path: root/Electric_Power_Distribution_System_Engineering_by_T._Gonen/ch3_1.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 3 : Application of Distribution Transformers"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.1  Page No : 118"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "# Variables\n",
      "S = 25.*(10**3);         #Rating of the transformer in VA\n",
      "#Values in per unit\n",
      "Rt = 0.014;         #Resismath.tance of Transformer\n",
      "Xt = 0.012;         #Reacmath.tance of transformer\n",
      "Vh = 7200;         #High Voltage End in V\n",
      "Vx = 120;         # Low Voltage End in V\n",
      "Rb = (Vh**2)/S;         #Base Value of Resismath.tance\n",
      "#Accroding to Lloyd's Formula\n",
      "\n",
      "# Calculations\n",
      "Zhx12 = (1.5*Rt)+(1j*1.2*Xt);         #Impedance referred to HV side when the winding x2x3 is shorted\n",
      "\n",
      "n = Vh/Vx;         #Turns Ratio\n",
      "\n",
      "Zhx13 = Rt+(1j*Xt);         #Use of Entire low voltage winding\n",
      "\n",
      "#Impedances of the required terms in pu\n",
      "A = (2*Zhx13)-Zhx12;\n",
      "B = ((2*Zhx12)-(2*Zhx13))/(n**2);\n",
      "C = B;\n",
      "\n",
      "#Angle of Impedances\n",
      "ta = math.degrees(math.atan(A.imag/A.real));\n",
      "tb = math.degrees(math.atan(B.imag/B.real));\n",
      "\n",
      "# Results\n",
      "print 'The Circuit impedances on the high voltage side is %g/_%g ohm'%(abs(A*Rb),ta)\n",
      "print 'Each of the Circuit impedances on the low voltage side is %g/_%g ohm'%(abs(B*Rb),tb)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Circuit impedances on the high voltage side is 24.6366/_53.9017 ohm\n",
        "Each of the Circuit impedances on the low voltage side is 0.0085248/_18.9246 ohm\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.2  Page No : 119"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from numpy import exp\n",
      "\n",
      "# Variables\n",
      "#Impedances from the previous example\n",
      "Zh = 24.6437*exp(1j*53.9*math.pi/180);\n",
      "Zl = 8.525*(10**-3)*exp(1j*18.9*math.pi/180);\n",
      "#Voltages\n",
      "Vh = 7200.;         #High End\n",
      "Vx = 120.;         # Low End\n",
      "S = 25.*1000;         #Transformer Rating in VA\n",
      "N = Vh/Vx;         #Turns Ratio\n",
      "\n",
      "# Calculations\n",
      "#R of service drop is zero         #Line to Neutral Currents\n",
      "IfLVn = Vx/(Zl+((1/(N**2))*Zh));         #Secondary Fault Current\n",
      "IfHVn = IfLVn/N;         #Primary Fault Current\n",
      "\n",
      "#R of service drop is zero         #Line to Line Currents\n",
      "Nl = Vh/(2*Vx);         #New Truns Ratio\n",
      "IfLVl = 2*Vx/((2*Zl)+((1/(Nl**2))*Zh));         #Secondary Fault Current\n",
      "IfHVl = IfLVl/Nl;         #Primary Fault Current\n",
      "\n",
      "# Results\n",
      "print 'a) The Magnitude of Line to Neutral Fault Currentson HV and LV when R of service drop\\\n",
      " is zero are %g A and %g A respectively'%(abs(IfHVn),abs(IfLVn))\n",
      "print 'b) The Magnitude of Line to Line Fault Currentson HV and LV when R of service drop is zero are\\\n",
      " %g A and %g A respectively'%(abs(IfHVl),abs(IfLVl))\n",
      "print 'c) The Minimum Allowable interrupting capacity for circuit breaker isconnected to the LV is %g A'%(abs(IfLVn))\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The Magnitude of Line to Neutral Fault Currentson HV and LV when R of service drop is zero are 136.353 A and 8181.2 A respectively\n",
        "b) The Magnitude of Line to Line Fault Currentson HV and LV when R of service drop is zero are 188.283 A and 5648.5 A respectively\n",
        "c) The Minimum Allowable interrupting capacity for circuit breaker isconnected to the LV is 8181.2 A\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.3  Page No : 121"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from sympy import solve,Symbol\n",
      "from numpy import exp\n",
      "\n",
      "# Variables\n",
      "Vx = 120.;         #Low End Voltage\n",
      "#When Service drop is Zero\n",
      "IfLVn = 8181.7*exp(-1*1j*34.3*math.pi/180);         #Line to Neutral Vault Current\n",
      "IfLVl = 5649*exp(-1*1j*40.6*math.pi/180);         #Line to Line Fault Current\n",
      "\n",
      "Ral4 = 2.58;         #        #4 AWG Aluminium Conductor Resismath.tance per mile\n",
      "Ralinf = 1.03;         #        #1/0 AWG Aluminium Conductor Resismath.tance per mile\n",
      "\n",
      "# Calculations\n",
      "#Impedances when Service drop is zero, suffix l denotes line to line\n",
      "#Suffix n denotes line to neutral\n",
      "Zl0 = (2*Vx)/IfLVl;\n",
      "Zn0 = (Vx)/IfLVn;\n",
      "\n",
      "#When there is R service drop\n",
      "#Magnitudes of Line to Line and Line to Earth fault currents are equal\n",
      "\n",
      "R = Symbol('R');         #Variable Value\n",
      "#Effective Impedances\n",
      "Zl = Zl0+(2*R);\n",
      "Zn = Zn0+(2*R);\n",
      "#Fault Currents\n",
      "Ifl = 2*Vx/Zl;\n",
      "Ifn = Vx/Zn;\n",
      "#print Ifl[1]\n",
      "#Magnitudes of Currents\n",
      "MIfl = abs(240.)/abs(Ifl.subs(R,3));\n",
      "MIfn = abs(Ifn.subs(R,2))/abs(Ifn.subs(R,3));\n",
      "DI = MIfl-MIfn;\n",
      "X = - 1.5781966 + 240*R   #DI.subs(R,2);         #Polynomial Equation to find 'R'\n",
      "R = solve(X)[0];         #Numerical Value\n",
      "\n",
      "#The Magnitude of R found is Wrong in the Textbook\n",
      "\n",
      "#Length of service drop cable\n",
      "SDL4 = R/Ral4;\n",
      "SDLinf = R/Ralinf;\n",
      "\n",
      "# Results\n",
      "print 'a) The Value of Service drop in the Cable is %g ohm'%(R)\n",
      "print 'b The Length of service drop cable for:'\n",
      "print 'i)         #4 AWG Conductor is %g miles'%(SDL4) \n",
      "print 'ii)         #1/0 AWG Conductor is %g miles'%(SDLinf) \n",
      "\n",
      "#Length is printed in Miles\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The Value of Service drop in the Cable is 0.00657582 ohm\n",
        "b The Length of service drop cable for:\n",
        "i)         #4 AWG Conductor is 0.00254877 miles\n",
        "ii)         #1/0 AWG Conductor is 0.00638429 miles\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.4  Page No : 122"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Variables\n",
      "#Transformer Ratings in kVA\n",
      "Sr1 = 250.; \n",
      "Sr2 = 500.;\n",
      "\n",
      "#percentage impedances\n",
      "Zr1 = 2.4;\n",
      "Zr2 = 3.1;\n",
      "\n",
      "# Calculations\n",
      "#Ratio of Maximum Loads\n",
      "R = Sr1*Zr2/(Sr2*Zr1);\n",
      "\n",
      "#If 500 kVA is chosen as the full load transformer, Transformer 1 becomes overloaded\n",
      "SL1 = Sr1;         #To Avoid OverLoading of transformer 1\n",
      "\n",
      "SL2 = SL1/R;         #Maximum Load on transformer 2\n",
      "\n",
      "Tl = SL1+SL2;         #Total Load without overloading\n",
      "\n",
      "# Results\n",
      "print 'The Maximum Load Carried without overloading any of the transformer is %g kVA'%(Tl)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The Maximum Load Carried without overloading any of the transformer is 637.097 kVA\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.5  Page No : 127"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "# Variables\n",
      "#Considering Van as reference voltage\n",
      "SL3phi = 200*(10**3);         #Load to be powered\n",
      "pf3 = 0.8;         #Power Factor of three phase load\n",
      "t3 = math.acos(pf3);         #Power FActor Angle for three phase load\n",
      "pf1 = 0.9;         #Power Factor of math.single phase load\n",
      "t1 = math.acos(pf1);         #Power Factor angle of math.single phase load\n",
      "SL1 = 80.*(10**3);         #Single Phase Light Load\n",
      "Vll = 240.;         #Secondary Voltage\n",
      "#Rating of Single Phase Transformers between individual lines\n",
      "Sbc = 100.*(10**3);\n",
      "Sab = 75.*(10**3);\n",
      "Sca = Sab;\n",
      "#Angles of Three phase voltages \n",
      "ta = 0.;\n",
      "tb = -120.;\n",
      "tc = 120.;\n",
      "#Angles of three  line currents\n",
      "tai = ta-t3;\n",
      "tbi = tb-t3;\n",
      "tci = tc-t3;\n",
      "\n",
      "# Calculations\n",
      "I = SL3phi/(math.sqrt(3)*Vll);         #Magnitude of Current\n",
      "#3 Phase Line Currents\n",
      "Ia3 = I*exp(1j*math.pi*tai/180);\n",
      "Ib3 = I*exp(1j*math.pi*tbi/180);\n",
      "Ic3 = I*exp(1j*math.pi*tci/180);\n",
      "\n",
      "MIbc = SL1/Vll;         #Magnitude Single Phase Current\n",
      "\n",
      "tbc = -90;         #Lagging Van         #Angle of Vbc\n",
      "tbci = tbc-t1;         #Angle of Current Ibc\n",
      "Ibc = MIbc*exp(1j*math.pi*tbci/180);\n",
      "\n",
      "#Load Currents\n",
      "Ia = Ia3;\n",
      "Ta = math.degrees(math.atan(Ia.imag/Ia.real));\n",
      "Ib = Ib3+Ibc;\n",
      "Tb = math.degrees(math.atan(Ib.imag/Ib.real));\n",
      "Ic = Ic3-Ibc;         #Current is wrong in the textbook\n",
      "Tc = math.degrees(math.atan(Ic.imag/Ic.real));\n",
      "\n",
      "#Current Flowing in the secondary winding of the transformers 1,2 and 3\n",
      "Iac = ((Ic/Sbc)-(Ia/Sab))/((1/Sab)+(1/Sbc)+(1/Sca));\n",
      "T1 = math.degrees(math.atan(Iac.imag/Iac.real));         #Angle of the above current\n",
      "Iba = ((Ia/Sca)-(Ib/Sbc))/((1/Sab)+(1/Sbc)+(1/Sca));\n",
      "T2 = math.degrees(math.atan(Iba.imag/Iba.real));         #Angle of the above current\n",
      "Icb = ((Ib/Sab)-(Ic/Sca))/((1/Sab)+(1/Sbc)+(1/Sca));\n",
      "T3 = math.degrees(math.atan(Icb.imag/Icb.real));         #Angle of the above current\n",
      "\n",
      "#Kilovoltampere Load on each transformer\n",
      "SLab = Vll*abs(Iba)/1000;\n",
      "SLbc = Vll*abs(Icb)/1000;\n",
      "SLca = Vll*abs(Iac)/1000;\n",
      "\n",
      "Vlls = Vll;         #Secondary Voltage\n",
      "Vllp = 7620;         #Primary Voltage\n",
      "n = Vllp/Vlls;         #Turns Ratio\n",
      "\n",
      "#Primary Currents of the transformer\n",
      "IAC = Iac/n;\n",
      "IBA = Iba/n;\n",
      "ICB = Icb/n;\n",
      "\n",
      "#Primary Current in each each phase wire\n",
      "IA = IAC-IBA;\n",
      "TA = math.degrees(math.atan(IA.imag/IA.real));        #Angle of the above current\n",
      "IB = IBA-ICB;\n",
      "TB = math.degrees(math.atan(IB.imag/IB.real));        #Angle of the above current\n",
      "IC = ICB-IAC;\n",
      "TC = math.degrees(math.atan(IC.imag/IC.real));        #Angle of the above current\n",
      "\n",
      "# Results\n",
      "print 'a The Line Currents flowing in secondary phase wire :'\n",
      "print 'A phase is %g/_%g A'%(abs(Ia),Ta)\n",
      "print 'B phase is %g/_%g A'%(abs(Ib),Tb)\n",
      "print 'C phase is %g/_%g A'%(abs(Ic),Tc)\n",
      "print 'b The Current flowing in secondary winding of each transformer:'\n",
      "print 'AC is %g/_%g A'%(abs(Iac),T1)\n",
      "print 'AB is %g/_%g A'%(abs(Iba),T2)\n",
      "print 'BC is %g/_%g A'%(abs(Icb),T3)\n",
      "print 'c The Load on Each Transformer is:'\n",
      "print '1 : %g kVA'%(SLca)\n",
      "print '2 : %g kVA'%(SLab)\n",
      "print '3 : %g kVA'%(SLbc)\n",
      "print 'd The Current flowing in primary winding of each transformer:'\n",
      "print 'AC is %g/_%g A'%(abs(IAC),T1)\n",
      "print 'AB is %g/_%g A'%(abs(IBA),T2)\n",
      "print 'BC is %g/_%g A'%(abs(ICB),T3)\n",
      "print 'e The Line Currents flowing in primary phase wire :'\n",
      "print 'A phase is %g/_%g A'%(abs(IA),TA)\n",
      "print 'B phase is %g/_%g A'%(abs(IB),TB)\n",
      "print 'C phase is %g/_%g A'%(abs(IC),TC)\n",
      "\n",
      "#Ic is calculation is wrong, the author has added Ibc instead of subtracting, so if you change - into + in line 45, you get the answer as in the textbook\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a The Line Currents flowing in secondary phase wire :\n",
        "A phase is 481.125/_-0.643501 A\n",
        "B phase is 787.293/_71.6504 A\n",
        "C phase is 787.977/_-72.7823 A\n",
        "b The Current flowing in secondary winding of each transformer:\n",
        "AC is 316/_-40.9814 A\n",
        "AB is 315.535/_39.7662 A\n",
        "BC is 545.454/_89.442 A\n",
        "c The Load on Each Transformer is:\n",
        "1 : 75.84 kVA\n",
        "2 : 75.7283 kVA\n",
        "3 : 130.909 kVA\n",
        "d The Current flowing in primary winding of each transformer:\n",
        "AC is 9.95276/_-40.9814 A\n",
        "AB is 9.9381/_39.7662 A\n",
        "BC is 17.1796/_89.442 A\n",
        "e The Line Currents flowing in primary phase wire :\n",
        "A phase is 15.1536/_-0.643501 A\n",
        "B phase is 24.7966/_71.6504 A\n",
        "C phase is 24.8182/_-72.7823 A\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.6  Page No : 140"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\n",
      "# Variables\n",
      "#Considering Van as reference voltage\n",
      "SL3phi = 100.*(10**3);         #Load to be powered\n",
      "pf3 = 0.8;         #Power Factor of three phase load\n",
      "t3 = math.acos(pf3);         #Power FActor Angle for three phase load\n",
      "pf1 = 0.9;         #Power Factor of math.single phase load\n",
      "t1 = math.acos(pf1);         #Power Factor angle of math.single phase load\n",
      "SL1 = 50.*(10**3);         #Single Phase Light Load\n",
      "Vll = 240.;         #Secondary Voltage\n",
      "#Angles of Three phase voltages \n",
      "ta = 0.;\n",
      "tb = -120.;\n",
      "tc = 120.;\n",
      "#Angles of three  line currents\n",
      "tai = ta-t3;\n",
      "tbi = tb-t3;\n",
      "tci = tc-t3;\n",
      "\n",
      "# Calculations\n",
      "I = SL3phi/(math.sqrt(3)*Vll);         #Magnitude of Current\n",
      "#3 Phase Line Currents\n",
      "Ia3 = I*exp(1j*math.pi*tai/180);\n",
      "Ib3 = I*exp(1j*math.pi*tbi/180);\n",
      "Ic3 = I*exp(1j*math.pi*tci/180);\n",
      "\n",
      "MI1 = SL1/Vll;         #Magnitude Single Phase Current\n",
      "\n",
      "t1v = 30;         #Leading Van         #Angle of Vbc\n",
      "t1i = t1v-t1;         #Angle of Current Ibc\n",
      "I1 = MI1*exp(1j*math.pi*t1i/180);\n",
      "\n",
      "#Load Currents\n",
      "Ia = Ia3+I1;\n",
      "Ta = math.degrees(math.atan(Ia.imag/Ia.real));\n",
      "Ib = Ib3-I1;\n",
      "Tb = -180+(math.degrees(math.atan(Ib.imag/Ib.real)));\n",
      "Ic = Ic3;         #Current is wrong in the textbook\n",
      "Tc = math.degrees(math.atan(Ic.imag/Ic.real));\n",
      "\n",
      "#Current flowing in the secondary of the transformer\n",
      "Iba = Ia;\n",
      "T2 = math.degrees(math.atan(Iba.imag/Iba.real));         #Angle of the above current\n",
      "Icb = Ic;\n",
      "T3 = 180+(math.degrees(math.atan(Icb.imag/Icb.real)));         #Angle of the above current\n",
      "\n",
      "#Load on Each Transformer\n",
      "SLba = Vll*abs(Iba)/1000;\n",
      "SLcb = Vll*abs(Icb)/1000;\n",
      "\n",
      "Vlls = Vll;         #Secondary Voltage\n",
      "Vllp = 7620;         #Primary Voltage\n",
      "n = Vllp/Vlls;         #Turns Ratio\n",
      "\n",
      "#Primary Currents of the transformer\n",
      "IA = Iba/n;\n",
      "TA = math.degrees(math.atan(IA.imag/IA.real));        #Angle of the above current\n",
      "IB = Icb/n;\n",
      "TB = T3;        #Angle of the above current\n",
      "IN = IA+IB;         #Neutral Current\n",
      "TN = math.degrees(math.atan(IN.imag/IN.real));        #Angle of the above current\n",
      "\n",
      "# Results\n",
      "print 'a The Line Currents flowing in secondary phase wire :'\n",
      "print 'A phase is %g/_%g A'%(abs(Ia),Ta)\n",
      "print 'B phase is %g/_%g A'%(abs(Ib),Tb)\n",
      "print 'C phase is %g/_%g A'%(abs(Ic),Tc)\n",
      "print 'b The Current flowing in secondary winding of each transformer:'\n",
      "print 'AB is %g/_%g A'%(abs(Iba),T2)\n",
      "print 'BC is %g/_%g A'%(abs(Icb),T3)\n",
      "print 'c The Load on Each Transformer is:'\n",
      "print '1 : %g kVA'%(SLba)\n",
      "print '2 : %g kVA'%(SLcb)\n",
      "print 'd The Line Currents flowing in primary phase wire :'\n",
      "print 'AB is %g/_%g A'%(abs(IA),TA)\n",
      "print 'CB is %g/_%g A'%(abs(IB),TB)\n",
      "print 'The Neutral Current is %g/_%g'%(abs(IN),TN)\n",
      "\n",
      "#Note the mistake in the Textbook for the calulation for Neutral Current\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a The Line Currents flowing in secondary phase wire :\n",
        "A phase is 433.486/_13.3432 A\n",
        "B phase is 433.874/_-134.453 A\n",
        "C phase is 240.563/_-60.6435 A\n",
        "b The Current flowing in secondary winding of each transformer:\n",
        "AB is 433.486/_13.3432 A\n",
        "BC is 240.563/_119.356 A\n",
        "c The Load on Each Transformer is:\n",
        "1 : 104.037 kVA\n",
        "2 : 57.735 kVA\n",
        "d The Line Currents flowing in primary phase wire :\n",
        "AB is 13.6531/_13.3432 A\n",
        "CB is 7.57678/_119.356 A\n",
        "The Neutral Current is 13.6653/_45.5475\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.8  Page No : 152"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from numpy import exp\n",
      "\n",
      "\n",
      "# Variables\n",
      "Vll = 480.;         #Line to Line Voltage\n",
      "Vln = 277.;         #Line to neutral Voltage\n",
      "\n",
      "# Calculations\n",
      "#From the Phasor Diagram from the result file\n",
      "Vab = Vll*exp(1j*0);         #Vab is taken as reference\n",
      "Vabh = 50*Vab/100;\n",
      "VAB = 4160.;\n",
      "VABh = 50*VAB/100; \n",
      "VH1H2o = math.sqrt((VAB**2)-(VABh**2));\n",
      "VH1H2t = (VABh);\n",
      "Vx1x2o = 1*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
      "Vx2x3o = 2*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
      "VH2H3t = (VABh);\n",
      "Vx1x2t = Vabh;\n",
      "Vx2x3t = Vabh;\n",
      "\n",
      "# Results\n",
      "print 'a The Phasor diagram is shown in the result file attached to the code'\n",
      "print 'b) Vab is %g/_%g V'%(abs(Vab),Vab.imag/Vab.real)\n",
      "print 'c The Magnitudes of the following rated winding voltages'\n",
      "print 'i) The Voltage VH1H2 on transformer 1 : %g V'%(VH1H2o.real)\n",
      "print 'ii) The Voltage Vx1x2 on transformer 1 : %g V'%(Vx1x2o.real)\n",
      "print 'iii) The Voltage Vx2x3 on transformer 1 : %g V'%(Vx2x3o.real)\n",
      "print 'iv) The Voltage VH1H2 on transformer 2 : %g V'%(VH1H2t.real)\n",
      "print 'v) The Voltage VH2H3 on transformer 2 : %g V'%(VH2H3t.real)\n",
      "print 'vi) The Voltage Vx1x2 on transformer 2 : %g V'%(Vx1x2t.real)\n",
      "print 'vii) The Voltage Vx1x2 on transformer 2 : %g V'%(Vx2x3t.real)\n",
      "print 'd i NO ii NO iii YES'\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a The Phasor diagram is shown in the result file attached to the code\n",
        "b) Vab is 480/_0 V\n",
        "c The Magnitudes of the following rated winding voltages\n",
        "i) The Voltage VH1H2 on transformer 1 : 3602.67 V\n",
        "ii) The Voltage Vx1x2 on transformer 1 : 138.564 V\n",
        "iii) The Voltage Vx2x3 on transformer 1 : 277.128 V\n",
        "iv) The Voltage VH1H2 on transformer 2 : 2080 V\n",
        "v) The Voltage VH2H3 on transformer 2 : 2080 V\n",
        "vi) The Voltage Vx1x2 on transformer 2 : 240 V\n",
        "vii) The Voltage Vx1x2 on transformer 2 : 240 V\n",
        "d i NO ii NO iii YES\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.9  Page No : 154"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from numpy import exp\n",
      "\n",
      "# Variables\n",
      "R = 2.77;         #Resismath.tance of the balanced load\n",
      "#From Phasor Diagram in Result file\n",
      "Vab = 480*exp(1j*0);         #Reference Voltage\n",
      "MVn = abs(Vab)/math.sqrt(3);         #Magnitude of line to neutral voltages\n",
      "#Angles of Three phase voltages \n",
      "ta = -30.;\n",
      "tb = -150.;\n",
      "tc = 90.;\n",
      "\n",
      "# Calculations\n",
      "#Angles of Winding according to the Line Currents\n",
      "tx3x2 = 30;         #Leading\n",
      "tx1x2 = -30;         #Lagging\n",
      "\n",
      "I = MVn/R;         #Magnitude of current\n",
      "\n",
      "#Low Voltage Current Phasors\n",
      "Ia = I*exp(1j*math.pi*ta/180);\n",
      "Ib = I*exp(1j*math.pi*tb/180);\n",
      "Ic = I*exp(1j*math.pi*tc/180);\n",
      "pfT = math.cos(math.radians(ta-ta));         #Angle of Ia is same as phase voltage         #Resismath.tance load\n",
      "\n",
      "# Results\n",
      "print 'a The Low voltage current phasors are:'\n",
      "print 'A is %g/_%g A'%(abs(Ia),ta)\n",
      "print 'B is %g/_%g A'%(abs(Ib),tb)\n",
      "print 'C is %g/_%g A'%(abs(Ic),tc)\n",
      "print 'b The Phasor Diagram is the ''b'' diagram of in the result file'\n",
      "print 'c) The Power Factor of the Transformer is %g'%(pfT)\n",
      "print 'd) Power Factor as seen by winding x3x2 of transformer 2 is %g leading'%(math.cos(math.radians(tx3x2)))\n",
      "print 'e) Power Factor as seen by winding x1x2 of transformer 2 is %g lagging'%(math.cos(math.radians(tx1x2)))\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a The Low voltage current phasors are:\n",
        "A is 100.046/_-30 A\n",
        "B is 100.046/_-150 A\n",
        "C is 100.046/_90 A\n",
        "b The Phasor Diagram is the b diagram of in the result file\n",
        "c) The Power Factor of the Transformer is 1\n",
        "d) Power Factor as seen by winding x3x2 of transformer 2 is 0.866025 leading\n",
        "e) Power Factor as seen by winding x1x2 of transformer 2 is 0.866025 lagging\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.10  Page No : 156"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from numpy import exp\n",
      "\n",
      "#'o' and 't' represent transformer one and two respectively\n",
      "# Variables\n",
      "#Objective is to find the Factor which has to be multiplied to get VA rating\n",
      "Vll = 1.;         #Assumption Made\n",
      "#From the Phasor Diagram from the result file\n",
      "Vab = Vll*exp(1j*0);         #Vab is taken as reference\n",
      "Vabh = 50.*Vab/100;\n",
      "Vx1x2o = 1*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
      "Vx2x3o = 2*math.sqrt((Vab**2)-(Vabh**2))/3;\n",
      "Vx1x2t = Vabh;\n",
      "Vx2x3t = Vabh;\n",
      "\n",
      "#Let I be unity\n",
      "I = 1;\n",
      "\n",
      "# Calculations\n",
      "#VA Ratings of the respective windings\n",
      "Sx1x2o = Vx1x2o*I;\n",
      "Sx2x3o = Vx2x3o*I;\n",
      "Sx1x2t = Vx1x2t*I;\n",
      "Sx2x3t = Vx2x3t*I;\n",
      "\n",
      "#Total VA rating of transformer\n",
      "S1 = Sx1x2o+Sx2x3o;\n",
      "S2 = Sx1x2t+Sx2x3t;\n",
      "\n",
      "#Ratio of total rating to maximum rating\n",
      "Rt = (S1+S2)/(math.sqrt(3)*Vll*I);\n",
      "\n",
      "# Results\n",
      "print 'a) The voltampere raing of x1x2 of transformer 1 is %g*VI VA'%(Sx1x2o.real)\n",
      "print 'b) The voltampere raing of x1x2 of transformer 1 is %g*VI VA'%(Sx2x3o.real)\n",
      "print 'c) The Total Output from transformer 1 is %g*VI VA'%(S1.real)\n",
      "print 'd) The voltampere raing of x1x2 of transformer 2 is %g*VI VA'%(Sx1x2t.real)\n",
      "print 'e) The voltampere raing of x1x2 of transformer 2 is %g*VI VA'%(Sx2x3t.real)\n",
      "print 'f) The Total Output from transformer 2 is %g*VI VA'%(S2.real)\n",
      "print 'g) The Total Rating to the Maximum Continous Output is %g'%(Rt.real)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The voltampere raing of x1x2 of transformer 1 is 0.288675*VI VA\n",
        "b) The voltampere raing of x1x2 of transformer 1 is 0.57735*VI VA\n",
        "c) The Total Output from transformer 1 is 0.866025*VI VA\n",
        "d) The voltampere raing of x1x2 of transformer 2 is 0.5*VI VA\n",
        "e) The voltampere raing of x1x2 of transformer 2 is 0.5*VI VA\n",
        "f) The Total Output from transformer 2 is 1*VI VA\n",
        "g) The Total Rating to the Maximum Continous Output is 1.07735\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3.11  Page No : 157"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from numpy import exp\n",
      "\n",
      "# Variables\n",
      "#Per unit value\n",
      "Zt = 0.01+(1j*0.03);         #Transformer impedance\n",
      "\n",
      "Vll = 240.;         #Secondary Voltage\n",
      "\n",
      "Sl = 90.;         #Lighting Load\n",
      "pfl = 0.9;\n",
      "tl = math.acos(pfl);\n",
      "S = 25.;         #Balanced Load\n",
      "pf = 0.8;\n",
      "t = math.acos(pf);\n",
      "\n",
      "\n",
      "# Calculations\n",
      "def angle(y): \n",
      "    return math.degrees(math.atan(y.imag/y.real))\n",
      "\n",
      "tab = 30;         #Phase angle of Vab\n",
      "\n",
      "Il = Sl*1000/Vll;         #Magnitude of Light Load\n",
      "#Umath.sing the symmetrical - components theory\n",
      "Ia1 = Il*exp(1j*math.pi*(tab-tl)/180);\n",
      "Ta1 = angle(Ia1);         #Angle for the above current\n",
      "Ib1 = -1*Ia1;\n",
      "Ic1 = 0;         #Neutral Wire\n",
      "#Angles of three line to line voltages\n",
      "ta = 0;\n",
      "tb = -120;\n",
      "tc = 120;\n",
      "\n",
      "Ib = S*1000/(math.sqrt(3)*Vll);         #Magnitude of balanced load\n",
      "\n",
      "#Currents in Three phase load\n",
      "Ta2 = ta-t;\n",
      "Ia2 = Ib*exp(1j*math.pi*Ta2/180);\n",
      "Tb2 = tb-t;\n",
      "Ib2 = Ib*exp(1j*math.pi*Tb2/180);\n",
      "Tc2 = tc-t;\n",
      "Ic2 = Ib*exp(1j*math.pi*Tc2/180);\n",
      "\n",
      "#Currents in phase wire\n",
      "Ia = Ia1+Ia2;\n",
      "Ta = angle(Ia);         #Angle corresponding to the above angle\n",
      "Ib = Ib1+Ib2;\n",
      "Tb = angle(Ib);         #Angle corresponding to the above angle\n",
      "Ic = Ic1+Ic2;\n",
      "Tc = angle(Ic);         #Angle corresponding to the above angle\n",
      "\n",
      "#Transformer Loads\n",
      "ST1 = Vll*abs(Ia)/1000;\n",
      "T1 = 100;         #From the above value of Load, this transformer is chosen to meet the specific characteristic\n",
      "ST1pu = ST1/T1;         #Per unit Load\n",
      "ST2 = Vll*abs(Ic)/1000;\n",
      "T2 = 15;         #From the above value of Load, this transformer is chosen to meet the specific characteristic\n",
      "ST2pu = ST2/T2;         #Per unit Load\n",
      "\n",
      "#Transformer Power Factors\n",
      "pfT1 = math.cos(math.radians(tab-Ta));\n",
      "pfT2 = math.cos(math.radians(90-Tc));         #Vcb makes angle of 90\n",
      "\n",
      "Vh = 7200;         #High End Voltage\n",
      "n = Vh/Vll;         #Turns Ratio\n",
      "\n",
      "# The Primary Line Currents\n",
      "IA = Ia/n;\n",
      "IB = -1*Ic/n;\n",
      "IN = -1*(IA+IB);\n",
      "\n",
      "Ibase = T1*1000/Vll;         #Base Current\n",
      "Iapu = Ia/Ibase;\n",
      "Icpu = Ic/Ibase;\n",
      "\n",
      "#Phase Voltages\n",
      "Vab = Vll*exp(1j*math.pi*tab/180); \n",
      "Vbc = Vll*exp(-1*1j*math.pi*90/180);\n",
      "#Per Unit Voltages\n",
      "VANpu = (Vab/Vll)+(Iapu*Zt);\n",
      "VBNpu = (Vbc/Vll)-(Icpu*Zt);\n",
      "\n",
      "#Actual Voltages\n",
      "VAN = VANpu*Vh;\n",
      "VBN = VBNpu*Vh;\n",
      "\n",
      "# Results\n",
      "print 'a The Phasor Currents:'\n",
      "print 'Ia is %g/_%g A'%(abs(Ia),Ta)\n",
      "print 'Ib is %g/_%g A'%(abs(Ib),180+Tb)\n",
      "print 'Ic is %g/_%g A'%(abs(Ic),Tc)\n",
      "print 'b) Suitable ratings of the transformers are %g kVA and %g kVA'%(T1,T2)\n",
      "print 'c) The Per Unit kVA load on each transformer is %g pu and %g pu'%(ST1pu,ST2pu)\n",
      "print 'd) The power factor of output of each transformer is %g and %g both lagging'%(pfT1,pfT2)\n",
      "print 'e The phasor currents at the high voltage leads:'\n",
      "print 'IA is %g/_%g A'%(abs(IA),Ta)\n",
      "print 'IB is %g/_%g A'%(abs(IB),180+angle(IB))\n",
      "print 'IN is %g/_%g A'%(abs(IN),angle(IN))\n",
      "print 'f) VAN is %g/_%g V and VBN is %g/_%g V'%(abs(VAN),angle(VAN),abs(VBN),angle(VBN))\n",
      "\n",
      "#Highly Accuracy of Answers; Upto 5 decimal Places\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a The Phasor Currents:\n",
        "Ia is 428.052/_25.4972 A\n",
        "Ib is 428.229/_213.552 A\n",
        "Ic is 60.1407/_-60.6435 A\n",
        "b) Suitable ratings of the transformers are 100 kVA and 15 kVA\n",
        "c) The Per Unit kVA load on each transformer is 1.02732 pu and 0.96225 pu\n",
        "d) The power factor of output of each transformer is 0.996914 and -0.871586 both lagging\n",
        "e The phasor currents at the high voltage leads:\n",
        "IA is 14.2684/_25.4972 A\n",
        "IB is 2.00469/_119.356 A\n",
        "IN is 14.5415/_17.5913 A\n",
        "f) VAN is 7294.34/_31.6923 V and VBN is 7193.85/_-89.743 V\n"
       ]
      }
     ],
     "prompt_number": 20
    }
   ],
   "metadata": {}
  }
 ]
}