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diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_2.ipynb deleted file mode 100755 index 28433f9f..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_21-checkpoint_2.ipynb +++ /dev/null @@ -1,1641 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 21: D.c. machines</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 600;# no. of conductors\n",
- "c = 2;# for a wave winding\n",
- "p = 4;# no. of pairs\n",
- "n = 625/60;# in rev/sec\n",
- "Phi = 20E-3;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the generated e.m.f is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the generated e.m.f is 500.0 V "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the speed at which the machine must be driven to generate an e.m.f. of 240 V\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 50*16;# no. of conductors\n",
- "p = 1;# let no. of pairs\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 30E-3;# in Wb\n",
- "E = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- " #Rearranging gives, speed\n",
- "n = E*c/(2*p*Phi*Z)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the speed at which the machine must be driven is \",round(n,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the speed at which the machine must be driven is 10.0 rev/sec "
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 354</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the e.m.f. generated when running at 500 rev/min.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 1200;# no. of conductors\n",
- "p = 1;# let, no. of pairs\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 30E-3;# in Wb\n",
- "n = 500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 300.0 V "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 355</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Z = 1200;# no. of conductors\n",
- "p = 4;# let, no. of pairs\n",
- "c = 2;# for a wave winding\n",
- "Phi = 30E-3;# in Wb\n",
- "n = 500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #Generated e.m.f., E = 2*p*Phi*n*Z/c\n",
- "E = 2*p*Phi*n*Z/c\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 1200.0 V "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 355</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the change in the generated voltage when the field current is reduced by 20%, \n",
- "#assuming the flux is proportional to the field current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 150; # in Volts\n",
- "x = 0.2;\n",
- "\n",
- "#calculation:\n",
- "E2 = E1*(1- x)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Generated e.m.f. is \",round(E2,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Generated e.m.f. is 120.0 V "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 356</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the percentage increase in the flux per pole required to generate 250 V at 20 rev/s.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "n1 = 30;# in rev/sec\n",
- "E1 = 200;# in Volts\n",
- "n2 = 20;# in rev/sec\n",
- "E2 = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then\n",
- " # E proportional to phi*n\n",
- " # E1/E2 = Phi1*n1/(Phi2*n2)\n",
- " # let Phi2/Phi1 = Phi\n",
- "Phi = E2*n1/(E1*n2)\n",
- "Phi_inc = (Phi - 1)*100#/in percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n percentage increase in the flux per pole is \",round(Phi_inc,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " percentage increase in the flux per pole is 87.5 percent "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 7, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the terminal voltage of a generator\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.30;# in ohms\n",
- "Ia = 30;# in Amperes\n",
- "E = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #terminal voltage,\n",
- " #V = E - Ia*Ra\n",
- "V = E - Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n terminal voltage of a generator is \",round(V,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " terminal voltage of a generator is 191.0 V "
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the terminal voltage, and (b) the generated e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "RL = 60;# in ohms\n",
- "Ia = 8;# in Amperes\n",
- "Ra = 1;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #terminal voltage,\n",
- " #V = Ia*RL\n",
- "V = Ia*RL\n",
- " #Generated e.m.f., E\n",
- "E = V + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)terminal voltage is \",round(V,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)terminal voltage is 480.0 V \n",
- "\n",
- " (b)generated e.m.f. is 488.0 V "
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 357</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the generated e.m.f. when \n",
- "#(a) the speed increases to 25 rev/s and the pole flux remains unchanged, \n",
- "#(b) the speed remains at 20 rev/s and the pole flux is decreased to 0.08 Wb,\n",
- "#and (c) the speed increases to 24 rev/s and the pole flux is decreased to 0.07 Wb.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "E1 = 150;# in Volts\n",
- "n1 = 20;# in rev/sec\n",
- "Phi1 = 0.10;# in Wb\n",
- "n2 = 25;# in rev/sec\n",
- "Phi2 = 0.10;# in Wb\n",
- "n3 = 20;# in rev/sec\n",
- "Phi3 = 0.08;# in Wb\n",
- "n4 = 24;# in rev/sec\n",
- "Phi4 = 0.07;# in Wb\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E proportional to phi*w and since w = 2*pi*n, then\n",
- " # E proportional to phi*n\n",
- " # E1/E2 = Phi1*n1/(Phi2*n2)\n",
- "E2 = E1*Phi2*n2/(Phi1*n1)\n",
- "E3 = E1*Phi3*n3/(Phi1*n1)\n",
- "E4 = E1*Phi4*n4/(Phi1*n1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)the generated e.m.f is \",round(E2,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E3,2),\" V \"\n",
- "print \"\\n (c)generated e.m.f. is \",round(E4,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)the generated e.m.f is 187.5 V \n",
- "\n",
- " (b)generated e.m.f. is 120.0 V \n",
- "\n",
- " (c)generated e.m.f. is 126.0 V "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 359</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the terminal voltage, and (b) the e.m.f. generated in the armature.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ps = 20000;# in Watts\n",
- "Vs = 200;# in Volts\n",
- "Rs = 0.1;# in ohms\n",
- "Rf = 50;# in ohms\n",
- "Ra = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Load current, I\n",
- "Is = Ps/Vs\n",
- " #Volt drop in the cables to the load\n",
- "Vd = Is*Rs\n",
- " #Hence terminal voltage,\n",
- "V = Vs + Vd\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current\n",
- "Ia = If + Is\n",
- " #Generated e.m.f. E\n",
- "E = V + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)terminal voltage is \",round(V,2),\" V \"\n",
- "print \"\\n (b)generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)terminal voltage is 210.0 V \n",
- "\n",
- " (b)generated e.m.f. is 214.17 V "
- ]
- }
- ],
- "prompt_number": 9
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 361</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the e.m.f. generated.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Is = 80;# in amperes\n",
- "Vs = 200;# in Volts\n",
- "Rf = 40;# in ohms\n",
- "Rse = 0.02;# in ohms\n",
- "Ra = 0.04;# in ohms\n",
- "\n",
- "#calculation:\n",
- " #Volt drop in series winding\n",
- "Vse = Is*Rse\n",
- " #P.d. across the field winding = p.d. across armature\n",
- "V1 = Vs + Vse\n",
- " #Field current, If\n",
- "If = V1/Rf\n",
- " #Armature current\n",
- "Ia = If + Is\n",
- " #Generated e.m.f. E\n",
- "E = V1 + Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n generated e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " generated e.m.f. is 205.0 V "
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 363</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the efficiency of the generator at full load\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ps = 10000;# in Watt\n",
- "Pl = 600;# in Watt\n",
- "Ra = 0.75;# in ohms\n",
- "Rf = 125;# in ohms\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Output power Ps = V*I\n",
- " #from which, load current I\n",
- "I = Ps/V\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current\n",
- "Ia = If + I\n",
- " #Efficiency,\n",
- "eff = Ps*100/((V*I) + (Ia*Ia*Ra) + (If*V) + (Pl))# in Percent\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Efficiency is \",round(eff,2),\" percent \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Efficiency is 80.5 percent "
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 364</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the back e.m.f. when the armature current is 50 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.2;# in ohms\n",
- "V = 240;# in Volts\n",
- "Ia = 50;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #For a motor, V = E + Ia*Ra\n",
- "E = V - Ia*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n back e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " back e.m.f. is 230.0 V "
- ]
- }
- ],
- "prompt_number": 14
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 365</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate the e.m.f. generated when it is running: (a) as a generator giving 100 A, and\n",
- "#(b) as a motor taking 80 A.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ra = 0.25;# in ohms\n",
- "V = 300;# in Volts\n",
- "Ig = 100;# in Amperes\n",
- "Im = 80;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #As a generator, generated e.m.f.,\n",
- " # E = V + Ia*Ra\n",
- "Eg = V + Ig*Ra\n",
- " #For a motor, generated e.m.f. (or back e.m.f.),\n",
- " # E = V - Ia*Ra\n",
- "E = V - Im*Ra\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)As a generator, generated e.m.f. is \",round(Eg,2),\" V \"\n",
- "print \"\\n (b)back e.m.f. is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)As a generator, generated e.m.f. is 325.0 V \n",
- "\n",
- " (b)back e.m.f. is 280.0 V "
- ]
- }
- ],
- "prompt_number": 15
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the torque exerted when a current of 30 A flows in each armature conductor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 4;\n",
- "c = 2;# for a wave winding\n",
- "Phi = 25E-3;# Wb\n",
- "Z = 900;\n",
- "Ia = 30;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #torque T = p*Phi*Z*Ia/(pi*c)\n",
- "T = p*Phi*Z*Ia/(1*math.pi*c)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the torque exerted is 429.72 Nm "
- ]
- }
- ],
- "prompt_number": 16
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the torque\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "V = 350;# in Volts\n",
- "Ra = 0.5;# in ohms\n",
- "n = 15;# in rev/sec\n",
- "Ia = 60;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #Back e.m.f. E = V - Ia*Ra\n",
- "E = V - Ia*Ra\n",
- " #torque T = E*Ia/(2*n*pi)\n",
- "T = E*Ia/(2*n*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the torque exerted is 203.72 Nm "
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 366</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Calculate (a) the speed and (b) the torque developed when the armature current is 40 A\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "p = 1;# let\n",
- "c = 2*p;# for a lap winding\n",
- "Phi = 20E-3;# Wb\n",
- "Z = 500;\n",
- "V = 250;# in Volts\n",
- "Ra = 1;# in ohms\n",
- "Ia = 40;# in Amperes\n",
- "\n",
- "#calculation:\n",
- " #Back e.m.f. E = V - Ia*Ra\n",
- "E = V - Ia*Ra\n",
- " #E.m.f. E = 2*p*Phi*n*Z/c\n",
- " # rearrange,\n",
- "n = E*c/(2*p*Phi*Z)\n",
- " #torque T = E*Ia/(2*n*pi)\n",
- "T = E*Ia/(2*n*math.pi)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)speed n is \",round(n,2),\" rev/sec \"\n",
- "print \"\\n (b)the torque exerted is \",round(T,2),\" Nm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)speed n is 21.0 rev/sec \n",
- "\n",
- " (b)the torque exerted is 63.66 Nm "
- ]
- }
- ],
- "prompt_number": 18
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 18, page no. 367</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the armature current at this new value of torque.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T1 = 25;# in Nm\n",
- "T2 = 35;# in Nm\n",
- "Ia1 = 16;# in Amperes\n",
- "V = 100;# in Volts\n",
- "x = 0.15;\n",
- "\n",
- " #calculation:\n",
- " #the shaft torque T of a generator is proportional to (phi*Ia),\n",
- " #where Phi is the flux and Ia is the armature current. Thus, T = k*Phi*Ia, where k is a constant.\n",
- " #The torque at flux phi1 and armature current Ia1 is T1 = k*Phi1*Ia1.\n",
- " #similarly T2 = k*Phi2*Ia2\n",
- "\n",
- "Ia2 = T2*Ia1/(0.85*T1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n armature current at the new value of torque is \",round(Ia2,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " armature current at the new value of torque is 26.35 A "
- ]
- }
- ],
- "prompt_number": 19
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 19, page no. 367</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the efficiency of the generator and (b) the power loss in the generator.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = 12;# in Nm\n",
- "I = 15;# in Amperes\n",
- "V = 100;# in Volts\n",
- "n = 1500/60;# in rev/sec\n",
- "\n",
- "#calculation:\n",
- " #the efficiency of a generator = (output power/input power)*100 %\n",
- " #The output power is the electrical output, i.e. VI watts. \n",
- " #The input power to a generator is the mechanical power in the shaft driving the generator, \n",
- " #i.e. T*w or T(2*pi*n) watts, where T is the torque in Nm and n is speed of rotation in rev/s. Hence, for a generator \n",
- " #efficiency = V*I*100/(T*2*pi*n) %\n",
- "eff = V*I*100/(T*2*math.pi*n)# in Percent\n",
- " #The input power = output power + losses\n",
- " # hence, T*2*math.pi*n = V*I + losses\n",
- "Pl = T*2*math.pi*n - V*I\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) efficiency is \",round(eff,2),\" % \"\n",
- "print \"\\n (b) power loss is \",round(Pl,2),\" W \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) efficiency is 79.58 % \n",
- "\n",
- " (b) power loss is 384.96 W "
- ]
- }
- ],
- "prompt_number": 20
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 20, page no. 368</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the current in the armature, and (b) the back e.m.f.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Rf = 150;# in Ohms\n",
- "Ra = 0.4;# in Ohms\n",
- "I = 30;# in Amperes\n",
- "V = 240;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Field current If\n",
- "If = V/Rf\n",
- " #Supply current I = Ia + If\n",
- " #Hence armature current, Ia\n",
- "Ia = I - If\n",
- " #Back e.m.f. E = V -\u0004 Ia*Ra\n",
- "E = V - (Ia*Ra)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a) current in the armature is \",round(Ia,2),\" A \"\n",
- "print \"\\n (b) Back e.m.f. E is \",round(E,2),\" V \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a) current in the armature is 28.4 A \n",
- "\n",
- " (b) Back e.m.f. E is 228.64 V "
- ]
- }
- ],
- "prompt_number": 21
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 21, page no. 370</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the speed of the motor, assuming the flux remains constant.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 30;# in Amperes\n",
- "Ia2 = 45;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "n1 = 1350/60;# in Rev/sec\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #The relationship E proportional to (Phi*n) applies to both generators and motors. For a motor,\n",
- " #E = V - (Ia*Ra)\n",
- "E1 = V - (Ia1*Ra)\n",
- "E2 = V - (Ia2*Ra)\n",
- " #The relationship, E1/E2 = Phi1*n1/Phi2*n2, applies to both generators and motors.\n",
- " #Since the flux is constant, Phi1 = Phi2\n",
- "n2 = E2*n1/(E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n the speed of the motor is \",round(n2,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " the speed of the motor is 21.78 rev/sec "
- ]
- }
- ],
- "prompt_number": 22
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 22, page no. 370</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the maximum value of armature current if the flux is suddenly reduced by 10% and \n",
- "#(b) the steadystate value of the armature current at the new value of flux,\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 30;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "n = 800/60;# in Rev/sec\n",
- "V = 220;# in Volts\n",
- "x= 0.1;\n",
- "\n",
- "#calculation:\n",
- " #For a d.c. shunt-wound motor, E = V - (Ia*Ra),Hence initial generated e.m.f.,\n",
- "E1 = V - (Ia1*Ra)\n",
- " #The generated e.m.f. is also such that E proportional to (Phi*n) \n",
- " #so at the instant the flux is reduced, the speed has not had time to change, and\n",
- "E = E1*(1-x)\n",
- " #Hence, the voltage drop due to the armature resistance is\n",
- "Vd = V - E\n",
- " #The instantaneous value of the current is\n",
- "Ia = Vd/Ra\n",
- " #T proportional to (Phi*Ia), since the torque is constant,\n",
- " #Phi1*Ia1 = Phi2*Ia2, The flux 8 is reduced by 10%, hence\n",
- "Ia2 = Ia1/0.9\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)instantaneous value of the current \",round(Ia,2),\" A \"\n",
- "print \"\\n (b)steady state value of armature current, \",round(Ia2,2),\" A \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)instantaneous value of the current 82.0 A \n",
- "\n",
- " (b)steady state value of armature current, 33.33 A "
- ]
- }
- ],
- "prompt_number": 23
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 23, page no. 372</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the generated e.m.f. at this load. \n",
- "#(b) Calculate the speed of the motor when the load is changed such that the current is increased to 30 A.\n",
- "#Assume that this causes a doubling of the flux.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 15;# in Amperes\n",
- "Ia2 = 30;# in Amperes\n",
- "Rf = 0.3;# in ohms\n",
- "Ra = 0.2;# in ohm\n",
- "n1 = 24;# in Rev/sec\n",
- "V = 240;# in Volts\n",
- "x= 2;\n",
- "\n",
- "#calculation:\n",
- " #generated e.m.f., E, at initial load, is given by\n",
- "E1 = V - Ia1*(Ra + Rf)\n",
- " #When the current is increased to 30 A, the generated e.m.f. is given by:\n",
- "E2 = V - Ia2*(Ra + Rf)\n",
- " #E proportional to (Phi*n)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = E2*n1/(2*E1) \n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)generated e.m.f., E is \",round(E1,2),\" V \"\n",
- "print \"\\n (b)speed of motor, n2, \",round(n2,2),\" rev/sec \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)generated e.m.f., E is 232.5 V \n",
- "\n",
- " (b)speed of motor, n2, 11.61 rev/sec "
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 24, page no. 374</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the overall efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 80;# in Amperes\n",
- "C = 1500;# in Watt\n",
- "Rf = 40;# in ohms\n",
- "Ra = 0.2;# in ohm\n",
- "n = 1000/60;# in Rev/sec\n",
- "V = 320;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Field current, If\n",
- "If = V/Rf\n",
- " #Armature current Ia\n",
- "Ia = I - If\n",
- " #Efficiency\n",
- "eff = ((V*I - (Ia*Ia*Ra) - (If*V) - C)/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 80.09 %"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 25, page no. 374</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the maximum efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "I = 40;# in Amperes\n",
- "Rf = 0.05;# in ohms\n",
- "Ra = 0.15;# in ohm\n",
- "V = 250;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #However for a series motor, If = 0 and the Ia*Ia*Ra loss needs to be I*I*(\u0011Ra + Rf)\n",
- " #For maximum efficiency I*I*\u0011(Ra + Rf) = C\n",
- " #Efficiency\n",
- "eff = ((V*I - (2*I*I*(Ra + Rf)))/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 93.6"
- ]
- }
- ],
- "prompt_number": 25
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 26, page no. 375</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the current supplied to the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "T = 15;# in Nm\n",
- "n = 1200/60;# in rev/sec\n",
- "eff = 0.8;\n",
- "V = 200;# in Volts\n",
- "\n",
- "#calculation:\n",
- "I = T*2*math.pi*n/(V*eff)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n current supplied, I is \",round(I,2),\"A\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " current supplied, I is 11.78 A"
- ]
- }
- ],
- "prompt_number": 26
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 27, page no. 376</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the efficiency of the motor.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "R = 2;# in ohm\n",
- "n = 30;# in rev/sec\n",
- "I = 10;# in A\n",
- "C = 300;# in Watt\n",
- "V = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #Efficiency\n",
- "eff = ((V*I - (I*I*R) - C)/(V*I))*100 # in percent\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n efficiency is\",round(eff,2),\"%\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " efficiency is 87.5 %"
- ]
- }
- ],
- "prompt_number": 27
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 28, page no. 378</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) Determine the speed when the current is 60 A and a resistance of 0.5 ohmis connected in series with the armature, \n",
- "#the shunt field remaining constant. \n",
- "#(b) Determine the speed when the current is 60 A and the shunt field is reduced to 80% of its normal value \n",
- "#by increasing resistance in the field circuit.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 120;# in A\n",
- "Ia2 = 60;# in A\n",
- "Ra = 0.2;# in ohm\n",
- "n1 = 10;# in rev/sec\n",
- "R = 0.5;# in ohm\n",
- "x = 0.8;\n",
- "V = 500;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #back e.m.f. at Ia1\n",
- "E1 = V - Ia1*Ra\n",
- " #at Ia2\n",
- "E2 = V - Ia2*(Ra + R)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = n1*E2/E1\n",
- " #Back e.m.f. when Ia2\n",
- "E3 = V - Ia2*Ra\n",
- "n3 = n1*E3/(x*E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n (a)speed n2 is \",round(n2,2),\" rev/sec\"\n",
- "print \"\\n (b)speed n3 is \",round(n3,2),\" rev/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " (a)speed n2 is 9.62 rev/sec\n",
- "\n",
- " (b)speed n3 is 12.82 rev/sec"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 29, page no. 379</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the speed when developing full load torque but with a 0.2 ohm diverter in parallel with the field winding.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 90;# in Amperes\n",
- "Ra = 0.1;# in ohm\n",
- "Rse = 0.05;# in ohm\n",
- "Rd = 0.2;# in Ohm\n",
- "n1 = 15;# in rev/sec\n",
- "V = 300;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1\n",
- "E1 = V - Ia1*(Ra + Rse)\n",
- " #With the Rd diverter in parallel with Rse\n",
- " #equivalent resistance, Re\n",
- "Re = Rd*Rse/(Rd+Rse)\n",
- " #Torque, T proprtional to Ia*Phi and for full load torque, Ia1*Phi1 = Ia2*Phi2\n",
- " #Since flux is proportional to field current Phi1 proportional to Ta1 and Phi2 Proportional to I1\n",
- "I1 = (Ia1*Ia1*0.8)**0.5\n",
- " #By current division, current I1\n",
- "Ia2 = I1/(Rd/(Rd + Rse))\n",
- " #Hence e.m.f. E2\n",
- "E2 = V - Ia2*(Ra + Re)\n",
- " #E1/E2 = Phi1*n1/Phi2*n2\n",
- "n2 = E2*Ia1*n1/(I1*E1)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n speed n2 is \",round(n2,2),\" rev/sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " speed n2 is 16.74 rev/sec"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 30, page no. 380</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the resistance to be connected in series to reduce the speed to 600 rev/min with the same current.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "Ia1 = 25;# in Amperes\n",
- "Ra = 0.4;# in ohm\n",
- "Rse = 0.2;# in ohm\n",
- "n1 = 800/60;# in rev/sec\n",
- "n2 = 600/60;# in rev/sec\n",
- "V = 400;# in Volts\n",
- "\n",
- "#calculation:\n",
- " #e.m.f. E1\n",
- "E1 = V - Ia1*(Ra + Rse)\n",
- " #At n2, since the current is unchanged, the flux is unchanged.\n",
- " #E1/E2 = n1/n2\n",
- "E2 = E1*n2/n1\n",
- " #and E2 = V - Ia1(\u0011Ra + Rse + R)\n",
- "R = (V - E2)/Ia1 - Ra - Rse\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n Resistance is \",round(R,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " Resistance is 3.85 ohm"
- ]
- }
- ],
- "prompt_number": 30
- }
- ],
- "metadata": {}
- }
- ]
-}
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