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{
"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": {}
}
]
}
|
