{
 "metadata": {
  "name": "",
  "signature": "sha256:a93d445dad4ffd499630570fa7ced24b4b25ee06fcd5352ae47d0eb721a47db5"
 },
 "nbformat": 3,
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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "CHAPTER1 : WHAT MACHINES AND TRANSFORMERS HAVE IN COMMON"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E01 : Pg 16"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt\n",
      "horsepower=2.5  # rating of induction motor in horsepower at half load\n",
      "Vl=230. # terminal voltage of motor in volts\n",
      "Il=7.   # load current of motor in amperes\n",
      "pf=0.8 # power factor of the machine\n",
      "Pin=sqrt(3.)*Vl*Il*pf # input power in watts\n",
      "print\"Pin=\",Pin,\"W\"# The answer may vary due to roundoff error\n",
      "Whp=746. # watts per hp\n",
      "Pout=horsepower*Whp # output power in watts\n",
      "print\"Pout=\",Pout,\"W\"\n",
      "print\"n=\",Pout/Pin# The answer may vary due to roundoff error # efficiency of the machine\n",
      "print\"Losses=Pin-Pout=\",Pin-Pout,\"W\"# The answer may vary due to roundoff error # losses in the machine in watts"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Pin= 2230.88144015 W\n",
        "Pout= 1865.0 W\n",
        "n= 0.835992431707\n",
        "Losses=Pin-Pout= 365.881440149 W\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E02 : Pg 17"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# the below exmaple is an extension of Ex1_1.sce\n",
      "from math import sqrt \n",
      "Vl=230. # terminal voltage of machine in volts\n",
      "Il=7. # current drawn by machine in amperes\n",
      "pf=0.8 # power factor of machine\n",
      "Pin=sqrt(3.)*Vl*Il*pf # from Ex1_1 # input power in watts\n",
      "Losses=365. # in watts\n",
      "Pout=Pin-Losses # output power in watts\n",
      "Whp=746. # watts per hp\n",
      "print\"n=1-(Losses/Input)=\",1.-(Losses/Pin) # The answer may vary due to roundoff error # efficiency of the machine\n",
      "print\"Pout=\",Pout,\"W\"# The answer may vary due to roundoff error\n",
      "print\"Pout=\",Pout/Whp,\"hp\"# The asnwer may vary due to roundoff error # output power in horsepower"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "n=1-(Losses/Input)= 0.836387540175\n",
        "Pout= 1865.88144015 W\n",
        "Pout= 2.50118155516 hp\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E03 : Pg 17"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt,pi \n",
      "f=60. # frequency of voltage source in Hz\n",
      "x=1.9 # Steinmetz coefficient\n",
      "V=80. # applied sinusoidal voltage in volts\n",
      "t=100. # no of turns wound on a coil\n",
      "hc=500. # hysteresis coefficient \n",
      "w=2.*pi*f # angular frequency in rads/sec\n",
      "phimax=(sqrt(2.)*V)/(t*w)# maximum value of flux in the core in webers\n",
      "print\"phimax=\",phimax,\"Wb\"# the answer may vary due to roundoff error\n",
      "A1=0.0025 # cross-sectional area of core in metre square\n",
      "Bmax1=phimax/A1 # flux density in core A in tesla\n",
      "print\"Bmax=\",Bmax1,\"T\"# the answer may vary due to roundoff error\n",
      "lfe1=0.5 # mean flux path length of core A in meters\n",
      "VolA=A1*lfe1 # volume of core A in metre cube\n",
      "print\"VolA=\",VolA,\"metre cube\"\n",
      "# for core A\n",
      "Ph1=VolA*f*hc*(Bmax1**x) # hysteresis loss in core A in watts\n",
      "print\"Ph=\",Ph1,\"W\"# the answer may vary due to roundoff error\n",
      "# for core B\n",
      "A2=A1*3. # cross sectional area of core B in metre square\n",
      "lfe2=0.866 # mean flux path length of core B in metres\n",
      "Bmax2=phimax/A2 # flux density in core B in tesla\n",
      "VolB=A2*lfe2 # volume of core B in metre cubes\n",
      "Ph2=VolB*f*hc*(Bmax2**x) # hysteresis loss of core B in watts\n",
      "print\"Ph=\",Ph2,\"W\"# the answer may vary due to roundoff error"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "phimax= 0.00300105438719 Wb\n",
        "Bmax= 1.20042175488 T\n",
        "VolA= 0.00125 metre cube\n",
        "Ph= 53.0597985532 W\n",
        "Ph= 34.1904136606 W\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E04 : Pg 18"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "V1=240. # voltage applied to a winding of transformer(three phase) in volts\n",
      "f1=60. # initial applied frequency in Hz\n",
      "f2=30. # reduced frequency in Hz\n",
      "Phe1=400. # core loss in watts at f1 frequency\n",
      "Phe2=169. # core losses in watts at f2 frequency\n",
      "print\"V2=\",(f2*V1)/f1,\"V\"# voltage at 30 Hz frequency\n",
      "print\"Ph+e/f=Ch+Ce*f\"# equation for claculating  hysteresis and eddy current loss coefficients\n",
      "#a=[1 f1;1 f2] # left hand side matix for the equation above\n",
      "#b=[Phe1/f1;Phe2/f2] # right hand side matrix for the equation above\n",
      "#c=inv(a)*b\n",
      "Ch=4.6#c(1,:)# hysteresis loss coefficient in W/Hz\n",
      "Ce=0.0344#c(2,:)# eddy current loss coefficient in W/(Hz*Hz)\n",
      "print\"Ph=\",Ch*f1,\"W\"# ans may vary due to roundoff error # hysteresis loss in watts at 60 Hz\n",
      "print\"Pe=\",round(Ce*f1*f1),\"W\"# ans may vary due to roundoff error # eddy current loss at 60 Hz in watts"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "V2= 120.0 V\n",
        "Ph+e/f=Ch+Ce*f\n",
        "Ph= 276.0 W\n",
        "Pe= 124.0 W\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E05 : Pg 20"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt \n",
      "Pk=75. # core loss of transfomer in watts\n",
      "R=0.048 # internal resistance in ohms\n",
      "V2=240.#  secondary voltage in volts\n",
      "I2=sqrt(Pk/R)# secondary current in amperes\n",
      "print\"I2=\",round(I2),\"A\"# ans may vary due to roundoff error\n",
      "print\"|S|=V2*I2=\",round(V2*I2),\"VA\"# The answer in the textbook is wrong # output volt ampere of transformer"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "I2= 40.0 A\n",
        "|S|=V2*I2= 9487.0 VA\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E06 : Pg 22"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "sfl=1746 # speed at full load in rev/min\n",
      "snl=1799.5 # speed at no load in rev/min\n",
      "print\"Voltage Regulation=\",round((snl-sfl)/sfl,5) # the ans may vary due to round of error"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Voltage Regulation= 0.03064\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example E07 : Pg 22"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "Vnl=27.3 # no load voltage in volts\n",
      "Vfl1=24.   # full load voltage at power factor 1 in volts\n",
      "print\"(Vnl-Vfl/Vfl)=\",(Vnl-Vfl1)/Vfl1# ans may vary due to roundoff error\n",
      "Vfl2=22.1 # full load voltage at power factor 0.7 in volts\n",
      "print\"Voltage Regulation=\",round((Vnl-Vfl2)/Vfl1,4)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(Vnl-Vfl/Vfl)= 0.1375\n",
        "Voltage Regulation= 0.2167\n"
       ]
      }
     ],
     "prompt_number": 7
    }
   ],
   "metadata": {}
  }
 ]
}