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| author | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
|---|---|---|
| committer | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
| commit | fffcc90da91b66ee607066d410b57f34024bd1de (patch) | |
| tree | 7b8011d61013305e0bf7794a275706abd1fdb0d3 /Basic_Electrical_Engineering/Chapter16.ipynb | |
| parent | 299711403e92ffa94a643fbd960c6f879639302c (diff) | |
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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 |