Question

A certain natural gas has the following volumetric analysis: 65 percent \(\mathrm{CH}_{4}, 8\) percent \(\mathrm{H}_{2}, 18\) percent \(\mathrm{N}_{2}\) 3 percent \(\mathrm{O}_{2},\) and 6 percent \(\mathrm{CO}_{2}\). This gas is now burned completely with the stoichiometric amount of dry air. What is the air- fuel ratio for this combustion process?

Short answer

Answer: The air-fuel ratio for the complete combustion of the given natural gas with the stoichiometric amount of dry air is approximately 2.83.

Step-by-step solution

1. Identify the combustion reactions

We need to identify the combustion reactions for each of the components of the natural gas: - CH\(_4\) + 2O\(_2\) → CO\(_2\) + 2H\(_2\)O - H\(_2\) + 1/2O\(_2\) → H\(_2\)O - As for the N\(_2\) and O\(_2\), they do not take part in the combustion and are considered inert. - CO\(_2\) is a product of the combustion of other components.

2. Calculate the amounts of components in the natural gas

Now that we have the components and their respective volumes, we can calculate the amounts. Let's assume we have 100mL of gas (just for convenience, as percentages will be equal to the actual volumes): - CH\(_4\): 65mL - H\(_2\): 8mL - N\(_2\): 18mL - O\(_2\): 3mL - CO\(_2\): 6mL

3. Calculate the oxygen (O\(_2\)) required for complete combustion of the gas

From the reactions, we can observe the following stoichiometric relationships for O\(_2\) usage: - 1 mol of CH\(_4\) requires 2 mol of O\(_2\) - 1 mol of H\(_2\) requires 1/2 mol of O\(_2\) Let's calculate the moles of each component and find out how much O\(_2\) is required: - For CH\(_4\): - n\(_{CH_4}\) = \(\frac{65}{22.4}\) (mol) - 2 × n\(_{CH_4}\) = \(\frac{130}{22.4}\) (mol of O\(_2\)) - For H\(_2\): - n\(_{H_2}\) = \(\frac{8}{22.4}\) (mol) - 1/2 × n\(_{H_2}\) = \(\frac{4}{22.4}\) (mol of O\(_2\)) Now we sum up the O\(_2\) required for both the components: - O\(_2\) required = \(\frac{130+4}{22.4}\) mol

4. Calculate the amount of air needed

As air consists of 21% of oxygen (O\(_2\)) and 79% of nitrogen (N\(_2\)) by volume, we can determine the amount of air needed: O\(_2\) required = \(\frac{134}{22.4}\) mol Air = \(\frac{\frac{134}{22.4}}{0.21}\) mol (since 21% of air is oxygen)

5. Determine the air-fuel ratio

Now we have the amount of fuel (100 mL) and the amount of air required for complete combustion. The air-fuel ratio can be determined as: AFR = \(\frac{\text{Air}}{\text{Fuel}}\) AFR = \(\frac{\frac{\frac{134}{22.4}}{0.21}}{100}\) Hence, the air-fuel ratio for this combustion process is approximately \(2.83\).

Key concepts

Stoichiometry in Combustion

Stoichiometry is the mathematical relationship between reactants and products in a chemical reaction. In the context of combustion, it deals with the precise calculation of how much fuel can react with a given amount of oxygen. This balance ensures there is just enough oxygen to completely burn the fuel, which is essential for efficient energy production and minimizing pollutants.

When analyzing combustion processes, a stoichiometric mixture means that fuel and air are mixed in the perfect proportions according to the chemical equation of the reaction. For instance, methane (CH_4) requires two molecules of oxygen (O_2) to produce one molecule of carbon dioxide (CO_2) and two molecules of water (H_2O). If we have 65 percent methane in our fuel, then we need exactly double that amount of oxygen for complete combustion.

Understanding stoichiometry is critical in applications like engine design, where the air-fuel ratio affects performance and emissions. Ensuring that the stoichiometric balance is maintained optimizes energy use and reduces the production of harmful byproducts such as carbon monoxide.

Combustion Reactions

Combustion reactions are exothermic processes where a fuel reacts with an oxidant, releasing energy in the form of heat and light. In the scenario given, we are working with a natural gas mixture that includes components like methane (CH_4), hydrogen (H_2), and other inert gases like nitrogen (N_2) and carbon dioxide (CO_2).

For methane and hydrogen, which are the main combustible components, the combustion reactions can be represented as:

• CH_4 + 2O_2 → CO_2 + 2H_2O
• H_2 + 1/2O_2 → H_2O

The nitrogen (N_2) and the oxygen (O_2) in the gas mixture do not combust and are considered inert. During combustion, they act as diluents, reducing the likelihood of knock in engines and contributing to the control of combustion temperatures.

It's important to note that not every part of the gas combusting contributes to providing energy; carbon dioxide (CO_2), for example, is a product of the combustion and does not react further under normal conditions.

Thermochemistry of Combustion

Thermochemistry is the study of energy changes during chemical reactions. In the case of combustion, it focuses on how the chemical energy stored in fuels is transformed into heat. This field gives us insights into the energy yield from different fuels and helps us understand the efficiency of combustion processes.

During combustion, the energy content of the fuel is released primarily as heat. The amount of heat produced is determined by the calorific value of the fuel and its stoichiometry in the reaction. For example, methane has a high calorific value, meaning it releases a large amount of heat per unit of mass when it is burned, which can be calculated using its specific energy values and the amount combusted.

In a controlled environment like an engine or a furnace, we desire a combustion process that releases heat in a way that is efficient and manageable. If combustion is incomplete due to an incorrect air-fuel ratio, energy is lost, and pollutants are created, demonstrating the importance of using stoichiometry to calculate the exact air-fuel ratio necessary for complete combustion.