{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 13: Fuel and Combustions"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 1, Page No: 350"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"C = 84 # Percentage\n",
"S = 1.5 # Percentage\n",
"N = 0.6 # Percentage\n",
"H = 5.5 # Percentage\n",
"O = 8.4 # Percentage\n",
"\n",
"# Solution\n",
"GCV = (8080 * C + 34500 * (H - O / 8) + 2240 * S) / 100\n",
"LCV = (GCV - 9 * H / 100 * 587)\n",
"print \"Gross Calorific Value\", int(GCV), \"kcal / kg\"\n",
"print \"Net Calorific Value\", \"{:.2f}\".format(LCV), \"kcal / kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Gross Calorific Value 8356 kcal / kg\n",
"Net Calorific Value 8065.48 kcal / kg\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 2, Page No: 350"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"C = 90 # Percentage\n",
"O = 3.0 # Percentage\n",
"S = 0.5 # Percentage\n",
"N = 0.5 # Percentage\n",
"ash = 2.5 # Percentage\n",
"LCV = 8490.5 # kcal / kg\n",
"\n",
"# Solution\n",
"print \"HCV = LCV + 9 * H / 100 * 587\"\n",
"print \"HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)\"\n",
"H = (8490.5 - 7754.8) / (345 - 52.8)\n",
"H = 4.575\n",
"print \"The precentage of H is\", H, \"%\"\n",
"HCV = LCV + 52.8 * H\n",
"print \"Higeher calorific value of coal\", \"{:.1f}\".format(HCV), \"kcal / kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"HCV = LCV + 9 * H / 100 * 587\n",
"HCV = 1/100 * (8080 * C + 34500 * (H - O / 8) + 2240 * N)\n",
"The precentage of H is 4.575 %\n",
"Higeher calorific value of coal 8732.1 kcal / kg\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 3, Page No: 351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"x = 0.72 # g\n",
"W = 250 # g\n",
"w = 150 # g\n",
"t1 = 27.3 # C\n",
"t2 = 29.1 # C\n",
"\n",
"# Solution\n",
"HCV = ((W + w) * (t2 - t1)) / x\n",
"HCV *= 4185.0 / 10 ** 6\n",
"print \"HCV of fuel is\", \"{:.3f}\".format(HCV), \"KJ / Kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"HCV of fuel is 4.185 KJ / Kg\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 4, Page No: 351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"x = 0.84 # g\n",
"W = 1060 # g\n",
"w = 135 # g\n",
"delta_t = 2.5 # C\n",
"\n",
"# Solution\n",
"HCV = ((W + w) * delta_t) / x\n",
"print \"HCV of fuel is\", \"{:.2f}\".format(HCV), \"kcal / kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"HCV of fuel is 3556.55 kcal / kg\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 5, Page No: 351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"V = 0.1 # m ^ 3\n",
"W = 25 # kg\n",
"t1 = 20 # C\n",
"t2 = 33 # C\n",
"m = 0.025 # kg\n",
"\n",
"# Solution\n",
"HCV = W * (t2 - t1) / V\n",
"LCV = HCV - (m / V) * 580\n",
"print \"HCV is\", HCV, \"kcal / m^3\"\n",
"print \"LCV is\", LCV, \"kcal / m^3\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"HCV is 3250.0 kcal / m^3\n",
"LCV is 3105.0 kcal / m^3\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 6, Page No: 351"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"w1 = 2.5 # g\n",
"w2 = 2.415 # g\n",
"r = 1.528 # g\n",
"ma = 0.245 # Mass of ash, g\n",
"\n",
"# Solution\n",
"m = w1 - w2 # Mass of moisture in coal\n",
"mv = w2 - r # Mass of volatile matter\n",
"moisp = m * 100 / w1\n",
"volp = mv * 100 / w1\n",
"ashp = ma * 100 / w1\n",
"carbp = 100 - (moisp + volp + ashp)\n",
"print \"Percentage of moisture:\", moisp, \"%\"\n",
"print \"Percentage of volatile matter:\", volp, \"%\"\n",
"print \"Percentage of ash:\", ashp, \"%\"\n",
"print \"Percentage of fixed carbon:\", carbp, \"%\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Percentage of moisture: 3.4 %\n",
"Percentage of volatile matter: 35.48 %\n",
"Percentage of ash: 9.8 %\n",
"Percentage of fixed carbon: 51.32 %\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 7, Page No: 352"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"wt_coke = 2 # kg\n",
"\n",
"# Solution\n",
"wt_O = 2 * 32 / 12.0\n",
"wt_air = wt_O * 100 / 23.2\n",
"Vol_air = wt_air / 28.94 * 22.4\n",
"print \"Volume of air needed for the complete combustion of 2kg coke\",\n",
"print \"is\", \"{:.3f}\".format(Vol_air), \"litres at NTP\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Volume of air needed for the complete combustion of 2kg coke is 17.793 litres at NTP\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 8, Page No: 352"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"C = 86 # Percent\n",
"H = 4 # Percent\n",
"N = 1.3 # Percent\n",
"S = 3 # Percent\n",
"O = 4 # Percent\n",
"Ash = 1.7 # Percent\n",
"wt = 500 # g\n",
"\n",
"# Solution\n",
"wt_C = C / 100.0\n",
"wt_S = S / 100.0\n",
"wt_H = H / 100.0\n",
"wt_O = O / 100.0\n",
"\n",
"print \"Nitrogen and ash are incombustible, so they do not require oxygen.\"\n",
"wt_O_C = 32 / 12.0 * wt_C\n",
"wt_O_S = 32 / 32.0 * wt_S\n",
"wt_O_H = 32 / 4.0 * wt_H\n",
"\n",
"Twt_O = wt_O_H + wt_O_S + wt_O_C\n",
"wt_O_needed = Twt_O - wt_O\n",
"wt_air = (100.0 / 23.0 * wt_O_needed) * 500 / 1000.0\n",
"print \"Minimum Wt. of air required by 500g of fuel\", \"{:.2f}\".format(wt_air), \"kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Nitrogen and ash are incombustible, so they do not require oxygen.\n",
"Minimum Wt. of air required by 500g of fuel 5.66 kg\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 9, Page No: 353"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"wt_C = 3 # kg\n",
"\n",
"# Solution\n",
"wt_air = wt_C * 32 * 100 / 12.0 / 23.0\n",
"vol_air = wt_air * 1000 * 22.4 / 28.94\n",
"\n",
"print \"H2(g) + 1/2 O2(g) --> H20(l)\"\n",
"print \" 1 0.5 1\\t\\t(By Vol.)\"\n",
"print \"CO(g) + 1/2 O2(g) --> CO2(g)\"\n",
"print \" 1 0.5 1\\t\\t(By Vol.)\"\n",
"print \"CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)\"\n",
"print \" 1 2 1\\t\\t(By Vol.)\"\n",
"\n",
"print \"Weight of air for the combustion of 3kg carbon\",\n",
"print \"{:.3f}\".format(wt_air), \"kg\"\n",
"print \"Vol. pf air required for combustion of 3kg carbon\",\n",
"print \"{:.3e}\".format(vol_air), \"L\",\n",
"print \"or\", \"{:.2f}\".format(vol_air / 1000), \"m^3\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"H2(g) + 1/2 O2(g) --> H20(l)\n",
" 1 0.5 1\t\t(By Vol.)\n",
"CO(g) + 1/2 O2(g) --> CO2(g)\n",
" 1 0.5 1\t\t(By Vol.)\n",
"CH4(g) + 2 O2(g) --> CO2(g) + 2H2O(l)\n",
" 1 2 1\t\t(By Vol.)\n",
"Weight of air for the combustion of 3kg carbon 34.783 kg\n",
"Vol. pf air required for combustion of 3kg carbon 2.692e+04 L or 26.92 m^3\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 10, Page No: 353"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"H = 0.30 # m^3\n",
"CO = 0.10 # m^3\n",
"CH4 = 0.04 # m^3\n",
"N2 = 0.56 # m^3\n",
"\n",
"# Soution\n",
"vol_oxygen = H * 0.5 + CO * 0.5 + CH4 * 2\n",
"vol_air = vol_oxygen * 100 / 21\n",
"print \"Volumer of air required for complete combustion of 1 m^3 of\",\n",
"print \"producer gas:\", \"{:.3f}\".format(vol_air), \"m^3\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Volumer of air required for complete combustion of 1 m^3 of producer gas: 1.333 m^3\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 11, Page No: 354"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"H = 15.4 # Percentage\n",
"C = 84.6 # Percentage\n",
"wt_fuel = 1 # kg\n",
"wt_C = 0.846 # kg\n",
"wt_H = 0.154 # kg\n",
"\n",
"# Solution\n",
"print \"The combustion reactions are,\"\n",
"print \"C + O2 --> CO2\"\n",
"print \"12 32 \\t(by Weight)\"\n",
"print \"2H2 + O2 --> H20\"\n",
"print \" 4 32\\t(by Weight)\"\n",
"\n",
"wt_O = 32 / 12.0 * wt_C\n",
"wt_O_H = 32 / 4.0 * wt_H\n",
"Twt_O = wt_O + wt_O_H\n",
"print \"Because 32 gm of O2 occupies a volume of 22.4 liters at NTP\"\n",
"print \"3.488 * 1000 gm of O2 will occupy\",\n",
"print \"{:.1f}\".format(22.4 / 32 * Twt_O * 1000), \"liters\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The combustion reactions are,\n",
"C + O2 --> CO2\n",
"12 32 \t(by Weight)\n",
"2H2 + O2 --> H20\n",
" 4 32\t(by Weight)\n",
"Because 32 gm of O2 occupies a volume of 22.4 liters at NTP\n",
"3.488 * 1000 gm of O2 will occupy 2441.6 liters\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 12, Page No: 354"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"C = 750 # g\n",
"H = 52 # g\n",
"O = 121 # g\n",
"N = 32 # g\n",
"ash = 45 # g\n",
"\n",
"# Solution\n",
"min_wt_air = (C * 32 / 12. + H * 16 / 2. - O) * 100 / 23.\n",
"HCV = 1 / 1000. * (8080 * C + 34500 * (H - O / 8.) + 2240 * 0)\n",
"LCV = HCV - 0.09 * H * 587 / 10.0\n",
"\n",
"print \"HCV is\", int(HCV), \"kcal/kg\"\n",
"print \"LCV is\", int(LCV), \"kcal/kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"HCV is 7332 kcal/kg\n",
"LCV is 7057 kcal/kg\n"
]
}
],
"prompt_number": 25
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 13, Page No: 355"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"C = 81 # Percent\n",
"H = 8 # Percent\n",
"N = 2 # Percent\n",
"O = 5 # Percent\n",
"\n",
"# Solution\n",
"print \"In 1kg coal,\"\n",
"\n",
"wt_C = C * 10\n",
"wt_H = H * 10\n",
"wt_N = N * 10\n",
"wt_O = O * 10\n",
"wt_ash = 100 - (wt_O + wt_N + wt_H + wt_C)\n",
"\n",
"wt_air = ((wt_C * 32 / 12. + wt_H * 16 / 2. - wt_O) * 100 / 23.) / 1000.\n",
"\n",
"print \"Weight of air required for complete combustion of 10kg coal\",\n",
"print \"=\", \"{:.2f}\".format(wt_air * 10), \"kg\"\n",
"\n",
"HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.))\n",
"LCV = HCV - 0.09 * H * 587\n",
"\n",
"print \"HCV is\", int(HCV), \"kcal/kg\"\n",
"print \"LCV is\", int(LCV), \"kcal/kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"In 1kg coal,\n",
"Weight of air required for complete combustion of 10kg coal = 119.57 kg\n",
"HCV is 9089 kcal/kg\n",
"LCV is 8666 kcal/kg\n"
]
}
],
"prompt_number": 27
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Problem: 14, Page No: 355"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"# Variables\n",
"C = 80 # Percent\n",
"H = 7 # Percent\n",
"N = 2.1 # Percent\n",
"O = 3 # Percent\n",
"S = 3.5 # Percent\n",
"Ash = 4.4 # Percent\n",
"\n",
"# Solution\n",
"HCV = 1 / 100. * (8080 * C + 34500 * (H - O / 8.) + 2240 * S)\n",
"LCV = HCV - 0.09 * H * 587\n",
"\n",
"print \"HCV is\", int(HCV), \"kcal/kg\"\n",
"print \"LCV is\", int(LCV), \"kcal/kg\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"HCV is 8828 kcal/kg\n",
"LCV is 8458 kcal/kg\n"
]
}
],
"prompt_number": 30
}
],
"metadata": {}
}
]
}