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+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 14: ALTERNATORS AND SYNCHRONOUS MOTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.1,Page number: 433"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the number of poles on the generator if the frequency of the generated voltage is decreased.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "f=60 #Frequency of ac-generator(in Hertz)\n",
+ "P=6 #Number of poles\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Ns=(120*f)/P\n",
+ "f=20\n",
+ "P=(120*f)/Ns\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The speed of rotation of the generator is %d rpm.\" %(Ns)\n",
+ "print \"If the frequency is reduced to 20Hz,the required number of poles is %d.\" %(P)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The speed of rotation of the generator is 1200 rpm.\n",
+ "If the frequency is reduced to 20Hz,the required number of poles is 2.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.2,Page number: 441\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the distribution factor for a machine.\"\"\"\n",
+ "\n",
+ "from math import radians,degrees,sin\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "slots=9 #Number of slots\n",
+ "slot_angle=radians(180/9) #Slot angle(in radians)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "q_a=120.0/degrees(slot_angle)\n",
+ "k_d_a=sin(q_a*(slot_angle/2.0))/(q_a*sin(slot_angle/2.0))\n",
+ "q_b=60.0/degrees(slot_angle)\n",
+ "k_d_b=sin(q_b*(slot_angle/2.0))/(q_b*sin(slot_angle/2.0))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The distribution factor for a machine for a three phase winding with 120 degrees phase group is %.3f.\" %(k_d_a)\n",
+ "print \"(b)The distribution factor for a machine for a three phase winding with 60 degrees phase group is %.3f.\" %(k_d_b)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The distribution factor for a machine for a three phase winding with 120 degrees phase group is 0.831.\n",
+ "(b)The distribution factor for a machine for a three phase winding with 60 degrees phase group is 0.960.\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.3,Page number: 441"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the speed,the generated emf per phase,and the line emf of a three-phase alternator.\"\"\"\n",
+ "\n",
+ "from math import radians,degrees,sqrt,sin\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "phase=3 #Number of phases\n",
+ "f=50 #Frequency rating(in Hertz)\n",
+ "P=20 #Number of poles \n",
+ "slots=180 #Number of slots on the stator\n",
+ "cond_per_slot=8 #Number of conductors per slot\n",
+ "flux=25e-03 #Flux per pole(in Weber) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Z=slots*cond_per_slot\n",
+ "T=(Z/2)/phase\n",
+ "Ns=(120*f)/P\n",
+ "k_p=1\n",
+ "slots_per_pole=slots/P\n",
+ "slot_angle=radians(180/slots_per_pole)\n",
+ "q=slots_per_pole/phase\n",
+ "k_d=sin(q*(slot_angle/2))/(q*sin(slot_angle/2))\n",
+ "E=4.44*f*flux*T*k_p*k_d\n",
+ "line_emf=sqrt(3.0)*E\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The speed of the alternator is %d rpm.\" %(round(Ns,0))\n",
+ "print \"(b)The rms value of generated EMF per phase is %.2f V.\" %(E)\n",
+ "print \"(c)The line EMF is %.2f V.\" %(line_emf)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The speed of the alternator is 300 rpm.\n",
+ "(b)The rms value of generated EMF per phase is 1278.45 V.\n",
+ "(c)The line EMF is 2214.34 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.4,Page number: 446\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the voltage regulation for full-load for a three-phase alternator.\"\"\"\n",
+ "\n",
+ "from math import sqrt,cos,acos \n",
+ "\n",
+ "#Variable Declaration:\n",
+ "P=600e06 #Power rating of the alternator(in VA) \n",
+ "V_L=22e03 #Rated terminal voltage(in Volts) \n",
+ "sync_imp=0.16 #Synchronous impedance per phase(in Ohms)\n",
+ "res_phase=0.014 #Resistance per phase(in Ohms) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "I_L=P/(sqrt(3)*V_L)\n",
+ "Iph=I_L\n",
+ "V=V_L/sqrt(3)\n",
+ "Vz=Iph*sync_imp\n",
+ "theta=acos(res_phase/sync_imp)\n",
+ "pf=0.8\n",
+ "phi=acos(pf)\n",
+ "alpha=theta-phi\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "vol_reg_a=(E-V)/V\n",
+ "pf=1\n",
+ "phi=acos(pf)\n",
+ "alpha=theta-phi\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "vol_reg_b=(E-V)/V\n",
+ "pf=0.8\n",
+ "phi=acos(pf)\n",
+ "alpha=theta+phi\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "vol_reg_c=(E-V)/V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)For a power factor of 0.8 lagging:\"\n",
+ "print \"The voltage regulation for full load is %.2f percent.\" %(vol_reg_a*100)\n",
+ "print \"(b)For unity power factor:\"\n",
+ "print \"The voltage regulation for full load is %.2f percent.\" %(vol_reg_b*100)\n",
+ "print \"(c)For a power factor of 0.8 leading:\" \n",
+ "print \"The voltage regulation for full load is %.2f percent.\" %(vol_reg_c*100)\n",
+ "print \"\\nNote:The voltage regulation for leading power-factor load is negative.\"\n",
+ "print \"It means that on removing the load, the terminal voltage decreases.\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)For a power factor of 0.8 lagging:\n",
+ "The voltage regulation for full load is 14.20 percent.\n",
+ "(b)For unity power factor:\n",
+ "The voltage regulation for full load is 3.64 percent.\n",
+ "(c)For a power factor of 0.8 leading:\n",
+ "The voltage regulation for full load is -8.90 percent.\n",
+ "\n",
+ "Note:The voltage regulation for leading power-factor load is negative.\n",
+ "It means that on removing the load, the terminal voltage decreases.\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.5,Page number: 450\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the synchronous reactance per phase and the voltage regulation for a three-phase star-connected alternator.\"\"\"\n",
+ "\n",
+ "from math import acos,sqrt,cos\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_L=900 #Open-circuit voltage(line to line)(in Volts)\n",
+ "V_L_rated=3.3e03 #Rated voltage of the alternator(in Volts)\n",
+ "I_f=100 #Full-load current(in Amperes)\n",
+ "R=0.9 #Armature Resistance(in Ohm/phase)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "V_oc=V_L/sqrt(3)\n",
+ "I_sc=I_f\n",
+ "sync_imp=V_oc/I_sc\n",
+ "sync_rea=sqrt((sync_imp*sync_imp)-(R*R))\n",
+ "theta=acos(R/sync_imp)\n",
+ "V=V_L_rated/sqrt(3)\n",
+ "Vz=I_f*sync_imp\n",
+ "pf=0.8\n",
+ "phi=acos(pf)\n",
+ "alpha=theta-phi\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "vol_reg_a=(E-V)/V\n",
+ "pf=0.8\n",
+ "phi=acos(pf)\n",
+ "alpha=theta+phi\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "vol_reg_b=(E-V)/V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The synchronous reactance per phase is %.3f Ohms.\" %(sync_rea)\n",
+ "print \"(a) For a power factor of 0.8 lagging:\"\n",
+ "print \"The voltage regulation for full load is %.2f percent\" %(vol_reg_a*100)\n",
+ "print \"(b) For a power factor of 0.8 leading:\"\n",
+ "print \"The voltage regulation for full load is %.2f percent.\" %(vol_reg_b*100)\n",
+ "print \"\\nNote: The voltage regulation for leading power-factor load is negative.\"\n",
+ "print \"It means that on removing the load, the terminal voltage decreases.\"\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The synchronous reactance per phase is 5.118 Ohms.\n",
+ "(a) For a power factor of 0.8 lagging:\n",
+ "The voltage regulation for full load is 21.34 percent\n",
+ "(b) For a power factor of 0.8 leading:\n",
+ "The voltage regulation for full load is -9.03 percent.\n",
+ "\n",
+ "Note: The voltage regulation for leading power-factor load is negative.\n",
+ "It means that on removing the load, the terminal voltage decreases.\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.6,Page number: 455\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the angle of retard of a synchronous motor.\"\"\"\n",
+ "\n",
+ "from math import sqrt,acos,cos,pi,asin,sin,degrees\n",
+ "from cmath import phase\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Po=9e03 #Power rating of synchronous motor(in Watts) \n",
+ "V_L=400 #Voltage rating of synchronous motor(in Watts)\n",
+ "Zs=0.4+3*1j #Synchronous impedance per phase(in Ohms) \n",
+ "pf=0.8 #Power factor(leading) \n",
+ "effi=0.9 #Efficiency of the motor\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Pin=Po/effi\n",
+ "I_L=Pin/(sqrt(3)*V_L*pf)\n",
+ "I=I_L\n",
+ "phi=acos(pf)\n",
+ "mod_Zs=abs(Zs)\n",
+ "theta=phase(Zs)\n",
+ "V=V_L/sqrt(3)\n",
+ "Er=I*mod_Zs\n",
+ "E=sqrt((V*V)+(Er*Er)+(2*V*Er*cos(pi-(theta+phi))))\n",
+ "E_L=sqrt(3)*E\n",
+ "angle_retard=asin((Er*sin(theta+phi))/E)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The angle of retard of the rotor is %.2f degrees.\" %(degrees(angle_retard))\n",
+ "print \"The excitation emf E to which the motor has to be excited to give a full-load output at 0.8 leading power factor is %.2f V.\" %(E_L)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle of retard of the rotor is 10.47 degrees.\n",
+ "The excitation emf E to which the motor has to be excited to give a full-load output at 0.8 leading power factor is 453.81 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.7,Page number: 459"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the line emf generated by the alternator.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "S=24.0 #Number of slots in the alternator\n",
+ "C=12.0 #Number of conductors per slot\n",
+ "flux=0.1 #Flux per pole(in Weber)\n",
+ "P=4.0 #Number of poles in the alternator\n",
+ "Ns=1500.0 #Synchronous speed of the alternator(in rpm)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Zph=S*C/3.0\n",
+ "T=Zph/2.0\n",
+ "S_pole=S/P\n",
+ "slot_ang=180.0/S_pole\n",
+ "q=S_pole/3.0\n",
+ "kd=sin((q*radians(slot_ang)/2))/(q*sin(radians(slot_ang/2)))\n",
+ "f=(P*Ns)/120.0\n",
+ "kp=1.0\n",
+ "E=4.44*flux*f*T*kp*kd\n",
+ "E_L=sqrt(3.0)*E\n",
+ " \n",
+ " \n",
+ "#Result:\n",
+ "print \"The line emf generated when the alternator runs at 1500 rpm is %.2f V.\" %(E_L)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The line emf generated when the alternator runs at 1500 rpm is 1782.78 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.8,Page number: 460"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the net emf induced in the 6 coil in series constituting the alternator winding.\"\"\"\n",
+ "\n",
+ "from math import radians,sin\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "q=6.0 #Number of slots per pole per phase\n",
+ "angle=30.0 #Electrical angle between two consecutive slotss(in degrees) \n",
+ "e=10.0 #Emf of each coil(in Volts) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "kd=sin(q*(radians(angle/2.0)))/(q*sin(radians(angle/2.0)))\n",
+ "arith_sum=6*e\n",
+ "Er=kd*arith_sum\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The net emf induced in the six coils in series is %.3f V.\" %(round(Er,3)) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The net emf induced in the six coils in series is 38.637 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.9,Page number: 460"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the number of poles and the current rating of an alternator.\"\"\"\n",
+ "\n",
+ "from math import sqrt,pi\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "N=120.0 #Speed of the alternator(in rpm)\n",
+ "f=50.0 #Frequency of the alternator(in Hertz)\n",
+ "VA=100e06 #VA rating of the alternator(in Volt-Ampere)\n",
+ "pf=1.0 #Power factor\n",
+ "V_L=11e03 #Line voltage(in Volts)\n",
+ "effi=0.97 #Efficiency of the alternator \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "P=(120.0*f)/N\n",
+ "Po=VA*pf\n",
+ "I_L=VA/(sqrt(3.0)*V_L)\n",
+ "Pin=Po/effi\n",
+ "tor=Pin/(2*pi*N/60.0)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The number of poles is %d.\" %(P)\n",
+ "print \"(b)The current rating is %.2f A.\" %(I_L)\n",
+ "print \"(c)The input power is %.2f MW.\" %(round((Pin/1000000),2))\n",
+ "print \"(d)The prime-mover torque applied to the genrator shaft is %e Nm.\" %(tor)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The number of poles is 50.\n",
+ "(b)The current rating is 5248.64 A.\n",
+ "(c)The input power is 103.09 MW.\n",
+ "(d)The prime-mover torque applied to the genrator shaft is 8.203863e+06 Nm.\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.10,Page number: 460"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the percentage regulation for a load.\"\"\"\n",
+ "\n",
+ "from math import atan,radians,acos\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_L=11e03 #Line voltage of the star-connected alternator(in Volts)\n",
+ "VA=800e03 #Power rating of the alternator(in Volt-Amperes)\n",
+ "Rs=1.5 #Resistance per phase(in Ohms)\n",
+ "Xs=25.0 #Synchronous reactance(in Ohms)\n",
+ "pf=0.8 #Leading power factor\n",
+ "Po=600e03 #Output power(in Watts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Vph=V_L/sqrt(3.0)\n",
+ "V=Vph\n",
+ "Iph=Po/(sqrt(3.0)*V_L*pf)\n",
+ "Zs=sqrt((Rs*Rs)+(Xs*Xs))\n",
+ "theta=atan(Xs/Rs)\n",
+ "Vz=Iph*Zs\n",
+ "pf_ang=acos(pf)\n",
+ "alpha=theta+pf_ang\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "reg=(E-V)/V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The percentage regulation is %.3f per cent.\" %(reg*100)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage regulation is -7.641 per cent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.11,Page number: 461"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the full-load voltage regulation of the alternator.\"\"\"\n",
+ "\n",
+ "from math import atan,acos\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_L=6e03 #Line voltage of the alternator(in Volts)\n",
+ "VA=6000e03 #Power rating of the alternator(in Volt-Amperes)\n",
+ "R=0.2 #Winding resistance per phase(in Ohms)\n",
+ "pf=0.8 #Lagging power factor\n",
+ "V_L_OC=480.0 #Line voltage in open-circuit test(in Volts)\n",
+ "I_f_OC=10.0 #Field current in open-circuit test(in Amperes)\n",
+ "I_L_SC=105.0 #Line current in short-circuit test(in Amperes)\n",
+ "I_f_SC=5.0 #Field current in short-circuit test(in Amperes) \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "\"\"\"In the short-circuit test,the currents are small compared to the full-load current.\"\"\"\n",
+ "V=V_L/sqrt(3.0)\n",
+ "I=VA/(3*V)\n",
+ "V_ph_OC=V_L_OC/sqrt(3.0)\n",
+ "\"\"\"Since the field current of 5 A gives an armature current of 105 A,a field current of 10 A will give an armature \n",
+ " current of (105*2)=210 A.\"\"\"\n",
+ "I_ph=I_L_SC*2\n",
+ "Zs=V_ph_OC/I_ph\n",
+ "Xs=sqrt((Zs*Zs)-(R*R))\n",
+ "theta=atan(Xs/R)\n",
+ "Vz=I*Zs\n",
+ "pf_ang=acos(pf)\n",
+ "alpha=theta-pf_ang\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "reg=(E-V)/V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The full-load voltage regulation of the alternator at 0.8 lagging power factor is %.2f per cent.\" %(reg*100)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The full-load voltage regulation of the alternator at 0.8 lagging power factor is 16.73 per cent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.12,Page number: 462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the terminal voltage of the generator.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "f=50.0 #Frequency rating of the generator(in Hertz)\n",
+ "Z_p=96.0 #Number of conductors per phase\n",
+ "flux=0.1 #Flux per pole(in Webers)\n",
+ "Xs=5.0 #Synchronous reactance per phase(in Ohms)\n",
+ "kd=0.96 #Distribution factor for the stator winding\n",
+ "Z_L=10.0 #Load impedance(in Ohms)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Zs=1j*Xs\n",
+ "kp=1.0\n",
+ "T=Z_p/2.0\n",
+ "E=4.44*f*flux*kp*kd*T\n",
+ "V=E/(1+(Zs/Z_L))\n",
+ "V_L=sqrt(3.0)*abs(V)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The terminal voltage of the generator is %.2f V.\" %(V_L)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The terminal voltage of the generator is 1584.79 V.\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.13,Page number: 462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the percentage change of voltage.\"\"\"\n",
+ "\n",
+ "from math import atan,radians,acos\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_L=6.6e03 #Line voltage of the star-connected alternator(in Volts)\n",
+ "VA=1500e03 #Power rating of the alternator(in Volt-Amperes)\n",
+ "Rs=0.5 #Resistance per phase(in Ohms)\n",
+ "Xs=5.0 #Synchronous reactance(in Ohms)\n",
+ "pf=0.8 #Lagging power factor\n",
+ "\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Vph=V_L/sqrt(3.0)\n",
+ "V=Vph\n",
+ "Iph=VA/(3.0*Vph)\n",
+ "Zs=sqrt((Rs*Rs)+(Xs*Xs))\n",
+ "theta=atan(Xs/Rs)\n",
+ "Vz=Iph*Zs\n",
+ "pf_ang=acos(pf)\n",
+ "alpha=theta-pf_ang\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "reg=(E-V)/V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The percentage change in voltage is %.3f per cent.\" %(reg*100)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The percentage change in voltage is 12.432 per cent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.14,Page number: 463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the terminal voltage and load regulation.\"\"\"\n",
+ "\n",
+ "from math import atan,radians,acos,asin\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_L=6599.0 #Line voltage of the star-connected alternator(in Volts)\n",
+ "Rs=0.5 #Resistance per phase(in Ohms)\n",
+ "Xs=5.0 #Synchronous reactance(in Ohms)\n",
+ "I=130.0 #Full-load current(in Amperes)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "E=V_L/sqrt(3.0)\n",
+ "Zs=Rs+1j*Xs\n",
+ "theta=atan(Xs/Rs)\n",
+ "Vz=I*abs(Zs)\n",
+ "pf=0.8\n",
+ "pf_ang=acos(pf)\n",
+ "alpha=theta-pf_ang\n",
+ "tor_ang=asin((Vz*sin(alpha))/E)\n",
+ "V_a=(E*cos(tor_ang))-(Vz*cos(alpha))\n",
+ "reg_a=(E-V_a)/V_a\n",
+ "pf=0.6\n",
+ "pf_ang=acos(pf)\n",
+ "alpha=theta+pf_ang\n",
+ "beta=pi-alpha\n",
+ "tor_ang=asin((Vz*sin(beta))/E)\n",
+ "V_b=(E*cos(tor_ang))+(Vz*cos(beta))\n",
+ "reg_b=(E-V_b)/V_b\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The terminal voltage when the power factor is 0.8 lagging is %.2f V.\" %(V_a)\n",
+ "print \" The load regulation is %.2f per cent.\" %(reg_a*100) \n",
+ "print \"(b)The terminal voltage when the power factor is 0.6 leading is %.2f V.\" %(V_b)\n",
+ "print \" The load regulation is %.2f per cent.\" %(reg_b*100)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The terminal voltage when the power factor is 0.8 lagging is 3337.45 V.\n",
+ " The load regulation is 14.16 per cent.\n",
+ "(b)The terminal voltage when the power factor is 0.6 leading is 4265.21 V.\n",
+ " The load regulation is -10.67 per cent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.15,Page number: 464"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the percentage load regulation.\"\"\"\n",
+ "\n",
+ "from math import acos,atan\n",
+ "\n",
+ "#Variable Declration:\n",
+ "VA=1.5e06 #Power rating of the synchronous generator(in Volt-Amperes)\n",
+ "V_L=11e03 #Line voltage(in Volts)\n",
+ "Rs=1.2 #Armature resistance(in Ohms)\n",
+ "Xs=25.0 #Synchronous reactance per phase(in Ohms)\n",
+ "P_L=1.4375e06 #Load to be delivered(in Volt-Amperes)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "V=V_L/sqrt(3.0)\n",
+ "I=P_L/(3*V)\n",
+ "Zs=sqrt((Rs*Rs)+(Xs*Xs))\n",
+ "Vz=I*Zs\n",
+ "pf=0.8\n",
+ "pf_ang=acos(pf)\n",
+ "theta=atan(Xs/Rs)\n",
+ "alpha=theta-pf_ang\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "reg_a=(E-V)/V\n",
+ "alpha=theta+pf_ang\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "reg_b=(E-V)/V\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The percentage load regulation for a power factor of 0.8 lagging is %.2f per cent.\" %(reg_a*100)\n",
+ "print \"(b)The percentage load regulation for a power factor of 0.8 leading is %.2f per cent.\" %(reg_b*100)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The percentage load regulation for a power factor of 0.8 lagging is 21.15 per cent.\n",
+ "(b)The percentage load regulation for a power factor of 0.8 leading is -13.12 per cent.\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.16,Page number: 465"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the torque angle of the alternator.\"\"\"\n",
+ "\n",
+ "from math import acos,degrees\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "R=0.5 #Effective resistance per phase of the alternator(in Ohms)\n",
+ "V_L=2200.0 #Line voltage of the alternator(in Volts)\n",
+ "I_FL=200.0 #Full-load current(in Amperes)\n",
+ "If=30.0 #Field current(in Amperes)\n",
+ "V_L_oc=1100.0 #Line-to-line voltage on open circuit(in Volts)\n",
+ "pf=0.8 #Lagging power factor \n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Zs=V_L_oc/(sqrt(3.0)*I_FL)\n",
+ "theta=acos(R/Zs)\n",
+ "Vz=I_FL*Zs\n",
+ "V=V_L/sqrt(3.0)\n",
+ "pf_ang=acos(pf)\n",
+ "alpha=theta-pf_ang\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "Po=V*I_FL*pf\n",
+ "tor_ang=theta-acos((Po+(V*V*cos(theta)/Zs))*(Zs/(E*V)))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The torque angle of the alternator is %.2f degrees.\" %(degrees(tor_ang))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The torque angle of the alternator is 14.35 degrees.\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.17,Page number: 465"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"\"Finding the synchronising power of the alternator.\"\"\"\n",
+ "\n",
+ "from math import acos\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "P=6.0 #Number of poles\n",
+ "VA=3e06 #Power rating of the alternator(in Volt-Amperes)\n",
+ "Ns=1000.0 #Speed of operation of the alternator(in rpm)\n",
+ "V_L=3.3e03 #Line voltage of the load(in Volts)\n",
+ "pf=0.8 #Lagging power factor\n",
+ "\n",
+ "#Calculations:\n",
+ "V=V_L/sqrt(3.0)\n",
+ "I=VA/(sqrt(3.0)*V_L)\n",
+ "IXs=0.25*V\n",
+ "Xs=IXs/I\n",
+ "rotor_dis=radians(1*(P/2.0))\n",
+ "Vz=I*Xs\n",
+ "theta=pi/2\n",
+ "pf_ang=acos(0.8)\n",
+ "alpha=theta-pf_ang\n",
+ "E=sqrt((V*V)+(Vz*Vz)+(2*V*Vz*cos(alpha)))\n",
+ "Psy=(3*E*V*sin(rotor_dis))/(Xs)\n",
+ "tor=(3*Psy)/(2*pi*Ns/60)\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The synchronising power is %.2f kW.\" %(Psy/1000)\n",
+ "print \"(b)The synchronising torque per mechanical degree is %.2f kNm.\" %(tor/1000)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The synchronising power is 733.08 kW.\n",
+ "(b)The synchronising torque per mechanical degree is 21.00 kNm.\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.18,Page number: 466"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the armature current and power factor.\"\"\"\n",
+ "\n",
+ "from math import radians,degrees\n",
+ "from cmath import rect,phase\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_L=2300.0 #Line voltage of the winding(in Volts)\n",
+ "f=50.0 #Operating frequency of the synchronous motor(in Hertz)\n",
+ "Psh=205.0 #Power delivered by the motor(in Horse Power)\n",
+ "ang=15.0 #Power angle(in degrees)\n",
+ "Xs=11.0 #Synchronous reactance(in Ohms)\n",
+ "effi=0.90 #Efficiency of the motor\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "V=V_L/sqrt(3.0)\n",
+ "Psh=Psh*746.0\n",
+ "Pd=Psh/effi\n",
+ "E=(Pd*Xs)/(3.0*V*sin(radians(ang)))\n",
+ "I=(rect(V,0)-rect(E,radians(-ang)))/(1j*Xs)\n",
+ "pf=cos(phase(I))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The excitation voltage per phase is %.2f V per phase.\" %(E)\n",
+ "print \"(b)The armature current is %.2f A at a phase angle of %.2f degrees.\" %(abs(I),degrees(phase(I)))\n",
+ "print \"(c)The power factor is %.4f leading.\" %(pf)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The excitation voltage per phase is 1812.83 V per phase.\n",
+ "(b)The armature current is 57.44 A at a phase angle of 42.05 degrees.\n",
+ "(c)The power factor is 0.7426 leading.\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.19,Page number: 466"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the line current and the power factor.\"\"\"\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_L=6600.0 #Line voltage(in Volts)\n",
+ "Xs=20.0 #Synchronous reactance per phase(in Ohms)\n",
+ "Pin=915e03 #Power consumed by the motor(in Watts)\n",
+ "E_L=8942.0 #Induced line emf per phase(in Volts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "V=V_L/sqrt(3.0)\n",
+ "E=E_L/sqrt(3.0)\n",
+ "I_cos=Pin/(sqrt(3.0)*V_L)\n",
+ "BN=Xs*I_cos\n",
+ "NA=sqrt((E*E)-(BN*BN))\n",
+ "NO=NA-V\n",
+ "Er=sqrt((NO*NO)+(BN*BN))\n",
+ "I_L=Er/Xs\n",
+ "pf=I_cos/I_L\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The line current is %.2f A.\" %(I_L)\n",
+ "print \"(b)The power factor is %.4f leading.\" %(pf) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The line current is 97.05 A.\n",
+ "(b)The power factor is 0.8247 leading.\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.20,Page number: 467"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the power factor when the load on the motor increases.\"\"\"\n",
+ "\n",
+ "from math import acos,asin\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "V_L=6600.0 #Line voltage(in Volts)\n",
+ "Xs=15.0 #Synchronous reactance per phase(in Ohms)\n",
+ "Pin=500e03 #Power consumed by the motor(in Watts)\n",
+ "pf=0.8 #Leading power factor\n",
+ "Pin_new=800e03 #New Input power(in Watts)\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "V=V_L/sqrt(3.0)\n",
+ "Zs=Xs\n",
+ "I_L=Pin/(sqrt(3.0)*V_L*pf)\n",
+ "Er=I_L*Zs\n",
+ "pf_ang=acos(pf)\n",
+ "alpha=(pi/2)-pf_ang\n",
+ "E=sqrt((V*V)+(Er*Er)+(2*V*Er*cos(alpha)))\n",
+ "I1_cos=Pin_new/(sqrt(3.0)*V_L)\n",
+ "sin_tor=(Pin_new*Xs)/(E*V)\n",
+ "AN=V*sin_tor\n",
+ "NB=E-(V*cos(asin(sin_tor)))\n",
+ "Er1=sqrt((AN*AN)+(NB*NB))\n",
+ "I1=Er1/Xs\n",
+ "pf=I1_cos/I1\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"The power factor when the load on the motor increases is %.4f leading.\" %(pf)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The power factor when the load on the motor increases is 0.3229 leading.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14.21,Page number: 469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Question:\n",
+ "\"\"\"Finding the power factor and kVA rating of the synchronous motor.\"\"\"\n",
+ "\n",
+ "from math import atan,acos\n",
+ "\n",
+ "#Variable Declaration:\n",
+ "Si=500e03 #Load supplied by the three-phase system(in Volt-Amperes)\n",
+ "pf_i=0.5 #Lagging power factor of three-phase system\n",
+ "Po=100.0 #Load supplied by the synchronous motor(in Horse-Power)\n",
+ "pf_t=0.9 #Overall lagging power factor \n",
+ "effi=0.87 #Efficiency of the synchronous motor\n",
+ "\n",
+ "\n",
+ "#Calculations:\n",
+ "Pi=Si*pf_i\n",
+ "Qi=Si*sin(acos(pf_i))\n",
+ "Ps=(Po*746)/effi\n",
+ "Pt=Pi+Ps\n",
+ "St=Pt/pf_t\n",
+ "Qt=St*sin(acos(pf_t))\n",
+ "Qs=Qt-Qi\n",
+ "pf_ang=atan(Qs/Ps)\n",
+ "pf_s=cos(pf_ang)\n",
+ "Ss=sqrt((Ps*Ps)+(Qs*Qs))\n",
+ "\n",
+ "\n",
+ "#Result:\n",
+ "print \"(a)The power faactor of the synchronous motor is %.2f leading.\" %(pf_s)\n",
+ "print \"(b)The kVA rating of the synchronous motor is %.2f kVA.\" %(Ss/1000.0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)The power faactor of the synchronous motor is 0.30 leading.\n",
+ "(b)The kVA rating of the synchronous motor is 283.67 kVA.\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file