/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 99 An ideal gas with constant speci... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An ideal gas with constant specific heat ratio \(k\) enters a nozzle operating at steady state at pressure \(p_{1}\), temperature \(T_{1}\), and velocity \(\mathrm{V}_{1} .\) The air expands isentropically to a pressure of \(p_{2}\) (a) Develop an expression for the velocity at the exit, \(\mathrm{V}_{2}\), in terms of \(k, R, \mathrm{~V}_{1}, T_{1}, p_{1}\), and \(p_{2}\), only. (b) For \(\mathrm{V}_{1}=0, T_{1}=1000 \mathrm{~K}\), plot \(\mathrm{V}_{2}\) versus \(p_{2} / p_{1}\) for selected values of \(k\) ranging from \(1.2\) to \(1.4\).

Short Answer

Expert verified
The exit velocity \( \mathrm{V}_2 \) is \( \mathrm{V}_2 = \sqrt{ 2 \left( \frac{kR}{k-1} \right) T_1 \left(1 - \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \right) + \mathrm{V}_1^2 } \).Plotting \( \mathrm{V}_2 \) vs \( \frac{p_2}{p_1} \) for \( V_1 = 0 \) and \( T_1 = 1000 \ensuremath{K} \), use \( k = 1.2 \) to \( k = 1.4 \).

Step by step solution

01

Apply the Energy Equation

For a steady flow process, the energy equation for the nozzle is given by \[ h_1 + \frac{1}{2} \mathrm{V}_1^2 = h_2 + \frac{1}{2} \mathrm{V}_2^2 \]where \(h\) is the specific enthalpy. Since the process is isentropic, we can use \( h = c_p T\).
02

Substitute the enthalpy terms

Using \( h = c_p T \), the energy equation becomes \[ c_p T_1 + \frac{1}{2} \mathrm{V}_1^2 = c_p T_2 + \frac{1}{2} \mathrm{V}_2^2 \]. Since the process is isentropic, relate the temperatures as part of the isentropic relations.
03

Use the Isentropic Relations

For an isentropic process, the following relation holds: \[ \frac{T_2}{T_1} = \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \] Substituting for \( T_2 \) gives: \[ c_p T_1 + \frac{1}{2} \mathrm{V}_1^2 = c_p T_1 \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} + \frac{1}{2} \mathrm{V}_2^2 \].
04

Express \( T_1 \)

Rearrange the equation to solve for \( \mathrm{V}_2 \): \[ \mathrm{V}_2 = \sqrt{ 2c_p T_1 \left(1 - \left(\frac{p_2}{p_1}\right)^{\frac{k-1}{k}} \right) + \mathrm{V}_1^2 }. \]To simplify further, use \( c_p = \frac{kR}{k-1} \).
05

Substitute \( c_p \) with \( \frac{kR}{k-1} \)

Substitute \( c_p \) in the equation: \[ \mathrm{V}_2 = \sqrt{ 2 \left( \frac{kR}{k-1} \right) T_1 \left(1 - \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \right) + \mathrm{V}_1^2 }. \]
06

Plot \( \mathrm{V}_2 - \frac{p_2}{p_1} \) for different \( k \)

Given \( V_1 = 0 \) and \( T_1 = 1000 \) K, use different values of \( k \) (from 1.2 to 1.4) to plot \( V_2 \) versus \( \frac{p_2}{p_1} \) using the derived formula: \[ \mathrm{V}_2 = \sqrt{ 2 \left( \frac{kR}{k-1} \right) 1000 \left(1 - \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \right) }. \]

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 91Ó°ÊÓ!

Key Concepts

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

nozzle flow
When air or any ideal gas flows through a nozzle, its pressure, temperature, and velocity change. A nozzle is a device designed to control the speed and direction of this gas flow. In many practical applications, understanding how these parameters transform is crucial for performance and efficiency.
Throughout nozzle flow, we apply thermodynamic principles to predict changes in state properties. For steady-state operations, where conditions no longer change with time, specific equations help solve for these state properties. This forms the backbone of working with nozzles in aerospace, mechanical engineering, and other fields.
steady state
Steady state refers to a condition where variables (like velocity, pressure, and temperature in a nozzle) do not change with time. This simplification allows us to focus on spatial changes. For a nozzle operating at steady state, we leverage the energy equation to understand how kinetic and internal energies transform.
In other words, while the gas flows through the nozzle, its properties might vary from entry to exit, but at any point in time, these properties remain constant. This assumption is crucial as it allows us to use mathematical relations, such as the steady-state energy equation for our calculations.
isentropic process
An isentropic process is a reversible adiabatic process, meaning no heat is transferred in or out of the system, and entropy remains constant. In our nozzle flow scenario, assuming an isentropic expansion simplifies the analysis by using specific straightforward relationships.
For instance, for an ideal gas undergoing an isentropic process, the relation between temperatures and pressures is given by:
\[ \frac{T_2}{T_1} = \left( \frac{p_2}{p_1} \right)^{\frac{k-1}{k}} \],
where \(T_1\) and \(T_2\) are the temperatures at entry and exit, and \(p_1\) and \(p_2\) are the corresponding pressures. This relationship helps us to calculate the temperature at the nozzle's exit, given the entry conditions and pressure changes.
specific enthalpy
Specific enthalpy (h) is a measure of energy content per unit mass in a thermodynamic system. For an ideal gas, it's related to temperature and specific heat at constant pressure (\(c_p\)), given by \( h = c_p T\).
In nozzle flow analysis, we use specific enthalpy in the energy equation to determine how much kinetic energy change occurs due to the gas expansion: \[ h_1 + \frac{1}{2} \mathrm{V}_1^2 = h_2 + \frac{1}{2} \mathrm{V}_2^2 \]
By substituting appropriate values and isentropic relations, this equation enables us to solve for the exit velocity of the gas.
ideal gas
An ideal gas is a theoretical gas composed of many randomly moving point particles that interact only through elastic collisions. The ideal gas law, given by\( PV = nRT \), relates pressure \( P \), volume \( V \), amount of gas \( n \), gas constant \( R \), and temperature \( T \).
In our nozzle flow problem, assuming air behaves as an ideal gas allows us to utilize simple mathematical relations and specific calculations. For instance, using constants and specific heat ratios \( k \) helps derive and simplify expressions for temperature, pressure, and velocity changes in an isentropic process.
Remember, while real gases may exhibit different behaviors under extreme conditions, the ideal gas assumption holds robust and simplifies many practical problems efficiently.

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

How might you explain the entropy production concept in terms a child would understand?

Ammonia enters a counterflow heat exchanger at \(-20^{\circ} \mathrm{C}\), with a quality of \(35 \%\), and leaves as saturated vapor at \(-20^{\circ} \mathrm{C}\). Air at \(300 \mathrm{~K}, 1\) atm enters the heat exchanger in a separate stream with a flow rate of \(4 \mathrm{~kg} / \mathrm{s}\) and exits at \(285 \mathrm{~K}, 0.98\) atm. The heat exchanger is at steady state, and there is no appreciable heat transfer from its outer surface. Neglecting kinetic and potential energy effects, determine the mass flow rate of the ammonia, in \(\mathrm{kg} / \mathrm{s}\), and the rate of entropy production within the heat exchanger, in \(\mathrm{kW} / \mathrm{K}\).

The theoretical steam rate is the quantity of steam required to produce a unit amount of work in an ideal turbine. The Theoretical Steam Rate Tables published by The American Society of Mechanical Engineers give the theoretical steam rate in lb per \(\mathrm{kW} \cdot \mathrm{h}\). To determine the actual steam rate, the theoretical steam rate is divided by the isentropic turbine efficiency. Why is the steam rate a significant quantity? Discuss how the steam rate is used in practice.

Steam enters a turbine operating at steady state at a pressure of \(3 \mathrm{MPa}\), a temperature of \(400^{\circ} \mathrm{C}\), and a velocity of \(160 \mathrm{~m} / \mathrm{s}\). Saturated vapor exits at \(100^{\circ} \mathrm{C}\), with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). Heat transfer from the turbine to its surroundings takes place at the rate of \(30 \mathrm{~kJ}\) per kg of steam at a location where the average surface temperature is \(350 \mathrm{~K}\). (a) For a control volume including only the turbine and its contents, determine the work developed, in \(\mathrm{kJ}\), and the rate at which entropy is produced, in \(\mathrm{kJ} / \mathrm{K}\), each per \(\mathrm{kg}\) of steam flowing. (b) The steam turbine of part (a) is located in a factory where the ambient temperature is \(27^{\circ} \mathrm{C}\). Determine the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of steam flowing, for an enlarged control volume that includes the turbine and enough of its immediate surroundings so that heat transfer takes place from the control volume at the ambient temperature. Explain why the entropy production value of part (b) differs from that calculated in part (a).

Air enters a compressor operating at steady state at 1 bar, \(22^{\circ} \mathrm{C}\) with a volumetric flow rate of \(1 \mathrm{~m}^{3} / \mathrm{min}\) and is compressed to 4 bar, \(177^{\circ} \mathrm{C}\). The power input is \(3.5 \mathrm{~kW}\). Employing the ideal gas model and ignoring kinetic and potential energy effects, obtain the following results: (a) For a control volume enclosing the compressor only, determine the heat transfer rate, in \(\mathrm{kW}\), and the change in specific entropy from inlet to exit, in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\). What additional information would be required to evaluate the rate of entropy production? (b) Calculate the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for an enlarged control volume enclosing the compressor and a portion of its immediate surroundings so that heat transfer occurs at the ambient temperature, \(22^{\circ} \mathrm{C}\).

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.