{ "metadata": { "name": "", "signature": "sha256:ffa252e0f1b2a360d9cd4d249675a66968869a58480fb1a0e3c0c3fd2fa6c2ba" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "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": {} } ] }