/*! 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 114 An insulated, rigid tank whose v... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 114

An insulated, rigid tank whose volume is \(0.5 \mathrm{~m}^{3}\) is connected by a valve to a large vessel holding steam at 40 bar, \(500^{\circ} \mathrm{C}\). The tank is initially evacuated. The valve is opened only as long as required to fill the tank with steam to a pressure of 20 bar. Determine (a) the final temperature of the steam in the tank, in \({ }^{\circ} \mathrm{C}\), (b) the final mass of the steam in the tank, in \(\mathrm{kg}\), and (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

Short Answer

Expert verified
Final temperature: \( 595^{\circ} \mathrm{C} \). Mass of steam: \( 5 \mathrm{kg} \). Entropy produced: \( 38.245 \mathrm{kJ}/ \mathrm{K} \).

Step by step solution

01

Identify known values

Given:The volume of the tank, \( V = 0.5 \mathrm{m}^{3} \)Initial state of the tank: evacuated (no initial mass)Pressure of the large vessel: \( P_{1} = 40 \mathrm{bar} \)Temperature of the large vessel: \( T_{1} = 500^{\circ} \mathrm{C} \)Final pressure in the tank: \( P_{2} = 20 \mathrm{bar} \)
02

Use the steam tables to find specific volume and entropy for given conditions

Using the steam tables, locate the properties of steam at: \[ P_{1} = 40 \mathrm{bar}, \ T_{1} = 500^{\circ} \mathrm{C} \]From the steam tables:Specific volume: \( v_{1} = 0.06251 \mathrm{m}^{3}/ \mathrm{kg} \)Specific entropy: \( s_{1} = 7.649 \mathrm{kJ}/ \mathrm{kg} \mathrm{K} \)
03

Use ideal gas or steam properties to find the final temperature in the tank

Since the tank is insulated and rigid, no heat is transferred (adiabatic process) and no work is done. Therefore, use the fact that specific entropy is constant:\[ s_{1} = s_{2} \]Assume the final specific entropy from steam tables at \(20 \mathrm{bar} \). From the steam tables:At \(20 \mathrm{bar} \), find temperature \(T_{2}\) that matches specific entropy \(s_{2} = s_{1} = 7.649 \mathrm{kJ}/ \mathrm{kg} \mathrm{K} \)\[ T_{2} \approx 595^{\circ} \mathrm{C} \]
04

Determine the final mass of the steam in the tank

Using the specific volume at \( T_{2} = 595^{\circ} \mathrm{C} \) and \( P_{2} = 20 \mathrm{bar}, \ v_{2} \approx 0.1 \mathrm{m}^{3}/ \mathrm{kg} \):Tank volume: \( V = 0.5 \mathrm{m}^{3} \)Final mass of steam: \[ m = \frac{V}{v_{2}} = \frac{0.5 \mathrm{m}^{3}}{0.1 \mathrm{m}^{3}/ \mathrm{kg}} = 5 \mathrm{kg} \]
05

Calculate the entropy produced

Since the initial state is evacuated and the final state properties are known, the entropy change is:\[ \Delta S = ms_2 - ms_1 \]Therefore, with initial entropy \( s_{initial} = 0 \):Use the specific entropy values:\[ \Delta S = 5 \mathrm{kg} \times s_{2} \]Where \( s_{2} = s_{1} = 7.649 \mathrm{kJ}/ \mathrm{kg} \mathrm{K} \)So total entropy produced:\[ \Delta S = 5 \times 7.649 = 38.245 \mathrm{kJ}/ \mathrm{K} \]

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.

Adiabatic Process
An adiabatic process is one where no heat exchange occurs between the system and its surroundings. In other words, the process is thermally insulated. In this exercise, the rigid tank is insulated, meaning no heat transfer happens while steam fills the tank. This is crucial because it implies that the changes in the system, such as temperature and pressure, are achieved without any heat gain or loss. For an adiabatic process, the specific entropy remains constant. This means the entropy at the initial state before the steam enters the tank is equal to the entropy at the final state after the steam fills the tank.
Steam Tables
Steam tables are valuable tools that provide the properties of water and steam at different pressures and temperatures. They include values such as specific volume, specific entropy, specific enthalpy, and saturation temperature. For this problem, steam tables help us determine the properties of steam at given conditions. For instance, at 40 bar and 500掳C, we use the steam tables to find the specific volume and specific entropy of steam. These tables help students and engineers quickly find necessary thermodynamic properties to solve problems related to steam and water.
Specific Entropy
Specific entropy is a measure of the energy disorderedness per unit mass in a system. In this context, it helps us understand the energy changes inside the tank. Given that the tank is insulated (thus an adiabatic process), the specific entropy of the steam remains constant when moving from one state to another. Here, the specific entropy at 40 bar and 500掳C is used to find the final temperature in the tank at 20 bar by ensuring the specific entropy remains the same. This consistency helps identify the temperature corresponding to the specific entropy value in the steam tables.
Insulated System
An insulated system is one that does not allow any heat transfer to or from the surroundings. Insulation is critical in ensuring that the energy change within the process is due solely to work done or changes in internal energy. In this problem, the rigid tank is insulated, signifying an adiabatic process. As a result, even when the steam enters and fills the tank, no heat is exchanged with the environment. This characteristic simplifies the calculations since it implies specific entropy remains unchanged during the process.
Rigid Tank
A rigid tank means the volume remains constant regardless of changes in pressure or temperature within the tank. Unlike systems where expansion or compression can occur (leading to work done by or on the system), a rigid tank maintains its volume. Here, the rigid nature of the tank ensures that when 0.5 m鲁 of steam is introduced at 20 bar, we can directly relate this to specific volume and derive the mass. Knowing that the tank is rigid and insulated simplifies the process of determining the final states of pressure and temperature within the system.

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 \(\mathrm{kg}\) of propane initially at 8 bar and \(50^{\circ} \mathrm{C}\) undergoes a process to 3 bar, \(20^{\circ} \mathrm{C}\) while being rapidly expanded in a piston-cylinder assembly. Heat transfer between the propane and its surroundings occurs at an average temperature of \(35^{\circ} \mathrm{C}\). The work done by the propane is measured as \(42.4 \mathrm{~kJ}\). Kinetic and potential energy effects can be ignored. Determine whether it is possible for the work measurement to be correct.

One lb of water contained in a piston-cylinder assembly, initially saturated vapor at \(1 \mathrm{~atm}\), is condensed at constant pressure to saturated liquid. Evaluate the heat transfer, in Btu, and the entropy production, in Btu/ \({ }^{\circ} \mathrm{R}\), for (a) the water as the system. (b) an enlarged system consisting of the water and enough of the nearby surroundings that heat transfer occurs only at the ambient temperature, \(80^{\circ} \mathrm{F}\). Assume the state of the nearby surroundings does not change during the process of the water, and ignore kinetic and potential energy.

Air within a piston-cylinder assembly, initially at \(30 \mathrm{lbf} /\) in. \({ }^{2}, 510^{\circ} \mathrm{R}\), and a volume of \(6 \mathrm{ft}^{3}\), is compressed isentropically to a final volume of \(1.2 \mathrm{ft}^{3}\). Assuming the ideal gas model with \(k=1.4\) for the air, determine the (a) mass, in lb, (b) final pressure, in lbf/in. \({ }^{2}\), (c) final temperature, in \({ }^{\circ} \mathrm{R}\), and (d) work, in Btu.

Air at \(1 \mathrm{~atm}, 520^{\circ} \mathrm{R}\) enters a compressor operating at steady state and is compressed adiabatically to 3 atm. The isentropic compressor efficiency is \(80 \%\). Employing the ideal gas model with \(k=1.4\) for the air, determine for the compressor (a) the power input, in Btu per lb of air flowing, and (b) the amount of entropy produced, in Btu/ \(/ \mathrm{R}\) per lb of air flowing. Ignore kinetic and potential energy effects.

Air at \(400 \mathrm{kPa}, 970 \mathrm{~K}\) enters a turbine operating at steady state and exits at \(100 \mathrm{kPa}, 670 \mathrm{~K}\). Heat transfer from the turbine occurs at an average outer surface temperature of \(315 \mathrm{~K}\) at the rate of \(30 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing. Kinetic and potential energy effects are negligible. For air as an ideal gas with \(c_{p}=1.1 \mathrm{~kJ} /\) \(\mathrm{kg} \cdot \mathrm{K}\), determine (a) the rate power is developed, in kJ per \(\mathrm{kg}\) of air flowing, and (b) the rate of entropy production within the turbine, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of air flowing.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Particle Model of Matter

Read Explanation

Solid State Physics

Read Explanation

Nuclear Physics

Read Explanation

Electricity

Read Explanation

Scientific Method Physics

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.