{
"metadata": {
"name": "",
"signature": "sha256:090249f83a260ba2b6a1983f6426576be55ba0e881a9fae7d9c5cbc674e22e36"
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"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": {}
}
]
}