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Chapter, Combustion of Fuels, In This Chapter., 8.1 TYPES OF FUELS, Fuels are classified into three main groups as:, 8.1 Types of Fuels, 8.2 Calorific Values of Fuels, (a) Solid fuels,, 8.3 Theoretical Determination of, CV of Fuel, (b) Liquid fuels and, (c) Gaseous fuels., 8.4 Combustion of Fuels, Solid Fuels, 8.5 Combustion Equation for, Hydrocarbon Fuel, 8.6 Determination of Minimum Air, Required for Combustion, 8.7 Conversion of Volumetric Analysis to, Mass Analysis, Wood, peat, coal and coke and various solid, wastes from processes like sugarcane crushing,, municipal waste etc. fall in this group., The analysis of solid fuel is generally carried out in, two ways., 8.8 Determination of Air Supplied when, Volumetric Analysis of Dry Flue Gases, is Known, (a) Proximate Analysis is the determination of the, percentages of fixed carbon, volatile matter,, moisture and ash in fuel. It is easier to perform and, is sufficient for commercial, 8.9 Determination of Excess Air Supplied, purposes., 8.10 Determination of % of Carbon in Fuel, Burning to CO and CO2., 8.11 Determination of Minimum Quantity of, Air Supplied to Gaseous Fuels, (b) Ultimate Analysis. The percentage of each, constituent element in the fuel such as C, H, S, O,,, N, and ash is determined in this method. This, analysis is very useful in combustion process, study and for finding the composition of the, products of combustion., 8.12 Determination of Excess Air Supplied, m³ per m3 in Gaseous Fuels, 8.13 Flue Gas Analysis., (8.1), Scanned by CamScanner

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8.2, I.C. ENGINES, Liquid Fuels, In all liquid fuels, the basic combustible elements are carbon and hydrogen. Most of the, liquid fuels are obtained from petroleum. Petroleum is a mixture of many different hydrocarbons., These hydrocarbons are grouped into four general categories :, 1. Paraffins (C, H2n+ 2), 2. Olefins (C, H2„), 4. Aromatics (C, H4n-2), 3. Naphthenes (C, H3„), The commonly used liquid fuels for power generation are petrol, paraffins, diesel oil and, heavy fuel oil., The advantages of liquid fuels over solid fuels are listed below :, 1. Higher calorific value., 3. Cleanliness of the surrounding., 5. Elimination of wear and tear of grate., 7. Easy starting and stopping., 2. Economy in, 4. Easy control of combustion., 6. Easy handling and supply., space., Gaseous Fuels, The, gaseous fuels of engineering importance are natural gas, coal gas, town gas, coke oven, gas, blast furnace gas and producer gas. Some of these gases are obtained as byproducts in, engineering industries and therefore they can be used most economically., many, The advantages of gaseous fuels over solid fuels are :, 1. The gas can be produced at one point and can be easily distributed over a wide area, through pipe lines., 2. Remote and instant control of combustion is possible., 3. Smoke and ash disposal difficulties are removed., 4. Cleanliness., 5. Low grade solid fuels can be successfully used by gasification., 6. Complete combustion without pollution is possible due to uniform mixing of air and, fuel., 8.2, CALORIFIC VALUES OF FUELS, The calorific value of solid or liquid fuel is defined as the heat evolved by the complete, combustion of unit mass of fuel. The calorific value of, gaseous, fuel is, expressed as heat developed, by the complete combustion of one cubic metre of gas at S.T.P., All fuels containing hydrogen will produce water vapour during the process of combustion., If the products of combustion containing water vapour are cooled back to the initial temperature, (room temperature), then all water vapour formed will be condensed evolving their latent heat,, producing the maximum amount of heat per kg of fuel. This heat is known as higher calorific value, of fuel., In most of the combustion processes, the cooling of burned gases to room temperature is not, possible and therefore the water vapour formed is not condensed, but carried to the atmosphere., The amount of latent heat carried depends upon the pressure at which evaporation takes place and, the quantity of water vapour formed. It is difficult to decide the pressure of evaporation. The, Scanned by CamScanner

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COMBUSTION OF FUELS, 8.3, practice is to assume that the evaporation takes place at a saturation temperature of 15 C, common, The latent heat of water vapour at this saturation temperature is 2460 kl/kg. The lower calorific value, of the fuel is given by, L.C.V. =, H.C.V. - Heat carried by water vapour formed per kg of fuel burned, %3D Н.С.V.- m, х 2460 in kJ/kg, here m, is the mass of water vapour formed per kg of fuel burned., .(8.1), 83. THEORETICAL DETERMINATION OF CALORIFIC VALUE OF FUEL, The calorific values of fuel constituents which are present in most of the fuels are listed below:, Calorific value on mass basis (kJ/kg), Calorific value on volume basis, (k]/m2), Fuel, Higher, Lower, Higher, Lower, Hydrogen, 143000, 121000, 12200, C to CO, 10200, C to CO2, CO to CO2, 35000, 10250, Methane, 55700, 50100, 37800, S to SO2, 9160, Petrol, 43120, Paraffin, 46500, Fuel oil, 45500, Natural, gas, 37000, 34500, Producer gas, 6080, 5750, Assume that the fuel contains C, H, O and S percent of carbon, hydrogen, oxygen and, sulphur respectively by mass. It is always assumed that the oxygen in the fuel is in combination, with hydrogen and it is not available in free condition. This fact reduces the mass of hydrogen, available for combustion. The mass of hydrogen combined with oxygen is, as 8 kg of oxygen are, 8, required per kg of hydrogen for complete combustion. Therefore, the free hydrogen available for, combustion is H-), 8, The higher calorific value of fuel is given by, 1, 35000 C+ 143000 H-, 100, 9160 S kJ/kg, 8., Н. С.V.%3D, .(8.2), where C, H, O and S are the percentages of carbon, hydrogen, oxygen and sulphur., The above equation gives H.C.V. The L.C.V. is given by, 9, L.C. V.= H. C. V:-, -x Hx 2460 kJ/kg., ..(8.3), 100, The methods used for determining the actual calorific values of fuels are discussed in, articles 8.14 and 8.15., Scanned by CamScanner, | || |

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8.4, I.C. ENGINES, 8.4, COMBUSTION OF FUELS, All fuels contain combustible elements such as C, H, and S which readily combine with, oxygen and evolve heat during combustion. It is always necessary to supply sufficient air for the, complete combustion of fuels and so it is necessary in the design of combustion equipments to, calculate the mass of air required for the complete consumption of the fuel. If the air supplied is, insufficient (less than required for complete combustion), then some of the carbon will be burned, to CO instead of CO, and this will reduce the amount of heat produced during the combustion in, addition to the pollution caused by CO., The following chemical equations are used to calculate the amount of oxygen required, and, the amount of gases produced by the combustion of fuel,, (a) C+ O, = CO,, using the molecular weights of the elements, 12 kg of carbon combines with 32 kg of oxygen and, forms 44 kg of CO2, 8, 11, 1 kg of C+ kg of O2, 3, - kg of CO,, .., 3, (b) 2 C+ 0, =2CO, 24 kg of carbon combines with 32 kg of oxygen and forms 56 kg of carbon monoxide., 4, 1 kg of C+- kg of O,, 3, = kg of CO, 3, (c) 2 CO+ O2 =2 CO,, 56 kg of CO combines with 32 kg of O, and forms 88 kg of CO,., 4, 11, 1 kg of CO+ kg of O2, 7, kg of CO,., (d) 2 H, + O, =2 H,0, 4 kg of H, combines with 32 kg of O, and forms 36 kg of water, 1 kg of H, kg of O, =9 kg of H,O, (e) S+ O2 = SO,, + 8, 32 kg of sulphur combines with 32 kg of O, and forms 64 kg of SO,, 1 kg of S+1 kg of O, =2 kg of SO,, 8.5, COMBUSTION EQUATIONS FOR HYDROCARBON FUEL, The combustion equation for any hydrocarbon fuel can be prepared as follows :, Assume hydrocarbon fuel is Cn H2n+ 2', then, xC, H2 + 2 + yO2 = pCO, + qH,O, ..(8.4), where x, y, p and q are constants and are to be calculated., Equating masses of C, H and O of the two sides of the equation,, x. n= p, p= x. 11., Scanned by CamScanner

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COMBUSTION OF FUELS, 8.5, x (21n + 2) = 29, (21+2)=x (n+ 1)., %3D, 2y = 2p+q, y =(2p+q) = 2.xn+ xn+ x] = (3n+ 1), 2, Substituting these values in the equation 8.4, x C„H2n + 2 (1+ 1) H,O, + (3n + 1) O, =xnCO, +x, 2n+2, Зп+1, C„H2n+ 2, - 02 =nCO, +(n+1) H,O, Assuming n=10 (as a particular case), 31, C10 H22 +, O2 =10 CO, + 11 H,0, 2, 2 C10 H22 + 310, =20 CO, + 22 H,O, or, Such an equation can be prepared for any given hydrocarbon fuel., DETERMINATION OF MINIMUM AIR REQUIRED PER KG OF, LIQUID FUEL FOR COMPLETE COMBUSTION, 8.6, If the content of C, H, , O, and S in the fuel by mass are C, H, O and S percent respectively,, then the mass of oxygen required for complete combustion is given by :, 1, Oxygen per kg of fuel for complete combustion =-, 100 3, 8, C+8 H-, Air contains 23% of oxygen by mass and therefore kg of air required per kg of fuel for, complete combustion is given by, 8, C+8 H-, 1, 100, +S, 100 3, 23, 1, 8, C+8 H-, 3, .. (8.5), 23, 8, 8.7, CONVERSION OF VOLUMETRIC ANALYSIS TO, MASS ANALYSIS AND VICE-VERSA, Generally, the volumetric analysis of dry exhaust gases can be determined easily by Orsat, apparatus. It is always necessary to convert the volumetric analysis into mass analysis in order to, find the mass of air supplied., Assume that the exhaust gases contain C, % of CO, C, % of CO2, O% of O2, and N% of N, by, volume as obtained by Orsat apparatus then these volume percentage can be converted into mass, percentages, as shown in the following table., Scanned by CamScanner