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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": {} + } + ] +} |