{
"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": {}
}
]
}