{
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"name": "",
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"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 pi, sqrt, sin, cos\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": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4, Page 38"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.5, Page 39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.6, Page 39"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.7, Page 40"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.8, Page 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.9, Page 58"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.10, Page 60"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.11, Page 61"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, cos, sin\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": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.12, Page 62"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 21
},
{
"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 pi, sqrt, sin, cos\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 pi, sqrt, sin, cos\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 pi, sqrt, sin, cos\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 pi, sqrt, sin, cos, atan, tan\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": 32
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.18, Page 82"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.19, Page 82"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos, acos, tan, atan\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": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.20, Page 84"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos, atan, tan, acos\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": 29
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.21, Page 97"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 30
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.22, Page 97"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from math import pi, sqrt, sin, cos\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": 31
}
],
"metadata": {}
}
]
}