Question

Air enters a one-inlet, one-exit control volume at 8 bar, \(600 \mathrm{~K}\), and \(40 \mathrm{~m} / \mathrm{s}\) through a flow area of \(20 \mathrm{~cm}^{2}\). At the exit, the pressure is 2 bar, the temperature is \(400 \mathrm{~K}\), and the velocity is \(350 \mathrm{~m} / \mathrm{s}\). The air behaves as an ideal gas. For steady-state operation, determine (a) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\). (b) the exit flow area, in \(\mathrm{cm}^{2}\).

Short answer

The mass flow rate is 0.371 kg/s. The exit flow area is 6.088 cm^2.

Step-by-step solution

Given Data

Identify and write down the given data from the problem. Inlet conditions: Pressure, \( P_1 = 8 \text{ bar} \) Temperature, \( T_1 = 600 \text{ K} \) Velocity, \( V_1 = 40 \text{ m/s} \) Flow area, \( A_1 = 20 \text{ cm}^2 = 0.002 \text{ m}^2 \). Exit conditions: Pressure, \( P_2 = 2 \text{ bar} \) Temperature, \( T_2 = 400 \text{ K} \) Velocity, \( V_2 = 350 \text{ m/s} \)

Convert pressures

Convert the pressures from bar to Pascals for standard units. \[ P_1 = 8 \text{ bar} = 8 \times 10^5 \text{ Pa} \] \[ P_2 = 2 \text{ bar} = 2 \times 10^5 \text{ Pa} \]

Use Ideal Gas Equation

Use the ideal gas equation to find the densities of the air at the inlet and the exit. The ideal gas equation is \( \rho = \frac{P}{RT} \), where \( R = 287 \text{ J/kg.K} \) (for air). \[ \rho_1 = \frac{ P_1 }{ R T_1 } = \frac{ 8 \times 10^5 }{ 287 \times 600 } = 4.641 \text{ kg/m}^3 \] \[ \rho_2 = \frac{ P_2 }{ R T_2 } = \frac{ 2 \times 10^5 }{ 287 \times 400 } = 1.742 \text{ kg/m}^3 \]

Calculate Mass Flow Rate

Use the continuity equation to calculate the mass flow rate. The mass flow rate \( \textdot{m} \) is given by \( \textdot{m} = \rho A V \). At the inlet: \[ \textdot{m} = \rho_1 A_1 V_1 = 4.641 \times 0.002 \times 40 = 0.371 \text{ kg/s} \]

Calculate Exit Flow Area

Rearrange the continuity equation to solve for the exit flow area \( A_2 \). \[ A_2 = \frac{ \textdot{m} }{ \rho_2 V_2 } = \frac{ 0.371 }{ 1.742 \times 350 } = 0.0006088 \text{ m}^2 = 6.088 \text{ cm}^2 \]

Key concepts

Ideal Gas Equation

The ideal gas equation is a fundamental concept in thermodynamics that relates the pressure, volume, and temperature of an ideal gas. This relation is given by the equation:
\[ PV = nRT \] \
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. However, for problems involving mass instead of moles, we often use the form:
\[ \rho = \frac{P}{RT} \]
where \( \rho \) is the density of the gas.
This equation is fundamental when dealing with gases in engineering and physics. It allows us to find the density of the air at different states, which is crucial for other calculations like mass flow rate. Remember, the specific gas constant \( R \) for air is 287 J/(kg·K).
In our problem, we found the densities at the inlet and the exit using this equation, which helped us proceed with mass flow rate and flow area calculations.

Continuity Equation

The continuity equation is a principle derived from the conservation of mass. It states that for any control volume, the amount of mass entering must equal the amount of mass exiting for a steady-state flow. Mathematically, it can be written as:
\[ \text{mass rate in} = \text{mass rate out} \]
For fluids, the mass flow rate \( \textdot{m} \) is the product of density \( \rho \), flow area \( A \), and velocity \( V \):
\[ \textdot{m} = \rho A V \]
This equation is useful for determining various properties at different points in a flow system. In our exercise, we used the continuity equation to first find the mass flow rate at the inlet. Then, by knowing the mass flow at the exit and densities at both points, we determined the exit flow area. It's crucial to ensure all units are consistent when performing these calculations.

Steady-State Operation

Steady-state operation is an important concept that simplifies the analysis of systems by assuming that all properties are constant over time. This means that variables such as pressure, temperature, and flow rates do not change as time progresses.

In the context of our problem, assuming steady-state operation means the mass entering the control volume equals the mass exiting it at any given moment. This assumption helps to apply the continuity equation straightforwardly.

• Inlet mass flow rate = Exit mass flow rate
• No accumulation of mass within the control volume
• Simplifies calculations and assumptions for engineering problems

This helps us to focus only on the given data without worrying about transient changes. For processes involving gases or fluids, steady-state assumptions are often valid and lead to simpler, yet accurate, solutions.
Understanding steady-state operation allows engineers to make critical assumptions that are both practical and accurate for many real-world systems.