/*! 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 25 A quantity of liquid water under... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 25

A quantity of liquid water undergoes a process from \(80^{\circ} \mathrm{C}\), 5 MPa to saturated liquid at \(40^{\circ} \mathrm{C}\). Determine the change in specific entropy, in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\), using (a) Tables A-2 and A-5. (b) saturated liquid data only from Table A-2. (c) the incompressible liquid model with a constant specific heat from Table A-19. (d) \(I T\).

Short Answer

Expert verified
Use tables to find entropy values at given temperatures and states; calculate changes based on differences. For part (c), utilize specific heat and temperature ratio.

Step by step solution

01

Given Data

Identify the given data and all initial conditions: Temperature at initial state: \( T_1 = 80^{\text{o}} \text{C} \) Pressure at initial state: \( P_1 = 5 \text{ MPa} \) Final state: Saturated liquid at \( T_2 = 40^{\text{o}} \text{C} \)
02

Use Tables A-2 and A-5 for Part (a)

Locate the specific entropy values in the tables. From Table A-2: Specific entropy at \( 80^{\text{o}} \text{C} \), \( s_1 \) From Table A-5: Specific entropy of the saturated liquid at \( 40^{\text{o}} \text{C} \), \( s_{f2} \) Calculate the change: \[ \triangle s = s_{f2} - s_1 \]
03

Calculate Using Saturated Liquid Data from Table A-2 for Part (b)

From Table A-2, find specific entropies: At initial state, \( s_1 \)At final state, \( s_{f2} \) for saturated liquid at \( 40^{\text{o}} \text{C} \)Calculate the change: \[ \triangle s = s_{f2} - s_1 \]
04

Incompressible Liquid Model with Constant Specific Heat for Part (c)

Use the formula: \[ \triangle s = c_p \times \text{ln} \frac{T_2}{T_1} \]Where: \( c_p \) is the specific heat at constant pressure. From Table A-19, locate \( c_p \) for water. Using the temperatures: Convert temperatures to Kelvin for calculation: \( T_1 = 80+273 = 353 \text{ K} \)\( T_2 = 40+273 = 313 \text{ K} \)Substitute and compute \( \triangle s \):\[ \triangle s = c_p \times \text{ln} \frac{313}{353} \]
05

Use Ideal Gas Tables for Part (d)

Locate the specific entropy values for water in the ideal gas tables (if applicable).Generally, for liquid or real substances, ideal gas assumptions might not apply directly.However, if ideal gas tables are given, find values analogous to the initial and final conditions and calculate the change similarly.

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.

Specific Entropy
Specific entropy is a thermodynamic property that measures the disorder or randomness in a substance per unit mass. It is most commonly denoted by the symbol 's' and has units of \( \text{kJ} / \text{kg} \cdot \text{K} \). Understanding specific entropy is crucial for solving problems related to heat and work in thermodynamic processes. When comparing states in thermodynamics, we often look at the change in specific entropy as an indicator of energy transformations and efficiency. For example, in the given exercise, we calculate the specific entropy at different states to find out how the system changes from one state to another.
Thermodynamic Tables
Thermodynamic tables are essential tools used to find the properties of substances at various states. These tables provide data for temperature, pressure, specific volume, specific entropy, and specific enthalpy, among others. In the exercise, we use tables A-2 and A-5 to find specific entropy values for water at given conditions.
By referencing these tables, we can accurately determine properties such as specific entropy without needing complex calculations. This helps in solving parts (a) and (b) of the exercise where knowing the exact state properties enables us to find the entropy change accurately. Using thermodynamic tables simplifies the process and provides reliable data for further computations.
Incompressible Liquid Model
The incompressible liquid model is an approximation used in thermodynamics to simplify computations for liquids. This model assumes that the liquid鈥檚 volume does not change significantly under pressure variations. It is particularly useful for substances like water. In our exercise, we apply this model to determine the change in specific entropy, treating water as incompressible.
We use the formula \[ \triangle s = c_p \times \text{ln} \frac{T_2}{T_1} \] where \( c_p \) is the specific heat at constant pressure. Tables like A-19 provide the value of \( c_p \) for water. This model is handy when exact table data might not be contextually applicable or available, allowing for a simpler, yet effective, estimation of entropy changes.
Constant Specific Heat
Constant specific heat refers to the assumption that the specific heat capacity of a substance does not change with temperature. This simplification is often used in thermodynamic calculations to make the integration of temperature-dependent properties easier. In the given exercise, for part (c), we use the constant specific heat of water to calculate the change in entropy.
With this assumption, we can apply the formula \[ \triangle s = c_p \times \text{ln} \frac{T_2}{T_1} \] where \( T_1 \) and \( T_2 \) are the initial and final temperatures converted to Kelvin, respectively. The specific heat \( c_p \) is obtained from standard tables. While this model may introduce some approximations, it simplifies the math significantly and is often sufficiently accurate for engineering purposes.

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

A system consists of \(2 \mathrm{~m}^{3}\) of hydrogen gas \(\left(\mathrm{H}_{2}\right)\), initially at \(35^{\circ} \mathrm{C}, 215 \mathrm{kPa}\), contained in a closed rigid tank. Energy is transferred to the system from a reservoir at \(300^{\circ} \mathrm{C}\) until the temperature of the hydrogen is \(160^{\circ} \mathrm{C}\). The temperature at the system boundary where heat transfer occurs is \(300^{\circ} \mathrm{C}\). Modeling the hydrogen as an ideal gas, determine the heat transfer, in \(\mathrm{kJ}\), the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\), and the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\). For the reservoir, determine the change in entropy, in \(\mathrm{kJ} / \mathrm{K}\). Why do these two entropy changes differ?

At steady state, work at a rate of \(25 \mathrm{~kW}\) is done by a paddle wheel on a slurry contained within a closed, rigid tank. Heat transfer from the tank occurs at a temperature of \(250^{\circ} \mathrm{C}\) to surroundings that, away from the immediate vicinity of the tank, are at \(27^{\circ} \mathrm{C}\). Determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), (a) for the tank and its contents as the system. (b) for an enlarged system including the tank and enough of the nearby surroundings for the heat transfer to occur at \(27^{\circ} \mathrm{C}\).

Ammonia enters a valve as a saturated liquid at 7 bar with a mass flow rate of \(0.06 \mathrm{~kg} / \mathrm{min}\) and is steadily throttled to a pressure of 1 bar. Determine the rate of entropy production in \(\mathrm{kW} / \mathrm{K}\). If the valve were replaced by a power-recovery turbine operating at steady state, determine the maximum theoretical power that could be developed, in \(\mathrm{kW}\). In each case, ignore heat transfer with the surroundings and changes in kinetic and potential energy. Would you recommend using such a turbine?

An insulated, rigid tank is divided into two compartments by a frictionless, thermally conducting piston. One compartment initially contains \(1 \mathrm{~m}^{3}\) of saturated water vapor at \(4 \mathrm{MPa}\) and the other compartment contains \(1 \mathrm{~m}^{3}\) of water vapor at \(20 \mathrm{MPa}, 800^{\circ} \mathrm{C}\). The piston is released and equilibrium is attained, with the piston experiencing no change of state. For the water as the system, determine (a) the final pressure, in MPa. (b) the final temperature, in \({ }^{\circ} \mathrm{C}\). (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

An electric motor operating at steady state draws a current of 10 amp with a voltage of \(220 \mathrm{~V}\). The output shaft rotates at 1000 RPM with a torque of \(16 \mathrm{~N} \cdot \mathrm{m}\) applied to an external load. The rate of heat transfer from the motor to its surroundings is related to the surface temperature \(T_{\mathrm{b}}\) and the ambient temperature \(T_{0}\) by \(\mathrm{hA}\left(T_{\mathrm{b}}-T_{0}\right)\), where \(\mathrm{h}=100 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}, \mathrm{A}=0.195 \mathrm{~m}^{2}\), and \(T_{0}=293 \mathrm{~K}\). Energy transfers are considered positive in the directions indicated by the arrows on Fig. P6.51. (a) Determine the temperature \(T_{\mathrm{b}}\), in \(\mathrm{K}\). (b) For the motor as the system, determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\). (c) If the system boundary is located to take in enough of the nearby surroundings for heat transfer to take place at temperature \(T_{0}\), determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for the enlarged system.
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Physical Quantities And Units

Read Explanation

Waves Physics

Read Explanation

Particle Model of Matter

Read Explanation

Famous Physicists

Read Explanation

Electromagnetism

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.