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diff --git a/Electrical_Power-i__by_M.L._Anand/Chapter2.ipynb b/Electrical_Power-i__by_M.L._Anand/Chapter2.ipynb new file mode 100755 index 00000000..4d0e3eeb --- /dev/null +++ b/Electrical_Power-i__by_M.L._Anand/Chapter2.ipynb @@ -0,0 +1,626 @@ +{ + "metadata": { + "name": "", + "signature": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "chapter 2, Layout of transmission systems" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.1 : page 32" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols\n", + "from __future__ import division\n", + "a=symbols('a')\n", + "#Given Data :\n", + "CableCost=20+400*a #in Rs./meter (a=cross section in cm**2)\n", + "Cable_cost=(20+400*a)*1000 #in Rs./km\n", + "l=1 #in Km\n", + "P=1 #in MW\n", + "V=11 #in KV\n", + "cosfi=0.8 #powerfactor\n", + "h=3000 #hours\n", + "i=10 #in %\n", + "E_cost=15 #in paisa/kwh\n", + "rho=1.75*10**-6 #sp. resistance in ohm-cm\n", + "C1=CableCost*1000 #in Rs./km\n", + "R=rho*l*10**3/(a*10**-2) #in ohm\n", + "Ifl=(P*10**6)/(V*10**3*cosfi) #in Ampere\n", + "Ploss=2*Ifl**2*R #in Watts\n", + "Annual_cost=Ploss*10**-3*h*E_cost/100 #in Rs.\n", + "AnnualCost2=400*10**3*a*i/100 #in Rs.\n", + "annually_energy_lost = 40000/a # Rs.\n", + "#comparing Annual_cost = AnnualCost2\n", + "a=Annual_cost/AnnualCost2*a**2\n", + "print \"Most economical cross section area, a = \",round(a,4), \"cm2 \"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Most economical cross section area, a = 0.0508 cm2 \n" + ] + } + ], + "prompt_number": 245 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.2 : page 33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#Given Data :\n", + "Pt=30*10**6 #in watt\n", + "V=220*10**3 #in volt\n", + "l=275*10**3 #in meter\n", + "R=0.173 #in ohm/km\n", + "Eta=90 #in %\n", + "density=8.9 #copper density\n", + "Loss=100-Eta #in %\n", + "cosfi=0.8 #powerfactor\n", + "print \"3-phase 3 wire :\"\n", + "IL=Pt/(sqrt(3)*V*cosfi) #in Ampere\n", + "LineLosses=(Loss/100)*Pt #in watts\n", + "rho=R*10**-4/(1*10**3) #in ohm-meter\n", + "a=3*IL**2*rho*l/(LineLosses) #in m**2\n", + "Volume=3*a*l #in m**3\n", + "Cu_weight=Volume*density #in Tones\n", + "print \"Weight of copper = %0.f Tones\" %Cu_weight\n", + "print \"Single phase 2 wire :\"\n", + "IL=Pt/(V*cosfi) #in Ampere\n", + "a=2*IL**2*rho*l/(LineLosses) #in m**2\n", + "Volume=2*a*l #in m**3\n", + "Cu_weight=Volume*density #in Tones\n", + "print \"Weight of copper = %0.1f Tones\" %Cu_weight\n", + "#Note : answer is not accurate in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "3-phase 3 wire :\n", + "Weight of copper = 338 Tones\n", + "Single phase 2 wire :\n", + "Weight of copper = 451.1 Tones\n" + ] + } + ], + "prompt_number": 246 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.3 : page 35" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols\n", + "from __future__ import division\n", + "from math import sqrt\n", + "a=symbols('a')\n", + "#Given Data :\n", + "l=1 #in Km\n", + "l=l*10**5 #in cm\n", + "I=200 #in Ampere\n", + "CableCost=50*a #in Rs./meter (a=cross section in cm**2)\n", + "E_cost=5 #in paisa/kwh\n", + "i=10 #in %\n", + "rho=1.72*10**-6 #resistivity in ohm-cm\n", + "R=rho*l/a #in ohm\n", + "Eloss=2*I**2*R*24*365/1000 #in kwh\n", + "AnnualCost2=(E_cost/100)*2*I**2*rho*l*24*365/1000/a #in Rs.\n", + "C1=CableCost*1000 #in Rs./km\n", + "AnnualCharges=C1*i/100 #in Rs.\n", + "a=sqrt(AnnualCost2/AnnualCharges*a**2) # cm2\n", + "print \"Most economical cross section area a = %0.3f cm2\" %a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Most economical cross section area a = 1.098 cm2\n" + ] + } + ], + "prompt_number": 247 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.4 : page 36" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols\n", + "from __future__ import division\n", + "from math import sqrt\n", + "a=symbols('a')\n", + "#Given Data : \n", + "l=4*10**5 #in cm\n", + "VL=30 #in KV\n", + "LineCost=40000*a+7500 #in Rs/km\n", + "i=8 #in %\n", + "E_cost=4 #in paisa/kwh\n", + "rho=1.72*10**-6 #in ohm-cm\n", + "R=rho*l/a #in ohm\n", + "P1=3*10**6 #in watt\n", + "h1=10 #in hours\n", + "cosfi1=0.8#unitless\n", + "I1=P1/(sqrt(3)*VL*10**3*cosfi1) #in Ampere\n", + "P2=1.5*10**6 #in watt\n", + "h2=6 #in hours\n", + "cosfi2=0.9#unitless\n", + "I2=P2/(sqrt(3)*VL*10**3*cosfi2) #in Ampere\\\n", + "P3=0.5*10**6 #in watt\n", + "h3=8 #in hours\n", + "cosfi3=0.9 #unitless\n", + "I3=P3/(sqrt(3)*VL*10**3*cosfi3) #in Ampere\n", + "Etot=3*(I1**2*h1+I2**2*h2+I3**2*h3)*R*365/1000 #in kwh\n", + "Ccost_line=40000*a*4 #in Rs.\n", + "AnnualCharges=Ccost_line*i/100 #in Rs.\n", + "AnnualCost2=(E_cost/100)*Etot #in Rs.\n", + "a=sqrt(AnnualCost2/AnnualCharges*a**2) # cm2\n", + "print \"Most economical cross section area a = %0.2f cm2\" %a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Most economical cross section area a = 0.37 cm2\n" + ] + } + ], + "prompt_number": 248 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.5 : page 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols\n", + "from __future__ import division\n", + "from math import sqrt\n", + "a, l=symbols('a l')\n", + "#Given Data :\n", + "P=5*10**6 #in watt\n", + "VL=33*10**3 #in volt\n", + "cosfi=0.8#unitless\n", + "LineCost=31250*a+4000 #in Rs/km\n", + "rho=10**-6 #in ohm-cm\n", + "i=8 #in %\n", + "E_cost=4 #in paisa/kwh\n", + "IL=P/(sqrt(3)*VL*cosfi) #in Ampere\n", + "Line_length=l*10**5 #in cm\n", + "R=rho*l*10**5/a #in ohm\n", + "E_lost=3*IL**2*R*365/1000 #in kwh\n", + "Ccost_line=31250*a*l #in Rs.\n", + "AnnualCharges=Ccost_line*i/100 #in Rs.\n", + "a=sqrt(E_lost/AnnualCharges*a**2)\n", + "print \"Most economical cross section area a = %0.2f cm2\" %a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Most economical cross section area a = 0.72 cm2\n" + ] + } + ], + "prompt_number": 249 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.6 : page 38" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given Data :\n", + "P=50*10**6 #in watt\n", + "VL=220*10**3 #in volt\n", + "cosfi=0.8#unitless\n", + "Eta=90 #in %\n", + "l=200*10**3 #in meter\n", + "rho=1.75*10**-8 #in ohm-cm\n", + "W=P*(100-Eta)/100 #in Wats(Line losses)\n", + "#Part (i) : 3 phase 3 wire with Cu condutor\n", + "gravity=8.9 #specific gravity\n", + "IL=P/(sqrt(3)*VL*cosfi) #in Ampere\n", + "a=3*IL**2*rho*l/W #in m**2\n", + "Vol3=3*a*l #volume of 3 lines(in m**3)\n", + "CuWeight=Vol3*gravity #in Tones\n", + "print \"Weight of copper = %0.2f Tones\" %CuWeight \n", + "#Part (ii) : When Al conductor is used.\n", + "gravity=2.7 #specific gravity\n", + "rho=3*10**-8 #in ohm-meter\n", + "a=3*IL**2*rho*(l/W) #in m**2\n", + "Vol=3*a*l #volume of 3 lines(in m**3)\n", + "AlWeight=Vol*gravity #in Tones\n", + "print \"Weight of Alluminium = %0.3f Tones\" %AlWeight " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Weight of copper = 301.69 Tones\n", + "Weight of Alluminium = 156.896 Tones\n" + ] + } + ], + "prompt_number": 250 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.7 : page 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols\n", + "from __future__ import division\n", + "from math import sqrt\n", + "P2, V, cosfi, R = symbols('P2 V cosfi R')\n", + "#Given Data :\n", + "Vcon=V #in volt\n", + "pf=cosfi #unitless\n", + "Rcon=R #in ohm\n", + "P1=15*10**6 #in watt\n", + "I1=P1/(V*cosfi) #in Ampere\n", + "W1=2*I1**2*Rcon #in Wats(Line losses)\n", + "Lloss_percent=W1/P1*100 #in % eqn(1)\n", + "\n", + "I2=P2/(V*cosfi*sqrt(3)) #in Ampere\n", + "W2=3*I2**2*Rcon #in Wats(Line losses)\n", + "Lloss_percent=W2/P2*100 #in % eqn(2)\n", + "# Equating eqn(1) & eqn(2)\n", + "P2=W1/P1*100/W2/100*P2**2 # W\n", + "P2/=10**6 # MW\n", + "print \"3 phase load = %0.f MW \" %P2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "3 phase load = 30 MW \n" + ] + } + ], + "prompt_number": 251 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.8 : page 40" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given Data :\n", + "from sympy import symbols\n", + "P, V, rho, l, a = symbols('P V rho l a')\n", + "I=P/V #in Ampere\n", + "R=rho*l/a\n", + "a=2*I**2*2*rho*l/W\n", + "K=a*V**2\n", + "#(i) When V=220 volts\n", + "V1=220 #in volts\n", + "vol1=K/V1**2\n", + "#(ii) When V=500 volts\n", + "V2=500 #in volts\\\n", + "vol2=K/V2**2\n", + "saving=(vol1-vol2)/vol1*100 # in % \n", + "print \"Saving in copper =\",round(saving,2),\"%\"\n", + "# Answer not accurate in the textbook." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Saving in copper = 80.64 %\n" + ] + } + ], + "prompt_number": 252 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.9 : page 41" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given Data :\n", + "P=30*10**6 #in watts\n", + "V=220*10**3 #in Volt\n", + "l=250*10**3 #in meter\n", + "Eta=85 #in %\n", + "rho=3*10**-8 #in ohm-meter\n", + "cosfi=0.8 #power factor\n", + "W=P*(100-Eta)/100 #in watts\n", + "I=P/(sqrt(3)*V*cosfi) #in Ampere\n", + "a=3*I**2*rho*l/W #in m**2\n", + "Volume=3*a*l #in m**3\n", + "print \"Volume of the conductor material = %0.1f m3\" %Volume\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Volume of the conductor material = 36.3 m3\n" + ] + } + ], + "prompt_number": 253 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.10 : page 42" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given Data :\n", + "P=20*10**6 #in VA\n", + "cosfi=0.75 #power factor\n", + "P=20*10**6*cosfi #in watts\n", + "V=33*10**3 #in Volt\n", + "l=20*10**3 #in meter\n", + "Eta=85 #in %\n", + "rho=3*10**-8 #in ohm-meter\n", + "W=P*(100-Eta)/100 #in watts\n", + "#For single phase system :\n", + "I=P/(V*cosfi) #in Ampere\n", + "a1=2*I**2*rho*l/W #in m**2\n", + "V1=2*a1*l #in m**3\n", + "print \"For single phase system :\\nVolume of the conductor material = %0.3f m3\" %(V1)\n", + "#For 3 phase 3 wire system :\n", + "I=P/(sqrt(3)*V*cosfi) #in Ampere\n", + "a2=3*I**2*rho*l/W #in m**2\n", + "V2=3*a2*l #in m**3\n", + "print \"For three phase 3-wire system :\\nVolume of the conductor material = %0.2f m3\" %V2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "For single phase system :\n", + "Volume of the conductor material = 7.836 m3\n", + "For three phase 3-wire system :\n", + "Volume of the conductor material = 5.88 m3\n" + ] + } + ], + "prompt_number": 254 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.11 : page 43" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols\n", + "from __future__ import division\n", + "from math import sqrt\n", + "a=symbols('a')#Given Data :\n", + "l=1*10**3 #in meter\n", + "rho=1.85*10**-6 #in ohm-cm\n", + "R=rho*l*100/a #in ohm\n", + "IL=300 #in Ampere\n", + "Rate=10 #in Rs/kwh\n", + "Eloss=2*IL**2*R*365*24/1000 #in kwh\n", + "CableCost=100*a #in Rs/meter : a=cross sectional area(in cm**2)\n", + "i=10 #in %\n", + "AnnualCost=Eloss/Rate #in Rs\n", + "Ccost=100*a*l #in Rs\n", + "Ccost=Ccost*i/100 # with effet of depreciation\n", + "#a=sqrt(29170.8/10000)\n", + "a=sqrt(AnnualCost/Ccost*a**2) # cm2\n", + "print \"Most economical cross section = %0.3f cm2\" %a" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Most economical cross section = 1.708 cm2\n" + ] + } + ], + "prompt_number": 255 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.12 : page 44 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols\n", + "pf, Vcon, Rcon = symbols('cosfi V R')\n", + "#Given Data :\n", + "#Part (i) : single phase system\n", + "P1=5*10**6 #in watt\n", + "I1=P1/(Vcon*pf) #in Ampere\n", + "W1=2*I1**2*Rcon #in Watts(Line losses)\n", + "Lloss_percent=W1/P1*100 #in % eqn(1)\n", + "#Part (ii) : 3 phase 3 wire system\n", + "P2 = symbols('P2') # MW\n", + "I2=P2/(Vcon*pf*sqrt(3)) #in Ampere\n", + "W2=3*I2**2*Rcon #in Watts(Line losses)\n", + "Lloss_percent=W2/P2*100 #in % eqn(2)\n", + "# Equating two eqn\n", + "P2=(W1/P1*100)/(W2*100)*P2**2\n", + "P2/=10**6 # MW\n", + "print \"3 phase load = %0.f MW \" %P2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "3 phase load = 10 MW \n" + ] + } + ], + "prompt_number": 256 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exa 2.13 : page 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#Given Data :\n", + "from sympy import symbols\n", + "P, V, rho, l, a = symbols('P V rho l a')\n", + "I=P/V #in Ampere\n", + "R=rho*l/a\n", + "a=2*I**2*2*rho*l/W\n", + "K=a*V**2\n", + "#(i) When V=200 volts\n", + "V1=200 #in volts\n", + "vol1=K/V1**2\n", + "#(ii) When V=600 volts\n", + "V2=600 #in volts\\\n", + "vol2=K/V2**2\n", + "saving=(vol1-vol2)/vol1*100 # in % \n", + "print \"Saving in copper =\",math.floor(saving),\"%\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Saving in copper = 88.0 %\n" + ] + } + ], + "prompt_number": 257 + } + ], + "metadata": {} + } + ] +} |