PKI6=11%Internal Combustion Engines/ch1.ipynb{
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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1 : Introduction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.1 Page no : 7"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\t\t\n",
"#Input data\n",
"BHP = 51.\t\t\t\t\t#Brake horse power in h.p\n",
"N = 1000.\t\t\t\t\t#Speed in r.p.m\n",
"FHP = 17.\t\t\t\t\t#Friction horse power in h.p\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"IHP = (BHP+FHP)\t\t\t\t\t#Indicated Horse power in h.p\n",
"mn = (BHP/IHP)*100\t\t\t\t\t#Mechanical efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Mechanical efficiency of the engine is %i percent'%(mn)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Mechanical efficiency of the engine is 75 percent\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2 Page no : 10"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Data taken from Ex.No.1\n",
"BHP = 51.\t\t\t\t\t#Brake horse power in h.p\n",
"N = 1000.\t\t\t\t\t#Speed in r.p.m\n",
"FHP = 17.\t\t\t\t\t#Friction horse power in h.p\n",
"\t\t\t\t\t\n",
"#Input data\n",
"O1 = BHP/2\t\t\t\t\t#Half of b.h.p output in h.p\n",
"O2 = 10\t\t\t\t\t#Brake horse power in h.p\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"#Case(i)\n",
"IHP1 = (O1+FHP)\t\t\t\t\t#Indicated Horse power in h.p\n",
"mn1 = (O1/IHP1)*100\t\t\t\t\t#Mechanical efficiency in percent\n",
"\n",
"#Case(ii)\n",
"IHP2 = (O2+FHP)\t\t\t\t\t#Indicated Horse power in h.p\n",
"mn2 = (O2/IHP2)*100\t\t\t\t\t#Mechanical efficiency in percent\n",
"\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Mechanical efficiency of the engine when it delivers \\\n",
"\\na) Half the b.h.p output is %3.0f percent \\\n",
"\\nb) 10 b.h.p is %3.0f percent'%(mn1,mn2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Mechanical efficiency of the engine when it delivers \n",
"a) Half the b.h.p output is 60 percent \n",
"b) 10 b.h.p is 37 percent\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3 Page no : 12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"Fc = 220.\t\t\t\t\t#Fuel consumption in gm/(b.h.p*hr)\n",
"CV = 10600.\t\t\t\t\t#Calorific value in kcal/kg\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"hf = (Fc/1000)*CV\t\t\t\t\t#Heat supplied in kcal/hr\n",
"O = 632.\t\t\t\t\t\n",
"#Output in terms of kcal/hr\n",
"bn = (O/hf)*100\t\t\t\t\t#Brake thermal efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Brake thermal efficiency is %3.1f percent'%(bn)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Brake thermal efficiency is 27.1 percent\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.4 Page no : 12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"IHP = 45.\t\t\t\t\t#Indicated horse power in h.p\n",
"Fc = 13.\t\t\t\t\t#Fuel consumption in litres/hr\n",
"g = 0.8\t\t\t\t\t#Specific gravity of oil\n",
"nm = 80.\t\t\t\t\t#Mechanical efficiency in percent\n",
"CV = 10000.\t\t\t\t\t#Calorific value of fuel in kcal/kg\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"BHP = (IHP*nm)/100\t\t\t\t\t#Brake horse power in h.p\n",
"hi = (Fc*g*CV)\t\t\t\t\t#Heat supplied in kcal/hour\n",
"In = ((IHP*4500*60)/(427*hi))*100\t\t\t\t\t#Indicated thermal efficiency in percent\n",
"Bn = (In*(nm/100))\t\t\t\t\t#Brake thermal efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Indicated thermal efficiency is %3.2f percent \\\n",
"\\nBrake thermal efficiency is %3.2f percent'%(In,Bn)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Indicated thermal efficiency is 27.36 percent \n",
"Brake thermal efficiency is 21.89 percent\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.5 Page no : 13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"BHP = 15.\t\t\t\t\t#Brake horse power in h.p\n",
"In = 28.\t\t\t\t\t#Indicated thermal efficiency in percent\n",
"mn = 75.\t\t\t\t\t#Mechanical efficiency in percent\n",
"CV = 10000.\t\t\t\t\t#Calorific value of fuel in kcal/kg\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Bn = ((In/100)*(mn/100))*100\t\t\t\t\t#Brake thermal efficiency in percent\n",
"I = (BHP/(Bn/100))*((4500.*60)/427)\t\t\t\t\t#Input in kcal/hr\n",
"Fc = (I/CV)\t\t\t\t\t#Fuel consumption in kg/hr\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Fuel consumption of the engine is %3.2f kg/hr'%(Fc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Fuel consumption of the engine is 4.52 kg/hr\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}PKI~zz%Internal Combustion Engines/ch3.ipynb{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3 : Air Standard Cycles"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.1 Page no : 19"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"p = [1,8]\t\t\t\t\t#Pressure at the beginning and end of compression in kg/m**3\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"r = (p[1]/p[0])**(1/g)\t\t\t\t\t#Compression ratio\n",
"n = (1-(1/r)**(g-1))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Air standard efficiency of an engine working on the Otto cycle is %3.1f percent'%(n)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Air standard efficiency of an engine working on the Otto cycle is 44.8 percent\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.2 Page no : 24"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"D = 0.25\t\t\t\t\t#Bore in m\n",
"L = 0.45\t\t\t\t\t#Stroke in m\n",
"Cv = 5.\t\t\t\t\t#Clearance volume in litres\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"IHP = 32.\t\t\t\t\t#Indicated Horse power in h.p\n",
"m = 14.\t\t\t\t\t#Gas consumption in m**3/hr\n",
"CV = 4000.\t\t\t\t\t#Calorific value of gas in kcal/m**3\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = (3.14/4)*D**2*L\t\t\t\t\t#Stroke volume in m**3\n",
"Vc = Cv/1000\t\t\t\t\t#Clearance volume in m**3\n",
"r = (Vs+Vc)/Vc\t\t\t\t\t#Compression ratio\n",
"na = (1-(1/r)**(g-1))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"q = (m*CV)/60\t\t\t\t\t#Heat supplied in kcal/min\n",
"aI = (IHP*4500)/427\t\t\t\t\t#Heat equivalent of I.H.P in kcal/min\n",
"itn = (aI/q)*100\t\t\t\t\t#Indicated thermal efficiency in percent\n",
"rn = (itn/na)*100\t\t\t\t\t#Relative efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The air standard efficiency is %3.1f percent \\\n",
"\\nIndicated thermal efficiency is %3.1f percent \\\n",
"\\nRelative efficiency is %3.1f percent'%(na,itn,rn)\n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The air standard efficiency is 49.1 percent \n",
"Indicated thermal efficiency is 36.1 percent \n",
"Relative efficiency is 73.6 percent\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.3 Page no : 26"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"r = 6.\t\t\t\t\t#Compression ratio\n",
"It = 0.6\t\t\t\t\t#Indicated thermal efficiency ratio\n",
"CV = 10000.\t\t\t\t\t#Calorific value in kcal/kg\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"an = (1-(1/r)**(g-1))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"In = (It*(an/100))\t\t\t\t\t#Indicated thermal efficiency \n",
"SFC = ((4500*60)/(427*CV*In))\t\t\t\t\t#Specific fuel consumption in kg/I.H.P.hr\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Specific fuel consumption is %3.3f kg/I.H.P. hr'%(SFC)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Specific fuel consumption is 0.206 kg/I.H.P. hr\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.4 Page no : 27"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"T = [100.+273,473.+273]\t\t\t\t\t#Temperatures at the beginning and at the end of adiabatic compression in K\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"an = (1-(T[0]/T[1]))*100\t\t\t\t\t#Air standard efficiency in percent \n",
"r = (T[1]/T[0])**(1/(g-1))\t\t\t\t\t#Compression ratio\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The compression ratio is %3.2f \\\n",
"\\nAir standard efficiency is %i percent'%(r,an)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The compression ratio is 5.66 \n",
"Air standard efficiency is 50 percent\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.5 Page no : 29"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"T1 = 45.+273\t\t\t\t\t#Temperature at the beginning of compression in K\n",
"p1 = 1.\t\t\t\t\t#Pressure at the beginning of compression in kg/cm**2\n",
"T2 = 325.+273\t\t\t\t\t#Temperature at the end of compression in K\n",
"T3 = 1500.+273\t\t\t\t\t#Temperature at the end of consmath.tant volume heat addition in K\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"r = (T2/T1)**(1/(g-1))\t\t\t\t\t#Compression ratio\n",
"an = (1-(1/r)**(g-1))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"p2 = (p1*r**g)\t\t\t\t\t#Pressure at the end of compression in kg/cm**2\n",
"p3 = (p2*(T3/T2))\t\t\t\t\t#Pressure at the end of consmath.tant volume heat addition in kg/cm**2\n",
"p4 = p3/p2\t\t\t\t\t#Pressure at the end of adiabatic expansion in kg/cm**2\n",
"T4 = T3/r**(g-1)\t\t\t\t\t#Temperature at the end of adiabatic expansion in K\n",
"t4 = T4-273\t\t\t\t\t#Temperature at the end of adiabatic expansion in degree C\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The air standard efficiency is %3.1f percent \\\n",
"\\nTemperature at the end of adiabatic expansion is %i degree C \\\n",
"\\nPressure at the end of adiabatic expansion is %3.0f kg/cm**2'%(an,t4,p4)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The air standard efficiency is 46.8 percent \n",
"Temperature at the end of adiabatic expansion is 669 degree C \n",
"Pressure at the end of adiabatic expansion is 3 kg/cm**2\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.6 Page no : 34"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"T1 = 40.+273\t\t\t\t\t#Temperature at the beginning of compression in K\n",
"p1 = 1.\t\t\t\t\t#Pressure at the beginning of compression in kg/cm**2\n",
"p2 = 15.\t\t\t\t\t#Pressure at the end of adabatic compression in kg/cm**2\n",
"T3 = 2000.+273\t\t\t\t\t#Maximum temperature during the cycle in K\n",
"Cv = 0.17\t\t\t\t\t#Specific heat at consmath.tant volume in kJ/kg.K\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"T2 = T1*(p2/p1)**((g-1)/g)\t\t\t\t\t#Temperature at the end of adabatic compression in K\n",
"na = (1-(T1/T2))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"q = (Cv*(T3-T2))\t\t\t\t\t#Heat added in kcal/kg of air\n",
"W = ((na/100)*q)\t\t\t\t\t#Workdone per kg of air in kcal\n",
"W1 = (4.28*W)\t\t\t\t\t#Workdone per kg of air in kg.m\n",
"p3 = (p2*(T3/T2))\t\t\t\t\t#Pressure at the end of consmath.tant volume heat addition in kg/cm**2\n",
"p4 = (p3*p1)/p2\t\t\t\t\t#Pressure at the end of adiabatic expansion in kg/cm**2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) The heat supplied is %3.0f kcal/kg of air \\\n",
"\\nb) The workdone is %i kcal/kg of air \\\n",
"\\nc) The pressure at the end of adiabatic expansion is %3.2f kg/cm**2'%(q,W,p4)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) The heat supplied is 271 kcal/kg of air \n",
"b) The workdone is 146 kcal/kg of air \n",
"c) The pressure at the end of adiabatic expansion is 3.35 kg/cm**2\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.7 Page no : 38"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"r = 16.\t\t\t\t\t#Compression ratio\n",
"k = 5.\t\t\t\t\t#Cut off takes place at 5% of the stroke\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"c = (((k/100)*(r-1))+1)\t\t\t\t\t#Cut off ratio\n",
"na = (1-((1/r**(g-1))*((c**g-1)/(g*(c-1)))))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The air standard efficiency is %3.1f percent'%(na)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The air standard efficiency is 62.6 percent\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.8 Page no : 38"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"p1 = 1.05\t\t\t\t\t#Inlet pressure in kg/cm**2\n",
"T1 = 15.+273\t\t\t\t\t#Inlet temperature in K\n",
"p2 = 33.4\t\t\t\t\t#Pressure at the end of adiabatic compression in kg/cm**2\n",
"r = 5.\t\t\t\t\t#The ratio of expansion\n",
"Cp = 0.238\t\t\t\t\t#Specific heat at consmath.tant pressure in kJ/kg.K\n",
"Cv = 0.17\t\t\t\t\t#Specific heat at consmath.tant volume in kJ/kg.K\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"r1 = (p2/p1)**(1/g)\t\t\t\t\t#Compression ratio\n",
"k = r1/r\t\t\t\t\t#Cutoff ratio\n",
"T2 = (p2/p1)**((g-1)/g)*T1\t\t\t\t\t#Temperature at the end of adiabatic compression in K\n",
"T3 = T2*k\t\t\t\t\t#Temperature at the end of consmath.tant pressure heat addition in K\n",
"T4 = T3*(1/r)**(g-1)\t\t\t\t\t#Temperature at the end of adiabatic expansion in K\n",
"qa = (Cp*(T3-T2))\t\t\t\t\t#Heat added in kcal/kg of air\n",
"qre = (Cv*(T4-T1))\t\t\t\t\t#Heat rejected in kcal/kg of air\n",
"nt = ((qa-qre)/qa)*100\t\t\t\t\t#Ideal thermal efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The ideal thermal efficiency is %3.1f percent'%(nt)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The ideal thermal efficiency is 54.5 percent\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.9 Page no : 41"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\t\t\t\t\t\n",
"#Input data\n",
"p1 = 1.\t\t\t\t\t#Pressure at the end of suction stroke in kg/cm**2\n",
"T1 = 30.+273\t\t\t\t\t#Temperature at the end of suction stroke in kg/cm**2\n",
"T3 = 1500.+273\t\t\t\t\t#Maximum temperature during the cycle in K\n",
"r = 16.\t\t\t\t\t#Compression ratio\n",
"Cp = 0.24\t\t\t\t\t#Specific heat at consmath.tant pressure in kJ/kg.K\n",
"Cv = 0.17\t\t\t\t\t#Specific heat at consmath.tant volume in kJ/kg.K\n",
"g = 1.41\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"T2 = T1*r**(g-1)\t\t\t\t\t#Temperature at the end of adiabatic compression in K\n",
"s = (((T3/T2)-1)/(r-1))*100\t\t\t\t\t#Percentage of the stroke at which cut off occurs\n",
"r1 = (r/(T3/T2))\t\t\t\t\t#Expansion ratio\n",
"T4 = T3/(r1)**(g-1)\t\t\t\t\t#Temperature at the end of adiabatic expansion in K\n",
"qa = (Cp*(T3-T2))\t\t\t\t\t#Heat added in kcal/kg of air\n",
"qre = (Cv*(T4-T1))\t\t\t\t\t#Heat rejected in kcal/kg of air\n",
"nt = ((qa-qre)/qa)*100\t\t\t\t\t#Air standard efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) The percentage of stroke at which cut off takes place is %3.2f percent \\\n",
"\\nb) The temperature at the end of expansion stroke is %3.0f K \\\n",
"\\nc) The theoretical efficiency is %3.0f percent'%(s,T4,nt) \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) The percentage of stroke at which cut off takes place is 5.85 percent \n",
"b) The temperature at the end of expansion stroke is 737 K \n",
"c) The theoretical efficiency is 63 percent\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.10 Page no : 46"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"p1 = 1.\t\t\t\t\t#Pressure at the beginning of compression in kg/cm**2\n",
"T1 = 80.+273\t\t\t\t\t#Temperature at the beginning of compression in K\n",
"r = 14.\t\t\t\t\t#Compression ratio\n",
"p4 = 2.7\t\t\t\t\t#Pressure at the end of expansion in kg/cm**2\n",
"Cp = 0.24\t\t\t\t\t#Specific heat at consmath.tant pressure in kJ/kg.K\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"p2 = p1*r**g\t\t\t\t\t#Pressure at the end of compression in kg/cm**2\n",
"s = (((r*(p4/p2)**(1/g))-1)/(r-1))*100\t\t\t\t\t#Percentage of stroke when the fuel is cut off in percent\n",
"T2 = (T1*(p2/p1))/r\t\t\t\t\t#Temperature at the end of compression in K\n",
"T3 = (T2*r*(p4/p2)**(1/g))\t\t\t\t\t#Temperature at the end of adiabatic expansion in K\n",
"q = (Cp*(T3-T2))\t\t\t\t\t#Heat supplied in kcal/kg\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) The maximum pressure attained during the cycle is %3.1f kg/cm**2 \\\n",
"\\nb) The percentage of working stroke at which the heat supply to the working fluid ceases is %3.2f percent \\\n",
"\\nc) The heat received per kg of woring substance during the cycle is %3.0f kcal/kg'%(p2,s,q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) The maximum pressure attained during the cycle is 40.2 kg/cm**2 \n",
"b) The percentage of working stroke at which the heat supply to the working fluid ceases is 7.95 percent \n",
"c) The heat received per kg of woring substance during the cycle is 251 kcal/kg\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.11 Page no : 47"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"# variables\n",
"d = 0.25\t\t\t\t\t#Diameter of the cylinder in m\n",
"L = 0.35\t\t\t\t\t#Stroke in m\n",
"Cv = 1500.\t\t\t\t\t#Clearance volume in c.c\n",
"s = 5.\t\t\t\t\t#cut off ratio takes place at 5 percent of stroke\n",
"a = 1.4\t\t\t\t\t#Explosion ratio\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats for air\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = (3.14/4)*d**2*L\t\t\t\t\t#Stroke volume in m**3\n",
"r = (Vs*10**6+Cv)/Cv\t\t\t\t\t#Compression ratio\n",
"k = (Cv+((s/100)*Vs*10**6))/Cv\t\t\t\t\t#Cut off ratio\n",
"na = (1-((1/(r**(g-1)))*((a*k**g-1)/((a-1)+a*g*(k-1)))))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The air standard efficiency of the engine is %3.1f percent'%(na)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The air standard efficiency of the engine is 60.7 percent\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.12 Page no : 48"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 0.2\t\t\t\t\t#Diameter of the cylinder in m\n",
"L = 0.4\t\t\t\t\t#Stroke in m\n",
"r = 13.5\t\t\t\t\t#Compression ratio\n",
"a = 1.42\t\t\t\t\t#Explosion ratio\n",
"s = 5.1\t\t\t\t\t#Cut off occurs at 5.1 percent of the stroke\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats for air\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = (3.14/4)*d**2*L*10**-6\t\t\t\t\t#Stroke volume in c.c\n",
"Vc = Vs/r\t\t\t\t\t#Clearance volume in c.c\n",
"k = (((s/100)*Vs)+Vc)/Vc\t\t\t\t\t#Cut off ratio\n",
"ASE = (1-((1/(r**(g-1)))*((a*k**g-1)/((a-1)+a*g*(k-1)))))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The air standard efficiency of the engine is %3.1f percent'%(ASE)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The air standard efficiency of the engine is 61.4 percent\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.13 Page no : 53"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"x = [2./3,1./3]\t\t\t\t\t#The dual cycle atkes two-thirds of its total heat supply at consmath.tant volume and one-third at consmath.tant pressure\n",
"r = 13.\t\t\t\t\t#Compression ratio\n",
"p3 = 43.\t\t\t\t\t#Maximum pressure of the cycle in kg/cm**2\n",
"p1 = 1.\t\t\t\t\t#Pressure at intake in kg/cm**2\n",
"T1 = 15.+273\t\t\t\t\t#Intake temperature in K\n",
"Cp = 0.24\t\t\t\t\t#Specific heat at consmath.tant pressure in kJ/kg.K\n",
"Cv = 0.17\t\t\t\t\t#Specific heat at consmath.tant volume in kJ/kg.K\n",
"g = 1.41\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"T2 = T1*r**(g-1)\t\t\t\t\t#Temperature at the end of compression in K\n",
"p2 = (p1*r**g)\t\t\t\t\t#Pressure at the end of compression in kg/cm**2\n",
"T3 = T2*p3/p2\t\t\t\t\t#Temperature at the end of consmath.tant volume heat addition in K\n",
"q23 = Cv*(T3-T2)\t\t\t\t\t#Heat added at consmath.tant volume in kcal/kg\n",
"q34 = (1./2)*q23\t\t\t\t\t#Heat added at consmath.tant pressure in kcal/kg\n",
"T4 = (q34/Cp)+T3\t\t\t\t\t#Temperature at the end of consmath.tant pressure heat supply in K\n",
"T5 = (T4*((p1*(T4/T3))/r)**(g-1))\t\t\t\t\t#Temperature at the end of expansion in K\n",
"na = (1-((Cv*(T5-T1))/((Cv*(T3-T2))+(Cp*(T4-T3)))))*100\t\t\t\t\t#Efficiency in percent\n",
"T = [T1-273,T2-273,T3-273,T4-273,T5-273]\t\t\t\t\t#Temperature at the five cardinal points in degree C\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) The temperature at the five cardinal points of the cycle are : point 1 is %3.0f degree C \\\n",
"\\npoint 2 is %3.0f degree C \\\n",
"\\npoint 3 is %3.0f degree C \\\n",
"\\npoint 4 is %3.1f degree C \\\n",
"\\npoint 5 is %3.0f degree C \\\n",
"\\nb) The ideal thermal efficiency of the cycle is %3.1f percent'%(T[0],T[1],T[2],T[3],T[4],na)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) The temperature at the five cardinal points of the cycle are : point 1 is 15 degree C \n",
"point 2 is 551 degree C \n",
"point 3 is 680 degree C \n",
"point 4 is 725.0 degree C \n",
"point 5 is 82 degree C \n",
"b) The ideal thermal efficiency of the cycle is 65.0 percent\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.14 Page no : 56"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"p1 = 1.\t\t\t\t\t#Pressure at intake in kg/cm**2\n",
"T1 = 100.+273\t\t\t\t\t#Intake temperature in K\n",
"r = 10.\t\t\t\t\t#Compression ratio\n",
"p3 = 70.\t\t\t\t\t#Maximum pressure of the cycle in kg/cm**2\n",
"q = 400.\t\t\t\t\t#Amount of heat added in kcal/kg of air\n",
"Cp = 0.24\t\t\t\t\t#Specific heat at consmath.tant pressure in kJ/kg.K\n",
"Cv = 0.17\t\t\t\t\t#Specific heat at consmath.tant volume in kJ/kg.K\n",
"g = 1.41\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"T2 = (T1*r**(g-1))\t\t\t\t\t#Temperature at the end of compression in K\n",
"p2 = (p1*r**g)\t\t\t\t\t#Pressure at the end of compression in kg/cm**2\n",
"T3 = T2*(p3/p2)\t\t\t\t\t#Temperature at the end of consmath.tant volume heat addition in K\n",
"qv = (Cv*(T3-T2))\t\t\t\t\t#Heat added at consmath.tant volume in kcal/kg\n",
"qp = (q-qv)\t\t\t\t\t#Heat added at consmath.tant pressure in kcal/kg\n",
"T4 = (qp/Cp)+T3\t\t\t\t\t#Temperature at the end of consmath.tant pressure heat supply in K\n",
"k = (T4/T3)\t\t\t\t\t#Cut off ratio\n",
"T5 = T4/(r/k)**(g-1)\t\t\t\t\t#Temperature at the end of expansion in K\n",
"qv2 = Cv*(T5-T1)\t\t\t\t\t#Heat added at consmath.tant volume in kcal/kg\n",
"W = q-qv2\t\t\t\t\t#Workdone in kcal/kg of air\n",
"na = (W/q)*100\t\t\t\t\t#Air standard efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The temperature at the five cardinal points of the cycle are : \\\n",
"\\npoint 1 is %3.0f K \\\n",
"\\npoint 2 is %3.0f K \\\n",
"\\npoint 3 is %3.0f K \\\n",
"\\npoint 4 is %3.0f K \\\n",
"\\npoint 5 is %3.0f K \\\n",
"\\nThe air standard efficiency of the engine is %3.1f percent'%(T1,T2,T3,T4,T5,na)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The temperature at the five cardinal points of the cycle are : \n",
"point 1 is 373 K \n",
"point 2 is 959 K \n",
"point 3 is 2611 K \n",
"point 4 is 3107 K \n",
"point 5 is 1298 K \n",
"The air standard efficiency of the engine is 60.7 percent\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.15 Page no : 59"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"r = 12.\t\t\t\t\t#Compression ratio\n",
"p1 = 0.955\t\t\t\t\t#Pressure at the start of compression in kg/cm**2\n",
"T1 = 85.+273\t\t\t\t\t#Temperature at the start of compression in K\n",
"p3 = 55.\t\t\t\t\t#Maximum pressure of the cycle in kg/cm**2\n",
"x = (1./30)\t\t\t\t\t#Consmath.tant pressure heat reception contnues for 1/30 of the stroke\n",
"Cp = 0.238\t\t\t\t\t#Specific heat at consmath.tant pressure in kJ/kg.K\n",
"Cv = 0.17\t\t\t\t\t#Specific heat at consmath.tant volume in kJ/kg.K\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"T2 = T1*r**(g-1)\t\t\t\t\t#Temperature at the end of compression in K\n",
"p2 = (p1*r**g)\t\t\t\t\t#Pressure at the end of compression in kg/cm**2\n",
"T3 = T2*(p3/p2)\t\t\t\t\t#Temperature at the end of consmath.tant volume heat addition in K\n",
"T4 = (T3*((p1+x*(r-1))/p1))\t\t\t\t\t#Temperature at the end of consmath.tant pressure heat supply in K\n",
"T5 = T4*((p1+x*(r-1))/r)**(g-1)\t\t\t\t\t#Temperature at the end of expansion in K\n",
"qs = (Cv*(T3-T2))+(Cp*(T4-T3))\t\t\t\t\t#Heat supplied in kcal/kg of air\n",
"qre = (Cv*(T5-T1))\t\t\t\t\t#Heat rejected in kcal/kg of air\n",
"W = (qs-qre)\t\t\t\t\t#Workdone in kcal/kg of air\n",
"an = ((qs-qre)/qs)*100\t\t\t\t\t#Air standard efficiency in percent\n",
"\n",
"\t\t\t\t\t#Ouptut\n",
"print 'The wordone per kg of air is %3.2f kcal \\\n",
"\\nThe ideal thermal efficiency is %3.1f percent'%(W,an)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The wordone per kg of air is 178.24 kcal \n",
"The ideal thermal efficiency is 62.6 percent\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.16 Page no : 63"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"p1 = 1.\t\t\t\t\t#Pressure of air intake in kg/cm**2\n",
"T1 = 50.+273\t\t\t\t\t#Temperature of air intake in K\n",
"v = (1./14)\t\t\t\t\t#Volume compresses by it adiabatically of its original volume\n",
"r = (1/v)\t\t\t\t\t#Compression ratio\n",
"Cp = 0.237\t\t\t\t\t#Specific heat at consmath.tant pressure in kJ/kg.K\n",
"Cv = 0.169\t\t\t\t\t#Specific heat at consmath.tant volume in kJ/kg.K\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats for air\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"T2 = T1*r**(g-1)\t\t\t\t\t#Temperature at the end of compression in K\n",
"p2 = (p1*r**g)\t\t\t\t\t#Pressure at the end of compression in kg/cm**2\n",
"p3 = x*p2\t\t\t\t\t#Pressure at the end of the heat addition at consmath.tant volume in kg/cm**2\n",
"T3 = T2*(p3/p2)\t\t\t\t\t#Temperature at the end of consmath.tant volume heat addition in K\n",
"T4 = (T3*x)\t\t\t\t\t#Temperature at the end of consmath.tant pressure heat supply in K\n",
"T5 = T4/(r/x)**(g-1)\t\t\t\t\t#Temperature at the end of expansion in K\n",
"qs = (Cv*(T3-T2))+(Cp*(T4-T3))\t\t\t\t\t#Heat supplied in kcal/kg of air\n",
"qre = (Cv*(T5-T1))\t\t\t\t\t#Heat rejected in kcal/kg of air\n",
"na = ((qs-qre)/qs)*100\t\t\t\t\t#Air standard efficiency in percent\n",
"T = [T1-273,T2-273,T3-273,T4-273,T5-273]\t\t\t\t\t#Temperature at the five key points in degree C\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) The temperature at the five key points of the cycle are : \\\n",
"\\npoint 1 is %3.0f K = %3.0f degree C \\\n",
"\\npoint 2 is %3.0f K = %3.0f degree C \\\n",
"\\npoint 3 is %3.0f K = %3.0f degree C \\\n",
"\\npoint 4 is %3.0f K = %3.0f degree C \\\n",
"\\npoint 5 is %3.0f K = %3.0f degree C \\\n",
"\\nb) The ideal thermal efficiency of the cycle is %3.2f percent'%(T1,T[0],T2,T[1],T3,T[2],T4,T[3],T5,T[4],na)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) The temperature at the five key points of the cycle are : \n",
"point 1 is 323 K = 50 degree C \n",
"point 2 is 928 K = 655 degree C \n",
"point 3 is 31 K = -242 degree C \n",
"point 4 is 1 K = -272 degree C \n",
"point 5 is 0 K = -273 degree C \n",
"b) The ideal thermal efficiency of the cycle is 65.62 percent\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.18 Page no : 68"
]
},
{
"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#Six cylinder engine\n",
"r = 5.\t\t\t\t\t#Compression ratio\n",
"Vc = 110.\t\t\t\t\t#Clearance volume in c.c\n",
"a = 0.66\t\t\t\t\t#Efficiency ratio referred to the air standard cycle\n",
"N = 2400.\t\t\t\t\t#Speed in r.p.m\n",
"m = 9.9\t\t\t\t\t#Mass of petrol in kg\n",
"CV = 10600.\t\t\t\t\t#Calorific value of fuel in kcal/kg\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = (r*Vc-Vc)\t\t\t\t\t#Swept Volume in c.c\n",
"na = (1-(1/r)**(g-1))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"nt = (na/100)*a\t\t\t\t\t#Thermal efficiency \n",
"IHP = (nt*CV*m*427)/(4500*60)\t\t\t\t\t#Indicated Horse Power in h.p\n",
"pm = (((IHP/n)*4500*100*2)/(Vs*N))\t\t\t\t\t#Average indicated mean effective pressure in kg/cm**2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The average indicated mean effective pressure in each cylinder is %3.3f kg/cm**2'%(pm)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The average indicated mean effective pressure in each cylinder is 7.386 kg/cm**2\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.19 Page no : 68"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4.\t\t\t\t\t#Four cylinder engine\n",
"BHP = 40.\t\t\t\t\t#Brake horse power in h.p\n",
"N = 3000.\t\t\t\t\t#Speed in r.p.m\n",
"nm = 70.\t\t\t\t\t#Mechanical efficiency in percent\n",
"pm = 13.5\t\t\t\t\t#Indicated mean effective pressure in kg/cm**2\n",
"\t\t\t\t\t#Bore is equal to stroke\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"#case(i)\n",
"d1 = ((BHP*100*4500*n*2)/(n*(nm/100)*pm*N*3.14))**(1/3)\t\t\t\t\t#Cylinder bore or stroke length in cm\n",
"\n",
"#Case(ii)\n",
"d2 = ((BHP*100*4500*n)/(n*(nm/100)*pm*N*3.14))**(1/3)*10\t\t\t\t\t#Cylinder bore or stroke length in cm\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The cylinder sizes for a bore equal to stroke of a four cylinder in case of \\\n",
"\\ni)Four stroke engine is %3.1f cm \\\n",
"\\nii)Two stroke engine is %3.0f mm'%(d1,d2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The cylinder sizes for a bore equal to stroke of a four cylinder in case of \n",
"i)Four stroke engine is 1.0 cm \n",
"ii)Two stroke engine is 10 mm\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.20 Page no : 72"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"T = [50.+273,345.+273]\t\t\t\t\t#Temperatures at the beginning and end of compression in K\n",
"g = 1.4\t\t\t\t\t#ratio of specific heats \n",
"IHP = 25.\t\t\t\t\t#Indicated horse power in h.p\n",
"m = 5.44\t\t\t\t\t#Mass of fuel consumed per hour in kg\n",
"CV = 10300.\t\t\t\t\t#Calorific value in kcal/kg\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"na = (1-(T[0]/T[1]))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"r = (T[1]/T[0])**(1/(g-1))\t\t\t\t\t#Compression ratio\n",
"qIHP = (IHP*4500)/427\t\t\t\t\t#Heat equivalent of I.H.P in kcal/min\n",
"q = (m*CV)/60\t\t\t\t\t#Heat supplied per minute in kcal/min\n",
"Ith = (qIHP/q)*100\t\t\t\t\t#Indicated thermal efficiency in percent\n",
"nr = (Ith/na)*100\t\t\t\t\t#Efficiency ratio\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The air standard efficiency is %3.1f percent \\\n",
"\\nThe compression ratio is %3.2f \\\n",
"\\nIndicated thermal efficiency is %3.1f percent \\\n",
"\\nEfficiency ratio is %3.1f percent'%(na,r,Ith,nr)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The air standard efficiency is 47.7 percent \n",
"The compression ratio is 5.06 \n",
"Indicated thermal efficiency is 28.2 percent \n",
"Efficiency ratio is 59.1 percent\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.21 Page no : 74"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"CV = 10000.\t\t\t\t\t#Calorific value of petrol in kcal/kg\n",
"pe = [30.,70.]\t\t\t\t\t#Percentage of compression strokes in percent\n",
"p = [1.33,2.66]\t\t\t\t\t#Pressures in the cylinder corresponding to the compression strokes in kg/cm**2\n",
"n = 1.33\t\t\t\t\t#Polytropic consmath.tant\n",
"rn = 50.\t\t\t\t\t#Relative efficiency in percent\n",
"g = 1.4\t\t\t\t\t#ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"v = (p[1]/p[0])**(1/n)\t\t\t\t\t#Ratio of specific volumes\n",
"r = ((pe[1]/100)*v-(pe[0]/100))/((pe[1]/100)-((pe[0]/100)*v))\t\t\t\t\t#Compression ratio\n",
"na = (1-(1/r)**(g-1))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"ith = (rn*na)/100\t\t\t\t\t#Indicated thermal efficiency in percent\n",
"q = (4500*60)/(427*(ith/100))\t\t\t\t\t#Heat supplied in kcal/i.h.p.hr\n",
"Sc = (q/CV)\t\t\t\t\t#Specific consumption in kg/i.h.p.hr\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Compression ratio is %3.2f \\\n",
"\\nSpecific consumption is %3.3f kg/i.h.p.hr'%(r,Sc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Compression ratio is 4.51 \n",
"Specific consumption is 0.279 kg/i.h.p.hr\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.22 Page no : 76"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4.\t\t\t\t\t#Four cylinder four stroke\n",
"d = 7.5\t\t\t\t\t#Bore in cm\n",
"L = 8.75\t\t\t\t\t#Stroke in cm\n",
"r = 6.\t\t\t\t\t#Compression ratio\n",
"n1 = 55.\t\t\t\t\t#Efficiency in percent\n",
"g = 1.4\t\t\t\t\t#ratio of specific heats\n",
"N = 2400.\t\t\t\t\t#Speed in r.p.m\n",
"pm = 7.\t\t\t\t\t#Brake mean effective pressure in kg/cm**2\n",
"m = 9.\t\t\t\t\t#Mass of fuel per hour in kg\n",
"CV = 10500.\t\t\t\t\t#Calorific Value in kcal/kg\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"an = (1-(1/r)**(g-1))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"In = (an*n1)/100\t\t\t\t\t#Indicated thermal efficiency in percent. In textbook, answer is wrong\n",
"BHP = (pm*(3.14/4)*d**2*(L/100)*(N/2)*n)\t\t\t\t\t#Brake horse power in kg.m/min\n",
"Bth = ((BHP*60)/(427*CV*m))*100\t\t\t\t\t#Brake thermal efficiency in percent\n",
"nm = (Bth/In)*100\t\t\t\t\t#Mechanical efficiency in percent\n",
"Sc = ((4500*60)/(427*(Bth/100)*CV))\t\t\t\t\t#Specific consumption in g/i.h.p.hr\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Indicated thermal efficiency is %3.1f percent \\\n",
"\\nBrake thermal efficiency is %3.1f percent \\\n",
"\\nMechanical efficiency is %3.1f percent \\\n",
"\\nSpecific fuel consumption is %3.3f kg/i.h.p.hr'%(In,Bth,nm,Sc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Indicated thermal efficiency is 28.1 percent \n",
"Brake thermal efficiency is 19.3 percent \n",
"Mechanical efficiency is 68.6 percent \n",
"Specific fuel consumption is 0.312 kg/i.h.p.hr\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.26 Page no : 79"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"r = 7.\t\t\t\t\t#Compression ratio\n",
"v = 1.\t\t\t\t\t#Specific heat at consmath.tant volume increases by 1 percent\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"e = (1-(1/r**(g-1)))\t\t\t\t\t#Air standard efficiency\n",
"dee = -(((1-e)*(g-1)*math.log(r)*(v/100))/e)*100\t\t\t\t\t#Change in efficiency to the original efficiency\n",
"x = -(dee)\t\t\t\t\t#For Output purpose\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Percentage change is efficiency is %3.2f percent i.e. a decrease of %3.2f percent'%(dee,x)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage change is efficiency is -0.66 percent i.e. a decrease of 0.66 percent\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.30 Page no : 84"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"N = 210.\t\t\t\t\t#Speed in r.p.m\n",
"d = 0.3\t\t\t\t\t#Diameter of the piston in m\n",
"L = 0.4\t\t\t\t\t#Stroke in m\n",
"v = 2.5\t\t\t\t\t#Clearance volume is 2.5% of the swept volume. But in textbook it is given wrong as 25%\n",
"CO = 19.7\t\t\t\t\t#Percentage of CO gas\n",
"H2 = 28.8\t\t\t\t\t#Percentage of H2 gas\n",
"CO2 = 14.4\t\t\t\t\t#Percentage of CO2 gas\n",
"N2 = 37.1\t\t\t\t\t#Percentage of N2 gas\n",
"x = 0.875\t\t\t\t\t#Total mixture at N.T.P admitted per suction stroke is 0.875 of the total volume behind the piston at the end of the stroke\n",
"tn = 35.\t\t\t\t\t#Thermal efficiency in percent\n",
"CVH2 = 13200.\t\t\t\t\t#Calorific value of H2 per kg in kcal\n",
"CVC = 2540.\t\t\t\t\t#Calorific value of carbon burning from CO to CO2 in kcal/kg\n",
"de = 1.293\t\t\t\t\t#Density of air in kg/m**3\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mH2 = 2.\t\t\t\t\t#Molecular weight of H2\n",
"mCO = 28.\t\t\t\t\t#Molecular weight of CO\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"a = ((100./21)*((CO2/100)+((CO/2)/100)))\t\t\t\t\t#Air per cu.m of gas in cu.m\n",
"Vm = (a+1)\t\t\t\t\t#Volume of mixture per cu.m of gas in cu.m\n",
"Vs = ((3.14/4)*d**2*L)\t\t\t\t\t#Swept volume in cu.m\n",
"Vc = (Vs*v)/100\t\t\t\t\t#Clearance volume in cu.m\n",
"V = Vc+Vs\t\t\t\t\t#Total volume in cu.m\n",
"VC = V*x\t\t\t\t\t#Volume of charge admitted per stroke in cu.m\n",
"VM = VC*(N/2)\t\t\t\t\t#Charge volume per minute in cu.m\n",
"VG = (VM/Vm)\t\t\t\t\t#cu.m of gas per minute\n",
"vH2 = (VG*(H2/100))\t\t\t\t\t#Volume of H2 per minute in cu.m\n",
"vCO = (VG*(CO/100))\t\t\t\t\t#Volume of CO per minute in cu.m\n",
"CVH2cum = (mH2*CVH2)/(vH2*1000)\t\t\t\t\t#Calorific value of H2 per cu.m in kcal\n",
"CVCO = (CVC*(2*mC)/(2*mCO))\t\t\t\t\t#Calorific value of CO per kg in kcal\n",
"CVCOcum = (mCO*CVCO)/(vH2*1000)\t\t\t\t\t#Calorific value of CO per cu.m in kcal\n",
"qH2 = (16.09*CVH2cum)\t\t\t\t\t#Heat in charge due to H2 in kcal\n",
"qCO = (11*CVCOcum)\t\t\t\t\t#Heat in charge due to CO in kcal\n",
"qt = (qH2+qCO)\t\t\t\t\t#Heat supplied per minute in kcal\n",
"qu = (qt*(tn/100))\t\t\t\t\t#Heat utilised in kcal\n",
"hp = (qu*427)/4500\t\t\t\t\t#H.P developed\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Maximum horse power that can be developed is %3.1f H.P'%(hp)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum horse power that can be developed is 71.0 H.P\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.31 Page no : 84"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"vCH4 = 65.\t\t\t\t\t#Composition by volume of CH4 \n",
"vH2 = 2.\t\t\t\t\t#Composition by volume of H2\n",
"vN2 = 2.\t\t\t\t\t#Composition by volume of N2\n",
"vCO2 = 31.\t\t\t\t\t#Composition by volume of CO2\n",
"O2 = 5.3\t\t\t\t\t#Composition of O2 in dry exhaust gases when analysed in orsat apparatus\n",
"N2 = 83.\t\t\t\t\t#Composition of N2 in dry exhaust gases when analysed in orsat apparatus\n",
"CO = 0.3\t\t\t\t\t#Composition of CO in dry exhaust gases when analysed in orsat apparatus\n",
"CO2 = 11.4\t\t\t\t\t#Composition of CO2 in dry exhaust gases when analysed in orsat apparatus\n",
"an = 79.\t\t\t\t\t#Air contains 79% by volume of nitrogen \n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"a = (100./(100-an))*(((vCH4/100)*2)+((vN2/100)*(1./2)))\t\t\t\t\t#Total air required for complete combustion of 1 cu.m of gas in cu.m\n",
"xCO = (CO/2)\t\t\t\t\t#O2 required to burn the CO in cu.m\n",
"xCO2 = CO\t\t\t\t\t#CO2 formed in cu.m \n",
"tO2 = O2-xCO\t\t\t\t\t#Total O2 in cu.m\n",
"tN2 = N2\t\t\t\t\t#Total N2 in cu.m\n",
"tCO2 = CO2+xCO2\t\t\t\t\t#Total CO2 in cu.m\n",
"T = tO2+tN2+tCO2\t\t\t\t\t#Total mixture in cu.m\n",
"pCO2 = (tCO2*100)/T\t\t\t\t\t#Percentage of CO2 in percent\n",
"mm = (a*100)\t\t\t\t\t#Minimum air supply required for complete combustion of 100 cu.m of the gas in cu.m\n",
"an2 = (an/100)*mm\t\t\t\t\t#N2 for this air in cu.m\n",
"tn2 = (an2+vN2)\t\t\t\t\t#Total N2 in cu.m\n",
"v = (((vCH4+vCO2)*100)/pCO2)-(vCH4+vCO2+tn2)\t\t\t\t\t#Increase in air supply for reduction in percentage of CO2 in cu.m\n",
"pea = (v*100)/mm\t\t\t\t\t#Percentage of excess air. In textbook it is given wrong as 26.7 percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) the air-fuel ratio by volume to give complete combustion is %3.3f \\\n",
"\\nb) the percentage of excess air actually used in the test is %3.1f percent'%(a,pea)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) the air-fuel ratio by volume to give complete combustion is 6.238 \n",
"b) the percentage of excess air actually used in the test is 36.6 percent\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.32 Page no : 86"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"Vs = 9.45\t\t\t\t\t#Swept volume in litres\n",
"Vc = 2.32\t\t\t\t\t#Clearance volume in litres\n",
"m = 4.25\t\t\t\t\t#Consumption of gas per hour in cu.m\n",
"N = 165.\t\t\t\t\t#Speed in r.p.m\n",
"bhp = 5.62\t\t\t\t\t#Brake horse power in h.p\n",
"nm = 73.4\t\t\t\t\t#Mechanical efficiency in percent\n",
"CV = 3500.\t\t\t\t\t#Calorific value in kcal per cubic meter\n",
"vn = 0.87\t\t\t\t\t#Volumetric efficiency\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"tV = (Vs+Vc)*1000\t\t\t\t\t#Total volume in c.c\n",
"rc = (tV/Vc)\t\t\t\t\t#Compression ratio\n",
"na = (1-(1/rc**(g-1)))*100\t\t\t\t\t#Air math.radians(numpy.arcmath.tan(ard efficiency in percent\n",
"W = (bhp*4500)/427\t\t\t\t\t#Workdone per minute in kcal\n",
"Iw = (W/(nm/100))\t\t\t\t\t#Indicated work in kcal/min\n",
"q = (m/60)*CV\t\t\t\t\t#Heat supplied in kcal/min\n",
"ith = (Iw/q)*100\t\t\t\t\t#Indicated thermal efficiency in percent\n",
"rn = (ith/na)*100\t\t\t\t\t#Relative efficiency in percent\n",
"Vm = (Vs*1000)*vn\t\t\t\t\t#Volume of mixture taken in per stroke in c.c\n",
"Vg = (m*2*10**6)/(60*N)\t\t\t\t\t#Volume of gas taken in per stroke in c.c\n",
"Va = (Vm-Vg)\t\t\t\t\t#Volume of air taken in per stroke in c.c\n",
"agr = (Va/Vg)\t\t\t\t\t#Air gas ratio\n",
"CVc = (CV/(agr+1))\t\t\t\t\t#Calorific value of charge in kcal\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Ratio of air to gas used is %3.2f \\\n",
"\\nCalorific value of 1 cu.m of the mixture in the cylinder is %3.1f kcal'%(agr,CVc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Ratio of air to gas used is 8.58 \n",
"Calorific value of 1 cu.m of the mixture in the cylinder is 365.5 kcal\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 3.33 Page no : 87"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 18.\t\t\t\t\t#Bore in cm\n",
"l = 37.5\t\t\t\t\t#Stroke in cm\n",
"N = 220.\t\t\t\t\t#Speed in r.p.m\n",
"#Mean effective pressure in kg/cm**2\n",
"#Firing\n",
"pp = 5.9\t\t\t\t\t#Positive loop\n",
"pn = 0.248\t\t\t\t\t#Negative loop\n",
"\t\t\t\t\t#Mismath.sing\n",
"nn = 0.432\t\t\t\t\t#Negative loop\n",
"bhp = 8.62\t\t\t\t\t#Brake horse power in h.p\n",
"ex = 100.\t\t\t\t\t#Explosions per minute\n",
"vg = 0.101\t\t\t\t\t#Gas used in cu.m per minute\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"tc = (N/2)\t\t\t\t\t#The number of cycles \n",
"nw = ex\t\t\t\t\t#Number of working cycles \n",
"nm = (tc-nw)\t\t\t\t\t#Number of mismath.sing cycles\n",
"ihp = ((l/100)*(3.14/4)*(d**2/4500))*((pp-pn)*(100-nn))\t\t\t\t\t#Net I.H.P in h.p\n",
"fhp = (ihp-bhp)\t\t\t\t\t#Friction horse power in h.p\n",
"W = ((pp-pn)*(3.14/4)*(d**2*(l/100)))\t\t\t\t\t#Workdone per firing done in kg.m\n",
"Wp = (nn*(3.14/4)*d**2*(l/100))\t\t\t\t\t#Workdone per pumping stroke in kg.m\n",
"n = ((fhp*4500)+(Wp*tc))/(W+Wp)\t\t\t\t\t#Number of strokes\n",
"gf = (vg/nw)\t\t\t\t\t#Gas per firing stroke in cu.m\n",
"gl = (n*gf)\t\t\t\t\t#Gas per minute at no load in cu.m\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Friction horse power of the engine is %3.2f Gas consumption at no load is %3.3f cu.m/min'%(fhp,gl)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Friction horse power of the engine is 3.31 Gas consumption at no load is 0.034 cu.m/min\n"
]
}
],
"prompt_number": 26
}
],
"metadata": {}
}
]
}PKII2al
l
%Internal Combustion Engines/ch4.ipynb{
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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 4 : Fuel Air Cycles And Real Cycles"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.1 Page no : 100"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 20.\t\t\t\t\t#Bore in cm\n",
"l = 38.\t\t\t\t\t#Stroke in cm\n",
"Vc = 900.\t\t\t\t\t#Clearance volume in c.c\n",
"p1 = 1.\t\t\t\t\t#Pressure at the start of compression stroke in kg/cm**2\n",
"T1 = 90.+273\t\t\t\t\t#Temperature at the start of compression stroke in K\n",
"x = 0.75\t\t\t\t\t#Piston travelled 0.75 of the compression stroke\n",
"n = 1.32\t\t\t\t\t#Compression curve index\n",
"wa = 0.0125\t\t\t\t\t#Weight of air in kg\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = (3.14/4)*d**2*l\t\t\t\t\t#Swept volume in c.c\n",
"V1 = (Vs+Vc)\t\t\t\t\t#Volume in c.c\n",
"V2 = (1-x)*Vs+Vc\t\t\t\t\t#Volume in c.c\n",
"p2 = p1*(V1/V2)**n\t\t\t\t\t#Pressure in kg/cm**2\n",
"T2 = (T1*(p2/p1)*(V2/V1))\t\t\t\t\t#Temperature in K\n",
"W = ((p1*V1-p2*V2)/(n-1))*10**-2\t\t\t\t\t#Workdone in kg.m\n",
"dI = wa*0.17*(T2-T1)\t\t\t\t\t#Change in internal energy in kcal\n",
"q = (dI+(W/427))\t\t\t\t\t#Heat in kcal\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'When the cylinder has travelled %3.2f of the compression stroke, \\\n",
"\\nThe volume is %3.0f c.c \\\n",
"\\nThe pressure is %3.2f kg/cm**2 \\\n",
"\\nTemperature is %3.0f K \\\n",
"\\nThe workdone on the gas is %3.2f kg.m \\\n",
"\\nChange in internal energy between the two points is %3.3f kcal'%(x,V2,p2,T2,W,dI)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"When the cylinder has travelled 0.75 of the compression stroke, \n",
"The volume is 3883 c.c \n",
"The pressure is 4.84 kg/cm**2 \n",
"Temperature is 532 K \n",
"The workdone on the gas is -186.85 kg.m \n",
"Change in internal energy between the two points is 0.359 kcal\n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}PKIک6AA%Internal Combustion Engines/ch5.ipynb{
"metadata": {
"name": "",
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"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": {}
}
]
}PKIԲ%Internal Combustion Engines/ch6.ipynb{
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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 6 : Combustion Charts"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.1 Page no : 123"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"r = 7.5\t\t\t\t\t#Compression ratio\n",
"\t\t\t\t\t#Data from combustion chart\n",
"p = [1,15.1,26.95,1.95]\t\t\t\t\t#Pressure of air fuel mixture in kg/cm**2\n",
"T = [60,460,1150,435]\t\t\t\t\t#Temperature of air fuel mixture in K\n",
"V = [16.98,2.264,2.264,16.98]\t\t\t\t\t#Volume in m**3/kg\n",
"U = [17,78.8,212,80]\t\t\t\t\t#Internal energy in kcal/kg\n",
"S = [0.07,0.07,0.22,0.22]\t\t\t\t\t#Entropy in kcal/kg.degree C\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"n = (((U[2]-U[3])-(U[1]-U[0]))/(U[2]-U[1]))*100\t\t\t\t\t#Thermal efficiency in percent\n",
"na = (1-(1/r)**(g-1))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Thermal efficiency is %3.1f percent \\\n",
"\\nAir standard efficiency is %3.1f percent'%(n,na)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thermal efficiency is 52.7 percent \n",
"Air standard efficiency is 55.3 percent\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.2 Page no : 127"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"\t\t\t\t\t#Data from combustion chart\n",
"p = [1,33,33,1]\t\t\t\t\t#Pressure of air fuel mixture in kg/cm**2\n",
"T = [65,600,1450,725]\t\t\t\t\t#Temperature of air fuel mixture in K\n",
"V = [16,1.23,3.45,16]\t\t\t\t\t#Volume in m**3/kg\n",
"U = [11.8,110,295,140]\t\t\t\t\t#Internal energy in kcal/kg\n",
"H = [22.7,150,395,225]\t\t\t\t\t#Enthalpy in kcal/kg\n",
"S = [0.068,0.068,0.264,0.264]\t\t\t\t\t#Entropy in kcal/kg.degree C\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"r = (V[0]/V[1])\t\t\t\t\t#Compression ratio\n",
"q = (H[2]-H[1])\t\t\t\t\t#Heat supplied in kcal/kg\n",
"qre = (U[3]-U[0])\t\t\t\t\t#Heat rejected in kcal/kg\n",
"nt = ((q-qre)/q)*100\t\t\t\t\t#Thermal efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a)Compression ratio is %3.0f \\\n",
"\\nb) Heat supplied to the cycle is %3.0f kcal/kg \\\n",
"\\nc) Heat rejected by the cycle is %3.2f kcal/kg \\\n",
"\\nd) Thermal efficiency is %3.2f percent'%(r,q,qre,nt)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a)Compression ratio is 13 \n",
"b) Heat supplied to the cycle is 245 kcal/kg \n",
"c) Heat rejected by the cycle is 128.20 kcal/kg \n",
"d) Thermal efficiency is 47.67 percent\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 6.3 Page no : 130"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"\t\t\t\t\t#Data from combustion chart\n",
"p = [1,51.5,77.25,77.25,3.75]\t\t\t\t\t#Pressure of air fuel mixture in kg/cm**2\n",
"T = [16,1,1,1.5,16]\t\t\t\t\t#Temperature of air fuel mixture in K\n",
"V = [65,745,1400,2200,1030]\t\t\t\t\t#Volume in m**3/kg\n",
"U = [14.7,135,275,475,197]\t\t\t\t\t#Internal energy in kcal/kg\n",
"H = [21.9,85,372,625,280]\t\t\t\t\t#Enthalpy in kcal/kg\n",
"S = [0.068,0.068,0.19,0.32,0.32]\t\t\t\t\t#Entropy in kcal/kg.degree C\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"nth = (((U[2]-U[1])+(H[3]-H[2])-(U[4]-U[0]))/((U[2]-U[1])+(H[3]-H[2])))*100\t\t\t\t\t#Thermal efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Thermal efficiency is %3.2f percent'%(nth)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thermal efficiency is 53.61 percent\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}PKI9t. %Internal Combustion Engines/ch8.ipynb{
"metadata": {
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{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chatper 8 : Combustion"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.1 Page no : 143"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"C = 88.6\t\t\t\t\t#Composition of C in percent\n",
"H2 = 11.4\t\t\t\t\t#Composition of H2 in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"w1 = (C/100)\t\t\t\t\t#Weight per kg of fuel of C in kg\n",
"w2 = (H2/100)\t\t\t\t\t#Weight per kg of fuel of H2 in kg\n",
"O1 = (8./3)\t\t\t\t\t#Oxygen required per kg of constituent for C in kg\n",
"O2 = 8.\t\t\t\t\t#Oxygen required per kg of constituent for H2 in kg\n",
"O11 = (w1*O1)\t\t\t\t\t#Oxygen required per kg of fuel for C in kg\n",
"O22 = (w2*O2)\t\t\t\t\t#Oxygen required per kg of fuel for H2 in kg\n",
"T = (O11+O22)\t\t\t\t\t#Total Oxygen required per kg of fuel in kg\n",
"P1 = (w1+O11)\t\t\t\t\t#Composition of CO2 in kg\n",
"P2 = (w2+O22)\t\t\t\t\t#Composition of H2O in kg\n",
"w = (T*(100./23))\t\t\t\t\t#Weight of air required in kg per kg of fuel\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The weight of air required for complete combustion of liquid fuel is %3.2f kg per kg of fuel \\\n",
"\\nThe composition of CO2 is %3.3f kg \\\n",
"\\nThe composition of H2O is %3.3f kg'%(w,P1,P2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The weight of air required for complete combustion of liquid fuel is 14.24 kg per kg of fuel \n",
"The composition of CO2 is 3.249 kg \n",
"The composition of H2O is 1.026 kg\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.2 Page no : 147"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"#Input data\n",
"C = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"H2 = 2.\t\t\t\t\t#Molecular weght of H2\n",
"O2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"\n",
"\n",
"#Calculations\n",
"C7H16 = (7*C+8*H2)\t\t\t\t\t#Molecular weight of C7H16\n",
"O2x = (11*O2)\t\t\t\t\t#Molecular weight of 22O2\n",
"wt = (O2x/C7H16)*(100/23.2)\t\t\t\t\t#Weight of air in kg per kg of fuel\n",
"#Now in actual experiment, we have\n",
"#1[C7H16] +x[O2] +...[N2] = a[CO2] +8[H2O] +a[O2] +...[N2]\n",
"#This is the new equation written in volumes. The volumes of CO2 and O2 being equal, with no CO present, and the usual assumption that all the hydrogen is burnt to H2O\n",
"#Now, if all the carbon is burnt, we must have 7 mols of CO2\n",
"x = (7+4+7)\t\t\t\t\t#Total number of mols from CO2,H2O and O2 terms respectively\n",
"W = ((x*O2)/100)*(100/23.2)\t\t\t\t\t#Weight of air in kg per kg of fuel\n",
"\n",
"\n",
"#Output\n",
"print 'The weight of air is %3.1f kg per kg of fuel which would just suffice for theoretically complete combustion \\\n",
"\\nThe ratio of air to fuel by weight as actually supplied is %3.1f kg of air per kg of fuel'%(wt,W)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The weight of air is 15.2 kg per kg of fuel which would just suffice for theoretically complete combustion \n",
"The ratio of air to fuel by weight as actually supplied is 24.8 kg of air per kg of fuel\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.3 Page no : 151"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"H = 15.\t\t\t\t\t#Percentage of Hydrogen by volume\n",
"CO = 25.\t\t\t\t\t#Percentage of carbon monoxide by volume\n",
"CH4 = 4.\t\t\t\t\t#Percentage of methane by volume\n",
"CO2 = 4.\t\t\t\t\t#Percentage of carbon dioxide by volume\n",
"O2 = 2.\t\t\t\t\t#Percentage of oxygen by volume. In textbook it is given wrong as 25\n",
"N2 = 50.\t\t\t\t\t#Percentage of nitrogen by volume\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"O21 = (H/100)*(1./2)\t\t\t\t\t#Volume of oxygen required in m**3 by 15 percent of H2\n",
"O22 = (CO/100)*(1./2)\t\t\t\t\t#Volume of oxygen required in m**3 by 25 percent of CO\n",
"CO21 = (CO/100)*1\t\t\t\t\t#Volume of CO2 produced in m**3 by 25 percent of CO\n",
"O23 = (CH4/100)*2\t\t\t\t\t#Volume of oxygen required in m**3 by 4 percent of CH4\n",
"CO22 = (CH4/100)*1\t\t\t\t\t#Volume of CO2 produced in m**3 by 4 percent of CH4\n",
"H201 = (CO/100)*2\t\t\t\t\t#Volume of H2O produced in m**3 by 4 percent of CH4\n",
"TO2 = (O21+O22+O23-(O2/100))\t\t\t\t\t#Total vol. of oxygen in m**3\n",
"wa = (TO2*(100./21))\t\t\t\t\t#Theoretical volume of air required in m**3\n",
"vN2 = (wa*(79./100))\t\t\t\t\t#Volume of N2 present in air in m**3\n",
"TvN2 = (vN2+(N2/100))\t\t\t\t\t#Total volume of N2 after combustion of 1 m**3 of fuel in m**3\n",
"xCO2 = (CO21+CO22)\t\t\t\t\t#CO2 produced due to combustion of fuel in m**3\n",
"TCO2 = (xCO2+(CO2/100))\t\t\t\t\t#Total volume of CO2 in the flue gas in m**3\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The air required for complete combustion of one m**3 of the fuel is %3.3f cu.m \\\n",
"\\nThe dry flue gas contains %3.3f cu.m volume of N2 and %3.2f cu.m volume of CO2'%(wa,TvN2,TCO2) \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The air required for complete combustion of one m**3 of the fuel is 1.238 cu.m \n",
"The dry flue gas contains 1.478 cu.m volume of N2 and 0.33 cu.m volume of CO2\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.4 Page no : 155"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"C = 88.1\t\t\t\t\t#Composition of C in percent\n",
"H2 = 10.7\t\t\t\t\t#Composition of H2 in percent\n",
"O2 = 1.2\t\t\t\t\t#Composition of O2 in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"w1 = (C/100)\t\t\t\t\t#Weight per kg of fuel of C in kg\n",
"w2 = (H2/100)\t\t\t\t\t#Weight per kg of fuel of H2 in kg\n",
"w3 = (O2/100)\t\t\t\t\t#Weight per kg of fuel of O2 in kg\n",
"O1 = (8/3)\t\t\t\t\t#Oxygen required per kg of constituent for C in kg\n",
"O2 = 8\t\t\t\t\t#Oxygen required per kg of constituent for H2 in kg\n",
"O11 = (w1*O1)\t\t\t\t\t#Oxygen required per kg of fuel for C in kg\n",
"O22 = (w2*O2)\t\t\t\t\t#Oxygen required per kg of fuel for H2 in kg\n",
"T = (O11+O22-w3)\t\t\t\t\t#Total Oxygen required per kg of fuel in kg\n",
"P1 = (w1+O11)\t\t\t\t\t#Composition of CO2 in kg\n",
"P2 = (w2+O22)\t\t\t\t\t#Composition of H2O in kg\n",
"w = (T*(100/23))\t\t\t\t\t#Weight of air required in kg per kg of fuel\n",
"wN2 = (w*(77/100))\t\t\t\t\t#Weight of N2 in 'w' kg of fuel in kg\n",
"T1 = (P1+P2+wN2)\t\t\t\t\t#Total weight of all products of combustion in kg\n",
"pCO2 = (P1/T1)*100\t\t\t\t\t#Percentage composition of CO2 by weight\n",
"pH2O = (P2/T1)*100\t\t\t\t\t#Percentage composition of H2O by weight\n",
"pN2 = (wN2/T1)*100\t\t\t\t\t#Percentage composition of N2 by weight\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The weight of air required to burn one kg of the fuel is %3.1f kg \\\n",
"\\nThe composition of products of combustion by weight is %3.2f percent of CO2, %3.2f percent of H2O and %3.2f percent of N2'%(w,pCO2,pH2O,pN2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The weight of air required to burn one kg of the fuel is 10.4 kg \n",
"The composition of products of combustion by weight is 73.29 percent of CO2, 26.71 percent of H2O and 0.00 percent of N2\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.5 Page no : 156"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"C = 85.\t\t\t\t\t#Composition of C in percent\n",
"H2 = 15.\t\t\t\t\t#Composition of H2 in percent\n",
"CV = 10600.\t\t\t\t\t#Calorific value in kcal/kg\n",
"eO2 = 60.\t\t\t\t\t#Percentage of air in excess\n",
"bhp = 240.\t\t\t\t\t#Brake horse power in h.p\n",
"nth = 30.\t\t\t\t\t#Thermal efficiency in percent\n",
"O2 = 23.\t\t\t\t\t#Percentage of oxygen contained in air by weight\n",
"wC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"wH2 = 2.\t\t\t\t\t#Molecular weght of H2\n",
"wO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"mma = (100./23)*(((C/100)*(wO2/wC))+((H2/100)*(wO2/(wH2*2))))\t\t\t\t\t#Minimum air in kg per kg oil\n",
"aa = ((100+eO2)/100)*mma\t\t\t\t\t#Actual air supplied in kg per kg oil\n",
"q = ((bhp*(4500./427))/(nth/100))\t\t\t\t\t#Heat supplied in kcal/min\n",
"mf = (q/CV)\t\t\t\t\t\t#Mass of fuel supplied in kg/min\n",
"ma = (aa*mf)\t\t\t\t\t#Mass of air supplied in kg/min\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The weight of air is %3.2f kg/min'%(ma)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The weight of air is 19.18 kg/min\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.6 Page no : 157"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"#C + O2 -> CO2\n",
"# 12 + 32 -> 44\n",
"C = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"O2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"CO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"N2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"wair = (O2/C)*(100./23)\t\t\t\t\t#Air required per kg of C in kg\n",
"wN2 = (O2/C)*(77./23)\t\t\t\t\t#N2 associated with the air in kg\n",
"pCO2 = (CO2/C)/CO2\t\t\t\t\t#Parts by volume/k for CO2\n",
"pN2 = (wN2/N2)\t\t\t\t\t#Parts by volume/k for N2\n",
"Tv = (pCO2+pN2)\t\t\t\t\t#Total parts by volume\n",
"ppCO2 = (pCO2/Tv)*100\t\t\t\t\t#Percentage volume of CO2\n",
"ppN2 = (pN2/Tv)*100\t\t\t\t\t#Percentage volume of N2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The volumetric analysis of the flue gas when pure carbon is burnt with \\\n",
" a minimum quantity of air is given by CO2 -> %3.1f percent N2 -> %3.1f percent'%(ppCO2,ppN2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The volumetric analysis of the flue gas when pure carbon is burnt with a minimum quantity of air is given by CO2 -> 20.7 percent N2 -> 79.3 percent\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.7 Page no : 157"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"C = 90.\t\t\t\t\t#Percentage composition of C \n",
"H2 = 3.3\t\t\t\t\t#Percentage composition of H2\n",
"O2 = 3\t\t\t\t\t#Percentage composition of O2\n",
"N2 = 0.8\t\t\t\t\t#Percentage composition of N2\n",
"S = 0.9\t\t\t\t\t#Percentage composition of S\n",
"Ash = 2.\t\t\t\t\t#Percentage composition of Ash\n",
"eO2 = 50.\t\t\t\t\t#Percentage of excess air\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mS = 32.\t\t\t\t\t#Molecular weight of sulphur\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mSO2 = 64.\t\t\t\t\t#Molecular weight of SO2\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"w1 = (C/100)\t\t\t\t\t#Weight per kg of fuel of C in kg\n",
"w2 = (H2/100)\t\t\t\t\t#Weight per kg of fuel of H2 in kg\n",
"w3 = (S/100)\t\t\t\t\t#Weight per kg of fuel of S in kg\n",
"O1 = (8./3)\t\t\t\t\t#Oxygen required per kg of constituent for C in kg\n",
"O2 = 8.\t\t\t\t\t#Oxygen required per kg of constituent for H2 in kg\n",
"O3 = 1.\t\t\t\t\t#Oxygen requred per kg of constituent for S in kg\n",
"O11 = (w1*O1)\t\t\t\t\t#Oxygen required per kg of fuel for C in kg\n",
"O22 = (w2*O2)\t\t\t\t\t#Oxygen required per kg of fuel for H2 in kg\n",
"O33 = (w3*O3)\t\t\t\t\t#Oxygen required per kg of fuel for S in kg\n",
"T = (O11+O22+O33-(O2/100))\t\t\t\t\t#Total Oxygen required per kg of fuel in kg\n",
"ma = (T*(100./23))\t\t\t\t\t#Minimum air required in kg\n",
"aN2 = (ma*((100+eO2)/100)*(77/100))\t\t\t\t\t#N2 in actual air supply in kg\n",
"TN2 = (aN2+(N2/100))\t\t\t\t\t#Total N2 in kg\n",
"wt = (ma*(eO2/100)*(23./100))\t\t\t\t\t#Weight of air due to excess O2 in kg\n",
"TSO2 = (w3*(mSO2/mS))\t\t\t\t\t#Total SO2 in kg\n",
"TCO2 = (w1*(mCO2/mC))\t\t\t\t\t#Total CO2 in kg\n",
"pCO2 = (TCO2/mCO2)\t\t\t\t\t#Parts by volume of CO2\n",
"pSO2 = (TSO2/mSO2)\t\t\t\t\t#Parts by volume of SO2\n",
"pO2 = (wt/mO2)\t\t\t\t\t#Parts by volume of O2\n",
"pN2 = (TN2/mN2)\t\t\t\t\t#Parts by volume of N2\n",
"Tv = (pCO2+pSO2+pN2+pO2)\t\t\t\t\t#Total parts by volume\n",
"ppCO2 = (pCO2/Tv)*100\t\t\t\t\t#Percentage volume of CO2\n",
"ppSO2 = (pSO2/Tv)*100\t\t\t\t\t#Percenatge volume of SO2\n",
"ppO2 = (pO2/Tv)*100\t\t\t\t\t#Percentage volume of O2\n",
"ppN2 = (pN2/Tv)*100\t\t\t\t\t#Percentage volume of N2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Percentage combustion of the dry flue gases by volume is CO2 %3.2f percent SO2 %3.2f percent O2 %3.1f percent N2 %3.2f percent'%(ppCO2,ppSO2,ppO2,ppN2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage combustion of the dry flue gases by volume is CO2 64.61 percent SO2 0.24 percent O2 34.9 percent N2 0.25 percent\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.8 Page no : 158"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"C = 85.\t\t\t\t\t#Composition of C in percent\n",
"H2 = 12.3\t\t\t\t\t#Composition of H2 in percent\n",
"i = 2.7\t\t\t\t\t#Incombustible residue composition in percent\n",
"ma = 25.\t\t\t\t\t#Mass of air supplied in kg of air per kg of fuel\n",
"pO2 = 23.\t\t\t\t\t#Percentage of oxygen in gemetric analysis of air\n",
"pN2 = 77.\t\t\t\t\t#Percentage of nitrogen in gemetric analysis of air\n",
"p = 1.03\t\t\t\t\t#Total pressure of the exhaust gases in kg/cm**2\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mH2 = 2.\t\t\t\t\t#Molecular weght of H2\n",
"mH2O = 18.\t\t\t\t\t#Molecular weight of H2O\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"xCO2 = ((C/100)*(mCO2/mC))\t\t\t\t\t#per kg of fuel, the products formed in kg\n",
"xH2O = ((H2/100)*((2*mH2O)/(2*mH2)))\t\t\t\t\t#per kg of fuel, the products formed in kg\n",
"xO2 = (((C/100)*(mO2/mC))+((H2/100)*(mO2/(2*mH2))))\t\t\t\t\t#Oxygen used in kg\n",
"xN2 = (pN2/pO2)*xO2\t\t\t\t\t#Associated nitrogen in kg\n",
"mma = (xO2+xN2)\t\t\t\t\t#Minimum air required in kg\n",
"ea = (ma-mma)\t\t\t\t\t#Excess air supplied in kg\n",
"XO2 = ((pO2/100)*ea)\t\t\t\t\t#Mass of O2 in excess air in kg\n",
"XN2 = ((pN2/100)*ea)\t\t\t\t\t#Mass of N2 in excess air in kg\n",
"wCO2 = xCO2/mCO2\t\t\t\t\t#Parts by volume for CO2\n",
"wO2 = XO2/mO2\t\t\t\t\t#Parts by volume for O2\n",
"wN2 = ((XN2+xN2)/mN2)\t\t\t\t\t#Parts by volume for N2\n",
"wH2O = (xH2O/mH2O)\t\t\t\t\t#Parts by volume for H2)\n",
"Tv = (wCO2+wO2+wN2+wH2O)\t\t\t\t\t#Total parts by volume\n",
"ppCO2 = (wCO2/Tv)*100\t\t\t\t\t#Percentage volume of CO2\n",
"ppO2 = (wO2/Tv)*100\t\t\t\t\t#Percentage volume of O2\n",
"ppN2 = (wN2/Tv)*100\t\t\t\t\t#Percentage volume of N2\n",
"ppH2O = (wH2O/Tv)*100\t\t\t\t\t#Percenatage volume of H2O\n",
"Tv1 = (wCO2+wO2+wN2)\t\t\t\t\t#Total parts by volume for dry products\n",
"pp1CO2 = (wCO2/Tv1)*100\t\t\t\t\t#Percentage volume of CO2 for dry analysis\n",
"pp1O2 = (wO2/Tv1)*100\t\t\t\t\t#Percentage volume of O2 for dry analysis\n",
"pp1N2 = (wN2/Tv1)*100\t\t\t\t\t#Percentage volume of N2 for dry analysis\n",
"papH2O = (ppH2O/100)*p\t\t\t\t\t#Partial pressure of H2O in kg/cm**2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The volumetric analysis for wet products gives in percent CO2 -> %3.1f O2 -> %3.1f N2 -> %3.1f H2O -> %3.1f \\\n",
"\\nThe volumetric analysis for dry products gives in percent CO2 -> %3.1f O2 -> %3.1f N2 -> %3.1f \\\n",
"\\nThe partial pressure of the vapour is %3.4f kg/cm**2'%(ppCO2,ppO2,ppN2,ppH2O,pp1CO2,pp1O2,pp1N2,papH2O) \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The volumetric analysis for wet products gives in percent CO2 -> 7.9 O2 -> 8.7 N2 -> 76.6 H2O -> 6.8 \n",
"The volumetric analysis for dry products gives in percent CO2 -> 8.5 O2 -> 9.3 N2 -> 82.2 \n",
"The partial pressure of the vapour is 0.0705 kg/cm**2\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.9 Page no : 162"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"CO2 = 15.\t\t\t\t\t#Volumetric analysis composition in percent\n",
"CO = 2.2\t\t\t\t\t#Volumetric analysis composition in percent\n",
"O2 = 1.6\t\t\t\t\t#Volumetric analysis composition in percent\n",
"N2 = 81.2\t\t\t\t\t#Volumetric analysis composition in percent\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mCO = 28.\t\t\t\t\t#Molecular weight of CO2\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"pCO2 = (CO2/100)*mCO2\t\t\t\t\t#Proportional weight for CO2\n",
"pCO = (CO/100)*mCO\t\t\t\t\t#Proportional weight for CO\n",
"pO2 = (O2/100)*mO2\t\t\t\t\t#Proportional weight for O2\n",
"pN2 = (N2/100)*mN2\t\t\t\t\t#Proportional weight for N2\n",
"T = (pCO2+pCO+pO2+pN2)\t\t\t\t\t#Total proportional weight\n",
"ppCO2 = (pCO2/T)*100\t\t\t\t\t#Weight per kg of exhaust gas for CO2\n",
"ppCO = (pCO/T)*100\t\t\t\t\t#Weight per kg of exhaust gas for CO\n",
"ppO2 = (pO2/T)*100\t\t\t\t\t#Weight per kg of exhaust gas for O2\n",
"ppN2 = (pN2/T)*100\t\t\t\t\t#Weight per kg of exhaust gas for N2\n",
"\n",
"print 'The analysis by weight is given by in percent CO2 -> %3.1f CO -> %3.1f O2 -> %3.1f N2 -> %3.1f'%(ppCO2,ppCO,ppO2,ppN2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The analysis by weight is given by in percent CO2 -> 21.7 CO -> 2.0 O2 -> 1.7 N2 -> 74.6\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.10 Page no : 167"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\n",
"# variables\n",
"CO2 = 10.9\t\t\t\t\t#Volumetric analysis composition in percent\n",
"CO = 1.\t\t\t\t\t#Volumetric analysis composition in percent\n",
"O2 = 7.1\t\t\t\t\t#Volumetric analysis composition in percent\n",
"N2 = 81.\t\t\t\t\t#Volumetric analysis composition in percent\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mCO = 28.\t\t\t\t\t#Molecular weight of CO\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"pCO2 = (CO2/100)*mCO2\t\t\t\t\t#Proportional weight for CO2\n",
"pCO = (CO/100)*mCO\t\t\t\t\t#Proportional weight for CO\n",
"pO2 = (O2/100)*mO2\t\t\t\t\t#Proportional weight for O2\n",
"pN2 = (N2/100)*mN2\t\t\t\t\t#Proportional weight for N2\n",
"T = (pCO2+pCO+pO2+pN2)\t\t\t\t\t#Total proportional weight\n",
"ppCO2 = (pCO2/T)*100\t\t\t\t\t#Weight per kg of exhaust gas for CO2\n",
"ppCO = (pCO/T)*100\t\t\t\t\t#Weight per kg of exhaust gas for CO\n",
"ppO2 = (pO2/T)*100\t\t\t\t\t#Weight per kg of exhaust gas for O2\n",
"ppN2 = (pN2/T)*100\t\t\t\t\t#Weight per kg of exhaust gas for N2\n",
"\n",
"print 'The gravimetric analysis is given by in percent CO2 -> %3.2f CO -> %3.2f O2 -> %3.2f N2 -> %3.2f'%(ppCO2,ppCO,ppO2,ppN2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The gravimetric analysis is given by in percent CO2 -> 15.97 CO -> 0.93 O2 -> 7.57 N2 -> 75.53\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.11 Page no : 168"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"CO2 = 10.\t\t\t\t\t#Volumetric analysis composition in percent\n",
"N2 = 80.\t\t\t\t\t#Volumetric analysis composition in percent\n",
"C = 80.\t\t\t\t\t#Carbon content of the fuel in percent\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"O2 = 100-(N2+CO2)\t\t\t\t\t#Volumetric analysis composition in percent\n",
"pCO2 = (CO2/100)*mCO2\t\t\t\t\t#Proportional weight for CO2\n",
"pO2 = (O2/100)*mO2\t\t\t\t\t#Proportional weight for O2\n",
"pN2 = (N2/100)*mN2\t\t\t\t\t#Proportional weight for N2\n",
"T = (pCO2+pO2+pN2)\t\t\t\t\t#Total proportional weight\n",
"ppCO2 = (pCO2/T)\t\t\t\t\t#Weight per kg of exhaust gas for CO2\n",
"ppO2 = (pO2/T)\t\t\t\t\t#Weight per kg of exhaust gas for O2\n",
"ppN2 = (pN2/T)\t\t\t\t\t#Weight per kg of exhaust gas for N2\n",
"wC = (ppCO2*(mC/mCO2))\t\t\t\t\t#Weight of carbon per kg of exhaust gases in kg\n",
"WC = ((C/100)/wC)\t\t\t\t\t#Weight of exhaust gases per kg of fuel burned in kg\n",
"wa = (WC-(ppCO2+ppO2+ppN2))\t\t\t\t\t#Weight of air supplied per kg fuel in kg\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Weight of air supplied per kg of fuel is %i kg'%(wa)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Weight of air supplied per kg of fuel is 19 kg\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.12 Page no : 168"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"CO2 = 12.\t\t\t\t\t#Volumetric analysis composition in percent\n",
"CO = 4.\t\t\t\t\t#Volumetric analysis composition in percent\n",
"N2 = 84.\t\t\t\t\t#Volumetric analysis composition in percent\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mCO = 28.\t\t\t\t\t#Molecular weight of CO\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mH2 = 2.\t\t\t\t\t#Molecular weight of H2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"pCO2 = (CO2/100)*mCO2\t\t\t\t\t#Proportional weight for CO2\n",
"pCO = (CO/100)*mCO\t\t\t\t\t#Proportional weight for CO\n",
"pN2 = (N2/100)*mN2\t\t\t\t\t#Proportional weight for N2\n",
"T = (pCO2+pCO+pN2)\t\t\t\t\t#Total proportional weight\n",
"ppCO2 = (pCO2/T)\t\t\t\t\t#Weight per kg of exhaust gas for CO2\n",
"ppCO = (pCO/T)\t\t\t\t\t#Weight per kg of exhaust gas for CO\n",
"ppN2 = (pN2/T)\t\t\t\t\t#Weight per kg of exhaust gas for N2\n",
"wC = ((ppCO2*(mC/mCO2))+(ppCO*(mC/mCO)))\t\t\t\t\t#Weight of carbon per kg of flue gases\n",
"pC = ((6*mC)/(6*mC+7*mH2))\t\t\t\t\t#Percentage by weight of carbon in C6H14\n",
"we = (pC/wC)\t\t\t\t\t#Weight of exhaust gases per kg of fuel in kg\n",
"wa = (we-(ppCO2+ppCO+ppN2))\t\t\t\t\t#Weight of air supplied per kg of fuel in kg\n",
"tw = ((100./23)*(((mO2/mC)*pC)+((mO2/(2*mH2))*0.163)))\t\t\t\t\t#Theoretical amount of air required for complete combustion of C6H14 in kg\n",
"exc = (wa-tw)\t\t\t\t\t#Excess air supplied per kg of fuel in kg\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Excess air supplied per kg of fuel is %3.1f kgdeficient)'%(exc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Excess air supplied per kg of fuel is -3.3 kgdeficient)\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.13 Page no : 171"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"C = 84.\t\t\t\t\t#Gravimetric analysis composition in percent\n",
"H2 = 12.\t\t\t\t\t#Gravimetric analysis composition in percent\n",
"S = 1.5\t\t\t\t\t#Gravimetric analysis composition in percent\n",
"O2 = 1.5\t\t\t\t\t#Gravimetric analysis composition in percent\n",
"ma = 20.\t\t\t\t\t#Mass of air in kg\n",
"pC = 4.\t\t\t\t\t#Percent of carbon in the fuel which is burnt to form CO\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mCO = 28.\t\t\t\t\t#Molecular weight of CO\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mH2 = 2.\t\t\t\t\t#Molecular weight of H2\n",
"mS = 32.\t\t\t\t\t#Molecular weight of S\n",
"mSO2 = 64.\t\t\t\t\t#Molecular weight of SO2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"mm = ((100./23)*((C/100)*(mO2/mC)+(H2/100)*(mO2/(2*mH2))+(S/100)*(mO2/mS)-(O2/100)))\t\t\t\t\t#Minimum air in kg/kg of fuel \n",
"#When 20 kg of air is supplied\n",
"xCO2 = ((C/100)*(mCO2/mC))\t\t\t\t\t#Mass of CO2 in kg\n",
"xSO2 = ((S/100)*(mSO2/mS))\t\t\t\t\t#Mass of SO2 in kg\n",
"xO2 = ((23/100)*(ma-mm))\t\t\t\t\t#Mass of O2 in kg\n",
"xN2 = ((77/100)*ma)\t\t\t\t\t#Mass of N2 in kg\n",
"nCO2 = (xCO2/mCO2)\t\t\t\t\t#Parts by volume of CO2\n",
"nSO2 = (xSO2/mSO2)\t\t\t\t\t#Parts by volume of SO2\n",
"nO2 = (xO2/mO2)\t\t\t\t\t#Parts by volume of O2\n",
"nN2 = (xN2/mN2)\t\t\t\t\t#Parts by volume of N2\n",
"T = (nCO2+nSO2+nO2+nN2)\t\t\t\t\t#Total parts by volume\n",
"pCO2 = (nCO2/T)*100\t\t\t\t\t#Percentage volume of CO2\n",
"pSO2 = (nSO2/T)*100\t\t\t\t\t#Percentage volume of SO2\n",
"pO2 = (nO2/T)*100\t\t\t\t\t#Percentage volume of O2\n",
"pN2 = (nN2/T)*100\t\t\t\t\t#Percentage volume of N2\n",
"\t\t\t\t\t#4% of available carbon is burnt to CO then per kg of fuel\n",
"yCO2 = ((C/100)/(1+(pC/100)))*(mCO2/mC)\t\t\t\t\t#Mass of CO2 in kg\n",
"yCO = (((C/100)-((C/100)/(1+(pC/100))))*(mCO/mC))\t\t\t\t\t#Mass of CO in kg\n",
"yO2 = ((C/100)*(mO2/mC))\t\t\t\t\t#Mass of O2 in kg\n",
"eO2 = (yO2-(((C/100)/(1+(pC/100)))*(mO2/mC)+(((C/100)-((C/100)/(1+(pC/100))))*(mO2/(2*mC)))))\n",
"nnCO2 = (yCO2/mCO2)\t\t\t\t\t#Parts by volume of CO2\n",
"nnCO = (yCO/mCO)\t\t\t\t\t#Parts by volume of CO\n",
"nnSO2 = (xSO2/mSO2)\t\t\t\t\t#Parts by volume of SO2\n",
"nnO2 = ((xO2+eO2)/mO2)\t\t\t\t\t#Parts by volume of O2\n",
"nnN2 = (xN2/mN2)\t\t\t\t\t#Parts by volume of N2\n",
"TT = (nnCO2+nnCO+nnSO2+nnO2+nnN2)\t\t\t\t\t#Total parts by volume\n",
"ppCO2 = (nnCO2/TT)*100\t\t\t\t\t#Percentage volume of CO2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Minimum weight of air required for complete combustion of 1 kg of the fuel is %3.1f kg/kg of fuel \\\n",
"\\nPercentage composition by volume when %i kg of air is supplied in percent) CO2 -> %3.1f SO2 -> %3.1f O2 -> %3.1f N2 -> %3.1f \\\n",
"\\nThe percentage volume of CO2 when %i percent of the carbon in the fuel is burnt to form CO is %3.1f percent'%(mm,ma,pCO2,pSO2,pO2,pN2,pC,ppCO2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Minimum weight of air required for complete combustion of 1 kg of the fuel is 13.9 kg/kg of fuel \n",
"Percentage composition by volume when 20 kg of air is supplied in percent) CO2 -> 99.3 SO2 -> 0.7 O2 -> 0.0 N2 -> 0.0 \n",
"The percentage volume of CO2 when 4 percent of the carbon in the fuel is burnt to form CO is 93.7 percent\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.14 Page no : 173"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"C = 85.\t\t\t\t\t#Composition by weight in percent\n",
"H2 = 14.\t\t\t\t\t#Composition by weight in percent\n",
"x = 50.\t\t\t\t\t#Percentage of excess air\n",
"Ta = [70.+273,500+273]\t\t\t\t\t#Temperature of air entering and leaving in K\n",
"Cp = 0.24\t\t\t\t\t#Mean specific heat of air in kJ/kg.K\n",
"qC = 8080.\t\t\t\t\t#Heat liberated in kcal/kg\n",
"qH2 = 34250.\t\t\t\t\t#Heat liberated in kcal/kg\n",
"a = 23.\t\t\t\t\t#Air contains 23% by weight of O2\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mCO = 28.\t\t\t\t\t#Molecular weight of CO\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mH2 = 2.\t\t\t\t\t#Molecular weight of H2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"mm = ((100/a)*((C/100)*(mO2/mC)+(H2/100)*(mO2/(2*mH2))))\t\t\t\t\t#Minimum air required in kg/kg of fuel\n",
"Q1 = ((C/100)*qC+(H2/100)*qH2)\t\t\t\t\t#Heat in kcal/kg fuel\n",
"ea = ((x/100)*mm)\t\t\t\t\t#Excess air supplied in kg/kg fuel\n",
"Q2 = ((mm/2)*Cp*(Ta[1]-Ta[0]))\t\t\t\t\t#Heat in kcal/kg fuel\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) Minimum quantity of air necessary for the complete combustion of 1 kg of fuel is %3.2f kg/kg of fuel \\\n",
"\\nb) Heat released per kg of fuel when the carbon is burnt to CO2 and hydrogen is burnt to H2O is %3.0f kcal/kg fuel \\\n",
"\\nc) Heat carried away by the excess air is %3.0f kcal/kg fuel'%(mm,Q1,Q2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) Minimum quantity of air necessary for the complete combustion of 1 kg of fuel is 14.72 kg/kg of fuel \n",
"b) Heat released per kg of fuel when the carbon is burnt to CO2 and hydrogen is burnt to H2O is 11663 kcal/kg fuel \n",
"c) Heat carried away by the excess air is 760 kcal/kg fuel\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.15 Page no : 173"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"CO = 17.\t\t\t\t\t#Percentage composition by volume\n",
"H2 = 53.4\t\t\t\t\t#Percentage composition by volume\n",
"CH2 = 28.8\t\t\t\t\t#Percentage composition by volume\n",
"O2 = 0.8\t\t\t\t\t#Percentage composition by volume\n",
"ea = 30.\t\t\t\t\t#Percentage of excess air\n",
"v = 1.\t\t\t\t\t#Volume in m**3\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"ta = ((100./21)*((CO/100)/2+(H2/100)/2+(CH2/100)*2-(O2/100)))\t\t\t\t\t#Theoretical air in m**3/m**3 of gas\n",
"aa = ((1+(ea/100))*ta)\t\t\t\t\t#Actual air in m**3/m**3 of gas\n",
"Vg = (v+aa)\t\t\t\t\t#Volume of gas air mixture in m**3/m**3 of gas\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Total quantity of air required is %3.2f m**3/m**3 of gas \\\n",
"\\nThe volume of gas air mixture is %3.2f m**3/m**3 of gas'%(ta,Vg)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Total quantity of air required is 4.38 m**3/m**3 of gas \n",
"The volume of gas air mixture is 6.70 m**3/m**3 of gas\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.16 Page no : 178"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"CH4 = 20.\t\t\t\t\t#Percentage volumetric analysis\n",
"C2H4 = 2.\t\t\t\t\t#Percentage volumetric analysis\n",
"H2 = 50.\t\t\t\t\t#Percentage volumetric analysis\n",
"CO = 16.\t\t\t\t\t#Percentage volumetric analysis\n",
"CO2 = 4.\t\t\t\t\t#Percentage volumetric analysis\n",
"O2 = 1.5\t\t\t\t\t#Percentage volumetric analysis\n",
"N2 = 6.5\t\t\t\t\t#Percentage volumetric analysis\n",
"v = 6.8\t\t\t\t\t#Volume of air supplied in m**3 per m**3 of coal gas\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"mmO2 = ((2*CH4)+(3*C2H4)+(H2/2)+(CO/2))-O2\t\t\t\t\t#Minimum moles of O2\n",
"mCO2 = (CH4+(2*C2H4)+CO+CO2)\t\t\t\t\t#Moles of CO2\n",
"mH2O = ((2*CH4)+(2*C2H4)+H2)\t\t\t\t\t#Moles of H2O\n",
"mN2 = (N2+(79./21)*mmO2)\t\t\t\t\t#Moles of N2\n",
"ma = ((100./21)*(mmO2/100))\t\t\t\t\t#Minimum air in m**3/m**3 of gas\n",
"ea = (v-ma)\t\t\t\t\t#Excess air in m**3/m**3 of gas\n",
"tm = (mCO2+mN2+ea)*2\t\t\t\t\t#Total moles of dry products per 100 moles of gas\n",
"pCO2 = (mCO2/tm)*100\t\t\t\t\t#Percentage of CO2 by volume in dry flue gases\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Minimum volume of air necessary for the complete combustion of 1 m**3 of coal gas is %3.2f m**3/m**3 of gas \\\n",
"\\nPercentage volume of CO2 in dry flue gases is %3.2f percent'%(ma,pCO2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Minimum volume of air necessary for the complete combustion of 1 m**3 of coal gas is 3.69 m**3/m**3 of gas \n",
"Percentage volume of CO2 in dry flue gases is 6.37 percent\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.17 Page no : 179"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"C = 86.\t\t\t\t\t#Percentage of carbon\n",
"H2 = 14.\t\t\t\t\t#Percentage of Hydrogen\n",
"ea = 20.\t\t\t\t\t#Percentage of excess air\n",
"O2 = 23.\t\t\t\t\t#Weight of oxygen in air in percent\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mCO = 28.\t\t\t\t\t#Molecular weight of CO\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mH2 = 2.\t\t\t\t\t#Molecular weight of H2\n",
"mH2O = 18.\t\t\t\t\t#Molecular weight of H2O\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"ma = ((100/O2)*((C/100)*(mO2/mC)+(H2/100)*(mO2/(2*mH2))))\t\t\t\t\t#Minimum weight of air required in kg/kg petrol\n",
"\t\t\t\t\t#Products of combustion by weight per kg- petrol \n",
"XCO2 = (C/100)*(mCO2/mC)\t\t\t\t\t#CO2 in kg\n",
"XH2O = (H2/100)*(mH2O/mH2)\t\t\t\t\t#H2O in kg\n",
"XO2 = (XCO2+XH2O-1)*(ea/100)\t\t\t\t\t#O2 in kg\n",
"XN2 = (ma*(1+(ea/100))*((100-O2)/100))\t\t\t\t\t#N2 in kg\n",
"XT = (XCO2+XH2O+XO2+XN2)\t\t\t\t\t#Total weight in kg\n",
"\t\t\t\t\t#Percentage analysis by weight\n",
"xCO2 = (XCO2/XT)*100\t\t\t\t\t#CO2\n",
"xH2O = (XH2O/XT)*100\t\t\t\t\t#H2O\n",
"xO2 = (XO2/XT)*100\t\t\t\t\t#O2\n",
"xN2 = (XN2/XT)*100\t\t\t\t\t#N2\n",
"\t\t\t\t\t#Percentage by weight to molecular weight\n",
"xxCO2 = (xCO2/mCO2)\t\t\t\t\t#CO2\n",
"xxH2O = (xH2O/mH2O)\t\t\t\t\t#H2O\n",
"xxO2 = (xO2/mO2)\t\t\t\t\t#O2\n",
"xxN2 = (xN2/mN2)\t\t\t\t\t#N2\n",
"xxt = (xxCO2+xxH2O+xxO2+xxN2)\t\t\t\t\t#Total percentage by weight to molecular weight\n",
"\t\t\t\t\t#Percentage by volume\n",
"pCO2 = (xxCO2/xxt)*100\t\t\t\t\t#CO2\n",
"pH2O = (xxH2O/xxt)*100\t\t\t\t\t#H2O\n",
"pO2 = (xxO2/xxt)*100\t\t\t\t\t#O2\n",
"pN2 = (xxN2/xxt)*100\t\t\t\t\t#N2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Volumetric composition of the products of combustion in percent) CO2 -> %3.1f H2O -> %3.1f O2 -> %3.2f N2 -> %3.2f'%(pCO2,pH2O,pO2,pN2)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Volumetric composition of the products of combustion in percent) CO2 -> 11.0 H2O -> 10.7 O2 -> 3.27 N2 -> 75.03\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.18 Page no : 179"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"bhp = 20.\t\t\t\t\t#Brake horse in h.p\n",
"N = 320.\t\t\t\t\t#Speed in r.p.m\n",
"C = 84.\t\t\t\t\t#Percentage of carbon\n",
"H2 = 16.\t\t\t\t\t#Percentage of hydrogen\n",
"CV = 10800.\t\t\t\t\t#Calorific value in kcal/kg\n",
"bth = 30.\t\t\t\t\t#Brake thermal efficiency in percent\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mCO = 28.\t\t\t\t\t#Molecular weight of CO\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mH2 = 2.\t\t\t\t\t#Molecular weight of H2\n",
"mH2O = 18.\t\t\t\t\t#Molecular weight of H2O\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"W = (bhp*4500)/427\t\t\t\t\t#Work done in kcal\n",
"Wc = (W*2)/N\t\t\t\t\t#Work done per cycle in kcal\n",
"qs = (Wc/(bth/100))\t\t\t\t\t#Heat supplied per cycle in kcal\n",
"wf = (qs/CV)\t\t\t\t\t#Weight of fuel used per cycle in kg\n",
"tO2 = ((C/100)*(mO2/mC)+(H2/100)*(mO2/(2*mH2)))\t\t\t\t\t#Total O2/kg fuel in kg\n",
"mw = (tO2/(23./100))\t\t\t\t\t#Minimum weight of air required in kg/kg fuel\n",
"aw = (mw*2)\t\t\t\t\t#Actual weight of air supplied in kg/kg fuel\n",
"wac = (aw*wf)\t\t\t\t\t#Wt. of air supplied/ cycle in kg. In textbook, it is given wrong as 0.1245 kg\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) the weight of fuel used per cycle is %3.6f kg \\\n",
"\\nb) the actual weight of air taken in per cycle is %3.4f kg'%(wf,wac)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) the weight of fuel used per cycle is 0.000407 kg \n",
"b) the actual weight of air taken in per cycle is 0.0124 kg\n"
]
}
],
"prompt_number": 18
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 8.19 Page no : 179"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"CO2 = 8.85\t\t\t\t\t#Percentage composition by volume\n",
"CO = 1.2\t\t\t\t\t#Percentage composition by volume\n",
"O2 = 6.8\t\t\t\t\t#Percentage composition by volume\n",
"N2 = 83.15\t\t\t\t\t#Percentage composition by volume\n",
"C = 84.\t\t\t\t\t#Percentage composition by weight\n",
"H2 = 14.\t\t\t\t\t#Percentage composition by weight\n",
"aO2 = 2.\t\t\t\t\t#Percentage composition by weight\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mCO2 = 44.\t\t\t\t\t#Molecular weight of CO2\n",
"mCO = 28.\t\t\t\t\t#Molecular weight of CO\n",
"mN2 = 28.\t\t\t\t\t#Molecular weight of N2\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mH2 = 2.\t\t\t\t\t#Molecular weight of H2\n",
"mH2O = 18.\t\t\t\t\t#Molecular weight of H2O\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"\t\t\t\t\t#O2 required per kg of fuel\n",
"xC = ((C/100)*(mO2/mC))\t\t\t\t\t#C\n",
"xH2 = ((H2/100)*(mO2/(2*mH2)))\t\t\t\t\t#H2\n",
"xO2 = -(aO2/100)\t\t\t\t\t#O2\n",
"ttO2 = (xC+xH2-xO2)\t\t\t\t\t#Theoretical total oxygen required in kg/kg fuel\n",
"twa = (ttO2/(23./100))\t\t\t\t\t#Theoretical weight of air in kg/kg fuel\n",
"\t\t\t\t\t#Conversion of volumetric analysis of the flue gas into a weight analysis\n",
"\t\t\t\t\t#Percenatge by volume * mol. wt\n",
"xxCO2 = (CO2*mCO2)\t\t\t\t\t#CO2\n",
"xxCO = (CO*mCO)\t\t\t\t\t#CO\n",
"xxO2 = (O2*mO2)\t\t\t\t\t#O2\n",
"xxN2 = (N2*mN2)\t\t\t\t\t#N2\n",
"xxt = (xxCO2+xxCO+xxO2+xxN2)\t\t\t\t\t#Total\n",
"\t\t\t\t\t#Percentage by weight\n",
"yCO2 = (xxCO2/xxt)*100\t\t\t\t\t#CO2\n",
"yCO = (xxCO/xxt)*100\t\t\t\t\t#CO\n",
"yO2 = (xxO2/xxt)*100\t\t\t\t\t#O2\n",
"yN2 = (xxN2/xxt)*100\t\t\t\t\t#N2\n",
"wcd = ((yCO2/100)*(mC/mCO2))+((yCO/100)*(mC/mCO))\t\t\t\t\t#Weight of carbon/ kg of dry flue gas in kg\n",
"wdf = ((C/100)/wcd)\t\t\t\t\t#Wt. of dry flue gas/kg fuel in kg\n",
"wxf = (wdf*(yO2/100))\t\t\t\t\t#Weight of excess O2/kg fuel in kg\n",
"weO2 = (wxf/(23./100))\t\t\t\t\t#Weight of excess air in kg/kg fuel\n",
"was = (twa+weO2)\t\t\t\t\t#Weight of air supplied/kg fuel in kg\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Weight of air supplied per kg fuel burnt is %3.2f kg'%(was)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Weight of air supplied per kg fuel burnt is 21.29 kg\n"
]
}
],
"prompt_number": 21
}
],
"metadata": {}
}
]
}PKI Zmm%Internal Combustion Engines/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": {}
}
]
}PKI *0*0&Internal Combustion Engines/ch11.ipynb{
"metadata": {
"name": "",
"signature": "sha256:4e8467c44185deda0f287a5fb2df2f33438b3eb316ab3092c4e75cf0fe34abd0"
},
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chatper 11 : Four Stroke Spark Ignition Engine"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.1 Page no : 239"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 0.0625\t\t\t\t\t#Diameter in m\n",
"L = 0.09\t\t\t\t\t#Stroke in m\n",
"nv = 0.75\t\t\t\t\t#Volumetric efficiency\n",
"p = 1.03\t\t\t\t\t#Pressure at N.T.P in kg/cm**2\n",
"T = 273\t\t\t\t\t#Temperature at N.T.P in K\n",
"R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/kg.degree C\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = ((3.14/4)*d**2*L)\t\t\t\t\t#Swept volume in cu.m\n",
"V = (nv*Vs)\t\t\t\t\t#Volume of charge at N.T.P in cu.m\n",
"w = (p*10**4*V)/(R*T)\t\t\t\t\t#Weight of the charge in kg/cycle\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The weight of the charge is %3.6f kg/cycle'%(w)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The weight of the charge is 0.000267 kg/cycle\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.2 Page no : 242"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 9.\t\t\t\t\t#Number of cylinder\n",
"d = 0.145\t\t\t\t\t#Bore in m\n",
"l = 0.19\t\t\t\t\t#Stroke in m\n",
"r = 5.9\t\t\t\t\t#Compression ratio\n",
"bhp = 460.\t\t\t\t\t#Brake horse power in B.H.P\n",
"N = 2000.\t\t\t\t\t#Speed in r.p.m\n",
"x = 20.\t\t\t\t\t#Percentage rich in mixture\n",
"CV = 11200.\t\t\t\t\t#Calorific value in kcal/kg\n",
"pC = 85.3\t\t\t\t\t#Percentage of carbon\n",
"pH2 = 14.7\t\t\t\t\t#Percentage of Hydrogen\n",
"nv = 70.\t\t\t\t\t#Volumetric efficiency in percent\n",
"T = 15.+273\t\t\t\t\t#Temperature in K\n",
"nm = 90.\t\t\t\t\t#Mechanical efficiency in percent\n",
"wO2 = 23.3\t\t\t\t\t#Percentage of oxygen by weight in air\n",
"da = 1.29\t\t\t\t\t#Density of air in kg/m**3\n",
"mC = 12.\t\t\t\t\t#Molecular weight of carbon\n",
"mO2 = 32.\t\t\t\t\t#Molecular weight of O2\n",
"mH2 = 2.\t\t\t\t\t#Molecular weight of H2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"hihp = ((bhp/(nm/100))*(4500./427))\t\t\t\t\t#Heat equivalent in kcal\n",
"Vs = ((3.14/4)*d**2*l*(N/2)*n)\t\t\t\t\t#Swept volume in c.m per min\n",
"cw = (Vs/da)\t\t\t\t\t#Charge weight of air per minute in kg\n",
"ma = (100/wO2)*((pC/100)*(mO2/mC)+(pH2/100)*(mO2/(2*mH2)))\t\t\t\t\t#Wt. of air required per kg of fuel in kg\n",
"mf = (cw/ma)\t\t\t\t\t#Minimum fuel inkg\n",
"ith = (hihp/(mf*(100+x)/100*CV))*100\t\t\t\t\t#Indicated thermal efficiciency in percent'\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Indicated thermal efficiency of the engine is %3.1f percent'%(ith)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Indicated thermal efficiency of the engine is 27.1 percent\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.3 Page no : 244"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 8.\t\t\t\t\t#Number of cylinders\n",
"d = 8.57\t\t\t\t\t#Bore in cm\n",
"l = 8.25\t\t\t\t\t#Stroke in cm\n",
"r = 7.\t\t\t\t\t#Compression ratio\n",
"N = 4000.\t\t\t\t\t#Speed in r.p.m\n",
"la = 53.35\t\t\t\t\t#Length of the arm in cm\n",
"t = 10.\t\t\t\t\t#Test duration in min\n",
"br = 40.8\t\t\t\t\t#Beam reading in kg\n",
"gas = 0.455\t\t\t\t\t#gasoline in kg. In textbook, it is given wrong as 4.55\n",
"CV = 11400.\t\t\t\t\t#Calorific value in kcal/kg\n",
"Ta = 21.+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",
"wa = 5.44\t\t\t\t\t#Rate of air in kg/min\n",
"J = 427.\t\t\t\t\t#Mechanical equivalent of heat in kg.m/kcal\n",
"R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/kg.K\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"bhp = (2*3.14*N*br*la)/(4500*100)\t\t\t\t\t#Brake horse power in B.H.P\n",
"pb = (bhp*4500)/((n/2)*(l/100)*(3.14/4)*d**2*N)\t\t\t\t\t#Brake mean effective pressure in kg/cm**2\n",
"bsfc = (gas*60)/bhp\t\t\t\t\t#Brake specific fuel consumption in kg/b.h.p.hr\n",
"bsac = ((wa*60)/bhp)\t\t\t\t\t#Brake specific fuel consumption in kg/b.h.p.hr\n",
"nb = ((bhp*4500)/(J*gas*CV))*100\t\t\t\t\t#Brake thermal efficiency in percent\n",
"Vd = ((3.14/4)*d**2*l)\t\t\t\t\t#Piston print lacement in c.c/cycle\n",
"Pd = (Vd/10**6)*(N/2)*n\t\t\t\t\t#Piston print lacement in m**3/min\n",
"Va = ((wa*R*Ta)/(pa*10**4))\t\t\t\t\t#Volume of air used in m**3/min\n",
"nv = (Va/Pd)*100\t\t\t\t\t#Volumetric efficiency in percent\n",
"af = (wa/gas)\t\t\t\t\t#Air fel ratio\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) the B.H.P delivered s %3.0f h.p \\\n",
"\\nb) the b.m.e.p is %3.1f kg/cm**2 \\\n",
"\\nc) the b.s.f.c is %3.3f kg/b.h.p.hr \\\n",
"\\nd) the brake specific air consumption is %3.3f kg/b.h.p.hr \\\n",
"\\ne) the brake thermal efficiency is %3.1f percent \\\n",
"\\nf) the volumetric efficiency is %3.0f percent \\\n",
"\\ng) the air fuel ratio is %3.2f'%(bhp,pb,bsfc,bsac,nb,nv,af)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) the B.H.P delivered s 122 h.p \n",
"b) the b.m.e.p is 7.2 kg/cm**2 \n",
"c) the b.s.f.c is 0.225 kg/b.h.p.hr \n",
"d) the brake specific air consumption is 2.686 kg/b.h.p.hr \n",
"e) the brake thermal efficiency is 24.7 percent \n",
"f) the volumetric efficiency is 60 percent \n",
"g) the air fuel ratio is 11.96\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.4 Page no : 246"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4.\t\t\t\t\t#Number of cylinders\n",
"N = 2000.\t\t\t\t\t#Speed in r.p.m\n",
"m = 13.15\t\t\t\t\t#Mass of fuel in kg/hour\n",
"Vd = 655.5\t\t\t\t\t#Displacement volume in c.c\n",
"da = 1.2\t\t\t\t\t#Density of air in kg/m**3\n",
"d = 12.7\t\t\t\t\t#Manometer depression in cm\n",
"\t\t\t\t\t#Qa = 0.231*math.sqrt(ha); Qa is the flow rate in cu.m/min and ha is the pressure difference in metres\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Qa = (0.231*math.sqrt(((d*1000)/da)/100))\t\t\t\t\t#Flow rate in cu.m/min\n",
"Wa = (Qa*da)\t\t\t\t\t#Weight of air in kg/min\n",
"Va = (Qa*(2/N)*(1/n))*10**6\t\t\t\t\t#Volume of air drawn in per cycle per cylinder in c.c\n",
"nv = (Va/Vd)*100\t\t\t\t\t#Volumetric efficiency in percent\n",
"af = (Wa/(m/60))\t\t\t\t\t#Air fuel ratio\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) the weight of air drawn is %3.3f kg/min \\\n",
"\\nb) volumetric efficiency taking air into account is %3.1f percent \\\n",
"\\nc) the air-fuel ratio is %i'%(Wa,nv,af)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) the weight of air drawn is 2.852 kg/min \n",
"b) volumetric efficiency taking air into account is 90.6 percent \n",
"c) the air-fuel ratio is 13\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.5 Page no : 248"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 10.\t\t\t\t\t#Diameter in cm\n",
"l = 15.\t\t\t\t\t#Stroke in cm\n",
"r = 6.\t\t\t\t\t#Compression ratio\n",
"ihp = 20.\t\t\t\t\t#Indicated horse power in h.p\n",
"N = 1000.\t\t\t\t\t#Speed in r.p.m\n",
"n = 4.\t\t\t\t\t#Number of cylinders\n",
"nt = 30.\t\t\t\t\t#Thermal efficiency in percent\n",
"CV = 10000.\t\t\t\t\t#Calorific value in kca/kg\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"Vs = ((3.14/4)*d**2*l)\t\t\t\t\t#Swept volume in c.c\n",
"Vc = (Vs/(r-1))\t\t\t\t\t#Clearance volume in c.c\n",
"na = (1-(1/r)**(g-1))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"pm = ((ihp*4500)/((l/100)*(3.14/4)*(d/100)**2*(N/2)*n))\t\t\t\t\t#Pressure in kg/cm**2\n",
"pc = (ihp*4500*60)/(427*(nt/100)*CV)\t\t\t\t\t#Petrol consumption in kg/hr\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Clearance volume is %3.1f c.c \\\n",
"\\nThe air standard efficiency is %3.1f percent \\\n",
"\\nPetrol consumption is %3.2f kg/hr'%(Vc,na,pc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Clearance volume is 235.5 c.c \n",
"The air standard efficiency is 51.2 percent \n",
"Petrol consumption is 4.22 kg/hr\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 11.6 Page no : 253"
]
},
{
"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",
"P = 62.\t\t\t\t\t#Power in HP\n",
"N = 3000.\t\t\t\t\t#Speed in r.p.m\n",
"nv = 85.\t\t\t\t\t#Volumetric efficiency in percent\n",
"nt = 25.\t\t\t\t\t#Thermal efficiency in percent\n",
"CV = 10500.\t\t\t\t\t#Calorific value in kcal/kg\n",
"af = 15.\t\t\t\t\t#Air fuel ratio\n",
"T = 273.\t\t\t\t\t#standard atmosphere temperature in K\n",
"p = 1.03\t\t\t\t\t#standard atmosphere pressure in kg/cm**2\n",
"R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/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",
"q = (P*4500)/(J*(nt/100))\t\t\t\t\t#Heat supplied in kcal/min\n",
"F = (q/CV)\t\t\t\t\t#Fuel supplied per minute in kg\n",
"Fc = (F/N)*(2/n)\t\t\t\t\t#Fuel supplied per cycle per cylinder in kg\n",
"wt = (af*Fc)\t\t\t\t\t#Weight of air supplied per cycle in kg\n",
"d = ((((wt)*R*T)/(p*10**4*(3.14/4)*(nv/100)))**(1./3))*100\t\t\t\t\t#Diameter in cm\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Cylinder bore = stroke = %3.2f cm'%(d)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Cylinder bore = stroke = 7.84 cm\n"
]
}
],
"prompt_number": 7
}
],
"metadata": {}
}
]
}PKI}@:&Internal Combustion Engines/ch14.ipynb{
"metadata": {
"name": "",
"signature": "sha256:ba05265a254158c2105561a6062dfe30873e9384b5393c6f2ca64b6f28ed143f"
},
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 14 : The Diesel Engine"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.1 Page no : 258"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"r = 14.3\t\t\t\t\t#Compression ratio\n",
"c = 5\t\t\t\t\t#Fuel cutoff in percent of stroke\n",
"w = 0.006\t\t\t\t\t#Weight of charge in kg\n",
"T4 = 912\t\t\t\t\t#Final temperature in degree C abs\n",
"q = 8300\t\t\t\t\t#Heat in kcal\n",
"x = [0.258,0.000048]\t\t\t\t\t#Temperature expression is 0.258T+0.000048T**2, where T is in degree C abs\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"v4 = 1. #Assuming clearance volume as unity\n",
"v1 = 1.665\t\t\t\t\t#v1 from fig. 14.2 on page no. 352\n",
"T1 = (T4*v1)/v4\t\t\t\t\t#Temperature in degree C abs\n",
"qp1 = (x[0]*T1+x[1]*T1**2)\t\t\t\t\t#constant pressure heat of mixture at temperature T1 in kcal/kg\n",
"qp4 = (x[0]*T4+x[1]*T4**2)\t\t\t\t\t#constant pressure heat of mixture at temperature T4 in kcal/kg\n",
"qre = (qp1-qp4)\t\t\t\t\t#Heat required by the mixture in kcal/kg\n",
"wf = (w*qre)/q\t\t\t\t\t#Weight of oil in kg\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The weight of oil that must be injected is %3.6f kg'%(wf)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The weight of oil that must be injected is 0.000164 kg\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.2 Page no : 261"
]
},
{
"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",
"p = 1.2\t\t\t\t\t#Induction pipe pressure in kg/cm**2\n",
"bp = 0.65\t\t\t\t\t#Exhaust back pressure in kg/cm**2\n",
"Tc = 87+273\t\t\t\t\t#Charge temperature in K\n",
"Te = 850+273\t\t\t\t\t#Exhaust temperature in K\n",
"T1 = 111+273\t\t\t\t\t#Temperature at the beginning of compression in K\n",
"g = 1.2\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Cw1 = ((bp*10**4)/Te)\t\t\t\t\t#specific heat in kJ/kg.K\n",
"Cw2 = ((p*10**4*(r-1))/Tc)\t\t\t\t\t#specific heat in kJ/kg.K\n",
"T3 = ((g*Te*Cw1+Cw2*Tc)/(Cw1*g+Cw2))\t\t\t\t\t#Temperature in K\n",
"t3 = T3-273\t\t\t\t\t#Temperature in degree C\n",
"rw = (Cw1/Cw2)\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The ratio of the mass residuals to fresh charge is %3.4f'%(rw)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The ratio of the mass residuals to fresh charge is 0.0134\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}PKIDK@@&Internal Combustion Engines/ch15.ipynb{
"metadata": {
"name": "",
"signature": "sha256:f62e01247c5e7505c14dabc00585a7fb2179310320c1005e1a7d9854022a73b1"
},
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 15 : Fuel Injection"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.1 Page no : 274"
]
},
{
"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\t#Number of cylinders\n",
"p = 720.\t\t\t\t\t#Horse power in h.p\n",
"N = 180.\t\t\t\t\t#Speed in r.p.m\n",
"f = 250.\t\t\t\t\t#Fuel rate in gm per horse power hour\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"w = (((f/1000)*p)/((N/2)*60*n))*1000\t\t\t\t\t#Weight of fuel per cycle in gm/cycle\n",
"\n",
"\t\t\t\t\t#Outptut\n",
"print 'The quantity of fuel to be injected per cylinder is %3.2f gm/cycle'%(w)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The quantity of fuel to be injected per cylinder is 5.56 gm/cycle\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.2 Page no : 278"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4\t\t\t\t\t#Number of cylinders\n",
"fc = 0.215\t\t\t\t\t#Brake specific fuel consumption in kg/B.H.P hour\n",
"BHP = 400\t\t\t\t\t#Brake horse power in B.H.P\n",
"N = 250\t\t\t\t\t#Speed in r.p.m\n",
"sg = 0.9\t\t\t\t\t#Specific gravity\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Fc = (fc*BHP)\t\t\t\t\t#Fuel consumption per hour in kg/hr\n",
"Fcy = (Fc/n)\t\t\t\t\t#Fuel consumption per cylinder in kg/hr\n",
"Fcyc = ((Fcy/(60*(N/2)))/(sg*1000))*10**6\t\t\t\t\t#Fuel consumption per cycle in kg. In textbook it is given wrong as 0.0287 instead of 3.185\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The quantity of fuel to be injected per cycle per cylinder is %3.3f c.c'%(Fcyc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The quantity of fuel to be injected per cycle per cylinder is 3.185 c.c\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.3 Page no : 279"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4.\t\t\t\t\t#Number of cylinders\n",
"p = 450.\t\t\t\t\t#Brake Horse power in B.H.P\n",
"N = 200.\t\t\t\t\t#Speed in r.p.m\n",
"f = 0.2\t\t\t\t\t#Fuel rate in kg per horse power hour\n",
"g = 0.9\t\t\t\t\t#Specific gravity of fuel\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"Fc = (p*f)\t\t\t\t\t#Fuel consumption per hour in kg/hr\n",
"Fcy = (Fc/n)\t\t\t\t\t#Fuel consumption per cylinder in kg/hr\n",
"Fcyc = (Fcy/(60*(N/2)))\t\t\t\t\t#Fuel consumption per cycle in kg\n",
"q = (Fcyc/(g*1000))*10**6\t\t\t\t\t#Quantity of fuel injected per cylinder per cycle in c.c\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The quantity of fuel to be injected per cycle per cylinder is %3.3f c.c'%(q)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The quantity of fuel to be injected per cycle per cylinder is 4.167 c.c\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.4 Page no : 284"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"#Data from problem 1\n",
"n = 6.\t\t\t\t\t#Number of cylinders\n",
"p = 720.\t\t\t\t\t#Horse power in h.p\n",
"N = 180.\t\t\t\t\t#Speed in r.p.m\n",
"f = 250.\t\t\t\t\t#Fuel rate in gm per horse power hour\n",
"\n",
"Vo = 20.\t\t\t\t\t#Volume of oil in the suction chamber in c.c\n",
"dp = 80.\t\t\t\t\t#Discharge pressure in kg/cm**2\n",
"voi = 6.\t\t\t\t\t#Volume of oil in the injector in c.c\n",
"g = 0.9\t\t\t\t\t#Specific gravity of oil\n",
"b = 78.8*10**-6\t\t\t\t\t#Coefficient of compressibility in cm**2/kg when pressure is taken as atmospheric\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"w = (((f/1000)*p)/((N/2)*60*n))*1000\t\t\t\t\t#Weight of fuel per cycle in gm/cycle\n",
"Va = (w/g)\t\t\t\t\t#Volume of air per cycle in c.c\n",
"V1 = (Vo+Va)\t\t\t\t\t#Initial volume in c.c\n",
"dV12 = (b*V1*dp)\t\t\t\t\t#Change in volume in c.c\n",
"\t\t\t\t\t#Assuming in accordance with average practice that s = 2d, nv = 0.94 and full load in this pump type x = 0.5\n",
"d = ((voi+dV12)/((3.14/4)*2*0.94*0.5))**(1./3)\t\t\t\t\t#Diameter in cm\n",
"l = (2*d)\t\t\t\t\t#Stroke in cm\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The diameter of the pump is %3.2f cm \\\n",
"\\nThe total stroke is %3.2f cm'%(d,l)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The diameter of the pump is 2.03 cm \n",
"The total stroke is 4.06 cm\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.5 Page no : 287"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"p = 110.\t\t\t\t\t#Oil pressure in kg/cm**2\n",
"pc = 25.\t\t\t\t\t#Pressure in the combustion chamber in kg/cm**2\n",
"q = 0.805\t\t\t\t\t#Velocity coefficient. In textbook it is given wrong as 9.805\n",
"d = 0.906\t\t\t\t\t#Specific gravity\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"v = (37.1*q*math.sqrt((p-pc)/d))\t\t\t\t\t#Velocity in m/s\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The velocity of injection is %3.0f m/s'%(v)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The velocity of injection is 289 m/s\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.6 Page no : 290"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"Vf = 6.2\t\t\t\t\t#Volume of fuel in c.c\n",
"l = 65\t\t\t\t\t#Length of fuel line in cm\n",
"di = 2.5\t\t\t\t\t#Inner diameter in mm\n",
"V = 2.75\t\t\t\t\t#Volume of fuel in the injector valve in c.c\n",
"Vd = 0.15\t\t\t\t\t#Volume of fuel to be delivered in c.c. In textbook it is given wrong as 0.047\n",
"p = 140\t\t\t\t\t#Pressure in kg/cm**2\n",
"pp = 1\t\t\t\t\t#Pump pressure in kg/cm**2\n",
"patm = 1.03\t\t\t\t\t#Atmospheric pressure in kg/cm**2\n",
"b = 78.8*10**-6\t\t\t\t\t#Coefficient of compressibility in cm**2/kg when pressure is taken as atmospheric\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"V1 = (Vf+(3.14/4)*(di/10)**2*l+V)\t\t\t\t\t#Initial volume in c.c\n",
"dV = ((b*V1*(p-pp)/patm))\t\t\t\t\t#Change in volume in c.c\n",
"d = (dV+Vd)\t\t\t\t\t#Total print lacement of the plunger in c.c\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The total print lacement of the plunger is %3.3f c.c'%(d)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total print lacement of the plunger is 0.279 c.c\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.7 Page no : 292"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"Vf = 6.75\t\t\t\t\t#Volume of fuel in c.c\n",
"l = 65\t\t\t\t\t#Length of fuel line in cm\n",
"di = 2.5\t\t\t\t\t#Inner diameter in mm\n",
"V = 2.45\t\t\t\t\t#Volume of fuel in the injector valve in c.c\n",
"Vd = 0.15\t\t\t\t\t#Volume of fuel to be delivered in c.c. \n",
"p = 150\t\t\t\t\t#Pressure in kg/cm**2\n",
"pp = 1\t\t\t\t\t#Pump pressure in kg/cm**2\n",
"patm = 1.03\t\t\t\t\t#Atmospheric pressure in kg/cm**2\n",
"b = 78.8*10**-6\t\t\t\t\t#Coefficient of compressibility in cm**2/kg when pressure is taken as atmospheric\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"V1 = (Vf+(3.14/4)*(di/10)**2*l+V)\t\t\t\t\t#Initial volume in c.c\n",
"dV = ((b*V1*(p-pp)/patm))\t\t\t\t\t#Change in volume in c.c\n",
"d = (dV+Vd)\t\t\t\t\t#Total print lacement of the plunger in c.c\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The total print lacement of the plunger is %3.3f c.c'%(d)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The total print lacement of the plunger is 0.291 c.c\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.8 Page no : 295"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"Vf = 6.75\t\t\t\t\t#Volume of fuel in c.c\n",
"l = 65\t\t\t\t\t#Length of fuel line in cm\n",
"di = 2.5\t\t\t\t\t#Inner diameter in mm\n",
"V = 2.45\t\t\t\t\t#Volume of fuel in the injector valve in c.c\n",
"Vd = 0.15\t\t\t\t\t#Volume of fuel to be delivered in c.c. \n",
"p = 150\t\t\t\t\t#Pressure in kg/cm**2\n",
"pp = 1\t\t\t\t\t#Pump pressure in kg/cm**2\n",
"patm = 1.03\t\t\t\t\t#Atmospheric pressure in kg/cm**2\n",
"b = 78.8*10**-6\t\t\t\t\t#Coefficient of compressibility in cm**2/kg when pressure is taken as atmospheric\n",
"dp = 0.75\t\t\t\t\t#Diameter of the plunger in cm\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"V1 = (Vf+(3.14/4)*(di/10)**2*l+V)\t\t\t\t\t#Initial volume in c.c\n",
"dV = ((b*V1*(p-pp)/patm))\t\t\t\t\t#Change in volume in c.c\n",
"d = (dV+Vd)\t\t\t\t\t#Total print lacement of the plunger in c.c\n",
"s = ((4/3.14)*(d/dp**2))*10\t\t\t\t\t#Stroke in mm\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The effective plunger stroke is %3.1f mm'%(s)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The effective plunger stroke is 6.6 mm\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.9 Page no : 298"
]
},
{
"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",
"p = 300.\t\t\t\t\t#Horse power in H.P\n",
"N = 1200.\t\t\t\t\t#Speed in r.p.m\n",
"f = 0.2\t\t\t\t\t#Fuel rate in kg per B.H.P hour\n",
"ip = 200.\t\t\t\t\t#Injection pressure in kg/cm**2\n",
"cp = 40.\t\t\t\t\t#Pressure in the combustion chamber in kg/cm**2\n",
"pic = 33.\t\t\t\t\t#Period of injection of the crank angle in degrees\n",
"g = 0.83\t\t\t\t\t#Specific gravity of fuel. In textbook, it is given wrong as 0.89\n",
"Cd = 0.9\t\t\t\t\t#Coefficient of discharge \n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"Fc = (p*f)\t\t\t\t\t#Fuel consumption per hour in kg/hr\n",
"Fcy = (Fc/n)\t\t\t\t\t#Fuel consumption per cylinder in kg/hr\n",
"Fcyc = (Fcy/(60*(N/2)))\t\t\t\t\t#Fuel consumption per cycle in kg\n",
"q = (Fcyc/(g*1000))*10**6\t\t\t\t\t#Quantity of fuel injected per cylinder per cycle in c.c\n",
"I = ((pic/360.)*(1/N)*60)\t\t\t\t\t#Injection period in sec\n",
"df = (g/1000)\t\t\t\t\t#Density of fuel in kg/m**3\n",
"v = math.sqrt(2*981*((ip-cp)/df))\t\t\t\t\t#Velocity of fuel through orifice in m/s\n",
"A = (q/(Cd*v*I))\t\t\t\t\t#Area of orifice in cm**2\n",
"d = math.sqrt(A/(3.14/4))*10\t\t\t\t\t#Diameter in mm\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The diameter of the math.single orifice injector is %3.2f mm'%(d)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The diameter of the math.single orifice injector is 0.73 mm\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 15.10 Page no : 302"
]
},
{
"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 = 11.5\t\t\t\t\t#Bore in cm\n",
"l = 14\t\t\t\t\t#Stroke in cm\n",
"af = 16\t\t\t\t\t#Air fuel ratio\n",
"pa = 1.03\t\t\t\t\t#Pressure of air intake in kg/cm**2\n",
"Ta = 24+273\t\t\t\t\t#Temperature of air intake in K\n",
"nv = 76.5\t\t\t\t\t#Volumetric efficiency in percent\n",
"R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/kg.K\n",
"N = 1500\t\t\t\t\t#Speed in r.p.m\n",
"ip = 125\t\t\t\t\t#Injection pressure in kg/cm**2\n",
"cp = 40\t\t\t\t\t#Compression pressure in kg/cm**2\n",
"q = 18.5\t\t\t\t\t#Fuel injection occupies 18.5 degrees of crenk travel\n",
"fsw = 760\t\t\t\t\t#Fuel specific weight in kg/m**2\n",
"dc = 0.94\t\t\t\t\t#Orifice discharge coefficient\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = ((3.14/4)*d**2*l)\t\t\t\t\t#Stroke volume in c.c\n",
"Va = (Vs*(nv/100))\t\t\t\t\t#Volume of air supplied in c.c\n",
"wa = ((pa*10**4*Va*10**-6)/(R*Ta))\t\t\t\t\t#Weight of air supplied per cylinder per cycle in kg\n",
"wf = (wa/af)\t\t\t\t\t#Weight of fuel injected per cylinder per cycle in kg\n",
"I = ((60*q)/(N*360))\t\t\t\t\t#Injection time per cycle in sec\n",
"F = (wf/I)\t\t\t\t\t#Fuel injected per cylinder per sec in kg/sec\n",
"Af = (F/(dc*math.sqrt(2*9.81*fsw*(ip-cp)*10**4)))\t\t\t\t\t#Area of orifice in sq.m\n",
"df = math.sqrt(Af/(3.14/4))*1000\t\t\t\t\t#Diameter of orifice in mm\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Maximum amount of fuel injected per cylinder per sec is %3.2f kg/sec \\\n",
"\\nDiameter of orifice is %3.3f mm'%(F,df)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Maximum amount of fuel injected per cylinder per sec is 0.04 kg/sec \n",
"Diameter of orifice is 0.694 mm\n"
]
}
],
"prompt_number": 11
}
],
"metadata": {}
}
]
}PKIGu&Internal Combustion Engines/ch16.ipynb{
"metadata": {
"name": "",
"signature": "sha256:188e5ca03e48e521630f0518a9d462567af55f5cfab6623682e9d579c84b2da8"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 16 : Combustion In Compresson Ignition Engines"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 16.1 Page no : 315"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"from numpy import array\n",
"\t\t\t\t\t\n",
"#Input data\n",
"s = 0.005\t\t\t\t\t#Delay in sec\n",
"d = 30.\t\t\t\t\t#Bore in cm\n",
"N = 600.\t\t\t\t\t#Speed in r.p.m\n",
"dx = array([10,15,20])\t\t\t\t\t#Bore diameters in cm\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"t = (s/d)*dx\t\t\t\t\t#Time of delay in sec. In textbook, t[1] is given wrong as 0.00025 sec instead of 0.0025 sec\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The delay time for %i cm diameter bore is %3.5f sec \\\n",
"\\nThe delay time for %i cm diameter bore is %3.5f sec \\\n",
"\\nThe delay time for %i cm diameter bore is %3.5f sec'%(dx[0],t[0],dx[1],t[1],dx[2],t[2])\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The delay time for 10 cm diameter bore is 0.00167 sec \n",
"The delay time for 15 cm diameter bore is 0.00250 sec \n",
"The delay time for 20 cm diameter bore is 0.00333 sec\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 16.2 Page no : 317"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = [15.,60.]\t\t\t\t\t#Bore in cm\n",
"N = [1600.,400.]\t\t\t\t\t#Speed in r.p.m respectively\n",
"q = 30.\t\t\t\t\t#Injection of oil occupies 30 degrees of crank travel in each case\n",
"pc = 30.\t\t\t\t\t#Compression pressure in kg/cm**2\n",
"d = 0.001\t\t\t\t\t#Delay time in sec\n",
"rp = 5.\t\t\t\t\t#Rapid combustion period is 5 degree of crank travel\n",
"pe = 60.\t\t\t\t\t#Compression pressure at the end of rapid compression in kg/cm**2\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"\t\t\t\t\t#For small engine\n",
"It1 = (60/N[0])*(q/360)\t\t\t\t\t#Injection time in sec\n",
"pf1 = ((d/It1)+(rp/pc))*100\t\t\t\t\t#Percent fuel\n",
"\t\t\t\t\t#For large engine\n",
"It2 = (60/N[1])*(q/360)\t\t\t\t\t#Injection time in sec\n",
"pf2 = ((d/It2)+(rp/pc))*100\t\t\t\t\t#Percent fuel\n",
"pr = (pc*(pf2/pf1))\t\t\t\t\t#Pressure rise in kg/cm**2\n",
"mp = (pc+pr)\t\t\t\t\t#Maximum pressure in kg/cm**2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Pressure in the large engine is %3.1f kg/cm**2'%(mp)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Pressure in the large engine is 45.2 kg/cm**2\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 16.3 Page no : 320"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4.\t\t\t\t\t#Number of cylinders\n",
"d = 105.\t\t\t\t\t#Bore in mm\n",
"l = 127.\t\t\t\t\t#Stroke in mm\n",
"BHP = 63.\t\t\t\t\t#Brake horse power in h.p\n",
"N = 1800.\t\t\t\t\t#Speed in r.p.m\n",
"t = 15.\t\t\t\t\t#Test time in min\n",
"mf = 2.75\t\t\t\t\t#Mass of fuel in kg\n",
"CV = 11000.\t\t\t\t\t#Calorific value in kcal/kg\n",
"af = 14.8\t\t\t\t\t#Air fuel ratio\n",
"v = 0.805\t\t\t\t\t#Specific volume in m**3/kg\n",
"nv = 80.\t\t\t\t\t#Volumetric efficiency in percent\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",
"bth = ((BHP*4500)/(J*(mf/t)*CV))*100\t\t\t\t\t#Brake thermal efficiency in percent\n",
"Vs = ((3.14/4)*(d/10)**2*(l/10))\t\t\t\t\t#Stroke volume in c.c\n",
"Vsw = (Vs*n*(N/2)*t)\t\t\t\t\t#Swept volume in c.c\n",
"Va = (Vsw*10**-6*(nv/100))\t\t\t\t\t#Volume of air sucked in m**3\n",
"wa = (Va/v)\t\t\t\t\t#Weight of air sucked in kg\n",
"wr = (af*mf)\t\t\t\t\t#Weight of air reqired in kg\n",
"pei = (wr/wa)*100\t\t\t\t\t#Percentage of air available for combustion\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Brake thermal efficiency is %3.1f percent \\\n",
"\\nThe percentage of air used for combustion is %i percent'%(bth,pei)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Brake thermal efficiency is 32.9 percent \n",
"The percentage of air used for combustion is 69 percent\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}PKI;EE&Internal Combustion Engines/ch18.ipynb{
"metadata": {
"name": "",
"signature": "sha256:0edb003d2406dbd42a27924dc90436706150dd4044705ed46aaf76f3c5aa44bf"
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"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 18 : Super Charging"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 18.1 Page no : 327"
]
},
{
"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 = 9.\t\t\t\t\t#Bore in cm\n",
"l = 10.\t\t\t\t\t#Stroke in cm\n",
"N = 2500.\t\t\t\t\t#Speed in r.p.m\n",
"Ta = 25.+273\t\t\t\t\t#Temperature of air entering the compressor in K\n",
"q = 16800.\t\t\t\t\t#Heat rate in kcal/hour\n",
"T = 60.+273\t\t\t\t\t#Temperature of air leaving the cooler in K\n",
"p = 1.6\t\t\t\t\t#Pressure of air leaving the cooler in kg/cm**2\n",
"t = 14.5\t\t\t\t\t#Engine torque in kg.m\n",
"nv = 75.\t\t\t\t\t#Volumetric efficiency in percent\n",
"nm = 74.\t\t\t\t\t#Mechanical efficiency in percent\n",
"R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/kg.K\n",
"Cp = 0.24\t\t\t\t\t#Specific heat at constant pressure n kcal/kg.K\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"BHP = (2*3.14*N*t)/4500\t\t\t\t\t#Brake horse power in B.H.P\n",
"IHP = (BHP/(nm/100))\t\t\t\t\t#Indicated horse power in I.H.P\n",
"pm = ((IHP*4500)/((l/100)*(3.14/4)*d**2*(N/2)*n))\t\t\t\t\t#Mean effective pressure in kg/cm**2\n",
"Vs = (n*(3.14/4)*(d/100)**2*(l/100)*(N/2))\t\t\t\t\t#Swept volume in m**3/min\n",
"Va = (Vs*(nv/100))\t\t\t\t\t#Aspirated Volume of air into engine in m**3/min\n",
"ma = (p*10**4*Va)/(R*T)\t\t\t\t\t#Aspirated mass flow into the engine in kg/min\n",
"mcdT = ((BHP*4500/427)/Cp)\t\t\t\t\t#Product of mass flow rate and change in temperature\n",
"msdT = ((q/60)/Cp)\t\t\t\t\t#Product of mass flow rate and change in temperature\n",
"x = (mcdT/msdT)\t\t\t\t\t#Ratio\n",
"T2 = ((Ta-(x*T)))/(1-x)\t\t\t\t\t#Temperature in K\n",
"mc = (msdT/(T2-T))\t\t\t\t\t#Air flow in kg/min\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) the mean effective pressure is %3.2f kg/cm**2 \\\n",
"\\nb) the air consumption is %3.3f kg/min \\\n",
"\\nc) the air flow into the compressor is %3.2f kg/min'%(pm,ma,mc)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) the mean effective pressure is 6.45 kg/cm**2 \n",
"b) the air consumption is 5.871 kg/min \n",
"c) the air flow into the compressor is 30.14 kg/min\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 18.2 Page no : 328"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"IMEP = 10.\t\t\t\t\t#Indicated mean effective pressure in kg/cm**2\n",
"x = 20.\t\t\t\t\t#Mixture strength 20% richer math.tan chemically correct\n",
"pIMEP = 0.41\t\t\t\t\t#Pumping Indicated mean effective pressure in kg/cm**2\n",
"p1 = 0.97\t\t\t\t\t#Charge pressure at the beginning of compression in kg/cm**2\n",
"T1 = 100.+273\t\t\t\t\t#Charge temperature at the beginning of compression in K\n",
"pm = 0.91\t\t\t\t\t#Mean pressure during the conduction stroke in kg/cm**2\n",
"bn = 70.\t\t\t\t\t#Blower adiabatic efficiency in percent\n",
"T2 = 50.\t\t\t\t\t#Temperature of the charge after delivery by the blower in degree C\n",
"dp = 0.07\t\t\t\t\t#Pressure drop in kg/cm**2\n",
"pi = 1.47\t\t\t\t\t#Charge pressure in the cylinder during the induction stroke in kg/cm**2\n",
"Ta = 15.+273\t\t\t\t\t#Atomspheric temperature in K\n",
"pa = 1.03\t\t\t\t\t#Atmospheric pressure in kg/cm**2\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"T2x = ((((pi/pa)**((g-1)/g)-1)/(bn/100))+1)*Ta+T2\t\t\t\t\t#Temperature in K\n",
"rIMEP = ((pi/pa)*(T1/T2x))\t\t\t\t\t#Ratio of I.M.E.P\n",
"gIMEP = (rIMEP*IMEP)\t\t\t\t\t#Gross I.M.E.P in kg/cm**2\n",
"nsIMEP = (gIMEP+(pi-pa))\t\t\t\t\t#Net I.M.E.P supercharged in kg/cm**2\n",
"nuIMEP = (IMEP-pIMEP)\t\t\t\t\t#Net I.M.E.P unsupercharged in kg/cm**2 \n",
"iIMEP = (nsIMEP-nuIMEP)\t\t\t\t\t#Increase in I.M.E.P in kg/cm**2\n",
"pei = (iIMEP*100)/nuIMEP\t\t\t\t\t#Percentage increase\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Percentage increase in the net I.M.E.P due to supercharging is %3.1f percent'%(pei)\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage increase in the net I.M.E.P due to supercharging is 49.9 percent\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 18.3 Page no : 331"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"l = 4.5\t\t\t\t\t#Capacity in litres\n",
"P = 20.\t\t\t\t\t#Power in H.P per m**3 of free air induced per minute\n",
"N = 1700.\t\t\t\t\t#Speed in r.p.m\n",
"nv = 75.\t\t\t\t\t#Volumetric efficiency in percent\n",
"Ta = 27.+273\t\t\t\t\t#Atomspheric temperature in K\n",
"pa = 1.03\t\t\t\t\t#Atmospheric pressure in kg/cm**2\n",
"pr = 1.75\t\t\t\t\t#Pressure ratio\n",
"ie = 70.\t\t\t\t\t#Isentropic efficiency in percent\n",
"nm = 75.\t\t\t\t\t#Mechanical efficiency in percent\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"nb = 80.\t\t\t\t\t#Efficiency of blower in percent\n",
"R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/kg.K\n",
"Cp = 0.24\t\t\t\t\t#Specific heat at constant pressure in kJ/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",
"Vs = (l/1000*(N/2))\t\t\t\t\t#Swept volume in m**3/min\n",
"uVs = ((nm/100)*Vs)\t\t\t\t\t#Unsupercharged swept volume in m**3/min\n",
"dp = (pr*pa)\t\t\t\t\t#Blower delivery pressure in kg/cm**2\n",
"Tc = (Ta*pr**((g-1)/g))\t\t\t\t\t#Temperature after isentropic compression in K\n",
"dT = (Ta+(Tc-Ta)/(ie/100))\t\t\t\t\t#Blow delivery temperature in K\n",
"Va = (Vs*(dp*Ta)/(pa*dT))\t\t\t\t\t#Equivalent volume at free air condition in m**3/min\n",
"iiv = (Va-uVs)\t\t\t\t\t#Increase in the induced volume in m**3/min\n",
"iIHP = (P*iiv)\t\t\t\t\t#ncrease in I.H.P \n",
"iBHP = (iIHP*(nm/100))\t\t\t\t\t#Increase in B.H.P\n",
"ma = (dp*10**4*Vs)/(R*dT)\t\t\t\t\t#Mass of air delivered by blower in kg/min\n",
"HP = (ma*Cp*(dT-Ta)*J)/(4500*(80./100))\t\t\t\t\t#H.P required for blower\n",
"nibhp = (iBHP-HP)\t\t\t\t\t#Net increse in engine b.h.p\n",
"pei = (nibhp/(P*uVs))*100\t\t\t\t\t#Percentage increase\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Percentage net increase in B.H.P is %3.1f percent'%(pei)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage net increase in B.H.P is 42.1 percent\n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}PKI0ώ&Internal Combustion Engines/ch19.ipynb{
"metadata": {
"name": "",
"signature": "sha256:2e4ee2364140e55a3d90d649c8582a33c5c2d95f57ec4b00aa0dee8989c40b8e"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 19 : Two Stroke Engines"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 19.3 Page no : 334"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 11.25\t\t\t\t\t#Bore in cm\n",
"l = 15.\t\t\t\t\t#Stroke in cm\n",
"r = 7.\t\t\t\t\t#Compression ratio\n",
"N = 1800.\t\t\t\t\t#Speed in r.p.m\n",
"a = 4.5\t\t\t\t\t#Air supply in kg/min\n",
"Ta = 72.+273\t\t\t\t\t#Temperature of air in K\n",
"af = 14.3\t\t\t\t\t#Air fuel ratio\n",
"ep = 1.\t\t\t\t\t#Exhaust pressure in kg/cm**2\n",
"R = 29.27\t\t\t\t\t#Characteristic gas constant in kg.m/kg.degree C\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vc = ((r/(r-1))*(3.14/4)*(d/100)**2*(l/100))\t\t\t\t\t#Swept volume in m**3\n",
"Wa = (Vc*N*ep*10**4)/(R*Ta)\t\t\t\t\t#Ideal air capacity in kg/min\n",
"sr = (a/Wa)\t\t\t\t\t#Scavenging ratio\n",
"sn = (1-math.exp(-sr))\t\t\t\t\t#Scavenging efficiency \n",
"nt = (sn/sr)\t\t\t\t\t#Trapping efficiency\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) Ideal air capacity is %3.2f kg/min \\\n",
"\\nb) Scavenging ratio is %3.2f \\\n",
"\\nc) Scavenging efficiency is %3.3f \\\n",
"\\nd) Trapping efficiency is %3.2f'%(Wa,sr,sn,nt)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) Ideal air capacity is 3.10 kg/min \n",
"b) Scavenging ratio is 1.45 \n",
"c) Scavenging efficiency is 0.766 \n",
"d) Trapping efficiency is 0.53\n"
]
}
],
"prompt_number": 3
}
],
"metadata": {}
}
]
}PKI슿<֢֢&Internal Combustion Engines/ch23.ipynb{
"metadata": {
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"signature": "sha256:0f0f5140414de1792e9d5844fd5e51c3893bcfd2190a5a84bfb8220b6aa3a6fc"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 23 : Testing of Engines"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.1 Page no : 340"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4.\t\t\t\t\t#Four cylinder engine\n",
"N = 1200.\t\t\t\t\t#Speed in r.p.m\n",
"BHP1 = 26.3\t\t\t\t\t#Brake horse power in B.H.P\n",
"T = 11.3\t\t\t\t\t#Average torque in kg\n",
"CV = 10000.\t\t\t\t\t#Calorific value of the fuel in kcal/kg\n",
"m = 270.\t\t\t\t\t#Flow rate in gm of petrol per B.H.P hour\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"BHP2 = (T*2*3.14*N)/4500\t\t\t\t\t#Average B.H.P on 3 cylinders \n",
"IHP = BHP1-BHP2\t\t\t\t\t#Average I.H.P of one cylinder\n",
"TIHP = (n*IHP)\t\t\t\t\t#Total I.H.P\n",
"p = ((m/1000)*BHP1)/TIHP\t\t\t\t\t#Petrol used in kg/I.H.P hr\n",
"nth = ((4500*60)/(427*p*CV))*100\t\t\t\t\t#Indicated Thermal efficiency in percent \n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Thermal efficiency is %3.1f percent'%(nth)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Thermal efficiency is 26.3 percent\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.2 Page no : 344"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"n = 4\t\t\t\t\t#Four cylinder engine\n",
"d = 0.1\t\t\t\t\t#Diameter of piston in m\n",
"l = 0.15\t\t\t\t\t#Stroke in m\n",
"RPM = 1600\t\t\t\t\t#Speed in r.p.m\n",
"ap = (5.76*10**-4)\t\t\t\t\t#Area of positive loop of indicator diagram in sq.m\n",
"an = (0.26*10**-4)\t\t\t\t\t#Area of negative loop of indicator diagram in sq.m\n",
"L = 0.055\t\t\t\t\t#Length of the indicator diagram in m\n",
"k = (3.5/10**-6)\t\t\t\t\t#Spring constant in kg/m**2 per m\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"NA = (ap-an)\t\t\t\t\t#Net area of the indicator diagram in sq.m\n",
"h = (NA/L)\t\t\t\t\t#Average height of diagram in m\n",
"Pm = (h*k)\t\t\t\t\t#Mean effective pressure in kg/m**2\n",
"IHP = (Pm*l*(3.14/4)*d**2*RPM*n)/4500\t\t\t\t\t#Indicated Horse Power\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Indicated horse power of a four cylinder two stroke petrol engine is %3.1f'%(IHP)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Indicated horse power of a four cylinder two stroke petrol engine is 58.6\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.3 Page no : 348"
]
},
{
"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 = 0.089\t\t\t\t\t#Bore in m\n",
"l = 0.1016\t\t\t\t\t#Stroke in m\n",
"vc = 3.183\t\t\t\t\t#Compression ratio\n",
"rn = 55\t\t\t\t\t#Relative efficiency in percent\n",
"m = 0.218\t\t\t\t\t#Petrol consumption in kg/hp.hr\n",
"Pm = (8.4/10**-4)\t\t\t\t\t#Indicated mean effective pressure in kg/m**2\n",
"N = 2500\t\t\t\t\t#Speed in r.p.m\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"an = (1-(1/(vc-1)))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"nth = (rn*an)/100\t\t\t\t\t#Thermal efficiency in percent\n",
"CV = ((4500*60)/(m*(nth/100)*427))\t\t\t\t\t#Calorific value in kcal/kg\n",
"IHP = ((Pm*(3.14/4)*d**2*l*N*n)/(4500*2))\t\t\t\t\t#Indicated horse power\n",
"p = (m*IHP)\t\t\t\t\t#Petrol consumption in kg/hour\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print '1) The calorific value of petrol is %i kcal/kg \\\n",
"\\n2) Corresponding petrol consumption is %3.1f kg/hour'%(CV,p)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"1) The calorific value of petrol is 9731 kcal/kg \n",
"2) Corresponding petrol consumption is 19.3 kg/hour\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.4 Page no : 349"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4.\t\t\t\t\t#Number of cylinders\n",
"d = 0.2\t\t\t\t\t#Bore in m\n",
"l = 0.3\t\t\t\t\t#Stroke in m\n",
"N = 300.\t\t\t\t\t#Speed in r.p.m\n",
"af = 5.\t\t\t\t\t#Air to fuel ratio by volume. In textbook it is given as 4 which is wrong\n",
"nv = 78.\t\t\t\t\t#Volumetric efficiency in percent\n",
"CV = 2200.\t\t\t\t\t#Calorific value in kcal/cu.m at N.T.P\n",
"bth = 23.\t\t\t\t\t#Brake thermal efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = ((3.14/4)*d**2*l)\t\t\t\t\t#Swept volume in cu.m\n",
"c = ((nv/100)*Vs)\t\t\t\t\t#Total charge per stroke in cu.m\n",
"Vg = ((c/af)*N)\t\t\t\t\t#Volume of gas used per min in cu.m at N.T.P\n",
"q = (CV*Vg)\t\t\t\t\t#Heat supplied in kcal/min\n",
"BHP = ((bth/100)*q)/(4500./427)\t\t\t\t\t#Brake horse power\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The volume of gas used per min is %3.3f cu.m at N.T.P \\\n",
"\\nB.H.P of engine is %3.1f'%(Vg,BHP)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The volume of gas used per min is 0.441 cu.m at N.T.P \n",
"B.H.P of engine is 21.2\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.5 Page no : 353"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 20.\t\t\t\t\t#Bore in cm\n",
"l = 37.5\t\t\t\t\t#Stroke in cm\n",
"r = 6.\t\t\t\t\t#Compression ratio\n",
"IPm = 5.\t\t\t\t\t#Indicated Mean effective pressure in kg/cm**2\n",
"ag = 6.\t\t\t\t\t#Air to gas ratio\n",
"CV = 3070.\t\t\t\t\t#Calorific value of gas in kcal/cu.m\n",
"T = 75.+273\t\t\t\t\t#Temperature in K\n",
"p = 0.975\t\t\t\t\t#Pressure in kg/cm**2\n",
"RPM = 240.\t\t\t\t\t#Speed in r.p.m\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = (3.14/4)*d**2*l\t\t\t\t\t#Stroke Volume in cu.m\n",
"Vg = (1/(r+1))*Vs\t\t\t\t\t#Volume of gas in cylinder in cu.m per cycle\n",
"x = (Vg*(p/1.03)*(273/T))\t\t\t\t\t#Volume at 'Vg' cu.m at 'p' kg/cm**2 and 'T' K are equivalent in cu.m\n",
"q = (CV*x)/10**6\t\t\t\t\t#Heat added in kcal per cycle\n",
"IHP = (IPm*(Vs/100)*(RPM/2))/4500\t\t\t\t\t#Indicated horse power\n",
"nth = ((IHP*4500)/(427*q*(RPM/2)))*100\t\t\t\t\t#Thermal efficiency in percent\n",
"na = (1-(1/r**(g-1)))*100\t\t\t\t\t#Air standard efficiency in percent\n",
"rn = (nth/na)*100\t\t\t\t\t#Relative effeciency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The thermal efficiency is %3.1f percent \\\n",
"\\nThe relative efficiency is %3.1f percent \\\n",
"\\nIndicated horese power is %3.1f H.P'%(nth,rn,IHP)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The thermal efficiency is 36.0 percent \n",
"The relative efficiency is 70.3 percent \n",
"Indicated horese power is 15.7 H.P\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.6 Page no : 356"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4\t\t\t\t\t#Number of cylinders\n",
"d = 6.25\t\t\t\t\t#Diametre in cm\n",
"l = 9.5\t\t\t\t\t#Stroke in cm\n",
"t = 678\t\t\t\t\t#Torque in kg.m\n",
"N = 3000\t\t\t\t\t#Speed in r.p.m\n",
"Vc = 60\t\t\t\t\t#Clearance volume in c.c\n",
"be = 0.5\t\t\t\t\t#Brake efficiency ratio based on the air standard cycle\n",
"CV = 10000\t\t\t\t\t#Calorific value in kcal/kg\n",
"g = 1.4\t\t\t\t\t#Ratio of specific heats\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"Vs = (3.14/4)*d**2*l\t\t\t\t\t#Stroke volume in c.c per cylinder\n",
"r = ((Vs+Vc)/Vc)\t\t\t\t\t#Compression ratio\n",
"na = (1-(1/r**(g-1)))\t\t\t\t\t#Air standard efficiency\n",
"bth = (be*na)*100\t\t\t\t\t#Brake thermal efficiency in percent\n",
"bhp = ((t/100)*2*3.14*N)/4500\t\t\t\t\t#B.H.P in H.P\n",
"q = (bhp*(4500/427))/(bth/100)\t\t\t\t\t#Heat supplied in kcal/min\n",
"F = (q*60)/CV\t\t\t\t\t#Fuel consumption in kg/hour\n",
"P = (bhp*4500*2*100)/(n*Vs*N)\t\t\t\t\t#pressure in kg/cm**2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'The fuel consumption is %3.2f kg/hour \\\n",
"\\nThe brake mean effective pressure is %3.2f kg/cm**2'%(F,P)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The fuel consumption is 5.95 kg/hour \n",
"The brake mean effective pressure is 6.47 kg/cm**2\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.7 Page no : 360"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 1.\t\t\t\t\t#Number of cylinders\n",
"t = 30.\t\t\t\t\t#Trail time in min\n",
"m = 5.6\t\t\t\t\t#Oil consumption in l\n",
"CV = 9980.\t\t\t\t\t#Calorific value of oil in kcal/kg\n",
"g = 0.8\t\t\t\t\t#Specific gravity of oil \n",
"a = 8.35\t\t\t\t\t#Average area of indicator diagram in sq.cm\n",
"l = 8.4\t\t\t\t\t#Length of the indicator diagram in cm\n",
"is1 = 5.5\t\t\t\t\t#Indicator spring scale\n",
"L = 147.5\t\t\t\t\t#Brake load in kg\n",
"sp = 20.\t\t\t\t\t#Spring balance reading in kg\n",
"d = 1.5\t\t\t\t\t#Effective brake wheel diameter in m\n",
"N = 200.\t\t\t\t\t#Speed in r.p.m\n",
"cyd = 30.\t\t\t\t\t#Cylinder diameter in cm\n",
"l1 = 45.\t\t\t\t\t#Stroke in cm\n",
"mw = 11.\t\t\t\t\t#Jacket cooling water in kg/min\n",
"Tc = 35.+273\t\t\t\t\t#Temperature rise of cooling water in K\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"mp = (a/l)*is1\t\t\t\t\t#Mean effective pressure\n",
"ihp = ((mp*(l1/100)*(3.14/4)*cyd**2*(N/2))/4500)\t\t\t\t\t#Indicated horse power in h.p\n",
"bhp = (L*3.14*d*N)/4500\t\t\t\t\t#Brake horse power in h.p\n",
"nm = (bhp/ihp)*100\t\t\t\t\t#Mechanical efficiency in percent\n",
"F = (m*(60/t)*g)\t\t\t\t\t#Fuel consumption in kg/hour\n",
"Fc = (F/bhp)\t\t\t\t\t#Specific fuel consumption in kg/B.H.P/hour\n",
"ith = ((ihp*(4500./427))/((F/60)*CV))*100\t\t\t\t\t#Indicated thermal efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) I.H.P is %3.1f \\\n",
"\\nb) B.H.P is %3.1f \\\n",
"\\nc) Mechanical efficiency is %3.1f percent \\\n",
"\\nd) Specific fuel consumption is %3.2f kg/B.H.P/hour \\\n",
"\\ne) Indicated thermal efficiency is %3.1f percent'%(ihp,bhp,nm,Fc,ith)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) I.H.P is 38.6 \n",
"b) B.H.P is 30.9 \n",
"c) Mechanical efficiency is 79.9 percent \n",
"d) Specific fuel consumption is 0.29 kg/B.H.P/hour \n",
"e) Indicated thermal efficiency is 27.3 percent\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.8 Page no : 361"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 15.\t\t\t\t\t#Diameter in cm. In textbook it is given wrong as 39\n",
"l = 45.\t\t\t\t\t#Stroke in cm\n",
"f = 9.5\t\t\t\t\t#Total fuel used in litres\n",
"CV = 10500.\t\t\t\t\t#Calorific value in kcal/kg\n",
"n = 12624.\t\t\t\t\t#Total no. of revolutions\n",
"imep = 7.24\t\t\t\t\t#Gross i.m.e.p in kg/cm**2\n",
"pimep = 0.34\t\t\t\t\t#Pumping i.m.e.p in kg/cm**2\n",
"L = 150.\t\t\t\t\t#Net load on brake in kg\n",
"db = 1.78\t\t\t\t\t#Diameter of the brake wheel drum in m\n",
"dr = 4.\t\t\t\t\t#Diameter of rope in cm\n",
"cw = 545.\t\t\t\t\t#Cooling water circulated in litres\n",
"Tc = 45.\t\t\t\t\t#Cooling water temperature rise in degree C\n",
"g = 0.8\t\t\t\t\t#Specific gravity of oil\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"ihp = ((imep-pimep)*(l/100)*3.14*d**2*n)/(4500*60)\t\t\t\t\t#I.H.P in h.p\n",
"q = (f*g*CV)/60\t\t\t\t\t#Heat supplied in kcal/min\n",
"bhp = (L*3.14*(db+(dr/100))*n)/(4500*60)\t\t\t\t\t#B.H.P in h.p\n",
"qbhp = (bhp*4500)/427\t\t\t\t\t#Heat equivalent of B.H.P in kcal/min\n",
"qw = (cw*Tc)/60\t\t\t\t\t#Heat lost to jacket cooling water in kcal/min\n",
"dq = (q-(qbhp+qw))\t\t\t\t\t#Heat unaccounted in kcal/min\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Heat supplied is %3.0f kcal/min \\\n",
"\\nHeat equivalent of B.H.P is %3.0f kcal/min \\\n",
"\\nHeat lost to jacket cooling water is %3.0f kcal/min \\\n",
"\\nHeat unaccounted is %3.0f kcal/min'%(q,qbhp,qw,dq)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat supplied is 1330 kcal/min \n",
"Heat equivalent of B.H.P is 422 kcal/min \n",
"Heat lost to jacket cooling water is 409 kcal/min \n",
"Heat unaccounted is 499 kcal/min\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.9 Page no : 364"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 27.\t\t\t\t\t#Diameter in cm\n",
"l = 45.\t\t\t\t\t#Stroke in cm\n",
"db = 1.62\t\t\t\t\t#Effective diameter of the brake in m\n",
"t = (38.*60+30)\t\t\t\t\t#Test duration in sec\n",
"CV = 4650.\t\t\t\t\t#Calorific value in kcal/m**3 at N.T.P\n",
"n = 8080.\t\t\t\t\t#Total no. of revolutions\n",
"en = 3230.\t\t\t\t\t#Total number of explosions\n",
"p = 5.75\t\t\t\t\t#Mean effective pressure in kg/cm**2\n",
"V = 7.7\t\t\t\t\t#Gas used in m**3\n",
"T = 15.+273\t\t\t\t\t#Atmospheric temperature in K\n",
"pg = 135.\t\t\t\t\t#pressure of gas in mm of water above atmospheric pressure\n",
"hb = 750.\t\t\t\t\t#Height of barometer in mm of Hg\n",
"L = 92.\t\t\t\t\t#Net load on brake in kg\n",
"w = 183.\t\t\t\t\t#Weigh of jacket cooling water in kg\n",
"Tc = 47.\t\t\t\t\t#Cooling water temperature rise in degree C\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"ihp = (p*(l/100)*(3.14/4)*d**2*en)/(4500*(t/60))\t\t\t\t\t#I.H.P in h.p\n",
"bhp = (L*3.14*db*n)/(4500*(t/60))\t\t\t\t\t#B.H.P in h.p\n",
"pa = (hb+(pg/13))\t\t\t\t\t#Pressure of gas supplied in mm of Hg\n",
"Vg = (V*(273/T)*(pa/760))\t\t\t\t\t#Volume of gas used at N.T.P in m**3\n",
"q = (Vg*CV)/(t/60)\t\t\t\t\t#Heat supplied per minute in kcal\n",
"qbhp = (bhp*4500)/427\t\t\t\t\t#Heat equivalent of B.H.P in kcal/min\n",
"qc = (w/(t/60))*Tc\t\t\t\t\t#Heat lost to jacket cooling water in kcal/min\n",
"qra = (q-(qbhp+qc))\t\t\t\t\t#Heat lost to exhaust, etc in kcal/min\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Heat supplied is %3.1f kcal/min \\\n",
"\\nHeat equivalent of B.H.P is %3.0f kcal/min \\\n",
"\\nHeat lost to jacket cooling water is %3.1f kcal/min \\\n",
"\\nHeat lost to exhaust radiation etc. is %3.1f kcal/min'%(q,qbhp,qc,qra)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat supplied is 882.0 kcal/min \n",
"Heat equivalent of B.H.P is 230 kcal/min \n",
"Heat lost to jacket cooling water is 223.4 kcal/min \n",
"Heat lost to exhaust radiation etc. is 428.6 kcal/min\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.10 Page no : 366"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 25.\t\t\t\t\t#Bore in cm\n",
"l = 50.\t\t\t\t\t#Stroke in cm\n",
"N = 240.\t\t\t\t\t#Speed in r.p.m\n",
"n = 100.\t\t\t\t\t#Number of times fires per minute\n",
"qc = 0.3\t\t\t\t\t#Quantity of coal gas used in cu.m per minute\n",
"h = 100.\t\t\t\t\t#Head in mm of water\n",
"bp = 1.03\t\t\t\t\t#Barometric pressure in kg/cm**2\n",
"T = 15.+273\t\t\t\t\t#Temperature in K\n",
"ma = 2.82\t\t\t\t\t#Mass of air used in kg per minute\n",
"R = 29.45\t\t\t\t\t#Characteristic gas constant in kg.m/kg.K\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"gp = (bp+(100/13.6)*(bp/76))\t\t\t\t\t#Gas pressure in kg/cm**2\n",
"Vc = (qc*(gp/bp)*(273/T))\t\t\t\t\t#Volume of coal gas at N.T.P in cu.m per minute\n",
"Vce = (Vc/n)\t\t\t\t\t#Volume of coal gas per explosion in cu.m at N.T.P\n",
"va = (ma*R*273)/(bp*10**4)\t\t\t\t\t#Volume of air taken in at N.T.P in cu.m per min\n",
"V = ((va-(((N/2)-n)*Vce))/(N/2))\t\t\t\t\t#Volume in cu.m\n",
"tV = (V+Vce)\t\t\t\t\t#Total volume of charge in cu.m at N.T.P\n",
"Vs = ((3.14/4)*(d**2*l)*10**-6)\t\t\t\t\t#Swept volume in cu.m\n",
"nv = (tV/Vs)*100\t\t\t\t\t#Volumetric efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'a) the charge of air per working cycle as measured at N.T.P is %3.5f cu.m \\\n",
"\\nb) the volumetric efficiency is %3.1f percent'%(tV,nv)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"a) the charge of air per working cycle as measured at N.T.P is 0.02094 cu.m \n",
"b) the volumetric efficiency is 85.4 percent\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.11 Page no : 368"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 18.\t\t\t\t\t#Diameter in cm\n",
"l = 24.\t\t\t\t\t#Stroke in cm\n",
"t = 30.\t\t\t\t\t#Duration of trail in min\n",
"r = 9000.\t\t\t\t\t#Total number of revolutins\n",
"e = 4445.\t\t\t\t\t#Total number of explosions\n",
"mep = 5.85\t\t\t\t\t#Mean effective pressure in kg/cm**2\n",
"Nl = 40.\t\t\t\t\t#Net load on brake wheel in kg\n",
"ed = 1.\t\t\t\t\t#Effective diameter of brake wheel in meter\n",
"tg = 2.3\t\t\t\t\t#Total gas used at N.T.P in m**3\n",
"CV = 4600.\t\t\t\t\t#Calorific value of gas in kcal/m**3 at N.T.P\n",
"ta = 36.\t\t\t\t\t#Total air used in m**3\n",
"pa = 720.\t\t\t\t\t#Pressure of air in mm of Hg\n",
"Ta = 18.+273\t\t\t\t\t#Temperature of air in K\n",
"da = 1.293\t\t\t\t\t#Density of air at N.T.P in kg/m**3\n",
"Te = 350.+273\t\t\t\t\t#Temperature of exhaust gases in K\n",
"Tr = 18.+273\t\t\t\t\t#Room temperature in K\n",
"Cp = 0.24\t\t\t\t\t#Specific heat of exhaust gases in kJ/kg.K\n",
"twc = 81.5\t\t\t\t\t#Total weight of cylinder jacket cooling water in kg\n",
"dT = 33.\t\t\t\t\t#Rise in temperature of jacket cooling water in degree C\n",
"R = 29.45\t\t\t\t\t#Characteristic gas constant in kg.m/kg.degree C\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"ihp = (mep*(l/100)*(3.14/4)*d**2*(e/t))/4500\t\t\t\t\t#Indicated horse power in h.p\n",
"bhp = (Nl*3.14*r*ed)/(4500*t)\t\t\t\t\t#Brake horse power in h.p\n",
"qs = (tg/t)*CV\t\t\t\t\t#Heat supplied at N.T.P in kcal\n",
"qbhp = (bhp*4500)/427\t\t\t\t\t#Heat equivalent of B.H.P in kcal/min\n",
"ql = (twc/t)*dT\t\t\t\t\t#Heat lost to cylinder jacket cooling water in kcal/min\n",
"VA = (ta*(273/Ta)*(pa/760))\t\t\t\t\t#Volume of air used at N.T.P in m**3\n",
"WA = (VA*da)/t\t\t\t\t\t#Weight of air used per min in kg\n",
"WG = (1.03*tg*10**4)/(R*273)\t\t\t\t\t#Weight of gas in kg\n",
"Wg = (WG/t)\t\t\t\t\t#Weight of gas per minute in kg\n",
"We = (WA+Wg)\t\t\t\t\t#Total weight of exhaust gases in kg\n",
"qle = (We*(Te-Tr)*Cp)\t\t\t\t\t#Heat lost of exhaust gases in kcal/min\n",
"qra = (qs-(qbhp+ql+qle))\t\t\t\t\t#Heat lost by radiation in kcal/min\n",
"nm = (bhp/ihp)*100\t\t\t\t\t#Mechanical efficiency in percent\n",
"ith = ((ihp*4500)/(427*qs))*100\t\t\t\t\t#Indicated thermal efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print ' HEAT BALANCE SHEET ------------------ \\\n",
"\\nHeat supplied per minute is %3.1f kcal/min \\\n",
"\\nHeat expenditure kcal per minute \\\n",
"\\n1.Heat equivalent of B.H.P is %3.1f \\\n",
"\\n2.Heat lost to jacket cooling water is %3.1f \\\n",
"\\n3.Heat lost in exhaust gases is %3.1f \\\n",
"\\n4.Heat lost by radiation, etc, is %3.1f \\\n",
"\\n-------- %3.1f --------'%(qs,qbhp,ql,qle,qra,qs)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" HEAT BALANCE SHEET ------------------ \n",
"Heat supplied per minute is 352.7 kcal/min \n",
"Heat expenditure kcal per minute \n",
"1.Heat equivalent of B.H.P is 88.2 \n",
"2.Heat lost to jacket cooling water is 89.7 \n",
"3.Heat lost in exhaust gases is 117.7 \n",
"4.Heat lost by radiation, etc, is 57.1 \n",
"-------- 352.7 --------\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.12 Page no : 372"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"gu = 0.135\t\t\t\t\t#Gas used in m**3/min at N.T.P\n",
"CV = 3990\t\t\t\t\t#Calorific value of gas in kcal/m**3 at N.T.P\n",
"dg = 0.64\t\t\t\t\t#Density of gas in kg/m**3 at N.T.P\n",
"au = 1.52\t\t\t\t\t#Air used in kg/min\n",
"C = 0.24\t\t\t\t\t#Specific heat of exhaust gases in kJ/kg.K\n",
"Te = 397+273\t\t\t\t\t#Temperature of exhaust gases in K\n",
"Tr = 17+273\t\t\t\t\t#Room temperature in K\n",
"cw = 6\t\t\t\t\t#Cooling water per minute in kg\n",
"rT = 27.5\t\t\t\t\t#Rise in temperature in degree C\n",
"ihp = 12.3\t\t\t\t\t#Indicated horse power in h.p\n",
"bhp = 10.2\t\t\t\t\t#Brake horse power in h.p\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"qs = (gu*CV*60)\t\t\t\t\t#Heat supplied in kcal/hour\n",
"qbhp = ((bhp*4500*60)/427)\t\t\t\t\t#Heat equivalent of B.H.P in kcal/hr\n",
"ql = (cw*60*rT)\t\t\t\t\t#Heat lost in jacket cooling water in kcal/hr\n",
"mg = (gu*dg)\t\t\t\t\t#Mass of gas used per minute in kg\n",
"me = (mg+au)\t\t\t\t\t#Mass of exhaust gases per minute in kg\n",
"qe = (me*C*(Te-Tr)*60)\t\t\t\t\t#Heat carried away by exhaust gases in kcal/hour\n",
"qun = (qs-(qbhp+ql+qe))\t\t\t\t\t#Heat unaccounted in kcal/hour\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Heat supplied is %3.0f kcal/hour \\\n",
"\\nHeat equivalent of B.H.P is %3.0f kcal/hr \\\n",
"\\nHeat lost in jacket cooling water is %3.0f kcal/hr \\\n",
"\\nHeat carried away by exhaust gases is %3.0f kcal/hour \\\n",
"\\nHeat unaccounted is %3.0f kcal/hour'%(qs,qbhp,ql,qe,qun)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat supplied is 32319 kcal/hour \n",
"Heat equivalent of B.H.P is 6450 kcal/hr \n",
"Heat lost in jacket cooling water is 9900 kcal/hr \n",
"Heat carried away by exhaust gases is 8790 kcal/hour \n",
"Heat unaccounted is 7179 kcal/hour\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.13 Page no : 372"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"n = 4.\t\t\t\t\t#Number of cylinders\n",
"r = 1.\t\t\t\t\t#Radius in metre\n",
"N = 1400.\t\t\t\t\t#Speed in r.p.m\n",
"bl = 14.5\t\t\t\t\t#Net brake load in kg\n",
"P = [9.8,10.1,10.3,10]\t\t\t\t\t#Loads on the brake in kg\n",
"d = 9.\t\t\t\t\t#Bore in cm\n",
"l = 12.\t\t\t\t\t#Stroke in cm\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"bhp = (bl*2*3.14*r*N)/4500\t\t\t\t\t#Brake horse power in h.p\n",
"bhp1 = (P[0]*2*3.14*r*N)/4500\t\t\t\t\t#Brake horse power in h.p\n",
"bhp2 = (P[1]*2*3.14*r*N)/4500\t\t\t\t\t#Brake horse power in h.p\n",
"bhp3 = (P[2]*2*3.14*r*N)/4500\t\t\t\t\t#Brake horse power in h.p\n",
"bhp4 = (P[3]*2*3.14*r*N)/4500\t\t\t\t\t#Brake horse power in h.p\n",
"ihp1 = bhp-bhp1\t\t\t\t\t#Indicated horse power in h.p\n",
"ihp2 = bhp-bhp2\t\t\t\t\t#Indicated horse power in h.p\n",
"ihp3 = bhp-bhp3\t\t\t\t\t#Indicated horse power in h.p\n",
"ihp4 = bhp-bhp4\t\t\t\t\t#Indicated horse power in h.p\n",
"ihp = (ihp1+ihp2+ihp3+ihp4)\t\t\t\t\t#Indicated horse power in h.p\n",
"nm = (bhp/ihp)*100\t\t\t\t\t#Mechanical efficiency in percent\n",
"pm = ((4500*bhp)/((l/100)*(3.14/4)*d**2*(N/2)))\t\t\t\t\t#Brake mean effective pressure in kg/cm**2\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'I.H.P is %3.1f h.p \\\n",
"\\nMechanical efficiency is %3.1f percent \\\n",
"\\nBrake mean effective pressure is %3.0f kg/cm**2'%(ihp,nm,pm)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"I.H.P is 34.8 h.p \n",
"Mechanical efficiency is 81.5 percent \n",
"Brake mean effective pressure is 24 kg/cm**2\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.14 Page no : 374"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"N = 350\t\t\t\t\t#Speed in r.p.m\n",
"L = 60\t\t\t\t\t#Net brake load in kg\n",
"mep = 2.75\t\t\t\t\t#Mean effective pressure in kg/cm**2\n",
"oc = 4.25\t\t\t\t\t#Oil consumption in kg/hour\n",
"jcw = 490\t\t\t\t\t#Jacket cooling water in kg/hour\n",
"Tw = [20+273,45+273]\t\t\t\t\t#Temperature of jacket water at inlet and outlet in K\n",
"au = 31.5\t\t\t\t\t#Air used per kg of oil in kg\n",
"Ta = 20+273\t\t\t\t\t#Temperature of air in the test room in K\n",
"Te = 390+273\t\t\t\t\t#Temperature of exhaust gases in K\n",
"d = 22\t\t\t\t\t#Cylinder diameter in cm\n",
"l = 28\t\t\t\t\t#Stroke in cm\n",
"bd = 1\t\t\t\t\t#Effective brake diameter in m\n",
"CV = 10500\t\t\t\t\t#Calorific value of oil in kcal/kg\n",
"pH2 = 15\t\t\t\t\t#Proportion of hydrogen in fuel oil in percent\n",
"C = 0.24\t\t\t\t\t#Mean specific heat of dry exhaust gases\n",
"Cs = 9.5\t\t\t\t\t#Specific heat of steam in kJ/kg.K\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"ibp = (mep*(l/100)*(3.14/4)*d**2*N)/4500\t\t\t\t\t#Indicated brake power in h.p\n",
"bhp = (L*3.14*N*bd)/4500\t\t\t\t\t#Brake horse power in h.p\n",
"qs = (oc*CV)/60\t\t\t\t\t#Heat supplied per minute in kcal\n",
"qbhp = (bhp*4500)/427\t\t\t\t\t#Heat equivalent of B.H.P in kcal/min\n",
"pqbhp = (qbhp/qs)*100\t\t\t\t\t#Percenatge of heat\n",
"ql = (jcw/60)*(Tw[1]-Tw[0])\t\t\t\t\t#Heat lost to cooling water in kcal/min\n",
"pql = (ql/qs)*100\t\t\t\t\t#Percenatge of heat\n",
"wH2O = (9*(pH2/100)*(oc/60))\t\t\t\t\t#Weight of H2O produced per kg of fuel burnt in kg/min\n",
"twe = (oc*(au+1))/60\t\t\t\t\t#Total weight of wet exhaust gases per minute in kg\n",
"twd = (twe-wH2O)\t\t\t\t\t#Weight of dry exhaust gases per minute in kg\n",
"qle = (twd*C*(Te-Ta))\t\t\t\t\t#Heat lost to dry exhaust gases/min in kcal\n",
"pqle = (qle/qs)*100\t\t\t\t\t#Percenatge of heat\n",
"qx = (100+538.9+0.5*(Te-373))\t\t\t\t\t#Heat in kcal/kg\n",
"qst = (wH2O*qx)\t\t\t\t\t#Heat to steam in kcal/min\n",
"pqst = (qst/qs)*100\t\t\t\t\t#Percenatge of heat\n",
"qra = (qs-(qbhp+ql+qle+qst))\t\t\t\t\t#Heat lost by radiation in kcal/min\n",
"pqra = (qra/qs)*100\t\t\t\t\t#Percenatge of heat\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print ' HEAT BALANCE SHEET ------------------ \\\n",
"\\nHeat supplied per minute is %3.0f kcal/min 100 percent \\\n",
"\\nHeat expenditure kcal per minute percent \\\n",
"\\n1.Heat equivalent of B.H.P is %3.1f %3.1f \\\n",
"\\n2.Heat lost to cooling water is %3.0f %3.1f \\\n",
"\\n3.Heat lost to dry exhaust gases is %3.1f %3.1f \\\n",
"\\n4.Heat lost of steam in exhaust gases is %3.0f %3.1f \\\n",
"\\n5.Heat lost by radiation, etc., is %3.0f %3.1f \\\n",
"\\n ---------- Total %3.0f %3.0f ------------------'\\\n",
"%(qs,qbhp,pqbhp,ql,pql,qle,pqle,qst,pqst,qra,pqra,qs,(pqbhp+pql+pqle+pqst+pqra))\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
" HEAT BALANCE SHEET ------------------ \n",
"Heat supplied per minute is 744 kcal/min 100 percent \n",
"Heat expenditure kcal per minute percent \n",
"1.Heat equivalent of B.H.P is 154.4 20.8 \n",
"2.Heat lost to cooling water is 200 26.9 \n",
"3.Heat lost to dry exhaust gases is 204.4 27.5 \n",
"4.Heat lost of steam in exhaust gases is 0 0.0 \n",
"5.Heat lost by radiation, etc., is 185 24.9 \n",
" ---------- Total 744 100 ------------------\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.15 Page no : 376"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"d = 20\t\t\t\t\t#Diameter in cm\n",
"l = 40\t\t\t\t\t#Stroke in cm\n",
"mep = 5.95\t\t\t\t\t#Mean effective pressure in kg/cm**2\n",
"bt = 41.5\t\t\t\t\t#Brake torque in kg.m\n",
"N = 250\t\t\t\t\t#Speed in r.p.m\n",
"oc = 4.2\t\t\t\t\t#Oil consumption in kg per hour\n",
"CV = 11300\t\t\t\t\t#Calorific value of fuel in kcal/kg\n",
"jcw = 4.5\t\t\t\t\t#Jacket cooling water in kg/min\n",
"rT = 45\t\t\t\t\t#Rise in temperature in degree C\n",
"au = 31\t\t\t\t\t#Air used in kg\n",
"Te = 420\t\t\t\t\t#Temperature of exhaust gases in degree C\n",
"Tr = 20\t\t\t\t\t#Room temperature in degree C\n",
"Cm = 0.24\t\t\t\t\t#Mean specific heat of exhaust gases in kJ/kg.K\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"ihp = (mep*(l/100)*(3.14/4)*d**2*(N/2))/4500\t\t\t\t\t#Indicated horse power in h.p\n",
"bhp = (bt*2*3.14*N)/4500\t\t\t\t\t#Brake horse power in h.p\n",
"q = (oc*CV)\t\t\t\t\t#Heat supplied in kcal/hour\n",
"qbhp = (bhp*4500*60)/427\t\t\t\t\t#Heat equivalent of B.H.P in kcal/hour\n",
"qfhp = ((ihp-bhp)*4500*60)/427\t\t\t\t\t#Heat equivalent F.H.P in kcal/hour\n",
"qc = (jcw*rT*60)\t\t\t\t\t#Heat lost in cooling water in kcal/hour\n",
"qe = (oc*32*Cm*(Te-Tr))\t\t\t\t\t#Heat lost in exhaust gases in kcal/hour\n",
"hu = (q-(qbhp+qfhp+qc+qe))\t\t\t\t\t#Heat unaccounted in kcal/hour\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Indicated horse power is %3.1f h.p \\\n",
"\\nBrake horse power is %3.2f h.p \\\n",
"\\nHeat supplied is %3.0f kcal/hour \\\n",
"\\nHeat equivalent of B.H.P is %3.0f kcal/hour \\\n",
"\\nHeat equivalent of F.H.P is %3.0f kcal/hour \\\n",
"\\nHeat lost in cooling water is %3.0f kcal/hour \\\n",
"\\nHeat lost in exhaust gases is %3.0f kcal/hour \\\n",
"\\nHeat unaccounted is %3.0f kcal/hour'%(ihp,bhp,q,qbhp,qfhp,qc,qe,hu)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Indicated horse power is 0.0 h.p \n",
"Brake horse power is 14.48 h.p \n",
"Heat supplied is 47460 kcal/hour \n",
"Heat equivalent of B.H.P is 9155 kcal/hour \n",
"Heat equivalent of F.H.P is -9155 kcal/hour \n",
"Heat lost in cooling water is 12150 kcal/hour \n",
"Heat lost in exhaust gases is 12902 kcal/hour \n",
"Heat unaccounted is 22408 kcal/hour\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.16 Page no : 381"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"ihp = 45.\t\t\t\t\t#Indicated horse power in h.p\n",
"bhp = 37.\t\t\t\t\t#Brake horse power in h.p\n",
"fu = 8.4\t\t\t\t\t#Fuel used in kg/hour\n",
"CV = 10000.\t\t\t\t\t#Calorific value in kcal/kg\n",
"Tc = [15.,70.]\t\t\t\t\t#Inlet and outlet temperatures of cylinders in degree C\n",
"cj = 7.\t\t\t\t\t#Rate of flow of cylinder jacket in kg/min\n",
"Tw = [15.,55.]\t\t\t\t\t#Inlet and outlet temperatures of water in degree C\n",
"rw = 12.5\t\t\t\t\t#Rate of water flow in kg per minute\n",
"Te = 82.\t\t\t\t\t#Final temperature of exhaust gases in degree C\n",
"Tr = 17.\t\t\t\t\t#Room temperature in degree C\n",
"af = 20.\t\t\t\t\t#Air fuel ratio\n",
"Cm = 0.24\t\t\t\t\t#Mean specific heat of exhaust gases in kJ/kg.K\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"q = (fu/60)*CV\t\t\t\t\t#Heat supplied in kcal/min\n",
"qbhp = (bhp*4500)/427\t\t\t\t\t#Heat equivalent of B.H.P in kcal/min\n",
"ql = (cj*(Tc[1]-Tc[0]))\t\t\t\t\t#Heat lost to cylinder jacket cooling water in kcal/min\n",
"qe = (rw*(Tw[1]-Tw[0]))\t\t\t\t\t#Heat lost by exhaust gases in kcal/min\n",
"qee = (Te-Tr)*Cm*(af+1)*fu/60\t\t\t\t\t#Heat of exhaust gas in kcal/min\n",
"te = (qe+qee)\t\t\t\t\t#Total heat lost to exhaust gases in kcal/min\n",
"hra = (q-(qbhp+ql+te))\t\t\t\t\t#Heat lost to radiation in kcal/min\n",
"ith = ((ihp*4500)/(427*q))*100\t\t\t\t\t#Indicated thermal efficiency in percent\n",
"bth = ((bhp*4500)/(427*q))*100\t\t\t\t\t#Brake thermal efficiency in percent\n",
"nm = (bhp/ihp)*100\t\t\t\t\t#Mechanical efficiency in percent\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Heat supplied is %3.0f kcal/min \\\n",
"\\nHeat equivalent of B.H.P is %3.0f kcal/min \\\n",
"\\nHeat lost to cylinder jacket cooling water is %3.0f kcal/min \\\n",
"\\nTotal heat lost to exhaust gases is %3.1f kcal/min \\\n",
"\\nHeat lost to radiation is %3.1f kcal/min \\\n",
"\\nIndicated thermal efficiency is %3.1f percent \\\n",
"\\nBrake thermal efficiency is %3.1f percent \\\n",
"\\nMechanical efficiency is %3.1f percent'%(q,qbhp,ql,te,hra,ith,bth,nm)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat supplied is 1400 kcal/min \n",
"Heat equivalent of B.H.P is 390 kcal/min \n",
"Heat lost to cylinder jacket cooling water is 385 kcal/min \n",
"Total heat lost to exhaust gases is 545.9 kcal/min \n",
"Heat lost to radiation is 79.2 kcal/min \n",
"Indicated thermal efficiency is 33.9 percent \n",
"Brake thermal efficiency is 27.9 percent \n",
"Mechanical efficiency is 82.2 percent\n"
]
}
],
"prompt_number": 26
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 23.17 Page no : 382"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math \n",
"\t\t\t\t\t\n",
"#Input data\n",
"Vs = 0.0015\t\t\t\t\t#Stroke volume in cu.m\n",
"rc = 5.5\t\t\t\t\t#Volume compression ratio\n",
"p2 = 8.\t\t\t\t\t#Pressure at the end of compression stroke in kg/cm**2\n",
"T2 = 350.+273\t\t\t\t\t#Temperature at the end of compression stroke in K\n",
"p3 = 25.\t\t\t\t\t#Pressure in kg/cm**2\n",
"x = (1./30)\t\t\t\t\t#Fraction of dismath.tance travelled by piston\n",
"pa = 1./16\t\t\t\t\t#Petrol air mixture ratio\n",
"R = 29.45\t\t\t\t\t#Characteristic gas constant in kg.m/kg degree C\n",
"CV = 10000.\t\t\t\t\t#Calorific value of fuel in kcal per kg\n",
"Cv = 0.23\t\t\t\t\t#Specific heat in kJ/kg.K\n",
"\n",
"\t\t\t\t\t\n",
"#Calculations\n",
"V2 = (Vs*10**6)/(rc-1)\t\t\t\t\t#Volume in c.c\n",
"V3 = (Vs*10**6)*x+V2\t\t\t\t\t#Volume in c.c\n",
"T3 = (T2*p3*V3)/(p2*V2)\t\t\t\t\t#Temperature in K\n",
"W = ((p3+p2)/2)*(V3-V2)\t\t\t\t\t#Workdone in kg.cm\n",
"mM = ((p2*V2)/(T2*R*100))\t\t\t\t\t#Mass of mixture present in kg\n",
"dE = (mM*Cv*(T3-T2))\t\t\t\t\t#Change in energy in kcal\n",
"q = (dE+(W/(427*100)))\t\t\t\t\t#Heat in kcal\n",
"qc = (1/(1+(1/pa)))*mM*CV \t\t\t\t\t#Heat in kcal\n",
"ql = (qc-q)/mM\t\t\t\t\t#Heat lost in kcal per kg of charge\n",
"\n",
"\t\t\t\t\t\n",
"#Output\n",
"print 'Heat lost per kg of charge during explosion is %3.0f kcal'%(ql)\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat lost per kg of charge during explosion is 203 kcal\n"
]
}
],
"prompt_number": 28
}
],
"metadata": {}
}
]
}PKI