Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For the isentropic flow of an ideal gas with constant specific heat ratio \(k\), the ratio of the temperature \(T^{*}\) to the stagnation temperature \(T_{\mathrm{o}}\) is \(T^{*} / T_{o}=2 /(k+1)\). Develop this relationship.

Short Answer

Expert verified
\( T^* / T_0 = \frac{2}{k+1} \).

Step by step solution

01

- Understanding the Isentropic Flow

In isentropic flow, the process is both adiabatic and reversible. This means there is no heat transfer and entropy remains constant.
02

- Ideal Gas Law

Recall the ideal gas law: \( PV = nRT \)
03

- Stagnation Temperature

The stagnation temperature is the temperature an ideal gas attains when brought to rest adiabatically. It includes both kinetic energy and internal energy.
04

- Applying Energy Conservation

Using the energy conservation principle for an ideal gas, the stagnation temperature \( T_0 \) can be expressed as:\[ T_0 = T + \frac{v^2}{2c_p} \]where: - \( T \) is the static temperature - \( v \) is the flow speed - \( c_p \) is the specific heat at constant pressure.
05

- Relating Temperature Ratio

For isentropic processes of an ideal gas with specific heat ratio \( k \), the temperature ratio can be simplified. The relationship between the static temperature \( T \) and the stagnation temperature \( T_0 \) is given by:\[ \frac{T}{T_0} = \left( \frac{2}{k+1} \right) \]
06

- Conclusion

Thus, the ratio of the temperature \( T^* \) of the flow to the stagnation temperature \( T_0 \) simplifies to \( T^* / T_{0} = \frac{2}{k+1} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Thermodynamics
Thermodynamics is the study of energy, heat, work, and how they interact with each other. The key concept in thermodynamics related to this problem is the isentropic process. In an isentropic process, entropy remains constant, meaning there is no loss or gain of energy due to friction or other forms of dissipation. In the context of isentropic flow, we are dealing with an ideal gas, where the process is both adiabatic (no heat transfer) and reversible. This means we can use thermodynamic relationships to determine various properties like temperature and pressure without loss of energy due to heat.
Stagnation Temperature
Stagnation temperature is a critical concept in fluid dynamics and thermodynamics. It is the temperature that a moving fluid would reach if it were brought to a complete stop adiabatically. This includes both the kinetic energy of the fluid and its internal energy. In mathematical terms, the stagnation temperature (\ref{T_0}) for an ideal gas can be expressed using the energy conservation principle as: \[ T_0 = T + \frac{v^2}{2c_p} \] where:
  • \ref{T} is the static temperature.
  • \ref{v} is the flow speed.
  • \ref{c_p} is the specific heat at constant pressure.
Understanding this helps in determining how temperature and energy are distributed in the flow.
Energy Conservation
The principle of energy conservation states that energy cannot be created or destroyed, only transferred or converted from one form to another. For the isentropic flow of an ideal gas, this principle allows us to relate different forms of energy within the system, like kinetic energy and thermal energy. In the context of the given problem, the stagnation temperature (\ref{T_0}) formula: \[ T_0 = T + \frac{v^2}{2c_p} \] illustrates how kinetic energy (from motion) and thermal energy (internal energy) contribute to the fluid's total energy when it is brought to rest adiabatically. This relationship helps in understanding the energy dynamics of the gas flow.
Ideal Gas Law
The ideal gas law is a fundamental equation in thermodynamics that relates the pressure (\ref{P}), volume (\ref{V}), and temperature (\ref{T}) of an ideal gas to the number of moles (\ref{n}) and the universal gas constant (\ref{R}). The equation is written as: \[ PV = nRT \] This law is crucial for understanding how gases behave under various conditions. In the context of the isentropic flow problem, the ideal gas law helps in determining how temperature changes due to changes in pressure and volume. This relationship is vital for further calculations, such as determining the temperature ratio (\ref{T/T_0}) for isentropic processes.
Specific Heat Ratio
The specific heat ratio (\ref{k}, also known as the adiabatic index) is a crucial parameter in thermodynamics, particularly for ideal gases. It is the ratio of specific heat at constant pressure (\ref{c_p}) to specific heat at constant volume (\ref{c_v}): \[ k = \frac{c_p}{c_v} \] This ratio is important for characterizing the thermodynamic behavior of gases under different processes. For the isentropic flow of an ideal gas, the specific heat ratio (\ref{k}) helps simplify the temperature ratio relationship. In the problem, the ratio of the temperature (\ref{T^*}) of the flow to the stagnation temperature (\ref{T_0}) is given by: \[ \frac{T^*}{T_0} = \frac{2}{k+1} \] This relationship helps understand how temperature and energy distribute in the gas flow during isentropic processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Air as an ideal gas with \(k=1.4\) undergoes a normal shock. The upstream conditions are \(p_{\mathrm{x}}=0.5\) bar, \(T_{\mathrm{x}}=280 \mathrm{~K}\), and \(M_{\mathrm{x}}=1.8 .\) Determine (a) the pressure \(p_{y}\), in bar. (b) the stagnation pressure \(p_{\mathrm{ox}}\), in bar. (c) the stagnation temperature \(T_{\mathrm{ax}}\), in \(\mathrm{K}\). (d) the change in specific entropy across the shock, in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\). (e) Plot the quantities of parts (a)-(d) versus \(M_{x}\) ranging from \(1.0\) to \(2.0\). All other upstream conditions remain the same.

A turboprop engine consists of a diffuser, compressor, combustor, turbine, and nozzle. The turbine drives a propeller as well as the compressor. Air enters the diffuser with a volumetric flow rate of \(83.7 \mathrm{~m}^{3} / \mathrm{s}\) at \(40 \mathrm{kPa}, 240 \mathrm{~K}\), and a velocity of \(180 \mathrm{~m} / \mathrm{s}\), and decelerates essentially to zero velocity. The compressor pressure ratio is 10 and the compressor has an isentropic efficiency of \(85 \%\). The turbine inlet temperature is \(1140 \mathrm{~K}\), and its isentropic efficiency is \(85 \%\). The turbine exit pressure is \(50 \mathrm{kPa}\). Flow through the diffuser and nozzle is isentropic. Using an air-standard analysis, determine (a) the power delivered to the propeller, in MW. (b) the velocity at the nozzle exit, in \(\mathrm{m} / \mathrm{s}\). Neglect kinetic energy except at the diffuser inlet and the nozzle exit.

Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each replaced with polytropic processes having \(n=1.3 .\) The compression ratio is 9 for the modified cycle. At the beginning of compression, \(p_{1}=1\) bar and \(T_{1}=300 \mathrm{~K}\). The maximum temperature during the cycle is \(2000 \mathrm{~K}\). Determine (a) the heat transfer and work per unit mass of air, in \(\mathrm{kJ} / \mathrm{kg}\), for each process in the modified cycle. (b) the thermal efficiency. \text { (c) the mean effective pressure, in bar. }

Air enters the compressor of a gas turbine at \(100 \mathrm{kPa}\), \(300 \mathrm{~K}\). The air is compressed in two stages to \(900 \mathrm{kPa}\), with intercooling to \(300 \mathrm{~K}\) between the stages at a pressure of \(300 \mathrm{kPa}\). The turbine inlet temperature is \(1480 \mathrm{~K}\) and the expansion occurs in two stages, with reheat to \(1420 \mathrm{~K}\) between the stages at a pressure of \(300 \mathrm{kPa}\). The compressor and turbine stage efficiencies are 84 and \(82 \%\), respectively. The net power developed is \(1.8 \mathrm{MW}\). Determine (a) the volumetric flow rate, in \(\mathrm{m}^{3} / \mathrm{s}\), at the inlet of each compressor stage. (b) the thermal efficiency of the cycle. (c) the back work ratio.

Air at \(22 \mathrm{kPa}, 220 \mathrm{~K}\), and \(250 \mathrm{~m} / \mathrm{s}\) enters a turbojet engine in flight at an altitude of \(10,000 \mathrm{~m}\). The pressure ratio across the compressor is \(12 .\) The turbine inlet temperature is \(1400 \mathrm{~K}\), and the pressure at the nozzle exit is \(22 \mathrm{kPa.}\) The diffuser and nozzle processes are isentropic, the compressor and turbine have isentropic efficiencies of 85 and \(88 \%\), respectively, and there is no pressure drop for flow through the combustor. On the basis of an air-standard analysis, determine (a) the pressures and temperatures at each principal state, in \(\mathrm{kPa}\) and \(\mathrm{K}\), respectively. (b) the velocity at the nozzle exit, in \(\mathrm{m} / \mathrm{s}\). Neglect kinetic energy except at the diffuser inlet and the nozzle exit.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free