path: root/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch9_1.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 9 : Distribution System Voltage Regulation"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.1  Page No : 468"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from numpy import exp\n",
      "\n",
      "# Variables\n",
      "#Base Value\n",
      "S3phib = 15;         #in MVA\n",
      "Vllst = 69;         #in kV\n",
      "Vllp = 13.2;         #in kV\n",
      "Vrrb = 120.;\n",
      "\n",
      "Ztpu = 1j*0.08;         #Transformer impedance per unit length\n",
      "VSTpuop = 1.05*exp(1j*0);         #Per Unit Maximum Subtransmission Voltage Off Peak\n",
      "VSTpup = 1.00*exp(1j*0);         #Per Unit Maximum Subtransmission Voltage Peak\n",
      "pftop = 0.95;         #Off Peak kilovoltageamperage power factor\n",
      "Sop = 0.25;         #Off Peak kilovoltageamperage\n",
      "pftp = 0.85;         #Off Peak kilovoltageamperage power factor\n",
      "Sp = 1.0;         #Off Peak kilovoltageamperage\n",
      "Regpu = 5./(8*100);         #Regulated percent volts for 32 steps\n",
      "K = 3.88*(10**-6);         #Drop Consmath.tant\n",
      "S = 4000.;         # Peak Load in kVA\n",
      "l = 10.;         #Length of the feeder\n",
      "#Case A\n",
      "Rset = 0.;\n",
      "Xset = 0.;\n",
      "Vpmax = 1.0417;\n",
      "BW = 0.0083;\n",
      "VRRpu = (Vpmax-BW);         #Setting of VRR\n",
      "VRR = (Vpmax-BW)*Vrrb;\n",
      "\n",
      "# Calculations\n",
      "#Case B\n",
      "IPpuop = (Sop/VSTpuop)*exp(1j*math.acos(pftop)*math.pi/180);         #No Load Primary Current at substation Off Peak\n",
      "VPpuop  = VSTpuop-(IPpuop*Ztpu);         #Highest Allowable Primary Voltage Off Peak\n",
      "IPpup = (Sp/VSTpup)*exp(-1*1j*math.acos(pftp)*math.pi/180);         #No Load Primary Current at substation Peak\n",
      "VPpup  = VSTpup-(IPpup*Ztpu);         #Highest Allowable Primary Voltage  Peak\n",
      "\n",
      "Step1 = (abs(VPpuop)-VRRpu)/Regpu;         #Off Peak Number Steps\n",
      "#To find the positive step value\n",
      "Step2 = -1*(abs(VPpup)-VRRpu)/Regpu;         # Peak Number Steps\n",
      "\n",
      "\n",
      "#Case C\n",
      "#At Peak Load Primary Voltages\n",
      "MaxVpp = 1.075;         #Max\n",
      "MinVpp = 1.000;         #Min\n",
      "\n",
      "TVDpu = K*S*l/2;         #Total Voltage Drop\n",
      "MinVPpu = VRRpu-TVDpu;\n",
      "\n",
      "#At Annual Peak Load Primary Voltages\n",
      "APMaxVPpu = MaxVpp-BW;         #Max\n",
      "APMinVPpu = MinVpp+BW;         #Min\n",
      "\n",
      "#At no load Load Primary Voltages\n",
      "NLMaxVPpu = Vpmax-BW;         #Max\n",
      "NLMinVPpu = APMinVPpu;         #Min\n",
      "\n",
      "# Results\n",
      "print 'a)The Setting of the VRR for the highest allowable primary voltage is %g V'%(VRR)\n",
      "print 'b The Maximum Number of Steps of buck and boost for:'\n",
      "print 'Off Peak : %g steps'%(math.ceil(Step1))\n",
      "print 'Peak : %g steps'%(math.ceil(Step2))\n",
      "print 'c) At Annual Load%( Significant Values on Voltage Curve'\n",
      "print 'The Total Voltage Drop is %g pu V'%(TVDpu)\n",
      "print 'The Minimum Feeder Voltage at the end of the feeder is %g'%(MinVPpu)\n",
      "print 'The Minimum and Maximum Primary Voltages at Peak Load is %g pu V and %g pu V'%(APMaxVPpu,APMinVPpu)\n",
      "print 'The Minimum and Maximum Primary Voltages at Peak Load is %g pu V and %g pu V'%(NLMaxVPpu,NLMinVPpu)\n",
      "\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a)The Setting of the VRR for the highest allowable primary voltage is 124.008 V\n",
        "b The Maximum Number of Steps of buck and boost for:\n",
        "Off Peak : 3 steps\n",
        "Peak : 5 steps\n",
        "c) At Annual Load%( Significant Values on Voltage Curve\n",
        "The Total Voltage Drop is 0.0776 pu V\n",
        "The Minimum Feeder Voltage at the end of the feeder is 0.9558\n",
        "The Minimum and Maximum Primary Voltages at Peak Load is 1.0667 pu V and 1.0083 pu V\n",
        "The Minimum and Maximum Primary Voltages at Peak Load is 1.0334 pu V and 1.0083 pu V\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.2  Page No : 472"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from sympy import Symbol,solve\n",
      "\n",
      "# Variables\n",
      "#Terms from previous example\n",
      "TVDpu = 0.0776;         #Total Voltage Drop\n",
      "VRRpu = 1.035;         #Setting Voltage of Regulator\n",
      "l = 10.;        #Length of the Feeder\n",
      "\n",
      "#Primary voltages for various cases\n",
      "VPpua = 1.01;\n",
      "VPpub = 1.00;\n",
      "\n",
      "s1 = Symbol('s1');         #Variable Value of Regulator length\n",
      "#Function to find the equation for the regulator dismath.tance\n",
      "def dist(y): \n",
      "    return (s1*(2-(s1/l))/l)-((VRRpu-y)/TVDpu)\n",
      "\n",
      "# Calculations\n",
      "#Different Cases\n",
      "Xa = dist(VPpua);\n",
      "Xb = dist(VPpub);\n",
      "\n",
      "s1a = solve(Xa);\n",
      "if((abs(l-s1a[0])+(l-s1a[0])) == 0):\n",
      "    s1a = s1a[1];\n",
      "else:\n",
      "    s1a = s1a[0];\n",
      "\n",
      "s1b = solve(Xb);\n",
      "if((abs(l-s1b[0])+(l-s1b[0])) == 0):\n",
      "    s1b = s1b[1];\n",
      "else:\n",
      "    s1b = s1b[0];\n",
      "\n",
      "# Results\n",
      "print 'a) The Regulator must be placed at %g miles from the start of the feeder'%(s1a)\n",
      "print 'b) The Regulator must be placed at %g miles from the start of the feeder'%(s1b)\n",
      "print 'c The Advantage of a over b is that it can compensate for future growth'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The Regulator must be placed at 1.76693 miles from the start of the feeder\n",
        "b) The Regulator must be placed at 2.59076 miles from the start of the feeder\n",
        "c The Advantage of a over b is that it can compensate for future growth\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.3  Page No : 473"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "# Variables\n",
      "l = 10.;         #Length of the feeder\n",
      "S3phi = 4000.;         #Annual Peak Load in kVA\n",
      "VPpu = 1.01;         #Primary Feeder Voltage\n",
      "s1 = 1.75;         # Dismath.tance of the Regulator\n",
      "Rmax = 10./100;         #Regulation Percent\n",
      "\n",
      "# Calculations\n",
      "S = S3phi*(1-(s1/l));         #Uniformly Distributed three phase load\n",
      "Sph = S/3;         #Single Phase Load\n",
      "\n",
      "Sreg = Rmax*Sph;         #Regulated Size\n",
      "\n",
      "# Results\n",
      "print 'The Calculated Circuit Kilovoltampere Size is %g kVA, And The corresponding Minimum kilovoltampere size \\\n",
      "of the regulator size can be found as 114.3 kVA'%(Sreg)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Calculated Circuit Kilovoltampere Size is 110 kVA, And The corresponding Minimum kilovoltampere size of the regulator size can be found as 114.3 kVA\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.4  Page No : 474"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "#To specify the best settings for regulation\n",
      "#Page 474\n",
      "\n",
      "# Variables\n",
      "s1 = 1.75;         #As Found in Example 2\n",
      "VRRpu = 1.035;         #As R and X are zero, the Settings turn out to produce this\n",
      "\n",
      "#Parameters of Distribution\n",
      "K = 3.88*(10**-6);\n",
      "S = 3300.;\n",
      "l = 10.;         #length of the line\n",
      "\n",
      "# Calculations\n",
      "VDpu = K*S*(l-s1)/2;         #Per unit voltage drop\n",
      "\n",
      "VP = VRRpu-VDpu;         #Primary Feeder Voltage\n",
      "\n",
      "#We Obtain VDs  =  K*(S3-((S3*s)/l))s+K(S*s/l)s/2;\n",
      "#We take various values of s and carry out the computation and hence form the table 9-4 given given in the result file\n",
      "\n",
      "#We Obtain from the voltage drop value for any give point s between the substation and the regulator  station as\n",
      "#VDs = I(r.math.cos(theta)+ del math.sin(theta))s*(1-(s/(2*l)))\n",
      "\n",
      "#We finally Get VDs  =  3.88 * (10**-6) * (3300-(3300s/8.25))s+3.88*(10**-6)*(3300s/8.25)*s/2\n",
      "\n",
      "# Results\n",
      "#Again for different values of s we get the table 9-5\n",
      "print 'a)The Best Settings for LDCs R and X, and for the VRR'\n",
      "print 'The best settings for LDC of the regulator are when settings for both R and X are zero and VRRpu  =  %g pu V'%(VRRpu)\n",
      "print 'b)The Voltage Drop occuring in the feeder portion between the regulating point and the end of the feeder is %g pu V'%(VDpu)\n",
      "print 'The Result Files give the Profiles and relevant information about the solution'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a)The Best Settings for LDCs R and X, and for the VRR\n",
        "The best settings for LDC of the regulator are when settings for both R and X are zero and VRRpu  =  1.035 pu V\n",
        "b)The Voltage Drop occuring in the feeder portion between the regulating point and the end of the feeder is 0.0528165 pu V\n",
        "The Result Files give the Profiles and relevant information about the solution\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.5  Page No : 478"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "#To determine the setting of the regulator so that the voltage criteria is met\n",
      "#Page 478\n",
      "\n",
      "# Variables\n",
      "l = 10.;         #Length of the feeder\n",
      "s1 = 1.75;\n",
      "ra = 0.386;\n",
      "xa = 0.4809;\n",
      "xd = 0.1802;\n",
      "xL = xa+xd;\n",
      "Vb = 120;\n",
      "pf = 0.85;         #Power Factor\n",
      "S = 1100.;         #Load in kVA\n",
      "Vln = 7.62;         #line to neutral voltage in kV\n",
      "Reff = ra*(l-s1)/2;\n",
      "Xeff = xL*(l-s1)/2;\n",
      "\n",
      "#Primary Ratings\n",
      "CTp = 150;         #Current Tranformer\n",
      "PTn = 63.5;         #POtential Transformer\n",
      "\n",
      "# Calculations\n",
      "#R Value of the dial\n",
      "Rset = (CTp/PTn)*Reff;\n",
      "Rsetpu = Rset/Vb;\n",
      "\n",
      "#X value of the dial\n",
      "Xset = (CTp/PTn)*Xeff;\n",
      "Xsetpu = Xset/Vb;\n",
      "\n",
      "VRP = 1.0138;         #Assumption Made on the Regulating Point\n",
      "#Output Voltage of the Regulator\n",
      "Vreg = VRP+((S/Vln)*((Rset*pf)+(Xset*math.sin(math.radians(math.acos(pf)))))/(CTp*Vb)); \n",
      "\n",
      "# Results\n",
      "print 'a) The Regulating Voltage is %g pu V'%(Vreg)\n",
      "print 'As per Table 9-6; the primary voltage criteria are met by using the R and X settings'\n",
      "print 'b The Voltage Profiles are given in the result file attached'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The Regulating Voltage is 1.03994 pu V\n",
        "As per Table 9-6; the primary voltage criteria are met by using the R and X settings\n",
        "b The Voltage Profiles are given in the result file attached\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.6  Page No : 480"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "# Variables\n",
      "#From Problems 4 and 5 the co-effcients are obtained\n",
      "VRRpu = 1.035;\n",
      "Vreg4 = 1.0337;\n",
      "Vreg5 = 1.0666;\n",
      "VRP4 = 1.0337;\n",
      "VRP5 = 1.0138;\n",
      "Vmin = 1.010;         #For s =  1.75\n",
      "\n",
      "# Calculations\n",
      "#Steps\n",
      "Buck4 = (VRRpu-VRP4)/(0.00625);\n",
      "Buck5 = (VRRpu-VRP5)/(0.00625);\n",
      "Boost4 = (Vreg4-Vmin)/(0.00625);\n",
      "Boost5 = (Vreg5-Vmin)/(0.00625);\n",
      "\n",
      "# Results\n",
      "print 'a) The Number of steps of buck and number is steps of boost in example 9-4 is %g and %g respectively'%(Buck4,Boost4)\n",
      "print 'b) The Number of steps of buck and number is steps of boost in example 9-5 is %g and %g respectively'%(Buck5,Boost5)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The Number of steps of buck and number is steps of boost in example 9-4 is 0.208 and 3.792 respectively\n",
        "b) The Number of steps of buck and number is steps of boost in example 9-5 is 3.392 and 9.056 respectively\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.8  Page No : 482"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "# Variables\n",
      "l = 3.;        #Length of the line\n",
      "Vlc = 2450.;         #Regulated Voltage\n",
      "Vcp = 12800.;         #Primary of customer transformer\n",
      "        #Base Values\n",
      "Vlbp = 2400.;         #Primary Bus Voltage of Customer's Bus(Low Voltage)\n",
      "Vlbs = 4160.;        #Secondary Bus Voltage of Customer's Bus\n",
      "Sb = 5000.;         #Power in kVA\n",
      "r = 0.3;         #Line Resismath.tance per mile\n",
      "x = 0.8;         #Line Reacmath.tance per mile\n",
      "Vhbp = 7390.;        #Primary Bus Voltage of High Voltage Bus\n",
      "Vhbs = 12800.;        #Secondary Bus Voltage of High Voltage Bus\n",
      "PTn = 63.5;         #Potential Transformer Turns Ratio\n",
      "CTp = 250.;        #Current Transformer Turns Ratio\n",
      "VRP = Vlc/Vlbp;        #Voltage at RP \n",
      "Vll = Vhbs/1000;         #Line Voltage\n",
      "VBsec = Vcp/(math.sqrt(3)*PTn);         #Secondary Reading of the Customer Transformer\n",
      "\n",
      "# Calculations\n",
      "VRRset = VRP*VBsec;         #Setting of the voltage-setting dial of VRR\n",
      "\n",
      "Zb = (Vll**2)*1000/Sb;         #Applicable Impedance Base\n",
      "Ztpu = 0.05*1j;        #Transformer Impedance per unit\n",
      "Zt = Ztpu*Zb;        #Transformer Impedance\n",
      "\n",
      "        #Effective Resismath.tances and Reacmath.tances\n",
      "Reff = (r*l)+Zt.real;\n",
      "Xeff = (x*l)+Zt.imag;\n",
      "\n",
      "Rset = CTp*Reff/PTn;         #X Dial Setting of LDCs\n",
      "Xset = CTp*Xeff/PTn;         #X Dial Setting of LDCs\n",
      "\n",
      "# Results\n",
      "print 'a) The Necessary Setting of the voltage-setting dial of the VRR of each single phase regulator in use is %g V'%(VRRset)\n",
      "print 'b) R and X dial settings of LDS is %g ohm and %g ohm respectively'%(Rset,Xset)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The Necessary Setting of the voltage-setting dial of the VRR of each single phase regulator in use is 118.804 V\n",
        "b) R and X dial settings of LDS is 3.54331 ohm and 15.8992 ohm respectively\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.9  Page No : 484"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "#To Determine the Design Parameters of a Distributed System\n",
      "#Page 484\n",
      "\n",
      "# Variables\n",
      "VPpu = 1.035;         #Primary Feeder Voltage per unit\n",
      "TVDpu = 0.0776;         #Total Voltage Drop of Feeder\n",
      "Vll = 13.2;         #Line Voltage in kV\n",
      "Vlpuqsw = 1;        #New Voltage at the End of the Feeder due to Qsw at annual peak load\n",
      "XL = 0.661;         #Inductive Reacmath.tance per mile\n",
      "Pl = 3400;         #Real Power\n",
      "Ql = 2100;         #Reactive Power\n",
      "l = 10.;         #Length of the Feeder in Miles\n",
      "Lf = 0.4;         #Load Factor\n",
      "CR = 0.27;         #Total Capacitor Compensation Ratio For the Above Load Factor\n",
      "QNSW = CR*Ql;         #Required Size of the Nonswitched capacitor bank\n",
      "s = 2*l/3;         #Loacation of Nonswitched capacitor bank for Optimum Result\n",
      "VRpu = QNSW*(2*XL*l/3)/(1000*(Vll**2));         #Per Unit Voltage Rise\n",
      "VDspu = TVDpu*s*(2-(s/l))/l;         #Voltage drop for the uniformaly distributed load\n",
      "\n",
      "VSpu = VPpu-VDspu;        #Feeder Voltage at 2l/3 dismath.tance\n",
      "\n",
      "nVSpu = VSpu+VRpu;         #New Voltage Rise when there is a fixed capacitor bank\n",
      "\n",
      "Vlpu = VPpu-TVDpu;         #When No Capcacitor bank is there, the voltage at the end of the feeder\n",
      "\n",
      "nVlpu = Vlpu+VRpu;         #When Capcacitor bank is there, the voltage at the end of the feeder\n",
      "VRpuqsw = Vlpuqsw-nVlpu;         #Required Voltage Rise\n",
      "\n",
      "Q3phisw = 1000*(Vll**2)*VRpuqsw/(XL*l);         #Required Size of the Capacitor Bank\n",
      "\n",
      "# Calculations\n",
      "#Let us assume the 15 single phase standard 50 kVAr Capacitor Units  =  750 kVAr\n",
      "\n",
      "SQ3phisw = 750;         #Selected Capacitor Bank\n",
      "\n",
      "RVRlpu = VRpuqsw*SQ3phisw/Q3phisw;         #Resulmath.tant Voltage Rises at dismath.tance l\n",
      "RVRspu = RVRlpu*s/l;         #Resulmath.tant Voltage Rises at dismath.tance s\n",
      "\n",
      "#At Peak Load when both the Non-Switched and Switched Capacitor Banks are on\n",
      "\n",
      "PVspu = nVSpu+RVRspu;         #Voltage at s\n",
      "PVlpu = nVlpu+RVRlpu;         #Voltage at l\n",
      "\n",
      "# Results\n",
      "print 'a) The NSW Capacitor Rating is %g kVAr, Which means 2 100kVAr Capacitor Banks per phase'%(QNSW)\n",
      "print 'b) Voltage Rise per unit is %g pu V'%(VRpu)\n",
      "print 'i When the No Capacitor Bank is Installed '\n",
      "print 'Voltage at %g miles is %g pu V'%(s,VSpu)\n",
      "print 'Voltage at %g miles is %g pu V'%(l,Vlpu)\n",
      "print 'ii When the Fixed Capacitor Bank is Installed '\n",
      "print 'Voltage at %g miles is %g pu V'%(s,nVSpu)\n",
      "print 'Voltage at %g miles is %g pu V'%(l,nVlpu)\n",
      "print 'c) At Annual Peak Load, The Size of Capacitor Bank Required is %g'%(Q3phisw)\n",
      "print 'The Voltage Rise at the Annual Load for the Required Capacitor Bank is %g pu V'%(VRpuqsw)\n",
      "\n",
      "#Note That The Calculations are highly accurate, Rounding of Terms hasn't be emplyed\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The NSW Capacitor Rating is 567 kVAr, Which means 2 100kVAr Capacitor Banks per phase\n",
        "b) Voltage Rise per unit is 0.0143399 pu V\n",
        "i When the No Capacitor Bank is Installed \n",
        "Voltage at 6.66667 miles is 0.966022 pu V\n",
        "Voltage at 10 miles is 0.9574 pu V\n",
        "ii When the Fixed Capacitor Bank is Installed \n",
        "Voltage at 6.66667 miles is 0.980362 pu V\n",
        "Voltage at 10 miles is 0.97174 pu V\n",
        "c) At Annual Peak Load, The Size of Capacitor Bank Required is 744.939\n",
        "The Voltage Rise at the Annual Load for the Required Capacitor Bank is 0.0282601 pu V\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.10  Page No : 488"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from numpy import exp\n",
      "#To Determine the proper 3 phase capacitor bank\n",
      "#Page 488\n",
      "\n",
      "# Variables\n",
      "V = 12.8;         #Voltage in kV\n",
      "xl = 0.8;         #Reacmath.tance per unit length\n",
      "l = 3;         #Dismath.tance of the line\n",
      "Xl = xl*l;         #Effective Reacmath.tance of the the Line\n",
      "pf = 0.8;         #Initial Power Factor\n",
      "pfn = 0.88;         #New Improved Power Factor\n",
      "Qcu = 150;         #Capacity of each unit available\n",
      "XT = 1.6384;         #Reacmath.tance of the transformer\n",
      "\n",
      "# Calculations\n",
      "S3phi = 5000*exp(1j*math.pi*math.acos(pf)/180);         #Presently existing Load\n",
      "\n",
      "#For New Load Real part of the Load doesn't Change\n",
      "\n",
      "QLnew = (S3phi.real)*math.degrees(math.atan(math.acos(pfn)));         #The Required VAr\n",
      "\n",
      "S3phinew  = math.sqrt(((S3phi.real)**2)+(QLnew**2));         #New Apparent Power\n",
      "\n",
      "Qc = (S3phi.imag)-QLnew;         #Minimum Size of capacitor bank;\n",
      "\n",
      "N = math.ceil(Qc/Qcu);         #Number of Units Required\n",
      "Qcn = N*Qcu;         #Required VAr\n",
      "\n",
      "XL = Xl+XT;         #Total Reacmath.tance\n",
      "\n",
      "VRpu = Qcn*XL/(1000*(V**2));         #Voltage Rise Per unit\n",
      "\n",
      "# Results\n",
      "print 'The The Voltage Rise found out is %g pu V, which is greater than the voltage rise criterion.Hence %g Capacitor units\\\n",
      " of %g kVAr must be installed'%(VRpu,N,Qcu)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The The Voltage Rise found out is -3.2425 pu V, which is greater than the voltage rise criterion.Hence -877 Capacitor units of 150 kVAr must be installed\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.11  Page No : 493"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "Skva = 6.3*(10**3);         #Starting kVA per HP of the Motor\n",
      "HPmotor = 10.;         #Power Rating\n",
      "Vll = 7.2*(10**3);         #Line Voltage\n",
      "I3phi = 1438.;         #Fault Current\n",
      "\n",
      "# Calculations\n",
      "Sstart = Skva*HPmotor;         #Starting kVA\n",
      "VDIP = 120*Sstart/(I3phi*Vll);         #Voltage Dip\n",
      "\n",
      "# Results\n",
      "print 'a) The Voltage dip due to the motor start is %g V'%(VDIP)\n",
      "print 'b) From the Permissible voltage flicker limit curve%( The Voltage dip of 0.73 Vwith a frequency of\\\n",
      " 15 times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The Voltage dip due to the motor start is 0.730181 V\n",
        "b) From the Permissible voltage flicker limit curve%( The Voltage dip of 0.73 Vwith a frequency of 15 times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers\n"
       ]
      }
     ],
     "prompt_number": 20
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.12  Page No : 495"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "Skva = 5.6*(10**3);         #Starting kVA per HP of the Motor\n",
      "HPmotor = 100.;         #Power Rating\n",
      "Vll = 12.47*(10**3);         #Line Voltage\n",
      "I3phi = 1765.;         #Fault Current\n",
      "\n",
      "# Calculations\n",
      "Sstart = Skva*HPmotor;         #Starting kVA\n",
      "VDIP = 69.36*Sstart/(I3phi*Vll);         #Voltage Dip\n",
      "\n",
      "# Results\n",
      "print 'a) The Voltage dip due to the motor start is %g V'%(VDIP)\n",
      "print 'b) From the Permissible voltage flicker limit curve, The Voltage dip of 1.72 Vwith a frequency of three\\\n",
      " times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The Voltage dip due to the motor start is 1.76476 V\n",
        "b) From the Permissible voltage flicker limit curve, The Voltage dip of 1.72 Vwith a frequency of three times per hour is in the satisfactory flicker zone and therefore is not objectionable to the immediate customers\n"
       ]
      }
     ],
     "prompt_number": 21
    }
   ],
   "metadata": {}
  }
 ]
}