Combustion Equation-Thermodynamics-Lecture Slides, Slides of Thermodynamics

Download Combustion Equation-Thermodynamics-Lecture Slides and more Slides Thermodynamics in PDF only on Docsity! 1 1. Applications of the Combustion Equation • (1) Stoichiometric proportions for finding the correct air supply rate for a fuel (2) Composition of the combustion products is useful during the design, commissioning and routine maintenance of a boiler installation • On-site measurements of flue gas composition and temperature are used as a basis for calculating the efficiency of the boiler at routine maintenance intervals. docsity.com 2 2. Combustion Air Requirements: Gaseous Fuels • Calculating the air required for gaseous fuels combustion is most convenient to work on a volumetric basis. • The stoichiometric combustion reaction of methane is : CH4 + 2O2 → CO2 + 2H2O which shows that each volume (normally 1 m3) of methane requires 2 volumes of oxygen to complete its combustion. docsity.com 5 • It can be seen that the complete combustion of one volume of methane will require (2+7.52=9.52) volumes of air, so the stoichiometric air-to-fuel (A/F) ratio for methane is 9.52. • In practice it is impossible to obtain complete combustion under stoichiometric conditions. Incomplete combustion is a waste of energy and it leads to the formation of carbon monoxide, an extremely toxic gas, in the products. docsity.com 6 • Excess air is expressed as a percentage increase over the stoichiometric requirement and is defined by: • Excess air will always reduce the efficiency of a combustion system. / / 100% / actual A F ratio stoichiometric A F ratio stoichiometric A F ratio   docsity.com 7 • It is sometimes convenient to use term excess air ratio, defined as: • Where sub-stoichiometric (fuel-rich) air- to-fuel ratios may be encountered, for instance, in the primary combustion zone of a low-NOX burner, the equivalence ratio is often quoted. This is given by: / / actual A F ratio stoichiometric A F ratio / / stoichiometric A F ratio actual A F ratio docsity.com 10 • Considering the combustion of methane with 20% excess air, the excess air (0.2×9.52) of 1.9 volumes will appear in the flue gases as (0.21×1.9)=0.4 volumes of oxygen and (1.9-0.4)=1.5 volumes of nitrogen. • The complete composition will be: constituent vol/vol methane CO2 1 O2 0.4 N2 9.02 H2O 2 giving a total product volume of 12.42 (wet) or 10.42 (dry). docsity.com 11 • The resulting composition of the flue gases, expressed as percentage by volume, is: Constituent % vol (dry) % vol (wet) CO2 9.6 8.1 O2 3.8 3.2 N2 86.6 72.6 H2O – 16.1 docsity.com 12 • Example 1: A gas consists of 70% propane (C3H8) and 30% butane (C4H10) by volume. Find: (a) The stoichiometric air-to-fuel ratio and (b) The percentage excess air present if a dry analysis of the combustion products shows 9% CO2 (assume complete combustion). • Solution: The combustion reactions for propane and butane are: 3 8 2 2 2 2 2 5 18.8 3 4 18.8C H O N CO H O N     4 10 2 2 2 2 26.5 24.5 4 5 24.5C H O N CO H O N     docsity.com 15 4. Combustion Air Requirements-Solid and Liquid Fuels • The way in which the combustion equation is used reflects the available information on the analysis of the solid or liquid fuels. This takes the form of an element-by-element analysis (referred to as an ultimate analysis) which gives the percentage by mass of each element present in the fuel. • An example of an ultimate analysis of a liquid fuel (oil) might be : Component % by mass Carbon (C) 86 Hydrogen(H2) 14 docsity.com 16 • Each constituent is considered separately via its own combustion equation. For the carbon: C + O2 → CO2 12kg 32kg 44kg or for 1 kg of fuel • So each kg of oil requires 2.29 kg oxygen for combustion of its carbon and produces 3.15 kg CO2 as product. 32 44 0.86 0.86 0.86 (kg) 12 12     docsity.com 17 • Similarly H2 + ½ O2 → H2O 2kg 16kg 18kg or per kg of fuel • In order to burn the hydrogen content of the oil 1.12 kg oxygen are needed and 1.26 kg water is formed. 16 18 0.14 0.14 0.14 (kg) 2 2     docsity.com 20 5. Combustion Products-Solid and Liquid Fuels • The stoichiometric combustion products from combustion of the oil are: CO2 3.15 kg H2O 1.26 kg N2 11.23 kg • The combustion products would normally be needed as a volume percentage, so the reverse operation to that which was performed for air above is required. docsity.com 21 • Hence if we require a dry volume percentage of the above products the following tabular procedure is convenient: Component Mass/kg fuel kmoles/kg fuel mole fraction CO2 3.15 0.151 N2 11.23 0.849 • The stoichiometric combustion products are thus 15.1% CO2 and 84.9% N2. 3.15 0.0716 44  11.23 0.4011 28 0.4727  docsity.com 22 • Solid fuels, and many liquid fuels, contain compounds of sulfur. For the purposes of stoichiometric calculations this is assumed to burn to sulfur dioxide: S + O2 → SO2 • In reality a mixture of sulfur dioxide and sulfur trioxide (SO3) is produced, but it is conventional to assume combustion to SO2 when calculating air requirements. docsity.com 25 • Solution: Lay out the calculation on a tabular basis using 1 kg coal: Mass (per kg) O2 Required Products Carbon 0.9 Hydrogen 0.03 Sulfur 0.005 Oxygen 0.025 -0.025 - Nitrogen 0.01 - 0.01 Ash 0.03 - - 32 0.9 2.4 12   16 0.03 0.24 2   32 0.005 0.005 32   64 0.005 0.01 32   18 0.03 0.27 2   44 0.9 3.3 12   docsity.com 26 • (a) Oxygen required to burn 1 kg coal = 2.4 + 0.24 + 0.005 - 0.025 = 2.62 kg. Air required = Actual air supplied = 11.25 × 1.2 = 13.5 kg Assuming a density for air of 1.2 kg/m3, the flow rate will be: 2.62 11.25 kg 0.233  350013.5 1.56 m /s 1.2 3600    docsity.com 27 • (b) To get the %CO2 in the combustion products we need to know the amounts of oxygen and nitrogen in the flue gases. Air supplied = 13.5 kg per kg coal, of which oxygen is 13.5 × 0.233 = 3.14 kg, and nitrogen 13.5 – 3.14 = 10.36 kg. • The combustion products will thus contain: 3.14 – 2.62 = 0.52 kg O2 and 10.36 + 0.01 = 10.37 kg N2. docsity.com 30 • If pure carbon is burnt, the only combustion product is carbon dioxide, so each molecule of oxygen in the combustion air becomes a molecule of carbon dioxide in the flue gas. This means that the stoichiometric combustion of carbon will produce 21% by volume CO2. • If we consider for the moment that hydrocarbon fuels consist only of carbon and hydrogen, as the carbon:hydrogen ratio of the fuel decreases the stoichiometric air-to-fuel ratio will increase. This is because 1 kg carbon requires 32/12=2.67 kg of oxygen for complete combustion but 1 kg hydrogen requires 8 kg oxygen. docsity.com 31 • The percentage CO2 in the flue gases will fall as the carbon:hydrogen ratio in the fuel decreases as (1) less carbon dioxide will be produced per kilogram of fuel and (2) the increased air requirement means that the carbon dioxide produced will be diluted by the extra nitrogen in the flue gas. • This effect is illustrated in Table 2.1 (next slide). The carbon:hydrogen ratio in fuels lie between the limits of 75:25 (methane) to around 95:5 (high carbon coals). docsity.com 32 Table 2.1 Carbon dioxide concentraton in flue gases C : H (by mass) Satoichiometric %CO2 100 0 95 5 90 10 85 15 80 20 75 25 70 30 65 35 21.00 18.67 16.62 14.81 13.19 11.73 10.42 9.23 docsity.com 35 • Next Slide (Fig. 2.2) A plot of the percentage CO2 in the flue gases over a range of values of excess air for C:H ratios ranging from 75:25 to 95:5. The curves for fuels with higher C:H ratios lie above those for fuels with a lower value of this ratio. docsity.com CO, in products (dry %) 0 50 100 150 200 excess air (%) Figure 2.2 Carbon dioxide in combustion products docsity.com 37 • Next Slide (Fig. 2.3) The relationship between the percentage oxygen in the flue gas and the excess air is very similar for a wide range of fuels. This is different from the CO2 curves. docsity.com 40 7. Sub-stoichiometric Combustion • There are circumstances in which localized fuel-rich combustion can take place, such as where combustion of the fuel is a two- stage process with secondary air added downstream of the primary combustion zone. • The mechanism of combustion of a fuel with less than the stoichiometric air requirement consists of the following sequence of events: (1) The available oxygen firstly burns all the hydrogen in the fuel to water vapor. (2) All the carbon in the fuel is then burned to carbon monoxide. (3) The remaining oxygen is consumed by burning carbon monoxide to carbon dioxide. docsity.com 41 • Next slide (Fig. 2.4) It can be seen that as the air supply falls below the stoichiometric requirement the percentage of carbon monoxide in the flue gas increases very quickly. docsity.com Air fuel ratio Figure 2.4 Sub-stiochiometric combustion of natural gas docsity.com 45 • Burning the carbon to CO will produce 3 volumes of CO and use up 1.5 volumes of oxygen, leaving (2.75- 1.5)=1.25 volumes of oxygen for further combustion. • Next reaction is CO + ½ O2 → CO2 So 1.25 volumes oxygen can burn 2.5 volumes of carbon monoxide, producing 2.5 volumes of carbon dioxide. • The remaining carbon monoxide is therefore (3-2.5)=0.5 volume. docsity.com 46 • The products of combustion are thus: N2 17.87 volumes CO 0.5 CO2 2.5 H2O 4.0 Total 24.87 • Giving the percentage compositions: Wet(%) Dry(%) N2 71.9 85.6 CO 2.0 2.4 CO2 10.0 12.0 H2O 16.1 - docsity.com