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diff --git a/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch10.ipynb b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch10.ipynb new file mode 100755 index 00000000..88974ba4 --- /dev/null +++ b/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch10.ipynb @@ -0,0 +1,410 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:bc905e996740356ca2e367534c643c16115b03b37f0c3f1fca3b88fd2929cfdd" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 10 : Distribution System Protection" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.1 Page No : 542" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Variables\n", + "#For Recloser\n", + "InstT = 0.03; #From Curve A #instaneous Time\n", + "TimeD = 0.17; #From Curve B #Time Delay\n", + "#For Relay\n", + "PickU = 0.42; #From Curve C #Pick Up \n", + "Reset = (1./10)*60; #Assuming a 60 s reset time for the relay with number 10 time dial setting\n", + "RecloserOT = 1; #Assumed Recloser Open Time\n", + "\n", + "RelayCTI = InstT/PickU; #Relay Closing Travel during instanmath.taneous operation\n", + "RelayRTI = (-1)*RecloserOT/Reset; #Relay Reset Travel during instanmath.taneuos\n", + "\n", + "# Calculations\n", + "RelayCTD = TimeD/PickU;\n", + "RelayRTD = (-1)*RecloserOT/Reset; #Relay Reset Travel during trip\n", + "NetRelayTravel = RelayCTD-RelayRTD;\n", + "\n", + "# Results\n", + "print 'During instanmath.taneous Operation'\n", + "print '|Relay Closing Travel| < |Relay Rest Travel|'\n", + "print '|%g percent| < |%g percent|'%(RelayCTI*100,RelayRTI*100)\n", + "\n", + "print 'During the Delayed Tripping Operations'\n", + "print 'Total Relay Travel is from:'\n", + "print '%g percent to %g percent to %g percent'%(RelayCTD*100,RelayRTD*100,RelayCTD*100)\n", + "print 'Since this Net Total Relay Travel is less than 100 percent the desired recloser to relay coordination is accomplished'\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "During instanmath.taneous Operation\n", + "|Relay Closing Travel| < |Relay Rest Travel|\n", + "|7.14286 percent| < |-16.6667 percent|\n", + "During the Delayed Tripping Operations\n", + "Total Relay Travel is from:\n", + "40.4762 percent to -16.6667 percent to 40.4762 percent\n", + "Since this Net Total Relay Travel is less than 100 percent the desired recloser to relay coordination is accomplished\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.2 Page No : 555" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "Vln = 7200.; #Line to Neutral Voltage\n", + "Vll = 12470.; #Line to Line Voltage\n", + "Z1sys = 0.7199+(1j*3.4619); #system impedance to the regulated 12.47kV bus\n", + "ZGsys = 0.6191+(1j*3.3397); #system impedance to ground\n", + "l = 2; #Dismath.tance of the Fault from the substation\n", + " #From Table 10-7; Various Paramters Can Be found out\n", + "z0a = 0.1122+(1j*0.4789);\n", + "z011 = (-0.0385-(1j*0.0996));\n", + "z1 = 0.0580+(1j*0.1208);\n", + "C = 5.28; #Cable consmath.tant\n", + "\n", + "# Calculations\n", + "Z0ckt = 2*(z0a+z011)*C; #Zero Sequence Impedance\n", + "Z1ckt = 2*z1*C; #Positive Sequence Impedance\n", + "ZGckt = ((2*Z1ckt)+Z0ckt)/3; #Impedance to ground of line\n", + " #Note That the calculation of the above term is wrong in the text book\n", + "\n", + "Z1 = Z1sys+Z1ckt; #Total Positive Sequence\n", + "ZG = ZGsys+ZGckt; #Total impedance to ground\n", + "\n", + "If3phi = Vln/abs(Z1); #Three Phase Fault at point 10\n", + "IfLL = 0.866*If3phi; #Line to Line Fault at point 10\n", + "IfLG = Vln/(abs(ZG)); #Single Line to Ground Fault\n", + "\n", + "# Results\n", + "print 'a The Zero and Postive sequence impedance of the line to point 10 are:',\n", + "print (Z0ckt),\n", + "print (Z1ckt)\n", + "print 'b The impedance to ground of the line to point 10',\n", + "print (ZGckt)\n", + "print 'c The Total positive sequence impedance including system impedance is',\n", + "print (Z1)\n", + "print 'd The Total Impedance to ground to point 10 including system impedance is',\n", + "print (ZG)\n", + "print 'All the Above impedances are in ohm'\n", + "print 'e) The Three phase fault current at point 10 is %g A'%(If3phi)\n", + "print 'f) The line to line fault current at point 10 is %g A'%(IfLL)\n", + "print 'g) The Line to Ground Current at point 10 is %g A'%(IfLG)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a The Zero and Postive sequence impedance of the line to point 10 are: (0.778272+4.005408j) (0.61248+1.275648j)\n", + "b The impedance to ground of the line to point 10 (0.667744+2.185568j)\n", + "c The Total positive sequence impedance including system impedance is (1.33238+4.737548j)\n", + "d The Total Impedance to ground to point 10 including system impedance is (1.286844+5.525268j)\n", + "All the Above impedances are in ohm\n", + "e) The Three phase fault current at point 10 is 1463.02 A\n", + "f) The line to line fault current at point 10 is 1266.97 A\n", + "g) The Line to Ground Current at point 10 is 1269.14 A\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.3 Page No : 562" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "from numpy import exp\n", + "\n", + "# Variables\n", + "St = 5*(10**6); #Capacity of Transformer\n", + "Zt = 1j*0.065; #Transformer Reacmath.tance\n", + "SB3phi = 1*(10**6); #3 Phase Power Base\n", + "VBLL = 69*(10**3); #Line to line voltage\n", + "VBLLn = 12.47*(10**3); #Line To line voltage\n", + "Vf = 1; #Per Unit Value of Voltage\n", + "Zb = (VBLL**2)/SB3phi; #Base Impedance\n", + "\n", + "#Zckt and Zf and Zt are Zero for Bus 1\n", + "#Zckt and Zf are Zero for Bus 2\n", + "#Power Generation of the system\n", + "SMax = 600*(10**6); #Maximum\n", + "SMin = 360*(10**6); #Minimum\n", + "\n", + "# Calculations\n", + "Xt = 0.065; #Transformer Reacmath.tance in per unit\n", + "MZsysmax = (VBLL**2)/SMax; #System Impedance at Maximum Power Generation\n", + "Ib = SB3phi/(math.sqrt(3)*VBLL); #Base Current\n", + "Zsysmaxpu = MZsysmax*exp(1j*math.pi*90/180)/Zb; #System Impedance Phasor\n", + "#Three Phase Fault Current\n", + "If3phimaxpu1 = abs(Vf/(Zsysmaxpu));\n", + "If3phimax1 = If3phimaxpu1*Ib;\n", + "Sf3phimax1 = math.sqrt(3)*VBLL*If3phimax1/1000000;\n", + "\n", + "#Line to Line Fault Current\n", + "IfLLmax1 = 0.866*If3phimax1;\n", + "SfLLmax1 = VBLL*IfLLmax1/1000000;\n", + "\n", + "#Line to Ground Fault\n", + "IfLGmaxpu1 = abs(3*Vf/((2*Zsysmaxpu)));\n", + "IfLGmax1 = IfLGmaxpu1*Ib;\n", + "SfLGmax1 = VBLL*IfLGmax1/(1000000*math.sqrt(3));\n", + "\n", + "Stn = SB3phi; #Numreical Value is Equal\n", + "Ztn = Zt*(Stn/St); #New Per Unit Transformer Reacmath.tance\n", + "#New Base Values\n", + "Zbn = (VBLLn**2)/SB3phi;\n", + "Ibn = Stn/(math.sqrt(3)*VBLLn);\n", + "\n", + "#Three Phase Fault Current\n", + "If3phimaxpu2 = abs(Vf/(Zsysmaxpu+Ztn));\n", + "If3phimax2 = If3phimaxpu2*Ibn;\n", + "Sf3phimax2 = math.sqrt(3)*VBLLn*If3phimax2/1000000;\n", + "\n", + "#Line to Line Fault Current\n", + "IfLLmax2 = 0.866*If3phimax2;\n", + "SfLLmax2 = VBLLn*IfLLmax2/1000000;\n", + "\n", + "#Line to Ground Fault\n", + "IfLGmaxpu2 = abs(3*Vf/((2*Zsysmaxpu)+(3*Ztn)));\n", + "IfLGmax2 = IfLGmaxpu2*Ibn;\n", + "SfLGmax2 = VBLLn*IfLGmax2/(1000000*math.sqrt(3));\n", + "\n", + "#Minimum Power Generation\n", + "MZsysmin = (VBLL**2)/SMin; #System Impedance at Maximum Power Generation\n", + "Ib = SB3phi/(math.sqrt(3)*VBLL); #Base Current\n", + "Zsysminpu = MZsysmin*exp(1j*math.pi*90/180)/Zb; #System Impedance Phasor\n", + "#Three Phase Fault Current\n", + "If3phiminpu1 = abs(Vf/(Zsysminpu));\n", + "If3phimin1 = If3phiminpu1*Ib;\n", + "Sf3phimin1 = math.sqrt(3)*VBLL*If3phimin1/1000000;\n", + "\n", + "#Line to Line Fault Current\n", + "IfLLmin1 = 0.866*If3phimin1;\n", + "SfLLmin1 = VBLL*IfLLmin1/1000000;\n", + "\n", + "#Line to Ground Fault\n", + "IfLGminpu1 = abs(3*Vf/((2*Zsysminpu)));\n", + "IfLGmin1 = IfLGminpu1*Ib;\n", + "SfLGmin1 = VBLL*IfLGmin1/(1000000*math.sqrt(3));\n", + "\n", + "Stn = SB3phi; #Numreical Value is Equal\n", + "Ztn = Zt*(Stn/St); #New Per Unit Transformer Reacmath.tance\n", + "#New Base Values\n", + "Zbn = (VBLLn**2)/SB3phi;\n", + "Ibn = Stn/(math.sqrt(3)*VBLLn);\n", + "\n", + "#Three Phase Fault Current\n", + "If3phiminpu2 = abs(Vf/(Zsysminpu+Ztn));\n", + "If3phimin2 = If3phiminpu2*Ibn;\n", + "Sf3phimin2 = math.sqrt(3)*VBLLn*If3phimin2/1000000;\n", + "\n", + "#Line to Line Fault Current\n", + "IfLLmin2 = 0.866*If3phimin2;\n", + "SfLLmin2 = VBLLn*IfLLmin2/1000000;\n", + "\n", + "#Line to Ground Fault\n", + "IfLGminpu2 = abs(3*Vf/((2*Zsysminpu)+(3*Ztn)));\n", + "IfLGmin2 = IfLGminpu2*Ibn;\n", + "SfLGmin2 = VBLLn*IfLGmin2/(1000000*math.sqrt(3));\n", + "\n", + "# Results\n", + "print 'a For Maximum Power Generation:'\n", + "print 'Bus 1'\n", + "print '3 phase fault current is %g A and %g MVA'%(If3phimax1,Sf3phimax1)\n", + "print 'Line to Line fault current is %g A and %g MVA'%(IfLLmax1,SfLLmax1)\n", + "print 'Line to ground fault current is %g A and %g MVA'%(IfLGmax1,SfLGmax1)\n", + "print 'Bus 2'\n", + "print '3 phase fault current is %g A and %g MVA'%(If3phimax2,Sf3phimax2)\n", + "print 'Line to Line fault current is %g A and %g MVA'%(IfLLmax2,SfLLmax2)\n", + "print 'Line to ground fault current is %g A and %g MVA'%(IfLGmax2,SfLGmax2)\n", + "print 'b For Minimum Power Generation:'\n", + "print 'Bus 1'\n", + "print '3 phase fault current is %g A and %g MVA'%(If3phimin1,Sf3phimin1)\n", + "print 'Line to Line fault current is %g A and %g MVA'%(IfLLmin1,SfLLmin1)\n", + "print 'Line to ground fault current is %g A and %g MVA'%(IfLGmin1,SfLGmin1)\n", + "print 'Bus 2'\n", + "print '3 phase fault current is %g A and %g MVA'%(If3phimin2,Sf3phimin2)\n", + "print 'Line to Line fault current is %g A and %g MVA'%(IfLLmin2,SfLLmin2)\n", + "print 'Line to ground fault current is %g A and %g MVA'%(IfLGmin2,SfLGmin2)\n", + "\n", + "#Note that 0.00166666666 is not rounded as 0.0017\n", + "#Hence you find all the answers close by\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a For Maximum Power Generation:\n", + "Bus 1\n", + "3 phase fault current is 5691.02 A and 680.143 MVA\n", + "Line to Line fault current is 4928.43 A and 340.061 MVA\n", + "Line to ground fault current is 8536.54 A and 340.071 MVA\n", + "Bus 2\n", + "3 phase fault current is 31490 A and 680.143 MVA\n", + "Line to Line fault current is 27270.4 A and 340.061 MVA\n", + "Line to ground fault current is 47235 A and 340.071 MVA\n", + "b For Minimum Power Generation:\n", + "Bus 1\n", + "3 phase fault current is 3064.4 A and 366.231 MVA\n", + "Line to Line fault current is 2653.77 A and 183.11 MVA\n", + "Line to ground fault current is 4596.6 A and 183.115 MVA\n", + "Bus 2\n", + "3 phase fault current is 16956.2 A and 366.231 MVA\n", + "Line to Line fault current is 14684 A and 183.11 MVA\n", + "Line to ground fault current is 25434.3 A and 183.115 MVA\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 10.4 Page No : 572" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "# Variables\n", + "#Percent Impedances of the substation transformer\n", + "Rtp = 1.;\n", + "Ztp = 7.;\n", + "Xtp = math.sqrt((Ztp**2)-(Rtp**2)); \n", + "Ztpu = Rtp+(1j*Xtp); #Transformer Impedance\n", + "Vll = 12.47; #Line to Line voltage in kV\n", + "Vln = 7.2; #Line to Neutral Voltage\n", + "V = 240.; #Secondary Voltage\n", + "St = 7500.; #Rating of the transformer in kVA\n", + "Sts = 100.; #Rating of Secondary Transformer\n", + "Ztp = Ztpu*((Vll**2)*10/St);\n", + "SSC = complex(.466,0.0293);\n", + "#From Table 10-7\n", + "Z1 = 0.0870+(1j*0.1812);\n", + "Z0 = complex(0.1653,0.4878);\n", + "\n", + "ZG = ((2*Z1)+Z0)/3; #Impedance to Ground\n", + "\n", + "Zsys = 0 ; #Assumption Made\n", + "Zeq = Zsys+Ztp+ZG; #Equivalent Impedance of the Primary\n", + "\n", + "PZ2 = Zeq*((V/(Vln*1000))**2); #Primary Impedance reffered to secondary\n", + "\n", + "# Calculations\n", + "#Distribution Tranformer Parameters\n", + "Rts = 1;\n", + "Zts = 1.9;\n", + "Xts = math.sqrt((Zts**2)-(Rts**2));\n", + "Ztspu = complex(Rts,Xts);\n", + "\n", + "Zts = Ztspu*((V/1000)**2)*10/Sts; #Distribution Transformer Reacmath.tance\n", + "\n", + "Z1SL = (60./1000)*SSC; #Impedance for 60 feet\n", + "\n", + "Zeq1 = PZ2+Zts+Z1SL; #Total Impedance to the fault in secondary\n", + "\n", + "IfLL = V/abs(Zeq1); #Fault Current At the secondary fault point F\n", + "\n", + "\n", + "# Results\n", + "print 'a The Impedance of the substation in ohms',\n", + "print (Ztp)\n", + "print 'b The Positive And Zero Sequence Impedances in ohms',\n", + "print (Z1),\n", + "print (Z0)\n", + "print 'c The Line to Ground impedance in the primary system in ohms',\n", + "print (ZG)\n", + "print 'd The Total Impedance through the primary in ohms',\n", + "print (Zeq)\n", + "print 'e The Total Primary Impedance referred to the secondary in ohms',\n", + "print (PZ2)\n", + "print 'f The Distribution transformer impedance in ohms',\n", + "print (Zts)\n", + "print 'g the Impedance of the secondary cable in ohms',\n", + "print (Z1SL)\n", + "print 'h The Total Impedance to the fault in ohms',\n", + "print (Zeq1)\n", + "print 'i) The Single Phase line to line fault for the 120/240 V three-wire service in amperes is %g A'%(IfLL)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "a The Impedance of the substation in ohms (0.207334533333+1.43645578359j)\n", + "b The Positive And Zero Sequence Impedances in ohms (0.087+0.1812j) (0.1653+0.4878j)\n", + "c The Line to Ground impedance in the primary system in ohms (0.1131+0.2834j)\n", + "d The Total Impedance through the primary in ohms (0.320434533333+1.71985578359j)\n", + "e The Total Primary Impedance referred to the secondary in ohms (0.00035603837037+0.00191095087065j)\n", + "f The Distribution transformer impedance in ohms (0.00576+0.00930556478673j)\n", + "g the Impedance of the secondary cable in ohms (0.02796+0.001758j)\n", + "h The Total Impedance to the fault in ohms (0.0340760383704+0.0129745156574j)\n", + "i) The Single Phase line to line fault for the 120/240 V three-wire service in amperes is 6582.1 A\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
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