{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 17: FRACTIONAL HORSE POWER MOTORS" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 17.1,Page number: 570\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Question:\n", "\"\"\"Finding the slip and efficiency of induction motor.\"\"\"\n", "\n", "#Variable Declaration:\n", "f=50 #Frequency rating of the induction motor(in Hertz) \n", "P=4 #Number of poles in the induction motor \n", "N=1410 #Speed of the motor(in rpm)\n", "Po=375 #Output Power(in Watts)\n", "V=230 #Voltage rating of the induction motor(in Volts) \n", "I=2.9 #Input current(in Amperes)\n", "pf=0.71 #Power factor(lagging) \n", "\n", "\n", "#Calculations:\n", "Ns=(120.0*f)/P\n", "slip=(Ns-N)/Ns\n", "\n", "\n", "#Result:\n", "print \"Slip is %.2f percent.\" %(slip*100)\n", "Pin=V*I*pf\n", "efficiency=Po/Pin\n", "print \"The efficiency is %.2f percent.\" %(efficiency*100)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Slip is 6.00 percent.\n", "The efficiency is 79.19 percent.\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 17.2,Page number: 570\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Question:\n", "\n", "\"\"\"Finding the currents and the power factor in the induction motor.\"\"\"\n", "\n", "from cmath import phase,rect,polar\n", "from math import radians,degrees,cos\n", "\n", "#Variable Declartion:\n", "V=rect(230,0) #Voltage rating of the split-phase induction motor(in Volts) \n", "Z_M=5+ 12*1j #Impedance of the main winding(in Ohms)\n", "Z_A=12+ 5*1j #Start-winding impedance(in Ohms)\n", "\n", "\n", "#Calculations:\n", "mod_Z_M=abs(Z_M)\n", "mod_Z_A=abs(Z_A)\n", "phi_M=phase(Z_M)\n", "phi_A=phase(Z_A)\n", "I_M=V/Z_M\n", "mod_I_M=abs(I_M)\n", "phi_I_M=degrees(phase(I_M))\n", "I_A=V/Z_A\n", "mod_I_A=abs(I_A)\n", "phi_I_A=degrees(phase(I_A))\n", "I_L=I_M+I_A\n", "mod_I_L=abs(I_L)\n", "phi_I_L=degrees(phase(I_L))\n", "phi=phi_I_A-phi_I_M\n", "pf=cos(radians(phi_I_L))\n", "\n", "\n", "#Result:\n", "print \"(a)The current in the main winding is %.2f A at a phase angle of %.2f degrees.\" %(mod_I_M,phi_I_M)\n", "print \"(b)The current in the starting winding is %.2f A at a phase angle of %.2f degrees.\" %(mod_I_A,phi_I_A)\n", "print \"(c)The line current is %.2f A at a phase angle of %.2f degrees.\" %(mod_I_L,phi_I_L)\n", "print \"(d)The phase displacement between the two winding currents is %.2f degrees.\" %(phi)\n", "print \"(e)The power factor is %.4f lagging.\" %(pf) " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)The current in the main winding is 17.69 A at a phase angle of -67.38 degrees.\n", "(b)The current in the starting winding is 17.69 A at a phase angle of -22.62 degrees.\n", "(c)The line current is 32.72 A at a phase angle of -45.00 degrees.\n", "(d)The phase displacement between the two winding currents is 44.76 degrees.\n", "(e)The power factor is 0.7071 lagging.\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 17.3,Page number: 571\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Question:\n", "\"\"\"Finding the capacitance in series with the auxiliary winding to maximize starting torque.\"\"\"\n", "\n", "from math import radians,degrees,atan,pi,tan\n", "\n", "#Variable Declaration:\n", "X_M=20 #Inductive reactance of the main winding(in Ohm)\n", "R_M=2 #Resistance of the main winding(in Ohm)\n", "X_A=5 #Inductive reactance of the auxiliary winding(in Ohm)\n", "R_A=25 #Resistance of the auxiliary winding(in Ohm)\n", "f=50 #Frequency rating of the split-phase induction motor(in Hertz) \n", "\n", "\n", "#Calculations:\n", "angle_M=atan(X_M/R_M)\n", "angle_A=degrees(angle_M)-90\n", "Xc=X_A-(R_A*tan(radians(angle_A)))\n", "\n", "\n", "#Result:\n", "C=1/(2*pi*f*Xc)\n", "print \"The value of capacitance connected in series with the auxiliary winding to obtain maximum starting torque is %e F.\" %(C)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The value of capacitance connected in series with the auxiliary winding to obtain maximum starting torque is 4.244132e-04 F.\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 17.4,Page number: 576\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Question:\n", "\"\"\"Finding the resolution and shaft speed of a stepper motor.\"\"\"\n", "\n", "#Variable Declaration:\n", "beta=2.5 #Step-angle of a stepper motor(in degrees)\n", "step_freq=3600 #Stepping frequency(in pps)\n", "\n", "\n", "#Calculations:\n", "res=360/beta\n", "number_steps=res*25\n", "shaft_speed=(beta*step_freq)/360\n", "\n", "\n", "#Result:\n", "print \"(a)The resolution is %d steps per revolution.\" %(res)\n", "print \"(b)The number of steps required for the shaft to make 25 revolutions=%d.\" %(number_steps)\n", "print \"(c)The shaft speed is %.2f rps.\" %(shaft_speed)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)The resolution is 144 steps per revolution.\n", "(b)The number of steps required for the shaft to make 25 revolutions=3600.\n", "(c)The shaft speed is 25.00 rps.\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 17.5,Page number:577\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Question:\n", "\"\"\"Finding the number of stator and rotor poles in a VR motor.\"\"\"\n", "\n", "#Variable Declaration:\n", "m=3 #Number of phases\n", "beta=15 #Step angle(in degrees)\n", "\n", "\n", "#Calculations:\n", "Nr=360/(m*beta)\n", "Ns1=(Nr*360)/(360-(beta*Nr))\n", "Ns2=(Nr*360)/(360+(beta*Nr))\n", "\n", "\n", "#Result:\n", "print \"(a) The number of rotor poles is %d.\" %(Nr)\n", "print \"(b)\"\n", "print \" Case 1: Ns>Nr\"\n", "print \" The number of stator poles is %d. \\n\" %(Ns1)\n", "print \" Case 2: NsNr\n", " The number of stator poles is 12. \n", "\n", " Case 2: Ns