path: root/Internal_Combustion_Engines_by_H._B._Keswani/ch5.ipynb

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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 5 : Variable Specific Heat"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.2  Page no :  105"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "r = 8.\t\t\t\t\t#Compression ratio\n",
      "n = 1.41\t\t\t\t\t#Adiabatic index of the medium\n",
      "cv = 0.17\t\t\t\t\t#Mean Specific heat at consmath.tant volume in kcal/kg/degree C\n",
      "x = 2.\t\t\t\t\t#Percentage with which spcific heat at consmath.tant volume increases\n",
      "R = 29.3\t\t\t\t\t#Characteristic gas consmath.tant in mkg/kg/degree C\n",
      "J = 427.\t\t\t\t\t#Mechanical equivalent of heat in kg.m/kcal\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "e = (1-(1/r**(n-1)))\t\t\t\t\t#Air standard efficiency neglecting the variation in specific heat\n",
      "debye = ((x/100)*((1-e)/e)*(R/(J*cv))*math.log(r))*100\t\t\t\t\t#Ratio of de and e in percent\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The change in air standard efficiency of the cycle is %3.3f percent'%(debye)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The change in air standard efficiency of the cycle is 1.247 percent\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.3  Page no :  106"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from scipy.integrate import quad\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "\t\t\t\t\t#Cv = 0.125+0.000005T where Cv is Specific heat at consmath.tant volume and T is the temperature in K\n",
      "R = 28.9\t\t\t\t\t#Characteristic gas consmath.tant in mkg/kg/degree C\n",
      "T = [100+273,50+273]\t\t\t\t\t#Temperature in K\n",
      "J = 427\t\t\t\t\t#Mechanical equivalent of heat in kg.m/kcal\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "def f(x):\n",
      "\treturn  0.125+(0.00005*x)\n",
      "\t\n",
      "I = J*quad(f,303,373)[0]\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The work done is %i m.kg/kg of gas'%(I)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The work done is 4241 m.kg/kg of gas\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.4  Page no :  109"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "af = 25.\t\t\t\t\t#Air fuel ratio\n",
      "cv = [0.17,0.00004]\t\t\t\t\t#Cv = 0.17+0.00004T where Cv is Specific heat at constant volume and T is the temperature in K\n",
      "r = 14.\t\t\t\t\t#Compression ratio\n",
      "p1 = 1.\t\t\t\t\t#Pressure at the beginning of compression in kg/cm**2\n",
      "T1 = 153.+273\t\t\t\t\t#Temperature at the beginning of compression in K\n",
      "CV = 10000.\t\t\t\t\t#Heating value of fuel in kcal/kg\n",
      "n = 1.35\t\t\t\t\t#Adiabatic constant\n",
      "R = 29.\t\t\t\t\t#Characteristic gas constant in mkg/kg.K\n",
      "J = 427.\t\t\t\t\t#Mechanical equivalent of heat in kg.m/kcal\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "T2 = (T1*r**(n-1))\t\t\t\t\t#Temperature at the end of compression in K\n",
      "a = (cv[1]/2)\t\t\t\t\t#For solving T3\n",
      "b = cv[0]+(R/J)\t\t\t\t\t#For solving T3\n",
      "c = (-T2*cv[0])-((cv[1]/2)*T2**2)-((R/J)*T2)-(CV/(af+1))\t\t\t\t\t#Foe solving T3\n",
      "T3 = (-b+math.sqrt(b**2-(4*a*c)))/(2*a)\t\t\t\t\t#Soving for T3 in K\n",
      "pc = (((T3/T2)-1)/(r-1))*100\t\t\t\t\t#Percentage cut off\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The percentage of stroke at which the constant pressure combustion stops is %i percent'%(pc)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The percentage of stroke at which the constant pressure combustion stops is 9 percent\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.5  Page no :  113"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "af = 25.\t\t\t\t\t#Air fuel ratio\n",
      "CV = 10000.\t\t\t\t\t#Calorific value in kcal/kg\n",
      "cv = [0.17,0.00004]\t\t\t\t\t#Cv = 0.17+0.00004T where Cv is Specific heat at constant volume and T is the temperature in K\n",
      "r = 14.\t\t\t\t\t#Compression ratio\n",
      "T2 = 800.+273\t\t\t\t\t#Temperature at the end of compression in K\n",
      "R = 29.\t\t\t\t\t#Characteristic gas constant in mkg/kg/degree C\n",
      "J = 427.\t\t\t\t\t#Mechanical equivalent of heat in kg.m/kcal\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "CVm = (CV/(af+1))\t\t\t\t\t#Calorific value of mixture in kcal/kg\n",
      "cpv = (R/J)\t\t\t\t\t#Difference in mean specific heats in kcal/kg mol.K\n",
      "a = (cv[1]/2)\t\t\t\t\t#For solving T3\n",
      "b = cpv+cv[0]\t\t\t\t\t#For solving T3\n",
      "c = (-T2*(cpv+cv[0]))-((cv[1]/2)*T2**2)-CVm\t\t\t\t\t#Foe solving T3\n",
      "T3 = (-b+math.sqrt(b**2-(4*a*c)))/(2*a)\t\t\t\t\t#Soving for T3 in K\n",
      "s = ((T3/T2)/(r-1))*100\t\t\t\t\t#Percentage of the stroke\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The percentage of the stroke at which the combustion will be complete is %3.2f percent'%(s)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The percentage of the stroke at which the combustion will be complete is 16.70 percent\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.6  Page no :  115"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from scipy.integrate import quad\n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "T = [500,2000]\t\t\t\t\t#Change in temperature in K\n",
      "x = [11.515,-172,1530]\t\t\t\t\t#Cp = 11.515-172/math.sqrt(T)+1530/T in kcal/kg mole.K\n",
      "mO2 = 32\t\t\t\t\t#Molecular weight of oxygen\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "\n",
      "def f(T):\n",
      "\treturn (x[0]+(x[1]/math.sqrt(T))+(x[2]/T))\n",
      "\t\n",
      "I = -quad(f,T[1],T[0])[0]\t\t\t\t\t#Integration\n",
      "dh = (I/mO2)\t\t\t\t\t#Change in enthalpy in kcal/kg\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The change in enthalpy is %3.1f kcal/kg'%(dh)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The change in enthalpy is 365.7 kcal/kg\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.7  Page no :  117"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "r = 14.\t\t\t\t\t#Compression ratio\n",
      "s = 5.\t\t\t\t\t#Fuel injection stops at 5% stroke after inner head centre\n",
      "pm = 50.\t\t\t\t\t#Maximum pressure in kg/cm**2\n",
      "p4 = 1.\t\t\t\t\t#Pressure at the end of suction stroke in kg/cm**2\n",
      "T4 = 90.+273\t\t\t\t\t#Temperature at the end of suction stroke in K\n",
      "R = 29.3\t\t\t\t\t#Characteristic gas constant in mkg/kg/degree C\n",
      "cv = [0.171,0.00003]\t\t\t\t\t#Cv = 0.171+0.00003T where Cv is Specific heat at constant volume and T is the temperature in K\n",
      "J = 427.\t\t\t\t\t#Mechanical equivalent of heat in kg.m/kcal\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "a = (R/J)+cv[0]\t\t\t\t\t#a value in kcal/kg.mole.K\n",
      "g = (a+cv[1]*T4)/(cv[0]+cv[1]*T4)\t\t\t\t\t#Adiabatic index of compression\n",
      "z = 1.3\t\t\t\t\t#Rounding off 'z' value to one decimal.\n",
      "T5 = (T4*r**(z-1))\t\t\t\t\t#Temperature in K\n",
      "p5 = (p4*r**g)\t\t\t\t\t#Pressure in kg/cm**2\n",
      "T1 = T5*(pm/p5)\t\t\t\t\t#Tmperature in K\n",
      "T2 = (T1*(1+(s/100)*(r-1)))\t\t\t\t\t#Temperature in K\n",
      "T3 = (T2*((1+(s/100)*(r-1))/r)**(g-1))\t\t\t\t\t#Temperature in K\n",
      "p3 = (p4*(T3/T4))\t\t\t\t\t#Pressure in kg/cm**2\n",
      "def f1(T):\n",
      "\treturn  cv[0]+(cv[1]*T)\n",
      "I1 = quad(f1,T5,T1)[0]\n",
      "\n",
      "def f2(T):\n",
      "\treturn (a+(cv[1]*T))\n",
      "\t\n",
      "I2 = quad(f2,T1,T2)[0]\t\t\t\t\t#I2 answer is given wrong in the textbook\n",
      "qs = (I1+I2)\t\t\t\t\t#Heat supplied per kg of air in kcal/kg\n",
      "\n",
      "def f3(T):\n",
      "\treturn  a+(cv[1]*T)\n",
      "\t\n",
      "qre = quad(f3,T4,T3)[0]\t\t\t\t\t#Heat required per kg of air in kcal/kg\n",
      "\n",
      "nth = ((qs-qre)/qs)*100\t\t\t\t\t#Thermal efficiency in percent\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'The tempertautes and pressures at salient points of the cycle are :  T1  =  %3.0f K  \\\n",
      "\\np1  =  %3.1f kg/cm**2  \\\n",
      "\\nT2  =  %3.0f K  \\\n",
      "\\np2  =  %3.1f kg/cm**2  \\\n",
      "\\nT3  =  %3.0f K  \\\n",
      "\\np3  =  %3.1f kg/cm**2  \\\n",
      "\\nT4  =  %3.0f K  \\\n",
      "\\np4  =  %3.1f kg/cm**2  \\\n",
      "\\nT5  =  %3.0f K  \\\n",
      "\\np5  =  %3.1f kg/cm**2  \\\n",
      "\\nHeat supplied per kg of air is %3.1f kcal/kg  \\\n",
      "\\nThe thermal efficiency of the cycle is %3.1f percent'%(T1,pm,T2,pm,T3,p3,T4,p4,T5,p5,qs,nth)\n",
      "\n",
      "#Textbook answers are given wrong\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The tempertautes and pressures at salient points of the cycle are :  T1  =  1057 K  \n",
        "p1  =  50.0 kg/cm**2  \n",
        "T2  =  1745 K  \n",
        "p2  =  50.0 kg/cm**2  \n",
        "T3  =  779 K  \n",
        "p3  =  2.1 kg/cm**2  \n",
        "T4  =  363 K  \n",
        "p4  =  1.0 kg/cm**2  \n",
        "T5  =  801 K  \n",
        "p5  =  37.9 kg/cm**2  \n",
        "Heat supplied per kg of air is 244.5 kcal/kg  \n",
        "The thermal efficiency of the cycle is 56.4 percent\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.8  Page no :  119"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "from scipy.integrate import quad\n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "r = 14.\t\t\t\t\t#Compression ratio\n",
      "c = 5.\t\t\t\t\t#Cut off takes place at 5% of the stroke\n",
      "p1 = 1.\t\t\t\t\t#Pressure at the beginning of compression in kg/cm**2. In texbook, it is given wrong as 10\n",
      "T1 = 90.+273\t\t\t\t\t#Temperature at the beginning of compression in K\n",
      "p3 = 50.\t\t\t\t\t#Maximum pressure in kg/cm**2\n",
      "R = 29.3\t\t\t\t\t#Characteristic gas constant in mkg/kg/degree C\n",
      "cv = [0.171,0.00003]\t\t\t\t\t#Cv = 0.171+0.00003T where Cv is Specific heat at constant volume and T is the temperature in K\n",
      "g1 = 1.4\t\t\t\t\t#Ratio of specific heats\n",
      "J = 427.\t\t\t\t\t#Mechanical equivalent of heat in kg.m/kcal\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "T2x = (T1*r**(g1-1))\t\t\t\t\t#Temperature in K\n",
      "def  f1(T):\n",
      "\treturn  cv[0]+(cv[1]*T)\n",
      "\n",
      "I1 = quad(f1,T1,T2x)[0]\n",
      "\n",
      "Cv = (1/(T2x-T1))*I1\t\t\t\t\t#Mean value of Cv in kJ/kg.K\n",
      "Cp = (Cv+(R/J))\t\t\t\t\t#Mean value of Cp in kJ/kg.K\n",
      "g = 1.35\t\t\t\t\t#(Cp/Cv) value and rounded off to 2 decimal places for calculation purpose. Ratio of specific heats\n",
      "T2 = (T1*r**(g-1))\t\t\t\t\t#Temperature in K\n",
      "I2 = quad(f1,T1,T2)[0]\n",
      "CV = (1/(T2-T1))*I2\t\t\t\t\t#Maen value of Cv in kJ/kg.K\n",
      "CP = (Cv+(R/J))\t\t\t\t\t#Mean value of Cp in kJ/kg.K\n",
      "g2 = 1.36\t\t\t\t\t#(Cp/Cv) value and rounded off to 2 decimal places for calculation purpose.Ratio of specific heats\n",
      "T2a = (T1*r**(g2-1))\t\t\t\t\t#Temperature in K\n",
      "p2 = (p1*r*(T2a/T1))\t\t\t\t\t#Pressure in kg/cm**2\n",
      "T3 = (T2a*(p3/p2))\t\t\t\t\t#Temperature in K\n",
      "T4 = (((r-1)*(c/100))+1)*T3\t\t\t\t\t#Temperature in K\n",
      "g3 = 1.3\t\t\t\t\t#Assuming gamma as 1.3 for process 4-5\n",
      "T5 = (T4/(r/(((r-1)*(c/100))+1))**(g3-1))\t\t\t\t\t#Temperature in K\n",
      "cV = cv[0]+(cv[1]/2)*(T5+T4)\t\t\t\t\t#Mean value of Cv in kJ/kg.K\n",
      "cP = cV+(R/J)\t\t\t\t\t#Mean value of Cp in kJ/kg.K\n",
      "g4 = (cP/cV)\t\t\t\t\t#Ratio of specific heats\n",
      "T5a = (T4/(r/(((r-1)*(c/100))+1))**(g4-1))\n",
      "I3 = quad(f1,T2a,T3)[0]\n",
      "\n",
      "def f2(T):\n",
      "\treturn cv[0]+(R/J)+(cv[1]*T)\n",
      "I4 = quad(f2,T3,T4)[0]\t\t\t\t\t#Textbook answer is wrong\n",
      "q = I3+I4\t\t\t\t\t#Heat supplied per kg of working substance in kcal/kg\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'a) Temperatures at all the points of the cycle are:  \\\n",
      "\\nT1  =  %i K  T2  =  %3.0f K  T3  =  %3.0f K  T4  =  %3.0f K  T5  =  %i K  \\\n",
      "\\nb) heat supplied per kg of the working substance is %3.1f kcal/kg'%(T1,T2a,T3,T4,T5a,q)\n",
      "\t\t\t\t\t#Textbook answer is wrong\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "a) Temperatures at all the points of the cycle are:  \n",
        "T1  =  363 K  T2  =  939 K  T3  =  1296 K  T4  =  2139 K  T5  =  1097 K  \n",
        "b) heat supplied per kg of the working substance is 318.5 kcal/kg\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5.10  Page no :  119"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math \n",
      "\t\t\t\t\t\n",
      "#Input data\n",
      "r = 20.\t\t\t\t\t#Compression ratio\n",
      "c = 5.\t\t\t\t\t#Cut off at 5%\n",
      "dc = 1.\t\t\t\t\t#Specific heat at constant volume increases by 1%\n",
      "Cv = 0.171\t\t\t\t\t#pecific heat at constant volume in kJ/kg.K\n",
      "R = 29.3\t\t\t\t\t#Characteristic gas constant in mkg/kg/degree C\n",
      "k = 1.95\t\t\t\t\t#k can be obtained from relation de/e = -dcv/cv*(1-e/e)*(g-1)*((1/g)+ln(r)-(k**g*lnk)/(k**g-1))\n",
      "J = 427.\t\t\t\t\t#Mechanical equivalent of heat in kg.m/kcal\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Calculations\n",
      "g = (R/(J*Cv))+1\t\t\t\t\t#Ratio of specific heats\n",
      "e = (1-((1/g)*(1/r**(g-1))*((k**g-1)/(k-1))))\t\t\t\t\t#Air standard efficiency of the cycle\n",
      "dee = ((-(dc/100)*((1-e)/e)*(g-1)*((1/g)+math.log(r)-((k**g*math.log(k))/(k**g-1))))*100)\t\t\t\t\t#Change in efficiency due to 1% change in cv\n",
      "\n",
      "\t\t\t\t\t\n",
      "#Output\n",
      "print 'Percentage change in air standard efficiency is %3.3f percent This indicates that there is a decrease in efficiency'%(dee)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Percentage change in air standard efficiency is -0.564 percent This indicates that there is a decrease in efficiency\n"
       ]
      }
     ],
     "prompt_number": 8
    }
   ],
   "metadata": {}
  }
 ]
}