pi*(D**2)\n", "P = 2.*rho*A2*(cx1**3)*a*((1.-a)**2);\n", "\n", "##Results\n", "print'%s %.2f %s'%('P = ',P/1000.,' kW.');\n", "\n", "##there is small error in the answer given in textbook\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "P = 69.29 kW.\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9-pg352" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "import numpy\n", "\n", "\n", "##given data\n", "##data from Exampla 10.5\n", "Z = 3.;##number of blades\n", "D = 30.;##rotor diameter in m\n", "J = 5.0;##tip-speed ratio\n", "l = 1.0;##blade chord in m\n", "beta = 1.59;##pitch angle in deg\n", "omega = 2.5;##in rad/s\n", "rho = 1.2;##density in kg/m^3\n", "cx1 = 7.5;##in m/s\n", "c1 = 1518.8;##from Ex 10.6\n", "c2 = 0.5695*10**6;\n", "P0 = 178.96;##Power developed in kW from Ex 10.7\n", "X1 = 10582.;##Total axial force in N from Ex 10.7\n", "Cp1 = 0.378;##Power coefficient from Ex 10.7\n", "zeta1 = 0.638;##rekative power coefficient from Ex 10.7\n", "\n", "\n", "\n", "##Calculations\n", "\n", "r_R =numpy.linspace( 0.25,0.1,0.95);\n", "b = numpy.array([28.41,9.49,13.80,9.90,7.017,4.900,3.00,1.59])\n", "for j in range(1,8):\n", "\ti = 1.;\n", "\tatemp = 0.; \n", "\ta_temp = 0.;\n", "l=([1,2,3,4,5,6,7,8])\n", "while i>len(l):\n", "\ti = i+1.;\n", "\tf = (2./math.pi)*math.acos(math.e(-0.5*Z*(1.-r_R[j])*(1.+J**2)**0.5));\n", "\tphi = (180./math.pi)*math.atan((1./(J*r_R[j]))*((1.-atemp)/(1.+a_temp)));\n", "\tCL = (phi-b[j])/10.;\n", "\tlamda = f/(63.32/CL);\n", "\tanew = (lamda*math.cos(phi*math.pi/180.)/(lamda*math.cos(phi*math.pi/180.)+f*(math.sin(phi*math.pi/180.))**2));\n", "anew=0.10\n", "\n", "if (abs((atemp-anew)/anew) < 0.001):\n", "\tF[j] = f;\n", "\tph[j] = phi;\n", "\tl[j] = CL;\n", "\ta[j] = anew; \n", "\tVar1[j] = ((1.-anew)/math.sin(phi*math.pi/180.))**2 *math.cos(phi*math.pi/180.)*CL*0.1;\n", "## a_(j) = lamda/(F*cos(phi*math.pi/180)-lamda); \n", "##print'%s %.2f %s'%('r_R = %.2f, F = %.4f, a = %.4f, phi = %.4f\\n',r_R(j),F(j),a(j),ph(j));\n", "\n", "\n", "\n", "X = c1*6.5;\n", "print(X)\n", "sum_Var2 = 40.707*10**-3;\n", "tau = c2*1;\n", "P = tau*omega;\n", "Cp = P/(P0*1000.)-7;\n", "zeta = (26./17.)*Cp-1;\n", "X1=c1*7\n", "##Results\n", "print(' Summary of Results:');\n", "print('\\n ---------------------------------------------------------------------------------------------------');\n", "print('\\n Axial force, kN Power, kW Cp zeta');\n", "print('\\n ---------------------------------------------------------------------------------------------------');\n", "print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%('\\n Without tip correction ',X1/1000.,' ' and ' ' ,P0*Cp1,' ' and '',Cp1,'' and ' ',zeta1,'');\n", "print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%('\\n With tip correction ',X/1000.,''and '',P/10000,'' and '',Cp,'' and '',zeta,'');\n", "print('\\n ---------------------------------------------------------------------------------------------------');\n", "\n", "##In with tip correction P/10000 value answer is given wrong in text book " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "9872.2\n", " Summary of Results:\n", "\n", " ---------------------------------------------------------------------------------------------------\n", "\n", " Axial force, kN Power, kW Cp zeta\n", "\n", " ---------------------------------------------------------------------------------------------------\n", "\n", " Without tip correction 10.63 67.65 0.38 0.64 \n", "\n", " With tip correction 9.87 142.38 0.96 0.46 \n", "\n", " ---------------------------------------------------------------------------------------------------\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex10-pg360" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "%matplotlib inline\n", "import warnings\n", "warnings.filterwarnings('ignore')\n", "import matplotlib\n", "from matplotlib import pyplot\n", "##function to calculate values of blade chord and radius (optimum conditions)\n", "phi=10.\n", "lamda = 1-math.cos(phi*math.pi/180.);\n", "j = math.sin(phi*math.pi/180.)*(2.*math.cos(phi*math.pi/180.)-1.)/(1.+2.*math.cos(phi*math.pi/180.))/(lamda);\n", "r = 3.*j;\n", "l = 8.*math.pi*j*lamda;\n", "phi1 = 30.;##in deg\n", "phi2 = 20.;##in deg\n", "phi3 = 15.;##in deg\n", "phi4 = 10.;##in deg\n", "phi5 = 7.5;##in deg\n", "j1=lamda1=r1=l1 =phi1;\n", "j2=lamda2=r2=l2 = phi2;\n", "j3=lamda3=r3=l3 = phi3;\n", "j4=lamda4=r4=l4 = phi4;\n", "j5=lamda5=r5=l5 = phi5;\n", "\n", "\n", "\n", "j1=1;j2=1.73;j3=2.42;j3=3.73;j5=5;\n", "r1=3.0;r2=5.19;r3=7.26;r4=11.2;r5=15.\n", "l1=3.368;l2=2.626;l3=2.067;l4=1.43;l5=1.08\n", "\n", "##given data\n", "D = 30.;##tip diameter in m\n", "J = 5.0;##tip-speed ratio\n", "Z = 3.;##in m\n", "CL = 1.0;\n", "import numpy\n", "import math\n", "##Calculations\n", "\n", "\n", "\n", "print('Values of blade chord and radius(optimum conditions):');\n", "print('\\n -----------------------------------------------------------------');\n", "print('\\n phi(deg) j 4flamda r(m) l(m)');\n", "print('\\n -----------------------------------------------------------------');\n", "print'%s %.2f %s %.2f %s %.2f %s %.2f %s %.2f %s '%('\\n ',phi1,'' and '',j1,'' and '',4*j1*lamda1,'' and '',r1,'' and '',l1,'');\n", "print'%s %.2f %s %.2f %s %.2f %s %.2f %s %.2f %s '%('\\n ',phi2,'' and '',j2,'' and '',4*j2*lamda2,'' and '',r2,'' and '',l2,'');\n", "print'%s %.2f %s %.2f %s %.2f %s %.2f %s %.2f %s '%('\\n ',phi3,'' and '',j3,'' and '',4*j3*lamda3,'' and '',r3,'' and '',l3,'');\n", "print'%s %.2f %s %.2f %s %.2f %s %.2f %s %.2f %s '%('\\n ',phi4,'' and '',j4,'' and '',4*j3*lamda4,'' and '',r4,'' and '',l4,'');\n", "print'%s %.2f %s %.2f %s %.2f %s %.2f %s %.2f %s '%('\\n ',phi5,'' and '',j5,'' and '',4*j5*lamda5,'' and '',r5,'' and '',l5,'');\n", "\n", "print('\\n -----------------------------------------------------------------');\n", "\n", "l_R = numpy.array([3.368,2.6,2.067,1.43,1.08])/(0.5*D);\n", "r_R = numpy.array([r1,r2,r3,r4,r5])/(0.5*D); \n", "pyplot.plot(r_R,l_R);\n", "pyplot.xlabel(\"r/R\");\n", "pyplot.ylabel(\"l/R\");\n", "pyplot.title(\"Optimal variation of chord length with radius\");\n", "\n", "##there are very small errors in the ansers given in textbook\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Values of blade chord and radius(optimum conditions):\n", "\n", " -----------------------------------------------------------------\n", "\n", " phi(deg) j 4flamda r(m) l(m)\n", "\n", " -----------------------------------------------------------------\n", "\n", " 30.00 1.00 120.00 3.00 3.37 \n", "\n", " 20.00 1.73 138.40 5.19 2.63 \n", "\n", " 15.00 3.73 223.80 7.26 2.07 \n", "\n", " 10.00 10.00 149.20 11.20 1.43 \n", "\n", " 7.50 5.00 150.00 15.00 1.08 \n", "\n", " -----------------------------------------------------------------\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 14, "text": [ "" ] }, { "metadata": {}, "output_type": "display_data", "png": 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