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Air enters a one-inlet, one-exit control volume at 6 bar, \(500 \mathrm{~K}\), and \(30 \mathrm{~m} / \mathrm{s}\) through a flow area of \(28 \mathrm{~cm}^{2}\). At the exit, the pressure is 3 bar, the temperature is \(456.5 \mathrm{~K}\), and the velocity is \(300 \mathrm{~m} / \mathrm{s}\). 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
Mass flow rate: 0.1163 kg/s. Exit flow area: 0.000135 m² or 13.5 cm².

Step by step solution

01

Determine the density at the inlet

Use the ideal gas law to find the density at the inlet. The ideal gas law is given by \[ \rho_1 = \frac{P_1}{R T_1} \] where \( P_1 = 6 \, \text{bar} = 600,000 \, \text{Pa} \), \( T_1 = 500 \, \text{K} \), and \( R = 287 \, \text{J} / \text{kg} \cdot \text{K} \) for air.
02

Calculate the mass flow rate at the inlet

The mass flow rate \( \dot{m} \) is given by \( \dot{m} = \rho_1 A_1 v_1 \). Using \( \rho_1 \) from Step 1, \( A_1 = 28 \, \text{cm}^2 = 0.0028 \, \text{m}^2 \), and \( v_1 = 30 \, \text{m/s} \), substitute these values to find the mass flow rate.
03

Determine the density at the exit

Use the ideal gas law to find the density at the exit. The ideal gas law is given by \[ \rho_2 = \frac{P_2}{R T_2} \] where \( P_2 = 3 \, \text{bar} = 300,000 \, \text{Pa} \), \( T_2 = 456.5 \, \text{K} \), and \( R = 287 \, \text{J} / \text{kg} \cdot \text{K} \).
04

Calculate the exit flow area

Use the mass flow rate expression: \( \dot{m} = \rho_2 A_2 v_2 \). With \( \dot{m} \) from Step 2, \( \rho_2 \) from Step 3, and \( v_2 = 300 \, \text{m/s} \), solve for the exit flow area \( A_2 \) by rearranging the formula to \( A_2 = \frac{\dot{m}}{\rho_2 v_2} \).

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Key Concepts

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

Mass Flow Rate
Understanding mass flow rate is crucial when analyzing fluid dynamics in a system. The mass flow rate, denoted by \( \dot{m} \), represents the mass of a substance that passes through a given surface per unit time. It is measured in kilograms per second (\( \mathrm{kg} / \mathrm{s} \)). To calculate the mass flow rate, we use the formula: \[ \dot{m} = \rho Av \] Here:
  • \( \rho \) is the fluid density (mass per unit volume)
  • \( A \) is the flow area (surface area through which the fluid moves)
  • \( v \) is the velocity of the fluid (speed of fluid flow)
For example, if air enters a one-inlet, one-exit control volume with certain pressure, temperature, and velocity, we can use the ideal gas law to find the density. Then, incorporating the known flow area and velocity, we derive the mass flow rate at the inlet using the formula above.
Density Calculation
Density calculation is a fundamental step in determining properties of gases and fluids. The density of an ideal gas can be obtained using the ideal gas law: \[ \rho = \frac{P}{RT} \] Here,
  • \( P \) is the absolute pressure of the gas
  • \( R \) is the specific gas constant
  • \( T \) is the absolute temperature
In the context of the original exercise, we were given the pressure and temperature of air both at the inlet and exit. Using the specific gas constant for air, which is 287 J/(kg·K), we substituted these values into the ideal gas equation to find the density at both points. At the inlet, for example, we had: \[ P_1 = 600,000 \, \text{Pa}, \ T_1 = 500 \, \text{K}, \ R = 287 \, \text{J} / \text{kg} \cdot \text{K} \] Substituting these in, we calculated the inlet density \( \rho_1 \). Repeating the process with the exit conditions, we found the exit density \( \rho_2 \). Knowing the density at both points is essential for determining flow rates and other key properties.
Flow Area
The flow area of a system, represented as \( A \), is the cross-sectional area through which the fluid flows. It is usually measured in square meters (m²) or square centimeters (cm²). Flow area is a critical factor in calculating the mass flow rate and other flow characteristics. The formula to relate the mass flow rate, density, velocity, and flow area is: \[ \dot{m} = \rho Av \] When solving for the flow area when other variables are known, we rearrange the formula: \[ A = \frac{\dot{m}}{\rho v} \] To find the exit flow area in the original exercise, we used the mass flow rate calculated at the inlet, along with the exit density and velocity. Substituting these values back into the equation provided us with the flow area at the exit. This process allows us to better understand how the fluid dynamics change as the fluid passes through different parts of a system.

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Most popular questions from this chapter

Hot combustion gases, modeled as air behaving as an ideal gas, enter a turbine at \(145 \mathrm{lbf} / \mathrm{in}^{2}, 2700^{\circ} \mathrm{R}\) with a mass flow rate of \(0.22 \mathrm{lb} / \mathrm{s}\) and exit at \(29 \mathrm{lbf} / \mathrm{in}^{2}\) and \(1620^{\circ} \mathrm{R}\). If heat transfer from the turbine to its surroundings occurs at a rate of \(14 \mathrm{Btu} / \mathrm{s}\), determine the power output of the turbine, in hp.

Helium gas flows through a well-insulated nozzle at steady state. The temperature and velocity at the inlet are \(550^{\circ} \mathrm{R}\) and \(150 \mathrm{ft} / \mathrm{s}\), respectively. At the exit, the temperature is \(400^{\circ} \mathrm{R}\) and the pressure is \(40 \mathrm{lbf} /\) in. \({ }^{2}\) The area of the exit is \(0.0085 \mathrm{ft}^{2}\). Using the ideal gas model with \(k=1.67\), and neglecting potential energy effects, determine the mass flow rate, in \(\mathrm{lb} / \mathrm{s}\), through the nozzle.

Air enters a compressor operating at steady state at \(14.7\) \(\mathrm{lbf} / \mathrm{in} .^{2}\) and \(60^{\circ} \mathrm{F}\) and is compressed to a pressure of \(150 \mathrm{lbf} / / \mathrm{in} .^{2}\) As the air passes through the compressor, it is cooled at a rate of 10 Btu per lb of air flowing by water circulated through the compressor casing. The volumetric flow rate of the air at the inlet is \(5000 \mathrm{ft}^{3} / \mathrm{min}\), and the power input to the compressor is 700 hp. The air behaves as an ideal gas, there is no stray heat transfer, and kinetic and potential effects are negligible. Determine (a) the mass flow rate of the air, lb/s, and (b) the temperature of the air at the compressor exit, in \({ }^{\circ} \mathrm{F}\).

Refrigerant \(134 \mathrm{a}\) enters a well-insulated nozzle at \(200 \mathrm{lbf} / \mathrm{in} .{ }^{2}, 220^{\circ} \mathrm{F}\), with a velocity of \(120 \mathrm{ft} / \mathrm{s}\) and exits at 20 lbf/in. \({ }^{2}\) with a velocity of \(1500 \mathrm{ft} / \mathrm{s}\). For steady-state operation, and neglecting potential energy effects, determine the exit temperature, in \({ }^{\circ} \mathrm{F}\).

Refrigerant \(134 \mathrm{a}\) enters an air conditioner compressor at 4 bar, \(20^{\circ} \mathrm{C}\), and is compressed at steady state to 12 bar, \(80^{\circ} \mathrm{C}\). The volumetric flow rate of the refrigerant entering is \(4 \mathrm{~m}^{3} /\) \(\min\). The work input to the compressor is \(60 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in \(\mathrm{kW}\).

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