Question

Hydrogen (H_) is burned with 100 percent excess air that enters the combustion chamber at \(80^{\circ} \mathrm{F}\) 14.5 psia, and 60 percent relative humidity. Assuming complete combustion, determine \((a)\) the air-fuel ratio and (b) the volume flow rate of air required to burn the hydrogen at a rate of \(40 \mathrm{lbm} / \mathrm{h}\).

Short answer

Question: Determine (a) the air-fuel ratio and (b) the volume flow rate of air required to burn hydrogen at a rate of 40 lbm/h with 100% excess air, given that the air entering the combustion chamber is at 80°F and 14.5 psia. Answer: (a) The air-fuel ratio is 4.762 moles of the air per mole of hydrogen. (b) The volume flow rate of air required to burn the hydrogen at a rate of 40 lbm/h is 0.535 m³/s.

Step-by-step solution

Find the stoichiometric equation for the complete combustion of hydrogen

The complete combustion of hydrogen gas with oxygen from the air results in water vapor as a product. The stoichiometric equation for the complete combustion of hydrogen is: $$ \mathrm{H_2} + \frac{1}{2}\mathrm{O_2} \longrightarrow \mathrm{H_2O} $$

Calculate the stoichiometric air-fuel ratio

Air is composed of approximately 21% oxygen (O₂) and 79% nitrogen (N₂) by volume. To find the stoichiometric air-fuel ratio, we need to determine how many moles of air are required to provide one mole of oxygen: $$ \frac{1 \ \mathrm{mole \ O_2}}{0.21} = 4.761 \ \mathrm{moles \ air} $$ One mole of hydrogen needs half mole of oxygen, so the stoichiometric air-fuel ratio is: $$ \frac{4.761 \ \mathrm{moles \ air}}{2} = 2.381 \ \mathrm{moles \ air} \ (\mathrm{per \ mole \ H_2} ) $$

Determine the actual air-fuel ratio considering the 100% excess air

Since there is 100% excess air, the actual air-fuel ratio is twice the stoichiometric air-fuel ratio: $$ \mathrm{Actual \ air-fuel \ ratio} = 2 \times 2.381 = 4.762 $$ (a) The air-fuel ratio is 4.762 moles of the air per mole of hydrogen.

Find the required mass flow rate of air

With the air-fuel ratio, we can find the mass of air needed to combust 40 lbm/h of hydrogen. The molar mass of hydrogen (H₂) is 2 g/mol (1 g/mol for each hydrogen atom, and there are two atoms in a molecule of hydrogen gas). The molar mass of air is approximately 29 g/mol (0.21 * 32 g/mol for O₂ + 0.79 * 28 g/mol for N₂). The mass flow rate of air required to burn the hydrogen at a given rate can be found using the air-fuel ratio: $$ \mathrm{Mass \ flow \ rate \ of \ air } = \mathrm{Air-fuel \ ratio }\times \frac{\mathrm{Mass \ flow \ rate \ of \ H_2}}{\mathrm{Molar \ mass \ of \ H_2}}\times \mathrm{Molar \ mass \ of \ air} \\ = 4.762 \times \frac{40 \ \mathrm{lbm/h}}{(32.2 \ \mathrm{lbm}/\mathrm{kg}) \cdot (2 \ \mathrm{g}/\mathrm{mol})} \times (29 \ \mathrm{g}/\mathrm{mol}) \\ = 179.1 \ \mathrm{ lbm/h } $$

Find the volume flow rate of air

To find the volume flow rate of air required, we can use the Ideal Gas Law: $$ P V = n R T $$ where P is the pressure of the air (14.5 psia = 14.5 lbm/ft² / (32.2 ft/s² × 144 in²/ ft²) = 99,437.9 Pa), V is the volume of air, n is the number of moles of air, R is the specific gas constant for air (R = 8.314 J/(mol·K) / 29 g/mol · 1000 g/kg = 286.9 J/(kg·K)), and T is the temperature in Kelvin (T = (80 °F - 32) × 5/9 + 273.15 = 300.37 K). We can rearrange the Ideal Gas Law to find the volume flow rate of air (V): $$ V = \frac{n R T}{P} \\ V = \frac{179.1 \ \mathrm{ lbm/h} \times \frac{1 \ \mathrm{kg}}{32.2 \ \mathrm{lbm}} \times \frac{1000 \ \mathrm{g}}{1 \ \mathrm{kg}} \times \frac{1 \ \mathrm{mol \ air}}{29 \ \mathrm{g}} \times 286.9 \ \mathrm{J/mol \cdot K} \times 300.37 \ \mathrm{K}}{99,437.9 \ \mathrm{Pa} \times \frac{3600 \ \mathrm{s}}{1 \ \mathrm{h}}} \\ V = 0.535 \ \mathrm{m^3/s} $$ (b) The volume flow rate of air required to burn the hydrogen at a rate of 40 lbm/h is 0.535 m³/s.

Key concepts

Understanding Air-Fuel Ratio

The air-fuel ratio is a critical parameter in combustion processes
and refers to the proportion of air to fuel mixed together for combustion.
For efficient combustion to occur, there must be just the right amount of air
to combine completely with the fuel. This perfect mix is termed the 'stoichiometric air-fuel ratio',
and it ensures that all the fuel reacts with oxygen, leaving no excess fuel or oxygen.

In the context of hydrogen combustion, the stoichiometric ratio is the number of moles of air
needed to supply enough oxygen for each mole of hydrogen. This exact mixture ensures that
each hydrogen atom finds an oxygen atom to react with, preventing the creation of pollutants
like carbon monoxide or unreacted hydrogen in the exhaust. However, in many real-world applications,
such as engines or furnaces, there is typically an excess of air supplied known as 'excess air'.
This excess air is provided to ensure complete combustion since actual conditions may vary, or
to control the temperature of the process. In the provided exercise, the combustion takes place
with 100 percent excess air. This means the actual air used is double the stoichiometric air-fuel ratio,
which helps ensure that all the hydrogen fuel burns completely, producing only water vapor.

Combustion of Hydrogen Examined

Hydrogen is a clean-burning fuel, and when it combusts,
the only product is water vapor. The combustion of hydrogen involves a chemical reaction
where hydrogen molecules (H₂) react with oxygen (O₂) from the air to form water (H₂O).
The reaction is exothermic, releasing energy that can be harnessed for power.

Unlike hydrocarbon fuels, hydrogen does not produce carbon dioxide (CO₂) during combustion,
which makes it an attractive fuel option for reducing greenhouse gas emissions. For hydrogen
to combust efficiently, it must be mixed with air in the right proportions, as determined
based on the stoichiometric air-fuel ratio. As revealed in the exercise solution, when dealing
with an air supply that has a known percent of relative humidity, as is common for air at
any given temperature and pressure, the stoichiometric calculations can still be accurately
conducted by considering the presence of water vapor and its influence on the overall composition
of the air-fuel mixture.

The Ideal Gas Law in Action

The Ideal Gas Law is a fundamental principle
in chemistry and physics that relates the pressure, volume, and temperature of an ideal gas
with the amount of gas present. The law is represented by the equation:
\[ PV = nRT \]
where P is the pressure of the gas, V is its volume, n is the number of moles, R is the universal
gas constant, and T is the temperature in Kelvin. One of the essential applications
of this law is calculating the volume flow rate of gases, as demonstrated in the solution
to our exercise.

By rearranging the equation to solve for V, and knowing the mass flow rate and molar mass
of the incoming air, we can find the volume flow rate needed for combustion. In practical terms,
engineers and scientists use the Ideal Gas Law to design and calculate the parameters for fuel combustion
systems, including engines and industrial furnaces. It's also crucial for environmental calculations,
such as determining the dispersion of pollutants in the atmosphere. For students, grasping the Ideal
Gas Law is vital, as it serves as a gateway to understanding a range of more complex thermodynamic
processes.