/*! 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 17 A closed system consisting of \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 17

A closed system consisting of \(2 \mathrm{~kg}\) of a gas undergoes a process during which the relation between pressure and volume is \(p V^{m}=\) constant. The process begins with \(p_{1}=150 \mathrm{kPa}\), \(V_{1}=0.3 \mathrm{~m}^{3}\) and ends with \(p_{2}=450 \mathrm{kPa}, V_{2}=0.137 \mathrm{~m}^{3}\) Determine (a) the value of \(n\) and (b) the specific volume at states 1 and 2, each in \(\mathrm{m}^{3} / \mathrm{kg}\). (c) Sketch the process on pressurevolume coordinates.

Short Answer

Expert verified
(a) m 鈮� 1.402 (b) Specific volumes are 0.15 m鲁/kg and 0.0685 m鲁/kg (c) Sketch: non-linear decreasing trend.

Step by step solution

01

Identify the given information

From the problem, we have the initial and final conditions: \(p_{1} = 150 \, \text{kPa}\), \(V_{1} = 0.3 \, \text{m}^{3}\), \(p_{2} = 450 \, \text{kPa}\), \(V_{2} = 0.137 \, \text{m}^{3}\). The relation between pressure and volume is given as \(pV^{m} = \text{constant}\). We are to find the value of \(m\), the specific volume at states 1 and 2, and sketch the process.
02

Determine the value of m

Use the relation \(pV^{m} = \text{constant}\): Since \(p_{1} V_{1}^{m} = p_{2} V_{2}^{m}\):taking the natural logarithm of both sides:\(\ln (p_{1}) + m \ln (V_{1}) = \ln (p_{2}) + m \ln (V_{2})\)Solving for \(m\):\(m = \frac{\ln \left(\frac{p_{2}}{p_{1}}\right)}{\ln \left(\frac{V_{1}}{V_{2}}\right)}\)Substitute given values:\(m = \frac{\ln \left(\frac{450}{150}\right)}{\ln \left(\frac{0.3}{0.137}\right)}\) Which simplifies to:\(m = \frac{\ln (3)}{\ln \left(\frac{0.3}{0.137}\right)}\)\(m \approx \frac{\ln (3)}{\ln (2.189)}\)\(m \approx \frac{1.0986}{0.7833}\)\(m \approx 1.402\)
03

Calculate specific volume at State 1

Specific volume \(v_{1}\) is the volume per unit mass, given by:\(v_{1} = \frac{V_{1}}{m}\)Substitute the values:\(v_{1} = \frac{0.3 \, \text{m}^{3}}{2 \, \text{kg}}\)\(v_{1} = 0.15 \, \text{m}^{3}/\text{kg}\)
04

Calculate specific volume at State 2

Specific volume \(v_{2}\) is given by:\(v_{2} = \frac{V_{2}}{m}\)Substitute the values:\(v_{2} = \frac{0.137 \, \text{m}^{3}}{2 \, \text{kg}}\)\(v_{2} = 0.0685 \, \text{m}^{3}/\text{kg}\)
05

Sketch the process on pressure-volume coordinates

To sketch the process, plot the initial and final states on a graph with pressure (p) on the y-axis and volume (V) on the x-axis. Mark the initial condition \((150 \, \text{kPa}, 0.3 \, \text{m}^{3})\) and the final condition \((450 \, \text{kPa}, 0.137 \, \text{m}^{3})\). Since the process follows the relation \(pV^{m} = \text{constant}\) with \(m \approx 1.402\), the curve will be a non-linear decreasing trend.

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 91影视!

Key Concepts

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

closed system
In thermodynamics, a closed system is one where no mass enters or leaves the system, but energy in the form of work or heat can be exchanged with the surroundings. For the given exercise, our system is the gas within the specified volume and its container. Because we are not adding or removing any gas, the system remains constant in mass. Understanding this concept is essential for solving problems where the properties of the system need to be calculated without the mass changing. Remember, because energy transfer does not involve a mass transfer, we focus on properties like pressure, volume, and temperature.
pressure-volume relationship
When studying the behavior of gases, particularly within a closed system, the relationship between pressure and volume becomes crucial. In the given problem, this relationship is expressed as, \[ p V^m = \text{constant} \]This implies that as the pressure changes, the volume must change in such a way that the product of pressure and volume to the power of m remains constant. For example, if the volume decreases, the pressure increases correspondingly to maintain this equilibrium. This form is particularly common in processes like adiabatic compression or expansion, where no heat is exchanged. To find the exponent m, we derived it using natural logarithms to simplify the equation: \[ m = \frac{\ln(\frac{p_2}{p_1})}{\ln(\frac{V_1}{V_2})} \]
specific volume
Specific volume is an important property in thermodynamics, defined as the volume per unit mass. In the exercise, we calculate the specific volume at both the initial and final states. It is mathematically expressed as: \[ v = \frac{V}{m} \]where V is the volume and m is the mass. In the given problem, we first find the specific volume at state 1: \[ v_1 = \frac{0.3 \text{ m}^3}{2 \text{ kg}} = 0.15 \text{ m}^3/ \text{kg} \]Then, for state 2: \[ v_2 = \frac{0.137 \text{ m}^3}{2 \text{ kg}} = 0.0685 \text{ m}^3/ \text{kg} \]Specific volume is essential for characterizing the state of the gas, particularly in processes like heating, cooling, and compression within a closed system. It can help determine other properties and understand how the gas behaves under different conditions.

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

One thousand \(\mathrm{kg}\) of natural gas at 100 bar and \(255 \mathrm{~K}\) is stored in a tank. If the pressure, \(p\), specific volume, \(v\), and temperature, \(T\), of the gas are related by the following expression $$ p=\left[\left(5.18 \times 10^{-3}\right) T /(v-0.002668)\right]-\left(8.91 \times 10^{-3}\right) / v^{2} $$ where \(v\) is in \(\mathrm{m}^{3} / \mathrm{kg}, T\) is in \(K\), and \(p\) is in bar, determine the volume of the tank, in \(\mathrm{m}^{3}\). Also, plot pressure versus specific volume for the isotherms \(T=250,500\), and \(1000 \mathrm{~K}\).

In a steam power station working on the ideal Rankine cycle with regeneration, steam enters the turbine at 150 bar, \(600^{\circ} \mathrm{C}\). One open feedwater heater is used in the plant. Some steam from the turbine enters the open feedwater heater at a pressure of 12 bar. The pressure in the condenser is \(0.1\) bar. Determine the rate of exergy input to the working fluid passing through the steam generator, in \(\mathrm{kJ} /\) \(\mathrm{kg}\) of steam entering the turbine. Let \(T_{0}=288 \mathrm{~K}\) and \(p_{0}=1\) bar. Also, determine the rate of exergy destruction in the open feedwater heater, in \(\mathrm{kJ} / \mathrm{kg}\) of steam entering the turbine.

A gas initially at \(p_{1}=1\) bar and occupying a volume of 1 liter is compressed within a piston-cylinder assembly to a final pressure \(p_{2}=4\) bar. (a) If the relationship between pressure and volume during the compression is \(p V=\) constant, determine the volume, in liters, at a pressure of 3 bar. Also plot the overall process on a graph of pressure versus volume. (b) Repeat for a linear pressure-volume relationship between the same end states.

In a steam power station working on the Rankine cycle with regeneration, steam enters the turbine at 150 bar, \(600^{\circ} \mathrm{C}\). One open feedwater heater is used in the plant. Some steam from the turbine enters the open feedwater heater at a pressure of 12 bar. The pressure in the condenser is \(0.1\) bar. The isentropic efficiencies at each turbine stage and pump is \(90 \%\). Determine the thermal efficiency of the cycle and the fraction of steam extracted from the turbine.

Water is the working fluid in an ideal Rankine cycle with reheat. Superheated vapor enters the turbine at \(8 \mathrm{MPa}\), \(440^{\circ} \mathrm{C}\), and the condenser pressure is \(8 \mathrm{kPa}\). Steam expands through the first- stage turbine to \(0.5 \mathrm{MPa}\) and then is reheated to \(440^{\circ} \mathrm{C}\). Determine for the cycle (a) the rate of heat addition, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of steam entering the first-stage turbine. (b) the thermal efficiency. (c) the rate of heat transfer from the working fluid passing through the condenser to the cooling water, in kJ per \(\mathrm{kg}\) of steam entering the first-stage turbine.
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Quantum Physics

Read Explanation

Physics of Motion

Read Explanation

Electromagnetism

Read Explanation

Medical Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation
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.