From efb9ead5d9758d5d0bed7a22069320b14f972e40 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Fri, 25 Jul 2014 12:37:07 +0530 Subject: adding books --- .../chapter_19-checkpoint_1.ipynb | 974 +++++++++++++++++++++ 1 file changed, 974 insertions(+) create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb') diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb new file mode 100755 index 00000000..8285878f --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_19-checkpoint_1.ipynb @@ -0,0 +1,974 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Chapter 19: Three phase systems

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

Example 1, page no. 299

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the system phase voltage, (b) the phase current and (c) the line current.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Vl = 415;# in Volts\n", + "Rp = 30;# in ohms\n", + "\n", + "#calculation:\n", + "Vp = Vl/(3**0.5)\n", + "Ip = Vp/Rp\n", + "Il = Ip\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)the system phase voltage is \",round(Vp,2),\" V\"\n", + "print \"\\n (b)phase current is \",round(Ip,2),\" A\"\n", + "print \"\\n (c)line current is \",round(Il,2),\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)the system phase voltage is 239.6 V\n", + "\n", + " (b)phase current is 7.99 A\n", + "\n", + " (c)line current is 7.99 A" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 2, page no. 299

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate the line voltage if the supply frequency is 50 Hz\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "R = 30;# in ohms\n", + "L = 0.1273;# in Henry\n", + "Ip = 5.08;# in Amperes\n", + "f = 50;# in Hz\n", + "\n", + "#calculation:\n", + "XL = 2*math.pi*f*L\n", + "Zp = (R*R + XL*XL)**0.5\n", + "Il = Ip\n", + "Vp = Ip*Zp\n", + "Vl = Vp*(3**0.5)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)line voltage is \",round(Vl,2),\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)line voltage is 439.89 V" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 4, page no. 301

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the current in each line and (b) the current in the neutral conductor.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "V = 415;# in Volts\n", + "PR = 24000;# in Watt\n", + "Py = 18000;# in Watt\n", + "Pb = 12000;# in Watt\n", + "VR = 240;# in Volts\n", + "Vy = 240;# in Volts\n", + "Vb = 240;# in Volts\n", + "\n", + "#calculation:\n", + " #For a star-connected system VL = Vp*(3**0.5)\n", + "Vp = V/(3**0.5)\n", + "phir = 90*math.pi/180\n", + "phiy = 330*math.pi/180\n", + "phib = 210*math.pi/180\n", + " # I = P/V for a resistive load\n", + "IR = PR/VR\n", + "Iy = Py/Vy\n", + "Ib = Pb/Vb\n", + "Inh = IR*math.cos(phir) + Ib*math.cos(phib) + Iy*math.cos(phiy)\n", + "Inv = IR*math.sin(phir) + Ib*math.sin(phib) + Iy*math.sin(phiy)\n", + "In = (Inh**2 + Inv**2)**0.5\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)cuurnt in R line is \",round(IR,2),\" A, cuurnt in Y line is \",round(Iy,2),\" A \"\n", + "print \"and cuurnt in B line is \",round(Ib,2),\" A\"\n", + "print \"\\n (b)cuurnt in neutral line is \",round(In,2),\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)cuurnt in R line is 100.0 A, cuurnt in Y line is 75.0 A and cuurnt in B line is 50.0 A\n", + "\n", + " (b)cuurnt in neutral line is 43.3 A" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 5, page no. 302

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the phase current, and (b) the line current.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "R = 30;# in ohms\n", + "L = 0.1273;# in Henry\n", + "VL = 440;# in Volts\n", + "f = 50;# in Hz\n", + "\n", + "#calculation:\n", + "XL = 2*math.pi*f*L\n", + "Zp = (R*R + XL*XL)**0.5\n", + "Vp = VL\n", + " #Phase current\n", + "Ip = Vp/Zp\n", + " #For a delta connection,\n", + "IL = Ip*(3**0.5)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)the phase current \",round(Ip,2),\" A\"\n", + "print \"\\n (b)line current \",round(IL,2),\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)the phase current 8.8 A\n", + "\n", + " (b)line current 15.24 A" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 6, page no. 302

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the capacitance of each of the capacitors.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "IL = 15;# in Amperes\n", + "VL = 415;# in Volts\n", + "f = 50;# in Hz\n", + "\n", + "#calculation:\n", + " #For a delta connection\n", + "Ip = IL/(3**0.5)#phase current\n", + "Vp = VL\n", + " #Capacitive reactance per phase\n", + "Xc = Vp/Ip\n", + "C = 1/(2*math.pi*f*Xc)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n capacitance is \",round(C*1E6,2),\"uF\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " capacitance is 66.43 uF" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 7, page no. 303

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate for each connection (a) the line and phase voltages and (b) the phase and line currents.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "R = 3;# in ohms\n", + "XL = 4;# in ohms\n", + "VL = 415;# in Volts\n", + "\n", + "#calculation:\n", + " #For a star connection:\n", + " #IL = Ip\n", + " #VL = Vp*(3**0.5)\n", + "VLs = VL\n", + "Vps = VLs/(3**0.5)\n", + " #Impedance per phase,\n", + "Zp = (R*R + XL*XL)**0.5\n", + "Ips = Vps/Zp\n", + "ILs = Ips\n", + " #For a delta connection:\n", + " #VL = Vp\n", + " #IL = Ip*(3**0.5)\n", + "VLd = VL\n", + "Vpd = VLd\n", + "Ipd = Vpd/Zp\n", + "ILd = Ipd*(3**0.5)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)the line voltage for star connection is \",round(VLs,2),\" V \"\n", + "print \"and the phase voltage for star connection is \",round(Vps,2),\" V \"\n", + "print \"and the line voltage for delta connection is \",round(VLd,2),\" V \"\n", + "print \"and the phase voltage for delta connection is \",round(Vpd,2),\" V\"\n", + "print \"\\n (b)the line current for star connection is \",round(ILs,2),\" A \"\n", + "print \"and the phase current for star connection is \",round(Ips,2),\" A \"\n", + "print \"and the line current for delta connection is \",round(ILd,2),\" A \"\n", + "print \"and the phase current for delta connection is \",round(Ipd,2),\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)the line voltage for star connection is 415.0 V \n", + "and the phase voltage for star connection is 239.6 V \n", + "and the line voltage for delta connection is 415.0 V \n", + "and the phase voltage for delta connection is 415.0 V\n", + "\n", + " (b)the line current for star connection is 47.92 A \n", + "and the phase current for star connection is 47.92 A \n", + "and the line current for delta connection is 143.76 A \n", + "and the phase current for delta connection is 83.0 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 8, page no. 304

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the total power dissipated by the resistors.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Rp = 12;# in ohms\n", + "VL = 415;# in Volts\n", + "\n", + "#calculation:\n", + " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n", + "Vp = VL/(3**0.5)# since the resistors are star-connected\n", + " #Phase current, Ip\n", + "Zp = Rp\n", + "Ip = Vp/Zp\n", + " #For a star connection\n", + "IL = Ip\n", + " # For a purely resistive load, the power factor cos(phi) = 1\n", + "pf = 1\n", + "P = VL*IL*(3**0.5)*pf\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)total power dissipated by the resistors is \",round(P,2),\" W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)total power dissipated by the resistors is 14352.08 W" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 9, page no. 304

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the power factor of the system.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "P = 5000;# in Watts\n", + "IL = 8.6;# in amperes\n", + "VL = 400;# in Volts\n", + "\n", + "#calculation:\n", + " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n", + "pf = P/(VL*IL*(3**0.5))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n power factor is \",round(pf,3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " power factor is 0.839" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 10, page no. 304

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the total power dissipated in each case.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "R = 10;# in ohms\n", + "L = 0.042;# in Henry\n", + "VL = 415;# in Volts\n", + "f = 50;# in Hz\n", + "\n", + "#calculation:\n", + " #For a star connection:\n", + " #IL = Ip\n", + " #VL = Vp*(3**0.5)\n", + "XL = 2*math.pi*f*L\n", + "Zp = (R*R + XL*XL)**0.5\n", + "VLs = VL\n", + "Vps = VLs/(3**0.5)\n", + " #Impedance per phase,\n", + "Ips = Vps/Zp\n", + "ILs = Ips\n", + " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n", + "pfs = R/Zp\n", + "Ps = VLs*ILs*(3**0.5)*pfs\n", + "\n", + " #For a delta connection:\n", + " #VL = Vp\n", + " #IL = Ip*(3**0.5)\n", + "VLd = VL\n", + "Vpd = VLd\n", + "Ipd = Vpd/Zp\n", + "ILd = Ipd*(3**0.5)\n", + " #Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n", + "pfd = R/Zp\n", + "Pd = VLd*ILd*(3**0.5)*pfd\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n total power dissipated in star is \",round(Ps,2),\" W and in delta is \",round(Pd,2),\" W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " total power dissipated in star is 6283.29 W and in delta is 18849.88 W" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 11, page no. 305

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the power input, (b) the line current and (c) the phase current\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Po = 12750;# in Watts\n", + "pf = 0.77;# power factor\n", + "eff = 0.85;\n", + "VL = 415;# in Volts\n", + "\n", + "#calculation:\n", + " #eff = power_out/power_in\n", + "Pi = Po/eff\n", + " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n", + "IL = Pi/(VL*(3**0.5)*pf)# line current\n", + " #For a delta connection:\n", + " #IL = Ip*(3**0.5)\n", + "Ip = IL/(3**0.5)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)power input is \",round(Pi,2),\" W\"\n", + "print \"\\n (b)line current is \",round(IL,2),\" A\"\n", + "print \"\\n (c)phase current is \",round(Ip,2),\" A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)power input is 15000.0 W\n", + "\n", + " (b)line current is 27.1 A\n", + "\n", + " (c)phase current is 15.65 A" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 13, page no. 308

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "R = 30;# in ohms\n", + "XL = 40;# in ohms\n", + "VL = 400;# in Volts\n", + "\n", + "#calculation:\n", + "Zp = (R*R + XL*XL)**0.5\n", + " #a delta-connected load\n", + "Vp = VL\n", + " #Phase current\n", + "Ip = Vp/Zp\n", + "IL = Ip*(3**0.5)\n", + " #Alternator output power is equal to the power dissipated by the load.\n", + " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n", + "pf = R/Zp\n", + "P = VL*IL*(3**0.5)*pf\n", + " #Alternator output kVA,\n", + "S = VL*IL*(3**0.5)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)the current supplied by the alternator is \",round(IL,2),\" A\"\n", + "print \"\\n (b)output power is \",round(P/1000,2),\"KW and kVA of the alternator is \",round(S/1000,2),\"kVA\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)the current supplied by the alternator is 13.86 A\n", + "\n", + " (b)output power is 5.76 KW and kVA of the alternator is 9.6 kVA" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 14, page no. 308

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate (a) the phase current, (b) the line current, (c) the total power dissipated and \n", + "#(d) the kVA rating of the load. Draw the complete phasor diagram for the load.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "R = 30;# in ohms\n", + "C = 80E-6;# in Farads\n", + "f = 50;# in Hz\n", + "VL = 400;# in Volts\n", + "\n", + "#calculation:\n", + " #Capacitive reactance\n", + "Xc = 1/(2*math.pi*f*C)\n", + "Zp = (R*R + Xc*Xc)**0.5\n", + "pf = R/Zp\n", + " #a delta-connected load\n", + "Vp = VL\n", + " #Phase current\n", + "Ip = Vp/Zp\n", + "IL = Ip*(3**0.5)\n", + " #Power P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n", + "P = VL*IL*(3**0.5)*pf\n", + " #Alternator output kVA,\n", + "S = VL*IL*(3**0.5)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)the phase current is \",round(Ip,2),\" A\"\n", + "print \"\\n (b)the line current is \",round(IL,2),\" A\"\n", + "print \"\\n (c) power is \",round(P/1000,2),\"kW\"\n", + "print \"\\n (d)kVA of the alternator is \", round(S/1000,2),\"kVA\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)the phase current is 8.03 A\n", + "\n", + " (b)the line current is 13.9 A\n", + "\n", + " (c) power is 5.8 kW\n", + "\n", + " (d)kVA of the alternator is 9.63 kVA" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 15, page no. 309

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the total power input and (b) the load power factor.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Pi1 = 8000;# in Watts\n", + "Pi2 = 4000;# in Watts\n", + "\n", + "#calculation:\n", + " #Total input power\n", + "Pi = Pi1 + Pi2\n", + "phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n", + " #Power factor\n", + "pf = math.cos(phi)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)power input is \",round(Pi,2),\"W\"\n", + "print \"\\n (b)power factor is \",round(pf,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)power input is 12000.0 W\n", + "\n", + " (b)power factor is 0.87" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 16, page no. 310

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the readings of each wattmeter.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Pi = 12000;# in Watts\n", + "pf = 0.6;# power factor\n", + "\n", + "#calculation:\n", + " #If the two wattmeters indicate Pi1 and Pi2 respectively\n", + " # Pit = Pi1 + Pi2\n", + "Pit = Pi\n", + " # Pid = Pi1 - Pi2\n", + " #power factor = 0.6 = cos(phi)\n", + "phi = math.acos(pf)\n", + "Pid = Pit*math.tan(phi)/(3**0.5)\n", + " #Hence wattmeter 1 reads\n", + "Pi1 = (Pid + Pit)/2\n", + " #wattmeter 2 reads\n", + "Pi2 = Pit - Pi1\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n reading in each wattameter are \",round(Pi1,2),\"W and \",round(Pi2,2),\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " reading in each wattameter are 10618.8 W and 1381.2 W" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 17, page no. 310

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine (a) the input power and (b) the load power factor.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Pi1 = 10000;# in Watts\n", + "Pi2 = -3000;# in Watts\n", + "\n", + "#calculation:\n", + " #Total input power\n", + "Pi = Pi1 + Pi2\n", + "phi = math.atan((Pi1 - Pi2)*(3**0.5)/(Pi1 + Pi2))\n", + " #Power factor\n", + "pf = math.cos(phi)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)power input is \",round(Pi,2),\"W\"\n", + "print \"\\n (b)power factor is \",round(pf,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)power input is 7000.0 W\n", + "\n", + " (b)power factor is 0.3" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 18, page no. 311

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate for each connection the readings on each of two wattmeters connected to measure the power by the two-wattmeter method.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "R = 8; # in ohms\n", + "XL = 8; # in ohms\n", + "VL = 415; # in Volts\n", + "\n", + "#calculation:\n", + "#For a star connection:\n", + "#IL = Ip\n", + "#VL = Vp*(3**0.5)\n", + "VLs = VL\n", + "Vps = VLs/(3**0.5)\n", + "#Impedance per phase,\n", + "Zp = (R*R + XL*XL)**0.5\n", + "Ips = Vps/Zp\n", + "ILs = Ips\n", + "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n", + "pf = R/Zp\n", + "Ps = VLs*ILs*(3**0.5)*pf\n", + "#If wattmeter readings are P1 and P2 then P1 + P2 = Pst\n", + "Pst = Ps\n", + "# Pid = Pi1 - Pi2\n", + "phi = math.acos(pf)\n", + "Psd = Pst*math.tan(phi)/(3**0.5)\n", + "#Hence wattmeter 1 reads\n", + "Ps1 = (Psd + Pst)/2\n", + "#wattmeter 2 reads\n", + "Ps2 = Pst - Ps1\n", + "\n", + "#For a delta connection:\n", + "#VL = Vp\n", + "#IL = Ip*(3**0.5)\n", + "VLd = VL\n", + "Vpd = VLd\n", + "Ipd = Vpd/Zp\n", + "ILd = Ipd*(3**0.5)\n", + "#Power dissipated, P = VL*IL*(3**0.5)*cos(phi) or P = 3*Ip*Ip*Rp)\n", + "Pd = VLd*ILd*(3**0.5)*pf\n", + "#If wattmeter readings are P1 and P2 then P1 + P2 = Pdt\n", + "Pdt = Pd\n", + "# Pid = Pi1 - Pi2\n", + "Pdd = Pdt*math.tan(phi)/(3**0.5)\n", + "#Hence wattmeter 1 reads\n", + "Pd1 = (Pdd + Pdt)/2\n", + "#wattmeter 2 reads\n", + "Pd2 = Pdt - Pd1\n", + "\n", + "#results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"(a)When the coils are star-connected the wattmeter readings are\", round(Ps1,2),\"W and \",round(Ps2,2),\"W\"\n", + "print \"(b)When the coils are delta-connected the wattmeter readings are are\", round(Pd1,2),\"W and\", round(Pd2,2),\"W\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "(a)When the coils are star-connected the wattmeter readings are 8489.35 W and 2274.71 W\n", + "(b)When the coils are delta-connected the wattmeter readings are are 25468.05 W and 6824.14 W" + ] + } + ], + "prompt_number": 17 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit