From 36a03d6d76bac315dba73b2ba9555c7e3fe0234f Mon Sep 17 00:00:00 2001 From: nice Date: Thu, 9 Oct 2014 18:07:00 +0530 Subject: updated books --- BASIC_ELECTRICAL_ENGINEERING_/Chapter16.ipynb | 1654 +++++++++++++++++++++++++ 1 file changed, 1654 insertions(+) create mode 100755 BASIC_ELECTRICAL_ENGINEERING_/Chapter16.ipynb (limited to 'BASIC_ELECTRICAL_ENGINEERING_/Chapter16.ipynb') diff --git a/BASIC_ELECTRICAL_ENGINEERING_/Chapter16.ipynb b/BASIC_ELECTRICAL_ENGINEERING_/Chapter16.ipynb new file mode 100755 index 00000000..7078e036 --- /dev/null +++ b/BASIC_ELECTRICAL_ENGINEERING_/Chapter16.ipynb @@ -0,0 +1,1654 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 16: DC MACHINES " + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.1,Page number: 518" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the terminal voltage,output current and total power generated by a dc generator.\"\"\" \n", + "\n", + "#Variable Declaration:\n", + "e=2.1 #Average emf generated in each conductor(in Volts) \n", + "full_load_I=200.0 #Full-load current(in Amperes) \n", + "Z=480.0 #Number of conductors in armature\n", + "P=8.0 #Number of poles\n", + "\n", + "\n", + "#Calculations:\n", + "A_1=P\n", + "E_1=e*(Z/A_1)\n", + "I_L_1=full_load_I*A_1\n", + "Po_1=E_1*I_L_1\n", + "A_2=2\n", + "E_2=e*(Z/A_2)\n", + "I_L_2=full_load_I*A_2\n", + "Po_2=E_2*I_L_2\n", + "\n", + "\n", + "#Result:\n", + "print(\"(a) When the armature is lap wound \\n \")\n", + "print \"The terminal voltage on no load is %.2f V\" %(E_1)\n", + "print \"The output current on full load is %.2f A\" %(I_L_1) \n", + "print \"The total power generated on full load is %e W\" %(Po_1)\n", + "print(\"\\n(b)When the armature is wave wound \\n \")\n", + "print \"The terminal voltage on no load is %.2f V\" %(E_2)\n", + "print \"The output current on full load is %.2f A\" %(I_L_2) \n", + "print \"The total power generated on full load is %e W\" %(Po_2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a) When the armature is lap wound \n", + " \n", + "The terminal voltage on no load is 126.00 V\n", + "The output current on full load is 1600.00 A\n", + "The total power generated on full load is 2.016000e+05 W\n", + "\n", + "(b)When the armature is wave wound \n", + " \n", + "The terminal voltage on no load is 504.00 V\n", + "The output current on full load is 400.00 A\n", + "The total power generated on full load is 2.016000e+05 W\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.2,Page number: 520" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the emf induced in the armature of a dc generator. \"\"\"\n", + "\n", + "#Variable Declaration:\n", + "slots=65.0 #Number of slots\n", + "cond_per_slot=12.0 #Number of conductors per slot\n", + "A=4.0 #Number of parallel paths\n", + "P=4.0 #Number of poles\n", + "flux=0.02 #Flux per pole(in Webers) \n", + "N=1200.0 #Speed of operation of the dc generator(in rpm)\n", + "\n", + "\n", + "#Calculations:\n", + "Z=slots*cond_per_slot\n", + "E=(flux*Z*N*P)/(60*A)\n", + "\n", + "\n", + "#Result:\n", + "print \"The emf induced in the armature is %.2f V\" %(E)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emf induced in the armature is 312.00 V\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.3,Page number: 520" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\" Finding the induced emf in a dc machine.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "N1=500.0 #Initial speed of operation of dc machine(in rpm) \n", + "E1=180.0 #Induced emf at 500rpm(in Volts)\n", + "N2=600.0 #New speed of operation(in rpm)\n", + "\n", + "\n", + "#Calculations:\n", + "E2=(N2/N1)*E1\n", + "\n", + "\n", + "#Result:\n", + "print \"The induced emf when the machine runs at 600 rpm is %.2f V\" %(E2) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The induced emf when the machine runs at 600 rpm is 216.00 V\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.4,Page number: 520" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"\"Finding the percentage increase in the field flux in a dc generator.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "N1=750.0 #Initial speed of operation of dc machine(in rpm)\n", + "E1=220.0 #Induced emf at 750 rpm(in Volts) \n", + "E2_a=250.0 #New emf(in Volts) \n", + "\n", + "\n", + "#Calculations and Result:\n", + "N2_a=(E2_a/E1)*N1\n", + "E2_b=250.0\n", + "N2_b=600.0\n", + "flux_ratio=(E2_b/E1)*(N1/N2_b)\n", + " \n", + " \n", + "#Result:\n", + "print \"(a)The speed at which the induced emf is 250V(assuming the flux to be constant) is %d rpm \" %(round(N2_a,0))\n", + "print \"(b)The required percentage increase in the field flux so that the induced emf is 250V,while the speed is only 600rpm is %d percent\" %(round(((flux_ratio-1)*100),0))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The speed at which the induced emf is 250V(assuming the flux to be constant) is 852 rpm \n", + "(b)The required percentage increase in the field flux so that the induced emf is 250V,while the speed is only 600rpm is 42 percent\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.5,Page number: 525 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\" Finding the emf induced in the armature.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "V=440.0 #Load voltage(in Volts) \n", + "Rsh=110.0 #Resistance of shunt field coil(in Ohms)\n", + "Ra=0.02 #Armature resistance(in Ohms)\n", + "I_L=496.0 #Load current(in Amperes)\n", + "\n", + "\n", + "#Calculations:\n", + "Ish=V/Rsh\n", + "Ia=I_L+Ish\n", + "Eg=V+(Ia*Ra)\n", + "\n", + "\n", + "#Result:\n", + "print \"The emf induced in the armature is %.2f V\" %(Eg)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emf induced in the armature is 450.00 V\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.6,Page number: 525 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\" Finding the total armature current and the generated emf.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "no_of_lamps=100.0 #Number of lamps\n", + "P=60.0 #Power rating of each lamp(in Watts) \n", + "V=200 #Voltage rating of each lamp(in Volts)\n", + "Ra=0.2 #Armature resistance(in Ohms)\n", + "Rsh=50 #Shunt field resistance(in Ohms)\n", + "Poles=4.0 #Number of poles\n", + "no_of_brushes=2.0 #Number of brushes\n", + "brush_drop_per_brush=1.0 #Brush drop at each brush(in Volts)\n", + "\n", + "\n", + "#Calculations:\n", + "I1=P/V\n", + "I_L=no_of_lamps*I1\n", + "Ish=V/Rsh\n", + "Ia=Ish+I_L\n", + "A=Poles\n", + "Ic=Ia/A\n", + "brush_drop=no_of_brushes*brush_drop_per_brush\n", + "Eg=V+(Ia*Ra)+brush_drop\n", + "\n", + "\n", + "#Result:\n", + "print \"The total armature current is %.2f A\" %(Ia)\n", + "print \"The current per path is %.2f A\" %(Ic)\n", + "print \"The generated emf is %.2f V\" %(Eg)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total armature current is 34.00 A\n", + "The current per path is 8.50 A\n", + "The generated emf is 208.80 V\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.7,Page number: 525" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the emf generated in a compound-wound dc generator.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "V=250.0 #Load voltage(in Volts)\n", + "Rsh=130.0 #Shunt field resistance(in Ohms)\n", + "Ra=0.1 #Armature resistance(in Ohms)\n", + "Rse=0.1 #Series field resistance(in Ohms)\n", + "I_L=100.0 #Load current(in Amperes)\n", + "no_of_brushes=2 #Number of brushes \n", + "brush_drop_per_brush=1.0 #Brush drop at each brush(in Volts)\n", + "\n", + "\n", + "#Calculations:\n", + "Ise=I_L\n", + "Vse=Ise*Rse\n", + "Vsh=V+Vse\n", + "Ish=Vsh/Rsh\n", + "Ia=I_L+Ish\n", + "brush_drop=no_of_brushes*brush_drop_per_brush\n", + "Eg=V+Vse+(Ia*Ra)+brush_drop\n", + "\n", + "\n", + "#Result:\n", + "print \"The emf generated is %.2f V.\" %(Eg)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emf generated is 272.20 V.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.8,Page number: 528 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the emf generated,the copper losses,and efficiency of a shunt generator.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "Po=30e03 #Full-load output power(in Watts)\n", + "V=200.0 #Terminal voltage(in Volts)\n", + "Ra=0.05 #Armature resistance(in Ohms)\n", + "Rsh=50.0 #Shunt field resistance(in Ohms)\n", + "loss=1000.0 #Friction losses(in Watts) \n", + "\n", + "\n", + "#Calculations:\n", + "I_L=Po/V\n", + "Ish=V/Rsh\n", + "Ia=Ish+I_L\n", + "Eg=V+(Ia*Ra)\n", + "copper_loss=(pow(Ish,2)*Rsh)+(pow(Ia,2)*Ra)\n", + "effi=Po/(Po+copper_loss+loss)\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The emf generated is %.2f V\" %(Eg)\n", + "print \"(b)The copper loss is %.2f W\" %(copper_loss)\n", + "print \"(c)The efficiency is %.2f percent\" %(effi*100)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The emf generated is 207.70 V\n", + "(b)The copper loss is 1985.80 W\n", + "(c)The efficiency is 90.95 percent\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.9,Page number:529 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding armature resistance and the load-current corresponding to maximum efficiency of dc shunt generator.\"\"\"\n", + "\n", + "from math import sqrt,pow\n", + "\n", + "#Variable Declaration:\n", + "V=210.0 #Full-load voltage(in Volts)\n", + "I_L=195.0 #Full-load current(in Amperes)\n", + "Rsh=52.5 #Shunt field resistance(in Ohms)\n", + "effi=0.90 #Full-load efficiency\n", + "stray_loss=710.0 #Stray losses(in Watts)\n", + "\n", + "\n", + "#Calculations:\n", + "Po=V*I_L\n", + "Pin=Po/effi\n", + "total_loss=Pin-Po\n", + "Ish=V/Rsh\n", + "Ia=I_L+Ish\n", + "sh_copp_loss=pow(Ish,2)*Rsh\n", + "const_loss=sh_copp_loss+stray_loss\n", + "arma_copp_loss=total_loss-const_loss\n", + "Ra=arma_copp_loss/(pow(Ia,2))\n", + "Ia_max_effi=sqrt(const_loss/Ra)\n", + "I_L_max_effi=Ia_max_effi-Ish\n", + "\n", + "\n", + "#Result:\n", + "print \"The armature resistance is %.5f ohm \" %(Ra)\n", + "print \"The load current corresponding to maximum efficiency is %.2f A\" %(I_L_max_effi)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The armature resistance is 0.07576 ohm \n", + "The load current corresponding to maximum efficiency is 139.04 A\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.10,Page number: 534 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the number of series turns required per pole for a level-compounded generator.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "I_full_load=100.0 #Full-load current(in Amperes)\n", + "sh_turns=1500.0 #Number of turns in the shunt winding \n", + "Ish_no_load=4.0 #Shunt current at no-load(in Amperes) \n", + "Ish_full_load=6.0 #Shunt current at full-load(in Amperes)\n", + "\n", + "\n", + "#Calculations:\n", + "At_no_load=Ish_no_load*sh_turns\n", + "At_full_load=Ish_full_load*sh_turns\n", + "At_series=At_full_load-At_no_load\n", + "Nse=At_series/I_full_load\n", + "\n", + "\n", + "#Result:\n", + "print \"The number of series turns required per pole is %d \" %(Nse)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The number of series turns required per pole is 30 \n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.11,Page number: 536" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the back emf generated in a dc shunt motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "I_L=41.0 #Full-load current(in Amperes)\n", + "V=250.0 #Full-load voltage(in Volts) \n", + "Ra=0.1 #Armature resistance(in Ohms) \n", + "Rsh=250.0 #Shunt field resistance(in Ohms)\n", + "\n", + "\n", + "#Calculations:\n", + "Ish=V/Rsh\n", + "Ia=I_L-Ish\n", + "Eb=V-(Ia*Ra)\n", + "\n", + "\n", + "#Result:\n", + "print \"The back emf generated in the motor is %.2f V\" %(Eb)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The back emf generated in the motor is 246.00 V\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.12,Page number: 536 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the speed of a dc motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "A=2.0 #Number of parallel paths\n", + "P=4.0 #Number of poles \n", + "Z=888.0 #Number of conductors \n", + "flux=23e-03 #Flux per pole(in Webers) \n", + "Ia=50.0 #Armature current(in Amperes)\n", + "Ra=0.28 #Armature resistance(in Ohms) \n", + "V=440.0 #Rated voltage(in Volts)\n", + "\n", + "\n", + "#Calculations:\n", + "Eb=V-(Ia*Ra)\n", + "N=(60*A*Eb)/(flux*Z*P)\n", + "\n", + "\n", + "#Result:\n", + "print \"The speed of the motor is %d rpm\" %(round(N,0)) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of the motor is 626 rpm\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.13,Page number: 536" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the speed of a dc motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "V1=460.0 #Initial supply voltage(in Volts)\n", + "N1=900.0 #Speed of motor at 460-V(in rpm)\n", + "V2=200.0 #Final supply voltage(in Volts)\n", + "\n", + "\n", + "#Calculations:\n", + "kflux=V1/N1\n", + "N2=V2/(0.7*kflux)\n", + "\n", + "\n", + "#Result:\n", + "ans_N2=\"The approximate speed of the motor when the motor is connected across a 200V supply is %d rpm\" %(round(N2,0))\n", + "print(ans_N2)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The approximate speed of the motor when the motor is connected across a 200V supply is 559 rpm\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.14,Page number: 537" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the speed and the gross torque developed by the armature of a dc motor.\"\"\"\n", + "\n", + "from math import pi\n", + "\n", + "#Variable Declaration:\n", + "V=480.0 #Rated voltage(in Volts) \n", + "Ia=110.0 #Armature current at rated voltage(in Amperes)\n", + "Ra=0.2 #Armature resistance(in Ohms) \n", + "flux=50e-03 #Flux per pole(in Webers) \n", + "A=6.0 #Number of parallel paths\n", + "P=6.0 #Number of poles \n", + "Z=864.0 #Number of conductors\n", + "\n", + "\n", + "#Calculations:\n", + "Eb=V-(Ia*Ra)\n", + "N=(60*A*Eb)/(flux*Z*P)\n", + "torque=((flux*Z)/(2*pi))*(P/A)*Ia\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The speed of the motor is %d rpm \" %(round(N,0))\n", + "print \"(b)The gross torque developed by the armature is %.2f Nm\" %(torque)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The speed of the motor is 636 rpm \n", + "(b)The gross torque developed by the armature is 756.30 Nm\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.15,Page number: 538 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the power generated in the armature winding of a dc generator.\"\"\"\n", + "\n", + "from math import pi \n", + "\n", + "#Variable Declaration:\n", + "N=900.0 #Operating speed of generator(in rpm)\n", + "torque=2e03 #Torque(in N-metre) \n", + "P_losses=8e03 #Power losses(in Watts) \n", + "\n", + "\n", + "#Calculations:\n", + "Pin=(2*pi*torque*N)/60.0\n", + "Pd=Pin-P_losses\n", + "\n", + "\n", + "#Result:\n", + "print \"The power generated in the armature winding is %e W\" %(Pd)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The power generated in the armature winding is 1.804956e+05 W\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.16,Page number: 540" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the speed of a series motor when the current changes.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "V=230.0 #Supply voltage(in Volts)\n", + "Ra=0.12 #Armature resistance(in Ohms) \n", + "Rse=0.03 #Series field resistance(in Ohms)\n", + "Ia1=110.0 #Current at 230 V(in Amperes)\n", + "flux1=24e-03 #Flux per pole at 110 A(in Webers)\n", + "N1=600.0 #Speed at 230 V(in rpm) \n", + "Ia2=50.0 #Armature current(in Amperes) \n", + "flux2=16e-03 #Flux per pole at 50 A(in Webers) \n", + "\n", + "\n", + "#Calculations:\n", + "Eb1=V-Ia1*(Ra+Rse)\n", + "k=Eb1/(N1*flux1)\n", + "Eb2=V-Ia2*(Ra+Rse)\n", + "N2=Eb2/(k*flux2)\n", + "\n", + "\n", + "#Result:\n", + "print \"The speed of the motor when the currenthas fallen to 50 A is %d rpm\" %(round(N2,0))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of the motor when the currenthas fallen to 50 A is 938 rpm\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.17,Page number: 540" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the current drawn by the machine.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "V=250.0 #Supply voltage(in Volts)\n", + "Ra=0.2 #Armature resistance(in Ohms)\n", + "\n", + "\n", + "#Calculations:\n", + "Eb_1=0\n", + "Ia_1=(V-Eb_1)/Ra\n", + "Eb_2=200\n", + "Ia_2=(V-Eb_2)/Ra\n", + "Eb_3=250\n", + "Ia_3=(V-Eb_3)/Ra\n", + "Eb_4=-250\n", + "Ia_4=(V-Eb_4)/Ra\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)When the machine is at rest,\"\n", + "print \"The current drawn by the machine is %.2f A\" %(Ia_1)\n", + "print \"(b)When the machine is generating an emf of 200V and is connected to the supply with correct polarities,\"\n", + "print \"The current drawn by the machine is %.2f A\" %(Ia_2)\n", + "print \"(c)When the machine is generating an emf of 250V and is connected to the supply with correct polarities,\"\n", + "print \"The current drawn by the machine is %.2f A\" %(Ia_3)\n", + "print \"(d)When the machine is generating an emf of 250V and is connected to the supply with reversed polarities,\"\n", + "print \"The current drawn by the machine is %.2f A\" %(Ia_4) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)When the machine is at rest,\n", + "The current drawn by the machine is 1250.00 A\n", + "(b)When the machine is generating an emf of 200V and is connected to the supply with correct polarities,\n", + "The current drawn by the machine is 250.00 A\n", + "(c)When the machine is generating an emf of 250V and is connected to the supply with correct polarities,\n", + "The current drawn by the machine is 0.00 A\n", + "(d)When the machine is generating an emf of 250V and is connected to the supply with reversed polarities,\n", + "The current drawn by the machine is 2500.00 A\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.18,Page number: 541" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the speed and the gross torque developed by the armature of a dc series motor.\"\"\"\n", + "\n", + "from math import pi\n", + "\n", + "#Variable Declaration:\n", + "P=6.0 #Number of poles\n", + "A=6.0 #Number of parallel paths \n", + "Z=864.0 #Number of conductors \n", + "flux=50e-03 #Flux per pole(in Webers)\n", + "Ia=110.0 #Armature current(in Amperes)\n", + "V=480.0 #Load voltage(in Volts)\n", + "Ra=0.18 #Armature resistance(in Ohms) \n", + "Rse=0.02 #Series field resistance(in Ohms)\n", + "\n", + "\n", + "#Calculations:\n", + "Eb=V-Ia*(Ra+Rse)\n", + "N=(60*A*Eb)/(flux*Z*P)\n", + "torque=(60*Eb*Ia)/(2*pi*N)\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The speed of the motor is %d rpm\" %(round(N,0))\n", + "print \"(b)The gross torque developed by the armature is %.2f Nm \" %(torque) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The speed of the motor is 636 rpm\n", + "(b)The gross torque developed by the armature is 756.30 Nm \n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.19,Page number: 541" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the series resistance to reduce the speed of a shunt motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "V=220.0 #Rated voltage of the motor(in Volts)\n", + "Ia=22.0 #Armature current(in Amperes)\n", + "Ra=0.45 #Armature resistance(in Ohms)\n", + "N1=700.0 #Initial speed of motor(in rpm)\n", + "N2=450.0 #Final speed of motor(in rpm)\n", + "\n", + "\n", + "#Calculations:\n", + "E1=V-(Ia*Ra)\n", + "E2=(N2/N1)*E1\n", + "R=((V-E2)/Ia)-Ra\n", + "\n", + "\n", + "#Result:\n", + "print \"The resistance that should be placed in series with the armature to reduce the speed to 450 rpm is %.3f ohm \" %(R)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance that should be placed in series with the armature to reduce the speed to 450 rpm is 3.411 ohm \n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.20,Page number: 541" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the speed of a dc series motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "V=230.0 #Rated voltage of the dc series motor(in Volts) \n", + "Ra=0.2 #Armature resistance(in Ohms) \n", + "Rse=0.1 #Series field resistance(in Ohms) \n", + "Ia1=40.0 #Line current at rated voltage(in Amperes)\n", + "N1=1000.0 #Speed of motor at rated voltage(in rpm)\n", + "Ia2=20.0 #Line current at 230 V(in Amperes)\n", + "\n", + "\n", + "#Calculations:\n", + "Eb1=V-Ia1*(Ra+Rse)\n", + "Eb2=V-Ia2*(Ra+Rse)\n", + "\"\"\" Eb=k*flux*N \"\"\"\n", + "N2=(Eb2*N1)/(Eb1*0.6)\n", + "\n", + "\n", + "#Result:\n", + "print \"The speed of the motor for a line current of 20A at 230V is %d rpm\" %(round(N2,0)) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of the motor for a line current of 20A at 230V is 1713 rpm\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.21,Page number: 543 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the terminal voltage of a dc shunt generator.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "P=4 #Number of poles\n", + "turns=260 #Number of turns in the armature winding\n", + "R=0.006 #Resistance of each turn of armature(in Ohms)\n", + "flux=0.08 #Useful flux per pole(in Webers)\n", + "I_L=55 #Load current(in Amperes)\n", + "N=1000 #Speed of the generator(in rpm)\n", + "\n", + "\n", + "#Calculations:\n", + "Z=2*turns\n", + "A=P\n", + "Eg=(flux*Z*N*P)/(60.0*A)\n", + "Rw=turns*R\n", + "R1=Rw/4.0\n", + "Ra=R1/4\n", + "V=Eg-(I_L*Ra)\n", + "\n", + "\n", + "#Result:\n", + "print \"The terminal voltage of the generator when it is running at 1000 rpm and supplying load current of 55 A is %.3f V.\" %(V)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The terminal voltage of the generator when it is running at 1000 rpm and supplying load current of 55 A is 687.971 V.\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.22,Page number: 544" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the armature current,the emf induced and the flux per pole for a dc shunt generator.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "P=8.0 #Number of poles\n", + "A=2.0 #Number of parallel paths\n", + "Z=778 #Number of conductors\n", + "V=250.0 #Load voltage(in Volts)\n", + "R_L=12.5 #Load resistance(in Ohms)\n", + "Ra=0.24 #Armature resistance(in Ohms)\n", + "Rsh=250.0 #Shunt field resistance(in Ohms)\n", + "N=500 #Speed of the dc shunt generator(in rpm) \n", + "\n", + "\n", + "#Calculations:\n", + "I_L=V/R_L\n", + "Ish=V/Rsh\n", + "Ia=I_L+Ish\n", + "Eg=V+(Ia*Ra)\n", + "flux=(60.0*A*Eg)/(Z*N*P)\n", + "\n", + "\n", + "#Result:\n", + "print \"The armature current is %.2f A.\" %(Ia)\n", + "print \"The emf induced is %.2f V.\" %(Eg)\n", + "print \"The flux per pole is %e Wb.\" %(flux)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The armature current is 21.00 A.\n", + "The emf induced is 255.04 V.\n", + "The flux per pole is 9.834447e-03 Wb.\n" + ] + } + ], + "prompt_number": 43 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.23,Page number: 544" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the percentage reduction in speed of dynamo.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "Po_1=500e03 #Initial power output(in Watts)\n", + "Po_2=250e03 #Final power output(in Watts)\n", + "V=500.0 #Constant excitation voltage(in Volts)\n", + "Ra=0.015 #Resistance between the terminals of dynamo(in Ohms)\n", + "\n", + "\n", + "#Calculations:\n", + "Ia_1=Po_1/V\n", + "E1=V+(Ia_1*Ra)\n", + "Ia_2=Po_2/V\n", + "E2=V+(Ia_2*Ra)\n", + "\"\"\" Since excitation emf remains constant in the two cases,we have \n", + "\n", + " E=(flux*Z*N*P)/(60*A) where E=emf generated;Z=number of conductors;N=speed of motor(in rpm);P=number of poles;A=number of parallel paths;\n", + " flux=useful flux per pole(in Wb).\n", + " \n", + " N=KE, where K is a constant.\n", + " \n", + " Hence,fractional reduction in speed is given as,\n", + " \n", + " (N1-N2)/N1=((K*(E1-E2))/(K*E1)). \"\"\"\n", + "\n", + "fract=((E1-E2)/E1)*100\n", + "\n", + "\n", + "#Result:\n", + "print \"The percentage reduction in speed of the dynamo is %.3f percent.\" %(fract) " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The percentage reduction in speed of the dynamo is 1.456 percent.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.24,Page number: 545 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the voltage between the far end of the feeder and the bus-bar of a dc series generator.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "Rt=0.3 #Resistance of transmission line(in Ohms) \n", + "I_L_1=160.0 #Load current in first case(in Amperes) \n", + "I_L_2=50.0 #Load current in second case(in Amperes)\n", + "\n", + "\n", + "#Calculations:\n", + "Vt_1=I_L_1*Rt\n", + "Vb_1=(50.0/200.0)*I_L_1\n", + "Vd_1=Vt_1-Vb_1\n", + "Vt_2=I_L_2*Rt\n", + "Vb_2=(50.0/200.0)*I_L_2\n", + "Vd_2=Vt_2-Vb_2\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The voltage between far end of the feeder and the bus-bar at a cusrrent of 160 A is %.2f V.\" %(Vd_1)\n", + "print \"(b)The voltage between far end of the feeder and the bus-bar at a cusrrent of 50 A is %.2f V.\" %(Vd_2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The voltage between far end of the feeder and the bus-bar at a cusrrent of 160 A is 8.00 V.\n", + "(b)The voltage between far end of the feeder and the bus-bar at a cusrrent of 50 A is 2.50 V.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.25,Page number: 545" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\" Finding the emf generated and the armature current in a dc long-shunt compound generator. \"\"\"\n", + "\n", + "#Variable Declaration:\n", + "I_L=50 #Load current(in Amperes)\n", + "V=500 #Terminal voltage(in Volts)\n", + "Ra=0.05 #Armature resistance(in Ohms)\n", + "Rse=0.03 #Series field resistance(in Ohms)\n", + "Rsh=250 #Shunt field resistance(in Ohms)\n", + "brush_drop=1.0 #Brush contact drop(in Volts) \n", + "\n", + "\n", + "#Calculations:\n", + "Ish=V/Rsh\n", + "Ia=Ish+I_L\n", + "Eg=V+(Ia*(Ra+Rse))+brush_drop\n", + "\n", + "\n", + "#Result:\n", + "print \"The armature current is %.2f A.\" %(Ia)\n", + "print \"The emf generated is %.2f V\" %(Eg)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The armature current is 52.00 A.\n", + "The emf generated is 505.16 V\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.26,Page number: 545" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\" Finding the voltage and the power generated by a dc generator.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "P=8 #Number of poles\n", + "Z=500 #Number of conductors on the armature\n", + "flux =0.02 #Magnetic flux per pole(in Webers)\n", + "N=1800 #Speed of the generator(in rpm)\n", + "I=5.0 #Allowable current per path(in Amperes)\n", + "\n", + "\n", + "#Calculations:\n", + "A_1=2\n", + "Eg_1=(flux*Z*N*P)/(60*A_1)\n", + "A_2=P\n", + "Eg_2=(flux*Z*N*P)/(60*A_2)\n", + "Ia_1=A_1*I\n", + "Pd_1=Eg_1*Ia_1\n", + "Ia_2=A_2*I\n", + "Pd_2=Eg_2*Ia_2\n", + "\n", + "\n", + "#Result:\n", + "print \"When the armature is wave wound:\"\n", + "print \"(a)The generated voltage is %.2f V.\" %(Eg_1)\n", + "print \"(b)The kW generated by the machine is %.2f kW.\" %(Pd_1/1000.0)\n", + "print \"\\nWhen the armature is lap wound:\"\n", + "print \"(a)The generated voltage is %.2f V.\" %(Eg_2)\n", + "print \"(b)The kW generated by the machine is %.2f kW.\" %(Pd_2/1000.0)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "When the armature is wave wound:\n", + "(a)The generated voltage is 1200.00 V.\n", + "(b)The kW generated by the machine is 12.00 kW.\n", + "\n", + "When the armature is lap wound:\n", + "(a)The generated voltage is 300.00 V.\n", + "(b)The kW generated by the machine is 12.00 kW.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.27,Page number: 546" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the emf generated and the copper losses in a dc shunt generator.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "V=250.0 #Terminal voltage(in Volts)\n", + "I_L=195 #Load current(in Amperes)\n", + "Ra=0.02 #Armature resistance(in Ohms)\n", + "Rsh=50.0 #Shunt-field resistance(in Ohms)\n", + "loss=950.0 #Iron and frictional losses(in Watts)\n", + "\n", + "\n", + "#Calculations:\n", + "Ish=V/Rsh\n", + "Ia=I_L+Ish\n", + "Eg=V+(Ia*Ra)\n", + "copp=(Ia*Ia*Ra)+(V*Ish)\n", + "Po=V*I_L\n", + "tot_loss=copp+loss\n", + "Pin=Po+tot_loss\n", + "Pe=Pin-loss\n", + "mech_effi=Pe/Pin\n", + "ele_effi=Po/Pe\n", + "comm_effi=Po/Pin\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The emf generated is %.2f V.\" %(Eg)\n", + "print \"(b)The copper losses is %.2f W.\" %(copp)\n", + "print \"(c)The output of the prime mover is %.3f kW.\" %(Pin/1000.0)\n", + "print \"(d)The commercial efficiency is %.2f.\\n The mechanical efficiency is %.2f.\" %((comm_effi*100),(mech_effi*100))\n", + "print \" The electrical efficiency is %.2f.\" %(ele_effi*100) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The emf generated is 254.00 V.\n", + "(b)The copper losses is 2050.00 W.\n", + "(c)The output of the prime mover is 51.750 kW.\n", + "(d)The commercial efficiency is 94.20.\n", + " The mechanical efficiency is 98.16.\n", + " The electrical efficiency is 95.96.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.28,Page number: 547" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the back emf generated by a dc shunt motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "V=250.0 #Terminal Voltage(in Volts)\n", + "I_L1=2.0 #No-load current(in Amperes)\n", + "N1=1000.0 #No-load speed(in rpm)\n", + "Ra=0.2 #Armature resistance(in Ohms)\n", + "Rsh=250.0 #Field resistance(in Ohms)\n", + "I_L2=51.0 #Current after loading(in Amperes)\n", + "\n", + "\n", + "#Calculations:\n", + "Ish=V/Rsh\n", + "Ia1=I_L1-Ish\n", + "E1=V-(Ia1*Ra)\n", + "Ia2=I_L2-Ish\n", + "E2=V-(Ia2*Ra)\n", + "\"\"\"As the motor is shunt-wound,the flux remains constant.The emf generated is directly proportional to the speed.\"\"\"\n", + "N2=(E2/E1)*N1\n", + "speed_drop=(N1-N2)/N1\n", + "\n", + "\n", + "#Result:\n", + "print \"(a)The back emf generated at no-load is %.3f V.\" %(E1)\n", + "print \"(b)On loading,\\n The back emf generated is %.2f V.\\n The speed of the motor is %d rpm.\" %(E2,round(N2,0))\n", + "print \" The percentage speed drop is %.3f percent.\" %(speed_drop*100)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The back emf generated at no-load is 249.800 V.\n", + "(b)On loading,\n", + " The back emf generated is 240.00 V.\n", + " The speed of the motor is 961 rpm.\n", + " The percentage speed drop is 3.923 percent.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.29,Page number: 547" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the value of starting resistance for a shunt motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "Po=14920.0 #Output power(in Watts)\n", + "V=240.0 #Supply voltage(in Volts)\n", + "Ra=0.25 #Armature resistance(in Ohms)\n", + "effi=0.86 #Efficiency at full-load\n", + "\n", + "\n", + "#Calculations:\n", + "Pin=Po/effi\n", + "I_L=Pin/V\n", + "Ist=1.5*I_L\n", + "Rt=V/Ist\n", + "Rst=Rt-Ra\n", + "\n", + "\n", + "#Result:\n", + "print \"The starting resistance for the shunt motor is %.3f Ohms.\" %(Rst)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The starting resistance for the shunt motor is 1.963 Ohms.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.30,Page number: 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the speed and efficiency of a dc shunt motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "I_L1=4.0 #No-load current(in Amperes)\n", + "N1=1000 #No-load speed(in rpm)\n", + "V=500.0 #Voltage rating of the dc shunt motor(in Volts)\n", + "Ra=0.2 #Armature resistance(in Ohms)\n", + "Ish=1.0 #Field current(in Amperes)\n", + "I_L2=100.0 #Full-load current(in Amperes)\n", + "\n", + "\n", + "#Calculations:\n", + "Ia1=I_L1-Ish\n", + "E1=V-(Ia1*Ra)\n", + "Ia2=I_L2-Ish\n", + "E2=V-(Ia2*Ra)\n", + "\"\"\" For a shunt motor,the flux remains constant and hence E is directly proportional to speed of the motor(N).\n", + " \n", + " E=kN where k is a constant. \"\"\"\n", + "N2=(E2/E1)*N1\n", + "\"\"\"At no-load,the power taken by the motor mainly meets the constant losses(iron and frictional losses).\"\"\"\n", + "Pc=V*I_L1\n", + "\"\"\"On loading,the copper loss in shunt field winding is negligible compared to the copper loss in armature winding.\"\"\"\n", + "Pv=Ia2*Ia2*Ra\n", + "Pin=V*I_L2\n", + "effi=(Pin-(Pv+Pc))/Pin\n", + "\n", + "\n", + "#Result:\n", + "print \"The speed of the dc shunt motor on loading is %d rpm.\" %(round(N2,0))\n", + "print \"The efficiency of the motor is %.2f percent.\" %(effi*100)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of the dc shunt motor on loading is 962 rpm.\n", + "The efficiency of the motor is 92.08 percent.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.31,Page number: 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the speed of a dc generator running as a shunt motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "Ra=0.02 #Armature resistance(in Ohms)\n", + "Rsh=50.0 #Shunt-field resistance(in Ohms)\n", + "V=250.0 #Terminal voltage(in Volts)\n", + "Po=50e03 #Output power(in Watts)\n", + "N1=500.0 #Speed of the dc generator(in rpm)\n", + "\n", + "\n", + "#Calculations:\n", + "Ish=V/Rsh\n", + "\"\"\" When working as a generator,the machine supplies a load of 50 kW at 250 V. \"\"\"\n", + "I_L=Po/V\n", + "Ia1=I_L+Ish\n", + "E1=V+(Ia1*Ra)\n", + "\"\"\"When working as a motor,the machine takes a power of 50 kW at 250 V. \"\"\"\n", + "Ia2=I_L-Ish\n", + "E2=V-(Ia2*Ra)\n", + "N2=(E2/E1)*N1\n", + "\"\"\" NOTE: The field current and the flux per pole is same in both cases.\"\"\"\n", + "\n", + "#Result:\n", + "print \"The speed of the machine running as a shunt motor is %d rpm.\" %(round(N2,0)) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The speed of the machine running as a shunt motor is 484 rpm.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.32,Page number: 548" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Question:\n", + "\"\"\"Finding the applied voltage and the current to run the motor.\"\"\"\n", + "\n", + "#Variable Declaration:\n", + "Ra=0.6 #Resistance of the armature(in Ohms)\n", + "Rse=0.4 #Series field resistance(in Ohms)\n", + "Ia1=20.0 #Initial armature current(in Amperes)\n", + "V1=400.0 #Initial terminal voltage(in Volts)\n", + "N1=250.0 #Initial speed of the motor(in rpm)\n", + "N2=350.0 #Final speed of the motor(in rpm)\n", + "\n", + "\n", + "#Calculations:\n", + "\"\"\" In a series motor,torque is directly proprtional to the square of the armature current(Ia).\n", + " \n", + " Given: Torque is directly proprtional to the square of the speed(N).\n", + " \n", + " Therefore, Ia is directly proportional to N. Ia=kN where k is a constant. \"\"\"\n", + "Ia2=(N2/N1)*Ia1\n", + "E1=V1-(Ia1*(Ra+Rse))\n", + "\"\"\"In a series motor,as the flux is directly proportional to Ia,the back emf is proportional to (Ia*N). \"\"\"\n", + "E2=E1*((Ia2*N2)/(Ia1*N1))\n", + "V2=E2+(Ia2*(Ra+Rse))\n", + "\n", + "\n", + "#Result:\n", + "print \"The applied voltage is %.2f V and the current is %.2f A to run the motor at 350 rpm.\" %(V2,Ia2) \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The applied voltage is 772.80 V and the current is 28.00 A to run the motor at 350 rpm.\n" + ] + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit