From c8733e4b6b4bffcddf7eb45ff1c72ccc837aa3af Mon Sep 17 00:00:00 2001 From: Jovina Dsouza Date: Tue, 22 Jul 2014 00:00:04 +0530 Subject: adding book --- .../chapter_22-checkpoint.ipynb | 791 +++++++++++++++++++++ 1 file changed, 791 insertions(+) create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb') diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb new file mode 100755 index 00000000..4bad2773 --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_22-checkpoint.ipynb @@ -0,0 +1,791 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Chapter 22: Three-phase induction motors

" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 1, page no. 389

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the synchronous speed of the motor in rev/min.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "f = 50;# in Hz\n", + "p = 2/2;# number of pairs of poles\n", + "\n", + "#calculation:\n", + " #ns is the synchronous speed, \n", + " #f is the frequency in hertz of the supply to the stator and \n", + " #p is the number of pairs of poles.\n", + "ns = f/p\n", + "nsrpm = ns*60\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\nsynchronous speed of the motor is \",nsrpm,\" rev/min\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "synchronous speed of the motor is 3000.0 rev/min" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 2, page no. 389

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the number of poles.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "f = 60;# in Hz\n", + "ns = 900/60;# in rev/sec\n", + "\n", + "#calculation:\n", + " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n", + " #p is the number of pairs of poles.\n", + "p = f/ns\n", + "np = p*2\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\nnumber of poles is \", round(np,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "number of poles is 8.0" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 3, page no. 390

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the frequency of the supply voltage.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "p = 2/2;# number of pairs of poles\n", + "ns = 6000/60;# in rev/sec\n", + "\n", + "#calculation:\n", + " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n", + " #\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"frequency is \",f,\" Hz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "frequency is 100.0 Hz" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 4, page no. 391

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the synchronous speed and (b) the slip at full load.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "p = 4/2;# number of pairs of poles\n", + "f = 50;# in Hz\n", + "nr = 1455/60;# in rev/sec\n", + "\n", + "#calculation:\n", + " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n", + " #p is the number of pairs of poles.\n", + "ns = f/p\n", + " #The slip, s\n", + "s = ((ns - nr)/ns)*100# in percent\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n", + "print \"\\n(b) slip is \",s,\" percent\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(a) synchronous speed is 25.0 rev/sec\n", + "\n", + "(b) slip is 3.0 percent" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 5, page no. 392

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the synchronous speed,\n", + "#(b) the speed of the rotor and (c) the frequency of the induced e.m.f.\u2019s in the rotor\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "p = 2/2;# number of pairs of poles\n", + "f = 60;# in Hz\n", + "s = 0.02;# slip\n", + "\n", + "#calculation:\n", + " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n", + " #p is the number of pairs of poles.\n", + "ns = f/p\n", + " #The the rotor runs at\n", + "nr = ns*(1 - s)\n", + " #frequency of the e.m.f. induced in the rotor bars is\n", + "fr = ns - nr\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(a) synchronous speed is \",ns,\" rev/sec\"\n", + "print \"\\n(b) rotor speed is \",nr,\" rev/sec\"\n", + "print \"\\n(c) frequency of the e.m.f. induced in the rotor bars is is \",fr,\" Hz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(a) synchronous speed is 60.0 rev/sec\n", + "\n", + "(b) rotor speed is 58.8 rev/sec\n", + "\n", + "(c) frequency of the e.m.f. induced in the rotor bars is is 1.2 Hz" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 6, page no. 392

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the synchronous speed.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "f = 50;# in Hz\n", + "nr = 1200/60;# in rev/min\n", + "s = 0.04;# slip\n", + "\n", + "#calculation:\n", + " #the synchronous speed.\n", + "ns = nr/(1 - s)\n", + "nsrpm = ns*60\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n synchronous speed is \",nsrpm,\" rev/min\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " synchronous speed is 1250.0 rev/min" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 7, page no. 394

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the slip, and (b) the rotor speed.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "p = 8/2;# number of pairs of poles\n", + "f = 50;# in Hz\n", + "fr = 3;# in Hz\n", + "\n", + "#calculation:\n", + " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n", + " #p is the number of pairs of poles.\n", + "ns = f/p\n", + " #fr = s*f\n", + "s = (fr/f)\n", + " #the rotor speed.\n", + "nr = ns*(1 - s)\n", + "nrrpm = nr*60\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(a) slip is \",s*100,\" percent\"\n", + "print \"\\n (b) rotor speed is \",nrrpm,\" rev/min\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(a) slip is 6.0 percent\n", + "\n", + " (b) rotor speed is 705.0 rev/min" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 8, page no. 396

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the rotor copper loss, \n", + "#(b) the total mechanical power developed by the rotor,\n", + "#(c) the output power of the motor if friction and windage losses are 750 W, and \n", + "#(d) the efficiency of the motor, neglecting rotor iron loss.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Psi = 32000;# in Watts\n", + "Psl = 1200;# in Watts\n", + "s = 0.05;# slip\n", + "Pfl = 750;# in Watts\n", + "\n", + "#calculation:\n", + " #Input power to rotor = stator input power -\u0006 stator losses\n", + "Pi = Psi - Psl\n", + " #slip = rotor copper loss/rotor input\n", + "Pl = s*Pi\n", + " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n", + "Pr = Pi - Pl\n", + " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n", + "Po = Pr - Pfl\n", + " #Efficiency of induction motor = (output power/input power)*100\n", + "eff = (Po/Psi)*100# in percent\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n", + "print \"\\n(b) Total mechanical power developed by the rotor is \",Pr,\" W\"\n", + "print \"\\n(c) Output power of motor is \",Po,\" Watt\"\n", + "print \"\\n(d) efficiency of induction motor is \",round(eff,2),\" percent\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(a) rotor copper loss is 1540.0 Watt\n", + "\n", + "(b) Total mechanical power developed by the rotor is 29260.0 W\n", + "\n", + "(c) Output power of motor is 28510.0 Watt\n", + "\n", + "(d) efficiency of induction motor is 89.09 percent" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 9, page no. 397

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the rotor copper loss, and (b) the efficiency of the motor.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Psi = 32000;# in Watts\n", + "Psl = 1200;# in Watts\n", + "Pfl = 750;# in Watts\n", + "x = 0.35;\n", + "\n", + "#calculation:\n", + " #The slip, s\n", + "s = 1-x\n", + " #Input power to rotor = stator input power - stator losses\n", + "Pi = Psi - Psl\n", + " #slip = rotor copper loss/rotor input\n", + "Pl = s*Pi\n", + " #Total mechanical power developed by the rotor = rotor input power -\u0006 rotor losses\n", + "Pr = Pi - Pl\n", + " #Output power of motor = power developed by the rotor -\u0006 friction and windage losses\n", + "Po = Pr - Pfl\n", + " #Efficiency of induction motor = (output power/input power)*100\n", + "eff = (Po/Psi)*100# in percent\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(a) rotor copper loss is \",Pl,\" Watt\"\n", + "print \"\\n(b) efficiency of induction motor is \",round(eff,2),\" percent\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(a) rotor copper loss is 20020.0 Watt\n", + "\n", + "(b) efficiency of induction motor is 31.34 percent" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 10, page no. 398

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate (a) the synchronous speed, (b) the slip, \n", + "#(c) the full load torque, (d) the power output if mechanical losses amount to 770 W, \n", + "#(e) the maximum torque, (f) the speed at which maximum torque occurs,\n", + "#and (g) the starting torque.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "V = 415;# in Volts\n", + "f = 50 ;# in Hz\n", + "nr = 24;# in rev/sec\n", + "p = 4/2;# no. of pole pairs\n", + "R2 = 0.35;# in Ohms\n", + "X2 = 3.5;# in Ohms\n", + "tr = 0.85;# turn ratio N2/N1\n", + "Pl = 770;# in Watt\n", + "m = 3;# no. of phases\n", + "\n", + "#calculation:\n", + " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n", + " #p is the number of pairs of poles.\n", + "ns = f/p\n", + " #The slip, s\n", + "s = ((ns - nr)/ns)*100# in percent\n", + " #Phase voltage, E1 = V/(3**0.5)\n", + "E1 = V/(3**0.5)\n", + " #Full load torque\n", + "T = (m*(tr**2)/(2*math.pi*ns))*((s/100)*E1*E1*R2/(R2*R2 + (X2*(s/100))**2))\n", + " #Output power, including friction losses\n", + "Pm = 2*math.pi*nr*T\n", + " #power output\n", + "Po = Pm - Pl\n", + " #Maximum torque occurs when R2 = Xr = 0.35 ohm\n", + " #Slip \n", + "sm = R2/X2\n", + " #maximum torque, Tm \n", + "Tm = (m*(tr**2)/(2*math.pi*ns))*(sm*E1*E1*R2/(R2*R2 + (X2*sm)**2))\n", + " #speed at which maximum torque occurs\n", + "nrm = ns*(1 - sm)\n", + "nrmrpm = nrm*60\n", + " #At the start, i.e., at standstill, slip, s=1\n", + "ss = 1\n", + " #starting torque\n", + "Ts = (m*(tr**2)/(2*math.pi*ns))*(ss*E1*E1*R2/(R2*R2 + (X2*ss)**2))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(a)Synchronous speed is \",round(ns,2),\" rev/sec\"\n", + "print \"\\n(b)Slip is \",round(s,2),\" percent\"\n", + "print \"\\n(c)Full load torque is \",round(T,2),\" Nm\"\n", + "print \"\\n(d)power output is \",round(Po,2),\"W\"\n", + "print \"\\n(e)maximum torque is \",round(Tm,2),\" Nm\"\n", + "print \"\\n(f)speed at which maximum torque occurs is \",round(nrmrpm,2),\"rev/min\"\n", + "print \"\\n(g)starting torque is \",round(Ts,2),\" Nm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(a)Synchronous speed is 25.0 rev/sec\n", + "\n", + "(b)Slip is 4.0 percent\n", + "\n", + "(c)Full load torque is 78.05 Nm\n", + "\n", + "(d)power output is 10998.99 W\n", + "\n", + "(e)maximum torque is 113.17 Nm\n", + "\n", + "(f)speed at which maximum torque occurs is 1350.0 rev/min\n", + "\n", + "(g)starting torque is 22.41 Nm" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 11, page no. 400

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine for the induction motor in problem 10 at full load, \n", + "#(a) the rotor current, (b) the rotor copper loss, and (c) the starting current.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "V = 415;# in Volts\n", + "f = 50 ;# in Hz\n", + "nr = 24;# in rev/sec\n", + "p = 4/2;# no. of pole pairs\n", + "R2 = 0.35;# in Ohms\n", + "X2 = 3.5;# in Ohms\n", + "tr = 0.85;# turn ratio N2/N1\n", + "m = 3;# no. of phases\n", + "\n", + "#calculation:\n", + " #ns is the synchronous speed, f is the frequency in hertz of the supply to the stator and \n", + " #p is the number of pairs of poles.\n", + "ns = f/p\n", + " #The slip, s\n", + "s = ((ns - nr)/ns)*100# in percent\n", + " #Phase voltage, E1 = V/(3**0.5)\n", + "E1 = V/(3**0.5)\n", + " #rotor current,\n", + "Ir = (s/100)*E1*tr/((R2**2 + (X2*(s/100))**2)**0.5)\n", + " #Rotor copper loss \n", + "Pcl = m*R2*(Ir**2)\n", + " #starting current,\n", + "ss =1\n", + "I2 = ss*tr*E1/((R2**2 + (X2*ss)**2)**0.5)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(a)rotor current is \",round(Ir,2),\" A\"\n", + "print \"\\n(b)Total copper loss is \",round(Pcl,2),\" W\"\n", + "print \"\\n(c)starting current is \",round(I2,2),\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(a)rotor current is 21.61 A\n", + "\n", + "(b)Total copper loss is 490.37 W\n", + "\n", + "(c)starting current is 57.9 A" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 12, page no. 401

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the power input at full load, \n", + "#(b) the efficiency of the motor at full load and \n", + "#(c) the current taken from the supply at full load, if the motor runs at a power factor of 0.87 lagging.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "V = 415;# in Volts\n", + "Psl = 650;# in Watt\n", + "pf = 0.87;# power factor\n", + "\n", + "#calculation:\n", + "Pm = 11770;# watts from part (d), Problem 22.10\n", + "Pcl = 490.35;# watts, Rotor copper loss, from part (b), Problem 22.11\n", + " #Stator input power\n", + "P1 = Pm + Pcl + Psl\n", + "Po = 11000# watts, Net power output, from part (d), Problem 22.10\n", + " #efficiency = (output/input) *100\n", + "eff = (Po/P1)*100# in percent\n", + " #Power input, P1 = (3**0.5)*VL*IL*cos(phi)\u000e\n", + " # pf = cos(phi)\n", + " #supply current, IL\n", + "I = P1/((3**0.5)*V*pf)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(aStator input power is \",round(P1,2),\" W\"\n", + "print \"\\n(b)efficiency is \",round(eff,2),\" percent\"\n", + "print \"\\n(c)supply current is \",round(I,2),\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(aStator input power is 12910.35 W\n", + "\n", + "(b)efficiency is 85.2 percent\n", + "\n", + "(c)supply current is 20.64 A" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 13, page no. 401

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the resistance of the rotor winding required for maximum starting torque.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "V = 415;# in Volts\n", + "f = 50 ;# in Hz\n", + "nr = 24;# in rev/sec\n", + "p = 4/2;# no. of pole pairs\n", + "R2 = 0.35;# in Ohms\n", + "X2 = 3.5;# in Ohms\n", + "\n", + "#calculation:\n", + " #At the moment of starting, slip, \n", + "s = 1\n", + " #Maximum torque occurs when rotor reactance equals rotor resistance\n", + " #for maximum torque\n", + "R2 = s*X2\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\nresistance of the rotor is \",R2,\" Ohm\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "resistance of the rotor is 3.5 Ohm" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit