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diff --git a/Elements_of_Power_system/Chapter_2.ipynb b/Elements_of_Power_system/Chapter_2.ipynb new file mode 100755 index 00000000..528910b1 --- /dev/null +++ b/Elements_of_Power_system/Chapter_2.ipynb @@ -0,0 +1,273 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:58c36c7b9784275db9287ae8c79867925c020abc46926776ba50c0c166f1d15f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2 - SUPPLY SYSTEM"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E1 - Pg 40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate saving in copper\n",
+ "#Given data :\n",
+ "VL1=220.##Volts\n",
+ "VL2=400.##Volts\n",
+ "print '%s' %(\"We know, W=I**2*2*R=(P/VL)**2*2*rho*l/a\")#\n",
+ "print '%s ' %(\"a=(P/VL)**2*2*rho*l/(I**2*2*R)\")#\n",
+ "print '%s' %(\"v=2*(P/VL)**2*2*rho*l/(I1**2*2)*l\")#\n",
+ "saving=(2./(VL1)**2.-2./(VL2)**2.)/(2./(VL1)**2.)*100.##%\n",
+ "print '%s %.2f' %(\"% saving in copper : \",saving)#\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "We know, W=I**2*2*R=(P/VL)**2*2*rho*l/a\n",
+ "a=(P/VL)**2*2*rho*l/(I**2*2*R) \n",
+ "v=2*(P/VL)**2*2*rho*l/(I1**2*2)*l\n",
+ "% saving in copper : 69.75\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E2 - pg 51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate 3-phase four wire system material required\n",
+ "print '%s' %(\"Two wire dc system : \")#\n",
+ "print '%s' %(\"I1=P/V & W=2*I1**2*R1=2*P**2*rho*l/V**2/a1\")#\n",
+ "print '%s' %(\"Therefore, Volume required, v1 is 2*a1*l=4*P**2*rho*l**2/V**2/W\")#\n",
+ "print '%s ' %(\"Three phase four wire system : \")#\n",
+ "print '%s ' %(\"I2=P/3/Vas Power by each phase is P/3 & W=3*I1**2*R2=P**2*rho*l/3/V**2/a2\")#\n",
+ "print '%s ' %(\"Therefore, Volume required, v2 is 3.5*a2*l=3.5*P**2*rho*l**2/3/V**2/W\")#\n",
+ "v2BYv1=3.5/3./4.##\n",
+ "print '%s %.3f %s' %(\"For 3-phase four wire system material required is \",v2BYv1,\" times the material required in two wire system.\")#\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Two wire dc system : \n",
+ "I1=P/V & W=2*I1**2*R1=2*P**2*rho*l/V**2/a1\n",
+ "Therefore, Volume required, v1 is 2*a1*l=4*P**2*rho*l**2/V**2/W\n",
+ "Three phase four wire system : \n",
+ "I2=P/3/Vas Power by each phase is P/3 & W=3*I1**2*R2=P**2*rho*l/3/V**2/a2 \n",
+ "Therefore, Volume required, v2 is 3.5*a2*l=3.5*P**2*rho*l**2/3/V**2/W \n",
+ "For 3-phase four wire system material required is 0.292 times the material required in two wire system.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E3 - Pg 52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate Additional load that can be tranmitted by converting sigle to 3-phase line\n",
+ "import math\n",
+ "print '%s ' %(\"For single phase ac system, P1=V*I1*cosd(fi) watts & W1=2*I1**2*R watts\")#\n",
+ "print '%s ' %(\"Line losses=W1/P1*100=2*I1**2*R*100/V/I1/cosd(fi)\")#\n",
+ "print '%s ' %(\"For three phase ac system, P2=sqrt(3)*V*I2*cosd(fi) watts & W2=3*I2**2*R watts\")#\n",
+ "print '%s ' %(\"Line losses=W2/P2*100=3*I2**2*R*100/sqrt(3)/V/I2/cosd(fi)\")#\n",
+ "#on equating W1/P1*100.=W2/P2*100.\n",
+ "I2BYI1=2*math.sqrt(3.)/3.#\n",
+ "P1=100\n",
+ "P2=2*P1#\n",
+ "Add_load=P2-P1#\n",
+ "Percent_add_load=(P2-P1)/P1*100##%\n",
+ "print '%s %.2f' %(\"Additional load that can be tranmitted by converting sigle to 3-phase line in %\",Percent_add_load)#\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For single phase ac system, P1=V*I1*cosd(fi) watts & W1=2*I1**2*R watts \n",
+ "Line losses=W1/P1*100=2*I1**2*R*100/V/I1/cosd(fi) \n",
+ "For three phase ac system, P2=sqrt(3)*V*I2*cosd(fi) watts & W2=3*I2**2*R watts \n",
+ "Line losses=W2/P2*100=3*I2**2*R*100/sqrt(3)/V/I2/cosd(fi) \n",
+ "Additional load that can be tranmitted by converting sigle to 3-phase line in % 100.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E4 - Pg 53 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate Extra power that can be supplied\n",
+ "print '%s' %(\"For three wire dc system, line current I1=(VS-VL)/R & P1=2*VL*I1=2*VL*(VS-VL)/R\")#\n",
+ "print '%s' %(\"For four wire three phase ac system, line current I2=(VS-VL)/R & P2=3*VL*I2*pf=3*VL*(VS-VL)/R\")#\n",
+ "#P2=3/2*2*VL*(VS-VL)/R##It implies that P2=3/2*P1\n",
+ "P1=100#\n",
+ "P2=3./2.*P1#\n",
+ "Diff=P2-P1#\n",
+ "Percent_Diff=(Diff/P1*100)##%\n",
+ "print '%s %.2f' %(\"Extra power that can be supplied in %\",Percent_Diff)#\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For three wire dc system, line current I1=(VS-VL)/R & P1=2*VL*I1=2*VL*(VS-VL)/R\n",
+ "For four wire three phase ac system, line current I2=(VS-VL)/R & P2=3*VL*I2*pf=3*VL*(VS-VL)/R\n",
+ "Extra power that can be supplied in % 50.00\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E5 - Pg 53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Additional load\n",
+ "pf=0.9##power factor\n",
+ "print '%s' %(\"Three wire dc system : \")#\n",
+ "print '%s' %(\"P1=2*I1*V & %P1loss=2*I1**2*R/(2*I1*V)*100=100*I1*R/V\")#\n",
+ "print '%s' %(\"Three phase 4-wire ac system : \")#\n",
+ "print '%s' %(\"P2=3*I1**2*V*pf & %P2loss=3*I2**2*R/(3*I2*V*pf)*100=100*I12*R/pf/V\")#\n",
+ "#on equating P1loss=P2loss#\n",
+ "I2BYI1=100*pf/100##ratio\n",
+ "#P2=3*I2*V*pf\n",
+ "P2BYI1V=3.*pf*I2BYI1#\n",
+ "P2BYP1=P2BYI1V/2.#\n",
+ "#LoadIncrease=(P2-P1)*100/P1#\n",
+ "LoadIncrease=(P2BYP1-1.)*100.##%\n",
+ "print '%s %.2f' %(\"% Additional load : \",LoadIncrease)#\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Three wire dc system : \n",
+ "P1=2*I1*V & %P1loss=2*I1**2*R/(2*I1*V)*100=100*I1*R/V\n",
+ "Three phase 4-wire ac system : \n",
+ "P2=3*I1**2*V*pf & %P2loss=3*I2**2*R/(3*I2*V*pf)*100=100*I12*R/pf/V\n",
+ "% Additional load : 21.50\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example E6 - Pg 57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate Weight of copper required for 3 conductors of 100 km length\n",
+ "#Given data :\n",
+ "import math\n",
+ "Pin=100.##MW\n",
+ "VL=380.##kV\n",
+ "d=100.##km\n",
+ "R=0.045##ohm/cm**2/km\n",
+ "w=0.01##kg/cm**3\n",
+ "Eta=90.##efficiency %\n",
+ "cosfi=1.#\n",
+ "IL=Pin*10.**6./math.sqrt(3.)/VL/10.**3./cosfi##Ampere\n",
+ "W=Pin*(1-Eta/100)##MW\n",
+ "LineLoss=W*10**6/3##Watts/conductor\n",
+ "R1=LineLoss/IL**2##in ohm\n",
+ "R2=R1/d##resistance per conductor per km\n",
+ "a=R/R2##in cm**2\n",
+ "volume=a*d*1000##cm**3 per km run\n",
+ "weight=w*volume##kg per km run\n",
+ "w3=3.*d*weight##kg(weight of copper required for 3 conductors for 100 km)\n",
+ "print '%s %.2f' %(\"Weight of copper required for 3 conductors of 100 km length(in kg) : \",w3)#\n",
+ "#Answer in the book is not accurate.\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Weight of copper required for 3 conductors of 100 km length(in kg) : 9349.03\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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