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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

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

Step by step solution

01

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, \( V_2 = 350 \text{ m/s} \)
02

Convert pressures

Convert the pressures from bar to Pascals for standard units. \[ P_1 = 8 \text{ bar} = 8 \times 10^5 \text{ Pa} \] \[ P_2 = 2 \text{ bar} = 2 \times 10^5 \text{ Pa} \]
03

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). \[ \rho_1 = \frac{ P_1 }{ R T_1 } = \frac{ 8 \times 10^5 }{ 287 \times 600 } = 4.641 \text{ kg/m}^3 \] \[ \rho_2 = \frac{ P_2 }{ R T_2 } = \frac{ 2 \times 10^5 }{ 287 \times 400 } = 1.742 \text{ kg/m}^3 \]
04

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: \[ \textdot{m} = \rho_1 A_1 V_1 = 4.641 \times 0.002 \times 40 = 0.371 \text{ kg/s} \]
05

Calculate Exit Flow Area

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Pumped-hydraulic storage power plants use relatively inexpensive off-peak baseload electricity to pump water from a lower reservoir to a higher reservoir. During periods of peak demand, electricity is produced by discharging water from the upper to the lower reservoir through a hydraulic turbinegenerator. A single device normally plays the role of the pump during upper-reservoir charging and the turbine-generator during discharging. The ratio of the power developed during discharging to the power required for charging is typically much less than \(100 \%\). Write a report describing the features of the pump-turbines used for such applications and their size and cost. Include in your report a discussion of the economic feasibility of pumped-hydraulic storage power plants,

What practical measures can be taken by manufacturers to use energy resources more efficiently? List several specific opportunities, and discuss their potential impact on profitability and productivity.

Ten \(\mathrm{kg} / \mathrm{min}\) of cooling water circulates through a water jacket enclosing a housing filled with electronic components. At steady state, water enters the water jacket at \(22^{\circ} \mathrm{C}\) and exits with a negligible change in pressure at a temperature that cannot exceed \(26^{\circ} \mathrm{C}\). There is no significant energy transfer by heat from the outer surface of the water jacket to the surroundings, and kinetic and potential energy effects can be ignored. Determine the maximum electric power the electronic components can receive, in \(\mathrm{kW}\), for which the limit on the temperature of the exiting water is met.

Infiltration of outside air into a building through miscellaneous cracks around doors and windows can represent a significant load on the heating equipment. On a day when the outside temperature is \(-18^{\circ} \mathrm{C}, 0.042 \mathrm{~m}^{3} / \mathrm{s}\) of air enters through the cracks of a particular office building. In addition, door openings account for about \(.047 \mathrm{~m}^{3} / \mathrm{s}\) of outside air infiltration. The internal volume of the building is \(566 \mathrm{~m}^{3}\), and the inside temperature is \(22^{\circ} \mathrm{C}\). There is negligible pressure difference between the inside and the outside of the building. Assuming ideal gas behavior, determine at steady state the volumetric flow rate of air exiting the building through cracks and other openings, and the number of times per hour that the air within the building is changed due to infiltration.

Ammonia enters a heat exchanger operating at steady state as a superheated vapor at 14 bar, \(60^{\circ} \mathrm{C}\), where it is cooled and condensed to saturated liquid at 14 bar. The mass flow rate of the refrigerant is \(450 \mathrm{~kg} / \mathrm{h}\). A separate stream of air enters the heat exchanger at \(17^{\circ} \mathrm{C}, 1\) bar and exits at \(42^{\circ} \mathrm{C}, 1\) bar. Ignoring heat transfer from the outside of the heat exchanger and neglecting kinetic and potential energy effects, determine the mass flow rate of the air, in \(\mathrm{kg} / \mathrm{min}\).

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