Question

A gaseous fuel with a volumetric analysis of 45 percent \(\mathrm{CH}_{4}, 35\) percent \(\mathrm{H}_{2},\) and 20 percent \(\mathrm{N}_{2}\) is burned to completion with 130 percent theoretical air. Determine \((a)\) the air-fuel ratio and ( \(b\) ) the fraction of water vapor that would condense if the product gases were cooled to \(25^{\circ} \mathrm{C}\) at 1 atm.

Short answer

Answer: The air-fuel ratio is 4.93, and the fraction of water vapor that would condense at 25°C and 1 atm is 0.0313.

Step-by-step solution

Find the chemical equation for the combustion of the fuel

Given that the fuel consists of 45% CH4, 35% H2, and 20% N2, let's write the general combustion equation for the fuel: $$ (\text{CH}_{4})_{0.45}+(\text{H}_{2})_{0.35}+(\text{N}_{2})_{0.20}+\text{O}_{2}+\text{N}_2→\text{CO}_{2}+\text{H}_{2}\text{O}+\text{N}_{2} $$ Now, let's balance the equation: $$ (\text{CH}_{4})_{0.45}+(\text{H}_{2})_{0.35}+(\text{N}_{2})_{0.20}+2\text{O}_{2}+\text{N}_2→\text{CO}_{2}+2\text{H}_{2}\text{O}+\text{N}_{2} $$

Calculate the air-fuel ratio

To determine the air-fuel ratio, we need to calculate the amount of air required for complete combustion. Air is composed of 21% O2 and 79% N2 by volume. The theoretical amount of O2 required for the combustion is given by the stoichiometric equation found in Step 1. The moles of O2 required for the combustion of the fuel are as follows: $$ \begin{aligned} \text{moles of O}_{2} &= 0.45(\text{moles of CH}_{4}) + 0.35 (\text{moles of H}_{2}) \\ &= 0.45 + 0.35 \\ &= 0.80\,\text{moles of O}_{2} \end{aligned} $$ Since the combustion occurs with 130% theoretical air, the actual amount of air required is 1.3 times the theoretical amount: $$ \text{Actual moles of O}_{2} = 1.3 \times 0.80 = 1.04\,\text{moles of O}_{2} $$ Consequently, the moles of N2 in the air are: $$ \text{moles of N}_{2} = \frac{79}{21} \times 1.04 = 3.89 \, \text{moles of N}_{2} $$ Now, the air-fuel ratio can be calculated as: $$ \text{Air-Fuel Ratio} = \frac{\text{moles of O}_{2} + \text{moles of N}_{2}}{\text{moles of Fuel}} = \frac{1.04 + 3.89}{1} = 4.93 $$

Calculate the fraction of water vapor that would condense at 25°C and 1 atm

In the combustion products, we have 2 moles of water vapor. We will use the vapor pressure of water at 25°C (3.17 kPa) to find the fraction of water vapor that would condense: $$ \text{Mole Fraction of water vapor at 25°C} = \frac{\text{Vapor Pressure of Water at 25°C}}{\text{Total Pressure}} = \frac{3.17 \,\text{kPa}}{101.3 \,\text{kPa}} = 0.0313 $$ Thus, the fraction of H2O that would condense at 25°C and 1 atm is 0.0313. So, the answers to the exercise are: \((a)\) The air-fuel ratio is 4.93, and \((b)\) the fraction of water vapor that would condense at 25°C and 1 atm is 0.0313.

Key concepts

Understanding Air-Fuel Ratio

When discussing combustion, 'air-fuel ratio' (AFR) is a crucial concept. Simply put, this ratio indicates how much air is needed to burn a certain amount of fuel completely. Getting the balance just right is key; too much air can cool down the combustion temperature leading to incomplete burning, whereas too little air means not all fuel gets burned, causing pollution and wasting fuel.

In complete combustion, every molecule of fuel gets enough oxygen to produce carbon dioxide and water. For gaseous fuels, like the one in our exercise—a mix of methane (CH₄), hydrogen (H₂), and nitrogen (N₂)—calculating the precise AFR involves stoichiometry. We balance the chemical reaction to find out the exact amount of oxygen needed and use the composition of air to convert it to the volume of air required. The air-fuel ratio is critical for engineers and scientists to ensure efficient, economic, and environmental-friendly combustion processes.

The Theoretical Air Calculation Process

To elucidate the concept of 'theoretical air calculation,' let's start from the basics. The air we breathe is approximately 21% oxygen and 79% nitrogen by volume. The theoretical air is the precise amount of this air mixture that would be required to completely combust a fuel without leaving any excess oxygen and without any unburnt fuel.

In theoretical air calculation, we first establish the combustion equation for the given fuel blend. The next step is to balance this equation in terms of oxygen, as demonstrated in the exercise. We then account for the air composition to determine the volume of air needed. Remember that the theoretical air is an ideal quantity and, in practice, actual combustion systems operate on 'excess air' to assure complete combustion—hence the 130% figure in our exercise. This calculation is the cornerstone for combustion efficiency and is used widely in designing burners, engines, and industrial combustion systems.

Water Vapor Condensation in Combustion

Finally, the phenomenon of 'water vapor condensation' is significant in understanding product gases from combustion. When a hydrocarbon fuel burns, one of the by-products is water vapor. The temperature and pressure of the exhaust gases determine if this water vapor remains as a gas or condenses into liquid water.

During the cooling of product gases, as in our exercise example, some water vapor may condense at a specific temperature—25°C in this case. The fraction of water vapor that will condense at a given temperature is governed by the physical properties (like vapor pressure) of water and the current pressure conditions. Calculating this fraction is important for applications such as heat recovery in exhaust systems and in environmental control to understand how much water is being added to the atmosphere. It is also a factor in the dew point, which indicates the temperature at which water starts to condense out of a gas mixture.