{ "metadata": { "name": "", "signature": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 4, Electrical Features of lines - 1" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.1 : page 104" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from numpy import log, pi\n", "import math\n", "#Given Data :\n", "r=1.213/2 #in cm\n", "f=60 #in Hz\n", "ds=0.77888*r #in cm\n", "spacing=1.25 #in meter\n", "L=4*10**-7*log(spacing*100/ds) #in H/m\n", "L*=1000 # in H/km\n", "print \"Inductance = %0.2e H/km\" %L\n", "XL=2*pi*f*L #in ohm/km\n", "XL*=60 # ohm per 60 km\n", "print \"Inductive reactance for 60 km line = %0.1f ohm\" %XL" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inductance = 2.23e-03 H/km\n", "Inductive reactance for 60 km line = 50.5 ohm\n" ] } ], "prompt_number": 105 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.2 : page 105" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from math import log\n", "#Given Data : \n", "l=20 # km (length of line)\n", "d=2.8*100 #in cm(spacing)\n", "r=0.5*1.5 #in cm\n", "ds=0.77888*r #in cm\n", "L=0.2*log(d/ds) #in H/m/phase\n", "L*=l # mH for 20km line\n", "print \"Inductance per phase for a 20 km line = %0.2f mH\" %L" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inductance per phase for a 20 km line = 24.69 mH\n" ] } ], "prompt_number": 106 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.3 : page 105" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from math import log\n", "#Given Data :\n", "a=1.5 #in cm**2\n", "d=8 #in meter(spacing)\n", "r=39.8/2 #in mm\n", "l=1*10**5 #in cm\n", "rho=1.73*10**-6 #in ohm-cm\n", "R=rho*l/a #in ohm/km\n", "print \"Resistance of line = %0.4f ohms/km\" %R\n", "ds=0.77888*r #in cm\n", "L=0.2*log(d/(ds*10**-3)) #in mH/km/phase\n", "print \"Inductance per phase for a 1 km line = %0.3f mH/km\" %L" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Resistance of line = 0.1153 ohms/km\n", "Inductance per phase for a 1 km line = 1.249 mH/km\n" ] } ], "prompt_number": 107 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.4 : page 106" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "#Given Data :\n", "Cs=1/3 #in uF\n", "Cc=(0.6-Cs)/2 #in uF\n", "#Part (a) :\n", "C1=(3/2)*Cc+(1/2)*Cs #in uF(between any two conductor)\n", "print \"Capacitance between any two conductor = %0.3f uF\" %C1\n", "#Part (b) :\n", "C2=2*Cc+2*Cs/3\n", "print \"Capacitance between any shorted conductors = %0.3f uF\" %C2" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Capacitance between any two conductor = 0.367 uF\n", "Capacitance between any shorted conductors = 0.489 uF\n" ] } ], "prompt_number": 108 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.5 : page 107" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "#Given Data :\n", "d1=3 #in meter\n", "d2=3 #in meter\n", "d3=d1+d2 #in meter\n", "d=378 #in cm\n", "dia=2.5 #in cm\n", "r=dia/2 #in cm\n", "epsilon_o=8.854*10**-12 #constnt\n", "L=(0.5+2*log10(d/r))*10**-7 #in H/m\n", "L*=60*1000*1000 # mH per 60 km\n", "print \"Inductance for 60 km line %0.2f mH\" %L\n", "C=2*pi*epsilon_o/log(d/r) #in F/m\n", "C*=60*10**3*10**6 # uF per 60 km line\n", "print \"Capacitnce for 60 km line = %0.4f uF\" %C\n", "# Answer is not accurate in the textbook." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inductance for 60 km line 32.77 mH\n", "Capacitnce for 60 km line = 0.5844 uF\n" ] } ], "prompt_number": 109 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.6 : page 107" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from numpy import log10\n", "#Given Data : \n", "dinner=6 #in meter\n", "douter=12 #in meter\n", "d=(dinner**2*douter)**(1/3) #in meter\n", "r=2.8 #in meter\n", "ds=0.7788*r #in cm\n", "L=2*log10(d*100/ds) #in mH/phase/km\n", "L*=100 # mH per 100 km\n", "print \"Inductance for 100 km line %0.f mH\" %L" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inductance for 100 km line 508 mH\n" ] } ], "prompt_number": 110 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.7 : page 108" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from math import log\n", "#Given Data :\n", "dia=5 #in mm\n", "d=1.5 #in meter(spacing)\n", "r=dia/2 #in mm\n", "r=r*10**-3 #in meter\n", "epsilon_o=8.854*10**-12 #constnt\n", "C=pi*epsilon_o/log(d/r) #in Farad per meter\n", "C*=50*1000 # F per 50km line\n", "print \"Capacitance for 50 km line = %0.2e F\" %C\n", "#Note : answer is not accurate in the book. " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Capacitance for 50 km line = 2.17e-07 F\n" ] } ], "prompt_number": 111 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.8 : page 108" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from math import log\n", "#Given Data :\n", "d=300 #in cm(spacing)\n", "r=1 #in cm\n", "#Formula : L=10**-7*[mu_r+4*log10(d/r)] #in H/m\n", "#Part (i) : mu_r=1\n", "mu_r=1 #constant\n", "L=10**-4*(mu_r+4*log(d/r)) #in H/m\n", "L*=1000 # mh per km\n", "print \"Loop inductance per km for copper %0.2f mH\" %L\n", "#Part (ii) : mu_r=100\n", "mu_r=100 #constant\n", "L=10**-4*(mu_r+4*log(d/r)) #in H/m\n", "L*=1000 # mH per km\n", "print \"Loop inductance per km for steel = %0.2f mH\" %L\n", "# Answer not calculated completely and calculation mistake." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Loop inductance per km for copper 2.38 mH\n", "Loop inductance per km for steel = 12.28 mH\n" ] } ], "prompt_number": 112 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.9 : page 109" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from math import log\n", "#Given Data :\n", "d1=100 #in cm(spacing)\n", "d2=100 #in cm(spacing)\n", "d3=100 #in cm\n", "r=1 #in cm\n", "L=10**-7*(0.5+2*log((d1*d2*d3)**(1/3)/r)) #in H/m\n", "L=L*1000*1000 #in mH/km\n", "print \"Inductance per km = %0.2f mH\" %L\n", "#Note : Answer in the book is wrong due to calculation mistake.\n", "#Note : In the last line it should be multiply by 10**6 to convert from H/m to mH/km instead of 10**8." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inductance per km = 0.97 mH\n" ] } ], "prompt_number": 113 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.10 : page 109" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from numpy import log\n", "#Given Data : \n", "d1=2 #in cm\n", "d2=2.5 #in cm\n", "d3=4.5 #in cm\n", "r=1.24/2 #in cm\n", "L=10**-7*(0.5+2*log((d1*d2*d3)**(1/3)/r)) #in H/m\n", "L=L*1000*1000 #in mH/km\n", "print \"Inductance per km per phase = %0.2f mH\" %L\n", "#Note : Answer in the book is wrong(calculation mistake)." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inductance per km per phase = 0.35 mH\n" ] } ], "prompt_number": 114 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.11 : page 110" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from numpy import log\n", "#Given Data :\n", "r=0.75*10 #in mm\n", "d=1.5*10**3 #in mm\n", "ds=0.7788*r #in mm\n", "L=4*10**-7*log(d/ds) #in H/m\n", "L=L*10**6 #in mH/km\n", "print \"Inductance of line = %0.2f mH/km\" %L" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inductance of line = 2.22 mH/km\n" ] } ], "prompt_number": 115 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.12 : page 110" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from numpy import log\n", "#Given Data :\n", "d1=4*100 #in cm\n", "d2=5*100 #in cm\n", "d3=6*100 #in cm\n", "r=1 #in cm\n", "ds=0.7788*r #in cm\n", "L=(0.2*log((d1*d2*d3)**(1/3)/ds)) #in mH\n", "L*=10**3 # uH\n", "print \"Inductance per km = %0.2f uH\" %L\n", "#Note : answer in the book is wrong." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inductance per km = 1290.20 uH\n" ] } ], "prompt_number": 116 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.13 : page 110" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from numpy import log\n", "#Given Data :\n", "d=300 #in cm(spacing)\n", "r=1 #in cm\n", "epsilon_o=8.854*10**-12 #constnt\n", "C=pi*epsilon_o/log(d/r) #in Farad per meter\n", "C*=30*1000*10**6 # uF per 30k km line\n", "print \"Capacitance for 30 km line = %0.2f uF\" %C" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Capacitance for 30 km line = 0.15 uF\n" ] } ], "prompt_number": 117 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.14 : page 111" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from numpy import log\n", "#Given Data :\n", "d=2.5*100 #in cm(spacing)\n", "r=2/2 #in cm\n", "epsilon_o=8.854*10**-12 #constnt\n", "C=2*pi*epsilon_o/log(d/r) #in Farad per meter\n", "C*=10*1000*10**6 # uF per 10 km line\n", "print \"Capacitance for 10 km line = %0.2f uF\" %C\n", "#Note : answer given in the book is wrong but calculated is right." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Capacitance for 10 km line = 0.10 uF\n" ] } ], "prompt_number": 119 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 4.15 : page 111" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given Data :\n", "VL=33 #in KV\n", "f=50 #in hz\n", "d1=4 #in meter\n", "d2=4 #in meter\n", "d3=8 #in meter\n", "d=(d1*d2*d3)**(1/3) #in meter\n", "epsilon_o=8.854*10**-12 #constnt\n", "d=d*100 #in cm\n", "r=0.62 #in cm\n", "C=2*pi*epsilon_o/log(d/r) #in Farad per meter\n", "C*=50*1000*10**6 # uF per m line\n", "print \"Capacitance for 50 km line %0.3f uF\" %C\n", "Vp=VL/sqrt(3) #in KV\n", "Vp=Vp*10**3 #in volt\n", "Ic=2*pi*f*(C*10**-6)*Vp #in Ampere\n", "print \"The charging current = %0.2f Ampere\" %Ic" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Capacitance for 50 km line 0.415 uF\n", "The charging current = 2.48 Ampere\n" ] } ], "prompt_number": 120 } ], "metadata": {} } ] }