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authorkinitrupti2017-05-12 18:40:35 +0530
committerkinitrupti2017-05-12 18:40:35 +0530
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-{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- " Chapter 2:Dynamics of Electric Drives"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:2.1,Page no:16"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "\n",
- "#variable declaration\n",
- "Jo=0.2 # inertia of the motor in kg-m2\n",
- "a1=0.1 # reduction gear\n",
- "J1=10 # inertia of the load in kg-m2\n",
- "Tl1=10 # load torque\n",
- "v=1.5 # speed of the translational load \n",
- "M1=1000 # mass of the translational load\n",
- "N=1420 # speed of the motor\n",
- "n1=.9 # efficiency of the reduction gear\n",
- "n1_=0.85 # efficiency of the translational load and the motor\n",
- "F1=M1*9.81 # force of the translational load \n",
- "\n",
- "#Calculation\n",
- "Wm=N*math.pi/30 #angular speed\n",
- "J=Jo+a1**2*J1+ M1*(v/Wm)**2 # total equivalent moment of inertia\n",
- "Tl= a1*Tl1/n1+F1/n1_*(v/Wm) # total equivalent torque\n",
- "\n",
- "#Result\n",
- "print\"\\nEquivalent moment of inertia is :\",round(J,1),\"kg-m2\"\n",
- "print\"\\nEquivalent load torque :\",round(Tl,2),\"N-m\"\n",
- "\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "Equivalent moment of inertia is: 0.4 kg-m2\n",
- "\n",
- "Equivalent load torque : 117.53 N-m\n"
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:2.2,Page no:22"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import scipy\n",
- "from scipy import integrate\n",
- "import math\n",
- "\n",
- "# variable declaration\n",
- "J=10 #moment of inertia of the drive in kg-m2\n",
- "print(\"Passive load torque during steady state is :Tl=0.05*N in N-m\")\n",
- "print(\"And load torque : T=100-0.1*N in N-m \")\n",
- "print(\"load torque when the direction is reversed T=-100-0.1*N in N-m\")\n",
- "\n",
- "#Calculation\n",
- "print(\"T-Tl=0\")\n",
- "print(\"100-0.1*N-0.05*N=0\")\n",
- "N=100/0.15 #Required speed of the motor in rpm during steady state\n",
- "N2=-100/0.15 #During reversal speed is in opposite direction\n",
- "print(\"\\nJdWm/dt=-100-0.1*N-0.05*N during reversing\")\n",
- "print(\"dN/dt=30/(J*pi)*(-100-0.15*N)\")\n",
- "print(\"dN/dt=(-95.49-0.143*N)\")\n",
- "N1=N\n",
- "N2=N2*0.95 #for speed reversal \n",
- "x2 = lambda N: 1/(-95.49-0.143*N)\n",
- "t=integrate.quad(x2, round(N1), round(N2))\n",
- "\n",
- "#result\n",
- "print\"\\nHence Time of reversal is :\",round(t[0],2),\"s\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Passive load torque during steady state is :Tl=0.05*N in N-m\n",
- "And load torque : T=100-0.1*N in N-m \n",
- "load torque when the direction is reversed T=-100-0.1*N in N-m\n",
- "T-Tl=0\n",
- "100-0.1*N-0.05*N=0\n",
- "\n",
- "JdWm/dt=-100-0.1*N-0.05*N during reversing\n",
- "dN/dt=30/(J*pi)*(-100-0.15*N)\n",
- "dN/dt=(-95.49-0.143*N)\n",
- "\n",
- "Hence Time of reversal is : 25.51 s\n"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example no:2.3,Page no:27"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#variable declaration\n",
- "Tlh=1000 # load torque in N-m\n",
- "Tmax=700 # maximum motor torque\n",
- "Tll=200 # light load for the motor to regain its steady state\n",
- "Tmin=Tll # minimum torque\n",
- "t_h=10 # period for which a load torque of 1000 N-m is apllied in sec\n",
- "Jm=10 # moment of inertia of the motor in Kg-m2\n",
- "No=500 # no load speed in rpm\n",
- "Tr=500 # torque at a given no load speed in N-m\n",
- "\n",
- "#Calculation\n",
- "Wmo=No*2*math.pi/60 # angular no load speed in rad/sec\n",
- "s=0.05 # slip at a torque of 500 N-m\n",
- "Wmr=(1-s)*Wmo # angular speed at a torque of 500 N-m in rad/sec\n",
- "\n",
- "y=math.log((Tlh-Tmin)/(Tlh-Tmax))\n",
- "x=Tr/(Wmo-Wmr)\n",
- "\n",
- "J=x*t_h/y\n",
- "Jf=J-Jm\n",
- "\n",
- "#Result \n",
- "#answer in the book is wrong\n",
- "print\"\\n\\nMoment of inertia of the flywheel : \", round(Jf,1),\"Kg-m2\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- "Moment of inertia of the flywheel : 1937.2 Kg-m2\n"
- ]
- }
- ],
- "prompt_number": 12
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file