{ "metadata": { "name": "", "signature": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter8 - Underground cables and faults" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 8.1 - page 222" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import exp, pi\n", "#given data\n", "R=500 #in Mohm/Km\n", "R=R*10**6 #in ohm\n", "r1=2.5/2 #in cm\n", "r1=r1*10**-2 #in meter\n", "rho=4.5*10**16 #in ohm/cm\n", "rho=rho*10**-2 #in ohm/m\n", "l=1 #in Km\n", "l=l*1000 #in meter\n", "#Formula : R=(rho/(2*pi*l))*log(r2/r1)\n", "r2=(exp(R/(rho/(2*pi*l))))*r1 #in meter\n", "thickness=r2-r1 #in meter\n", "thickness=thickness*100 #in cm\n", "print \"Thickness of Insulation = %0.3f cm\" %thickness\n", "# Answer in the textbook is wrong due to accuracy." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Thickness of Insulation = 0.009 cm\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 8.2 - page 223" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import log10\n", "#given data\n", "d=1 #in cm\n", "d=d*10**-2 #in meter\n", "D=1.8 #in cm\n", "D=D*10**-2 #in meter\n", "epsilon_r=4 #permittivity of insulation\n", "C=0.024*epsilon_r/log10(D/d) #in uF/Km\n", "print \"Capacitance/km of the fibre = %0.3f uF\" %C" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Capacitance/km of the fibre = 0.376 uF\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 8.3 - page 223" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import log\n", "#given data\n", "V=33 #in KV\n", "d=1 #in cm\n", "D=4 #in cm\n", "#Part (a) :\n", "gmax=2*V/(d*log(D/d)) #in KV/cm\n", "print \"Maximum Stress = %0.1f KV/cm\" %gmax\n", "#Part (b) :\n", "gmin=2*V/(D*log(D/d)) #in KV/cm\n", "print \"Minimum Stress = %0.0f KV/cm\" %round(gmin)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Maximum Stress = 47.6 KV/cm\n", "Minimum Stress = 12 KV/cm\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 8.4 - page 224" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import sqrt\n", "#given data\n", "Vrms=66 #in KV\n", "gmax=40 #in KV/cm\n", "V=sqrt(2)*Vrms #in Volt\n", "#Part (a) : \n", "d=2*V/gmax #in cm\n", "print \"The most economical diameter = %0.3f cm\" %d\n", "#Part (b) : \n", "PeakVoltage=sqrt(2)*Vrms/sqrt(3) #in Volt\n", "V=PeakVoltage #in Volt\n", "d=2*V/gmax #in cm\n", "print \"The most economical diameter for 3 phase system = %0.1f m\" %d" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The most economical diameter = 4.667 cm\n", "The most economical diameter for 3 phase system = 2.7 m\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 8.5 - page 224" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from math import sqrt\n", "#given data\n", "d=2 #in cm\n", "D=2.5*2 #in cm\n", "d1=(5/4)*d #in cm\n", "d2=(5/3)*d #in cm\n", "gmax=40 #in KV/cm\n", "PeakVoltage=(gmax/2)*(d*log(d1/d)+d1*log(d2/d1)+d2*log(D/d2)) #in KV\n", "print \"The safe Working Potential = %0.1f KV\" %(PeakVoltage/sqrt(2))" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The safe Working Potential = 35.6 KV\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 8.6 - page 25" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from numpy import pi, sqrt\n", "#given data\n", "CN=0.4 #in uF\n", "V=33 #in KV\n", "VP=V/sqrt(3) #in KV\n", "f=25 #in Hz\n", "#Capacitance between 2 cores for 15 Km length\n", "CN_1=15*CN #in uF\n", "#Capacitance of each core to neutral\n", "CN=2*CN_1 #in uF\n", "#Charging current per phase\n", "I=2*pi*f*VP*1000*CN*10**-6 #in Ampere\n", "print \"Charging current per phase = %0.2f A\" %round(I)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Charging current per phase = 36.00 A\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 8.7 - page 225" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "from numpy import pi, sqrt\n", "#given data\n", "l=10 #in Km\n", "C=0.3 #in uF\n", "V=22 #in KV\n", "VP=V/sqrt(3) #in KV\n", "VP=VP*1000 #in Volt\n", "f=50 #in Hz\n", "Capacitance=C*l #in uF\n", "CN=2*Capacitance #in uF\n", "KVA_Taken=3*VP*2*pi*f*VP*CN*10/1000 #in KVA\n", "print \"KVA taken by the 10 Km cable = %0.3e KVA\" %KVA_Taken" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "KVA taken by the 10 Km cable = 9.123e+09 KVA\n" ] } ], "prompt_number": 21 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 8.8 - page 225" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#given data\n", "P=10 #in Ohm\n", "Q=80 #in Ohm\n", "S2=3400 #in Ohm\n", "S1=2400 #in Ohm\n", "X=P*(S2-S1)/(P+Q) #in Ohm\n", "LoopResistance=P*S2/Q #in Ohm\n", "ResistancePerKm=LoopResistance/10 #in Ohm\n", "Distance=X/ResistancePerKm #in Km\n", "print \"Distance of fault from testing end = %0.3f Km\" %Distance" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Distance of fault from testing end = 2.614 Km\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 8.9 - page 226" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#given data\n", "Resistance=1.6 #in ohm/Km\n", "l=1000 #in meter\n", "PbyQ=3 #unitless\n", "PplusQbyQ=4 #unitless\n", "LoopResistance=(Resistance/1000)*2*l #in Ohm\n", "X=(1/PplusQbyQ)*LoopResistance #in Ohm\n", "Distance=X/(Resistance/1000) #in meter\n", "print \"Distance of Fault from testing end = %0.2f meters\" %Distance" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Distance of Fault from testing end = 500.00 meters\n" ] } ], "prompt_number": 24 } ], "metadata": {} } ] }