From 6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- .../Chapter_3.ipynb | 368 +++++++++++++++++++++ 1 file changed, 368 insertions(+) create mode 100755 Elements_of_Power_system_by_J.B._Gupta/Chapter_3.ipynb (limited to 'Elements_of_Power_system_by_J.B._Gupta/Chapter_3.ipynb') diff --git a/Elements_of_Power_system_by_J.B._Gupta/Chapter_3.ipynb b/Elements_of_Power_system_by_J.B._Gupta/Chapter_3.ipynb new file mode 100755 index 00000000..81522bb9 --- /dev/null +++ b/Elements_of_Power_system_by_J.B._Gupta/Chapter_3.ipynb @@ -0,0 +1,368 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8ef5a6db25d3ca5636ebb9bdfef72dd38c306363688bdeed8ee7c50c70ef98c0" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 3 - TRANSMISSION LINES" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E1 - Pg 66" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate Weight of copper required,Alluminium required\n", + "#Given data :\n", + "import math\n", + "P=30.*10.**6.##W\n", + "pf=0.8##lagging power factor\n", + "VL=132.*1000.##V\n", + "l=120.*1000.##m\n", + "Eta=90./100.##Efficiency\n", + "rho_Cu=1.78*10.**-8.##ohm-m\n", + "D_Cu=8.9*10.**3.##kg/m**3\n", + "rho_Al=2.6*10.**-8.##ohm-m\n", + "D_Al=2.*10.**3.##kg/m**3\n", + "IL=P/(math.sqrt(3.)*VL*pf)##A\n", + "#W=3.*IL**2.*rho*l/a=(1-Eta)*P\n", + "a_Cu=(3.*IL**2.*rho_Cu*l)/(1.-Eta)/P##m**2\n", + "V_Cu=3.*a_Cu*l##m**3\n", + "Wt_Cu=V_Cu*D_Cu##kg\n", + "print '%s %.2f' %(\"Weight of copper required(kg)\",Wt_Cu)#\n", + "a_Al=(3.*IL**2.*rho_Al*l)/(1.-Eta)/P##m**2\n", + "V_Al=3.*a_Al*l## m**3\n", + "Wt_Al=V_Al*D_Al##kg\n", + "print '%s %.2f' %(\"Weight of Alluminium required(kg)\",Wt_Al)#\n", + "#Answer in the textbook is not accurate.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Weight of copper required(kg) 184114.15\n", + "Weight of Alluminium required(kg) 60433.88\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E2 - Pg 70" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate Most economical cross sectional area\n", + "#Given data :\n", + "import math\n", + "a=100.#\n", + "cost=90.*a+20.##Rs./m\n", + "i=10.##%(interest and depreciation)\n", + "l=2.##km\n", + "cost_E=4.##paise/unit\n", + "Im=250.##A\n", + "a=1.##cm**2\n", + "rho_c=0.173##ohm/km/cm**2\n", + "l2=1.*1000.##km\n", + "R=rho_c*l/a##ohm\n", + "W=2.*Im**2.*R##W\n", + "Eloss=W/1000.*365.*24./2.##per annum(kWh)\n", + "P3BYa=cost_E/100.*Eloss##Rs\n", + "Cc=90.*a*l*1000.##Rs(capital cost of feeder cable)\n", + "P2a=Cc*i/100.##Rs\n", + "#P2a=P3BYa##For most economical cross section\n", + "a=math.sqrt(P3BYa*a/(P2a/a))##cm**2\n", + "print '%s %.3f' %(\"Most economical cross sectional area in cm**2 : \",a)#\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Most economical cross sectional area in cm**2 : 0.649\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E3 - Pg 71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate Best current density\n", + "#Given data :\n", + "import math\n", + "t=2600.##hour\n", + "Con_Cost=3.##Rs/kg(conductor cost)\n", + "R=1.78*10.**-8.##ohm-m\n", + "D=6200.##kg/m**3\n", + "E_Cost=10./100.##Rs/unit(energy cost)\n", + "i=12.##%(interest and depreciation)\n", + "a=100.##mm**2 ##cross sectional area\n", + "W=a*1000.*D/1000./1000.##kg/km(Weight of conductor of 1km length)\n", + "cost=Con_Cost*W##Rs./km(cost of conductor of 1km length)\n", + "In_Dep=cost*i/100.##Rs(Annual interest and depreciation per conductor per km)\n", + "In_DepBYa=In_Dep/a#\n", + "I=100.##A\n", + "E_lost_aBY_Isqr=R*1000./10.**-6.*t/1000.##Energy lost/annum/km/conductor\n", + "E_lost_cost_aBY_Isqr=E_Cost*E_lost_aBY_Isqr##Rs/annum\n", + "#In_Dep=E_lost_cost##For most economical cross section\n", + "IBYa=math.sqrt((In_DepBYa))/(E_lost_cost_aBY_Isqr)##cm**2\n", + "print '%s %.2f' %(\"Best current density in A/mm**2 : \",IBYa)#\n", + "#Answer in the textbook is not accurate.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Best current density in A/mm**2 : 0.32\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E4 - Pg 71" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate Diameter of conductor,Most economical current density\n", + "#Given data :\n", + "import math\n", + "V=11.##kV\n", + "P=1500.##kW\n", + "pf=0.8##lagging power factor\n", + "t=300.*8.##hours\n", + "a=100.##cross section area\n", + "Cc=8000.+20000.*a#Rs/km\n", + "R=0.173/a##ohm/km\n", + "E_lost_cost=2./100.##Rs/unit\n", + "i=12.##%(interest and depreciation)\n", + "Cc_var=20000.*a#Rs/km(variable cost)\n", + "P2a=Cc_var*i/100.##Rs/km\n", + "P2=P2a/a#\n", + "I=P/math.sqrt(3.)/V/pf##A\n", + "W=3.*I**2.*R##W\n", + "E_loss=W/1000.*t##kWh\n", + "P3BYa=E_lost_cost*E_loss##Rs\n", + "#P2a=P3BYa##For most economical cross section\n", + "a=math.sqrt((P3BYa))/(P2)##cm**2\n", + "d=math.sqrt(4.*a/math.pi)##cm\n", + "dela=I/a;##A/cm**2\n", + "print '%s %.2f' %(\"Diameter of conductor in cm : \",d)#\n", + "print '%s %.2f' %(\"Most economical current density in A/cm**2 : \",dela)#\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Diameter of conductor in cm : 0.03\n", + "Most economical current density in A/cm**2 : 152057.18\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E5 - Pg 72" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Exa 3.5\n", + "#Given data :\n", + "import math\n", + "a=100.##cross section area\n", + "I=100.##Current\n", + "Cc=500.+2000.*a#Rs/km\n", + "i=12.##%(interest and depreciation)\n", + "E_lost_cost=5./100.##Rs/kWh\n", + "rho=1.78*10.**-8.##ohm-cm\n", + "load_factor=0.12#\n", + "Cc_var=2000.*a#Rs/km(variable cost)\n", + "P2a=Cc_var*i/100.##Rs/km\n", + "P2=P2a/a#\n", + "R_into_a=rho*1000./(10.**-4.)##ohm\n", + "W_into_a=I**2*R_into_a##W\n", + "E_loss_into_a=W_into_a*load_factor/1000.*8760.##kWh\n", + "P3BYIsqr=E_lost_cost*E_loss_into_a/I**2##Rs\n", + "#P2a=P3BYa##For most economical cross section\n", + "IBYa=math.sqrt((P2))/(P3BYIsqr)##cm**2\n", + "print '%s %.2f' %(\"Most economical current density in A/cm**2 : \",IBYa)#\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Most economical current density in A/cm**2 : 1655.89\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E6 - Pg 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Exa 3.6\n", + "#Given data :\n", + "import math\n", + "A=100.##cross section area\n", + "I=100.##Current\n", + "Cc=500.+2000.*A#Rs/km\n", + "load_factor=0.12#\n", + "i=12.##%(depreciation)\n", + "E_lost_cost=0.05##Rs/kWh\n", + "R=0.17/A##ohm/km\n", + "Cc_var=2000.*A#Rs/km(variable cost)\n", + "P2A=Cc_var*i/100.##Rs/km\n", + "P2=P2A/A#\n", + "R_into_A=R*A##ohm\n", + "W_into_A_BY_Isqr=R_into_A##W\n", + "E_loss_into_A_BY_Isqr=W_into_A_BY_Isqr*load_factor/1000.*8760.##kWh\n", + "P3BYIsqr=E_lost_cost*E_loss_into_A_BY_Isqr##Rs\n", + "#P2a=P3BYa##For most economical cross section\n", + "IBYa=math.sqrt((P2))/(P3BYIsqr)##cm**2\n", + "print '%s %.2f' %(\"Most economical current density in A/cm**2 : \",IBYa)#\n", + "#Answer in the textbook is wrong.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Most economical current density in A/cm**2 : 1733.81\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E7 - Pg 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#calculate Most economical cross sectional area\n", + "#Given data :\n", + "import math\n", + "P1=1000.##kW\n", + "pf1=0.8##\n", + "t1=10.##hours\n", + "P2=500.##kW\n", + "pf2=0.9##\n", + "t2=8.##hours\n", + "P3=100.##kW\n", + "pf3=1.##\n", + "t3=6.##hours\n", + "a=100.##cross section area\n", + "I=100.##Current\n", + "L=100.##length in km\n", + "CcBYL=(8000.*a+1500.)#Rs/km(variable cost)\n", + "i=10.##%(depreciation)\n", + "E_lost_cost=80./100.##Rs/kWh\n", + "rho=1.72*10.**-6.##ohm-cm\n", + "Cc_varBYL=8000.*a*i/100.#Rs/km(variable cost)\n", + "I1=P1*1000./math.sqrt(3.)/10000./pf1##A\n", + "I2=P2*1000./math.sqrt(3.)/10000./pf2##A\n", + "I3=P3*1000./math.sqrt(3.)/10000./pf3##A\n", + "R_into_a_BY_L=rho*1000.*100.##ohm\n", + "W_into_A_BY_Isqr=R_into_a_BY_L##W\n", + "E_loss_into_A_BY_L=3*R_into_a_BY_L*(I1**2*t1+I2**2*t2+I3**2*t3)*365./1000.##kWh\n", + "E_loss_cost_into_A_BY_L=E_loss_into_A_BY_L*E_lost_cost##Rs\n", + "#Cc_var=E_loss_cost##For most economical cross section\n", + "a=math.sqrt((E_loss_cost_into_A_BY_L))/(Cc_varBYL/a)##cm**2\n", + "print '%s %.2f' %(\"Most economical cross sectional area in cm**2 : \",a)#\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Most economical cross sectional area in cm**2 : 0.12\n" + ] + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit