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diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter15.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter15.ipynb new file mode 100755 index 00000000..f49bd253 --- /dev/null +++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter15.ipynb @@ -0,0 +1,957 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 15: INDUCTION MOTORS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.1,Page number: 480" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the synchronous speed of an induction motor and the frequency of rotor currents at standstill.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "P=6.0 #Number of poles \n", + "f=50.0 #Operating frequency of the induction motor(in Hertz)\n", + "s_no_load=0.01 #Slip at no-load \n", + "s_full_load=0.03 #Slip at full-load\n", + "\n", + "\n", + "#Calculations:\n", + "Ns=(120.0*f)/P\n", + "N_no_load=Ns*(1-s_no_load)\n", + "N_full_load=Ns*(1-s_full_load)\n", + "s_standstill=1.0\n", + "fr_standstill=s_standstill*f\n", + "fr_full_load=s_full_load*f\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The synchronous speed is %d rpm.\" %(Ns) \n", + "print \"(b)The no-load speed is %d rpm.\" %(N_no_load) \n", + "print \"(c)The full-load speed is %d rpm.\" %(N_full_load) \n", + "print \"(d)The frequency of rotor-currents at standstill is %.2f Hz.\" %(fr_standstill)\n", + "print \"(e)The frequency of rotor-currents at full-load is %.2f Hz.\" %(fr_full_load)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The synchronous speed is 1000 rpm.\n", + "(b)The no-load speed is 990 rpm.\n", + "(c)The full-load speed is 970 rpm.\n", + "(d)The frequency of rotor-currents at standstill is 50.00 Hz.\n", + "(e)The frequency of rotor-currents at full-load is 1.50 Hz.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.2,Page number: 480" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the frequency of rotor-current in an induction motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "P=12.0 #Number of poles \n", + "f=50.0 #Operating frequency of induction motor(in Hertz)\n", + "N=485.0 #Speed of the motor(in rpm) \n", + "\n", + "\n", + "#Calculations:\n", + "Ns=(120.0*f)/P\n", + "s=(Ns-N)/Ns\n", + "fr=s*f\n", + "\n", + "\n", + "#Result:\n", + "print \"The frequency of the rotor-currents is %.2f Hz.\" %(fr) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of the rotor-currents is 1.50 Hz.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.3,Page number: 481 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the full-load slip and speed of an induction motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "fr=2.0 #Frequency of thr rotoremf at full-load(in Hertz) \n", + "f=50.0 #Frequency of the supply(in Hertz)\n", + "P=6.0 #Number of poles \n", + "\n", + "\n", + "#Calculations:\n", + "s=fr/f\n", + "Ns=(120*f)/P\n", + "N=Ns*(1-s)\n", + "\n", + "\n", + "#Result:\n", + "print \"The slip at full-load is %.2f percent.\" %(s*100)\n", + "print \"The full-load speed is %d rpm.\" %(N) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The slip at full-load is 4.00 percent.\n", + "The full-load speed is 960 rpm.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.4,Page number: 481 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the synchronous speed and the rotor frequency of an induction motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "f=50.0 #Frequency of the supply(in Hertz) \n", + "P=4.0 #Number of poles\n", + "\n", + "\n", + "#Calculations:\n", + "Ns=(120*f)/P\n", + "s1=0.04\n", + "N1=Ns*(1-s1) \n", + "N2=600\n", + "s2=(Ns-N2)/Ns\n", + "fr=s2*f\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The synchronous speed is %d rpm.\" %(Ns)\n", + "print \"(b)The speed of the rotor when the slip is 0.04 is %d rpm.\" %(N1)\n", + "print \"(c)The rotor frequency when the speed of the rotor is 600 rpm is %.2f Hz.\" %(fr)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The synchronous speed is 1500 rpm.\n", + "(b)The speed of the rotor when the slip is 0.04 is 1440 rpm.\n", + "(c)The rotor frequency when the speed of the rotor is 600 rpm is 30.00 Hz.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.5,Page number: 484 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the rotor current of a three-phase induction motor.\"\"\"\n", + "\n", + "from math import sqrt,pow,acos,radians,degrees\n", + "\n", + "#Variable Declaration:\n", + "E_L=100.0 #Induced EMF(in Volts)\n", + "R2=0.05 #Resistance of rotor windings(in Ohms) \n", + "X_20=0.1 #Standstill reactance of rotor windings(in Ohms)\n", + "\n", + "\n", + "#Calculations:\n", + "E_20=E_L/sqrt(3.0)\n", + "s_a=0.04\n", + "E2_a=s_a*E_20\n", + "Z2_a=sqrt((R2*R2)+pow((s_a*X_20),2))\n", + "I2_a=E2_a/Z2_a\n", + "angle_a=acos(R2/Z2_a)\n", + "s_b=1.0\n", + "E2_b=s_b*E_20\n", + "Z2_b=sqrt((R2*R2)+pow((s_b*X_20),2))\n", + "I2_b=E2_b/Z2_b\n", + "angle_b=acos(R2/Z2_b)\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)At 4% slip : \"\n", + "print \" The rotor current is %.2f A.\" %(round(I2_a,2))\n", + "print \" The phase difference between the rotor voltage and the rotor current is %.2f degrees.\" %(round(degrees(angle_a),2))\n", + "print \"\\n(b)At 100% slip :\"\n", + "print \" The rotor current is %.2f A.\" %(round(I2_b,2))\n", + "print \" The phase difference between the rotor voltage and the rotor current is %.2f degrees.\" %(round(degrees(angle_b),2))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)At 4% slip : \n", + " The rotor current is 46.04 A.\n", + " The phase difference between the rotor voltage and the rotor current is 4.57 degrees.\n", + "\n", + "(b)At 100% slip :\n", + " The rotor current is 516.40 A.\n", + " The phase difference between the rotor voltage and the rotor current is 63.43 degrees.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.6,Page number: 485" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the developed power,air-gap power,rotor copper loss,and stator loss in an induction motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "HP=746.0 #Value of Horse-Power(in Watts) \n", + "f=50.0 #Operating frequency of the induction motor(in Hertz)\n", + "N=1470.0 #Speed of the motor(in rpm)\n", + "P=4.0 #Number of poles\n", + "phase=3.0 #Number of phases \n", + "effi=87.5e-02 #Efficiency of the motor at full-load \n", + "\n", + "\n", + "#Calculations:\n", + "Po=5.0*HP\n", + "Pin=Po/effi\n", + "total_loss=Pin-Po\n", + "mech_loss=0.05*total_loss\n", + "elec_loss=total_loss-mech_loss\n", + "dev_pow=Po+mech_loss\n", + "Ns=(120*f)/P\n", + "s=(Ns-N)/Ns\n", + "Pg=dev_pow/(1-s)\n", + "P_R=s*Pg\n", + "P_S=Pin-Pg\n", + "\n", + "\n", + "#Result:\n", + "print \"The developed power is %.2f W.\" %(round(dev_pow,2))\n", + "print \"The air-gap power is %.2f W.\" %(round(Pg,2))\n", + "print \"The rotor copper loss is %.2f W.\" %(round(P_R,2))\n", + "print \"The stator loss is %.2f W.\" %(round(P_S,2))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The developed power is 3756.64 W.\n", + "The air-gap power is 3833.31 W.\n", + "The rotor copper loss is 76.67 W.\n", + "The stator loss is 429.55 W.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.7,Page number: 490 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the rotor speed,the stator current,the power factor and the efficiency of an induction motor. \"\"\"\n", + "\n", + "from math import sqrt,cos,degrees\n", + "from cmath import rect,phase\n", + "\n", + "#Variable Declaration:\n", + "V_L=400.0 #Operating voltage of the induction motor(in Volts)\n", + "P=4.0 #Number of poles\n", + "f=50.0 #Operating frequency of the motor(in Hertz)\n", + "s=0.02 #Slip at rated load \n", + "rot_loss=0.34e03 #Rotational losses(in Watts)\n", + "R1=0.641 #Impedance per phase on the stator side(in Ohms) \n", + "X1=1.106 #Impedance per phase on the stator side(in Ohms) \n", + "R2_eq=0.332 #Impedance per phase on the stator side(in Ohms)\n", + "X_20_eq=0.464 #Impedance per phase on the stator side(in Ohms) \n", + "Xg=26.3 #Impedance per phase on the stator side(in Ohms)\n", + "\n", + "\n", + "#Calculations:\n", + "V1_mod=V_L/sqrt(3)\n", + "V1=rect(V1_mod,0)\n", + "Ns=(120*f)/P\n", + "N=Ns*(1-s)\n", + "V_Th=V1*((Xg/(R1+(X1+Xg)*1j))*1j)\n", + "Z_Th=((Xg*(R1+X1*1j))/(R1+(X1+Xg)*1j))*1j\n", + "R_L_eq=((1-s)/s)*R2_eq\n", + "I1=V_Th/(Z_Th +(R2_eq+X_20_eq*1j)+R_L_eq)\n", + "angle=phase(I1)\n", + "pf=cos(angle)\n", + "Po=(3*abs(I1)*abs(I1)*R_L_eq)-rot_loss\n", + "Pi=3*abs(V1)*abs(I1)*cos(phase(I1))\n", + "effi=(Po/Pi)*100\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The rotor speed is %d rpm.\" %(round(N,0))\n", + "print \"(b)The stator current is %.2f A at a phase angle of %.2f degrees.\" %(abs(I1),degrees(phase(I1))) \n", + "print \"(c)The power factor is %.3f lagging.\" %(pf)\n", + "print \"(d)The output power is %.2f W.\" %(Po)\n", + "print \" The input power is %.2f W.\" %(Pi)\n", + "print \"(e)The efficiency of the motor is %.2f percent.\" %(effi)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The rotor speed is 1470 rpm.\n", + "(b)The stator current is 12.84 A at a phase angle of -3.78 degrees.\n", + "(c)The power factor is 0.998 lagging.\n", + "(d)The output power is 7703.00 W.\n", + " The input power is 8874.78 W.\n", + "(e)The efficiency of the motor is 86.80 percent.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.8,Page number: 495" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the standstill rotor reactance of an induction motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "f=50.0 #Operating frequency of induction motor(in Hertz)\n", + "P=6.0 #Number of poles\n", + "N=940.0 #Speed of motor(in rpm)\n", + "R2=0.1 #Resistance per phase(in Ohms)\n", + "\n", + "\n", + "#Calculations:\n", + "Ns=(120*f)/P\n", + "s=(Ns-N)/Ns\n", + "X_20=R2/s\n", + "\n", + "\n", + "#Result:\n", + "print \"The standstill rotor resistance is %.3f Ohms.\" %(X_20)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The standstill rotor resistance is 1.667 Ohms.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.9,Page number: 498" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the full-load slip of an induction motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "P_motor=6.0 #Number of poles in the induction motor\n", + "N_motor=960.0 #Full-load speed of the induction motor(in rpm)\n", + "P_alt=4.0 #Number of poles in the alternator\n", + "N_alt=1500.0 #Speed of the alternator(in rpm)\n", + "\n", + "\n", + "#Calculations:\n", + "f=(N_alt*P_alt)/120.0\n", + "Ns=(120.0*f)/P_motor\n", + "s=(Ns-N_motor)/Ns\n", + "\n", + "\n", + "#Result:\n", + "print \"The full-load slip of the motor is %.2f percent.\" %(s*100)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The full-load slip of the motor is 4.00 percent.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.10,Page number: 498" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the rotor input,motor input and the efficiency of an induction motor.\"\"\"\n", + "\n", + "from math import pi\n", + "\n", + "#Variable Declaration:\n", + "P=4.0 #Number of poles in the induction motor\n", + "useful_tor=160.0 #Useful torque(in Newton-metre)\n", + "s=0.05 #Slip\n", + "P_S=1000.0 #Stator losses(in Watts)\n", + "Pm=500.0 #Frictional and windage losses(in Watts)\n", + "f=50.0 #Frequency of induction motor(in Hertz)\n", + "#Calculations:\n", + "Ns=(120.0*f)/P\n", + "N=(1-s)*Ns\n", + "Po=(2*pi*useful_tor*N)/60.0\n", + "Pd=Po+Pm\n", + "Pg=Pd/(1-s)\n", + "Pin=Pg+P_S\n", + "effi=Po/Pin\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The rotor input is %.4f kW.\" %(Pg/1000.0)\n", + "print \"(b)The motor input is %.4f W.\" %(Pin)\n", + "print \"(c)The efficiency is %.4f percent.\" %(effi*100)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The rotor input is 25.6591 kW.\n", + "(b)The motor input is 26659.0570 W.\n", + "(c)The efficiency is 89.5609 percent.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.11,Page number: 499" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the slip of an induction motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "effi=0.9 #Efficiency of the induction motor\n", + "Po=50e03 #Load driven by the motor(in Watts)\n", + "\n", + "\n", + "#Calculations:\n", + "Pin=Po/effi\n", + "P_tot=Pin-Po\n", + "\"\"\"The no-load losses comprise of the stator iron loss(Pi) and mechanical losses(Pm)\n", + " (since the stator and rotor copper losses are negligible).These two losses are independent of the load.\n", + " \n", + " Given,the mechanical loss,Pm=(no-load loss)/3.0=((Pi+Pm)/3.0).\n", + " \n", + " Therefore,Pm=(Pi/2.0).\n", + " \n", + " Total loss=(Stator copper loss)+(Stator iron loss)+(Rotor copper loss)+(Mechanical loss).\"\"\"\n", + "Pi=(2.0/7.0)*P_tot\n", + "P_R=Pi\n", + "Pm=Pi/2.0\n", + "Pd=Po+Pm\n", + "Pg=Pd+P_R\n", + "s=P_R/Pg\n", + "\n", + "\n", + "#Result:\n", + "print \"The slip of the induction motor is %.2f percent.\" %(s*100)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The slip of the induction motor is 3.03 percent.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.12,Page number: 499" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the rotor current in an induction motor.\"\"\"\n", + "\n", + "from math import sqrt\n", + "\n", + "#Variable Declaration:\n", + "V2=100.0 #Induced emf between slip ring terminals(in Volts)\n", + "R2=0.4 #Resistance per phase of star-connected rotor windings(in Ohms)\n", + "s=0.04 #Slip of rotor\n", + "\n", + "\n", + "#Calculations:\n", + "E20=V2/sqrt(3.0)\n", + "\"\"\"The rotor reactance X2=(s*X20) is negligible for small values of sand hence cam be ignored.\"\"\"\n", + "I2=(s*E20)/R2\n", + "\n", + "\n", + "#Result:\n", + "print \"The rotor current is %.3f A.\" %(I2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rotor current is 5.774 A.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.13,Page number: 499" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the number of poles and the slip of an induction motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "N=285.0 #Full-load speed of an induction motor(in rpm)\n", + "f=50.0 #Frequency of supply(in Hertz)\n", + "P_R=250.0 #Original rotor losses(in Watts)\n", + "\n", + "\n", + "#Calculations:\n", + "P=(120.0*f)/N\n", + "\"There has to be even number of poles,such that Ns>N.Thus,the actual number of poles is 20.\"\"\"\n", + "P=round((P-1),0)\n", + "Ns=(120.0*f)/P\n", + "s=(Ns-N)/Ns\n", + "\"\"\"For small values of s,the reactance of (s*X20) is much smaller than the resistance R2,hence\n", + " torque is directly proportional to (s/R2).\n", + " \n", + " It means that the to keep the torque same,the (s/R2) ratio should remain the same.If R2 is doubled,then s also has to be \n", + " doubled.\"\"\" \n", + "s_new=2*s\n", + "\"\"\"Since the full-load current remains the same,on doubling the rotor resistance,the copper loss(I*I*R) is also doubled.\"\"\"\n", + "P_R_new=2*P_R\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The number of poles is %d.\" %(P)\n", + "print \"(b)The slip is %.2f percent.\" %(s*100)\n", + "print \"(c)The slip for full-load torque if the rotor resistance is doubled is %.2f percent.\" %(s_new*100)\n", + "print \"(d)The rotor copper losses with added rotor resistance is %.2f W.\" %(P_R_new)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The number of poles is 20.\n", + "(b)The slip is 5.00 percent.\n", + "(c)The slip for full-load torque if the rotor resistance is doubled is 10.00 percent.\n", + "(d)The rotor copper losses with added rotor resistance is 500.00 W.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.14,Page number: 500" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the slip and the power output of an induction motor when external resistances are inserted in each rotor phase.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "s=0.02 #Full-load slip of the induction motor\n", + "Po=500 #Power rating of the motor(in HorsePower)\n", + "R2=0.25 #Resistance per phase of the rotor(in Ohms)\n", + "\n", + "\n", + "#Calculations:\n", + "R2_new=2.0+R2\n", + "s_new=(s/R2)*(R2_new)\n", + "\"\"\"If Ns is the synchronous speed of the motor,then the speed of the rotor before inserting external resistance,\n", + " N=(1-s)*Ns;\n", + " \n", + " The speed of the rotor after inserting external resistance is N_new=(1-s_new)*Ns;\"\"\"\n", + "\n", + "Po_new=((1-s_new)/(1-s))*Po\n", + " \n", + "\n", + "#Result:\n", + "print \"The new slip is %.2f percent.\" %(s_new*100)\n", + "print \"The new power output is %.2f HP.\" %(Po_new)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The new slip is 18.00 percent.\n", + "The new power output is 418.37 HP.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.15,Page number: 500" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the full-load speed of an induction motor.\"\"\"\n", + "\n", + "from sympy import *;\n", + "\n", + "#Variable Declaration:\n", + "P=4.0 #Number of poles in the induction motor\n", + "f=50.0 #Rated frequency of the induction motor(in Hertz)\n", + "\n", + "\n", + "#Calculations:\n", + "\"\"\" Starting torque(tor_st)=1.6*(tor_fl) where tor_fl=full-load torque;\n", + " Maximum torque(tor_max)=2.0*(tor_fl);\n", + " \n", + " tor_st/tor_max=0.8;\n", + " \n", + " (tor_st/tor_max)=(2*s_m)/((s_m*s_m)+1))\n", + " \n", + " (s_m*s_m)-(2.5*s_m)+1=0 is a quadratic equation whose roots are 2 and 0.5 \n", + " \n", + " s_m has to be less than 1. Therefore, s_m=0.5; \n", + " \n", + " Similarly,\n", + " (tor_fl)/(tor_max)=0.5; \n", + " \n", + " (tor_fl)/(tor_max)=(2*s_fl*s_m)/((s_m*s_m)+(s_fl*s_fl))\n", + " \n", + " Substituting s_m=0.5,we get a quadratic equation,\n", + " \n", + " (s_fl*s_fl)-(s_fl)+0.125=0 whose roots are 0.8535,0.1465.\n", + " \n", + " s_fl=0.1465 as s_fl should be less than s_m. \"\"\"\n", + "\"\"\"Finding the roots:\"\"\"\n", + "s_m,s_fl= symbols('s_m s_fl')\n", + "a=solve(s_m**2-(2.5*s_m)+1.0,s_m)\n", + "b=solve(s_fl**2-s_fl+0.125, s_fl)\n", + "Ns=(120*f)/P\n", + "Nfl=Ns*(1-b[0])\n", + "Nm=Ns*(1-a[0])\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The full-load speed is %.2f rpm.\" %(Nfl)\n", + "print \"(b)The speed at maximum torque is %.2f rpm.\" %(Nm)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The full-load speed is 1280.33 rpm.\n", + "(b)The speed at maximum torque is 750.00 rpm.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.16,Page number: 501" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the rotor copper loss and the gross torque for an induction motor.\"\"\"\n", + "\n", + "from math import pi\n", + "\n", + "#Variable Declaration:\n", + "s=0.04 #Full-load slip\n", + "P=4.0 #Number of poles\n", + "f=50.0 #Frequency of the induction motor(in Hertz)\n", + "Po=18.65e03 #Power output(in Watts)\n", + "\n", + "\n", + "#Calculations:\n", + "Pm=0.025*Po\n", + "Pd=Pm+Po\n", + "P_R=Pd*(s/(1-s))\n", + "Pg=P_R/s\n", + "Ns=(120*f)/P\n", + "N=Ns*(1-s)\n", + "sh_tor=Po/(2*pi*(N/60.0))\n", + "gross_tor=Pd/(2*pi*(N/60.0))\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The rotor copper loss is %.2f W.\" %(P_R)\n", + "print \"(b)The rotor input is %.2f W.\" %(Pg)\n", + "print \"(c)The output(shaft) torque is %.2f Nm.\" %(sh_tor)\n", + "print \"(d)The gross torque is %.2f Nm.\" %(gross_tor)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The rotor copper loss is 796.51 W.\n", + "(b)The rotor input is 19912.76 W.\n", + "(c)The output(shaft) torque is 123.68 Nm.\n", + "(d)The gross torque is 126.77 Nm.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.17,Page number: 501" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the rotor current and the rotor power factor for an induction motor.\"\"\"\n", + "\n", + "from math import sqrt,pow,cos,atan\n", + "\n", + "#Variable Declaration:\n", + "P=4.0 #Number of poles in the induction motor\n", + "E1=1100.0 #Line voltage(in Volts) \n", + "f=50.0 #Operating frequency of the motor(in Hertz)\n", + "K=1.0/3.8 #Transformation ratio\n", + "R2=0.012 #Rotor resistance per phase(in Ohms)\n", + "X20=0.25 #Rotor stanstill reactance per phase(in Ohms) \n", + "N=1440.0 #Full-load speed of the motor(in rpm)\n", + "\n", + "\n", + "#Calculations:\n", + "Ns=(120.0*f)/P\n", + "s=(Ns-N)/Ns\n", + "E20=K*E1\n", + "Z20=sqrt((R2*R2)+(X20*X20))\n", + "Z2=sqrt((R2*R2)+(s*X20*s*X20))\n", + "I20=E20/Z20\n", + "pf_20=cos(atan(X20/R2))\n", + "I2=(s*E20)/Z2\n", + "pf=R2/Z2\n", + "I1=100.0/sqrt(3.0)\n", + "I_20=I1/K\n", + "Z2_rot=E20/I_20\n", + "r=sqrt((Z2_rot*Z2_rot)-(X20*X20))-R2\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The rotor current at starting with slip-rings shorted is %.2f A.\" %(I20)\n", + "print \"(b)The rotor power factor at starting with slip-rings shorted is %.5f,lagging.\" %(pf_20)\n", + "print \"(c)The rotor current while running at full load with slip-rings shorted is %.3f A.\" %(I2)\n", + "print \"(d)The rotor power factor while running at full-load with slip-rings shorted is %.5f,lagging.\" %(pf)\n", + "print \"(e)The external rotor resistance is %.4f Ohms.\" %(r)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The rotor current at starting with slip-rings shorted is 1156.56 A.\n", + "(b)The rotor power factor at starting with slip-rings shorted is 0.04794,lagging.\n", + "(c)The rotor current while running at full load with slip-rings shorted is 741.266 A.\n", + "(d)The rotor power factor while running at full-load with slip-rings shorted is 0.76822,lagging.\n", + "(e)The external rotor resistance is 1.2835 Ohms.\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
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