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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 2 : Direct Current Machines"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.4  Page No : 92"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "P = 2.         #number of poles\n",
      "Z = 400.        #number of conducters\n",
      "n = 300.        #speed in rpm\n",
      "E = 200.         #voltage of generator\n",
      "A = 2.           #number of parallel paths\n",
      "N = 1200.        #number of turns in each field coil\n",
      "\n",
      "# Calculations and Results\n",
      "phi = (E*60*A)/(Z*n*P)   #flux at the end of 0.15sec\n",
      "t = 0.15                     #time\n",
      "print \"magnitude of flux at the end of 15sec is %f wb\"%(phi)\n",
      "e = N*(phi/t)\n",
      "print \"induced emf in the field coil =  %d volts\"%(e)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "magnitude of flux at the end of 15sec is 0.100000 wb\n",
        "induced emf in the field coil =  800 volts\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.6  Page No : 93"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "P = 8.            #number of poles\n",
      "A = 8.            #number of parallel paths in the armature\n",
      "Z = 960.          #number of conductors\n",
      "N = 400.          #speed in rpm\n",
      "phi = 0.04       #flux per pole\n",
      "\n",
      "# Calculations\n",
      "E = (phi*Z*N*P)/(60*A)        #emf generated onopen circuit condition\n",
      "\n",
      "# Results\n",
      "print \"emf generated on open circuit condition, E = %d volts\"%(E)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "emf generated on open circuit condition, E = 256 volts\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.7  Page No : 97"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "E = 180.;                #induced emf at 500rpm\n",
      "N = 500.;                #speed in rpm\n",
      "\n",
      "# Calculations and Results\n",
      "K1 = (E/N)\n",
      "print \"K1 = %f\"%(K1)\n",
      "E1 = (K1*600)           #induced emf at 600rpm\n",
      "print \" induced emf at 600rpm is = %d V\"%(E1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "K1 = 0.360000\n",
        " induced emf at 600rpm is = 216 V\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.8  Page No : 97"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "E1 = 220.;           #induced emf at N1 speed in volts\n",
      "N1 = 750.;          # speed \n",
      "K1 = (E1/N1)\n",
      "E2 = 250.;          #induced emf at speed N2\n",
      "N2 = E2/K1\n",
      "print \"speed at induced emf of 250V  = %d rpm\"%(N2)\n",
      "print (\"when induced emf is 250V and speed 700 rpm\")\n",
      "E3 = 250.;            #induced emf at N3 speed\n",
      "N3 = 700.;            #speed\n",
      "ratio = (E3*N1)/(E1*N3)\n",
      "Pi = (ratio-1)*100\n",
      "print \"percentage increase in flux is %f percent\"%(Pi)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "speed at induced emf of 250V  = 852 rpm\n",
        "when induced emf is 250V and speed 700 rpm\n",
        "percentage increase in flux is 21.753247 percent\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.9  Page No : 98"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "E = 200.      #emf induced\n",
      "I = 15.        #armature current\n",
      "n = 1200.          #speed in rpm\n",
      "\n",
      "# Calculations and Results\n",
      "omega = (2*3.14*n)/60;\n",
      "print \"omega = %f \"%(omega)\n",
      "T = (E*I)/omega;\n",
      "print \"electromagnetic torque = %f Nm\"%(T)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "omega = 125.600000 \n",
        "electromagnetic torque = 23.885350 Nm\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.10  Page No : 98"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "n = 10.;             #number of turns in 1 coil\n",
      "l = 0.2;            \n",
      "d = 0.2;          #diameter in metres\n",
      "B = 1.;              #uniform magnetic field density in weber per m**2\n",
      "N = 1500.;            #speed in rpm\n",
      "\n",
      "# Calculations and Results\n",
      "r = (d/2);           #radius in metres\n",
      "E = (B*l*((2*3.14*N)/60)*r*2*n);\n",
      "print \"total induced emf = %f V\"%(E)\n",
      "R = 4;                #total resistance in ohms\n",
      "I = E/R;\n",
      "print \"The current through the armature coil when connected to the load, I = %f A\"%(I)\n",
      "T = (E*I)/((2*3.14*N)/60)\n",
      "print \"torque = %f Nm\"%(T)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "total induced emf = 62.800000 V\n",
        "The current through the armature coil when connected to the load, I = 15.700000 A\n",
        "torque = 6.280000 Nm\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.11  Page No : 99"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "V = 230.;            #armature voltage supply in volts\n",
      "Ia = 12.;            #armature current in amperes\n",
      "Ra = 0.8;            #armature resistance in ohms\n",
      "N = 100.;               #speed in radian per second\n",
      "\n",
      "# Calculations and Results\n",
      "E = (V-(Ia*Ra))\n",
      "print \"induced emf, E = %fV\"%(E)\n",
      "Te = (E*Ia)/N\n",
      "print \"the electromagnetic torque = %fNm\"%(Te)\n",
      "Pi = V*Ia\n",
      "print \"electrical input to the armature, Pinput =  %dW\"%(Pi)\n",
      "Pd = Te*N\n",
      "print \"mechanical developed = %fW\"%(Pd)\n",
      "loss = (Ia**2*Ra)\n",
      "print \"armature copper loss = %fW\"%(loss)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "induced emf, E = 220.400000V\n",
        "the electromagnetic torque = 26.448000Nm\n",
        "electrical input to the armature, Pinput =  2760W\n",
        "mechanical developed = 2644.800000W\n",
        "armature copper loss = 115.200000W\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.13  Page No : 101"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "P = 50000.;            #power delivered in watts\n",
      "V = 250.;              #voltage in volts\n",
      "Ra = 0.02;            #armature resistance in ohms\n",
      "Rf = 50.;              #field resistance in ohms\n",
      "\n",
      "# Calculations and Results\n",
      "If = V/Rf             #field current in amperes\n",
      "Ng = 400.;             #speed in generating condition in rpm\n",
      "print \"field current, If = %dA\"%(If)\n",
      "Il = P/V               #load current in amperes\n",
      "print \"Load current, If = %dA\"%(Il)\n",
      "Ia = If+Il             #armature current in amperes\n",
      "print \"Aramture current, If = %dA\"%(Ia)\n",
      "Eg = (V+(Ia*Ra))\n",
      "print (\"At motor condition\")\n",
      "Ia = (Il-If)\n",
      "print \"Aramture current, If = %dA\"%(Ia)\n",
      "Em = (V-(Ia*Ra))\n",
      "print \"Em = %fV\"%(Em)\n",
      "Nm = (Ng*Em)/Eg\n",
      "print \"Speed of the motor = %drpm\"%(Nm)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "field current, If = 5A\n",
        "Load current, If = 200A\n",
        "Aramture current, If = 205A\n",
        "At motor condition\n",
        "Aramture current, If = 195A\n",
        "Em = 246.100000V\n",
        "Speed of the motor = 387rpm\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.14  Page No : 101"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "V = 250.;                #voltage supply in volts\n",
      "Ra = 0.12;              #armature resistance in ohms\n",
      "Rf = 100.;                #field resistance in ohms\n",
      "Il = 80.;                #load current in amperes\n",
      "\n",
      "# Calculations and Results\n",
      "If = V/Rf                \n",
      "print \"Field current, If = %f\"%(If)\n",
      "print (\"When machine is generating\")\n",
      "Ia = Il+If\n",
      "Eg = (V+(Ia*Ra))\n",
      "print \"Ia = %fA\"%(Ia)\n",
      "print \"Eg = %fV\"%(Eg)\n",
      "print (\"When machine is motoring\")\n",
      "Ia = Il-If\n",
      "Em = (V-(Ia*Ra))\n",
      "print \"Ia = %fA\"%(Ia)\n",
      "print \"Eg = %fV\"%(Em)\n",
      "ratio = Eg/Em\n",
      "print \"Ratio of speeds = %f\"%(ratio)\n",
      "\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Field current, If = 2.500000\n",
        "When machine is generating\n",
        "Ia = 82.500000A\n",
        "Eg = 259.900000V\n",
        "When machine is motoring\n",
        "Ia = 77.500000A\n",
        "Eg = 240.700000V\n",
        "Ratio of speeds = 1.079767\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.15  Page No : 102"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "V = 550.;            #voltage supply in volts\n",
      "P = 16.;            #number of poles\n",
      "N = 150.;             #speed in rpm\n",
      "Z = 2500.;            #number of armature conductors\n",
      "A = 16.;              \n",
      "Power = 1500000.;        #power in watt\n",
      "Cl = 25000.;              #full-load copper loss\n",
      "B = 0.9;                #flux density in the pole\n",
      "\n",
      "# Calculations and Results\n",
      "Ia = Power/V\n",
      "print \"Full load current = %fA\"%(Ia)\n",
      "Ra = Cl/(Ia**2)\n",
      "print \"Ra = %fohms\"%(Ra)\n",
      "E = V+(Ia*Ra)\n",
      "print \"Induced emf = %fvolts\"%(E)\n",
      "phi = (E*60*A)/(Z*N*P)\n",
      "print \"flux density = %fWb/m**2\"%(B)\n",
      "print \"flux = %fWb\"%(phi)\n",
      "area = (phi/B)\n",
      "print \" Area of pole shoe = %fcm**2\"%(area*10000)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Full load current = 2727.272727A\n",
        "Ra = 0.003361ohms\n",
        "Induced emf = 559.166667volts\n",
        "flux density = 0.900000Wb/m**2\n",
        "flux = 0.089467Wb\n",
        " Area of pole shoe = 994.074074cm**2\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.16  Page No : 103"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "Cd = 0.76;            #commutator diameter in metres\n",
      "Cr = .38;              #commutator radius in metres\n",
      "bw = 1.5*10**(-2);        #brush width in metres\n",
      "N = 600.;                #speed in rpm\n",
      "n = 10.;                #speed in rps\n",
      "\n",
      "# Calculations and Results\n",
      "V = Cr*(2*3.14*n);        \n",
      "print \"peripheral speed of commutator, V = %fm/sec\"%(V);\n",
      "Tc = bw/V;\n",
      "print \"Time of commutation = %fseconds\"%(Tc)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "peripheral speed of commutator, V = 23.864000m/sec\n",
        "Time of commutation = 0.000629seconds\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.17  Page No : 123"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "V = 240.;            #supply voltage in volts\n",
      "N = 800.;            #speed in rpm\n",
      "Ia = 2.;              #armeture current in amperes\n",
      "Ra = 0.4;            #armature resistance in ohms\n",
      "Rf = 160.;            #field resistance in ohms\n",
      "Il1 = 30.;            #line current in amperes\n",
      "\n",
      "# Calculations and Results\n",
      "E = V-(Ia*Ra);       #induced emf in volts\n",
      "print (\"At no-load\")\n",
      "print \"E = %fV\"%(E)\n",
      "If = V/Rf;             #field current in amperes\n",
      "print \"If = %fA\"%(If)\n",
      "K1 = E/(If*N);\n",
      "print \"K1 = %f\"%(K1)\n",
      "print (\"At a load of 30A\")\n",
      "Ia1 = (Il1-If);\n",
      "E1 = V-(Ia1*Ra);\n",
      "N1 = 950;               #speed in rpm\n",
      "If1 = E1/(K1*N1);\n",
      "print \"If1 = %fA\"%(If1);\n",
      "Rr = V/If1;\n",
      "R = (Rr-Rf);\n",
      "print \"Extra resistance required in the field circuit, R = %fohms\"%(R)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "At no-load\n",
        "E = 239.200000V\n",
        "If = 1.500000A\n",
        "K1 = 0.199333\n",
        "At a load of 30A\n",
        "If1 = 1.207182A\n",
        "Extra resistance required in the field circuit, R = 38.810149ohms\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.18  Page No : 124"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "V = 230.;                    #voltage supply in volts\n",
      "Ia = 20.;                    #armature current in amperes\n",
      "Ra = 0.5;                    #armature resistance in ohms\n",
      "\n",
      "# Calculations and Results\n",
      "E = V-(Ia*Ra);\n",
      "print \"E = %dV\"%(E)\n",
      "print (\"when extra resistance is added in the armature circuit,the speed is halved\")\n",
      "E2 = E/2;\n",
      "R = ((V-E2)/Ia)-Ra;\n",
      "print (\"The load torque is conmath.atant\")\n",
      "print \"extra resistance in the armature circui, R = %fohms\"%(R)\n",
      "print (\"The load torque directly proportional to square of speed\")\n",
      "print (\"if N is halfed, Iais one-fourthed\")\n",
      "Ia2 = Ia/4;\n",
      "R = ((V-E2)/Ia2)-Ra;\n",
      "print \"extra resistance in the armature circui, R = %fohms\"%(R)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "E = 220V\n",
        "when extra resistance is added in the armature circuit,the speed is halved\n",
        "The load torque is conmath.atant\n",
        "extra resistance in the armature circui, R = 5.500000ohms\n",
        "The load torque directly proportional to square of speed\n",
        "if N is halfed, Iais one-fourthed\n",
        "extra resistance in the armature circui, R = 23.500000ohms\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.19  Page No : 125"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "V = 250.;                    #voltage supply in volts\n",
      "Ia = 50.;                    #armature current in amperes\n",
      "Ra = 0.3;                    #armature resistance in ohms\n",
      "N = 1000.;\n",
      "\n",
      "# Calculations and Results\n",
      "E = V-(Ia*Ra);\n",
      "print \"E = %dV\"%(E)\n",
      "print (\"when extra resistance is added in the armature circuit when the speed is 800rpm\")\n",
      "N2 = 800.;\n",
      "E2 = (E*N2)/N;\n",
      "print \"E at 800rpm = %dV\"%(E2)\n",
      "R = ((V-E2)/Ia)-Ra;\n",
      "print \"extra resistance in the armature circui, R = %fohms\"%(R)\n",
      "print (\"if load is halfed,Ia will be halfed\")\n",
      "Ia2 = Ia/2;\n",
      "E1 = V-(Ia2*(Ra+R));\n",
      "print \"E1 = %dV\"%(E1)\n",
      "N1 = (N2*E1)/E2;\n",
      "print \"N1 = %frpm\"%(N1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "E = 235V\n",
        "when extra resistance is added in the armature circuit when the speed is 800rpm\n",
        "E at 800rpm = 188V\n",
        "extra resistance in the armature circui, R = 0.940000ohms\n",
        "if load is halfed,Ia will be halfed\n",
        "E1 = 219V\n",
        "N1 = 931.914894rpm\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.20  Page No : 125"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "Il = 5.;            #current in amperes al no-load\n",
      "V = 250.;            #voltage in volts\n",
      "Rf = 250.;            #field resistance in ohms\n",
      "\n",
      "# Calculations and Results\n",
      "If1 = V/Rf;            #field current in amperes\n",
      "Ia1 = Il-If1;            #armature current\n",
      "Ra = 0.2;                    #armature resistance in ohms\n",
      "print (\"at a load current of 50A\")\n",
      "Il2 = 50;                #load current in amperes\n",
      "#armature reaction weakens by 3percent\n",
      "If2 = 0.97;                    #current in amperes\n",
      "Ia2 = Il2-If2;\n",
      "N1 = 1000; \n",
      "E1 = (V-(Ia1*Ra));\n",
      "E2 = (V-(Ia2*Ra));\n",
      "N2 = (N1*E2)/(0.97*E1);\n",
      "print \"N2 = %frpm\"%(N2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "at a load current of 50A\n",
        "N2 = 993.670467rpm\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.21  Page No : 126"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "P = 4;                  #pole\n",
      "V = 500;                #shunt motor  in volts\n",
      "Ia = 60;                    #armature current in amperes\n",
      "Ra = 0.2;                   #armature resistance in ohms\n",
      "\n",
      "# Calculations and Results\n",
      "E = V-(Ia*Ra)-2;\n",
      "print \"voltage drop across each brush = %fV\"%(E)\n",
      "phi = 0.03;                 #flux per pole in Wb\n",
      "Z = 720.;                    #total armature current in volts\n",
      "A = 2;\n",
      "N = (E*60*A)/(phi*Z*P)\n",
      "print \"full load speed of the motor = %frpm\"%(N)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "voltage drop across each brush = 486.000000V\n",
        "full load speed of the motor = 675.000000rpm\n"
       ]
      }
     ],
     "prompt_number": 19
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.22  Page No : 126"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "# Given Data\n",
      "V = 440;                #primary voltage in volts\n",
      "Ia = 50;                #armature current in amperes\n",
      "Ra = 0.2;                #armature resistance in ohms\n",
      "N = 600;                  #speed in rpm\n",
      "E = V-(Ia*Ra);           #emf induced  in volts before adding extra resistance\n",
      "#E = K*phi*N = K1*Ia*N\n",
      "K1 = E/(Ia*N);\n",
      "\n",
      "# Calculations and Results\n",
      "#we have the relation T = Kt1*Ia**2, T1 = Kt1*Ia1**2\n",
      "#when torque is half, say torque be T1\n",
      "#T1 = T/2. r = T/T1\n",
      "r = 2;\n",
      "Ia1 = math.sqrt(Ia**2/r);\n",
      "print \"Ia1 = %fA\"%(Ia1);\n",
      "#extra resistance R is introduced in the circuit\n",
      "N1 = 400;\n",
      "E1 = (K1*Ia1*N1);\n",
      "R = ((V-E1)/Ia1)-Ra;\n",
      "print \"value of extra resistance added = %fohms\"%(R)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Ia1 = 35.355339A\n",
        "value of extra resistance added = 6.511746ohms\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.23  Page No : 127"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Given Data\n",
      "V = 200.;                        #voltage in volts\n",
      "Ia = 20.;                        #armature current in amperes\n",
      "Ra = 0.5;                        #armature resistance in ohms\n",
      "Rse = 0.2;                        #field winding resistance in ohms\n",
      "\n",
      "# Calculations and Results\n",
      "E = V-(Ia*(Ra+Rse));\n",
      "print \"In first case, E = %fV\"%(E)\n",
      "#E = k*phi*N\n",
      "N = 1000;                        #speed in rpm\n",
      "Kphi = E/N;   \n",
      "#a resistance R is connected in parallel with the series field which is called diverter\n",
      "print (\"when resistace R is added and new conditions\")\n",
      "I = 20;                        #total current flowing\n",
      "#current is equally devided between series field and diverter\n",
      "Ise2 = I/2;\n",
      "#flux at 10A current is 20percent of flux at 20A current\n",
      "p = 0.70;                #percentage of flux\n",
      "Kpih1 = p*Kphi;\n",
      "E1 = (V-((Ia*Ra)+(Ise2*Rse)));\n",
      "print \"Induced emf = %fV\"%(E1)\n",
      "#new speed is N1\n",
      "N1 = E1/(p*Kphi)\n",
      "print \"N1 = %frpm\"%(N1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "In first case, E = 186.000000V\n",
        "when resistace R is added and new conditions\n",
        "Induced emf = 188.000000V\n",
        "N1 = 1443.932412rpm\n"
       ]
      }
     ],
     "prompt_number": 22
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.24  Page No : 128"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "V = 200.;                             #motor runs in volts\n",
      "Ia = 15.;                                  #current taken in amperes\n",
      "Ra = 1.;                                #motor resistance in ohms\n",
      "\n",
      "# Calculations and Results\n",
      "E1 = V-(Ia*Ra);\n",
      "print \"resistance when 1ohm = %fV\"%(E1)\n",
      "R = 5;                                   #resistance \n",
      "E2 = V-(Ia*(Ra+R))\n",
      "print \"resistance when 5ohms connected in series = %fV\"%(E2)\n",
      "N1 = 800;                                #speed of motor in rpm\n",
      "N2 = N1*(E2/E1);\n",
      "print \"speed at which motor will run when resistance is 5ohms = %frpm\"%(N2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "resistance when 1ohm = 185.000000V\n",
        "resistance when 5ohms connected in series = 110.000000V\n",
        "speed at which motor will run when resistance is 5ohms = 475.675676rpm\n"
       ]
      }
     ],
     "prompt_number": 23
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.25  Page No : 135"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "P = 8.;                          #pole\n",
      "Z = 107.;                        #generator with slots\n",
      "Ia = 1000.;                    #current containing in amperes\n",
      "Bag = 0.32;                      #gap flux density in Wb/m**2\n",
      "lg = 0.012;                          #interpole air gap in meters\n",
      "pi = 3.14;\n",
      "\n",
      "# Calculations\n",
      "Mu = (4*pi*10**-7)\n",
      "AT = (((Ia*Z)/(2*P))+((Bag*lg)/Mu));\n",
      "\n",
      "# Results\n",
      "print \"current for each commutating pole = %f\"%(AT)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "current for each commutating pole = 9744.824841\n"
       ]
      }
     ],
     "prompt_number": 24
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.26  Page No : 135"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "Bag = 0.3;                                  #flux density in the interpole air gap in Wb/m**2\n",
      "\n",
      "# Calculations and Results\n",
      "Ia = 200000./200;                        #armature current in amperes\n",
      "print \"Armature current = %f\"%(Ia)\n",
      "Z = 540.;                          #Number of armature conductors\n",
      "Zt = 540./2;                            #Number armature winding turns \n",
      "print \"Number armature winding turns = %f\"%(Zt)\n",
      "A = 6.;                             #the winding lap\n",
      "Ap = Zt/A;                        #Number of armature turns per parallel path\n",
      "print \"Number of armature turns per parallel path = %f\"%(Ap)\n",
      "P = 6;                              #pole\n",
      "Np = ((Ia*Ap)/P);\n",
      "print \"Number of armature ampere turns per pole = %f\"%(Np)\n",
      "lg = 0.01;                              #inter pole air gap in meters\n",
      "pi = 3.14;\n",
      "Mu = (4*pi*10**-7)\n",
      "Nipg = ((Bag*lg)/Mu);                          #Air gap\n",
      "print \"ampere turns for the air gap = %f\"%(Nipg)\n",
      "NipI = (Np+Nipg);                                #total interpole ampere\n",
      "print \"Total interpole ampere turns = %f\"%(NipI)\n",
      "Nip = (NipI/Ia);\n",
      "print \"Number of turns needed on each commutating pole = %f\"%(Nip)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Armature current = 1000.000000\n",
        "Number armature winding turns = 270.000000\n",
        "Number of armature turns per parallel path = 45.000000\n",
        "Number of armature ampere turns per pole = 7500.000000\n",
        "ampere turns for the air gap = 2388.535032\n",
        "Total interpole ampere turns = 9888.535032\n",
        "Number of turns needed on each commutating pole = 9.888535\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.27  Page No : 128"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "N = 960.;                           #speed in rpm\n",
      "F = 23.;                            #effictive load in kgf\n",
      "\n",
      "# Calculations and Results\n",
      "r = 45./2;                               #radius of the drum\n",
      "print \"radius of the drum = %fcm\"%(r)\n",
      "pi = 3.14;\n",
      "OP = (2*pi*N*F*r*9.81)/(60*100);\n",
      "print \"output power = %fW\"%(OP)\n",
      "\n",
      "Vi = 230.;                  #motor input in volts\n",
      "Ci = 28.;                       #input current in amperes\n",
      "IP = (Vi*Ci);\n",
      "print \"input power  = %fW\"%(IP)\n",
      "Effi = (OP/IP)*100;\n",
      "print \"Efficiency of the motor = %fpercent\"%(Effi)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "radius of the drum = 22.500000cm\n",
        "output power = 5101.043040W\n",
        "input power  = 6440.000000W\n",
        "Efficiency of the motor = 79.208743percent\n"
       ]
      }
     ],
     "prompt_number": 26
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.29  Page No : 145"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "I = 440.;                      #input at no-load in watt\n",
      "V = 220.;                        #voltage in volts\n",
      "Ic = I/V;                      #input current at no-load in amperes\n",
      "i = 1;                    #input current in amperes\n",
      "A = 2;                      #current in amperes\n",
      "C = A-i;                    #armature current at no-load in amperes\n",
      "L = I-((((C)**2)*0.5)+(V*C));                #iron,friction and windage losses in watt\n",
      "a = 40;                         #motor current in amperes\n",
      "OP = (V*a);\n",
      "Ra = 0.5;\n",
      "\n",
      "# Calculations and Results\n",
      "Effi = (OP*100)/(OP+(((a+i)**2)*Ra)+(V*i)+L)\n",
      "print \"Efficiency as a generator when delivering 40A at 220V = %fpercent\"%(Effi)\n",
      "Eff = ((OP-(((a-i)**2)*Ra)-(V*C)-L)/OP)*100;\n",
      "print \"Efficiency as a motor when taking 40A from at 220V = %fpercent\"%(Eff)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Efficiency as a generator when delivering 40A at 220V = 87.301587percent\n",
        "Efficiency as a motor when taking 40A from at 220V = 86.363636percent\n"
       ]
      }
     ],
     "prompt_number": 27
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.30  Page No : 147"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "V = 400.;                            #motor in volts\n",
      "Rf = 200.;                            #field resistance in ohms\n",
      "If = V/Rf;                          #current in amperes\n",
      "i = 5;                             #current at no load in amperes\n",
      "IP = V*i;                            #motor input at no load\n",
      "Ia = 3;                             #aramture current in amperes\n",
      "Ra = 0.5;                           #armature resistance in ohms\n",
      "\n",
      "# Calculations and Results\n",
      "L = IP-(((Ia)**2)*Ra)-(V*If);                     #iron,friction and windage in losses in watt\n",
      "print \"iron, friction and windage in losses = %fW\"%(L)\n",
      "At = 50.;                                        #armature total current in amperes\n",
      "A = At-2;                                       #armature current in amperes\n",
      "Ls = (((A)**2)*Ra)+(V*If)+L;                      #Losses\n",
      "Eff = (((V*At)-Ls)/(V*At))*100;\n",
      "print \"Efficiency of full load = %fpercent\"%(Eff)\n",
      "#flux is consmath.tant\n",
      "E1 = V-(Ia*Ra);                           #induced emf in the armature at no load\n",
      "E2 = V-(A*Ra);                             #induced emf in the armature at full load\n",
      "# math.since N1/N2 = E1/E2\n",
      "percentload = (1-(E2/E1))*100;\n",
      "print \"Percentage change in speed from no load to full load = %fpercent\"%(percentload)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "iron, friction and windage in losses = 1195.500000W\n",
        "Efficiency of full load = 84.262500percent\n",
        "Percentage change in speed from no load to full load = 5.646173percent\n"
       ]
      }
     ],
     "prompt_number": 29
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.31  Page No : 148"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "Ra = 0.5;                #armature resistance in ohms\n",
      "Rf = 750.;               #field circuit resistance in ohms\n",
      "V = 500.;                       #voltage in volts\n",
      "\n",
      "# Calculations\n",
      "If = V/Rf;                          #current in amperes \n",
      "l = 3.;                          #line current in amperes\n",
      "i = 2.33;                          #current in motor in amperes\n",
      "I = 0.67;                         #current i amperes\n",
      "L = (V*l)-(((i)**2)*Ra)-(V*I);                         #Iron,friction and windage losses\n",
      "O = 20.;                               #generator \n",
      "OP = (O*1000)/V;                #output current of the generator under loaded condition in amperes\n",
      "Ia = I+OP;            #output in amperes\n",
      "Effi = (O*1000*100)/((O*1000)+(((Ia)**2)*Ra)+(V*I)+L);\n",
      "\n",
      "# Results\n",
      "print \"efficiency of the machine = %fpercent\"%(Effi)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "efficiency of the machine = 89.588435percent\n"
       ]
      }
     ],
     "prompt_number": 30
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.32  Page No : 149"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "Ig = 25.;               #current of generator in amperes\n",
      "I = 30.;                   #current in motor in amperes\n",
      "Il = I-Ig;              #current in amperes\n",
      "Ra = 0.25;                #resistance in ohms\n",
      "Gl = ((Ig)**2)*Ra;                #loss in generator in watt\n",
      "M = ((I)**2)*Ra;                    #loss in motor in watt\n",
      "T = Gl+M;                   #total loss in watt\n",
      "V = 100.;             #voltage in volts\n",
      "P = V*Il;               #power supplied from mains in watt\n",
      "L = P-T;                  #iron,friction and windages losses in the two machines in ohms\n",
      "l = L/2;                   #iron,friction and windages losses in each machines in ohms\n",
      "IP = I*V;                    #input\n",
      "\n",
      "# Calculations and Results\n",
      "Eff = ((IP-M-l)/IP)*100;\n",
      "print \"Efficiency of the motor = %fpercent\"%(Eff)\n",
      "OP = Ig*V;                 #output\n",
      "Effi = ((OP)/(OP+Gl+l))*100;\n",
      "print \"Efficiency of the generator = %fpercent\"%(Effi)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Efficiency of the motor = 90.520833percent\n",
        "Efficiency of the generator = 92.059839percent\n"
       ]
      }
     ],
     "prompt_number": 31
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.33  Page No : 150"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "V = 440.;                    #voltage in volts\n",
      "P = 200.*1000;               #power in watt\n",
      "Ig = P/V;              #rated current of each machine in amperes\n",
      "\n",
      "# Calculations\n",
      "#assume losses to be equal\n",
      "I = 90;              #addition currnet supply\n",
      "Effi = math.sqrt(Ig/(Ig+I))*100;\n",
      "\n",
      "# Results\n",
      "print \"approximate efficiency = %fpercent\"%(Effi)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "approximate efficiency = 91.363261percent\n"
       ]
      }
     ],
     "prompt_number": 32
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2.34  Page No : 150"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "\n",
      "# Given Data\n",
      "Ig = 2000.;                             #output current of generator in amperes\n",
      "I = 380.;                               #Input current from supply mains in amperes\n",
      "\n",
      "# Calculations and Results\n",
      "Effi = math.sqrt(Ig/(Ig+I))*100;                  #Efficiency of generator assuming equal efficiencies of the two machines\n",
      "print \"Efficiences of the generator at full load assuming equal efficiencies = %fpercent\"%(Effi)\n",
      "S = 22.;                            #Shunt field current of generator\n",
      "G = Ig+S;                        #Armature current of generator in amperes\n",
      "R = 0.01;                               #resistance of the armature circuit of each machine in ohms\n",
      "Gc = ((G)**2)*R;                          #copper loss in arrmature circuit of generator in W\n",
      "V = 500.;                                #Voltage in volts\n",
      "L = V*S;                              #loss in the field circuit of the generator in W\n",
      "T = Ig+I;                            #total current suuply in amperes\n",
      "Sf = 17.;                                        #shunt field current of motor in amperes\n",
      "A = T-Sf;                              #armature current in motor in amperes\n",
      "Lc = ((A)**2)*R;                        #loss in armature circuit of motor in amperes\n",
      "Lf = V*Sf;                                 #loss in the shunt field circuit of motor in W\n",
      "Tin = V*I;                      #total input to motor and generator in W\n",
      "Ml = Tin-(Gc+L+Lc+Lf);                     #iron,friction and windage loss in both machines in W\n",
      "Me = Ml/2;                                   #iron,friction and windage loss in each machine in W\n",
      "p = 1000.;                     #power in kW\n",
      "OP = (Ig*V)/p;                        #full load output of the generator\n",
      "Eff = (p*100)/(p+((Gc+L+Me)/1000));\n",
      "print \"Efficiency of the generator at full load = %fpercent\"%(Eff)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Efficiences of the generator at full load assuming equal efficiencies = 91.669850percent\n",
        "Efficiency of the generator at full load = 91.846461percent\n"
       ]
      }
     ],
     "prompt_number": 33
    }
   ],
   "metadata": {}
  }
 ]
}