Combustion and Stoichiometry PDF

Combustion and Stoichiometry Dr. R. Arun Prasath Professor Department of Green Energy Technology Pondicherry University Puducherry – 605014 Email: [email protected] [email protected] R. Arun Prasath...

Combustion and Stoichiometry Dr. R. Arun Prasath Professor Department of Green Energy Technology Pondicherry University Puducherry – 605014 Email: [email protected] [email protected] R. Arun Prasath GETY514 2023 1  COMBUSTION Combustion (or) burning is a chemical process, an exothermic reaction between a substance (the fuel) and a gas (the oxidizer), usually O2, to release thermal energy (heat), electromagnetic energy (light), mechanical energy (noise) and electrical energy ( free ions and electrons ). In a complete combustion reaction, a fuel rapidly oxidized with oxygen to produce heat and light. Complete combustion of a fuel is possible only in the presence of an adequate supply of oxygen For example: CH4 + 2 O2 → CO2 + 2 H2O + Heat ( +light/noise/ions ) Fuel Gas R. Arun Prasath GETY514 2023 2  COMBUSTION BASICS Oxygen is one of the most common elements on earth making up 21% of air. Rapid fuel oxidation results in large amounts of heat. Most of the 79% of air (that is not oxygen) is nitrogen, with traces of other elements. Nitrogen is considered to be a temperature reducing dilutant that must be present to obtain the oxygen required for combustion. Nitrogen reduces combustion efficiency by absorbing heat from the combustion of fuels and diluting the flue gases Supply of pure oxygen for combustion…. economically and practically.. R. Arun Prasath GETY514 2023 3  COMBUSTION AND ENTHALPY The energy change of a system during a chemical reaction is due to a change in state and a change in chemical composition: When the existing chemical bonds are broken and new ones are formed during a combustion process, usually a large amount of sensible energy is absorbed or released. Example: Bond breaking/formation in Hydrogen with oxygen….? R. Arun Prasath GETY514 2023 4  COMBUSTION AND ENTHALPY Enthalpy of reaction hR : The difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction. The enthalpy of formation hf : The amount of energy absorbed or released as the component is formed from its stable elements during a steady-flow process at a specified state. Enthalpy of combustion hC : It is the enthalpy of reaction for combustion processes. It represents the amount of heat released during a steady-flow combustion process when 1 kmol (or 1 kg) of fuel is burned completely at a specified temperature and pressure. R. Arun Prasath GETY514 2023 5  COMBUSTION AND ENTHALPY Combustion is an oxidation reaction that produces heat, and it is therefore always exothermic. All chemical reactions first break bonds and then make new ones to form new materials Combustion is chemical reaction. The heat transfer in combustion reaction are classified based on the value of Qnet positive or negative? Exothermic Endothermic Exothermic combustion reactions Endothermic combustion reactions R. Arun Prasath GETY514 2023 6  COMBUSTION AND ENTHALPY Heating value: The amount of heat released when a fuel is burned completely in a steady-flow process and the products are returned to the state of the reactants. The heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel. Solid fuel and liquid: cal/g, Kcal/Kg, BTU/lb, J/Kg Gaseous fuels: cal/cm3, Kcal/m3, BTU/Ft3, J/m3 Lower heating value (LHV): H20 in the products is in the vapour form (NET) Higher heating value (HHV): H20 in the products is in the liquid form (GROSS) at the atmospheric room temp condition the steam gets condensed into water and latent heat is evolved. The latent heat of water is 587Kcal/Kg R. Arun Prasath GETY514 2023 7  COMBUSTION AND ENTHALPY Net calorific value= Gross calorific value - Latent heat of condensation of water vapour produced LCV = HCV – mass of hydrogen X 9 X latent heat of steam 1 part by mass of hydrogen produces 9 parts by mass of water ½ H2 + O2 H2O 2g 18g 16g 1g 9g 8g 9H/100 g = mass of H2O from 1 g of fuel = 0.09 H g The latent heat of steam is 587 KCal/Kg of water vapour formed at room temperature R. Arun Prasath GETY514 2023 8  COMBUSTION AND ENTHALPY STP stands for Standard Temperature and Pressure. STP is set as 0°C /273K and 100 kPa or 1 bar. Standard Lab Conditions (SLC) is considered to be 25°C and 100 kPa. At SLC the molar Gas Volume is 24.8 L mol-1 NTP stands for Normal Temperature and Pressure. NTP is set at 101.325 kPa [ 1atm] but uses 20°C /293K as the temperature R. Arun Prasath GETY514 2023 9  CALORIFIC VALUE BY DULONG FORMULA The calorific value of fuels is determined theoretically by Dulong formula or I.A. Davies formula HCV = 1/100 [8,080 C + 34,500 (H – O/8)+ 2,240 S] kcal/kg Where C, H, O, and S are the percentages of each elements R. Arun Prasath GETY514 2023 10  CALORIFIC VALUE BY DULONG FORMULA Exercises! Calculate the gross and net calorific value of a fuel having composition carbon = 79%, hydrogen =11%, sulphur =2% nitrogen =5% and remaining ash. Consider the following composition of fuel by weight %: C=88, O= 4, S=2, N=1 and remaining ash. The net calorific value of fuel found to be 9,500 Kcal/Kg. Calculate the % of H and GCV of fuel. R. Arun Prasath GETY514 2023 11 R. Arun Prasath GETY514 2023 12  CALORIFIC VALUE BY BOMB CALORIMETRIC Calculation: Let X = mass in g of fuel sample taken in crucible; W = mass of water in the calorimeter; w =water equivalent in g of calorimeter, stirrer, thermometer, bomb, etc.; t1 = initial temperature of water in calorimeter; t2 = final temperature of water in calorimeter; L = higher calorific value in fuel in cal/g. Heat liberated by burning fuel = X L Heat absorbed by water and apparatus, etc. = ( W + w )( t2 - t1 ) But heat liberated by fuel = heat absorbed by water, apparatus, etc. X L = ( W + w )( t2 - t1 ) or H.C.V of fuel (L) = (W + w)(t2 – t1)/X cal/g (or KCal/Kg) Note: The water equivalent is determined by burning a fuel of known calorific value and using the above equation. The fuels used for this purpose are benzoic acid (H.C.V = 6325 Kcal/Kg). R. Arun Prasath GETY514 2023 13  CALORIFIC VALUE BY BOMB CALORIMETRIC Heat Capacity of Calorimeter = cthermometer + cstirrer + ccontainer..... Heat Capacity of Calorimeter qc C= Heat Constant (e.g.. 900 J/0C) qc= CΔT Heat Capacity of Water qw = m x c x dt C = 4.184 J/goC R. Arun Prasath GETY514 2023 14  CALORIFIC VALUE BY BOMB CALORIMETRIC Exercises! Consider a fuel of 1 gm burned in a bomb calorimeter containing 1500 grams of water at an initial temp of 25 ⁰C. The final temp. after the combustion reaction of water found to be 28 ⁰C. The heat capacity of the calorimeter is 720J/⁰C. Calculate the heat of combustion of fuel in kJ Hints: Specific heat capacity of water = 4.18J/g ⁰C Heat capacity of the calorimeter = 720J/ ⁰C Heat absorbed by water = qw = m x c x dt = xxxJ Heat absorbed by calorimeter qc = C x dt Total heat released in combustion = qw +qc Consider 0.75 gm of dodecane (C12H26) burned in a bomb calorimeter containing 1500 grams of water at an initial temp of 25 ⁰C. The final temp. after the combustion reaction of water found to be 28 ⁰C. The heat capacity of the calorimeter is 720J/⁰C. Calculate the heat of combustion of fuel in kJ/mole Hints: Do all the above hints + convert to Kj/mole Dodecane Mwt = find out Heat of combustion of fuel in kJ/mole = (xkJ x Mwt g/mole) / 0.75 R. Arun Prasath GETY514 2023 15  CALORIFIC VALUE BY BOMB CALORIMETRIC Calculation with corrections Fuse wire/thread correction : During the experiment the heat liberated by the fuse wire is also included which actually has to be excluded from gross calorific value. Acid correction: During the experiment fuels containing S and N also get oxidized under high pressure and temperature to sulphuric and nitric acids respectively. S + 2H + 2O2 H2SO4 + Heat 2N + 2H + 3O2 2HNO3 + Heat Formation of these acid produce heat energy ( exothermic reactions). Hence these heat values have to be deducted from the gross calorific value. The correction for 1 mg of Sulphur is 2.25 Cal and if 1 ml of 0.1 N HNO3 is formed the heat produced is 1.43 Cal. Cooling Correction: For calculating correction, the time taken for cooling the calorimeter water from maximum to minimum temperature (room temperature) is noted and the rate of cooling is found out, which is multiplied with the time taken for cooling t0 Cooling correction = dt/minute x t (W + w)[(t 2 − t1 + cooling correction)] − (Acid + Fuse correction) C= X R. Arun Prasath GETY514 2023 16  CALORIFIC VALUE BY BOMB CALORIMETRIC Exercises! A sample of lignite coal contains 82% C, 11% H , 3% ash. When this fuel of 0.85 g burned in in the bomb calorimeter containing water of 1.5L, water equivalent of calorimeter is 700 g, rise in temp 2.5 ⁰C, cooling correction = 0.04 ⁰C, cotton thread correction = 20 cals with acid correction 15 cals. Calculate the gross and net calorific values of the fuel. Hints: (W + w)[(t 2 − t1 + cooling correction)] − (Acid + Fuse correction) C= X Net calorific value = Gross calorific value – 0.09H x 587 cal/g or..Kcal R. Arun Prasath GETY514 2023 17 CALORIFIC VALUE BY BOYS/JUNKER CALORIMETRIC For gases fuel Combustion Chamber R. Arun Prasath GETY514 2023 18  CALORIFIC VALUE BY BOYS/JUNKER CALORIMETRIC Components: Bunsen Burner: special type of Burner clamped at the bottom. It can be pulled out of the combustion chamber or pushed up in chamber during the carrying out combustion. Gasometer: It is employed to measure the volume of gas burning per unit time. This attached with manometer fitted with the thermometer so that pressure and temperature of the gas before burning can be read. Pressure controller: It can control the supply of quantity of gas at give pressure. Gas Calorimeter/ Combustion chamber: It is a vertical cylinder, which is surrounded by annular space for heating water and interchange coils. The entire is covered by an outer jacket in order to reduce the heat loss by radiation and convection. R. Arun Prasath GETY514 2023 19  CALORIFIC VALUE BY BOYS/JUNKER CALORIMETRIC Calculation: Volume of gas burn at STP in certain time (t)= V (m3) Mass of the cooling water used in time t = W (kg) Temperature of inlet water = T1 ( °C) Temperature of outlet water = T2 ( °C) Mass of steam condensed in time t in graduated cylinder in = m Higher calorific value of fuel = L Specific heat of water = S Heat absorbed by circulating water = W(T2-T1)×Specific heat of water (s) Heat produced by combustion of fuel = VL VL = W(T2-T1)×S HCV (L) = W(T2-T1)×S/V LCV = L - amount of water collected x latent heat V LCV = m x 587 L- Kcal/m3 V R. Arun Prasath GETY514 2023 20  CALORIFIC VALUE BY BOYS/JUNKER CALORIMETRIC Exercises! Consider the following data obtained in the Boy’s gas calorimeter experiment to determine the GCV and NCV of gases fuel. Volume of gaseous fuel combusted = 0.1 m3 at NTP Weight of water circulated for heat extraction = 50 Kg Temp. of inlet water = 25 ˚C Temp. of outlet water = 38 ˚C Weight of steam condensed = 0.025 kg Calculate the gross and net calorific value per m3 at NTP. Take the heat liberated in condensing water is 580 Kcal/Kg R. Arun Prasath GETY514 2023 21  3 T’s of Combustion (Principle) The objective of good combustion is to release all of the heat in the fuel. This is accomplished by controlling the “three T’s” of combustion which are – Temperature high enough to ignite and maintain ignition of the fuel, – Turbulence or intimate mixing of the fuel and oxygen, and – Time sufficient for complete combustion. R. Arun Prasath GETY514 2023 22 Combustion Stoichiometry R. Arun Prasath GETY514 2023 23  Combustion Stoichiometry Combustion Stoichiometry ( Theoretical ) If sufficient oxygen is available, a hydrocarbon fuel can be completely oxidized, the carbon is converted to carbon dioxide (CO2) and the hydrogen is converted to water (H2O). During combustion, each element reacts with oxygen to release heat : C + O2 -> CO2 + Heat H2+ 0.5 O2 -> H20 + Heat Pure oxygen…………. or air is mainly used for combustion. It contains 21 percent of Oxygen O2 and 79 percent of Nitrogen N2. The stoichiometric air/fuel ratio refers to the proportion of air and fuel present during a theoretical combustion. The heat released when the fuel burns completely is known as the heat of combustion R. Arun Prasath GETY514 2023 24  Combustion Stoichiometry General Equation for combustion with oxygen  m m Cn H m +  n + O2 → nCO2 + H 2O  4 2 For Methane CH 4 + 2O2 → CO 2 + 2 H 2O For Benzene C6 H 6 + 7.5O2 → 6CO 2 + 3H 2O For Propane…? 79/21 = 3.78 Combustion in Air (O2 = 21%, N2 = 79%) Cn H m + (O2 + 3.78 N 2 ) → CO 2 + H 2O + N 2  m m  m Cn H m +  n + (O2 + 3.78N 2 ) → nCO2 + H 2O + 3.78 n +  N 2  4 2  4 CH 4 + 2(O2 + 3.78 N 2 ) → CO 2 + 2 H 2O + 7.56 N 2 C6 H 6 + 7.5(O2 + 3.78 N 2 ) → 6CO 2 + 3H 2O + 28.35 N 2 For dodecane…? R. Arun Prasath GETY514 2023 25  Combustion Stoichiometry Air-Fuel (AF) mass ratio AF = m Air / m Fuel Where: m air = mass of air in the feed mixture m fuel = mass of fuel in the feed mixture Fuel-Air ratio: FA = m Fuel /m Air Air-Fuel molal ratio AFmole = nAir / nFuel Where: nair = moles of air in the feed mixture nfuel = moles of fuel in the feed mixture Rich mixture - more fuel than required (AF) mixture < (AF)stoich Lean mixture - more air than required (AF) mixture > (AF)stoich Most combustion systems operate under lean conditions. R. Arun Prasath GETY514 2023 26 Combustion Stoichiometry R. Arun Prasath GETY514 2023 27  Combustion Stoichiometry 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. Excess air is expressed as a percentage increase over the stoichiometric requirement and is defined by: actual A / F ratio − stoichiometric A / F ratio 100% stoichiometric A / F ratio R. Arun Prasath GETY514 2023 28  Combustion Stoichiometry Flue Gas Composition- calculation The composition of the stoichiometric combustion products of methane is: 1 volume CO2 7.52 volumes N2 2 volumes H2O Given a total product volume, per volume of fuel burned, of 10.52 if water is in the vapor phase, or 8.52 if the water is condensed to a liquid. The two cases are usually abbreviated to “wet” and “dry”. The proportion of carbon dioxide in this mixture is therefore 1  100% = 9.51% wet and 10.52 1  100% = 11.74% dry 8.52 R. Arun Prasath GETY514 2023 29 Combustion Stoichiometry R. Arun Prasath GETY514 2023 30 Combustion Stoichiometry R. Arun Prasath GETY514 2023 31  Combustion Stoichiometry General Equation for combustion with oxygen  m m Cn H m +  n + O2 → nCO2 + H 2O  4 2 79/21 = 3.78 Combustion in Air (O2 = 21%, N2 = 79%)  m m  m Cn H m +  n + (O2 + 3.78N 2 ) → nCO2 + H 2O + 3.78 n +  N 2  4 2  4 C + O2 O2 + Heat H2+ 0.5 O2 H2O + Heat S + O2 SO2 + Heat R. Arun Prasath GETY514 2023 32  Combustion Stoichiometry Practically Combustion ( Excess of Air) Due to fluctuations in fuel flow and the lack of perfect mixing between fuel and air in the combustion zone, excess air is required to achieve more complete combustion of the fuel. Without this extra air, the formation of partial products of combustion such as carbon monoxide and soot may occur. However, supplying too much excess air will decrease combustion efficiency and a balance between too much air and not enough air must be maintained. (air takes heat…!) R. Arun Prasath GETY514 2023 33  Combustion Stoichiometry ❑ Thermal NOx - Oxidation of atmospheric N2 at high temperatures N 2 + O2  2 NO NO + 12 O2  NO 2 - Formation of thermal NOx is favorable at higher temperature ❑ Fuel NOx - Oxidation of nitrogen compounds contained in the fuel ❑ Formation of CO - Incomplete Combustion - Dissociation of CO2 at high temperature CO2  CO + 12 O2 R. Arun Prasath GETY514 2023 34  Exercises! Consider burning of 1 kg of coal sample having the following composition, C =64%, H = 15%, O = 8% S = 5%, moisture = 11%. Calculate (a) the theoretical weight of air required for complete combustion of 1 kg of coal, (b) its volume in m3 at NTP, and (c) percentage composition by weight and volume of dry products of combustion. HINTS: MOLE METHOD or WEIGHT METHOD Mole: Wt/MWt of each element, e.g C =64/12 = x mole, same for other element H, S, and oxygen need to minus here. Consider combustion of 1 ton solid fuel which consist C =55%, H = 25%, O = 5% moisture = 10%. Calculate (a) the amount of energy released in MJ, (b) the amount of units produced in kWh- if 60% released energy converted to power, and (c) the amount of CO2 releases in atmosphere Consider 1 kmol of hydrogen H2 combusted with 6 kmol air. What is A/F ratio on a mole basis and what is the percent theoretical air? Calculate the theoretical air-fuel ratio on a mass and mole basis for the combustion of ethanol, C2H5OH. R. Arun Prasath GETY514 2023 35 Consider burning of 1 kg of coal sample having the following composition, C =74%, H= 18.5%, O2 = 3% S = 3%, moisture = 1.5%. Calculate (a) the theoretical weight of air required for complete combustion of 1 kg of coal, (b) its volume in m3 at NTP, and (c) percentage composition by weight and volume of dry products of combustion. Composition of dry products by volume By MOLE METHOD No. of Kmols of Oxygen CO2 = 0.06166 SO2 = 0.00094 C = 0.74/12 = 0.06166 H = 0.185/4 = 0.04625 N2 = 0.5138 x 79/100 = 0.4059 S = 0.03/32 = 0.00094 Total dry product in volume = 0.4685 O = 0.03/32 = - 0.00094 N2 = 86.64 % CO2 = 13.16 Total = 0.10791 Kmols of O2 for 1kg coal SO2 = 0.20% Total in kg of O2 = mol x Mwt of O2 Composition of dry products by Wt =0.10791 x 32 = 3.453 kg of O2 Theoretical air in kg = 3.453 x100/23 = CO2 = 0.06166 x 44 = 2.7130 15.0135 kg air SO2 = 0.00094 x 64 = 0.06016 N2 = 0.4059x 28 = 11.3652 Theoretical air in kmol = 0.10791 x 100/21 = Total dry product in Wt= 14.1383 0.5138 kmol N2 = 80.38% CO2 = 19.18% Volume of air = 0.5138 x 22.4 m3 (NTP) = SO2 = 0.42% 11.51 m3 R. Arun Prasath GETY514 2023 36 Consider burning of 1 kg of coal sample having the following composition, C =74%, H= 18.5%, O2 = 3% S = 3%, moisture = 1.5%. Calculate (a) the theoretical weight of air required for complete combustion of 1 kg of coal, (b) its volume in m3 at NTP, and (c) percentage composition by weight and volume of dry products of combustion. By WEIGHT METHOD Composition of dry product in Volume Wt of Oxygen required in kg CO2 = 2.713/44 = 0.06166 C = 0.74x32/12 = 1.9733 SO2 = 0.06/64 = 0.0009375 H = 0.185x16/2 = 1.48 N2 = 11.51/28 = 0.4110 S = 0.03x32/32 = 0.03 Total dry product in volume = 0.4735 O = 0.03 = - 0.03 N2 = 86.8 % Total oxygen = 3.453 kg CO2 = 13% SO2 = 0.19% Theoretical air in kg = 3.453 x100/23 = 15.013 kg air Composition of dry products by Wt 32 kg of Oxygen occupies 22.4 m3 CO2 =.74x44/12 = 2.7130 at NTP SO2 = 0.03x 64/32 = 0.06 N2 = 15.0135 - 3.453 = 11.56 So, 3.453 x22.4/32 = 2.4171 m3 Total dry product in Wt= 14.333 N2 = 80.6% Air = 2.4271 x100/21 = 11.51 m3 CO2 = 18.93% SO2 = 0.41% R. Arun Prasath GETY514 2023 37  Exercises! Consider combustion of 1 ton solid fuel which consist C =55%, H = 25%, O = 5% moisture = 10%. Calculate (a) the amount of energy released in MJ, (b) the amount of units produced in kWh- if 60% released energy converted to power, and (c) the amount of CO2 releases in atmosphere Hints Find how much C, H in moles, Wt/Mwt = mole Enthalpy of combustion of C, H, S Carbon to CO2 ~ 393 kJ/mol Carbon to CO ~110 kJ/mol Hydrogen to H2O ~ 286 kJ/mol Sulfur to SO2 ~ 296 kJ/mol Total in kJ kJ to MJ kWh = 3.6 MJ CO2 in moles of carbon …release xxx moles of CO2 In weight XXX x 44 = …xx kg R. Arun Prasath GETY514 2023 38 Exercises! R. Arun Prasath GETY514 2023 39  Assignment –I Compulsory Ignition: Concept, types, battery ignition and magneto ignition system, -components and its working. R. Arun Prasath GETY514 2023 40  Flue gases Flue gases are exhaust gases from combustion process occurs in a oven, furnace, boiler or steam generator at power plants via a pipe or channel. A mixture of gases produced by the burning of fuel or other materials in power stations / industrial plants and extracted via ducts H2O, CO2,CO, O2, N2 ………particulates, heavy metals. R. Arun Prasath GETY514 2023 41  Flue gases analysis. Why? ✓ Flue gas analysis tells about the extent of good/perfect/bad combustion i.e, are you achieving max efficient combustion or not? engine performance, environmental impact, …. Complete Combustion: CO2, H2O, and N2 Incomplete Combustion: CO2, H2O, N2, CO, NOx Combustion with Excess Oxygen: CO2, H2O, N2, and O2 Analysis methods: Chemical, Instruments, sensors… R. Arun Prasath GETY514 2023 42  Flue gas analysis-chemical methods Basic Principle: absorption of gases with chemicals leads for chemical absorption. i.e, Chemical reaction between absorbent and absorbate. The absorption depends upon the stoichiometry of the reaction and concentration of its reactants. R. Arun Prasath GETY514 2023 43  ORSAT APPARATUS- Construction Consists of a water-jacketed measuring burette, connected in series to a set of three absorption bulbs, each through a stop-cock. The other end is provided with a three-way stop-cock, the free end of which is further connected to a U-tube packed with glass wool (for avoiding the incoming of any smoke particles, etc.) The graduated burette is surrounded by a water-jacket to keep the temperature of the gas constant during the experiment. The lower end of the burette is connected to a water reservoir by means of a long rubber tubing. The absorption bulbs are usually filled with glass tubes, so that the surface area of contact between the gas and the solution is increased. R. Arun Prasath GETY514 2023 44 Working Principle – Orsat flue gases R. Arun Prasath GETY514 2023 45  Absorption in ORSAT APPARATUS The absorption bulbs have solutions for the absorption of CO2, O2 and CO respectively. First bulb has ‘potassium hydroxide solution (250g KOH in 500mL of boiled distilled water), and it absorbs only CO2. 2KOH + CO2 ↔ K2CO3+ H2O Second bulb has a solution of ‘alkaline pyrogallic acid’ (25g pyrogallic acid+200g KOH in 500 mL of distilled water) and it can absorb CO2 and O2. 2C6H3(OH)3(pyrogallol) + 2KOH(saturated alkaline)+ O2 ↔ 4H2O + 2C5H3OCOOK and a physical color change is observed. Brown from colourless absorbed. Third bulb contains ‘ammonical cuprous chloride’ (100g cuprous chloride + 125 mL liquor ammonia+375 mL of water) and it can absorb CO2, O2 and CO. 2CuCl + 2CO →(in NH3 solution)→ [CuCl(CO)]2 Hence, it is necessary that the flue gas is passed first through potassium hydroxide bulb, where CO2 is absorbed, then through alkaline pyrogallic acid bulb, when only O2 will be absorbed (because CO2 has already been removed) and finally through ammonical cuprous chloride bulb, where R. Arun only Prasath CO will GETY514 2023 be absorbed. 46 Working Principle – Orsat flue gases R. Arun Prasath GETY514 2023 47 Working Principle – Orsat flue gases R. Arun Prasath GETY514 2023 48 Working Principle – Orsat flue gases R. Arun Prasath GETY514 2023 49 Working Principle – Orsat flue gases R. Arun Prasath GETY514 2023 50 Working Principle – Orsat flue gases R. Arun Prasath GETY514 2023 51  Calculation Flue gas sample Volume = 100 ml Volume of CO2 = (100-a) ml = X; ‘a’ is the volume of flue gas absorbed in KOH bulb, X is the remaining flue gas measured in the burette. (The decrease in burette volume- reading the gives volume of CO2 in 100 mL of the flue gas sample taken) Volume of O2 = (X-b) ml = Y; ‘b’ is the volume of flue gas absorbed in pyrogallic bulb, Y is the remaining flue gas measured in the burette. (The decrease in burette reading from X to Y (gives volume of O2 of flue gas sample from X reading) Volume of CO = (Y-c) ml; = Z ‘c’ is the volume of flue gas absorbed in cuprous chloride bulb Z is the remaining flue gas measured in the burette. (The decrease in burette reading from Y to Z (gives volume of CO of flue gas sample from Y reading) Volume of N2 = Z- (a+b+c) = d Observation Table 100 ml = a+b+c+ d S.No Amount of CO2 Amount of O2 Amount of CO Amount of N2 1 a b c d R. Arun Prasath GETY514 2023 52  Dew Point in combustion (Steam) Basic Definition: The temperature at which the relative humidity reaches a 100% (saturation point) , that is the temperature in which saturation occurs is known as Dew Point temperature. i.e.- the temperature at which steam starts to condensed at its saturated partial pressure is called dew point temperature. Dew point or Dew point temperature of a gas (steam) is the temperature at which the water vapor starts to condensed to its liquid phase. Importance in Combustion: SOx, Nox corrosion as they from respective acids. The furnace walls get corroded -if the DEW point temperature is lower in the furnace/ boiler…- as steam get condensed and causes reactions with SO/NO oxides to produce acids. R. Arun Prasath GETY514 2023 53  Dew Point in combustion (Steam) Dew point is the temperature at which steam condenses Tdp Tdp = Tsat at Pw (Partial pressure of water alone) Mole fraction of water in the flue product = Moles of water/ Total mole of all product Total mole of all product = include steam, O2, CO2, SOx, N2 Partial pressure of water (Pw) = Mole fraction of water x given pressure (various) Tdp = Tsat at Pw refer from standard table (Saturation temperature and pressure data) R. Arun Prasath GETY514 2023 54  Dew Point in combustion (Steam) Saturation pressure at temperatures given in degree Celcius and pressure given in kiloPascals [kPa], bars, atmospheres [atm] and pounds per square inch [psi]: R. Arun Prasath GETY514 2023 55  Exercises! Liquid propane is burned with dry air. A volumetric analysis of the products of combustion yields the following volume percent composition on a dry basis: 8.6% CO2, 0.6% CO, 7.2% O2 and 83.6% N2. Determine the percent of theoretical air used in this combustion process. a C3H8 + b O2 + c N2 → 8.6 CO2 + 0.6 CO + d H2O + 7.2 O2 + 83.6 N2 C balance: 3a = 8.6 + 0.6 = 9.2 ⇒ a = 3.067 H2 balance: 4a = d ⇒ d = 12.267 N2 balance: c = 83.6 O2 balance: b = 8.6 + 0.6/2 + 12.267/2 + 7.2 = 22.234 Air-Fuel ratio = 22.234 + 83.6/3.067 = 34.51 Theoretical equation: 1C3H8 + 5 O2 + 18.8 N2 → 3 CO2 + 4 H2O + 18.8 N2 ⇒ theo. A-F ratio = 5 + 18.81/1 = 23.8 % theoretical air = 34.51/23.8 × 100 % = 145 % R. Arun Prasath GETY514 2023 56  Exercises! Liquid propane is burned with dry air. A volumetric analysis of the products of combustion yields the following volume percent composition on a dry basis: 7.5% CO2, 1.7% CO, 6.2% O2 and 84.6% N2. Determine the percent of theoretical air used in this combustion process. A hydrocarbon CxHy, is burnt with excess air. The Orsat analysis of the flue gas shows 10.81% CO2, 3.78% O2 and 85.40 N2. Calculate the atomic ratio of C:H in the hydrocarbon and the % excess air. R. Arun Prasath GETY514 2023 57  Exercises! Consider the hydrocarbon C8H18 combusted completely with 60% excess air. Calculate air fuel ratio in molar and find the dew-point of water assuming the flue products are at 1 atm. Hints: Write the stoichiometric and excess (actual) reaction and balance the same to calculate A/F ratio in molar Find the mole fraction of water and partial pressure of the water vapor. Compare with the standard Table of Saturation temperature and pressure data! R. Arun Prasath GETY514 2023 58 R. Arun Prasath GETY514 2023 59