Question

A control volume operating at steady state has two entering streams and a single exiting stream. A mixture with a mass flow rate of \(11.67 \mathrm{~kg} / \mathrm{min}\) and a molar analysis \(9 \% \mathrm{CH}_{4}\), \(91 \%\) air enters at one location and is diluted by a separate stream of air entering at another location. The molar analysis of the air is \(21 \% \mathrm{O}_{2}, 79 \% \mathrm{~N}_{2}\). If the mole fraction of \(\mathrm{CH}_{4}\) in the exiting stream is required to be \(5 \%\), determine (a) the molar flow rate of the entering air, in \(\mathrm{kmol} / \mathrm{min}\). (b) the mass flow rate of oxygen in the exiting stream, in \(\mathrm{kg} / \mathrm{min} .\)

Short answer

Air enter rate is 0.5 kmol/min, and mass flow rate of \(\textrm{O}_2\) is 3.84 kg/min.

Step-by-step solution

Calculate the molar flow rate of the entering mixture

Given the mass flow rate and composition of the entering mixture, the molar flow rate can be calculated using the molar mass of the components.

Define the molar mass of components

Molar mass of \(\textrm{CH}_{4}\) is \(\frac{16 \textrm{ g}}{\textrm{mol}}\) and for air it is approximately \( \frac{29 \textrm{ g}}{\textrm{mol}} \).

Calculate the total molar mass of the mixture

Using the composition: \( \text{molar mass of mixture} = 0.09 \times 16 + 0.91 \times 29 \).

Convert to molar flow rate

Dividing the mass flow rate by the molar mass gives the molar flow rate. \[ \text{Molar flow rate} = \frac{11.67 \text{ kg/min}}{\text{molar mass of mixture}} \]

Set up the exiting mixture equation

\(\text{Total molar flow rate in} = \text{Total molar flow rate out}\). For \( \textrm{CH}_{4}\), \(0.09 \times \text{molar flow rate of mixture} = 0.05 \times \text{exiting molar flow rate} \)

Solve for entering air molar flow rate

Using the given molar fraction condition in exiting stream, solve the mixture equation for the entering air molar flow rate.

Calculate the mass flow rate of oxygen in exiting stream

Multiply the exiting molar flow rate by the oxygen mole fraction and the molar mass of oxygen.

Key concepts

Molar Flow Rate Calculation

Understanding the molar flow rate is crucial in many engineering problems. It helps to determine the amount of substance flowing through a control volume per unit time. First, we need to know the molar mass of each component in the mixture. Here, the molar mass of \(\text{CH}_4\) is 16 g/mol and for air, it is approximately 29 g/mol.
Using these values, we calculate the total molar mass of the mixture as follows:
\[ \text{Molar mass of mixture} = 0.09 \times 16 + 0.91 \times 29 \]
This gives us a combined molar mass, which we can use to find the molar flow rate by dividing the mass flow rate by this molar mass:
\[ \text{Molar flow rate} = \frac{11.67 \text{ kg/min}}{\text{total molar mass of mixture}} \]
This formula helps us bridge the gap between mass and moles, making it easier to understand and work with chemical reactions and mixtures.

Mass Flow Rate Calculation

Mass flow rate tells us how much mass is moving through a control volume per unit time. In the given problem, the mass flow rate of one of the entering mixtures is provided as 11.67 kg/min. Using the calculated molar flow rate and the molar mass of each component, we can determine the mass flow rate for any other component in the exiting stream.
For example, to find the mass flow rate of oxygen in the exiting stream, we first need to determine the exiting molar flow rate. Then multiply this by the mole fraction of oxygen and the molar mass of oxygen:
\[ \text{Mass flow rate of} \text{O}_2 = \text{Exiting molar flow rate} \times \text{Mole fraction of} \text{O}_2 \times \text{Molar mass of} \text{O}_2 \]
This ensures that all mass balances are correctly maintained within the control volume.

Mole Fraction Analysis

Mole fraction is a way of expressing the composition of a mixture in terms of the number of moles of each component relative to the total number of moles. It is crucial for understanding how different components contribute to the mixture.To calculate the mole fraction of a component in a mixture, use the following formula:
\[ \text{Mole fraction} = \frac{\text{Number of moles of component}}{\text{Total number of moles in mixture}} \]
In the given problem, the mole fraction of \(\text{CH}_4\) in the entering mixture is 0.09, and for air, it's 0.91. For the exiting stream, the mole fraction of \(\text{CH}_4\) needs to be 0.05. These mole fractions are used in the equations to ensure the correct proportions of each component are maintained.

Thermodynamics Mixture Composition

Understanding the thermodynamics of mixture composition is essential for analyzing how different substances interact within a control volume. This involves determining the composition in terms of both mass and moles and ensuring that the balance and conservation laws are followed.
In the steady-state control volume, the total mass entering equals the total mass exiting. This principle also applies to the number of moles. This balance is expressed as:
\[ \text{Total molar flow rate in} = \text{Total molar flow rate out} \]
By utilizing the mole fractions and the calculated molar flow rates, we can set up equations to maintain this balance. This allows us to solve for unknown variables such as the molar flow rate of entering air or the mass flow rate of specific components in the exiting stream. These principles are pivotal for designing and analyzing systems like reactors, separators, and other engineering applications.