13.5 A fuel mixture with the molar analysis 40 % CH3 OH, 50 % C2 H5 OH, and 10 % N2 burns compl

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13.5 A fuel mixture with the molar analysis 40 % CH3 OH, 50...

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.

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.

Key Concepts

Air-Fuel Ratio Calculations

The air-fuel ratio is the mass or mole ratio of air to fuel, and it is a crucial parameter in combustion processes. It is determined by calculating the total amount of air (or oxygen and accompanying nitrogen) used in the combustion reaction and comparing it to the fuel used. This ratio can be expressed on both a molar basis (moles of air per mole of fuel) and a mass basis (mass of air per mass of fuel), providing different insights into the efficiency and design of combustion systems.

Air Composition

Air is not pure oxygen; it is a mixture of gases, primarily oxygen and nitrogen, with oxygen usually being about 21% by volume and nitrogen about 79%. In combustion calculations, air is often represented by a combination of oxygen (O2) and nitrogen (N2) in a fixed mole ratio, such as O2 + 3.76 N2, to accurately account for the inert gases present in the combustion process.

Fuel Mixtures and Molar Analysis

This concept involves interpreting the composition of a fuel mixture given in mole percentages. When a mixture contains several fuel components along with inert substances (like nitrogen), it is necessary to consider each component's contribution to the overall combustion reaction. The molar analysis helps in writing the reaction equation and calculating the precise amounts of reactants and products.

Balancing Chemical Equations

Balancing chemical equations is the process of ensuring that the number of atoms for each element is the same on both the reactant and product sides of a reaction. In combustion problems, this typically involves accounting for carbon, hydrogen, oxygen and any other species present in both fuel and oxidizer, including inert components like nitrogen.

Excess Air

Excess air refers to the additional amount of air supplied beyond the stoichiometric requirement to ensure that combustion is complete. It is often expressed as a percentage over the stoichiometric amount. Handling excess air in combustion calculations means accounting for surplus oxygen that appears in the products and adjusting the balance accordingly.

Stoichiometric Combustion

This concept involves determining the exact amount of oxygen required to completely oxidize all the fuel to its combustion products (typically CO2 and H2O), without leaving any excess fuel or oxygen. It requires balancing the atoms of each element (carbon, hydrogen, oxygen, nitrogen, etc.) to derive the stoichiometric coefficients in the reaction equation.

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