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diff --git a/Electrical_Machines_II/chapter_1_1.ipynb b/Electrical_Machines_II/chapter_1_1.ipynb new file mode 100755 index 00000000..0ea2a442 --- /dev/null +++ b/Electrical_Machines_II/chapter_1_1.ipynb @@ -0,0 +1,1297 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e4feee2966703522c4e009f16556422b345ba2813bb92ad5cc7b8ab8058895bb" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 1: Synchronous Machines\n" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1, Page 24" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#given data\n", + "slots=24;#no. of slotes\n", + "NoOfPhase=3;#no of phase\n", + "MotorSpeed=1450;#in rpm\n", + "N=1500;#Synchonous Speed in rpm\n", + "f=50;#in Hz\n", + "\n", + "#Calculations&Results\n", + "print \"As the winding is in single layer, each slot contains one coil slide only.\";\n", + "CoilSlidePerSlot=1;#coil slide per slot\n", + "CoilSlidePerCoil=2;#coil slide per Coil\n", + "TotalCoils=slots*CoilSlidePerSlot/CoilSlidePerCoil;#no. of coils\n", + "print \"Total no. of Coils : \",TotalCoils;\n", + "P=120*f/N;#no. of poles\n", + "print \"No. of poles : \",P;\n", + "PolesPitch=slots/P;#unitless\n", + "print \"Poles Pitch = \",PolesPitch;\n", + "print \"In case of single layer winding, the pole ppitch is generally taken in odd numbers only\"\n", + "print \"let pole pitch = 5(for short pitch winding)\";\n", + "PolesPitch=5;#for short pitch winding\n", + "print \"Coil Span or coil through = 1-6\";\n", + "CoilsPerPolePerPhase=TotalCoils/(P*NoOfPhase);#No. of Coils/Pole/Phase\n", + "print \"No. of Coils/Pole/Phase = \",CoilsPerPolePerPhase;\n", + "pair_of_poles=2;#no. of pair of poles\n", + "TotalElectricalDegree=360*pair_of_poles;#in degree \n", + "ElectricalDegreesPerSlot=TotalElectricalDegree/slots;#in degree electrical\n", + "print \"Electrical Degrees/Slot = \",ElectricalDegreesPerSlot;\n", + "Slots_required=120/ElectricalDegreesPerSlot;#No. of slotes required for proper phase displacement\n", + "print \"No. of slotes required for proper phase displacement = \",Slots_required;\n", + "print \"Winding Table is as follows :\";\n", + "print \"Coil No. Connection Lead from Coil Span Phase and Group No.\";\n", + "print \" 1 A1 1-6 A1\";\n", + "print \" 2 3-8 C4\";\n", + "print \" 3 B1 5-10 B1\";\n", + "print \" 4 7-12 A2\";\n", + "print \" 5 C1 9-14 C1\";\n", + "print \" 6 11-16 B2\";\n", + "print \" 7 13-18 A3\";\n", + "print \" 8 15-20 C2\";\n", + "print \" 9 17-22 B3\";\n", + "print \" 10 19-24 A4\";\n", + "print \" 11 21-2 C3\";\n", + "print \" 12 23-4 B4\";" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "As the winding is in single layer, each slot contains one coil slide only.\n", + "Total no. of Coils : 12\n", + "No. of poles : 4\n", + "Poles Pitch = 6\n", + "In case of single layer winding, the pole ppitch is generally taken in odd numbers only\n", + "let pole pitch = 5(for short pitch winding)\n", + "Coil Span or coil through = 1-6\n", + "No. of Coils/Pole/Phase = 1\n", + "Electrical Degrees/Slot = 30\n", + "No. of slotes required for proper phase displacement = 4\n", + "Winding Table is as follows :\n", + "Coil No. Connection Lead from Coil Span Phase and Group No.\n", + " 1 A1 1-6 A1\n", + " 2 3-8 C4\n", + " 3 B1 5-10 B1\n", + " 4 7-12 A2\n", + " 5 C1 9-14 C1\n", + " 6 11-16 B2\n", + " 7 13-18 A3\n", + " 8 15-20 C2\n", + " 9 17-22 B3\n", + " 10 19-24 A4\n", + " 11 21-2 C3\n", + " 12 23-4 B4\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2, Page 28" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#given data\n", + "slots=24;#no. of slotes\n", + "P=4;#no. of poles\n", + "CoilPitch=5;#unitless\n", + "NoOfPhase=3;#no of phase\n", + "\n", + "#Calculations&Results\n", + "print \"As the winding is in double layer, each slot contains 2 coils per slide.\";\n", + "CoilSlidePerSlot=2;#coil slide per slot\n", + "CoilSlidePerCoil=2;#coil slide per Coil\n", + "TotalCoils=slots*CoilSlidePerSlot/CoilSlidePerCoil;#no. of coils\n", + "print \"Total no. of Coils : \",TotalCoils;\n", + "PolesPitch=slots/P;#unitless\n", + "print \"Pole Pitch = \",PolesPitch;\n", + "CoilsPerPolePerPhase=TotalCoils/(P*NoOfPhase);#No. of Coils/Pole/Phase\n", + "print \"No. of Coils/Pole/Phase = \",CoilsPerPolePerPhase;\n", + "pair_of_poles=2;#no. of pair of poles\n", + "TotalElectricalDegree=360*pair_of_poles;#in degree \n", + "ElectricalDegreesPerSlot=TotalElectricalDegree/slots;#in degree electrical\n", + "print \"Electrical Degrees/Slot = \",ElectricalDegreesPerSlot;\n", + "Slots_required=120/ElectricalDegreesPerSlot;#No. of slotes required for proper phase displacement\n", + "print \"No. of slots required for proper phase displacement = \",Slots_required;\n", + "print \"ie. Phase A1 is brought out from slot no. = 1\";\n", + "print \"Phase B1 at slot no. = 1+4 = 5\";\n", + "print \"Phase C1 at slot no. = 5+4 = 9\";\n", + "print \"Col Connection - end to start to start\";\n", + "print \"Winding Table is as follows :\";\n", + "print \"Coil No. Connection Lead from Coil Span Phase and Group No.\";\n", + "print \" 1 A1 1-6 A1\";\n", + "print \" 2 2-7 \";\n", + "print \" 3 3-8 C4\";\n", + "print \" 4 4-9 \";\n", + "print \" 5 B1 5-10 B1\";\n", + "print \" 6 6-11 \";\n", + "print \" 7 7-12 A2\";\n", + "print \" 8 8-13 \";\n", + "print \" 9 C1 9-14 C1\";\n", + "print \" 10 10-15 \";\n", + "print \" 11 11-16 B2\";\n", + "print \" 12 12-17 \";\n", + "print \" 13 13-18 A3\";\n", + "print \" 14 14-19 \";\n", + "print \" 15 15-20 C2\";\n", + "print \" 16 16-21 \";\n", + "print \" 17 17-22 B3\";\n", + "print \" 18 18-23 \";\n", + "print \" 19 19-24 A4\";\n", + "print \" 20 20-1 \";\n", + "print \" 21 21-2 C3\";\n", + "print \" 22 22-3 \";\n", + "print \" 23 23-4 \";\n", + "print \" 24 24-5 B4\";" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "As the winding is in double layer, each slot contains 2 coils per slide.\n", + "Total no. of Coils : 24\n", + "Pole Pitch = 6\n", + "No. of Coils/Pole/Phase = 2\n", + "Electrical Degrees/Slot = 30\n", + "No. of slots required for proper phase displacement = 4\n", + "ie. Phase A1 is brought out from slot no. = 1\n", + "Phase B1 at slot no. = 1+4 = 5\n", + "Phase C1 at slot no. = 5+4 = 9\n", + "Col Connection - end to start to start\n", + "Winding Table is as follows :\n", + "Coil No. Connection Lead from Coil Span Phase and Group No.\n", + " 1 A1 1-6 A1\n", + " 2 2-7 \n", + " 3 3-8 C4\n", + " 4 4-9 \n", + " 5 B1 5-10 B1\n", + " 6 6-11 \n", + " 7 7-12 A2\n", + " 8 8-13 \n", + " 9 C1 9-14 C1\n", + " 10 10-15 \n", + " 11 11-16 B2\n", + " 12 12-17 \n", + " 13 13-18 A3\n", + " 14 14-19 \n", + " 15 15-20 C2\n", + " 16 16-21 \n", + " 17 17-22 B3\n", + " 18 18-23 \n", + " 19 19-24 A4\n", + " 20 20-1 \n", + " 21 21-2 C3\n", + " 22 22-3 \n", + " 23 23-4 \n", + " 24 24-5 B4\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3, Page 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "#given data\n", + "StatorSlots=36;#No. of stator slots\n", + "Poles=4;#No. of poles\n", + "#coilSpan=1:8;#unitless\n", + "\n", + "#Calculations&Results\n", + "SlotsPerPole=StatorSlots/Poles;# no. of slots per pole\n", + "print \"Slots/Pole = \",SlotsPerPole;\n", + "print \"Here the coil span falls short by, \",((2./9)*180),\" Degree\";\n", + "alfa=40*pi/180;#short pitch angle in degree\n", + "Kp=cos(alfa/2);#Coil span Factor\n", + "print \"Pitch Factor or coil span factor : \",round(Kp,1);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Slots/Pole = 9\n", + "Here the coil span falls short by, 40.0 Degree\n", + "Pitch Factor or coil span factor : 0.9\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4, Page 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import *\n", + "\n", + "#given data\n", + "NoOfPhase=3;#no of phase\n", + "P=16;#No. of pole alternator\n", + "Slots=144;#No. of slots\n", + "Conductors=10;#per slot\n", + "fi=0.03;#in Weber\n", + "N=375;#machine speed in rpm\n", + "\n", + "#Calculations\n", + "f=P*N/120;#in Hz\n", + "SlotsPerPole=Slots/P;#unitless\n", + "m=Slots/(P*NoOfPhase);#unitless\n", + "Beta=180/(SlotsPerPole);#in Degree\n", + "Kd=sin(m*Beta/2*pi/180)/(m*sin(Beta/2*pi/180));#unitless\n", + "TotalConductors=Conductors*Slots;#no. of conductors\n", + "TotalConductorsPerPhase=Conductors*Slots/NoOfPhase;#no. of conductors/phase\n", + "TurnsPerPhase=TotalConductorsPerPhase/2;#No. of turns per phase\n", + "EMFPerPhase=4.44*Kd*fi*f*TurnsPerPhase;#in Volt\n", + "LineVoltage=sqrt(3)*EMFPerPhase;#in Volt\n", + "\n", + "#Results\n", + "print \"Frequency in Hz : \",f;\n", + "print \"Phase Electromotive force in Volt : \",round(EMFPerPhase,1);\n", + "print \"Line Electromotive force in Volt : \",round(LineVoltage,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency in Hz : 50\n", + "Phase Electromotive force in Volt : 1534.1\n", + "Line Electromotive force in Volt : 2657.202\n" + ] + } + ], + "prompt_number": 76 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5, Page 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "NoOfPhase=3;#no of phase\n", + "SlotsPerPhase=3;#o. of slots\n", + "\n", + "#Calculations&Results\n", + "m=SlotsPerPhase;#no. of slots\n", + "SlotsPerPolePerPhase=SlotsPerPhase*NoOfPhase;#unitless\n", + "Beta=180./SlotsPerPolePerPhase;#in degree\n", + "print \"The phase difference between the induced emf in two coils lying in adjacent slots = \",(Beta),\" degree\";\n", + "Kb=sin((m*Beta*pi)/(2*180))/(m*sin((Beta*pi)/(2*180)));#unitless\n", + "print \"The breadth factor : \",round(Kb,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The phase difference between the induced emf in two coils lying in adjacent slots = 20.0 degree\n", + "The breadth factor : 0.96\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6, Page 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "StatorSlots=24#No. of stator slots\n", + "Poles=4;#No. of poles\n", + "SlotsPerPole=StatorSlots/Poles;# no. of slots per pole\n", + "#coilSpan=1:6;#unitless\n", + "\n", + "#Calculations&Results\n", + "print \"If the sides of the coil are placed in slots 1 and 7, then it is full pitched, \\nIf the coil slides are placed\" \\\n", + " + \" in 1 and 6 then it is short pitched and the distance equal to 5/6th of pole-pitch.\";\n", + "print \"Since it falls short by 1/6th of the pole-pitch, hence it is short by :\"\n", + "theta=180/SlotsPerPole;#in Degree\n", + "print (theta),\" Degree.\"\n", + "Kp=cos(theta/2*pi/180);#unitless\n", + "print \"Pitch-factor Kp : \",round(Kp,4);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "If the sides of the coil are placed in slots 1 and 7, then it is full pitched, \n", + "If the coil slides are placed in 1 and 6 then it is short pitched and the distance equal to 5/6th of pole-pitch.\n", + "Since it falls short by 1/6th of the pole-pitch, hence it is short by :\n", + "30 Degree.\n", + "Pitch-factor Kp : 0.9659\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7, Page 40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#given data\n", + "NoOfPhase=3;#no of phase\n", + "Eph=3300/sqrt(3);#in Volts\n", + "f=50;#in Hz\n", + "Poles=12;#No. of poles\n", + "StatorSlots=144#No. of stator slots\n", + "\n", + "#Calculations\n", + "SlotsPerPhase=StatorSlots/NoOfPhase;#no. of slots/phase\n", + "Conductors=5;#per slot\n", + "ConductorsPerphase=SlotsPerPhase*Conductors;#Conductors/Phase\n", + "S=ConductorsPerphase;#Conductors/phase\n", + "SlotsPerPolePerPhase=SlotsPerPhase/Poles;#no. of slots/phase\n", + "Kf=1.11;#Form Factor\n", + "Kb=0.96;#Breadth Factor\n", + "Kp=1;#For concentric winding\n", + "fi=Eph/(2*Kf*Kb*Kp*S*f);#in weber\n", + "\n", + "#Result\n", + "print \"The Flux per pole in weber : \",round(fi,4);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Flux per pole in weber : 0.0745\n" + ] + } + ], + "prompt_number": 93 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8, Page 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "Phase=3;#no. of phase\n", + "f=50;#in Hz\n", + "P=16;#No. of pole alternator\n", + "Slots=144.;#No. of slots\n", + "conductors=10;#conductors per slot\n", + "\n", + "#Calculations&Results\n", + "fi=2.48*10**-2;#in weber\n", + "n=Slots/P;#No. of slots/pole\n", + "Zr=Slots*conductors/Phase;#No. of conductors/Phase\n", + "T=Zr/2;#N. of turns/phase\n", + "Beta=180/n;#Angular displacement between slots in degree\n", + "m=n/Phase;#No. of slots/pole/Phase\n", + "Kd=sin(m*Beta/2*pi/180)/(m*sin(Beta/2*pi/180));#Distribution factor :unitless \n", + "print \"The coil span falls short of 2 slots i.e. \",(2*180/9),\" degree.\";\n", + "alfa=40;#short pitch angle in degree\n", + "Kp=(cos(alfa/2*pi/180));#Unitless\n", + "#Formula : f=P*N/120;#in Hz\n", + "N=120*f/P;#in rpm\n", + "Ep=4.44*Kd*Kp*fi*f*T;#in Volts\n", + "LineVoltage=sqrt(3)*Ep;#in Volts\n", + "print \"The speed is \",(N),\" in rpm\";\n", + "print \"The line emf is \",round(LineVoltage,2),\" Volts\";\n", + "#Note : Answer in the book is not accurate for last part due to rounding off errors" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The coil span falls short of 2 slots i.e. 40 degree.\n", + "The speed is 375 in rpm\n", + "The line emf is 2064.15 Volts\n" + ] + } + ], + "prompt_number": 109 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9, Page 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "RatedPower=100;#in KVA\n", + "RatedPower=100*1000;#in VA\n", + "VL=1040;#in Volt\n", + "Phase=3;#Machine phase\n", + "If=40;#in Ampere\n", + "Isc=200;#in Ampere\n", + "EL=1040;#in Volt\n", + "\n", + "#Calculations&Results\n", + "Eph=EL/sqrt(3);#in Volt\n", + "Zs=Eph/Isc;#in Ohm\n", + "Rs=0.2;#in Ohm\n", + "Xs=sqrt(Zs**2-Rs**2);#in Ohm\n", + "IL=19.25;#in Ampere\n", + "V=3000/sqrt(3);#in Volt\n", + "#At 0.8 power factor lagging\n", + "IRa=IL*0.2;#in Volt\n", + "IXs=IL*Xs;#in Volt\n", + "Vsin_fi=V*0.6;#in Volt\n", + "Vcos_fi=V*0.8;#in Volt\n", + "Eo=sqrt((Vcos_fi+IRa)**2+(Vsin_fi+IXs)**2);#in Volts\n", + "Regulation=((Eo-V)/V)*100;#in %\n", + "print \"Full load percentage regulation at a power factor of 0.8 lagging : \",round(Regulation,1);\n", + "\n", + "#At 0.8 power factor leading\n", + "Eo=sqrt((Vcos_fi+IRa)**2+(Vsin_fi-IXs)**2);#in Volts\n", + "Regulation=((Eo-V)/V)*100;#in %\n", + "print \"Full load percentage regulation at a power factor of 0.8 leading : \",round(Regulation,2);\n", + "print \"Negative regulation due to leading power factor.\";" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Full load percentage regulation at a power factor of 0.8 lagging : 2.2\n", + "Full load percentage regulation at a power factor of 0.8 leading : -1.78\n", + "Negative regulation due to leading power factor.\n" + ] + } + ], + "prompt_number": 110 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10, Page 60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "RatedPower=50;#in KVA\n", + "RatedPower=50*1000;#in VA\n", + "VL=173;#in Volts\n", + "Ra=0.1;#in Ohm\n", + "\n", + "#Calculations&Results\n", + "VP=VL/sqrt(3);#in Volts\n", + "print \"Some exciting curent on short circuit produces a current of 100 A.\";\n", + "OC_PhaseVoltage=100;#in Volt\n", + "SC_Current=100;#in Ampere\n", + "Zs=OC_PhaseVoltage/SC_Current;#n ohm\n", + "Xs=sqrt(Zs**2-Ra**2);#in Ohm\n", + "print \"Impedence of the alternator in Ohm : \",round(Xs,2);\n", + "V=400;#in Volts\n", + "I_FL=RatedPower/(sqrt(3)*V);#in Ampere\n", + "V=400/sqrt(3);#in Volts\n", + "Eo=sqrt((V+I_FL*Ra)**2+(I_FL*Xs)**2);#in Volts\n", + "Regulation=(Eo-V)*100/V;#in %\n", + "print \"Regulation at U.P.F. in % :\",round(Regulation,2);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Some exciting curent on short circuit produces a current of 100 A.\n", + "Impedence of the alternator in Ohm : 0.99\n", + "Regulation at U.P.F. in % : 7.71\n" + ] + } + ], + "prompt_number": 112 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11, Page 61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "OutputPower=500;#in KVA\n", + "OutputPower=500*1000;#in VA\n", + "VL=3300;#in Volts\n", + "Ra=0.3;#in Ohm\n", + "Xs=4;#in Ohm\n", + "PF=0.8;#Lagging Power factor\n", + "\n", + "#Calculations&Results\n", + "#Formula : outputPower=sqrt(3)*VL*IL\n", + "IL=OutputPower/(sqrt(3)*VL);#in Ampere\n", + "print \"For a star connected alternator, line current is equal to phase current. Therefore Ia=IL\";\n", + "Ia=IL;#in Ampere\n", + "#PF=cosd(fi)=0.8 and sind(fi)=0.6\n", + "cos_fi=0.8;#Power factor\n", + "sin_fi=0.6;#Unitless\n", + "VPerPhase=VL/sqrt(3);#in Volts\n", + "E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi+Ia*Xs)**2);#in Volts/Phase\n", + "Regulation=(E-VPerPhase)*100/VPerPhase;#in %\n", + "print \"Voltage Regulation at Full Load in % :\",round(Regulation,2);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For a star connected alternator, line current is equal to phase current. Therefore Ia=IL\n", + "Voltage Regulation at Full Load in % : 12.98\n" + ] + } + ], + "prompt_number": 113 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.12, Page 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "V=2000;#in Volt\n", + "Ia=100;#in Ampere\n", + "OC_Voltage=500;#in Volt\n", + "SC_Current=100;#in Ampere\n", + "\n", + "#Calculations&Results\n", + "Zs=OC_Voltage/SC_Current;#in ohm\n", + "Ra=0.8;#in Ohm\n", + "Xs=sqrt(Zs**2-Ra**2);#in Ohm\n", + "#formula : Induced EMF, E=sqrt((V*cos_fi+Ia*Ra)^2+(V*sin_fi+Ia*Xs)^2)\n", + "#Part (a) : at unity pf\n", + "cos_fi=1;#Unitless\n", + "sin_fi=0;#Unitless\n", + "E=sqrt((V*cos_fi+Ia*Ra)**2+(V*sin_fi+Ia*Xs)**2)\n", + "Regulation=(E-V)*100/V;#in %\n", + "print \"Regulation at U.P.F. in % :\",round(Regulation,2);\n", + "\n", + "#Part (b) : at 0.71 pf lagging\n", + "cos_fi=0.71;#Unitless\n", + "sin_fi=0.704;#Unitless\n", + "E=sqrt((V*cos_fi+Ia*Ra)**2+(V*sin_fi+Ia*Xs)**2)\n", + "Regulation=(E-V)*100/V;#in %\n", + "print \"Regulation at 0.71 pf lagging in % :\",round(Regulation,2);\n", + "\n", + "#Part (c) : at 0.8 pf leading\n", + "cos_fi=0.8;#Unitless\n", + "sin_fi=0.6;#Unitless\n", + "E=sqrt((V*cos_fi+Ia*Ra)**2+(V*sin_fi-Ia*Xs)**2)\n", + "Regulation=(E-V)*100/V;#in %\n", + "print \"Regulation at 0.8 pf leading in % :\",round(Regulation,2);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Regulation at U.P.F. in % : 6.89\n", + "Regulation at 0.71 pf lagging in % : 21.1\n", + "Regulation at 0.8 pf leading in % : -8.88\n" + ] + } + ], + "prompt_number": 114 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13, Page 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#given data\n", + "Ia=60;#in Ampere\n", + "\n", + "#Calculations&Results\n", + "print \"The value of synchronous impedence at this excitation :\" ;\n", + "OC_Voltage=900;#in Volt\n", + "SC_Current=150;#in Ampere\n", + "Zs=OC_Voltage/SC_Current;#in ohm\n", + "print \"Zs equals to \",(Zs),\" Ohm\";\n", + "print \"Internal Voltage drop when tthe load current is 60A=Ia*Zs=\",(Ia*Zs),\" Volts\";" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of synchronous impedence at this excitation :\n", + "Zs equals to 6 Ohm\n", + "Internal Voltage drop when tthe load current is 60A=Ia*Zs= 360 Volts\n" + ] + } + ], + "prompt_number": 118 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14, Page 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "V=6600;#in Volts\n", + "OutputPower=2000;#in KVA\n", + "\n", + "#Calculations\n", + "OutputPower=2000*1000;#in VA\n", + "#Formula : outputPower=sqrt(3)*VL*IL\n", + "IL=OutputPower/(sqrt(3)*V);#in Ampere\n", + "Ia=IL;#in Ampere\n", + "Ra=0.4;#in Ohm\n", + "Xs=4.5;#in Ohm\n", + "#PF=cosd(fi)=0.8 and sind(fi)=0.6\n", + "cos_fi=0.8;#Power factor\n", + "sin_fi=0.6;#Unitless\n", + "VPerPhase=V/sqrt(3);#in Volts\n", + "E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi+Ia*Xs)**2);#in Volts/Phase\n", + "Regulation=(E-VPerPhase)*100/VPerPhase;#in %\n", + "\n", + "#Result\n", + "print \"Percentage Change in Terminal Voltage :\",round(Regulation,1);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage Change in Terminal Voltage : 14.9\n" + ] + } + ], + "prompt_number": 119 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15, Page 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "OutputPower=1200;#in KVA\n", + "OutputPower=1200*1000;#in VA\n", + "V=3300;#in Volt\n", + "Ra=0.25;#in Ohm\n", + "\n", + "#Calculations&Results\n", + "#Formula : outputPower=sqrt(3)*VL*IL\n", + "IL=OutputPower/(sqrt(3)*V);#in Ampere\n", + "Ia=IL;#in Ampere\n", + "VPerPhase=V/sqrt(3);#in Volts\n", + "OC_Voltage=1100;#in Volt\n", + "SC_Current=200;#in Ampere\n", + "Zs=OC_Voltage/(sqrt(3)*SC_Current);#in ohmRa\n", + "Xs=sqrt(Zs**2-Ra**2);#in Ohm\n", + "#formula : Induced EMF, E=sqrt((V*cos_fi+Ia*Ra)^2+(V*sin_fi+Ia*Xs)^2)\n", + "\n", + "#Part (a) : For lagging pf load\n", + "cos_fi=0.8;#Unitless\n", + "sin_fi=0.6;#Unitless\n", + "E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi+Ia*Xs)**2);\n", + "Regulation=(E-VPerPhase)*100/VPerPhase;#in %\n", + "print \"Regulation at U.P.F. in % :\",round(Regulation,2);\n", + "\n", + "#Part (b) : For leading pf load\n", + "cos_fi=0.8;#Unitless\n", + "sin_fi=0.6;#Unitless\n", + "E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi-Ia*Xs)**2)\n", + "Regulation=(E-VPerPhase)*100/VPerPhase;#in %\n", + "print \"Regulation at 0.71 pf lagging in % :\",round(Regulation,2);\n", + "#Note: For leading power factor load, the regulation s negative. " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Regulation at U.P.F. in % : 25.9\n", + "Regulation at 0.71 pf lagging in % : -13.52\n" + ] + } + ], + "prompt_number": 120 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16, Page 64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "OutputPower=1500;#in KVA\n", + "OutputPower=1500*1000;#in VA\n", + "V=6600;#in Volt\n", + "Ra=0.4;#in Ohm\n", + "Xs=6;#in Ohm per phase\n", + "pf=0.8;#lagging power factor\n", + "\n", + "#Calculations&Results\n", + "#Formula : outputPower=sqrt(3)*VL*IL\n", + "Ia=OutputPower/(sqrt(3)*V);#in Ampere\n", + "VPerPhase=V/sqrt(3);#in Volts\n", + "#formula : Induced EMF, E=sqrt((V*cos_fi+Ia*Ra)^2+(V*sin_fi+Ia*Xs)^2)\n", + "cos_fi=0.8;#Unitless\n", + "sin_fi=0.6;#Unitless\n", + "E=sqrt((VPerPhase*cos_fi+Ia*Ra)**2+(VPerPhase*sin_fi+Ia*Xs)**2);#in volt\n", + "print \"Induced emf in volt : \",round(E,2);\n", + "print \"As excitation remains constant, E at 4364 volt remains constant.\";\n", + "E=4364;#in Volt\n", + "V=4743;#in Volts\n", + "TerminalVoltage=sqrt(3)*V;#in Volts\n", + "print \"Terminal voltage line to line in Volts : \",round(TerminalVoltage);\n", + "#Note ans of 1st part is wrong in the books" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Induced emf in volt : 4366.07\n", + "As excitation remains constant, E at 4364 volt remains constant.\n", + "Terminal voltage line to line in Volts : 8215.0\n" + ] + } + ], + "prompt_number": 123 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.17, Page 81" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "OutputPower=2500;#in KVA\n", + "OutputPower=2500*1000;#in VA\n", + "V=6600;#in Volt\n", + " \n", + "#Calculations\n", + "#For first load i.e. Lighting load : \n", + "KW1=2500;#in KWatts\n", + "KVAR1=0;#Kwatts\n", + "#For second load i.e. Motor load : \n", + "KW2=5000;#in KWatts\n", + "cos_fi=0.707;#unitless\n", + "sin_fi=0.707;#unitless\n", + "KVAR2=KW2*sin_fi/cos_fi;#Kwatts\n", + "#For total load\n", + "TotalKW=KW1+KW2;#in KWatts\n", + "TotalKVAR=KVAR1+KVAR2;#Kwatts \n", + "#For first Machine\n", + "KWm=4000;#in KWatts\n", + "cos_fi=0.8;#unitless\n", + "sin_fi=0.6;#unitless\n", + "KVARm=KWm*sin_fi/cos_fi;#Kwatts\n", + "#so, second machine will be supplying\n", + "KW=TotalKW-KWm;#in Kwatts\n", + "print \"KW output of second machine : \",KW;\n", + "KVAR=TotalKVAR-KVARm;#in KWatts\n", + "tan_fi=KVAR/KW;#unitless\n", + "fi=atan(tan_fi);#ib degree\n", + "#Power factor of other machine\n", + "pf=cos(fi);#unitless\n", + "\n", + "#Result\n", + "print \"Power factor of other machine : \",round(pf,4),\" lagging\";" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "KW output of second machine : 3500\n", + "Power factor of other machine : 0.8682 lagging\n" + ] + } + ], + "prompt_number": 126 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.18, Page 82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "\n", + "#Load1 : \n", + "KW1=500;#in KWatts\n", + "KVAR1=0;#Kwatts\n", + "\n", + "#Load2 : \n", + "KW2=1000;#in KWatts\n", + "pf=0.9;#lagging\n", + "cos_fi=0.9;#unitless\n", + "fi=acos(pf);\n", + "sin_fi=sin(fi);#unitless\n", + "KVAR2=KW2*sin_fi/cos_fi;#Kwatts\n", + "\n", + "#Load3 : \n", + "KW3=800;#in KWatts\n", + "pf=0.8;#lagging\n", + "cos_fi=0.8;#unitless\n", + "fi=acos(pf);\n", + "sin_fi=sin(fi);#unitless\n", + "KVAR3=KW3*sin_fi/cos_fi;#Kwatts\n", + "\n", + "#Load4 : \n", + "KW4=500;#in KWatts\n", + "pf=0.9;#lagging\n", + "cos_fi=0.9;#unitless\n", + "fi=acos(pf);\n", + "sin_fi=sin(fi);#unitless\n", + "KVAR4=-KW4*sin_fi/cos_fi;#Kwatts\n", + "#TOtalKW and TotalKVAR\n", + "TotalKW=KW1+KW2+KW3+KW4;#in KWatts\n", + "TotalKVAR=KVAR1+KVAR2+KVAR3+KVAR4;#in KWAtts\n", + "#For the first Machine :\n", + "KW=1500;#n Kwatts\n", + "cos_fi=0.95;#unitless\n", + "sin_fi=0.3123;#unitless\n", + "KVAR=KW*sin_fi/cos_fi;#Kwatts\n", + "\n", + "KW1=TotalKW-KW;#in KWatts\n", + "KVAR1=TotalKVAR-KVAR;#in Volts\n", + "print \"KW supplied by other machine : \",(TotalKW-KW);\n", + "print \"KVAR supplied by other machine : \",round(TotalKVAR-KVAR); #answer differs due to rounding off the digits\n", + "tan_fi=KVAR1/KW1;#unitless\n", + "#fi=atand(tan_fi);#in degree\n", + "cos_fi=cos(atan(tan_fi));#unitless\n", + "\n", + "#Result\n", + "print \"Power factor of the other machine : \",round(cos_fi,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "KW supplied by other machine : 1300\n", + "KVAR supplied by other machine : 349.0\n", + "Power factor of the other machine : 0.966\n" + ] + } + ], + "prompt_number": 129 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.19, Page 82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "\n", + "#Lighting Load : \n", + "MW1=20;#load in Mwatts\n", + "KW1=MW1*1000;#in KWatts\n", + "KVAR1=0;#Kwatts\n", + "\n", + "#Motor Load : \n", + "MW2=40;#load in Mwatts\n", + "KW2=MW2*1000;#in KWatts\n", + "pf=0.8;#unitless\n", + "cos_fi=0.8\n", + "fi=acos(pf);\n", + "sin_fi=sin(fi);#unitless\n", + "KVAR2=KW2*sin_fi/cos_fi;#Kwatts\n", + "\n", + "#For Total Load : \n", + "TotalKW=KW1+KW2;#load in Mwatts\n", + "TotalKVAR=KVAR1+KVAR2;#in KWatts\n", + "#For first machine : \n", + "MWm=32;#load in Mwatts\n", + "KWm=MWm*1000;#in KWatts\n", + "cos_fi=0.866;#unitless\n", + "fi=acos(cos_fi);\n", + "tan_fi=tan(fi);#unitless\n", + "KVARm=KWm*tan_fi;#in KWatts\n", + "#so, load supplied by the second machine\n", + "KW2=TotalKW-KWm;#in Kwatts\n", + "print \"Load of other machine,KW : \",(KW2);\n", + "KVAR2=TotalKVAR-KVARm;#in Kwatts\n", + "tan_fi=KVAR2/KW2;#unitless\n", + "fi=atan(tan_fi);#in degree\n", + "cos_fi=cos(atan(tan_fi));#unitless\n", + "\n", + "#Result\n", + "print \"Power factor of the other machine : \",round(cos_fi,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Load of other machine,KW : 28000\n", + "Power factor of the other machine : 0.925\n" + ] + } + ], + "prompt_number": 131 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.20, Page 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "cos_fi=0.8;#unitless\n", + "fi=acos(cos_fi);\n", + "tan_fi=tan(fi);#unitless\n", + "\n", + "#For Alternator A : \n", + "cos_fi_A=0.9;#unitless\n", + "fi_A=acos(cos_fi_A);\n", + "tan_fi_A=tan(fi_A);#unitless\n", + "#Formula : Active load, KW=V*I*cos_fi\n", + "#Formula : Reactive load, KVAR=V*I*sin_fi\n", + "ActiveLoad=8000;#in KW\n", + "ReactiveLoad=ActiveLoad*tan_fi;#in KVAR\n", + "\n", + "#For A:\n", + "ActiveLoadA=5000;#in KW\n", + "ReactiveLoadA=ActiveLoadA*tan_fi_A;#in KVAR\n", + "\n", + "#For B :\n", + "ActiveLoadB=ActiveLoad-ActiveLoadA;#in KW\n", + "ReactiveLoadB=ReactiveLoad-ReactiveLoadA;#in KVAR\n", + "tan_fi_B=ReactiveLoadB/ActiveLoadB;#unitless\n", + "fi_B=atan(tan_fi_B);#in degree\n", + "cos_fi=cos(atan(tan_fi_B));#unitless\n", + "\n", + "#Result\n", + "print \"Power factor of the other machine : \",round(cos_fi,3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power factor of the other machine : 0.642\n" + ] + } + ], + "prompt_number": 133 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.21, Page 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "V=6600;#in Volts\n", + "KW=6000.;#in KWatts\n", + "pf=0.8;#unitless\n", + "cos_fi=pf;#unitless\n", + "Eff=90.;#in %\n", + "\n", + "#Calculations&Results\n", + "#Part (a) : \n", + "KVA=KW/cos_fi;#in KVAR\n", + "print \"KVA rating of the alternator : \",(KVA),\" KVA\";\n", + "#Part (b) : \n", + "TotalRating=KVA;#in KVA\n", + "VA=TotalRating*1000;#in VA\n", + "I=VA/(sqrt(3)*KW);#in Ampere\n", + "print \"Current Rating in Ampere : \",round(I,2);\n", + "#Part (c) :\n", + "Input=KW/(Eff/100);#in KW\n", + "print \"Power Input(in KW) :\",round(Input,2);\n", + "Input=Input*1000/735.5;#in hp\n", + "print \"Power Input(in hp) :\",round(Input,2);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "KVA rating of the alternator : 7500.0 KVA\n", + "Current Rating in Ampere : 721.69\n", + "Power Input(in KW) : 6666.67\n", + "Power Input(in hp) : 9064.13\n" + ] + } + ], + "prompt_number": 139 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.22, Page 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import * \n", + "\n", + "#given data\n", + "Ecoil=8000;#in Volts\n", + "Icoil=418;#in Ampere\n", + "pf=80.;#in % lgging\n", + "\n", + "#Calculations&Results\n", + "pf=pf/100;#in fraction\n", + "cos_fi=pf;#unitless\n", + "#Part (i) : \n", + "EL=sqrt(3)*Ecoil;#in volt\n", + "print \"Line volts(in V): \",round(EL);\n", + "#Part (ii) : \n", + "IL=Icoil;#in Ampere\n", + "print \"Line Current in Ampere : \",IL;\n", + "#Part (iii) :\n", + "Rating=sqrt(3)*EL*IL/1000;#in KVA\n", + "print \"Rating (in KVA) :\",Rating;\n", + "#Part (iv) :\n", + "FullLoadPower=sqrt(3)*EL*IL*cos_fi/1000;#in KW\n", + "print \"Full Load Power in KW :\",FullLoadPower;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Line volts(in V): 13856.0\n", + "Line Current in Ampere : 418\n", + "Rating (in KVA) : 10032.0\n", + "Full Load Power in KW : 8025.6\n" + ] + } + ], + "prompt_number": 140 + } + ], + "metadata": {} + } + ] +}
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