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Consider the flow of air through a supersonic nozzle. The reservoir pressure and temperature are \(5 \mathrm{~atm}\) and \(500 \mathrm{~K}\), respectively. If the Mach number at the nozzle exit is 3, calculate the exit pressure, temperature, and density.

Short Answer

Expert verified
Exit pressure: 0.190 atm, Exit temperature: 178.57 K, Exit density: 0.372 kg/m^3.

Step by step solution

01

Gather Given Data

The problem provides the following data: Reservoir pressure \( P_0 = 5 \mathrm{~atm} \), Reservoir temperature \( T_0 = 500 \mathrm{~K} \), and Mach number at the nozzle exit \( M = 3 \).
02

Use Isentropic Flow Relations

For an isentropic flow through a nozzle, the relations between reservoir conditions and exit conditions are governed by the isentropic flow equations. The exit temperature \( T \), pressure \( P \), and density \( \rho \) can be calculated using the Mach number \( M \) and the specific heat ratio \( \gamma \) for air, which is \(1.4\).
03

Calculate Exit Temperature

The relationship for temperature is given by \[ \frac{T}{T_0} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{-1} \]Substituting the given values, \[ \frac{T}{500} = \left(1 + \frac{1.4 - 1}{2} \times 3^2 \right)^{-1} \] \[ \frac{T}{500} = \left(1 + 0.2 \times 9 \right)^{-1} \] \[ \frac{T}{500} = \left(1 + 1.8 \right)^{-1} = \left(2.8\right)^{-1} \]Therefore, \( T = 500 \times \frac{1}{2.8} \approx 178.57 \mathrm{~K} \).
04

Calculate Exit Pressure

The pressure relation is \[ \frac{P}{P_0} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{-\frac{\gamma}{\gamma - 1}} \]Substituting the known values, \[ \frac{P}{5} = \left(1 + \frac{0.4}{2} \times 3^2 \right)^{-\frac{1.4}{0.4}} \] \[ \frac{P}{5} = \left(2.8\right)^{-3.5} \]Therefore, \( P = 5 \times \left(2.8\right)^{-3.5} \approx 0.190 \mathrm{~atm} \).
05

Calculate Exit Density

The relationship for density is \[ \frac{\rho}{\rho_0} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{-\frac{1}{\gamma - 1}} \]Combining this with \( \frac{\rho_0}{\rho} = \frac{P_0}{P} \cdot \frac{T}{T_0} \), we get \[ \rho = \frac{P}{RT} \] where R is the specific gas constant for air \( R = 287 \mathrm{~J/kg} \cdot \mathrm{K} \).Therefore, \( \rho = \frac{0.190 \times 101325}{287 \times 178.57} \approx 0.372 \mathrm{~kg/m^3} \).

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Key Concepts

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

Isentropic Flow Equations
Isentropic flow equations are critical when analyzing fluid flows, such as air, through supersonic nozzles. These equations help us understand the relationships between various thermodynamic properties along a streamline when the process is adiabatic and no heat is transferred. It essentially means that the flow is extremely efficient, with minimal energy loss.

The isentropic relations connect properties like pressure, temperature, and density to the Mach number, which is a measure of the speed of the flow relative to the speed of sound. The specific heat ratio, or the adiabatic index of the gas (denoted by \( \gamma \)), plays a pivotal role in these equations. For air, this value is approximately 1.4. This ratio affects how pressure and temperature change during the flow.

The key formulas arising from isentropic flow assumptions include:
  • The relation for temperature: \[\frac{T}{T_0} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{-1}\]where \( T_0 \) is the reservoir temperature.
  • The relation for pressure:\[\frac{P}{P_0} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{-\frac{\gamma}{\gamma - 1}} \]where \( P_0 \) is the reservoir pressure.
  • The relation for density:\[\frac{\rho}{\rho_0} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{-\frac{1}{\gamma - 1}}\]These are used to derive the exit conditions from known reservoir conditions and the exit Mach number.
Mach Number
The Mach number is a dimensionless quantity used in fluid dynamics to compare the speed of a fluid to the speed of sound in that fluid. It is crucial for identifying flow regimes, be it subsonic, transonic, supersonic, or hypersonic.

In this exercise, the Mach number is 3 at the nozzle exit, which indicates supersonic flow—meaning that the air is moving three times faster than the speed of sound through the nozzle at the exit.

Understanding Mach numbers can help in predicting the behavior of air as it expands through a nozzle. In supersonic regimes, the nozzle design becomes highly complex and specific to ensure stability and efficiency of performance. The characteristic factor of supersonic flows is the decrease in pressure and temperature as the fluid velocity increases when it passes through the nozzle.

Engineers and scientists use Mach numbers to optimize nozzle shapes for desired flow characteristics, ensuring that jets and engines operate efficiently. For instance, at a Mach number of 3, different formulations like the isentropic flow equations are indispensable tools to determine downstream conditions such as exit pressure, temperature, and density.
Exit Pressure, Temperature, and Density
Calculating exit pressure, temperature, and density is an essential application of isentropic flow equations in fluid dynamics. As air flows through a nozzle and reaches supersonic speeds, these properties can be accurately predicted.

**Exit Temperature:** To determine the exit temperature \( T \), we use the formula:\[\frac{T}{T_0} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{-1}\]Given that the reservoir temperature \( T_0 = 500 \, \mathrm{K} \) and the Mach number \( M = 3 \), the exit temperature calculates to approximately 178.57 K, showcasing a significant drop as the air expands and accelerates.

**Exit Pressure:** The exit pressure \( P \) is calculated by:\[\frac{P}{P_0} = \left(1 + \frac{\gamma - 1}{2} M^2 \right)^{-\frac{\gamma}{\gamma - 1}} \]Using the initial condition \( P_0 = 5 \, \mathrm{atm} \), the exit pressure results in approximately 0.190 atm. This decrease illustrates the characteristic behavior of a gas in supersonic flow where the pressure drops as velocity increases.

**Exit Density:** Exit density \( \rho \) is derived from both the pressure and temperature: \[\rho = \frac{P}{RT} \]With known values and the specific gas constant \( R = 287 \, \mathrm{J/kg \cdot K} \), the exit density can be computed to be roughly 0.372 kg/m³. This lower density results from the gas expansion and acceleration within the nozzle.

Together, these calculations paint a comprehensive picture of how flowing air undergoes changes in its thermodynamic state due to an increase in velocity, governed by isentropic assumptions.

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Most popular questions from this chapter

An airplane is flying at a velocity of \(130 \mathrm{mi} / \mathrm{h}\) at a standard altitude of \(5000 \mathrm{ft}\). At a point on the wing, the pressure is \(1750.0 \mathrm{lb} / \mathrm{ft}^{2}\). Calculate the velocity at that point, assuming incompressible flow.

We wish to operate a low-speed subsonic wind tunnel so that the flow in the test section has a velocity of \(200 \mathrm{mi} / \mathrm{h}\). Consider two different types of wind tunnels (see figure below): \((a)\) a nozzle and a constant-area test section, where the flow at the exit of the test section simply dumps out to the surrounding atmosphere (that is, there is no diffuser); and ( \(b\) ) a conventional arrangement of nozzle, test section, and diffuser, where the flow at the exit of the diffuser dumps out to the surrounding atmosphere. For both wind tunnels \((a)\) and \((b)\), calculate the pressure differences across the entire wind tunnel required to operate them so as to have the given flow conditions in the test section. For tunnel \((a)\), the cross-sectional area of the entrance is \(20 \mathrm{ft}^{2}\), and the cross-sectional area of the test section is \(4 \mathrm{ft}^{2}\). For tunnel (b), a diffuser is added to \((a)\) with a diffuser exit area of \(18 \mathrm{ft}^{2}\). After completing your calculations, examine and compare your answers for tunnels \((a)\) and \((b)\). Which requires the smaller overall pressure difference? What does this say about the value of a diffuser in a subsonic wind tunnel?

A Pitot tube is mounted in the test section of a high-speed subsonic wind tunnel. The pressure and temperature of the airflow are 1 atm and \(270 \mathrm{~K}\), respectively. If the flow velocity is \(250 \mathrm{~m} / \mathrm{s}\), what is the pressure measured by the Pitot tube?

A high-speed subsonic Boeing 777 airliner is flying at a pressure altitude of \(12 \mathrm{~km}\). A Pitot tube on the vertical tail measures a pressure of \(2.96 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}\). At what Mach number is the airplane flying?

Consider an element of air in the standard atmosphere at a standard altitude of \(1000 \mathrm{~m}\). Assume that you somehow raise this element of air isentropically to a standard altitude of \(2000 \mathrm{~m}\), where the element now takes on the standard pressure at \(2000 \mathrm{~m}\). Calculate the density of this isentropically raised element of air and compare it with the density of its neighboring elements of air that all have a density equal to the standard density at \(2000 \mathrm{~m}\). What does this say about the stability of the atmosphere in this case? NOTE: The properties of the standard atmosphere are based on statics, that is, an element of fluid that is stationary, where the pressure change is dictated by the hydrostatic equation, Eq. (3.2). An isentropic process is not relevant to the establishment of the standard atmosphere. Indeed, a purpose of this question is to demonstrate that the changes in atmospheric properties with altitude are quite different from the changes corresponding to an isentropic process.

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