{ "metadata": { "name": "", "signature": "sha256:090249f83a260ba2b6a1983f6426576be55ba0e881a9fae7d9c5cbc674e22e36" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 11 : Belt, Rope and Chain Drives" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.1 Page No : 333" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "N1 = 150. \t\t\t#rpm\n", "d1 = 750. #mm\n", "d2 = 450. #mm\n", "d3 = 900. #mm\n", "d4 = 150. \t\t\t#mm\n", "\n", "#Solution:\n", "#Calculating the speed of the dynamo shaft when there is no slip\n", "N41 = N1*(d1*d3)/(d2*d4) \t\t\t#rpm\n", "#Calculating the speed of the dynamo shaft whne there is a slip of 2% at each drive\n", "s1 = 2.\n", "s2 = 2. \t\t\t#%\n", "N42 = N1*(d1*d3)/(d2*d4)*(1-s1/100)*(1-s2/100) \t\t\t#rpm\n", "\n", "#Results:\n", "print \" Speed of the dynamo shaft when there is no slip, N4 = %d rpm.\"%(N41)\n", "print \" Speed of the dynamo shaft when there is a slip of 2 %% at each drive, N4 = %d rpm.\"%(N42)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Speed of the dynamo shaft when there is no slip, N4 = 1500 rpm.\n", " Speed of the dynamo shaft when there is a slip of 2 % at each drive, N4 = 1440 rpm.\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.2 Page No : 334" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "d1 = 1.\n", "d2 = 2.25 \t\t\t#m\n", "N1 = 200. \t\t\t#rpm\n", "sigma1 = 1.4*10**6\n", "sigma2 = 0.5*10**6\n", "E = 100.*10**6 \t\t\t#N/m**2\n", "\n", "#Solution:\n", "#Calculating the speed of the driven pulley\n", "N21 = round(N1*(d1/d2),1) \t\t\t#rpm\n", "\n", "#Calculating the speed of the shaft considering creep\n", "N22 = N1*(d1/d2)*(E+math.sqrt(sigma2))/(E+math.sqrt(sigma1))\t\t\t#rpm\n", "#Calculating the speed lost by the driven pulley due to creep\n", "Nl = N21-N22 \t\t\t#Speed lost by the driven pulley due to creep rpm\n", "\n", "#Results:\n", "print \" Speed lost by the driven pulley due to creep = %.3f rpm.\"%(Nl)\n", "\n", "# note : answer is accurate. please check using calculator." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Speed lost by the driven pulley due to creep = 0.012 rpm.\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.3 Page No : 337" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import linalg\n", "from scipy.optimize import fsolve \n", "import math \n", "\n", "# Variables:\n", "N1 = 160. #rpm\n", "N3 = N1 #rpm\n", "N5 = N3 #rpm\n", "N2 = 60. #rpm\n", "N4 = 80. #rpm\n", "N6 = 100. \t\t\t#rpm\n", "x = 720.\n", "r1 = 40. \t\t\t#mm\n", "\n", "#Solution:\n", "#For a crossed belt:\n", "#Calcluating the radii of pulleys 2 3 4 5 and 6 \n", "r2 = r1*(N1/N2) \t\t\t#mm\n", "#For pulleys 3 and 4 r4 = r3*(N3/N4) or r3*(N3/N4)-r4 = 0\n", "#For a crossed belt drive r3+r4 = r1+r2\n", "A = [[N3/N4, -1],[ 1, 1]]\n", "B = [0, r1+r2]\n", "V = linalg.solve(A,B)\n", "r3 = V[0] \t\t\t#mm\n", "r4 = V[1] \t\t\t#mm\n", "#For pulleys 5 and 6 r6 = r5*(N5/N6) or r5*(N5/N6)-r6 = 0\n", "#For a crossed belt drive r5+r6 = r1+r2\n", "A = [[N5/N6, -1],[ 1, 1]]\n", "B = [0, r1+r2]\n", "V = linalg.solve(A,B)\n", "r5 = V[0] \t\t\t#mm\n", "r6 = V[1] \t\t\t#mm\n", "\n", "#Results:\n", "print \" For a crossed belt, r2 = %.1fmm;\"%(r2)\n", "print \" r3 = %.1f mm;\"%(r3)\n", "print \" r4 = %.1f mm;\"%(r4)\n", "print \" r5 = %.1f mm;\"%(r5)\n", "print \" r6 = %.1f mm.\"%(r6)\n", "#For an open belt:\n", "#Calcluating the radii of pulleys 2 3 4 5 and 6\n", "r2 = r1*(N1/N2) \t\t\t#mm\n", "#Calculating the length of belt for an open belt drive\n", "L = math.pi*(r1+r2)+(r2-r1)**2/x+2*x \t\t\t#mm\n", "#For pulleys 3 and 4 r4 = r3*(N3/N4) or r3*(N3/N4)-r4 = 0\n", "#Since L is constant for pulleys 3 and 4 pi*(r3+r4)+(r4-r3)**2/x+2*x-L = 0\n", "def f(a):\n", " r3 = a[0]\n", " r4 = a[1]\n", " y = [0,0]\n", " y[0] = r3*(N3/N4)-r4\n", " y[1] = math.pi*(r3+r4)+(r4-r3)**2/x+2*x-L\n", " return y\n", "\n", "z = fsolve(f,[1,1])\n", "r3 = z[0] \t\t\t#mm\n", "r4 = z[1] \t\t\t#mm\n", "#For pulleys 5 and 6 r6 = r5*(N5/N6) or r5*(N5/N6)-r6 = 0\n", "#Since L is constant for pulleys 5 and 6 pi*(r5+r6)+(r6-r5)**2/x+2*x-L = 0\n", "def f1(a):\n", " r5 = a[0]\n", " r6 = a[1]\n", " y = [0,0]\n", " y[0] = r5*(N5/N6)-r6\n", " y[1] = math.pi*(r5+r6)+(r6-r5)**2/x+2*x-L\n", " return y\n", "\n", "z = fsolve(f1,[1,1])\n", "r5 = z[0] \t\t\t#mm\n", "r6 = z[1] \t\t\t#mm\n", "\n", "#Results:\n", "print \" For an open belt, r2 = %.1fmm\"%(r2)\n", "print \" r3 = %.1f mm;\"%(r3)\n", "print \" r4 = %.1f mm;\"%(r4)\n", "print \" r5 = %d mm;\"%(round(r5,-1))\n", "print \" r6 = %d mm.\"%(r6)\n", "\n", "# note : answers are slightly different because of fsolve function and rounding off error." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " For a crossed belt, r2 = 106.7mm;\n", " r3 = 48.9 mm;\n", " r4 = 97.8 mm;\n", " r5 = 56.4 mm;\n", " r6 = 90.3 mm.\n", " For an open belt, r2 = 106.7mm\n", " r3 = 49.2 mm;\n", " r4 = 98.4 mm;\n", " r5 = 60 mm;\n", " r6 = 91 mm.\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.4 Page No : 342" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "d = 600./1000 \t\t\t#m\n", "N = 200. \t\t\t #rpm\n", "mu = 0.25\n", "theta = 160*math.pi/180 \t\t\t#radians\n", "T1 = 2500. \t\t\t#N\n", "\n", "#Solution:\n", "#Calcluating the velocity of the belt\n", "v = math.pi*d*N/60 \t\t\t#m/s\n", "#Calcluating the tension in the slack side of the belt\n", "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", "#Calcluating the power transmitted by the belt\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "\n", "#Results:\n", "print \" Power transmitted by the belt, P = %.2f kW.\"%(P)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Power transmitted by the belt, P = 7.89 kW.\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.5 Page No : 343" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "W = 9.*1000\n", "T1 = W \t\t\t#N\n", "d = 300./1000 \t#m\n", "N = 20. \t\t#rpm\n", "mu = 0.25\n", "\n", "#Solution:\n", "#Force required by the man:\n", "#Calculating the angle of contact\n", "theta = 2.5*2*math.pi \t\t\t#rad\n", "#Calculating the force required by the man\n", "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", "#Power to raise the casting:\n", "#Calculating the velocity of the rope\n", "v = math.pi*d*N/60 \t\t\t#m/s\n", "#Calculating the power to raise the casting\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "\n", "#Results:\n", "print \" Force required by the man, T2 = %.2f N.\"%(T2)\n", "print \" Power to raise the casting, P = %.3f kW.\"%(P)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Force required by the man, T2 = 177.33 N.\n", " Power to raise the casting, P = 2.772 kW.\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.6 Page No : 344" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "d1 = 450./1000 #mm\n", "r1 = d1/2 #m\n", "d2 = 200./1000 #m\n", "r2 = d2/2\n", "x = 1.95 \t\t\t#m\n", "N1 = 200. \t\t\t#rpm\n", "T1 = 1.*1000 \t\t\t#N\n", "mu = 0.25\n", "\n", "#Solution:\n", "#Calculating the speed of the belt\n", "v = math.pi*d1*N1/60 \t\t\t#m/s\n", "#Length of the belt:\n", "#Calculating the length of the crossed belt\n", "L = math.pi*(r1+r2)+2*x+(r1+r2)**2/x \t\t\t#m\n", "#Angle of contact between the belt and each pulley:\n", "#Calculating the angle alpha\n", "alpha = math.sin((r1+r2)/x)*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of contact between the belt and each pulley\n", "theta = (180+2*alpha)*math.pi/180 \t\t\t#radians\n", "#Power transmitted:\n", "#Calculating the tension in the slack side of the belt\n", "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", "#Calculating the power transmitted\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "\n", "#Results:\n", "print \" Length of the belt, L = %.3f m.\"%(L)\n", "print \" Angle of contact between the belt and each pulley, theta = %.3f rad.\"%(theta)\n", "print \" Power transmitted, P = %.2f kW.\"%(P)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Length of the belt, L = 4.975 m.\n", " Angle of contact between the belt and each pulley, theta = 3.473 rad.\n", " Power transmitted, P = 2.73 kW.\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.7 Page No : 347" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import linalg\n", "import math \n", "\n", "# Variables:\n", "N1 = 200.\n", "N2 = 300. \t\t\t#rpm\n", "P = 6.*1000 \t\t#W\n", "b = 100.\n", "t = 10. \t\t\t#mm\n", "x = 4.\n", "d2 = 0.5 \t\t\t#m\n", "mu = 0.3\n", "\n", "#Solution:\n", "#Stress in the belt for an open belt drive:\n", "#Calculating the diameter of the larger pulley\n", "d1 = d2*(N2/N1) \t\t\t#m\n", "#Calculating the velocity of the belt\n", "v = math.pi*d2*N2/60 \t\t\t#m/s\n", "#Calculating the angle alpha for an open belt drive\n", "alphao = math.sin((r1-r2)/x)*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of contact on the smaller pulley\n", "thetao = (180-2*alphao)*math.pi/180 \t\t\t#radians\n", "#Calculating the tensions in the belt\n", "#Ratio of the tensions in the belt T1/T2 = math.exp(mu*thetao) or T1-T2*math.exp(mu*thetao) = 0\n", "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", "A = [[1, -math.exp(mu*thetao)],[ 1, -1]]\n", "B = [0, P/v]\n", "V = linalg.solve(A,B)\n", "T1o = V[0] \t\t\t#N\n", "T2o = V[1] \t\t\t#N\n", "#Calculating the stress in the belt\n", "sigmao = T1o/(b*t) \t\t\t#MPa\n", "#Stress in the belt for a cross belt drive:\n", "#Calculating the angle alpha for a cross belt drive\n", "alphac = math.sin((d1+d2)/(2*x))*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of contact\n", "thetac = (180+2*alphac)*math.pi/180 \t\t\t#radians\n", "#Calculating the tensions in the belt \n", "#Ratio of the tensions in the belt T1/T2 = math.exp(mu*thetac) or T1-T2*math.exp(mu*thetac) = 0\n", "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", "A = [[1, -math.exp(mu*thetac)],[ 1, -1]]\n", "B = [0, P/v]\n", "V = linalg.solve(A,B)\n", "T1c = V[0] \t\t\t#N\n", "T2c = V[1] \t\t\t#N\n", "#Calculating the stress in the belt\n", "sigmac = T1c/(b*t) \t\t\t#MPa\n", "\n", "#Results:\n", "print \" Stress in the belt for an open belt drive, sigma = %.3f MPa.\"%(sigmao)\n", "print \" Stress in the belt for a cross belt drive, sigma = %.3f MPa.\"%(sigmac)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Stress in the belt for an open belt drive, sigma = 1.267 MPa.\n", " Stress in the belt for a cross belt drive, sigma = 1.184 MPa.\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.8 Page No : 348" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "P = 7.5*1000 \t\t#W\n", "d = 1.2 #m\n", "t = 10./1000 \t\t#m\n", "N = 250. \t\t\t#rpm\n", "theta = 165.*math.pi/180 \t\t\t#radians\n", "mu = 0.3\n", "sigma = 1.5*10**6 \t\t\t#N/m**2\n", "rho = 1.*10**3 \t\t\t#kg/m**3\n", "\n", "#Solution:\n", "#Calculating the velocity of the belt\n", "v = math.pi*d*N/60 \t\t\t#m/s\n", "#Calculating the tensions in the belt\n", "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", "#Ratio of tensions in the belt math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", "A = [[1, -1],[1, -math.exp(mu*theta)]]\n", "B = [P/v, 0]\n", "V = linalg.solve(A,B)\n", "T1 = V[0] \t\t\t#N\n", "T2 = V[1] \t\t\t#N\n", "#Calculating the width of the belt\n", "b = T1/(sigma*t-t*1*rho*v**2)*1000 \t\t\t#mm\n", "\n", "#Results:\n", "print \" Width of the belt, b = %.1f mm.\"%(b)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Width of the belt, b = 65.9 mm.\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.9 Page No : 349" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import linalg\n", "import math \n", "\n", "# Variables:\n", "t = 9.75/1000\n", "d1 = 300./1000\n", "x = 3. \t\t\t #m\n", "P = 15.*1000 \t\t\t#W\n", "N1 = 900.\n", "N2 = 300. \t\t\t#rpm\n", "rho = 1000. \t\t\t#kg/m**3\n", "sigma = 2.5*10**6 \t\t\t#N/m**2\n", "mu = 0.3\n", "\n", "#Solution:\n", "#Calculating the diameter of the driven pulley\n", "d2 = d1*(N1/N2) \t\t\t#m\n", "#Calculating the velocity of the belt\n", "v = math.pi*d1*N1/60 \t\t\t#m/s\n", "#Calculating the angle alpha for an open belt drive\n", "alpha = math.sin((d2-d1)/(2*x))*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of lap\n", "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", "#Calculating the tensions in the belt\n", "#Ratio of tensions math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", "A = [[1, -math.exp(mu*theta)],[ 1, -1]]\n", "B = [0, P/v]\n", "V = linalg.solve(A,B)\n", "T1 = V[0] \t\t\t#N\n", "T2 = V[1] \t\t\t#N\n", "#Calculating the width of the belt\n", "b = T1/(sigma*t-t*1*rho*v**2)*1000 \t\t\t#mm\n", "\n", "#Results:\n", "print \" Width of the belt, b = %d mm.\"%(b)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Width of the belt, b = 80 mm.\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.10 Page No : 350" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "theta = 120*math.pi/180 #degrees\t\t\t#radians\n", "b = 100./1000\n", "t = 6./1000 \t\t\t#m\n", "rho = 1000. \t\t\t#kg/m**3\n", "mu = 0.3\n", "sigma = 2*10**6 \t\t\t#N/m**2\n", "\n", "#Solution:\n", "#Speed of the belt for greatest power:\n", "#Calculating the maximum tension in the belt\n", "T = sigma*b*t \t\t\t#N\n", "#Calculating the mass of the belt per metre length\n", "l = 1. \t\t\t#m\n", "m = b*t*l*rho \t\t\t#kg/m\n", "#Calculating the speed of the belt for greatest power\n", "v = math.sqrt(T/(3*m)) \t\t\t#m/s\n", "#Greatest power which the belt can transmit\n", "#Calculating the centrifugal tension for maximum power to be transmitted\n", "TC = T/3 \t\t\t#N\n", "#Calculating the tension in the tight side of the belt\n", "T1 = T-TC \t\t\t#N\n", "#Calculating the tension in the slack side of the belt\n", "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", "#Calculating the greatest power which the belt can transmit\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "\n", "#Results:\n", "print \" Speed of the belt for greatest power, v = %.2f m/s.\"%(v)\n", "print \" Greatest power which the belt can transmit, P = %.2f kW.\"%(P)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Speed of the belt for greatest power, v = 25.82 m/s.\n", " Greatest power which the belt can transmit, P = 9.64 kW.\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.11 Page No : 351" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "d1 = 1.2 #m\n", "r1 = d1/2 #m\n", "d2 = 0.5 #m\n", "r2 = d2/2 #m\n", "x = 4. \t\t\t#m\n", "m = 0.9 \t\t\t#kg/m\n", "T = 2000. \t\t\t#N\n", "mu = 0.3\n", "N1 = 200. #rpm\n", "N2 = 450. \t\t\t#rpm\n", "\n", "#Solution:\n", "#Calculating the velocity of the belt\n", "v = math.pi*d1*N1/60 \t\t\t#m/s\n", "#Calculating the centrifugal tension\n", "TC = m*v**2 \t\t\t#N\n", "#Calculating the tension in the tight side of the belt\n", "T1 = T-TC \t\t\t#N\n", "#Calculating the angle alpha for an open belt drive\n", "alpha = math.sin((r1-r2)/x)*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of lap on the smaller pulley\n", "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", "#Calculating the tension in the slack side of the belt\n", "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", "#Calculating the torque on the shaft of larger pulley\n", "TL = (T1-T2)*r1 \t\t\t#N-m\n", "#Calculating the torque on the shaft of smaller pulley\n", "TS = (T1-T2)*r2 \t\t\t#N-m\n", "#Calculating the power transmitted\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "#Power lost in friction:\n", "#Calculating the input power\n", "P1 = TL*2*math.pi*N1/(60*1000) \t\t\t#kW\n", "#Calculating the output power\n", "P2 = TS*2*math.pi*N2/(60*1000) \t\t\t#kW\n", "#Calculating the power lost in friction\n", "Pf = P1-P2 \t\t\t#Power lost in friction kW\n", "#Calculating the efficiency of the drive\n", "eta = P2/P1*100 \t\t\t#%\n", "\n", "#Results:\n", "print \" Torque on the shaft of larger pulley, TL = %.1f N-m.\"%(TL)\n", "print \" Torque on the shaft of smaller pulley, TS = %d N-m.\"%(TS)\n", "print \" Power transmitted, P = %.2f kW.\"%(P)\n", "print \" Power lost in friction = %.2f kW.\"%(Pf)\n", "print \" Efficiency of the drive, eta = %.1f %%.\"%(eta)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Torque on the shaft of larger pulley, TL = 657.0 N-m.\n", " Torque on the shaft of smaller pulley, TS = 273 N-m.\n", " Power transmitted, P = 13.76 kW.\n", " Power lost in friction = 0.86 kW.\n", " Efficiency of the drive, eta = 93.8 %.\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.12 Page No : 353" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import linalg\n", "import math \n", "\n", "# Variables:\n", "T0 = 2000. \t\t\t#N\n", "mu0 = 0.3\n", "theta = 150.*math.pi/180 \t\t\t#radians\n", "r2 = 200./1000\n", "d2 = 2*r2 \t\t\t#m\n", "N2 = 500.\t\t\t#rpm\n", "\n", "#Solution:\n", "#Calculating the velocity of the belt\n", "v = math.pi*d2*N2/60 \t\t\t#m/s\n", "#Calculating the tensions in the belt\n", "#Initial tension T0 = (T1+T2)/2 or T1+T2 = 2*T0\n", "#Ratio of the tensions in the belt math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", "A = [[1, 1],[ 1 ,-math.exp(mu*theta)]]\n", "B = [2*T0, 0]\n", "V = linalg.solve(A,B)\n", "T1 = V[0] \t\t\t#N\n", "T2 = V[1] \t\t\t#N\n", "#Calculating the power transmitted\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "\n", "#Results:\n", "print \" Power transmitted, P = %.1f kW.\"%(P)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Power transmitted, P = 15.7 kW.\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.13 Page No : 354" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import linalg\n", "import math \n", "\n", "# Variables:\n", "x = 4.8\n", "d1 = 1.5 #m\n", "d2 = 1. \t\t\t#m\n", "T0 = 3.*1000 \t\t#N\n", "m = 1.5 \t\t\t#kg/m\n", "mu = 0.3\n", "N2 = 400. \t\t\t#rpm\n", "\n", "#Solution:\n", "#Calculating the velocity of the belt\n", "v = math.pi*d2*N2/60 \t\t\t#m/s\n", "#Calculating the centrifugal tension\n", "TC = m*v**2 \t\t\t#N\n", "#Calculating the angle alpha\n", "alpha = math.sin((d1-d2)/(2*x))*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of lap for the smaller pulley\n", "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", "#Calculating the tensions in the belt\n", "#Initial tension T0 = (T1+T2+2*TC)/2 or T1+T2 = 2*(T0-TC)\n", "#Ratio of tensions in the belt math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", "A = [[1, 1],[1, -math.exp(mu*theta)]]\n", "B = [2*(T0-TC), 0]\n", "V = linalg.solve(A,B)\n", "T1 = V[0] \t\t\t#N\n", "T2 = V[1] \t\t\t#N\n", "#Calculating the power transmitted\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "\n", "#Results:\n", "print \" Power transmitted, P = %.f kW.\"%(P)\n", "\n", "# answer are slightly different because of solve function" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Power transmitted, P = 42 kW.\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.14 Page No : 355" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "x = 1.2 #m\n", "d2 = 400./1000 #m\n", "t = 5./1000 #m\n", "b = 80./1000 \t\t#m\n", "N1 = 350. #rpm\n", "N2 = 140. \t\t\t#rpm\n", "mu = 0.3\n", "sigma = 1.4*10**6 \t\t\t#N/m**2\n", "\n", "#Solution:\n", "#Calculating the diameter of the driving pulley\n", "d1 = d2*(N2/N1) \t\t\t#m\n", "#Maximum power transmitted by the belting:\n", "#Refer Fig. 11.18\n", "#Calculating the angle alpha\n", "alpha = math.sin((d2-d1)/(2*x))*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of contact of the belt on the driving pulley\n", "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", "#Calculating the maximum tension to which the belt can be subjected\n", "T1 = sigma*b*t \t\t\t#N\n", "#Calculating the tension in the slack side of the belt\n", "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", "#Calculating the velocity of the belt\n", "v = math.pi*d1*N1/60 \t\t\t#m/s\n", "#Calculating the power transmitted\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "#Calculating the required initial belt tension\n", "T0 = (T1+T2)/2 \t\t\t#N\n", "\n", "#Results:\n", "print \" Diameter of the driving pulley, d1 = %.2f m.\"%(d1)\n", "print \" Maximum power transmitted by the belting, P = %.3f kW.\"%(P)\n", "print \" Required initial belt tension, T0 = %.1f N.\"%(T0)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Diameter of the driving pulley, d1 = 0.16 m.\n", " Maximum power transmitted by the belting, P = 0.963 kW.\n", " Required initial belt tension, T0 = 395.8 N.\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.15 Page No : 356" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import linalg\n", "import math \n", "\n", "# Variables:\n", "d2 = 240./1000 #m\n", "d1 = 600./1000 #m\n", "x = 3. \t\t\t #m\n", "P = 4.*1000 \t\t#W\n", "N2 = 300. \t\t\t#rpm\n", "mu = 0.3\n", "T1s = 10. \t\t\t#Safe working tension N/mm width\n", "\n", "#Solution:\n", "#Minimum width of the belt:\n", "#Calculating the velocity of the belt\n", "v = math.pi*d2*N2/60 \t\t\t#m/s\n", "#Calculating the angle alpha for an open belt drive\n", "alpha = math.sin((d1-d2)/(2*x))*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of lap on the smaller pulley\n", "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", "#Calculating the tensions in the belt\n", "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", "#Ratio of tensions math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", "A = [[1, -1],[ 1, -math.exp(mu*theta)]]\n", "B = [P/v, 0]\n", "V = linalg.solve(A, B)\n", "T1 = V[0] \t\t\t#N\n", "T2 = V[1] \t\t\t#N\n", "#Calculating the minimum width of the belt\n", "b = T1/T1s \t\t\t#mm\n", "#Calculating the initial belt tension\n", "T0 = (T1+T2)/2 \t\t\t#N\n", "#Calculating the length of the belt required\n", "L = math.pi/2*(d1+d2)+2*x+(d1-d2)**2/(4*x) \t\t\t#m\n", "\n", "#Results:\n", "print \" Minimum width of the belt, b = %.1f mm.\"%(b)\n", "print \" Initial belt tension, T0 = %.1f N.\"%(T0)\n", "print \" Length of the belt required, L = %.2f m.\"%(L)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Minimum width of the belt, b = 178.0 mm.\n", " Initial belt tension, T0 = 1249.5 N.\n", " Length of the belt required, L = 7.33 m.\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.16 Page No : 357" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import linalg\n", "import math\n", "\n", "# Variables:\n", "d1 = 400./1000 #m\n", "d2 = 250./1000 #m\n", "x = 2.\n", "mu = 0.4 \t\t\t#m\n", "T = 1200. \t\t\t#N\n", "v = 10. \t\t\t#m/s\n", "\n", "#Solution:\n", "#Power transmitted:\n", "#Calculating the angle alpha for an open belt drive\n", "alpha = math.sin((d1-d2)/(2*x))*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of contact\n", "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", "#Calculating the tension in the tight side of the belt\n", "T1 = T \t\t\t#Neglecting centrifugal tension N\n", "#Calculating the tension in the slack side of the belt\n", "T2 = T1/math.exp(mu*theta) \t\t\t#N\n", "#Calculating the power transmitted\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "\n", "#Results:\n", "print \" Power transmitted, P = %.2f kW.\"%(P)\n", "#Power transmitted when initial tension is increased by 10%:\n", "#Calculating the initial tension\n", "T0 = (T1+T2)/2 \t\t\t#N\n", "#Calculating the increased initial tension\n", "T0dash = T0+10./100*T0 \t\t\t#N\n", "#Calculating the corresponding tensions in the belt\n", "#We have T0dash = (T1+T2)/2 or T1+T2 = 2*T0dash\n", "#Ratio of the tensions math.log(T1/T2) = mu*theta or T1-T2*math.exp(mu*theta) = 0\n", "A = [[1, 1],[ 1, -math.exp(mu*theta)]]\n", "B = [2*T0dash, 0]\n", "V = linalg.solve(A,B)\n", "T1 = V[0] \t\t\t#N\n", "T2 = V[1] \t\t\t#N\n", "#Calculating the power transmitted\n", "P1 = (T1-T2)*v/1000 \t\t\t#kW\n", "#Power transmitted when coefficient of friction is increased by 10%:\n", "#Calculating the increased coefficient of friction\n", "mudash = mu+10./100*mu\n", "#Calculating the corresponding tensions in the belt\n", "#Ratio of the tensions math.log(T1/T2) = mudash*theta or T1-T2*math.exp(mudash*theta) = 0\n", "#Initial tension T0 = (T1+T2)/2 or T1+T2 = 2*T0\n", "A = [[1, -math.exp(mudash*theta)],[ 1, 1]]\n", "B = [0, 2*T0]\n", "V = linalg.solve(A,B)\n", "T1 = V[0] \t\t\t#N\n", "T2 = V[1] \t\t\t#N\n", "#Calculating the power transmitted\n", "P2 = (T1-T2)*v/1000 \t\t\t#kW\n", "\n", "#Results:\n", "if P1>P2:\n", " print \" Since the power transmitted by increasing the initial tension is more\\\n", " therefore in order to increase the power transmitted we shall adopt\\\n", " the method of increasing the initial tension.\"\n", "else:\n", " print \" Since the power transmitted by increasing the coefficient of friction is more,\\\n", " therefore in order to increase the power transmitted we shall adopt the method of increasing\\\n", " the coefficient of friction.\"\n", "#Percentage increase in power:\\\n", "#Calculating the percentage increase in power when the initial tension is increased\n", "I1 = (P1-P)/P*100. \t\t\t#Percentage increase in power when the initial tension is increased %\n", "#Calculating the percentage increase in power when coefficient of friction is increased\n", "I2 = (P2-P)/P*100. \t\t\t#Percentage increase in power when coefficient of friction is increased %\n", "\n", "\n", "#Results:\n", "print \" Percentage increase in power when the initial tension is increased = %.2f %%.\"%(I1)\n", "print \" Percentage increase in power when coefficient of friction is increased = %.1f %%.\"%(I2)\n", "\n", "#rounding off error" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Power transmitted, P = 8.48 kW.\n", " Since the power transmitted by increasing the initial tension is more therefore in order to increase the power transmitted we shall adopt the method of increasing the initial tension.\n", " Percentage increase in power when the initial tension is increased = 10.00 %.\n", " Percentage increase in power when coefficient of friction is increased = 7.6 %.\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.17 Page No : 362" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "beta = 30./2 \t\t\t#degrees\n", "alpha = 750.*10**-6 \t#mm**2\n", "mu = 0.12\n", "rho = 1.2*1000 \t\t\t#kg/m**3\n", "sigma = 7.*10**6 \t\t#N/m**2\n", "d = 300./1000 \t\t\t#m\n", "N = 1500. \t\t\t #rpm\n", "\n", "#Solution:\n", "#Power transmitted:\n", "#Calculating the velocity of the belt\n", "v = math.pi*d*N/60 \t\t\t#m/s\n", "#Calculating the mass of the belt per metre length\n", "l = 1 \t\t\t#m\n", "m = alpha*l*rho \t\t\t#kg/m\n", "#Calculating the centrifugal tension\n", "TC = m*v**2 \t\t\t#N\n", "#Calculating the maximum tension in the belt\n", "T = sigma*alpha \t\t\t#N\n", "#Calculating the tension in the tight side of the belt\n", "T1 = T-TC \t\t\t#N\n", "#Calculating the tension in the slack side of the belt\n", "theta = math.pi \t\t\t#Angle of contact radians\n", "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", "#Calculating the power transmitted\n", "P = (T1-T2)*v*2/1000 \t\t\t#kW\n", "#Shaft speed:\n", "#Calculating the belt speed for maximum power transmitted\n", "v1 = math.sqrt(T/(3*m)) \t\t\t#m/s\n", "#Calculating the shaft speed for maximum power transmitted\n", "N1 = v1*60/(math.pi*d) \t\t\t#rpm\n", "\n", "#Results:\n", "print \" Power transmitted, P = %.3f kW.\"%(P)\n", "print \" Shaft speed at which the power transmitted would be maximum, N1 = %d rpm.\"%(N1)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Power transmitted, P = 171.690 kW.\n", " Shaft speed at which the power transmitted would be maximum, N1 = 2807 rpm.\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.18 Page No : 363" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "beta = 30./2 \t\t\t#degrees\n", "t = 20./1000\n", "b = 20./1000 \t\t\t#m\n", "m = 0.35 \t\t\t #kg/m\n", "sigma = 1.4*10**6 \t\t\t#N/m**2\n", "theta = 140.*math.pi/180 \t\t\t#radians\n", "mu = 0.15\n", "\n", "#Solution:\n", "#Calculating the maximum tension in the belt\n", "T = sigma*b*t \t\t\t#N\n", "#Calculating the velocity of the belt for maximum power to be transmitted\n", "v = math.sqrt(T/(3*m)) \t\t\t#m/s\n", "#Calculating the centrifugal tension\n", "TC = T/3 \t\t\t#N\n", "#Calculating the tension in the tight side of the belt\n", "T1 = T-TC \t\t\t#N\n", "#Calculating the tension in the slack side of the belt\n", "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", "#Calculating the maximum power transmitted\n", "P = (T1-T2)*v/1000 \t\t\t#kW\n", "\n", "#Results:\n", "print \" Maximum power transmitted, P = %.2f kW.\"%(P)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Maximum power transmitted, P = 6.53 kW.\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.19 Page No : 363" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "P = 90. \t\t\t#kW\n", "N2 = 250.\n", "N1 = 750. \t\t\t#rpm\n", "d2 = 1.\n", "x = 1.75 \t\t\t#m\n", "v = 1600./60 \t\t\t#m/s\n", "a = 375.*10**-6 \t\t\t#m**2\n", "rho = 1000. \t\t\t#kg/m**3\n", "sigma = 2.5*10**6 \t\t\t#N/m**2\n", "beta = 35./2 \t\t\t#degrees\n", "mu = 0.25\n", "\n", "#Solution:\n", "#Calculating the diameter of the pulley on the motor shaft\n", "d1 = d2*(N2/N1) \t\t\t#m\n", "#Calculating the mass of the belt per metre length\n", "l = 1 \t\t\t #m\n", "m = a*l*rho \t\t\t#kg/m\n", "#Calculating the centrifugal tension\n", "TC = m*v**2 \t\t\t#N\n", "#Calculating the maximum tension in the belt\n", "T = sigma*a \t\t\t#N\n", "#Calculating the tension in the tight side of the belt\n", "T1 = T-TC \t\t\t #N\n", "#Refer Fig. 11.21\n", "#Calculating the angle alpha\n", "alpha = math.sin(math.radians((d2-d1)/(2*x)))*180/math.pi \t\t\t#degrees\n", "#Calculating the angle of lap on smaller pulley\n", "theta = (180-2*alpha)*math.pi/180 \t\t\t#radians\n", "#Calculating the tension in the slack side of the belt\n", "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", "#Number of V-belts:\n", "#Calculating the power transmitted per belt\n", "P1 = (T1-T2)*v/1000 \t\t\t#Power transmitted per belt kW\n", "#Calculating the number of V-belts\n", "n = P/16.086 \t\t\t#Number of V-belts\n", "#Calculating the length each of belt for an open belt drive\n", "L = math.pi/2*(d2+d1)+2*x+(d2-d1)**2/(4*x) \t\t\t#m\n", "\n", "#Results:\n", "print \" Number of V-belts = %.f.\"%(n)\n", "print \" Length of each belt, L = %.2f m.\"%(L)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Number of V-belts = 6.\n", " Length of each belt, L = 5.66 m.\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.20 Page No : 367" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "P = 600. \t\t\t#kW\n", "d = 4. \t\t\t #m\n", "N = 90. \t\t\t#rpm\n", "theta = 160.*math.pi/180 \t\t\t#radians\n", "beta = 45./2 \t\t\t#degrees\n", "mu = 0.28\n", "m = 1.5 \t\t\t#kg/m\n", "T = 2400. \t\t\t#N\n", "\n", "#Solution:\n", "#Calculating the velocity of the rope\n", "v = math.pi*d*N/60 \t\t\t#m/s\n", "#Calculating the centrifugal tension\n", "TC = m*v**2 \t\t\t#N\n", "#Calculating the tension in the tight side of the rope\n", "T1 = T-TC \t\t\t#N\n", "#Calculating the tension in the slack side of the belt\n", "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", "#Calculating the power transmitted per rope\n", "P1 = (T1-T2)*v/1000 \t\t\t#Power transmitted per rope kW\n", "#Calculating the number of ropes\n", "n = P/P1 \t\t\t#Number of ropes\n", "\n", "#Results:\n", "print \" Number of ropes required = %d.\"%(n+1)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Number of ropes required = 20.\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.21 Page No : 367" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "d = 3.6 \t\t\t#m\n", "n = 15. \t\t\t#Number of grooves\n", "beta = 45./2 \t\t#degrees\n", "theta = 170.*math.pi/180 \t\t\t#radians\n", "mu = 0.28\n", "T = 960. \t\t\t#N\n", "m = 1.5 \t\t\t#kg/m\n", "\n", "#Solution:\n", "#Speed of the pulley:\n", "#Calculating the velocity of the rope\n", "v = math.sqrt(T/(3*m)) \t\t\t#m/s\n", "#Calculating the speed of the pulley\n", "N = v*60/(math.pi*d) \t\t\t#rpm\n", "#Power transmitted\n", "#Calculating the centrifugal tension for maximum power\n", "TC = T/3 \t\t\t#N\n", "#Calculating the tension in the tight side of the rope\n", "T1 = T-TC \t\t\t#N\n", "#Calculating the tension in the slack side of the rope\n", "T2 = T1/math.exp(mu*theta*(1/math.sin(math.radians(beta)))) \t\t\t#N\n", "#Calculating the power transmitted per rope\n", "P1 = (T1-T2)*v/1000 \t\t\t#Power transmitted per rope kW\n", "#Calculating the total power transmitted\n", "P = P1*n \t\t\t#Total power transmitted kW\n", "\n", "#Results:\n", "print \" Speed of the pulley for maximum power, N = %.1f rpm.\"%(N)\n", "print \" Power transmitted = %.2f kW.\"%(P)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Speed of the pulley for maximum power, N = 77.5 rpm.\n", " Power transmitted = 124.22 kW.\n" ] } ], "prompt_number": 32 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.22 Page No : 368" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import linalg\n", "import math \n", "\n", "# Variables:\n", "PT = 24. \t\t\t#kW\n", "d = 400./1000 \t\t#m\n", "N = 110. \t\t\t#rpm\n", "beta = 45./2 \t\t#degrees\n", "theta = 160.*math.pi/180 \t\t\t#radians\n", "mu = 0.28\n", "n = 10.\n", "\n", "#Solution:\n", "#Initial tension:\n", "#Calculating the power transmitted per rope\n", "P = PT/n*1000 \t\t\t#W\n", "#Calculating the velocity of the rope\n", "v = math.pi*d*N/60 \t\t\t#m/s\n", "#Calculating the tensions in the rope\n", "#Power transmitted P = (T1-T2)*v or T1-T2 = P/v\n", "#Ratio of tensions math.log(T1/T2) = mu*theta*(1/math.sin(math.radians(beta)) or T1-T2*math.exp(mu*theta*(1/math.math.sin(math.radians(beta))) = 0\n", "A = [[1, -1],[ 1, -math.exp(mu*theta*(1/math.sin(math.radians(beta))))]]\n", "B = [P/v, 0]\n", "V = linalg.solve(A,B)\n", "T1 = V[0] \t\t\t#N\n", "T2 = V[1] \t\t\t#N\n", "#Calculating the initial tension in each rope\n", "T0 = (T1+T2)/2 \t\t\t#N\n", "#Diameter of each rope:\n", "#Calculating the girth of rope\n", "C = math.sqrt(T1/(122.*10**3-53.*v**2))*1000 \t\t\t#mm\n", "#Calculating the diameter of each rope\n", "d1 = C/math.pi \t\t\t#mm\n", "\n", "#Results:\n", "print \" Initial tension, T0 = %.2f N.\"%(T0)\n", "print \" Diameter of each rope, d1 = %.2f mm.\"%(d1)\n", "\n", "# rounding off error" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Initial tension, T0 = 676.00 N.\n", " Diameter of each rope, d1 = 31.56 mm.\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.23 Page No : 376" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables:\n", "N1 = 240. #rpm\n", "N2 = 120. \t\t\t#rpm\n", "T1 = 20.\n", "d2 = 600./1000 #m\n", "r2 = d2/2 #m\n", "x = 800./1000 \t\t#m\n", "\n", "#SOlution:\n", "#Calculating the number of teeth on the drive sprocket\n", "T2 = T1*(N1/N2)\n", "#Calculating the pitch of the chain\n", "p = r2*2*math.sin(math.radians(180/T2))*1000 \t\t\t#mm\n", "#Length of the chain:\n", "m = x*1000/p\n", "#Calculating the multiplying factor\n", "K = (T1+T2)/2+2*m+(1/math.sin(math.radians(180/T1))-1/math.sin(math.radians(180/T2)))**2/(4*m)\n", "#Calculating the length of the chain\n", "L = p*K/1000 \t\t\t#m\n", "\n", "#Results:\n", "print \" Number of teeth on the driven sprocket, T2 = %d.\"%(T2)\n", "print \" Pitch of the chain, p = %.1f mm.\"%(p)\n", "print \" Length of the chain, L = %.4f m.\"%(L)\n", "\n", "# rounding off error" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " Number of teeth on the driven sprocket, T2 = 40.\n", " Pitch of the chain, p = 47.1 mm.\n", " Length of the chain, L = 3.0402 m.\n" ] } ], "prompt_number": 23 } ], "metadata": {} } ] }