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| -rwxr-xr-x | Electrical_Machines_-_1_by_Tarlok_Singh/Chap2.ipynb | 2003 | ||||
| -rwxr-xr-x | Electrical_Machines_-_1_by_Tarlok_Singh/Chap3.ipynb | 1687 | ||||
| -rwxr-xr-x | Electrical_Machines_-_1_by_Tarlok_Singh/Chap4.ipynb | 155 | ||||
| -rw-r--r-- | Electrical_Machines_-_1_by_Tarlok_Singh/README.txt | 10 | ||||
| -rwxr-xr-x | Electrical_Machines_-_1_by_Tarlok_Singh/screenshots/ArmaturePowerC2.png | bin | 0 -> 107879 bytes | |||
| -rwxr-xr-x | Electrical_Machines_-_1_by_Tarlok_Singh/screenshots/WaveNLapWindingC2.png | bin | 0 -> 61876 bytes | |||
| -rwxr-xr-x | Electrical_Machines_-_1_by_Tarlok_Singh/screenshots/turnsPerPhaseC4.png | bin | 0 -> 78179 bytes |
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diff --git a/Electrical_Machines_-_1_by_Tarlok_Singh/Chap2.ipynb b/Electrical_Machines_-_1_by_Tarlok_Singh/Chap2.ipynb new file mode 100755 index 00000000..e53e4892 --- /dev/null +++ b/Electrical_Machines_-_1_by_Tarlok_Singh/Chap2.ipynb @@ -0,0 +1,2003 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter2 - DC Machines" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.3 page 60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "#Caption:Find the current per path of armature\n", + "\n", + "I_a=200 #rated armature current(in Amp)\n", + "P=12 #number of poles in machine\n", + "A_1=2 #number of parallel paths with wave winding\n", + "A_2=P #number of parallel paths with lap winding\n", + "I_1=I_a/A_1 #current per path in case of wave winding(in Amp)\n", + "print 'current per path in case of wave winding = %0.2f A'%I_1\n", + "I_2=I_a/A_2 #current per path in case of lap winding(in Amp)\n", + "print 'current per path in case of lap winding = %0.2f A'%I_2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "current per path in case of wave winding = 100.00 A\n", + "current per path in case of lap winding = 16.67 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.4 page 60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the induced emf in machine \n", + "N_1=500 #starting speed of machine(in rpm)\n", + "E_1=200 #emf of the machine at N_1 (in V)\n", + "N_2=600 #new speed of machine(in rpm)\n", + "E_2=E_1*N_2/N_1 #emf of the machine atN_2 (in V)\n", + "print 'induced emf while the machine is running at 600 rpm = %0.2f V '%E_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "induced emf while the machine is running at 600 rpm = 240.00 V \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.5 page 61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:calculate the generated emf\n", + "S_1=80 #Number of armature slots \n", + "S_2=6 #Number of conductor per slot\n", + "Z=S_1*S_2 #Number of armature conductors\n", + "F=50 #Flux per pole(in mWb)\n", + "F_1=F*10**-3 #(in Wb)\n", + "P=6 #Number of poles\n", + "A=P #Number of parallel paths\n", + "N=1200 #armature speed(in rpm)\n", + "E_g=F_1*Z*N*P/(60*A) #EMF generated in 6 pole lap wound dc generator (in Volts)\n", + "print 'EMF generated in 6 pole lap wound dc generator = %0.2f V'%E_g" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "EMF generated in 6 pole lap wound dc generator = 480.00 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.6 page 61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the voltage generated in 4 pole generator\n", + "\n", + "S_1=51 #Number of slots \n", + "S_2=20 #Number of conductor per slot\n", + "Z=S_1*S_2 #number of armature conductors\n", + "F=7 #flux per pole(in mWb) \n", + "F_1=F*10**-3 #flux per pole(in Wb)\n", + "P=4 #Number of poles\n", + "A=2 #Number of parallel paths (armature is wave wound)\n", + "N=1500 #speed of the machine(in rpm)\n", + "E_g=F_1*Z*N*P/(60*A) #generated emf\n", + "print 'voltage generated in machine = %0.2f V'%E_g" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage generated in machine = 357.00 V\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.7 page 61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Find out the speed for 6 pole machine \n", + "F=60 #flux per pole(in m Wb)\n", + "F_1=F*10**-3 #flux per pole(in Wb)\n", + "Z=480 #Number of armature conductors\n", + "P=6 #Number of poles\n", + "A=2 #Number of parallel paths(Armature wave wound)\n", + "E_g=320 #generated emf (in V)\n", + "N=E_g*60*A/(F_1*Z*P) #speed(in rpm)\n", + "print 'speed of the machine = %0.2f rpm '%N " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of the machine = 222.22 rpm \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.8 page 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import ceil\n", + "#Caption:calculate the suitable number of conductor per slot hence determine the actual value of flux\n", + "\n", + "F_1=0.05 #flux per pole(in Wb )\n", + "N=350 #speed(in rpm)\n", + "P=8 #no of poles\n", + "A=P #no of parallel path\n", + "E_g=240 #voltage generated (in V)\n", + "Z_1=E_g*60*A/(F_1*N*P) #total no of armature conductor required\n", + "C_s=ceil(Z_1/120) #number of conductor per slot\n", + "print 'number of conductor per slot =' ,C_s\n", + "A_s=120 #armature slots \n", + "Z_2=A_s*C_s #total conductors in armature slot\n", + "F_2=E_g*60*A/(N*Z_2*P) #Actual value of flux(in Wb)\n", + "print 'Actual value of flux = %0.4f Wb '%F_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "number of conductor per slot = 7.0\n", + "Actual value of flux = 0.0490 Wb \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.9 page 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the emf if the speed is 1500 rpm\n", + "\n", + "P=8 #Number of poles\n", + "F=2 #flux per pole (in mWb)\n", + "F_1=2*10**-2 #flux per pole (in Wb)\n", + "Z=960 #number of conductor\n", + "A=P #Number of parallel paths(lap winding)\n", + "N=1500 #speed (in rpm)\n", + "E_g=P*F*1e-2*Z*N/(60*A) #emf generated (in Volts)\n", + "print 'emf generated = %0.f V'%E_g" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emf generated = 480 V\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.10 page 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the increase of main field flux in percentage\n", + "\n", + "N_1=750 #speed of dc machine(in rpm)\n", + "E_1=220 #induced emf in dc machine when running at N_1\n", + "N_2=700 #speed of dc machine second time (in rpm)\n", + "E_2=250 #induced emf in dc machine when running at N_2\n", + "F=E_2*N_1/(E_1*N_2) \n", + "Inc=(F-1) \n", + "print 'increase in main field flux of the dc machine = %0.2f %%'%(Inc*100) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "increase in main field flux of the dc machine = 21.75 %\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.11 page 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#caption: a)find the emf generated in a 6 pole machine b)find speed at which machine generated 550 V emf\n", + "\n", + "F_1=0.06 #Flux per pole(in Wb)\n", + "N_1=250 #speed of the rotor(in rpm)\n", + "A=2 #number of parllel (paths armature wave wound)\n", + "P=6 #poles in machine\n", + "Z=664 #total conductor in machine\n", + "E_g=P*F_1*N_1*Z/(60*A) #emf generated\n", + "print 'emf generated in machine = %0.2f V'%E_g\n", + "E_2=550 #new emf generating machine(in V)\n", + "F_2=0.058 #flux per pole (in Wb) for generating E_2\n", + "N_2=60*E_2*A/(P*F_2*Z) #new speed at which machine generating E_2(in rpm)\n", + "print 'new speed of the rotor = %0.2f rpm '%N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emf generated in machine = 498.00 V\n", + "new speed of the rotor = 285.63 rpm \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.12 page 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption :determine the value of torque in Nw-m\n", + "\n", + "F=24 #flux per pole (in m Wb)\n", + "F_1=F*10**-3 #flux per pole (in Wb)\n", + "Z=760 #number of conductors in armature\n", + "P=4 #number of pole\n", + "A=2 #number of parallel paths\n", + "I_a=50 #armature cuurrent(in Amp)\n", + "T_a=0.159*F_1*Z*P*I_a/A #torque develope(in Nw-m)\n", + "print 'torque developed in machine = %0.f Nw-m '%T_a " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque developed in machine = 290 Nw-m \n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.13 page 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption: calculate the total torque in Nw-m\n", + "\n", + "P=6 #poles \n", + "A=P #number of parallel paths\n", + "S=60 #slots in motor\n", + "C_s=12 #conductor per slot\n", + "Z=S*C_s #total conductor in machine\n", + "I_a=50 #armature current(in Amp)\n", + "F_1=20#flux per pole(in m Wb)\n", + "F_2=F_1*10**-3 #flux per pole)(in Wb)\n", + "T=0.15924*F_2*Z*P*I_a/A #total torque (in Nw-m)\n", + "print 'total torque by motor = %0.2f Nw-m '%T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "total torque by motor = 114.65 Nw-m \n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.14 page 67" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the drop in speed when motor takes 51 Amp\n", + "\n", + "V=220 #supply voltage(in V)\n", + "R_sh=220 #shunt field resistance(in Ohm)\n", + "R_a=0.2 #armature resistance(in Ohm)\n", + "I_sh=V/R_sh #shunt field current(in Amp)\n", + "N_1=1200 #starting speed of the motor(in rpm)\n", + "I_1=5.4 #at N_1 speed current in motor(in Amp)\n", + "I_a1=I_1-I_sh #armature current at speed N_1(in Amp)\n", + "E_b1=V-I_a1*R_a #emf induced due to I_a1(in V)\n", + "I_2=51 #new current which motor taking(in Amp)\n", + "I_a2=I_2-I_sh #armature current at I_2(in Amp)\n", + "E_b2=V-I_a2*R_a #emf induced due to I_a2(in V)\n", + "N_2=E_b2*N_1/E_b1 #speed of the motor when taking I_2 current(in rpm)\n", + "N_r=ceil(N_1-N_2) #reduction in speed(in rpm)\n", + "print 'reduction in speed = %0.2f rpm '%N_r " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "reduction in speed = 50.00 rpm \n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.15 page 68" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a dc machine Calculate (a)induced emf (b)Electro magnetic torque (c)armature copper loss \n", + "\n", + "V=220 #voltage at the armature of dc motor\n", + "I_a=15 #current through armature(in Amp)\n", + "R_a=1 #armature resistance(in Ohm)\n", + "w=100 #speed of the machine(in radian/sec)\n", + "E=V-I_a*R_a #induced emf(in V)\n", + "print 'induced emf = %0.2f V '%E \n", + "T=E*I_a/w #electro magnentic torque developed(in Nw-m)\n", + "print 'electro magnentic torque developed = %0.2f Nw-m '%T \n", + "L=(I_a**2)*R_a #Armature copper loss(in Watt)\n", + "print 'Armature copper loss = %0.2f Watt' %L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "induced emf = 205.00 V \n", + "electro magnentic torque developed = 30.75 Nw-m \n", + "Armature copper loss = 225.00 Watt\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.16 page 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the electro magnetic torque\n", + "\n", + "E=250 #emf induced in dc machine(in V)\n", + "I_a=20 #current flowing through the armature(in Amp)\n", + "N=1500 #speed(in rpm)\n", + "T_e=0.1591*E*I_a*60/N #torque developed in machine(in Nw-m)\n", + "print 'electro magnetic torque developed in dc machine = %0.2f Nw-m '%T_e " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "electro magnetic torque developed in dc machine = 31.82 Nw-m \n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.17 page 69" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:calculate the gross torque in dc machine\n", + "\n", + "P=4 #number of poles \n", + "Z=1600 #number of armature conductor\n", + "F=0.027 #flux per pole(in Wb)\n", + "A=2 #number of parallel paths (wave wound)\n", + "I=75 #current in machine(in Amp)\n", + "N=1000 #speed of the motor(in rpm)\n", + "T=0.1591*P*F*Z*I/A #torque generate in machine(in Nw-m)\n", + "print 'Torque generated in machine = %0.2f Nw-m '%T " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Torque generated in machine = 1030.97 Nw-m \n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.18 page 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the value of back emf\n", + "\n", + "V=230 #applied voltage (in V)\n", + "R_a=0.1 #armature resistance(in Ohm)\n", + "I_a=60 #armature current (in Amp)\n", + "E_b=V-I_a*R_a #back emf(in Volts)\n", + "print 'back emf produced by machine = %0.2f V '%E_b " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "back emf produced by machine = 224.00 V \n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.19 page 75" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Capyion:find the change in back emf from no load to load\n", + "\n", + "V=220 #given voltage to machine(in V)\n", + "R_a=0.5 #armature circuit resistance(in ohm)\n", + "I_1=25 #full load armature current(in Amp)\n", + "I_2=5 #no load armature current(in Amp)\n", + "E_1=V-I_1*R_a #back emf at full load(in V)\n", + "E_2=V-I_2*R_a #back emf at no load(in V)\n", + "E=E_2-E_1 #change in back emf no load to load\n", + "print 'change in back emf from no load to load = %0.2f Volts '%E " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "change in back emf from no load to load = 10.00 Volts \n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.20 page 76\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption: Determine the back emf in dc shunt motor\n", + "\n", + "V=220 #voltage(in V)\n", + "R_a=0.7 #Armature resistance(in Ohm)\n", + "R_f=200 #field resistant(in Ohm)\n", + "P_1=8*10**3 #motor output power(in Watt)\n", + "P_2=8*10**3/0.8 #motor input power(in Watt)\n", + "I_m=P_2/V #motor input current(in Amp)\n", + "I_sh=V/R_f #shunt field current (in Amp)\n", + "I_a=I_m-I_sh #Armature current(in Amp)\n", + "E_b=V-I_a*R_a #Back emf (in V)\n", + "print 'Back emf produced in motor = %0.2f Volts '%E_b " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Back emf produced in motor = 188.95 Volts \n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:21 page 76" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption :Determine the total armature power developed when working as a)Generator b)motor\n", + "\n", + "P=25*10**3 #output generator power(in Watt)\n", + "V=250 #output generator voltage(in V)\n", + "R_f=125 #field resistance(in Ohm)\n", + "R_a=0.05 #armature resistance(in Ohm)\n", + "I_sh=V/R_f #shunt field current(in Amp)\n", + "#IN case of generator\n", + "I_o=P/V #output generator current(in Amp)\n", + "I_a1=I_o+I_sh #armature current for a generator(in Amp)\n", + "E_a1=250+I_a1*R_a #generated emf in armature(in V)\n", + "P_a1=E_a1*I_a1 #generated power in armature when working as a generator(in Watt)\n", + "P_g=P_a1*10**-3 #generated power in armature when working as a generator(in kW)\n", + "print 'power developed in armature when working as a generator = %0.2f kW '%P_g \n", + "#IN case of motor\n", + "I_in=P/V #motor input current(in Amp)\n", + "I_a2=I_in-I_sh #armature current for a motor(in Amp)\n", + "E_a2=250-I_a2*R_a #generated emf in armature when working as a motor(in V)\n", + "P_a2=E_a2*I_a2 #generated power in armature when working as a motor(in Watt)\n", + "P_m=P_a2*10**-3 #generated power in armature when working as a motor(in KW)\n", + "print 'power developed in armature when working as a motor = %0.2f kW '%P_m " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power developed in armature when working as a generator = 26.02 kW \n", + "power developed in armature when working as a motor = 24.02 kW \n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.22 page 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption :Calculated power developed in armature when machine running as (a)Generator (b)Motor\n", + "\n", + "V=250 #line voltage(in V)\n", + "R_sh=125 #shunt field resistance(in Ohm)\n", + "I_sh=V/R_sh #shunt field current(in Amp)\n", + "I_l=80 #line current(in Amp)\n", + "R_a=0.1 #armature resistance(in Ohm)\n", + "#As A Generator\n", + "I_a1=I_l+I_sh #armature current in generator(in Amp)\n", + "E_g=V+I_a1*R_a #generated emf(in V)\n", + "P_1=E_g*I_a1*10**-3 #power developed in armature (in KW)\n", + "print 'power developed in armature when machine running as a generator = %0.2f kW '%P_1 \n", + "#As A Motor\n", + "I_a2=I_l-I_sh #armature current in motor(in Amp)\n", + "E_b=V-I_a2*R_a #back emf in motor(in V)\n", + "P_2=E_b*I_a2*10**-3 #power developed in armature (in KW)\n", + "print 'power developed in armature when machine running as a Motor = %0.2f kW'%P_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power developed in armature when machine running as a generator = 21.17 kW \n", + "power developed in armature when machine running as a Motor = 18.89 kW\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.23 page 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculated speed of the motor when the current is 10 Amp\n", + "\n", + "V=230 #supply voltage(in V)\n", + "R_a=0.8 #armature resistance(in Ohm)\n", + "R_f=0.8#field resistance(in Ohm)\n", + "I_1=20 #dc motor taking current from supply(in Amp)\n", + "E_1=V-R_a*I_1 #emf generated due to I_1(in V)\n", + "N_1=600 #speed of the motor due to I_1\n", + "I_2=10 #current(in Amp) at which speed of the motor need to calculate\n", + "E_2=V-R_a*I_2 #emf generated due to I_2(in Volts)\n", + "N_2=E_2*I_1*N_1/(E_1*I_2) #speed of the motor when machine drawing 10(in Amp) current\n", + "print 'speed of the motor when machine drawing 10 Amp current = %0.0f rpm '%N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of the motor when machine drawing 10 Amp current = 1245 rpm \n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.24 page 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Percentage change in speed of a d.c. motor\n", + "\n", + "V=240 #supply voltage(in V)\n", + "R_a=0.5 #armature resistance(in Ohm)\n", + "I_1=100 #armature current (in Amp)\n", + "I_2=50 #changed armature current(in Amp)\n", + "E_1=V-R_a*I_1 #induced emf(in V)\n", + "E_2=V-R_a*I_2 #changed induced emf due to I_2\n", + "#flux per pole is constant\n", + "N_r=E_2/E_1 #ratio of speed in machine due to voltage change\n", + "N_rp=(N_r-1)*100 #Percentage change in speed of d.c. motor\n", + "print 'Percentage change in speed of d.c. motor =',round(N_rp ,2)\n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Percentage change in speed of d.c. motor = 13.16\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.25 page 100" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Find the speed at which motor will run when connected in series with a 4 Ohm resistance\n", + "\n", + "V=200 #supply voltage (in V)\n", + "R_m=1 #motor resistance b/w terminals(in Ohm)\n", + "I_1=15 #motor input current(in Amp)\n", + "N_1=800 #speed of the motor (in rpm)\n", + "E_1=V-(I_1*R_m) #back emf developed(in V)\n", + "R=4 #resistance connected in series with motor (in Ohm)\n", + "I_2=I_1 #when resistance of 4 Ohm connected in series with the motor ,motor input current is same\n", + "E_2=V-I_2*(R_m+R) #back emf developed when R connected in series with motor(in V)\n", + "N_2=E_2*N_1/E_1 #speed of the motor when it connected in series with a 4 Ohm resistance(in rpm)\n", + "print 'speed of the motor when it connected in series with a 4 Ohm resistance = %0.1f rpm '%N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of the motor when it connected in series with a 4 Ohm resistance = 540.5 rpm \n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.26 page 100" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine the speed when armature current is 75 Amp and the excitation is increased by 15 %\n", + "\n", + "V=220 #supply voltage(in V)\n", + "R_a=0.03 #armature resistance(in Ohm)\n", + "R_se=0.07 #field resistance(in Ohm)\n", + "I_a1=40 #armature current in first case(in Amp)\n", + "N_1=900 #motor running speed at 40 Amp armature current(in rpm)\n", + "E_1=V-I_a1*(R_a+R_se) #induced emf due to 40 Amp armature current (in V)\n", + "I_a2=75 #armature current in second case(in Amp)\n", + "E_2=V-I_a2*(R_a+R_se) #induced emf due to 75 Amp armature current (in V)\n", + "#Flux is F_1 when I_a1 and F_2 when I_a2. F_2=1.15*F_1 because Excitation is increased by 15% so F=(F_1/F_2)\n", + "F=1.15 #Ratio of F_2/F_1\n", + "N_2=ceil(N_1*E_2/(F*E_1)) #motor speed when armature current is 75 Amp and the excitation is increased by 15 %(in rpm)\n", + "print 'motor speed when armature current is 75 Amp and the excitation is increased by 15 %% = %0.2f rpm'%N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "motor speed when armature current is 75 Amp and the excitation is increased by 15 % = 770.00 rpm\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.27 page 101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate H.P. is being transmitted by the shaft of motor\n", + "\n", + "V=220 #supply voltage(in V)\n", + "N=900 #running speed of a shunt motor(in rpm)\n", + "T=1000 #torque exerted by motor (in Nw-m)\n", + "w=2*3.14159*N/60 #angular speed(in rad/sec)\n", + "P=w*T #power transmitted (in Watt)\n", + "H_p=P/735.5 #power transmitted in H.P.(metric)\n", + "print 'power transmitted by the shaft of motor = %0.1f H.P.(metric)' %H_p" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "power transmitted by the shaft of motor = 128.1 H.P.(metric)\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.28 page 102" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the torque\n", + "\n", + "P_1=70 #power transmitted by the shaft of a motor in H.P(metric)\n", + "P_2=P_1*735.5 #power (in Watts)\n", + "N=500 #speed of the motor(in rpm)\n", + "w=2*3.1416*N/60 #angular speed (in radian/sec)\n", + "T=P_2/w #torque in motor (in Nw-m)\n", + "print 'torque in motor = %0.1f Nw-m '%T " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque in motor = 983.3 Nw-m \n" + ] + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.29 page 102" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Find the speed when the armature takes 70 Amp \n", + "\n", + "V=400 #voltage for a shunt motor (in V)\n", + "R_a=0.2 #armature resistance(in Ohm)\n", + "I_a1=100 #starting armature current(in Amp)\n", + "N_1=1000 #speed of the motor when I_a1 armature current flows in armature\n", + "E_1=V-I_a1*R_a #emf induced at I_a1 armature current (in V)\n", + "I_a2=70 #changed armature current(in Amp)\n", + "E_2=V-I_a2*R_a #emf induced for I_a2 current(in V)\n", + "#flux is constant\n", + "N_2=ceil(E_2*N_1/E_1) #speed of the motor when 70 Amp current flowing through armature\n", + "print 'speed of the motor when 70 Amp current flowing through armature = %0.1f rpm'%N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of the motor when 70 Amp current flowing through armature = 1016.0 rpm\n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.30 page 103" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:resistance required to be connected in series to reduced speed of machine to 800 rpm (in Ohm)\n", + "\n", + "V=400 #voltage applied across the motor(in V)\n", + "R_sh=100 #shunt resistance of motor(in Ohm)\n", + "I=70 #total current flowing through motor(in Amp)\n", + "I_sh=V/R_sh #current flowing through the shunt resistance(in Amp)\n", + "I_a1=I-I_sh #current flowing through the armature(in Amp)\n", + "R_a1=0.03 #resistance of armature(in Ohm)\n", + "R_se=0.5 #series resistance with armature(in Ohm)\n", + "R_a=R_a1+R_se #total resistance in armature circuit(in Ohm)\n", + "E_1=V-I_a1*R_a #emf induced due to this R_a resistance(in V) \n", + "N_1=900 #given speed of the motor(in r.p.m.)\n", + "N_2=800 #desired speed of the motor(in r.p.m.)\n", + "E_2=E_1*(N_2/N_1) #emf induced due to armature resistance when motor spped is 800(in V)\n", + "R_a2=(400-E_2)/66 #resistance required to be connected in series(in Ohm)\n", + "print 'net resistance of armature which reduce speed of the machine to 800 rpm = %0.3f Ohm '%R_a2 \n", + "R=R_a2-R_a1 #additional resistance required to be connected in series to reduced speed of machine to 800 rpm (in Ohm)\n", + "print 'additional resistance required to be connected in series to reduced speed of machine to 800 rpm = %0.3f Ohm '%R " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "net resistance of armature which reduce speed of the machine to 800 rpm = 1.145 Ohm \n", + "additional resistance required to be connected in series to reduced speed of machine to 800 rpm = 1.115 Ohm \n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.31 page 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Motor speed at full load\n", + "\n", + "V=230 #supply voltage\n", + "I_a1=2 #no load current(in Amp)\n", + "N_1=1500 #speed of the motor at no load\n", + "R_a=0.3 #Armature resistance(in Ohm)\n", + "I_a2=50 #full load current(in Amp)\n", + "E_1=V-I_a1*R_a #emf generated at no load\n", + "E_2=V-I_a2*R_a #emf generated at full load\n", + "N_2=(E_2/E_1)*N_1 #full load speed (flux assumed constant)\n", + "print 'D.c. motor speed at full load when flux assumed constant = %0.2f rpm '%ceil(N_2) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "D.c. motor speed at full load when flux assumed constant = 1406.00 rpm \n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.32 page 104" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:calculate the speed of a d.c. shunt generator when it running as d.c. motor and taking 50 KW power at 250 volt\n", + "\n", + "#calculation when machine is running as generator\n", + "V=250 #applied voltage to d.c. shunt generator\n", + "P_1=50000 #power delivers by d.c. shunt generator at V_1\n", + "N_1=400 #generator running at V_1 ,P_1\n", + "R_a=0.02 #armature resistance(in Ohm)\n", + "R_sh=50 #field resistance(in Ohm)\n", + "I_l=P_1/V #load current(in Amp)\n", + "I_sh=V/R_sh #field current(in Amp)\n", + "I_a1=I_l+I_sh #armature current when machine working as a generator(in Amp)\n", + "C_d=1 #contact drop (in volt per brush)\n", + "E_1=V+I_a1*R_a+2*C_d #induced emf by machine when working as a generator(in V)\n", + "#calculation when machine is running as motor\n", + "I_a2=I_l-I_sh#armature current when machine working as a motor(in Amp)\n", + "E_2=V-I_a2*R_a-2*C_d #induced emf by machine when working as a motor(in V)\n", + "N_2=(E_2/E_1)*N_1 #speed of the machine when running as shunt motor(in r.p.m.)\n", + "print 'speed of the machine when running as shunt motor and taking 50 KW power at 250 volt = %0.2f r.p.m. '%N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of the machine when running as shunt motor and taking 50 KW power at 250 volt = 381.26 r.p.m. \n" + ] + } + ], + "prompt_number": 92 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.33 page 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the value of resistance to be connected in series with the armature to reduce the speed to 750 r.p.m.\n", + "\n", + "V=220 #applied voltage to shunt motor(in V)\n", + "I_a1=40 #armature current in first case(in Amp)\n", + "R_a=0.5 #armature circuit resistance(in Ohm)\n", + "N_1=900 #speed of the motor at I_a1 (in rpm)\n", + "E_b1=V-I_a1*R_a #emf generated in armature circuit due to I_a1(in V)\n", + "N_2=750 #desired motor speed (in rpm)\n", + "I_a2=30 #armature current in case of N_2 motor speed(in Amp)\n", + "E_b2=(N_2/N_1)*E_b1 #emf generated in second case when motor speed is N_2\n", + "#R resistance added in series with the armature circuit to reduced the speed of motor \n", + "R=(205-E_b2)/30 #resistance added in series with the armature circuit to reduced the speed of motor\n", + "print 'resistance added in series with the armature circuit to reduced the speed of motor = %0.2f Ohm '%R " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance added in series with the armature circuit to reduced the speed of motor = 1.28 Ohm \n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.34 page 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:what resistance should be placed in series with armature to reduced the speed of the motor to 700 rpm\n", + "\n", + "V=220 #applied voltage to shunt motor(in V)\n", + "R_a=0.3 #armature circuit resistance(in Ohm)\n", + "I_a=15 #armature current(in Amp)\n", + "E_b1=V-I_a*R_a #emf generated in armature circuit(in V)\n", + "N_1=1000 #motor speed when E_b1 generated(in rpm) \n", + "N_2=700 #desired motor speed (in rpm)\n", + "E_b2=(N_2/N_1)*E_b1 #emf generated in second case when motor speed is N_2\n", + "R=(215.5-E_b2)/15 #resistance added in series with the armature circuit to reduced the speed of motor (in Ohm)\n", + "print 'resistance added in series with the armature circuit to reduced the speed of motor = %0.2f Ohm '%R " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance added in series with the armature circuit to reduced the speed of motor = 4.31 Ohm \n" + ] + } + ], + "prompt_number": 65 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.35 page 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a shunt motor find the resistance required in series with the armature circuit to reduce the speed of motor by 50 percent\n", + "\n", + "I_a1=40 #Armature current(in Amp)\n", + "R_a=0.6 #Armature circuit resistance(in Ohm)\n", + "#T_1/T_2=I_a1/I_a2 \n", + "#T_1=T_2\n", + "V=220 \n", + "I_a2=I_a1 #current when speed reduced 50%\n", + "E_b1=V-(R_a*I_a1) #emf induced in circuit\n", + "N_r=0.5 #Ratio of speed (N_2/N_1)\n", + "#E_b2=V-I_a2*R\n", + "#R is a total armature resistance in the second case\n", + "#N_2/N_1=E_b2/E_b1\n", + "#N_r=(V-I_a2*R)/E_b1\n", + "R=(V-(N_r*E_b1))/I_a2-R_a #the resistance required in series with the armature circuit to reduce the speed of motor 50%\n", + "print 'the resistance required in series with the armature circuit to reduce the speed of motor by 50%% = %0.1f Ohm' % R" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the resistance required in series with the armature circuit to reduce the speed of motor by 50% = 2.4 Ohm\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.36 page 107" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the induced emf in a D.C. machine for speed of 600 rpm \n", + "#assuming the flux is constant\n", + "N_1=500 #primary speed of the motor(in rpm)\n", + "E_1=180 #induced emf in d.c. machine when running at N_1 (in V)\n", + "N_2=600 #secondary speed of the motor (in rpm)\n", + "E_2=(N_2/N_1)*E_1 #induced emf in d.c. machine when running at N_2 (in V)\n", + "print 'the emf induced in a D.C. machine when machine running at 600 rpm speed = %0.2f V '% E_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the emf induced in a D.C. machine when machine running at 600 rpm speed = 216.00 V \n" + ] + } + ], + "prompt_number": 76 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.37 page 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a dc machine calculate speed at which the induced emf will be 250 Volts\n", + "\n", + "E_1=220 #Primary emf(in Volts)\n", + "N_1=750 #Speed of the machine at 220 Volts\n", + "E_2=250 #Secondary emf(in Volts)\n", + "N_2=(E_2/E_1)*N_1 #Speed of the machine at which emf will be 250 Volts\n", + "print 'Speed of the machine at which emf will be 250 Volts = %0.f'%N_2 \n", + "N_3=700 #Speed of the machine when main field flux increase\n", + "E_3=250 #induced emf when flux increase(in Volts)\n", + "F_x=(E_3/E_2)*(N_2/N_3) #Ratio of flux when speed is N_3 and N_2\n", + "F=(F_x-1)*100 #Percentage change in flux for induced emf of 250 Volts and speed 700 rpm(in %)\n", + "print 'Percentage change in flux when induced emf 250 Volts and speed 700 rpm = %0.2f %%'% F " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Speed of the machine at which emf will be 250 Volts = 852\n", + "Percentage change in flux when induced emf 250 Volts and speed 700 rpm = 21.75 %\n" + ] + } + ], + "prompt_number": 78 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.38 page 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate induced emf when running at speed of 1380 rpm.\n", + "\n", + "P=4 #Poles in d.c. machine\n", + "Z=594 #number of conductor in d.c. machine\n", + "F=0.0075 #flux per pole(in Wb)\n", + "N=1380 #speed of the motor\n", + "A=2 #number of parallel paths \n", + "E=P*F*N*Z/(60*A) #emf generated in machine when running at speed of 1380 rpm.\n", + "print 'emf generated in machine when running at speed of 1380 rpm = %0.2f V '%ceil(E) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emf generated in machine when running at speed of 1380 rpm = 205.00 V \n" + ] + } + ], + "prompt_number": 79 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.39 page 109" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the speed and calculate the electro magnetic torque.\n", + "\n", + "V_1=230 #supply voltage(in V)\n", + "I_a1=100 #motor taking current from supply(in Amp)\n", + "N_1=600 #speed of the motor when I_a1 current taking from supply(in rpm)\n", + "R_a=0.12 #resistance of armature circuit(in Ohm)\n", + "R_f=0.03 #resistance of series winding(in Ohm)\n", + "R=R_a+R_f #total resistance(in Ohm)\n", + "I_a2=50 #desired current of the motor\n", + "E_1=V_1-I_a1*R #emf induced when current I_a1 flowing\n", + "E_2=V_1-I_a2*R #emf induced when current I_a2 flowing\n", + "N_2=(E_2/E_1)*(I_a1/I_a2)*N_1 #speed of the motor when 50 Amp current taking from supply(in rpm)\n", + "print 'speed of the motor when 50 Amp current taking from supply = %0.2f rpm '%N_2 \n", + "T_1=E_1*I_a1*60/(2*3.14*N_1) #electro-magnetic torque generated when motor running at 600 rpm(in Nw-m)\n", + "print 'electro-magnetic torque generatedwhen motor running at 600 rpm = %0.2f Nw-m '%T_1 \n", + "T_2=E_2*I_a2*60/(2*3.14*N_2) #electro-magnetic torque generated in second case(in Nw-m)\n", + "print 'electro-magnetic torque generated in second case = %0.2f Nw-m '%T_2\n", + "# Answer in the textbok are not accurate." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of the motor when 50 Amp current taking from supply = 1241.86 rpm \n", + "electro-magnetic torque generatedwhen motor running at 600 rpm = 342.36 Nw-m \n", + "electro-magnetic torque generated in second case = 85.59 Nw-m \n" + ] + } + ], + "prompt_number": 81 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.40 page 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:What resistance must be inserted in series with the armature to reduce the speed to 500 rpm\n", + "\n", + "V=250 #applied voltage to a shunt motor(in V)\n", + "I_a=20 #armature current(in Amp)\n", + "R_a=0.5 #armature resistance(in Ohm)\n", + "N_1=1000 #speed of the motor due to these readings(in rpm)\n", + "E_1=V-I_a*R_a #emf induced in machine(in V)\n", + "N_2=500 #desired speed of the motor(in rpm)\n", + "E_2=(N_2/N_1)*E_1 #emf in case of motor speed N_2\n", + "#R_1 additional resistance added to reduce the speed to 500 rpm\n", + "R_1=(V-E_2)/I_a-R_a #resistance applied in series with armature to reduce the speed to 500 rpm\n", + "print 'resistance applied in series with armature to reduce the speed to 500 rpm = %0.2f Ohm '%R_1 \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance applied in series with armature to reduce the speed to 500 rpm = 6.00 Ohm \n" + ] + } + ], + "prompt_number": 83 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.41 page 129" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine the torque in Kgm.,H.P. & efficiency of the motor.\n", + "\n", + "W_1=75 #load on one side of the brake(in Kg)\n", + "W_2=5 #load on other side of the brake(in Kg)\n", + "W=W_1-W_2 #effective force(in Kg)\n", + "R=1 #radius of the brake pulley (in m)\n", + "T=W*R #torque(in Kg-m)\n", + "N=1200 #speed of the small shunt motor(in rpm)\n", + "H=(2*3.14)*(N*T)/33000 #torque in H.P.\n", + "print 'torque = %0.2f H.P. '% H \n", + "H_P=735.5 #value of one H.P.\n", + "O_p=H*H_P #output power(in Watt)\n", + "I_c=80 #input current(in Amp)\n", + "V=250 #input voltage(in V)\n", + "I_p=I_c*V #input power(in Watt)\n", + "E=(O_p/I_p)*100 #efficiency of the motor\n", + "print 'efficiency of the motor = %0.2f %%'%E \n", + "# 2nd part Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "torque = 15.99 H.P. \n", + "efficiency of the motor = 58.79 %\n" + ] + } + ], + "prompt_number": 86 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.42 page 129\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine the torque in Kgm.,output in Watts & efficiency of the motor.\n", + "\n", + "W_1=40 #load on one side of the brake(in Kg)\n", + "W_2=5 #load on other side of the brake(in Kg)\n", + "W=W_1-W_2 #effective force(in Kg)\n", + "R=0.5 #radius of the brake pulley (in m)\n", + "T=W*R #torque(in Kg-m)\n", + "N=1500 #speed of the small shunt motor(in rpm)\n", + "O=(2*3.14)*(N*T)/60 #output (in Kg-m/sec)\n", + "O_p=O*9.81 #output (in watts)\n", + "print 'output = %0.2f watts'%ceil(O_p) \n", + "I_c=80 #input current(in Amp)\n", + "V=400 #input voltage(in V)\n", + "I_p=I_c*V #input power(in Watt)\n", + "E=(O_p/I_p)*100 #efficiency of the motor\n", + "print 'efficiency of the motor = %0.1f %%'%E " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output = 26953.00 watts\n", + "efficiency of the motor = 84.2 %\n" + ] + } + ], + "prompt_number": 88 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.43 page 129" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the efficiency of the motor\n", + "\n", + "W=10 #Spring reading (in Kg)\n", + "R=0.8 #brake arm length (in m)\n", + "T=W*R #torque (in kg-m)\n", + "N=1200 #speed of the small shunt motor(in rpm)\n", + "O=(2*3.14)*(N*T)/60 #output (in Kg-m/sec)\n", + "O_p=O*9.81 #output (in watts)\n", + "V=250 #input voltage(in V)\n", + "I=50 #input current(in Amp)\n", + "I_p=V*I #input power(in watts)\n", + "E=(O_p/I_p)*100 #efficiency of the motor\n", + "print 'efficiency of the motor = %0.2f %%'%E " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "efficiency of the motor = 78.86 %\n" + ] + } + ], + "prompt_number": 89 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.44 page 130" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Estimate the output and efficiency of a shunt motor when the input current is 20Amp and 100Amp\n", + "\n", + "V=500 #applied voltage(in V)\n", + "R_sh=500 #field resistance (in Ohm)\n", + "R_a=0.2 #armature resistance(in Ohm)\n", + "I_o=4 #no load current (in Amp)\n", + "I_sh=V/R_sh #field current(in Amp)\n", + "I_a=I_o-I_sh #armature current(in Amp)\n", + "L_cf=V*I_sh #copper losses in field(in watts)\n", + "L_ca=(I_a**2)*R_a #copper losses in armature(in watts)\n", + "L_tc=L_cf+L_ca #total copper losses\n", + "L_t=V*I_o #input power to motor on no load(in watts)\n", + "L_m=L_t-L_tc #iron & mechanical losses(in watts)\n", + "L_c=L_m+L_cf #constant losses(in watts)\n", + "#(a) when input current is 20 Amp\n", + "I_l1=20 #input current(in Amp)\n", + "I_a1=I_l1-I_sh #armature current when I_l1 input current(in Amp)\n", + "L_ca1=(I_a1**2)*R_a #copper losses in armature when I_l1 input current(in watts)\n", + "L_T1=L_c+L_ca1 #total losses at this load(in watts)\n", + "I_p=V*I_l1 #input power(in watts)\n", + "O_p1=I_p-L_T1 #output power when input current is 20 Amp(in watts)\n", + "print 'output power when input current is 20 Amp = %0.2f Watts '%O_p1 \n", + "E_1=(O_p1/I_p)*100 #efficiency of the shunt motor when input current is 20 Amp\n", + "print 'efficiency of the shunt motor when input current is 20 Amp = %0.3f %%'%E_1 \n", + "#(b) when input current is 100 Amp\n", + "I_l2=100 #input current(in Amp)\n", + "I_a2=I_l2-I_sh #armature current when I_l2 input current(in Amp)\n", + "L_ca2=(I_a2**2)*R_a #copper losses in armature when I_l2 input current(in watts)\n", + "L_T2=L_c+L_ca2 #total losses at this load(in watts)\n", + "I_p=V*I_l2 #input power(in watts)\n", + "O_p2=I_p-L_T2 #output power when input current is 100 Amp(in watts)\n", + "print 'output power when input current is 100 Amp = %0.2f Watts '%O_p2 \n", + "E_2=(O_p2/I_p)*100 #efficiency of the shunt motor when input current is 100 Amp\n", + "print 'efficiency of the shunt motor when input current is 100 Amp = %0.2f %%'%E_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output power when input current is 20 Amp = 7929.60 Watts \n", + "efficiency of the shunt motor when input current is 20 Amp = 79.296 %\n", + "output power when input current is 100 Amp = 46041.60 Watts \n", + "efficiency of the shunt motor when input current is 100 Amp = 92.08 %\n" + ] + } + ], + "prompt_number": 91 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.45 page 131" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the motor output, H.P. and efficiency when the total current taken from the mains is 35 Amp\n", + "\n", + "V=230 #applied voltage(in V)\n", + "R_sh=230 #field resistance (in Ohm)\n", + "R_a=0.3 #armature resistance(in Ohm)\n", + "I_o=2.5 #no load current (in Amp)\n", + "I_sh=V/R_sh #field current(in Amp)\n", + "I_a=I_o-I_sh #armature current(in Amp)\n", + "L_cf=V*I_sh #copper losses in field(in watts)\n", + "L_ca=(I_a**2)*R_a #copper losses in armature(in watts)\n", + "L_tc=L_cf+L_ca #total copper losses\n", + "L_t=V*I_o #input power to motor on no load(in watts)\n", + "L_m=L_t-L_tc #iron & mechanical losses(in watts)\n", + "L_c=L_m+L_cf #constant losses(in watts)\n", + "# when input current is 35 Amp(On load condition)\n", + "I_l=35 #input current(in Amp)\n", + "I_a1=I_l-I_sh #armature current when I_l input current(in Amp)\n", + "L_ca1=(I_a1**2)*R_a #copper losses in armature when I_l input current(in watts)\n", + "L_T1=L_c+L_ca1 #total losses at this load(in watts)\n", + "I_p=V*I_l #input power(in watts)\n", + "O_p=I_p-L_T1 #output power(in watts)\n", + "print 'output power of motor = %0.3f Watts '%O_p \n", + "Hp=O_p/746 #the shunt motor H.P\n", + "print 'the shunt motor H.P.= %0.2f'%Hp \n", + "E_1=(O_p/I_p)*100 #efficiency of the shunt motor when input current is 35 Amp\n", + "print 'efficiency of the shunt motor when input current is 35 Amp = %0.1f %%'%E_1 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output power of motor = 7128.875 Watts \n", + "the shunt motor H.P.= 9.56\n", + "efficiency of the shunt motor when input current is 35 Amp = 88.6 %\n" + ] + } + ], + "prompt_number": 95 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.46 page 132" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the full load motor output and its efficiency\n", + "\n", + "V=500 #applied voltage(in V)\n", + "R_sh=250 #field resistance (in Ohm)\n", + "R_a=0.2 #resistance of armature including brushes(in Ohm)\n", + "I_o=5 #no load current (in Amp)\n", + "I_sh=V/R_sh #shunt field current(in Amp)\n", + "I_a=I_o-I_sh #armature current(in Amp)\n", + "L_a=(I_a**2)*R_a #armature brush drop(in watts)\n", + "L_t=V*I_o #input to motor on no load(in watts)\n", + "L_c=L_t-L_a #constant losses(in watts)\n", + "#On full load condition\n", + "I_l=52 #input current(in Amp)\n", + "I_a1=I_l-I_sh #armature current when I_l1 input current(in Amp)\n", + "L_a1=(I_a1**2)*R_a #losses in armature when I_l1 input current(in watts)\n", + "L_T1=L_c+L_a1 #total losses at this load(in watts)\n", + "I_p=V*I_l #input power(in watts)\n", + "O_p=I_p-L_T1 #output power(in watts)\n", + "print 'output of the motor = %0.2f Watts '%O_p \n", + "Hp=O_p/735.5 #the shunt motor H.P\n", + "print 'the shunt motor H.P.= %0.2f' %Hp\n", + "E_1=(O_p/I_p)*100 #efficiency of the full load shunt motor when input current is 52 Amp\n", + "print 'efficiency of the full load shunt motor when input current is 52 Amp = %0.1f %%'%E_1 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "output of the motor = 23001.80 Watts \n", + "the shunt motor H.P.= 31.27\n", + "efficiency of the full load shunt motor when input current is 52 Amp = 88.5 %\n" + ] + } + ], + "prompt_number": 98 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.47 page 132" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine the input in Watts & efficiency of the generator\n", + "\n", + "O_p=50 #output of a machine (in KW)\n", + "O_p1=50*(10)**3 #output (in watts)\n", + "L_i=4000 #internal losses(in watts)\n", + "I_p=O_p1+L_i #input(in watts)\n", + "print 'input = %0.f Watts '%I_p \n", + "E=(O_p1/I_p)*100 #efficiency of the generator\n", + "print 'efficiency of the generator = %0.2f %%'% E " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "input = 54000 Watts \n", + "efficiency of the generator = 92.59 %\n" + ] + } + ], + "prompt_number": 100 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.48 page 133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine the internal losses,torque & efficiency of the motor\n", + "\n", + "V=240 #supply voltage(in V)\n", + "I=80 #motor taking current(in Amp)\n", + "Hp=20 #motor giving H.P.\n", + "I_p=V*I #input (in watts)\n", + "O_p=Hp*735.5 #output (in watts)\n", + "L_i=I_p-O_p #internal losses(in watts)\n", + "print 'internal losses in motor = %0.2f Watts '%L_i \n", + "N=1300 #motor speed (in rpm)\n", + "T=O_p*60/(2*3.14*N) #torque generated in motor(in Nw-m)\n", + "print 'torque generated in motor = %0.f Nw-m '% T \n", + "E=(O_p/I_p)*100 #efficiency of the motor\n", + "print 'efficiency of the motor = %0.1f %%' %E" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "internal losses in motor = 4490.00 Watts \n", + "torque generated in motor = 108 Nw-m \n", + "efficiency of the motor = 76.6 %\n" + ] + } + ], + "prompt_number": 103 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.49 page 133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Find the copper losses,iron& friction losses and commerical efficiency.\n", + "\n", + "V=250 #supply voltage(in V)\n", + "R_sh=100 #shunt field resistance(in Ohm)\n", + "R_a=0.1 #armature resistance(in Ohm)\n", + "I_sh=V/R_sh #field current(in Amp)\n", + "I_l=197.5 #motor taking current(in Amp)\n", + "I_a=I_l+I_sh #Armature current(in Amp)\n", + "E=V+I_a*R_a #E.M.F generated(in V)\n", + "Hp=80 #motor giving H.P.\n", + "P_m=Hp*735.5 #mechanical power input (in watts)\n", + "P_a=E*I_a #electrical power developed in armature(in watts)\n", + "L_i=P_m-P_a #iron & friction losses(in watts)\n", + "print 'iron & friction losses in motor = %0.2f Watts '%L_i \n", + "O_p=V*I_l #electrical power output\n", + "L_c=P_a-O_p #copper losses(in watts)\n", + "print 'copper losses in a shunt generator = %0.2f Watts '%L_c \n", + "L_t=L_i+L_c #Total losses(in Watts)\n", + "I_p=O_p+L_t #Input(in Watts)\n", + "E_f=(O_p/I_p)*100 #Commerical efficiency\n", + "print 'commerical efficiency = %0.2f %%'%E_f \n", + "# 3rd part Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "iron & friction losses in motor = 4840.00 Watts \n", + "copper losses in a shunt generator = 4625.00 Watts \n", + "commerical efficiency = 83.91 %\n" + ] + } + ], + "prompt_number": 105 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.50 page 134" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine Copper losses and total losses and Output and BHP of the motor and efficiency of the motor\n", + "\n", + "V=230 #supply voltage(in V)\n", + "I_l=200 #line current(in Amp)\n", + "I_p=V*I_l #input power (in watts)\n", + "R_sh=50 #shunt field resistance(in Ohm)\n", + "R_a=0.04 #armature resistance (in ohm)\n", + "I_sh=V/R_sh #shunt field currrent(in Amp)\n", + "I_a=I_l-I_sh #armature current(in Amp)\n", + "L_cf=V*I_sh #copper losses in the field(in watts)\n", + "L_ca=(I_a**2)*R_a #copper loses in armature(in watts)\n", + "L_ct=L_cf+L_ca #Total copper losses(in watts)\n", + "print 'copper losses of the motor = %0.2f Watts '%L_ct \n", + "L_s=1500 #Stray losses (in watts)\n", + "L_t=L_ct+L_s #total losses(in watts)\n", + "print 'total losses of the motor = %0.2f Watts '%L_t \n", + "O_p=I_p-L_t #output of the motor(in watts)\n", + "print 'Output of the motor = %0.2f Watts'%O_p \n", + "Bhp=O_p/735.5 #B.H.P. of the motor (in H.P.)\n", + "print 'B.H.P. of the motor = %0.3f H.P. '%Bhp \n", + "E=(O_p/I_p)*100 #efficiency of the motor(in %)\n", + "print 'efficiency of the motor = %0.2f %%'%E " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "copper losses of the motor = 2585.25 Watts \n", + "total losses of the motor = 4085.25 Watts \n", + "Output of the motor = 41914.75 Watts\n", + "B.H.P. of the motor = 56.988 H.P. \n", + "efficiency of the motor = 91.12 %\n" + ] + } + ], + "prompt_number": 107 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:2.51 page 135" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the speed and BHP of the motor\n", + "\n", + "V=250 #applied emf(in V)\n", + "R_sh=0.05 #field resistance (in Ohm)\n", + "R_a=0.1 #armature resistance(in Ohm)\n", + "I=80 #motor current(in Amp)\n", + "A_s=240 #armature slots \n", + "C_s=4 #number of conductor per slot\n", + "Z=A_s*C_s #total number of conductor\n", + "E_b=V-I*(R_a+R_sh) #Back emf(in V)\n", + "A=2 #number of parallel paths for wave wound\n", + "P=6 #poles\n", + "F=1.75 #flux per pole (in megalines)\n", + "F_1=1.75*10**-2 #flux per pole (in Wb)\n", + "N=E_b*60*A/(F_1*Z*P) #speed of the motor (in rpm)\n", + "print 'speed of the motor = %0.f rpm '%N \n", + "I_p=V*I #input to the motor(in watts)\n", + "L_c=(I**2)*(R_a+R_sh) #copper losses(in watts)\n", + "L_i=900 #iron and friction losses(in watts)\n", + "L_t=L_c+L_i #total losses(in watts)\n", + "O_p=I_p-L_t #output(in watts)\n", + "BHP=O_p/746 #B.H.P. of the motor\n", + "print 'B.H.P. of the motor = %0.1f H.P'% BHP " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "speed of the motor = 283 rpm \n", + "B.H.P. of the motor = 24.3 H.P\n" + ] + } + ], + "prompt_number": 111 + } + ], + "metadata": {} + } + ] +} diff --git a/Electrical_Machines_-_1_by_Tarlok_Singh/Chap3.ipynb b/Electrical_Machines_-_1_by_Tarlok_Singh/Chap3.ipynb new file mode 100755 index 00000000..cd64dbc4 --- /dev/null +++ b/Electrical_Machines_-_1_by_Tarlok_Singh/Chap3.ipynb @@ -0,0 +1,1687 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter3 - Transformer(Single Phase)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.1 page 167" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import ceil\n", + "#Caption:Calculate the number of turns on both primary and secondary winding in a single phase transformer\n", + "\n", + "E_1=500 #primary induced emf(in Volts)\n", + "E_2=250 #secondary induced emf(in Volts)\n", + "F=50 #supply frequency(in Hz)\n", + "B_max=1.2 #maximum flux density(in T)\n", + "a=0.009 #cross sectional area(in square meter)\n", + "F_x=B_max*a #maximum value of flux(in Wb)\n", + "N_1=ceil(E_1/(4.44*F*F_x)) #number of primary turns\n", + "print 'number of primary turns in transformer =',N_1 \n", + "N_2=ceil(E_2/(4.44*F*F_x)) #number of secondary turns\n", + "print 'number of secondary turns in transformer =',N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "number of primary turns in transformer = 209.0\n", + "number of secondary turns in transformer = 105.0\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.2 page 167" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a loss less transformer calculate \n", + "#(a)Number of turns on high voltage side (b)The primary current (c)The secondary current\n", + "\n", + "V_1=110 #primary voltage(in Volts)\n", + "V_2=220 #secondary voltage(in Volts)\n", + "N_1=130 #low voltage side turns\n", + "N_2=N_1*V_2/V_1 #high voltage side turns\n", + "print 'Number of turns on high voltage side =',N_2 \n", + "O_p=1 #Output of the transformer in KVA\n", + "I_2=O_p*1000/V_2 #Secondary current(in Amp)\n", + "I_1=O_p*1000/V_1 #Primary current(in Amp)\n", + "print 'Primary current = %0.2f Amp'%I_1\n", + "print 'Secondary current = %0.2f Amp'%I_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Number of turns on high voltage side = 260.0\n", + "Primary current = 9.09 Amp\n", + "Secondary current = 4.55 Amp\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.3 Page 168" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the secondary voltage and the volts per turn\n", + "\n", + "N_1=800 #Primary turns in a transformer\n", + "N_2=200 #Secondary turns in a transformer\n", + "V_1=100 #Primary voltage\n", + "V_2=V_1*(N_2/N_1) #Secondary voltage\n", + "print 'Secondary voltage = %0.2f Volts '%V_2 \n", + "V_n=V_1/N_1 #Volts per turn\n", + "print 'Volts per turn =',V_n " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Secondary voltage = 25.00 Volts \n", + "Volts per turn = 0.125\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.4 Page 169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the primary current in a single phase transformer\n", + "\n", + "V_1=230 #Primary voltage(in Volts)\n", + "V_2=460 #Secondary voltage(in Volts)\n", + "I_2=10 #secondary current(in Amp)\n", + "I_1=I_2*(V_2/V_1) #Primary current(in Amp)\n", + "print 'Primary current = %0.2f Amp'%I_1" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Primary current = 20.00 Amp\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.5 Page 169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate (a)the approximate values of the primary and secondary currents (b)the approimate number of primary turns\n", + "\n", + "V_1=6600 #Primary voltage(in Volts)\n", + "V_2=400 #Secondary voltage(in Volts)\n", + "N_2=100 #secondary turns\n", + "O_p=200 #output of a transformer in KVA\n", + "I_1=O_p*1000/V_1 #Full load primary current(in Amp)\n", + "print 'Full load primary current = %0.1f Amp'%I_1 \n", + "I_2=O_p*1000/V_2 #Full load secondary current(in Amp)\n", + "print 'Full load secondary current = %0.f Amp '%I_2 \n", + "N_1=N_2*(V_1/V_2) #Number of primary turns\n", + "print 'Number of primary turns =',N_1 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Full load primary current = 30.3 Amp\n", + "Full load secondary current = 500 Amp \n", + "Number of primary turns = 1650.0\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.6 page 169" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Find voltage in secondary terminal\n", + "V_1=100 #Primary voltage(in Volts)\n", + "N_1=20 #Number of primary turns\n", + "N_2=40 #Number of secondary turns\n", + "V_2=(N_2/N_1)*V_1 #voltage of the secondary terminal(in Volts)\n", + "print 'voltage of the secondary terminal = %0.2f V'%V_2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage of the secondary terminal = 200.00 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.7 page 170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the r.m.s value of the induced emf in the secondary coil.\n", + "\n", + "F_x=0.02 #Maximum value of flux(in Wb)\n", + "N_2=55 #Number of secondary turns\n", + "F=50 #Supply frequency(in Hz)\n", + "E_2=4.44*F*F_x*N_2 #r.m.s value of induced emf in the secondary (in Volts)\n", + "print 'r.m.s value of induced emf in the secondary = %0.1f Volts '%E_2 \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "r.m.s value of induced emf in the secondary = 244.2 Volts \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.8 page 170" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In single phase transformer Calculate (a)The maximum flux density in the core and (b)induced emf in the secondary\n", + "\n", + "N_1=80 #Primary turns\n", + "N_2=240 #secondary turns\n", + "f=50 #Supply frequency(in Hz)\n", + "E_1=240 #Supply voltage(in Volts)\n", + "F_max=E_1/(4.44*f*N_1) #Maximum value of the flux in the core\n", + "a=200 #Cross sectional area of core(in cm**2)\n", + "A=a*10**-4 #Cross sectional area of core(in m**2)\n", + "B_max=F_max/A #Peak value of flux density in the core(in T)\n", + "print 'Peak value of flux density in the core = %0.4f T '%B_max \n", + "E_2=E_1*(N_2/N_1) #Induced emf in the secondary winding(in Volts)\n", + "print 'Induced emf in the secondary winding = %0.f Volts '%E_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak value of flux density in the core = 0.6757 T \n", + "Induced emf in the secondary winding = 720 Volts \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.9 page 171" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate (a)Primary and secondary currents on full load (b)the maximum value of flux (c)the number of primary turns.\n", + "\n", + "O_p=200 #Rated output (in KVA)\n", + "V_1=3300 #Primary voltage (in Volts)\n", + "V_2=240 #Secondary voltage (in Volts)\n", + "N_2=100 #Secondary turns\n", + "f=50 #supply frequency(in Hz)\n", + "I_1=O_p*1000/V_1 #Primary current(in Amp)\n", + "print 'Primary current on full load = %0.2f Amp'%I_1 \n", + "I_2=O_p*1000/V_2 #secondary current(in Amp)\n", + "print 'secondary current on full load = %0.2f Amp '%I_2 \n", + "F_x=V_2/(4.44*f*N_2) #Maximum value of flux(in Wb)\n", + "print 'Maximum value of flux = %0.4f Wb '%F_x \n", + "N_1=N_2*(V_1/V_2) #Primary turns\n", + "print 'Primary turns =',N_1 \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Primary current on full load = 60.61 Amp\n", + "secondary current on full load = 833.33 Amp \n", + "Maximum value of flux = 0.0108 Wb \n", + "Primary turns = 1375.0\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.10 page 175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer primary side is open , Find (a)core loss, (b)loss component of current ,(c)Magnetising current.\n", + "\n", + "\n", + "O_p=50 #output (in KVA)\n", + "V_2=230 #Secondary voltage(in Volts)\n", + "P=187 #meter's power reading (in watt)\n", + "I_w=P/V_2 #Loss component of current(in Amp)\n", + "I_o=6.5 #meter's current reading (in Amp)\n", + "I_m=((I_o)**2-(I_w)**2) #Magnetising current(in Amp)\n", + "P=V_2*I_w #Core loss(in watt)\n", + "print 'Core loss = %0.2f Watt '%P \n", + "print 'Loss component of current = %0.3f Amp '%I_w \n", + "print 'Magnetising current = %0.2f Amp'%I_m " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Core loss = 187.00 Watt \n", + "Loss component of current = 0.813 Amp \n", + "Magnetising current = 41.59 Amp\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.11 page 175" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi, degrees, acos, sin, cos\n", + "#Caption:In a transformer Calculate \n", + "#(a)Magnetising component of no load current (b)the iron loss (c)the maximum value of flux in the core.\n", + "\n", + "V_1=460 #supply voltage(in Volts)\n", + "I_o=15 #No load current (in Amp)\n", + "p=degrees(acos(0.2)) #power angle(when power factor is 0.2)\n", + "n=sin(p*pi/180) \n", + "I_m=I_o*n #Magnetising component of no load current \n", + "print 'Magnetising component of no load current = %0.2f Amp '%I_m \n", + "L_ir=V_1*I_o*cos(p*pi/180)/1000 #the iron loss in transformer(in kWatt)\n", + "print 'the iron loss in transformer = %0.2f kW '%L_ir \n", + "E_1=V_1 #at no load condition\n", + "N_1=550 #primary winding\n", + "f=50 #supply frequency (in Hz)\n", + "F_m=E_1/(4.44*N_1*f) #the maximum value of flux in the core(in Wb)\n", + "print 'the maximum value of flux in the core = %0.4f Wb'%F_m " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetising component of no load current = 14.70 Amp \n", + "the iron loss in transformer = 1.38 kW \n", + "the maximum value of flux in the core = 0.0038 Wb\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.12 page 176" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer, Find (a)The magnetising current (b)The iron loss current \n", + "\n", + "p=degrees(acos(0.22)) #power angle(when power factor is 0.22)\n", + "I_o=0.3 #no load current(in Amp)\n", + "I_m=I_o*sin(p*pi/180) #Magnetising current(in Amp)\n", + "print 'Magnetising current = %0.4f Amp'%I_m \n", + "I_w=I_o*cos(p*pi/180) #Iron loss current (in Amp)\n", + "print 'Iron loss current = %0.3f Amp '%I_w " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Magnetising current = 0.2926 Amp\n", + "Iron loss current = 0.066 Amp \n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.13 page 177" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Find the active and reactive components of a single phase transformer.\n", + "\n", + "V_1=440 #supply Voltage(in Volts)\n", + "p=degrees(acos(0.3) )#power angle when power factor is 0.3\n", + "P_o=80 #power input to the hv winding(in Watt)\n", + "I_o=P_o/(V_1*cos(p*pi/180)) #No load current (in Amp)\n", + "I_w=I_o*cos(p*pi/180) #Active component of no load current (in Amp)\n", + "print 'Active component of no load current = %0.3f Amp '%I_w \n", + "I_m=I_o*sin(p*pi/180) #Reactive component of no load current (in Amp)\n", + "print 'Reactive component of no load current = %0.3f Amp '%I_m " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Active component of no load current = 0.182 Amp \n", + "Reactive component of no load current = 0.578 Amp \n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.14 page 181" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the current taken by the primary winding.\n", + "\n", + "V_1=6600 #Primary voltage(in Volts)\n", + "V_2=400 #Secondary voltage(in Volts)\n", + "I_o=0.7 #NO load current(in Amp)\n", + "p_o=0.24 #No load power factor\n", + "q_o=degrees(acos(p_o)) #power angle when no load power factor is 0.24\n", + "I_2=100 #Secondary current(in Amp)\n", + "p_2=0.8 #Secondary power factor\n", + "q_2=degrees(acos(p_2)) #power angle when secondary power factor is 0.8\n", + "K=V_2/V_1 #ratio of primary to secondary voltage\n", + "I_1=K*I_2 #primary current(in Amp)\n", + "Q=q_o-q_2 #resultant power angle\n", + "I=((I_o)**2 + (I_1)**2+2*I_o*I_1*cos(Q*pi/180))**(1/2) #Resultant current taken by the primary(in Amp)\n", + "print 'Resultant current taken by the primary = %0.2f Amp '%I " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resultant current taken by the primary = 6.62 Amp \n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.15 page 182" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the current taken by the primary winding.\n", + "\n", + "V_1=440 #Primary voltage(in Volts)\n", + "V_2=110 #Secondary voltage(in Volts)\n", + "I_o=5 #NO load current(in Amp)\n", + "p_o=0.2 #No load power factor\n", + "q_o=degrees(acos(p_o)) #power angle when no load power factor is 0.24\n", + "I_2=120 #Secondary current(in Amp)\n", + "p_2=0.8 #Secondary power factor\n", + "q_2=degrees(acos(p_2)) #power angle when secondary power factor is 0.8\n", + "K=V_2/V_1 #ratio of primary to secondary voltage\n", + "I_1=K*I_2 #primary current(in Amp)\n", + "Q=q_o-q_2 #resultant power angle\n", + "I=((I_o)**2 + (I_1)**2+2*I_o*I_1*cos(Q*pi/180))**(1/2) #Resultant current taken by the primary(in Amp)\n", + "print 'Resultant current taken by the primary = %0.2f Amp '%I " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resultant current taken by the primary = 33.90 Amp \n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.16 page 187" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate Equivalent resistance \n", + "\n", + "V_1=4400 #Primary voltage (in Volts)\n", + "V_2=220 #Secondary voltage (in Volts)\n", + "R_1=3.45 #Primary resistance (in Ohm)\n", + "X_1=5.2 #Primary reactances (in Ohm)\n", + "R_2=0.009 #secondary resistance (in Ohm)\n", + "X_2=0.015 #secondary reactance (in Ohm)\n", + "K=V_2/V_1 #voltage ratio\n", + "R_o1=R_1+(R_2/K**2) #Equivalent resistance as referred to primary(in Ohm)\n", + "print 'Equivalent resistance as referred to primary = %0.2f Ohm '%R_o1 \n", + "R_o2=R_2+(R_1*K**2) #Equivalent resistance as referred to secondary(in Ohm)\n", + "print 'Equivalent resistance as referred to secondary = %0.4f Ohm '%R_o2 \n", + "X_o1=X_1+(X_2/K**2) #Equivalent reactance as referred to primary(in Ohm)\n", + "print 'Equivalent reactance as referred to primary = %0.2f Ohm'%X_o1 \n", + "X_o2=X_2+X_1*(K**2) #Equivalent reactance as referred to secondary(in Ohm)\n", + "print 'Equivalent reactance as referred to secondary = %0.3f Ohm '%X_o2 \n", + "Z_o1=((R_o1)**2 + (X_o1)**2)**(1/2) #Equivalent impedence as referred to primary(in Ohm)\n", + "print 'Equivalent impedence as referred to primary = %0.2f Ohm'%Z_o1 \n", + "Z_o2=((R_o2)**2 + (X_o2)**2)**(1/2) #Equivalent impedence as referred to secondary(in Ohm)\n", + "print 'Equivalent impedence as referred to secondary = %0.3f Ohm'%Z_o2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent resistance as referred to primary = 7.05 Ohm \n", + "Equivalent resistance as referred to secondary = 0.0176 Ohm \n", + "Equivalent reactance as referred to primary = 11.20 Ohm\n", + "Equivalent reactance as referred to secondary = 0.028 Ohm \n", + "Equivalent impedence as referred to primary = 13.23 Ohm\n", + "Equivalent impedence as referred to secondary = 0.033 Ohm\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.17 page 190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the secondary terminal voltage at full load.\n", + "\n", + "V_1=2000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=400 #Secondary voltage at no load (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=5 #Primary resistance(in Ohm)\n", + "R_2=0.2 #Secondary resistance(in Ohm)\n", + "X_1=12 #Primary reactance(in Ohm)\n", + "X_2=0.48 #Secondary reactance(in Ohm)\n", + "R_o2=R_2 + (K**2)*R_1 #Equivalent resistance as referred to secondary(in Ohm)\n", + "X_o2=X_2 + (K**2)*X_1 #Equivalent reactance as referred to secondary(in Ohm)\n", + "O_p=10 #Rated output(i KVA)\n", + "I_2=O_p*1000/V_2 #Full load secondary current(in Amp)\n", + "p=0.8 #power factor\n", + "a=degrees(acos(p)) #power angle\n", + "V_d=I_2*R_o2*cos(a*pi/180) + I_2*X_o2*sin(a*pi/180) #Voltage drop (in Volts)\n", + "V=V_2-V_d #Secondary terminal voltage at full load (in Volts)\n", + "print 'Secondary terminal voltage at full load = %0.2f Volts '%V " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Secondary terminal voltage at full load = 377.60 Volts \n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.18 page 191" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the secondary terminal voltage at full load.\n", + "\n", + "V_1=11000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=400 #Secondary voltage at no load (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=4 #Primary resistance(in Ohm)\n", + "R_2=0.2 #Secondary resistance(in Ohm)\n", + "X_1=10 #Primary reactance(in Ohm)\n", + "X_2=0.4 #Secondary reactance(in Ohm)\n", + "R_o2=R_2 + (K**2)*R_1 #Equivalent resistance as referred to secondary(in Ohm)\n", + "X_o2=X_2 + (K**2)*X_1 #Equivalent reactance as referred to secondary(in Ohm)\n", + "O_p=10 #Rated output(in KVA)\n", + "I_2=O_p*1000/V_2 #Full load secondary current(in Amp)\n", + "Z_o2=((R_o2)**2 +(X_o2)**2)**(1/2) #Equivalent impedance as referred to secondary(in Ohm)\n", + "V=V_2-I_2*Z_o2 #Secondary terminal voltage at full load (in Volts)\n", + "print 'Secondary terminal voltage at full load = %0.3f Volts '%V " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Secondary terminal voltage at full load = 388.465 Volts \n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.19 page 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Determine equivalent resistance \n", + "\n", + "V_1=2000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=400 #Secondary voltage at no load (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=5.2 #Primary resistance(in Ohm)\n", + "R_2=0.2 #Secondary resistance(in Ohm)\n", + "X_1=12.5 #Primary reactance(in Ohm)\n", + "X_2=0.5 #Secondary reactance(in Ohm)\n", + "R_o2=R_2 + (K**2)*R_1 #Equivalent resistance as referred to secondary(in Ohm)\n", + "print 'Equivalent resistance as referred to secondary = %0.2f Ohm '%R_o2 \n", + "X_o2=X_2 + (K**2)*X_1 #Equivalent reactance as referred to secondary(in Ohm)\n", + "print 'Equivalent reactance as referred to secondary = %0.2f Ohm '%X_o2 \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent resistance as referred to secondary = 0.41 Ohm \n", + "Equivalent reactance as referred to secondary = 1.00 Ohm \n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.20 page 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine equivalent resistance and reactance \n", + "\n", + "V_1=2000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=200 #Secondary voltage at no load (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=2 #Primary resistance(in Ohm)\n", + "R_2=0.025 #Secondary resistance(in Ohm)\n", + "X_1=4 #Primary reactance(in Ohm)\n", + "X_2=0.04 #Secondary reactance(in Ohm)\n", + "R_eq12=(K**2)*R_1 #equivalent resistance of primary referred to secondary(in Ohm)\n", + "print 'equivalent resistance of primary referred to secondary = %0.2f Ohm '%R_eq12 \n", + "X_eq12=(K**2)*X_1 #equivalent reactance of primary referred to secondary(in Ohm)\n", + "print 'equivalent reactance of primary referred to secondary = %0.2f Ohm '%X_eq12 \n", + "R_e2=(K**2)*R_1 + R_2 #total resistance of primary referred to secondary(in Ohm)\n", + "print 'total resistance of primary referred to secondary = %0.3f Ohm '%R_e2 \n", + "X_e2=(K**2)*X_1 + X_2 #total reactance of primary referred to secondary(in Ohm)\n", + "print 'total reactance of primary referred to secondary = %0.3f Ohm '%X_e2 \n", + "R_eq21=R_2/(K**2) #equivalent resistance of secondary referred to primar(in Ohm)\n", + "print 'equivalent resistance of secondary referred to primary = %0.2f Ohm '%R_eq21 \n", + "X_eq21=X_2/(K**2) #equivalent reactance of secondary referred to primar(in Ohm)\n", + "print 'equivalent reactance of secondary referred to primary = %0.2f Ohm '%X_eq21 \n", + "R_e1=R_1 + R_2/(K**2) #total resistance of secondary referred to primary(in Ohm)\n", + "print 'total resistance of secondary referred to primary = %0.2f Ohm '%R_e1 \n", + "X_e1=X_1 + X_2/(K**2) #total reactance of secondary referred to primary(in Ohm)\n", + "print 'total reactance of secondary referred to primary = %0.2f Ohm '%X_e1 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "equivalent resistance of primary referred to secondary = 0.02 Ohm \n", + "equivalent reactance of primary referred to secondary = 0.04 Ohm \n", + "total resistance of primary referred to secondary = 0.045 Ohm \n", + "total reactance of primary referred to secondary = 0.080 Ohm \n", + "equivalent resistance of secondary referred to primary = 2.50 Ohm \n", + "equivalent reactance of secondary referred to primary = 4.00 Ohm \n", + "total resistance of secondary referred to primary = 4.50 Ohm \n", + "total reactance of secondary referred to primary = 8.00 Ohm \n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.21 Page 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Determine Primary resistance and total resistance of transformer referred to primary.\n", + "\n", + "V_1=2000 #Primary voltage at no load or full load(in Volts)\n", + "V_2=220 #Secondary voltage at no load (in Volts)\n", + "R_1=1.06 #Primary resistance(in Ohm)\n", + "R_2=0.013 #Secondary resistance(in Ohm)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_eq1=(K**2)*R_1 #Primary resistance referred to secondary(in Ohm)\n", + "print 'Primary resistance referred to secondary = %0.4f Ohm '%R_eq1 \n", + "R_eq2=R_2/(K**2) #secondary resistance referred to primary(in Ohm)\n", + "print 'secondary resistance referred to primary = %0.4f Ohm '%R_eq2 \n", + "R_e1=R_1 + R_eq2 #total resistance of transformer referred to primary(in Ohm)\n", + "print 'total resistance of transformer referred to primary = %0.4f Ohm '%R_e1 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Primary resistance referred to secondary = 0.0128 Ohm \n", + "secondary resistance referred to primary = 1.0744 Ohm \n", + "total resistance of transformer referred to primary = 2.1344 Ohm \n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.22 page 200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Find Full load regulation at a power factor (a) 0.8 lagging (b) unity (c) 0.8 leading.\n", + "\n", + "V_1=6600 #Primary voltage (in Volts)\n", + "V_2=250 #Secondary voltage (in Volts)\n", + "K=V_2/V_1 #Ratio of transformation\n", + "R_1=10 #Primary resistance (in Ohm)\n", + "R_2=0.02 #Secondary resistance (in Ohm)\n", + "X_o1=35 #Total leakage reactance referred to the primary winding(in Ohm)\n", + "R_o2=R_2 + R_1*(K**2) #Equivalent resistance reffered to the secondary(in Ohm)\n", + "X_o2=X_o1*(K**2) #Equivalent reactance reffered to the secondary(in Ohm)\n", + "O_p=40 #Rated output (in KVA)\n", + "I_2=O_p*1000/250 #Secondary current(in Amp)\n", + "P_1=0.8 #Power factor\n", + "q=(acos(P_1)) \n", + "V_r1=((I_2*R_o2*cos(q) + I_2*X_o2*sin(q))/V_2)*100 #Voltage regulation at 0.8 lagging power factor(in %)\n", + "print 'Voltage regulation at 0.8 lagging power factor = %0.3f %%'%V_r1 \n", + "V_r0=(I_2*R_o2/V_2)*100 #Voltage regulation at unity power factor(in %)\n", + "print 'Voltage regulation at unity power factor = %0.2f %%'%V_r0 \n", + "V_r2=((I_2*R_o2*cos(q) - I_2*X_o2*sin(q))/V_2)*100 #Voltage regulation at 0.8 leading power factor(in %)\n", + "print 'Voltage regulation at 0.8 leading power factor = %0.2f %%'%V_r2 \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage regulation at 0.8 lagging power factor = 3.687 %\n", + "Voltage regulation at unity power factor = 2.20 %\n", + "Voltage regulation at 0.8 leading power factor = -0.17 %\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.23 page 201" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the regulation of transformer\n", + "\n", + "#(I_2*R_a2/E_2)*100=1\n", + "Rs_d=1 #Percentage resistive drop\n", + "#(I_2*X_a2/E_2)*100=5\n", + "Re_d=5 #Percentage reactive drop\n", + "#power factor=cosd(q1)=0.8 (lagging)\n", + "q1=degrees(acos(0.8) )\n", + "#Voltage regulation= ((I_2*R_a2*cosd(q1)+I_2*X_a2*sind(q1))/100)*100\n", + "#V_r=(I_2*R_a2/E_2)*100*cosd(q1)+(I_2*X_a2/E_2)*100*sind(q1)\n", + "V_r1=Rs_d*cos(q1*pi/180)+Re_d*sin(q1*pi/180) #Voltage regulation when power factor is 0.8 lagging\n", + "print 'Voltage regulation when power factor is 0.8 lagging = %0.2f %%'%V_r1 \n", + "q2=-degrees(acos(0.8) )\n", + "#V_r2=(I_2*R_a2/E_2)*100*cosd(q2)+(I_2*X_a2/E_2)*100*sind(q2) #Voltage regulation when power factor is 0.8 leading\n", + "V_r2=Rs_d*cos(q2*pi/180)+Re_d*sin(q2*pi/180) #Voltage regulation when power factor is 0.8 leading\n", + "print 'Voltage regulation when power factor is 0.8 leading = %0.2f %%'%V_r2," + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage regulation when power factor is 0.8 lagging = 3.80 %\n", + "Voltage regulation when power factor is 0.8 leading = -2.20 %\n" + ] + } + ], + "prompt_number": 57 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.25 page 208" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the efficiency at unity power at Full load and Half load.\n", + "\n", + "P_c=400 #Full load copper loss(in Watts)\n", + "P_C=P_c/1000 #Full load copper loss(in KW)\n", + "P_i=350 #Full load iron loss(in Watts)\n", + "P_I=P_i/1000 #Full load iron loss(in KW)\n", + "P_f=1 #Power factor\n", + "KVA=25 #Rating of the transformer\n", + "O_p=KVA*P_f #Output at full load condition(in KW)\n", + "L_1=P_C+P_I #Losses at full load condition(in KW)\n", + "I_p=O_p+L_1 #Input at full load condition(in KW)\n", + "E_fL=(O_p/I_p)*100 #Efficiency in full load condition\n", + "print 'Efficiency of the tranformer at full load condition = %0.2f %%'%E_fL \n", + "#At half load condition\n", + "O_P=(1/2)*KVA*P_f #Output of the transformer at half load condition\n", + "L_2=((1/2)**2)*P_C+P_I #Losses at half load condition(in KW)\n", + "I_P=O_P+L_2 #Input at half load condition\n", + "E_hL=(O_P/I_P)*100 #Efficiency of the transformer at half load condition(in %)\n", + "print 'Efficiency of the transformer at half load condition = %0.2f %%'%E_hL " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the tranformer at full load condition = 97.09 %\n", + "Efficiency of the transformer at half load condition = 96.53 %\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.26 page 209" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the single phase transformer efficiency for 75% of the full load output at power factor unity and 0.8 lagging.\n", + "\n", + "P_i=1.5 #Core loss(in KW)\n", + "P_c=4.5 #Full load copper loss(in KW)\n", + "P_C=((3/4)**2)*P_c #Copper loss at 75% of full load(in KW)\n", + "P_t=P_i+P_C #Total loss at 75% of full load output(in KW)\n", + "KVA=300 #Rating of the transformer(in KVA)\n", + "P_f1=1 #power factor value when it is unity\n", + "P_f2=0.8 #power factor value when it is 0.8 lagging\n", + "O_p1=0.75*KVA*P_f1 #Output at 75% of full load and at unity power factor(in KW)\n", + "E_f1=(O_p1/(O_p1+P_t))*100 #Efficiency of the transformer for 75% of full load output at power factor unity(in %)\n", + "print 'Efficiency of the transformer for 75%% of full load output at power factor unity = %0.2f %%'%E_f1 \n", + "O_p2=0.75*KVA*P_f2 #Output at 75% of full load and at 0.8 lagging power factor(in KW)\n", + "E_f2=(O_p2/(O_p2+P_t))*100 #Efficiency of the transformer for 75% of full load output at power factor 0.8 lagging(in %)\n", + "print 'Efficiency of the transformer for 75%% of full load output at power factor 0.8 lagging = %0.2f %%' %E_f2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer for 75% of full load output at power factor unity = 98.24 %\n", + "Efficiency of the transformer for 75% of full load output at power factor 0.8 lagging = 97.81 %\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.27 page 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate efficiency at full load, unity power factor. \n", + "\n", + "P_f1=1 #power factor unity\n", + "P_f2=0.8 #power factor 0.8 lagging or leading\n", + "KVA=25 #Rating of the transformer(in KVA)\n", + "O_p1=KVA*P_f1 #Output at unity power factor(in KW)\n", + "P_c=400 #copper losses(in Watt)\n", + "P_C=P_c/1000 #copper losses(in KW)\n", + "P_i=320 #iron losses(in Watt)\n", + "P_I=P_i/1000 #iron losses(in KW)\n", + "P_T=P_I+P_C #total losses(in KW)\n", + "I_p1=O_p1+P_T #Input (in KW)\n", + "E_f1=(O_p1/I_p1)*100 #Efficiency of the transformer at full load and unity power factor(in %)\n", + "print 'Efficiency of the transformer at full load and unity power factor = %0.2f %%' %E_f1\n", + "O_p2=KVA*P_f2 #output at 0.8 lagging power factor(in KW)\n", + "I_p2=O_p2+P_T #input incase of 0.8 power factor(in KW)\n", + "E_f2=(O_p2/I_p2)*100 #Efficiency of the transformer at full load and 0.8 lagging power factor(in %)\n", + "print 'Efficiency of the transformer at full load and 0.8 lagging power factor = %0.2f %%' %E_f2\n", + "E_f3=E_f2 #At 0.8 leading power factor. since there is no change in input and output, so efficiency is unchanged\n", + "print 'At 0.8 leading power factor. since there is no change in input and output, so efficiency is unchanged = %0.2f %%' %E_f3" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer at full load and unity power factor = 97.20 %\n", + "Efficiency of the transformer at full load and 0.8 lagging power factor = 96.53 %\n", + "At 0.8 leading power factor. since there is no change in input and output, so efficiency is unchanged = 96.53 %\n" + ] + } + ], + "prompt_number": 59 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.28 page 210" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the efficiency of the transformer on Half load and Full load.\n", + "KVA=100 #rating of the(in KVA)\n", + "P_f=0.8 #power factor\n", + "O_p=(1/2)*KVA*P_f #Output at half load(in KW)\n", + "P_i=700 #iron loss (in Watt)\n", + "P_i1=P_i/1000 #iron loss at half and full load(in KW)\n", + "P_c=400 #copper losses (in Watt)\n", + "P_c1=((1/2)**2)*P_c/1000 #copper losses at half load condition (in KW)\n", + "P_t=P_c1+P_i1 #Total losses in half load condition(in KW)\n", + "E_f=(O_p/(O_p+P_t))*100 #Efficiency of the transformer on half load in percentage\n", + "print 'Efficiency of the transformer on half load = %0.2f %%' %E_f\n", + "O_P=KVA*P_f #Output in case of full load(in KW)\n", + "P_c2=P_c/1000 #Copper losss at full load condition(in KW)\n", + "P_T=P_c2+P_i1 #Total losses in full load condition(in KW)\n", + "E_F=(O_P/(O_P+P_T))*100 #Efficiency of the transformer on full load condition in percentage\n", + "print 'Efficiency of the transformer on full load condition = %0.2f %%' %E_F" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer on half load = 98.04 %\n", + "Efficiency of the transformer on full load condition = 98.64 %\n" + ] + } + ], + "prompt_number": 60 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.29 page 213" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the efficiency at full load unity power factor\n", + "\n", + "KVA=25 #Rating of the transformer(in KVA)\n", + "P_c=400 #Full load copper loss(in Watt)\n", + "P_c1=P_c/1000 #Full load copper loss(in KW)\n", + "P_i=350 #Iron loss(in Watt)\n", + "P_i1=P_i/1000 #Iron loss (in KW)\n", + "P_f=1 #Power factor unity\n", + "P_f1=0.8 #Power factor 0.8 lagging\n", + "O_p1=KVA*P_f #Output at full load and unity power factor(in KW)\n", + "P_t1=P_c1+P_i1 #Total losses at full load and unity power factor(in KW)\n", + "I_p1=O_p1+P_t1 #Input at full load and unity power factor(in KW)\n", + "E_f1=(O_p1/I_p1)*100 #Efficiency of the transformer at full load and unity power factor(in KW)\n", + "print 'Efficiency of the transformer at full load and unity power factor = %0.2f kW'%E_f1 \n", + "O_p2=(1/2)*KVA*P_f1 #Output At half full load, 0.8 power factor lag.(in KW)\n", + "P_c2=((1/2)**2)*P_c1 #Copper loss At half full load, 0.8 power factor lag.(in KW)\n", + "P_t2=P_c2+P_i1 #Total loss At half full load, 0.8 power factor lag.(in KW)\n", + "I_p2=O_p2+P_t2 #Input At half full load, 0.8 power factor lag.(in KW)\n", + "E_f2=(O_p2/I_p2)*100 #Efficiency of the transformer at half full load and 0.8 lagging power factor(in KW)\n", + "print 'Efficiency of the transformer at half full load and 0.8 lagging power factor = %0.2f kW'%E_f2 \n", + "#Maximum efficiency occurs when Copper loss = Iron loss,let the maximum efficiency occcur at x times full load\n", + "x=(P_i/P_c)**(1/2) \n", + "L=x*KVA #load in KVA corresponding to maximum efficiency\n", + "print 'the load for maximum efficiency = %0.2f kVA '%L " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer at full load and unity power factor = 97.09 kW\n", + "Efficiency of the transformer at half full load and 0.8 lagging power factor = 95.69 kW\n", + "the load for maximum efficiency = 23.39 kVA \n" + ] + } + ], + "prompt_number": 61 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.30 page 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the efficiency of full load current\n", + "\n", + "KVA=50 #Rating of the transformer(in KVA)\n", + "V_1=6600 #Primary voltage(in Volt)\n", + "V_2=200 #Secondary voltage(in VOlt)\n", + "I_1=KVA*1000/6600 #Full load primary current(in Amp)\n", + "P_f1=1 #power factor at unity\n", + "P_f2=0.8 #Power factor at 0.8\n", + "O_p1=KVA*P_f1 #Output at unity power factor(in KW)\n", + "P_i=650 #Iron loss(in Watt)\n", + "P_i1=P_i/1000 #Iron loss (in KW)\n", + "P_c=885 #Copper loss(in Watt)\n", + "P_c1=P_c/1000 #Copper loss(in KW)\n", + "I_p1=O_p1+P_c1+P_i1 #Input at unity power factor(in KW)\n", + "E_f1=(O_p1/I_p1)*100 #Efficiency of the transformer at unity power factor\n", + "print 'Efficiency of the transformer at unity power factor = %0.f %% '%E_f1 \n", + "O_p2=KVA*P_f2 #Output at 0.8 power factor(in KW) \n", + "#Maximum efficiency occurs when Copper loss = Iron loss,let the maximum efficiency occcur at x times full load\n", + "x=(P_i1/P_c1)**(1/2) \n", + "print 'the maximum efficiency occurs at the full load of = %0.3f'%x \n", + "O_P=x*KVA*P_f2 #Output at maximum efficiency(in KW)\n", + "E_F=(O_P/(O_P+P_i1+P_c1))*100 #Maximum Efficiency of the transformer at 0.8 power factor\n", + "print 'Maximum Efficiency of the transformer = %0.2f %%' %E_F\n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency of the transformer at unity power factor = 97 % \n", + "the maximum efficiency occurs at the full load of = 0.857\n", + "Maximum Efficiency of the transformer = 95.71 %\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.31 page 214" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In the single phase transformer,find the ratio of iron and copper loss such that maximum efficiency occurs at 75% of full load.\n", + "\n", + "x=75/100 #the value of load which is 75% of full load\n", + "P_r=x**2 #Ratio of the iron loss and copper loss for maximum efficiency\n", + "print 'Ratio of the iron and copper loss for the maximum efficiency =',P_r " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Ratio of the iron and copper loss for the maximum efficiency = 0.5625\n" + ] + } + ], + "prompt_number": 67 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.32 page 215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a single phase transformer Calculate the iron losses and the full load copper losses.\n", + "\n", + "KVA=400 #Rating of the transformer(in KVA)\n", + "E_f1=0.9877 #Transformer efficiency when delivering full load at 0.8 power factor\n", + "P_f1=0.8 #power factor at full load\n", + "P_f2=1 #power factor at half load\n", + "O_p1=KVA*P_f1 #Output on full load when power factor is 0.8(in KW)\n", + "I_p1=(O_p1/E_f1) #Input on full load when power factor is 0.8(in KW)\n", + "P_t1=I_p1-O_p1 #Total losses on full load when power factor is 0.8(in KW)\n", + "O_p2=(1/2)*KVA*P_f2 #Output at half load when power factor is 1\n", + "E_f2=0.9913 #Transformer efficiency when delivering half load at 1 power factor\n", + "I_p2=O_p2/E_f2 #Input at half load when power factor is 1\n", + "P_t2=I_p2-O_p2#Total losses at half load when power factor is 1\n", + "#P_t1=P_c+P_i\n", + "#P_t2=(1/4)P_c+P_i\n", + "P_c=(4/3)*(P_t1-P_t2) #Full load and copper losses\n", + "P_i=(1/3)*(4*P_t2-P_t1) #iron losses\n", + "print 'full load and copper losses = %0.3f kW'%P_c \n", + "print 'iron loss = %0.3f kW '%P_i " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "full load and copper losses = 2.973 kW\n", + "iron loss = 1.012 kW \n" + ] + } + ], + "prompt_number": 70 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.33 page 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import ceil\n", + "#Caption:In a transformer find all day efficiency\n", + "\n", + "KVA=15 #Rating of the transformer(in KVA)\n", + "E_f=0.98 #Efficiency of the transformer\n", + "P_F=1 #for unity power factor\n", + "O_P=KVA*P_F #Output of the transformer at unity power factor(in KW)\n", + "I_P=O_P/E_f #Input to the transformer(in KW)\n", + "P_T=I_P-O_P #Total losses(in KW)\n", + "#At Maximum efficiency\n", + "P_C=P_T/2 #copper loss for maximum efficiency(in KW)\n", + "P_I=P_C #iron losss at maximum efficiency copper loss=iron loss\n", + "L_1=2 #load for 12 hours (in KW)\n", + "L_2=12 #load for 6 hours (in KW)\n", + "L_3=18 #load for next 6 hours (in KW)\n", + "P_f1=0.5 #Power factor at L_1 load\n", + "P_f2=0.8 #Power factor at L_2 load\n", + "P_f3=0.9 #Power factor at L_3 load\n", + "T_1=12 #Time when L_1 working(in hours)\n", + "T_2=6 #Time when L_2 working(in hours)\n", + "T_3=6 #Time when L_3 working(in hours)\n", + "O_p1=L_1*T_1+L_2*T_2+L_3*T_3 #All day output(in KWh)\n", + "P_i1=P_I*24 #Iron losses for 24 hours(in KWh)\n", + "P_c1=T_1*P_C*((L_1/P_f1)/KVA)**2+T_2*P_C*((L_2/P_f2)/KVA)**2+T_3*P_C*((L_3/P_f3)/KVA)**2 #Copper loss for 24 hours(in KWh)\n", + "P_t=P_c1+P_i1 #Total losses of transformer for 24 hours(in KWh)\n", + "I_p1=O_p1+P_t #All day input(in KWh)\n", + "E_f1=(O_p1/I_p1)*100 #All day efficiency of transformer\n", + "print 'All day efficiency of transformer = %0.2f %% '%ceil(E_f1) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All day efficiency of transformer = 97.00 % \n" + ] + } + ], + "prompt_number": 73 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.34 page 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a transformer find all day efficiency\n", + "\n", + "KVA=1500 #Rating of the transformer(in KVA)\n", + "P_C=4.5 #copper loss for maximum efficiency(in KW)\n", + "P_I=3.2 #iron losss at maximum efficiency copper loss=iron loss\n", + "L_1=1200 #load for 6 hours (in KW)\n", + "L_2=900 #load for next 10 hours (in KW)\n", + "L_3=300 #load for next 4 hours (in KW)\n", + "L_4=0 #load for next 4 hours (in KW)\n", + "P_f1=0.8 #Power factor at L_1 load\n", + "P_f2=0.75 #Power factor at L_2 load\n", + "P_f3=0.8 #Power factor at L_3 load\n", + "P_f4=0 #Power factor at L_4 load\n", + "T_1=6 #Number of hours when L_1 working(in hours)\n", + "T_2=10 #Number of hours when L_2 working(in hours)\n", + "T_3=4 #Number of hours when L_3 working(in hours)\n", + "T_4=4 #Number of hours when L_4 working(in hours)\n", + "O_p1=L_1*T_1+L_2*T_2+L_3*T_3 #All day output(in KWh)\n", + "P_i1=P_I*24 #Iron losses for 24 hours(in KWh)\n", + "P_c1=T_1*P_C*((L_1/P_f1)/KVA)**2+T_2*P_C*((L_2/P_f2)/KVA)**2+T_3*P_C*((L_3/P_f3)/KVA)**2 #Copper loss for 24 hours(in KWh)\n", + "P_t=P_c1+P_i1 #Total losses of transformer for 24 hours(in KWh)\n", + "I_p1=O_p1+P_t #All day input(in KWh)\n", + "E_f1=(O_p1/I_p1)*100 #All day efficiency of transformer\n", + "print 'All day efficiency of transformer = %0.2f %%' %(E_f1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All day efficiency of transformer = 99.24 %\n" + ] + } + ], + "prompt_number": 74 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.35 page 218" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a transformer find all day efficiency\n", + "\n", + "KVA=15 #Rating of the transformer(in KVA)\n", + "P_c=0.35 #Full load copper loss (in KW)\n", + "P_i=0.25 #iron losss \n", + "P_I=P_i*24 #Iron loss per day\n", + "L_1=1/4 #load for 9 hours of full load\n", + "L_2=1 #load for 7 hours of full load\n", + "L_3=3/4 #load for next 6 hours of full load\n", + "P_f1=0.6 #Power factor at L_1 load\n", + "P_f2=0.8 #Power factor at L_2 load\n", + "P_f3=1 #Power factor at L_3 load\n", + "T_1=9 #Time when L_1 working(in hours)\n", + "T_2=7 #Time when L_2 working(in hours)\n", + "T_3=6 #Time when L_3 working(in hours)\n", + "P_c1=((1/4)**2)*P_c #Copper loss at 1/4 load\n", + "P_C1=9*P_c1 #Copper loss for 9 hours at 1/4 load\n", + "P_c2=P_c #Copper loss at full load\n", + "P_C2=7*P_c2 #Copper loss for 7 hours at full load\n", + "P_c3=((3/4)**2)*P_c #Copper loss at 3/4 load\n", + "P_C3=6*P_c3 #Copper loss for 6 hours at 3/4 load\n", + "P_C=P_C1+P_C2+P_C3 #Copper loss per day(in KW)\n", + "P_T=P_C+P_I #Iron loss per day(in KW)\n", + "O_P=L_1*KVA*P_f1*T_1+L_2*KVA*P_f2*T_2+L_3*KVA*P_f3*T_3 #Total output per day(in KWh)\n", + "I_P=O_P+P_T #Total input(in KWh)\n", + "E_F=(O_P/I_P)*100 #All day efficiency(in %)\n", + "print 'All day efficiency = %0.2f %%' %E_F" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "All day efficiency = 94.59 %\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.36 page 225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Calculate the economy of copper in auto transformer and current distribution in primary\n", + "\n", + "V_1=500 #Primary voltage\n", + "V_2=400 #Secondary voltage\n", + "I_2=100 #Secondary voltage\n", + "I_1=V_2*I_2/V_1 #Primary current\n", + "print 'Current in primary winding = %0.2f Amp '%I_1\n", + "K=V_2/V_1 #Transformer ratio\n", + "S=K*100 #Saving (in %)\n", + "print 'Economy of copper = %0.2f %%'%S" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current in primary winding = 80.00 Amp \n", + "Economy of copper = 80.00 %\n" + ] + } + ], + "prompt_number": 76 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.37 page 225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a auto transformer determine Transformation ratio secondary current primary current\n", + "\n", + "V_1=250 #Primary voltage(in Voltage)\n", + "V_2=125 #Secondary voltage(in Voltage)\n", + "K=V_2/V_1 #Transformation ratio\n", + "N_1=250 #Primary turns\n", + "print 'Transformation ratio =',K\n", + "P_f=1 #Unity power factor\n", + "L=5 #Value of load(in KW)\n", + "I_2=L*1000/(V_2*P_f) #Secondary current(in Amp)\n", + "print 'Secondary current = %0.2f Amp '%I_2 \n", + "I_1=K*I_2 #Primary current(in Amp)\n", + "print 'Primary current = %0.2f Amp '%I_1 \n", + "N_2=K*N_1 #Secondary turns\n", + "print 'number of turns across secondary winding =',N_2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transformation ratio = 0.5\n", + "Secondary current = 40.00 Amp \n", + "Primary current = 20.00 Amp \n", + "number of turns across secondary winding = 125.0\n" + ] + } + ], + "prompt_number": 77 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.38 page 29" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In an auto transformer determine current in different section and KVA output and power transferred\n", + "\n", + "KVA=10 \n", + "V_1=2400 #Voltage in first winding\n", + "V_2=240 #Voltage in second winding\n", + "I_1=KVA*1000/V_1 #Current rating of 2400 Volts winding\n", + "I_2=KVA*1000/V_2 #Current rating of 240 Volts winding\n", + "I_l=I_1+I_2 #Total load current\n", + "print 'Current rating of 2400 Volts winding = %0.2f Amp '%I_1 \n", + "print 'Current rating of 240 Volts winding = %0.2f Amp '%I_2 \n", + "print 'Total load current = %0.2f Amp '%I_l \n", + "KVA_r=V_1*I_l/1000 #KVA output rating\n", + "print 'KVA output rating = %0.2f KVA '%KVA_r \n", + "P_i=V_1*I_1 #power transferred inductively\n", + "P_c=V_1*I_2 #power transferred conductively\n", + "print 'power transferred inductively = %0.2f VA '%P_i \n", + "print 'power transferred conductively = %0.2f VA '%P_c \n", + "N_1=2640 #Number Primary winding in case of two winding transformer\n", + "N_2=2400 #Number Secondary winding in case of two winding transformer\n", + "K=N_1/N_2 #Transformer ratio\n", + "Saving=(1/K)*100 #Saving in copper\n", + "print 'Saving in copper = %0.2f %%'%Saving \n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current rating of 2400 Volts winding = 4.17 Amp \n", + "Current rating of 240 Volts winding = 41.67 Amp \n", + "Total load current = 45.83 Amp \n", + "KVA output rating = 110.00 KVA \n", + "power transferred inductively = 10000.00 VA \n", + "power transferred conductively = 100000.00 VA \n", + "Saving in copper = 90.91 %\n" + ] + } + ], + "prompt_number": 79 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:3.39 Page 230" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a transformer calculate Power output and power transformed and power conducted\n", + "KVA=25 \n", + "V_s=2200 #Source voltage\n", + "V_1=2200 #Voltage in first winding\n", + "V_2=220 #Voltage in second winding\n", + "I_1=KVA*1000/V_1 #Current rating of 2200 V winding\n", + "I_2=KVA*1000/V_2 #Current rating of 220 V winding\n", + "V_o=V_1+V_2 #Output voltage\n", + "I_l=I_1+I_2 #Input line current\n", + "I_o=I_2 #Output current of auto transformer\n", + "KVA_r=V_o*I_2/1000 #KVA rating\n", + "P_f1=0.8 #\n", + "P_o=KVA*P_f1 #Power output at full load and 0.8 power factor\n", + "KVA_t=V_1*I_1/1000 #KVA transformed \n", + "print 'KVA transformed = %0.2f kVA'%KVA_t\n", + "P_t=KVA_t*P_f1 #Power transformed(in KW)\n", + "print 'Power transformed = %0.2f kW'%P_t\n", + "KVA_c=V_s*I_o/1000 #KVA conducted(in KVA)\n", + "P_c=KVA_c*P_f1 #Power conducted(in KW)\n", + "print 'Power conducted = %0.2f kW'%P_c\n", + "#E_f=Output/(Output+Losses)\n", + "#Losses=((1/E_f)-1)*Output\n", + "E_f=0.9 #Efficiency of the two winding transformer\n", + "P_f2=0.85 #New power factor of the two einding transformer\n", + "O_p1=KVA_t*1000*P_f2 #Output of the two winding transformer\n", + "L=((1/E_f)-1)*O_p1 #losses in a 2-winding transformer\n", + "#Losses in auto transformer=losses in 2-winding transformer\n", + "O_P=KVA_r*1000*P_f2 #output of the auto transformer\n", + "E_F=(O_P/(O_P+L))*100 #Efficiency of the auto transformer(in %)\n", + "print 'Efficiency of the auto transformer = %0.2f %%'%E_F\n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "KVA transformed = 25.00 kVA\n", + "Power transformed = 20.00 kW\n", + "Power conducted = 200.00 kW\n", + "Efficiency of the auto transformer = 99.00 %\n" + ] + } + ], + "prompt_number": 81 + } + ], + "metadata": {} + } + ] +} diff --git a/Electrical_Machines_-_1_by_Tarlok_Singh/Chap4.ipynb b/Electrical_Machines_-_1_by_Tarlok_Singh/Chap4.ipynb new file mode 100755 index 00000000..5e5d9000 --- /dev/null +++ b/Electrical_Machines_-_1_by_Tarlok_Singh/Chap4.ipynb @@ -0,0 +1,155 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter4 - Three Phase Transformer" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:4.1 page 263" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from math import ceil\n", + "#Caption:Determines the turns per phase for the HV and LV winding of the 3 phase transformer.\n", + "\n", + "F_max=0.024 #Maximum flux (in weber)\n", + "f=50 #Supply frequency(in Hz)\n", + "E_1p=11000 #Primary phase voltage(in Volts)\n", + "N_1=ceil(E_1p/(4.44*F_max*f)) #Turns per phase on primary\n", + "print 'turns per phase for the H.V. winding of the 3 phase transformer =',N_1\n", + "E_2l=400 #Secondary line voltage(in Volts)\n", + "E_2p=E_2l/(3)**(1/2) #Secondary phase voltage(in Volts)\n", + "N_2=ceil(E_2p/(4.44*F_max*f)) #turns per phase for the L.V. winding of the 3 phase transformer\n", + "print 'turns per phase for the L.V. winding of the 3 phase transformer =',N_2\n", + "# Answer wrong in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "turns per phase for the H.V. winding of the 3 phase transformer = 2065.0\n", + "turns per phase for the L.V. winding of the 3 phase transformer = 44.0\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:4.2 page 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:In a three phase transformer Calculate the secondary line voltage at no load \n", + "\n", + "#when windings are star delta connected\n", + "V=3300 #the supply voltage(in Volts)\n", + "V_l1=V #Primary line voltage in star delta conection\n", + "N_1=420 #turns on the primary side of the transformer\n", + "N_2=36 #turns on the secondary side of the transformer\n", + "V_p1=V_l1/(3)**(1/2) #Primary phase voltage in star delta connection\n", + "V_p2=V_p1*(N_2/N_1) #Secondary phase voltage in star delta connection\n", + "V_l2=V_p2 #Secondary line voltage when windings are star delta connected(in Volts)\n", + "print 'Secondary line voltage when windings are star delta connected = %0.2f Volts '%V_l2 \n", + "#when windings are delta star connected\n", + "V_P1=V #Primary phase voltage(in Volts)\n", + "V_L1=V_P1 #Primary line voltage in delta star connection\n", + "V_P2=V_P1*(N_2/N_1) #Secondary phase voltage in delta star connection\n", + "V_L2=V_P2*(3)**(1/2) #Secondary line voltage when windings are delta star connected(in Volts)= \n", + "print 'Secondary line voltage when windings are delta star connected = %0.2f Volts '%V_L2 " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Secondary line voltage when windings are star delta connected = 163.31 Volts \n", + "Secondary line voltage when windings are delta star connected = 489.92 Volts \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exam:4.3 page 264" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Caption:Find the secondary no load voltage and primary secondary currents in a 3 phase transformer\n", + "\n", + "V_l=11000 #Input voltage(in Volts)\n", + "V_1ph=V_l/(3)**(1/2) #Phase voltage\n", + "KVA=50*10**(3) #\n", + "#KVA=((3)**(1/2))*V_l*I_l\n", + "I_1l=KVA/(((3)**(1/2))*V_l) #Line current\n", + "I_1ph=I_1l #Star system value of phase current\n", + "print 'Value of primary phase and line current = %0.2f Amp '%I_1ph \n", + "N_1=1000 #Primary turns\n", + "N_2=90 #Secondary turns\n", + "V_2ph=(N_2/N_1)*V_1ph #secondary phase voltage(in Volts)\n", + "#V_2ph=V_2l delta system\n", + "print 'Value of secondary phase and line voltage = %0.2f Volts '%V_2ph \n", + "I_2ph=(N_1/N_2)*I_1ph #secondary phase current(in Amp)\n", + "print 'Value of secondary phase current = %0.2f Amp '%I_2ph \n", + "I_2l=I_2ph*3**(1/2) #secondary line current(in Amp)\n", + "print 'Value of secondary line current = %0.2f Amp' %I_2l" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of primary phase and line current = 2.62 Amp \n", + "Value of secondary phase and line voltage = 571.58 Volts \n", + "Value of secondary phase current = 29.16 Amp \n", + "Value of secondary line current = 50.51 Amp\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +} diff --git a/Electrical_Machines_-_1_by_Tarlok_Singh/README.txt b/Electrical_Machines_-_1_by_Tarlok_Singh/README.txt new file mode 100644 index 00000000..2d0f162a --- /dev/null +++ b/Electrical_Machines_-_1_by_Tarlok_Singh/README.txt @@ -0,0 +1,10 @@ +Contributed By: Gaurav Mittal +Course: others +College/Institute/Organization: UKTU +Department/Designation: CS +Book Title: Electrical Machines - 1 +Author: Tarlok Singh +Publisher: S. K. Kataria & Sons, New Delhi +Year of publication: 2011 +Isbn: 978-81-85749-58-2 +Edition: 2
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