path: root/Internal_Combustion_Engines_by_H._B._Keswani/ch9.ipynb

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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 9 : Spark Ignition Engines"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.1  Page no :  188"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "d = 0.001\t\t\t\t\t#Diameter of the jet in m\n",
      "vd = 104.\t\t\t\t\t#Venturi depression in cm of water. In textbook, it is given as 10 cm\n",
      "Cd = 0.65\t\t\t\t\t#Coefficient of discharge \n",
      "g = 0.76\t\t\t\t\t#Specific gravity of petrol\n",
      "w = 1000.\t\t\t\t\t#Weight of water per one cu.m in kg\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "pa = (vd/100)*w\t\t\t\t\t#Venturi depression in kg/m**2\n",
      "dp = (g*w)\t\t\t\t\t#Density of petrol in kg/m**3\n",
      "wf = (((3.14*d**2)/4)*Cd*math.sqrt(2*9.81*dp*pa))/10**-3\t\t\t\t\t#Petrol discharge in gm/sec neglecting nozzle lip\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The weight of petrol discharged is %3.2f gm/sec'%(wf)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The weight of petrol discharged is 2.01 gm/sec\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.2  Page no :  193"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "d1 = 0.075\t\t\t\t\t#Throat diameter in m\n",
      "Ca = 0.93\t\t\t\t\t#Coefficient of air flow\n",
      "d2 = 0.005\t\t\t\t\t#Orifice diameter in m\n",
      "Cf = 0.68\t\t\t\t\t#Coefficient of fuel discharge\n",
      "ap = 1.\t\t\t\t\t#Approach factor\n",
      "dp = 0.15\t\t\t\t\t#Pressure drop in kg/cm**2\n",
      "da = 1.29\t\t\t\t\t#Density of air in kg/m**3\n",
      "df = 720\t\t\t\t\t#Density of fuel in kg/m**3\n",
      "\n",
      "\t\t\t\t\t#Calcultions\n",
      "w = (((3.14/4)*d1**2)/((3.14/4)*d2**2))*(Ca/Cf)*math.sqrt(da/df)\t\t\t\t\t#The air-fuel ratio neglecting the nozzle lip\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The air-fuel ratio neglecting the nozzle lip is %3.1f'%(w)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The air-fuel ratio neglecting the nozzle lip is 13.0\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.3  Page no :  193"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "af = 15.\t\t\t\t\t#Air fuel ratio\n",
      "dp = 753.\t\t\t\t\t#Density of petrol in kg/m**3\n",
      "da = 1.28\t\t\t\t\t#Density of air in kg/m**3\n",
      "C = [0.84,0.7]\t\t\t\t\t#Coefficient of discharge for air and fuel respectively\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "A = 1/(af*(C[1]/C[0])*math.sqrt(dp/da))\t\t\t\t\t#Ratio of areas\n",
      "d = math.sqrt(A)\t\t\t\t\t#Ratio of diamter of jet to diameter of venturi\n",
      "x = (1/d)\t\t\t\t\t#Reverse of ratio\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The ratio of diameter of jet to diameter of venturi is 1 : %3.1f'%(x)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The ratio of diameter of jet to diameter of venturi is 1 : 17.4\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.4  Page no :  194"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "D = [10.,12.]\t\t\t\t\t#Dimensions of four cylinder in 10 cm* 12 cm\n",
      "n = 4.\t\t\t\t\t#Four cylinder\n",
      "N = 2000.\t\t\t\t\t#Speed in r.p.m\n",
      "d = 0.03\t\t\t\t\t#Venturi throat in m\n",
      "nv = 70.\t\t\t\t\t#Volumetric efficiency of the engine in percent\n",
      "Ca = 0.8\t\t\t\t\t#Coefficient of air flow\n",
      "da = 1.29\t\t\t\t\t#Density of air in kg/m**3\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "Vs = (3.14/4)*(D[0]/100)**2*(D[1]/100)\t\t\t\t\t#Stroke volume of one cylinder in m**3\n",
      "Va = (Vs*n*(nv/100)*(N/2))\t\t\t\t\t#Volume of air drawn per minute in m**3\n",
      "w = (Va*da)/60\t\t\t\t\t#Weight of air drawn per sec\n",
      "dp = ((w/((3.14/4)*d**2*Ca))**2/(2*9.81*da))\t\t\t\t\t#Venturi depression in kg/cm**2\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The venturi depression is %3.1f kg/m**2'%(dp)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The venturi depression is 397.7 kg/m**2\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.5  Page no :  198"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "D = [8.25,11.5]\t\t\t\t\t#Dimensions of four cylinder in 8.25 cm* 11.5 cm\n",
      "n = 4.\t\t\t\t\t#Four cylinder\n",
      "N = 3000.\t\t\t\t\t#Speed in r.p.m\n",
      "v = 150.\t\t\t\t\t#Venturi depression in cm of water\n",
      "nv = 80.\t\t\t\t\t#Volumetric efficiency of the engine in percent\n",
      "af = 14.\t\t\t\t\t#Air fuel ratio\n",
      "Ca = 0.84\t\t\t\t\t#Coefficient of air flow\n",
      "Cf = 0.7\t\t\t\t\t#Coefficient of fuel orifice \n",
      "da = 1.29\t\t\t\t\t#Density of air in kg/m**3\n",
      "df = 700.\t\t\t\t\t#Density of fuel in kg/m**3\n",
      "dw = 1000.\t\t\t\t\t#Density of water in kg/m**3\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "Va = ((3.14/4)*(D[0]/100)**2*(D[1]/100)*n*(nv/100)*(N/(2*60)))\t\t\t\t\t#Maximum amount of air pasmath.sing through the venturi in m**3\n",
      "vd = (v/100)*dw\t\t\t\t\t#Venturi depression in kg/m**2\n",
      "va = (Ca*math.sqrt((2*9.81*vd)/da))\t\t\t\t\t#Velocity of air in m/s\n",
      "d = math.sqrt((Va/va)*(4/3.14))\t\t\t\t\t#Throat diameter of venturi in m\n",
      "Af = 1/(af*(va/Va)*math.sqrt(df/da)*(Cf/Ca))\t\t\t\t\t#Area of orifice in m**2\n",
      "df = math.sqrt((Af*4)/3.14)*1000\t\t\t\t\t#Diameter of orifice in mm\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The size of venturi is %i kg/m**2  \\\n",
      "\\nThe diameter of fuel orifice is %3.2f mm'%(vd,df)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The size of venturi is 1500 kg/m**2  \n",
        "The diameter of fuel orifice is 1.35 mm\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.6  Page no :  198"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "D = [7.5,10]\t\t\t\t\t#Dimensions of four cylinder in 7.5 cm diameter and 10 cm stroke\n",
      "n = 6.\t\t\t\t\t#Six cylinder\n",
      "pC = 84.\t\t\t\t\t#Percentage of carbon in volatile fuel\n",
      "pH2 = 16.\t\t\t\t\t#Percentage of hydrogen in volatile fuel\n",
      "dc = (38.5/1000)\t\t\t\t\t#Diameter of the throat of the choke tube in m\n",
      "N = 3000.\t\t\t\t\t#Speed in r.p.m\n",
      "nv = 0.8\t\t\t\t\t#Volumetric efficiency in ratio\n",
      "p = 0.914\t\t\t\t\t#Pressure at the throat of the choke tube in kg/cm**2\n",
      "T = 15.5+273\t\t\t\t\t#Temperature at the throat of the choke tube in K\n",
      "Ts = 273.\t\t\t\t\t#Temperature of 0 degree C in K\n",
      "ps = 1.027\t\t\t\t\t#Atmospheric pressure in kg/cm**2\n",
      "Ra = 29.27\t\t\t\t\t#Universal gas constant for air in kg.m/kg.K\n",
      "Rf = 9.9\t\t\t\t\t#Gas constant for fuel in kg.m/kg.K\n",
      "pO2 = 23.\t\t\t\t\t#Composition by weight of oxygen in air in percent\n",
      "pN2 = 77.\t\t\t\t\t#Composition by weight of nitogen in air in percent\n",
      "mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
      "mH2 = 2.\t\t\t\t\t#Molecular weight oh H2\n",
      "mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "Vm = ((3.14/4)*(D[0]/100)**2*(D[1]/100)*n*(N/2)*nv)\t\t\t\t\t#Volume of mixture supplied per sec in m**3\n",
      "qa = ((100/pO2)*(((pC/100)*(mO2/mC))+((pH2/100)*(mO2/(2*mH2)))))\t\t\t\t\t#Quantity of air required for complete combustion of fuel in kg\n",
      "vf = (Rf*Ts)/(ps*10**4)\t\t\t\t\t#Specific volume of volatile fuel in m**3/kg\n",
      "va = (Ra*Ts)/(ps*10**4)\t\t\t\t\t#Specific volume of air in m**3/kg\n",
      "wf = (Vm/(qa*va+vf))\t\t\t\t\t#Flow rate of fuel in kg/min\n",
      "Fc = (wf*60)\t\t\t\t\t#Fuel consumption in kg/hour\n",
      "da = (p*10**4)/(Ra*T)\t\t\t\t\t#Density of air at the throat of the choke in kg/m**3\n",
      "Va = ((qa*wf)/((3.14/4)*dc**2*da*60))\t\t\t\t\t#Speed of air at throat in m/s\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'a) The fuel consumption is %3.1f kg/hr  \\\n",
      "\\nb) The speed of the air through the choke is %3.1f m/s'%(Fc,Va)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) The fuel consumption is 0.0 kg/hr  \n",
        "b) The speed of the air through the choke is 0.0 m/s\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.7  Page no :  203"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "mf = 7.5\t\t\t\t\t#Consumption of petrol per hour\n",
      "gf = 0.75\t\t\t\t\t#Specific gravity of fuel\n",
      "Tf = 25+273\t\t\t\t\t#Temperature of fuel in K\n",
      "af = 15\t\t\t\t\t#Air fuel ratio\n",
      "dc = 22\t\t\t\t\t#diameter of choke tube in mm\n",
      "l = 4\t\t\t\t\t#Top of the jet is 4 mm above the petrol level in the float chamber\n",
      "Ca = 0.82\t\t\t\t\t#Coefficient of discharge for air \n",
      "Cf = 0.7\t\t\t\t\t#Coefficient of discharge for fuel\n",
      "R = 29.27\t\t\t\t\t#Characteristic gas constant for air in kg.m/kg.K\n",
      "p = 1.03\t\t\t\t\t#Atmospheric pressure in kg/cm**2\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "da = (p*10**4)/(R*Tf)\t\t\t\t\t#Density of air in kg/m**3 \n",
      "dp = (gf*1000)\t\t\t\t\t#Density of petrol in kg/cm**3\n",
      "dpa = ((af*mf*10**6)/(60*60*3.14*Ca*(dc/2)**2))**2/(2*9.81*da)\t\t\t\t\t#Change in pressure in kg/m**2\n",
      "df = math.sqrt((mf/(60*60*Cf*math.sqrt(2*9.81*dp*(dpa-((l/100)*dp)))))*(4/3.14))*1000\t\t\t\t\t#Diameter of the fuel jet in mm\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'Diameter of the jet of the carburettor is %3.2f mm'%(df) \n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Diameter of the jet of the carburettor is 1.22 mm\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.8  Page no :  207"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "td = 7.5\t\t\t\t\t#Throat diameter in cm\n",
      "Ca = 0.85\t\t\t\t\t#Coefficient of air flow\n",
      "fd = 0.5\t\t\t\t\t#Diameter of fuel orifice in cm\n",
      "Cd = 0.7\t\t\t\t\t#Coefficient of discharge\n",
      "l = 5\t\t\t\t\t#Nozzle lip in mm\n",
      "x = 1\t\t\t\t\t#Approach factor\n",
      "dpa = 0.15\t\t\t\t\t#Pressure drop in kg/cm**2\n",
      "da = 1.29\t\t\t\t\t#Density of air in kg/m**3\n",
      "dp = 720\t\t\t\t\t#Density of fuel in kg/m**3\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "afr1 = (((3.14*td**2)/(3.14*fd**2))*(Ca/Cd)*math.sqrt(da/dp))\t\t\t\t\t#Air fuel ratio\n",
      "afr2 = ((3.14*td**2)/(3.14*fd**2))*(Ca/Cd)*math.sqrt((da*dpa)/(dp*(dpa-((l/100)*(dp/10**6)))))\t\t\t\t\t#Air fuel ratio\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The air fuel ratio  \\\n",
      "\\na) neglecting nozzle lip is %3.2f  \\\n",
      "\\nb) nozzle lip is taken into account is %3.2f'%(afr1,afr2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The air fuel ratio  \n",
        "a) neglecting nozzle lip is 11.56  \n",
        "b) nozzle lip is taken into account is 11.56\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.9  Page no :  209"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "td = 7.5\t\t\t\t\t#Throat diameter in cm\n",
      "Ca = 0.85\t\t\t\t\t#Coefficient of air flow\n",
      "fd = 0.5\t\t\t\t\t#Diameter of fuel orifice in cm\n",
      "Cd = 0.7\t\t\t\t\t#Coefficient of discharge\n",
      "l = 5.\t\t\t\t\t#Nozzle lip in mm\n",
      "x = 1.\t\t\t\t\t#Approach factor\n",
      "dpa = 0.15\t\t\t\t\t#Pressure drop in kg/cm**2\n",
      "da = 1.29\t\t\t\t\t#Density of air in kg/m**3\n",
      "dp = 720\t\t\t\t\t#Density of fuel in kg/m**3\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "v = math.sqrt(2*9.81*(l/1000)*(dp/da))\t\t\t\t\t#Velocity of air in m/s\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'Velocity of air flow is %3.1f m/s'%(v)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Velocity of air flow is 7.4 m/s\n"
       ]
      }
     ],
     "prompt_number": 10
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.10  Page no :  212"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "x = 2.8\t\t\t\t\t#Height above the nozzle in mm\n",
      "va = 58\t\t\t\t\t#Velocity of air in m/s\n",
      "da = 1.28\t\t\t\t\t#Density of air in kg/m**3\n",
      "dp = 750\t\t\t\t\t#Density of petrol in kg/m**3\n",
      "An = 1.8\t\t\t\t\t#Area of cross section of nozzle in mm**2\n",
      "Cd = 0.6\t\t\t\t\t#Coefficient of discharge of nozzle \n",
      "Ca = 0.84\t\t\t\t\t#Coefficient of discharge of air\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "dpa = ((va/Ca)**2*(da/(2*9.81)))\t\t\t\t\t#Change in pressure in kg/m**2\n",
      "wf = ((An*10**-6)*Cd*math.sqrt(2*9.81*dp*(dpa-((x/1000)*dp))))\t\t\t\t\t#Petrol consumption in kg/sec\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'Petrol consumption is %3.4f kg/sec'%(wf)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Petrol consumption is 0.0023 kg/sec\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.11  Page no :  214"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "d = 0.155\t\t\t\t\t#Diameter of orifice in mm\n",
      "Cd = 0.94\t\t\t\t\t#Coefficient of discharge\n",
      "td = 3.18\t\t\t\t\t#Throat diameter in cm\n",
      "Ca = 0.84\t\t\t\t\t#Coefficient of discharge\n",
      "x = 29\t\t\t\t\t#Venturi depression\n",
      "dw = 0.89\t\t\t\t\t#Minimum depression of water in cm\n",
      "sa = 1.1\t\t\t\t\t#Specific weight of air in kg/m**3\n",
      "sg = 0.72\t\t\t\t\t#Specific gravity of petrol\n",
      "cyd = [7.75,10.75]\t\t\t\t\t#Cylinder dimensions in cm\n",
      "fc = 10.9\t\t\t\t\t#Fuel consumption in kg/hr\n",
      "N = 3200\t\t\t\t\t#Speed in r.p.m\n",
      "n = 4\t\t\t\t\t#Number of cylinders\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "w = (((3.14/4)*td**2)/((3.14/4)*d**2))*(Ca/Cd)*math.sqrt((sa/(sg*1000))*(x/(x-dw)))\t\t\t\t\t#Air fuel ratio\n",
      "Va = (3.14/4)*(td/100)**2*Ca*math.sqrt(2*9.81*sa*x*6)\t\t\t\t\t#Volume of air drawn/sec\n",
      "vn = (Va/((3.14/4)*(cyd[0]/100)**2*(cyd[1]/100)*n*(N/(2*60))))*100\t\t\t\t\t#Volumetric efficiency in percent\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'Air fue ratio is %3.1f  \\\n",
      "\\nVolumetric efficiency is %3.1f percent'%(w,vn)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Air fue ratio is 14.9  \n",
        "Volumetric efficiency is 77.5 percent\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.12  Page no :  219"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "af = 0.066\t\t\t\t\t#Air fuel ratio\n",
      "p = 0.83\t\t\t\t\t#Pressure at the venturi throat in kg/cm**2\n",
      "pd = 0.04\t\t\t\t\t#Pressure drop in kg/cm**2\n",
      "va = 245\t\t\t\t\t#Air flow at sea level in kg per hour\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "dpa = 1.03-p\t\t\t\t\t#Pressure at air cleaner in kg/cm**2\n",
      "d = (1.03-pd-dpa)\t\t\t\t\t#Throat pressure when the air cleaner is fitted in kg/cm**2\n",
      "naf = (af*math.sqrt((1.03-d)/dpa))\t\t\t\t\t#New air fuel ratio\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'a) Throat pressure when the air cleaner is fitted is %3.2f kg/cm**2  \\\n",
      "\\nb) New air fuel ratio is %3.4f'%(d,naf)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) Throat pressure when the air cleaner is fitted is 0.79 kg/cm**2  \n",
        "b) New air fuel ratio is 0.0723\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.13  Page no :  221"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "x = 3\t\t\t\t\t#Petrol smath.radians(numpy.arcmath.tan(s 3 mm below\n",
      "Ta = 15.5+273\t\t\t\t\t#Temperature of air in K\n",
      "pa = 1.027\t\t\t\t\t#Pressure of air in kg/cm**2\n",
      "R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/kg.K \n",
      "sg = 0.76\t\t\t\t\t#Specific gravity of fuel\n",
      "fc = 6.4\t\t\t\t\t#Consumption of fuel in kg/hour\n",
      "jd = 1.27\t\t\t\t\t#Jet diameter in mm\n",
      "Cd = 0.6\t\t\t\t\t#Nozzle discharge coefficienct\n",
      "Ca = 0.8\t\t\t\t\t#Discharge coefficient of air\n",
      "af = 0.066\t\t\t\t\t#Air fuel ratio\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "df = (sg*1000)\t\t\t\t\t#Density of fuel in kg/m**3\n",
      "da = (pa*10**4)/(R*Ta)\t\t\t\t\t#Density of air in kg/m**3\n",
      "va = Ca*math.sqrt((2*9.81*x*df)/(da*1000))\t\t\t\t\t#Critical velocity of air in m/s\n",
      "dpa = (((fc/(60*60))/((3.14/4)*(jd/1000)**2*Cd))**2/(2*9.81*df))+((x/1000)*df)\t\t\t\t\t#Drop in pressure in kg/m**3\n",
      "dpaw = (dpa/1000)*100\t\t\t\t\t#Drop in pressure in cm of water\n",
      "dj = math.sqrt(((fc/(3600*af))/(Ca*math.sqrt(2*9.81*da*dpa)))/(3.14/4))*1000\t\t\t\t\t#Effective throat diameter in mm\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'Critical air velocity is %3.2f m/sec  \\\n",
      "\\nEffective throat diameter of the venturi is %3.1f mm  \\\n",
      "\\nThe drop in pressure in the venturi is %3.2f cm of water'%(va,dj,dpaw)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Critical air velocity is 4.85 m/sec  \n",
        "Effective throat diameter of the venturi is 21.4 mm  \n",
        "The drop in pressure in the venturi is 36.73 cm of water\n"
       ]
      }
     ],
     "prompt_number": 14
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.14  Page no :  226"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "ma = 6.11\t\t\t\t\t#Flow rate of air in kg/min\n",
      "mf = 0.408\t\t\t\t\t#Flow arte of petrol in kg/min\n",
      "dp = 768\t\t\t\t\t#Density of petrol in kg/m**3\n",
      "Ta = 15.5+273\t\t\t\t\t#Temperature of air in K\n",
      "pa = 1.027\t\t\t\t\t#Pressure of air in kg/cm**2\n",
      "R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/kg.K \n",
      "va = 97.5\t\t\t\t\t#Speed of air in m/sec\n",
      "Cv = 0.84\t\t\t\t\t#Velocity coefficient\n",
      "g = 1.4\t\t\t\t\t#Ratio of specific heats \n",
      "x = 0.8\t\t\t\t\t#pressure at the venturi is 0.8 of the pressure drop at the choke\n",
      "Cd = 0.66\t\t\t\t\t#Coefficient of discharge\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "rp = (1-((va/Cv)**2/(((2*9.81*g)/(g-1))*R*Ta)))**(g/(g-1))\t\t\t\t\t#Pressure ratio\n",
      "p2 = (pa*rp)\t\t\t\t\t#Pressure in kg/cm**2\n",
      "T2 = (Ta*rp**((g-1)/g))\t\t\t\t\t#Temperature in K\n",
      "da = (p2/(R*T2))*10**4\t\t\t\t\t#Density in kg/m**3\n",
      "daa = math.sqrt((ma/(60*va*da))/(3.14/4))*1000\t\t\t\t\t#Throat diameter in mm\n",
      "df = math.sqrt((mf/(60*Cd*math.sqrt(2*9.81*dp*x*(pa-p2)*10**4)))/(3.14/4))*1000\t\t\t\t\t#Orifice diameter in mm\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'Throat diameter of the choke is %i mm  \\\n",
      "\\nThe orifice diameter is %3.2f mm'%(daa,df)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Throat diameter of the choke is 34 mm  \n",
        "The orifice diameter is 2.05 mm\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.15  Page no :  226"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "ma = 6.8\t\t\t\t\t#Mass flow rate of air in kg/min\n",
      "mf = 0.45\t\t\t\t\t#Mass flow rate of petrol in kg/min\n",
      "pa = 1.033\t\t\t\t\t#Pressure of air in kg/cm**2\n",
      "Ta = 20+273\t\t\t\t\t#Temperature of air in K\n",
      "va = 97.5\t\t\t\t\t#Velocity of air in m/s\n",
      "Cv = 0.8\t\t\t\t\t#Velocity coefficient\n",
      "g = 1.4\t\t\t\t\t#Ratio of specific heats \n",
      "R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/kg.K \n",
      "x = 0.75\t\t\t\t\t#pressure at the venturi is 0.8 of the pressure drop at the choke\n",
      "Cd = 0.65\t\t\t\t\t#Coefficient of discharge\n",
      "pw = 800\t\t\t\t\t#Weight of petrol in kg per cu.m\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "rp = (1-((va/Cv)**2/(((2*9.81*g)/(g-1))*R*Ta)))**(g/(g-1))\t\t\t\t\t#Pressure ratio\n",
      "p2 = (pa*rp)\t\t\t\t\t#Pressure in kg/cm**2\n",
      "T2 = (Ta*rp**((g-1)/g))\t\t\t\t\t#Temperature in K\n",
      "da = (p2/(R*T2))*10**4\t\t\t\t\t#Density in kg/m**3\n",
      "daa = math.sqrt((ma/(60*va*da))/(3.14/4))*100\t\t\t\t\t#Throat diameter in mm\n",
      "df = math.sqrt((mf/(60*Cd*math.sqrt(2*9.81*pw*x*(pa-p2)*10**4)))/(3.14/4))\t\t\t\t\t#Orifice diameter in mm\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'Throat diameter of the choke is %3.2f cm  \\\n",
      "\\nThe orifice diameter is %3.5f m'%(daa,df)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Throat diameter of the choke is 3.62 cm  \n",
        "The orifice diameter is 0.00213 m\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 9.16  Page no :  229"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "n = 6.\t\t\t\t\t#Number of cylinders\n",
      "d = 100.\t\t\t\t\t#Diameter in mm\n",
      "L = 100.\t\t\t\t\t#Stroke in mm\n",
      "N = 1500.\t\t\t\t\t#Speed in r.p.m\n",
      "ap = 13.5\t\t\t\t\t#Air fuel ratio\n",
      "Ta = 80.+273\t\t\t\t\t#Temperature of air in K\n",
      "x = (7./8)\t\t\t\t\t#Ratio of volume drawn\n",
      "nth = 22.\t\t\t\t\t#Thermal efficiency in percent\n",
      "p = 76.\t\t\t\t\t#Pressure in cm of mercury\n",
      "CV = 9000.\t\t\t\t\t#Calorific value of petrol in kcal/kg\n",
      "l = 1524.\t\t\t\t\t#Altitude in m\n",
      "dp = 2.54\t\t\t\t\t#Drop in pressure in cm of barometric reading\n",
      "lx = 274\t\t\t\t\t#Altitude rise in m\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "Vs = (3.14/4)*(d/10)**2*(L/10)\t\t\t\t\t#Swept volume in c.c\n",
      "Va = (x*Vs)\t\t\t\t\t#Volume of air drawn in per cylinder per stroke in c.c\n",
      "wa = (Va*10**-6*(N/2)*n)\t\t\t\t\t#Weight of air supplied to the engine per minute in kg\n",
      "wf = (wa/ap)\t\t\t\t\t#Weight of fuel consumed per minute in kg\n",
      "q = (wf*CV)\t\t\t\t\t#Heat supplied to the engine per minute in kcal\n",
      "P = (q*(nth/100)*427)/4500\t\t\t\t\t#Power developed at ground level in H.P\n",
      "db = (l/lx)*dp\t\t\t\t\t#Drop in barometric reading at an altitude of 1524 m in cm\n",
      "Pd = (P/p)*(p-db)\t\t\t\t\t#Power developed at 1524 m altitude in H.P\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'Power developed at the ground level is %i H.P  \\\n",
      "\\nPower developed at an altitude of %i m is %i H.P'%(P,l,Pd)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Power developed at the ground level is 43 H.P  \n",
        "Power developed at an altitude of 1524 m is 35 H.P\n"
       ]
      }
     ],
     "prompt_number": 17
    }
   ],
   "metadata": {}
  }
 ]
}