Question

13.5 A fuel mixture with the molar analysis \(40 \% \mathrm{CH}_{3} \mathrm{OH}, 50 \%\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), and \(10 \% \mathrm{~N}_{2}\) burns completely with \(33 \%\) excess air. Determine (a) the balanced reaction equation. (b) the air-fuel ratio, both on a molar and mass basis.

Short answer

(a) The balanced equation is provided in Step 6. (b) The air-fuel ratio is 14.967 (molar) and 12.185 (mass).

Step-by-step solution

Write the Molar Analysis of the Fuel Mixture

The given fuel mixture consists of: \[40\text{%} \text{CH}_3\text{OH}, 50\text{%} \text{C}_2\text{H}_5\text{OH}, \text{ and } 10\text{%} \text{N}_2\] Hence, the molar amounts are \(0.4 \text{CH}_3\text{OH}\), \(0.5 \text{C}_2\text{H}_5\text{OH}\), and \(0.1 \text{N}_2\).

Express the Combustion Reaction

The generic combustion reaction with oxygen is: \[a \text{CH}_3\text{OH} + b \text{C}_2\text{H}_5\text{OH} + c \text{N}_2 + d (\text{O}_2 + 3.76\text{N}_2) \rightarrow e \text{CO}_2 + f \text{H}_2\text{O} + g \text{N}_2\]

Setup the Stoichiometric Combustion Reactions

\(\text{CH}_3\text{OH} + 1.5\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)\( \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}\)

Account for Nitrogen in the Combustion

Nitrogen, \( \text{N}_2 \), does not participate in the chemical reaction but needs to be balanced as well. Therefore, it is found unreacted in the products.

Balance the Combustion Equation

The molar amounts of oxygen consumed: \(0.4 \times 1.5 = 0.6\) moles for \(\text{CH}_3\text{OH}\) and \(0.5 \times 3 = 1.5\) moles for \(\text{C}_2\text{H}_5\text{OH}\). Total oxygen required is \(0.6 + 1.5 = 2.1\) moles. With 33% excess air, the actual moles of \text{O}_2\rightarrow 2.1 \times 1.33 = 2.793 \ moles.

Write the Final Balanced Combustion Equation

The balanced equation is: \[0.4 \text{CH}_3\text{OH} + 0.5 \text{C}_2\text{H}_5\text{OH} + 0.1 \text{N}_2 + 2.793 (\text{O}_2 + 3.76 \text{N}_2) \rightarrow 0.4 \text{CO}_2 + 1 \text{CO}_2 + 0.8 \text{H}_2\text{O} + 1.5 \text{H}_2\text{O} + 2.793 \times 3.76 \text{N}_2\]

Calculate Air-Fuel Ratio (Molar Basis)

The moles of air used are \(2.793 + (2.793 \times 3.76)\). The moles of fuel is \(0.4 + 0.5\). Air-fuel ratio (molar) is given by: \[ \text{AFR}_{molar} = \frac{\text{moles of air}}{\text{moles of fuel}} = \frac{2.793 + (2.793 \times 3.76)}{0.9} = \frac{13.471}{0.9} = 14.967 \]

Calculate Air-Fuel Ratio (Mass Basis)

For mass basis, the amounts are converted using the molar masses: \(\text{CH}_3\text{OH}: 32 \text{g/mol}, \text{C}_2\text{H}_5\text{OH}: 46 \text{g/mol}, \text{Air}: 28.97 \text{g/mol}\). Total mass of fuel per mole is \(0.4 \times 32 + 0.5 \times 46\), mass of air is: \(13.471 \times 28.97\).

Final Air-Fuel Ratio (Mass Basis)

Calculating masses: Fuel mass \((0.4 \times 32 + 0.5 \times 46) = 32\) g/mole and Air mass \(13.471 \times 28.97 = 390.15\) g/mole. Thus, AFR (mass) is: \[ \text{AFR}_{mass} = \frac{390.15}{32} = 12.185 \]

Key concepts

balanced reaction equation

In chemistry, a balanced reaction equation ensures that the number of atoms for each element is the same on both the reactant and product sides of the equation. This is essential for the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

To balance a combustion reaction, we start by identifying and counting the moles of each element in the fuel and oxygen molecules initially present. We then adjust the coefficients of the products so that they contain the same number of each type of atom as the reactants. For example, in our given problem, we started with 0.4 moles of methanol (CH3OH) and 0.5 moles of ethanol (C2H5OH). By carefully balancing the oxygen (O2) and product molecules (CO2, H2O), we reached our final balanced equation: \[0.4 \text{CH}_3\text{OH} + 0.5 \text{C}_2\text{H}_5\text{OH} + 0.1 \text{N}_2 + 2.793 (\text{O}_2 + 3.76 \text{N}_2 ) \rightarrow 0.4 \text{CO}_2 + 1 \text{CO}_2 + 0.8 \text{H}_2\text{O} + 1.5 \text{H}_2\text{O} + 2.793 \times 3.76 \text{N}_2 \]

Balancing requires meticulous adjustments and cross-checking to ensure all atoms are accounted for properly.

air-fuel ratio

The air-fuel ratio (AFR) is a crucial concept in combustion analysis. It represents the ratio of the amount of air to the amount of fuel used during the combustion process. The AFR can be expressed both in terms of moles (molar basis) and mass (mass basis).

For our problem, we calculated the molar AFR by dividing the total moles of air used by the total moles of fuel. Given 2.793 moles of oxygen and the corresponding nitrogen, the total air moles were found to be 13.471. With 0.9 moles of fuel, the molar AFR was found to be 14.967.

When calculating the mass AFR, we accounted for the molecular masses of the fuels (32 g/mol for methanol and 46 g/mol for ethanol) and the air (28.97 g/mol). After converting the moles to mass, we determined that the mass AFR was 12.185.

Understanding the AFR helps in designing efficient combustion processes, ensuring optimal fuel usage and minimal pollutant emission.

stoichiometric combustion

Stoichiometric combustion refers to the ideal combustion process in which fuel burns completely with the exact amount of oxygen necessary to maintain a perfect balance – no excess air and no unburnt fuel.

The stoichiometric equation for methanol (CH3OH) and ethanol (C2H5OH) are given as: \[ \text{CH}_3\text{OH} + 1.5\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \text{ and } \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \]

However, in real-world applications, operating at stoichiometric conditions can often lead to higher temperatures which might be undesirable. It also means there's no margin for error, making it difficult to achieve in practice. Thus, stoichiometric combustion calculations primarily serve as a baseline or reference.

excess air

Excess air is the additional air supplied to a combustion process beyond the stoichiometric amount necessary for complete combustion. This concept is important because it ensures complete combustion of the fuel, reducing emissions of unburnt hydrocarbons and carbon monoxide.

In our exercise, we had 33% excess air, meaning that the actual moles of oxygen provided were 33% more than the theoretical requirement. If the stoichiometric moles of oxygen are calculated to be 2.1, then with excess air, the actual moles of oxygen become \(2.1 \times 1.33 = 2.793\).

While excess air ensures complete combustion, too much of it can lead to inefficiencies and energy losses as more heat is absorbed to raise the temperature of the extra air. Thus, careful management of excess air is essential for optimizing fuel combustion and efficiency.