Methane (CH4) is burned with air (79% N2 and 21% 02 by volume) at atmospheric pressure. The molar analysis of the flue gas yields CO, = 10.00%, O2 = 2.41%, CO = 0.52%, and N2 = 87.07%. Balance the combustion equation and determine the mass air-fuel ratio, the percentage of stoichiomet- ric air, and the percentage of excess air.
Ideal and Real Gases
Ideal gases obey conditions of the general gas laws under all states of pressure and temperature. Ideal gases are also named perfect gases. The attributes of ideal gases are as follows,
Gas Laws
Gas laws describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. The very first observations about the physical properties of gases was made by Robert Boyle in 1662. Later discoveries were made by Charles, Gay-Lussac, Avogadro, and others. Eventually, these observations were combined to produce the ideal gas law.
Gaseous State
It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed forms of matter studied are solids, gases and liquids. The best example of a substance that is present in different states is water. It is solid ice, gaseous vapor or steam and liquid water depending on the temperature and pressure conditions. This is due to the difference in the intermolecular forces and distances. The occurrence of three different phases is due to the difference in the two major forces, the force which tends to tightly hold molecules i.e., forces of attraction and the disruptive forces obtained from the thermal energy of molecules.
![**Methane Combustion with Air: Analysis and Calculations**
**Introduction:**
Methane (CH₄) undergoes combustion with air, which consists of 79% Nitrogen (N₂) and 21% Oxygen (O₂) by volume, at atmospheric pressure. The following information details the molar analysis of the flue gases produced and outlines the steps to balance the combustion equation. Furthermore, we will determine the mass air-fuel ratio, the percentage of stoichiometric air, and the percentage of excess air.
**Molar Analysis of Flue Gas:**
- CO₂: 10.00%
- O₂: 2.41%
- CO: 0.52%
- N₂: 87.07%
**Objectives:**
1. Balance the combustion equation.
2. Determine the mass air-fuel ratio.
3. Calculate the percentage of stoichiometric air.
4. Compute the percentage of excess air.
**Detailed Explanation:**
1. **Balancing the Combustion Equation:**
Ensure that the number of atoms for each element is the same on both sides of the equation. The general unbalanced combustion equation for methane is given by:
\[
CH₄ + a \cdot (O₂ + 3.76 \cdot N₂) \rightarrow b \cdot CO₂ + c \cdot H₂O + d \cdot O₂ + e \cdot CO + f \cdot N₂
\]
Here, \( a \) represents the moles of air used, and coefficients \( b, c, d, e, \) and \( f \) represent the moles of respective products.
2. **Determine the Mass Air-Fuel Ratio (AFR):**
The mass AFR is the ratio of the mass of air used to the mass of fuel burned. This ratio can be determined using balanced chemical equations and molar masses of the reactants and products.
3. **Calculate the Percentage of Stoichiometric Air:**
Stoichiometric air is the exact amount of oxygen needed for complete combustion of the fuel without any remaining oxygen or fuel. The percentage is a comparison between the actual air supplied to the stoichiometric air required.
4. **Compute the Percentage of Excess Air:**
Excess air is the additional air supplied beyond the stoichiometric requirement to ensure complete combustion. This percentage reflects the margin of air](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F719fb227-8848-4d2e-ac6c-7c3a609c5fdc%2F5a3716ed-a607-4dec-ad35-ce0895bfca81%2Fv217u17_processed.jpeg&w=3840&q=75)
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