{ "metadata": { "name": "", "signature": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 4, Fuel Air and Actual Cycles" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 4.10, page 340" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Initialisation of Variables\n", "r=7 #Compression Ratio\n", "t2=715 #Temperature at the end of isentropic compression in Kelvin\n", "t4=1610 #Temperature at the end of expansion in Kelvin\n", "#Calculations\n", "vr2=65.8 #From steam table\n", "u2=524.2 #From steam table\n", "vr4=5.69 #From steam table\n", "u4=1307.63 #From steam table\n", "vr1=r*vr2 \n", "t1=338 #From steam table\n", "u1=241.38 #From steam table\n", "vr3=vr4/r \n", "t3=2800 #From steam table\n", "u3=2462.5 #From steam table\n", "W=(u3-u2)-(u4-u1) #Work done\n", "Qa=(u3-u2) #Heat added\n", "eta=W/Qa #Cycle efficiency\n", "print \"The cycle work = %0.2f kJ/kg \" %W \n", "print \"The cycle efficiency = %0.2f %%\" %(eta*100) " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The cycle work = 872.05 kJ/kg \n", "The cycle efficiency = 44.99 %\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 4.2, page 342" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import log\n", "# Initialisation of Variables\n", "r=8 #Compression Ratio\n", "ga=1.4 #Degree of freedom for the gas\n", "Cvinc=1.1 #Increase of specific heat at constant volume in percentage\n", "#Calculations\n", "eta=1-1/(r**(ga-1)) #efficiency of otto cycle\n", "deta=(1-eta)*(ga-1)*log(r)*(Cvinc/100) #Change in efficiency\n", "etach=-deta/eta #Percentage change in efficiency of change in efficiency\n", "print \"The percentage change in the efficiency of otto cycle = %0.2f %%\"%(etach*100)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The percentage change in the efficiency of otto cycle = -0.71 %\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 4.3, page 345" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Initialisation of Variables\n", "r=7.0 #Compression Ratio\n", "ga=1.4 #Degree of freedom for the gas\n", "Cvinc=3.0 #Increase of specific heat at constant volume in percentage\n", "#Calculations\n", "eta=1-1/(r**(ga-1)) #efficiency of otto cycle\n", "deta=(1-eta)*(ga-1)*log(r)*(Cvinc/100) #Change in efficiency\n", "etach=-deta/eta #Percentage change in efficiency of change in efficiency\n", "print \"The percentage change in the efficiency of otto cycle = %0.2f %%\"%(etach*100)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The percentage change in the efficiency of otto cycle = -1.98 %\n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 4.4, page 349" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Initialisation of Variables\n", "r=18.0 #Compression Ratio\n", "co=5.0 #Cut off percent of stroke\n", "cv=0.71 #Mean specific heat for cycle in kJ/kg K\n", "R=0.285 #Charecteristic gas constant in kJ/kh K\n", "cvinc=2.0 #Percentage increase in mean specific heat of the cycle\n", "#Calculation\n", "rho=(co/100)*(r-1)+1 \n", "ga=1+(R/cv) \n", "eta=1-(1/(ga*(r**(ga-1))))*((rho**ga)-1)/(rho-1) #Efficiency of diesel cycle \n", "etach=-((1-eta)/eta)*(ga-1)*(log(r)-(((rho**ga)*log(rho))/((rho**ga)-1))+(1/ga))*(cvinc/100) #Variation in the air standard efficiency\n", "print \"The percentage change in the efficiency = %0.2f %%\"%(etach*100)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The percentage change in the efficiency = -1.15 %\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 4.5, page 351" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import sqrt\n", "# Initialisation of Variables\n", "r=7 #Compression ratio\n", "C=44000 #Calorific value of fuel used in kJ/kg\n", "afr=15 #Air fuel ratio\n", "t1=338 #Temperature of the charge at the end of the stroke in Kelvin\n", "p1=1 #Pressure of the charge at the end of the stroke in bar\n", "n=1.33 #Index of compression\n", "cv=0.71 #Specific heat constant at constant volume in kJ/kgK\n", "k=20*10**(-5) \n", "#Calculations\n", "p2=p1*(r)**n \n", "t2=(t1*p2)/(p1*r) \n", "ha=C/(afr+1) #Heat added per kg of charge in kJ\n", "t3=((-2*cv)+sqrt((4*cv*cv)+(4*k*((2*cv*t2)+(k*t2*t2)+(2*ha)))))/(2*k) \n", "p3=(p2*t3)/t2 #Max pressure for constant volume process in bar\n", "P3=p2*((ha/cv)+t2)/t2 #Max pressure for constant specific heat in bar\n", "print \"Max pressure in the cylinder = %0.2f bar \" %(p3)\n", "print \"Max pressure for constant specific heat = %0.2f bar\" %P3" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Max pressure in the cylinder = 65.52 bar \n", "Max pressure for constant specific heat = 93.52 bar\n" ] } ], "prompt_number": 26 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 4.6, page 356" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Initialisation of Variables\n", "r=10 #Compression ratio\n", "C=48000 #Calorific value of fuel used in kJ/kg\n", "afr=15 #Air fuel ratio\n", "t1=330 #Temperature of the charge at the end of the stroke in Kelvin\n", "p1=1 #Pressure of the charge at the end of the stroke in bar\n", "n=1.36 #Index of compression\n", "cv=0.7117 #Specific heat constant at constant volume in kJ/kgK\n", "k=2.1*10**(-4) \n", "#Calculations\n", "p2=p1*(r)**n \n", "t2=t1*((p2/p1)**((n-1)/n)) \n", "ha=C/(afr+1) #Heat added per kg of charge in kJ\n", "t3=((-2*cv)+sqrt((4*cv*cv)+(4*k*((2*cv*t2)+(k*t2*t2)+(2*ha)))))/(2*k) \n", "p3=(p2*t3)/t2 #Max pressure for constant volume process in bar\n", "P3=p2*((ha/cv)+t2)/t2 #Max pressure for constant specific heat in bar\n", "print \"Max pressure in the cylinder = %0.2f bar \" %(p3)\n", "print \"Max pressure for constant specific heat = %0.2f bar\" %P3" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Max pressure in the cylinder = 102.27 bar \n", "Max pressure for constant specific heat = 150.64 bar\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 4.7, page 360" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Initialisation of Variables\n", "r=15 #Compression ratio\n", "C=43000 #Calorific value of fuel used in kJ/kg\n", "afr=27 #Air fuel ratio\n", "t2=870 #Temperature of the charge at the end of the stroke in Kelvin\n", "cv=0.71 #Specific heat constant at constant volume in kJ/kgK\n", "R=0.287 #Gas constant in kJ/kgK\n", "k=20*10**(-5) \n", "#Calculations\n", "cp=cv+R #Specific heat at constant pressure\n", "ha=C/(afr+1) #Heat added per kg of charge in kJ\n", "t3=((-2*cp)+sqrt((4*cp*cp)+(4*k*((2*cp*t2)+(k*t2*t2)+(2*ha)))))/(2*k) \n", "co=((t3/t2)-1)/(r-1) #combustion occupies this amt of stroke \n", "print \"Percentage of the stroke when the combustion is completed is\",round((co*100),2)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Percentage of the stroke when the combustion is completed is 9.77\n" ] } ], "prompt_number": 28 } ], "metadata": {} } ] }