Combustion and Gasification Systems

Combustion and Gasification Systems

• The fundamental conversion process involves fuel and an oxidizer as reactants.
• Reactants and oxidizers undergo chemical reactions, producing fully oxidized products like CO2 and H2O, along with heat.
• Products typically include CO2, H2O, and nitrogen, where nitrogen doesn't participate in the reaction.
• Excess oxygen may also be present in the products if excess air is supplied during combustion.
• For a hydrocarbon fuel (CuHw), determining the air needed for full combustion is crucial.
• Complete combustion yields fully oxidized products (CO2 and H2O).
• Stoichiometrically balanced equations represent complete combustion.
• In practice, products may include carbon monoxide (CO) and uncombusted hydrocarbons, along with nitrogen and oxygen (when in excess).
• Nitrogen can lead to NOx compounds (N2O, NO2) and CO2 and H2O when under ideal conditions.
• In actual conditions, excess oxygen may be required for complete fuel burning.

Oxidizer and Combustion Reactions

• The amount of air/oxidizer needed to fully burn hydrocarbon fuels (CH4, C2H6, C3H8) must be determined.
• 'n' is the stoichiometric coefficient, depicting the moles of oxidizer needed for the reaction.
• The balanced equation is: CuHw + n(O2 + 3.76N2) -> yCO2 + zH2O + cN2 + dO2
• 'c' represents 3.76n (nitrogen moles), and 'd' represents excess oxygen moles.
• Five unknowns (n, y, z, c, d) need estimation to balance the equation stoichiometrically.
• Estimating these unknowns allows for the balanced equation of hydrocarbon combustion.
• The assumption is no excess O2.
• Theoretical air represents the air required for ideal combustion.
• Balancing atoms of each element yields four more equations (carbon, hydrogen, oxygen, nitrogen).
• These four equations, combined with the main equation, enable calculating the unknowns automatically.

Balancing Atoms in Combustion Reactions

• Balancing carbon atoms: 'u' (carbon in reactant) equals 'y' (carbon in product CO2) as only one C exists.
• Balancing hydrogen atoms: 'w' (hydrogen in reactant) results in '2z' (2 hydrogens per product H2O), giving z = w/2.
• Balancing oxygen atoms: '2n' (oxygen in reactant air) equates to '2y + z' (oxygen in products), leading to n = (y + z)/2.
• 'n' can be in terms of hydrocarbon fuel, e.g., n = u + w/4.
• Nitrogen balance: 'n' moles of incoming air are accompanied by 3.76 moles of N2, resulting in 2c (nitrogen in product), so c = 3.76n.
• This balance is achievable when excess air is used, but with value changes.
• Understanding combustion enables solving examples using hydrocarbon fuels.

Example: Ethane Combustion

• An industrial boiler with 3 mole percent O2 in flue gases is fueled by ethane (C2H6).
• Determine the operating air-to-fuel ratio.
• Molecular weight of air is 29, fuel (ethane) is 30; raw gas has 3% oxygen (0.03 mole fraction).
• Complete combustion assumes all fuel converts to CO2 and all hydrogen to H2O, with 3 mole percent oxygen also in the gas.
• No CO or NOx are formed; only CO2, H2O, and O2 are the resultants.
• Need to measure the amount of oxidizer necessary to burn fuel, releasing CO2, H2O, nitrogen, and oxygen.
• Balance the equation using theoretical air supplied.
• The equation: C2H6 + n(O2 + 3.76N2) --> 2CO2 + 3H2O + dO2 + 3.76nN2
• Balancing carbon and hydrogen: 2C becomes 2CO2 and 6H becomes 3H2O.
• Oxygen balance: '2n = (2x2) + 3 + 2d', results in 'd = n - 7/2'.
• Find 'n' using the mole fraction for oxygen.

Calculating Air-to-Fuel Ratio

• Mole fraction of oxygen = (moles of oxygen) / (total moles of gas mixture).
• From the combustion equation, oxygen is 'd'/ ('2 + 3 + d + 3.76n').
• 'd = n - 3.5'
• Substitute d: 0.03 = (n - 3.5) / (2 + 3 + (n - 3.5) + 3.76n)
• Solve for 'n', which comes to 4.136 moles of air.
• Theoretically, calculate air to fuel ratio, meaning stoichiometric calculations.
• Stoichiometric air-to-fuel ratio is mass of air/mass of fuel.
• This is the exact oxygen or air needed for burning hydrocarbon.
• Air mass is '4.136(1O2 + 3.76N2)' and fuel mass is '1 C2H6'.
• Value is 193 that is kilogram of air required per kilogram of fuel, for full combustion.
• Air/fuel ratio, number of moles of oxygen, air, or oxidizer required for burning process.

Combustion Equation and Ratios

• Develop combustion equation and determine air-to-fuel ratio.
• Ethane is the fuel, considering theoretical air.
• Consider 50% excess air.
• Two examples: theoretical air and 50% of theoretical air (excess).
• Molecular weight of air is 29, and fuel is 30.
• Complete combustion means all fuel converts to CO2 and all hydrogen to H2O.

Combustion of Ethane with Theoretical Air

• Ethane (C2H6) and air (O2 + 3.76N2) undergo a complete combustion process.
• The products are CO2, H2O, N2, and potentially O2.
• Theoretical air means that the product does not contain oxygen, all oxygen is consumed in the combustion
• The combustion equation represents the complete combustion of hydrocarbon fuel to produce CO2 and H2O.
• The balanced equation for complete combustion of ethane with theoretical air is C2H6 + 3.5(O2 + 3.76N2) -> 2CO2 + 3H2O + 3.5(3.76)N2.
• The coefficients in the balanced equation can be equated to determine the moles of each element.
• The number of moles of air (n) is calculated as y + z/2 or u + w/4, where y and u represents moles of CO2 and z and w represent Hydrogen.
• By simplifying the equation, the nitrogen in the product gas is approximately 13.16 moles.
• Theoretical air-fuel ratio is the mass of air divided by the mass of fuel (stoichiometric amount).
• Mass of air is 4.76 multiplied by the number of moles of air required for combustion, multiplied by the molecular weight of air (29).
• The molecular weight of ethane (C2H6) is 30.
• The theoretical air to fuel ratio for ethane is approximately 16.10 kg of air per kg of fuel.

Combustion of Ethane with Excess Air

• When 50% excess air is supplied, the amount of air is multiplied by 1.5 (150% of theoretical air).
• The balanced equation with 50% excess air is C2H6 + 1.5 * 3.5(O2 + 3.76N2) -> 2CO2 + 3H2O + 0.53.5O2 + 1.53.5(3.76)N2.
• The excess oxygen appears in the product gas (0.5 * 3.5 moles of O2).
• Nitrogen doesn't participate in the reaction; the total nitrogen in the product is 1.5 * 3.5 * 3.76 moles.
• The new air to fuel ratio with 50% excess air is approximately 24.15 kg of air per kg of fuel, which is higher than with theoretical air.

Calculating Percentage of Carbon Dioxide in Product Gas

• Ethane is combusted with 50% excess air at 25°C.
• Determination of air-fuel ratio and percentage of CO2 by volume in the product is required.
• Mole fraction of CO2 is calculated as moles of CO2 divided by total moles in the product gas (n total or n mix).
• The total moles in the product gas are obtained by summing up all product elements (CO2, H2O, O2, and N2).
• Steps for creating balanced stoichiometric equations are outlined
• The procedure for calculating the theoretical amount of required air is outlined
• The percentage of CO2 in the product gas is about 7.5 mole percent.