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

A system consisting of \(2 \mathrm{~kg}\) of ammonia undergoes a cycle composed of the following processes: Process 1-2: constant volume from \(p_{1}=10\) bar, \(x_{1}=0.6\) to saturated vapor Process 2-3: constant temperature to \(p_{3}=p_{1}, Q_{23}=+228 \mathrm{~kJ}\) Process 3-1: constant pressure Sketch the cycle on \(p-v\) and \(T-v\) diagrams. Neglecting kinetic and potential energy effects, determine the net work for the cycle and the heat transfer for each process, all in \(\mathrm{kJ}\).

Short Answer

Expert verified
Net work: Sum of works during each process. Heat transfers: Determine using first law of thermodynamics for each process. Sketches: Identify states and processes on \( p-v \) and \( T-v \) diagrams.

Step by step solution

01

- Identify State 1

At state 1, the pressure is given as \( p_1 = 10 \) bar and the quality of the ammonia is \( x_1 = 0.6 \). Using the ammonia property tables, find the specific volume \( v_1 \), internal energy \( u_1 \), and other relevant properties corresponding to these conditions.
02

- Determine State 2

For process 1-2, the volume is constant, and the ammonia reaches a saturated vapor state. Find the properties of the saturated vapor at 10 bar, specifically the specific volume \( v_2 \), the internal energy \( u_2 \), and the temperature \( T_2 \).
03

- Analyze Process 2-3 (Constant Temperature)

During process 2-3, the system undergoes a constant temperature process until the pressure returns to 10 bar. Use the specific volume from state 2 to find the properties at state 3. Also, ensure that the heat transfer \( Q_{23} = +228 \mathrm{kJ} \) is considered. Calculate the final internal energy \( u_3 \) at state 3.
04

- Determine State 3 Conditions

Since process 2-3 is at a constant temperature, use the relationship \( p_3 = p_1 \). Find the specific volume \( v_3 \) and internal energy \( u_3 \) at state 3.
05

- Analyze Process 3-1 (Constant Pressure)

For process 3-1, which is at constant pressure, use the specific volumes and internal energies already identified at states 1 and 3. Calculate the boundary work done during this process by using the area under the constant pressure line in the \( p-v \) diagram.
06

- Calculate Net Work for the Cycle

To find the net work for the cycle, sum the works done during each process from 1-2, 2-3, and 3-1. The work for process 1-2 is zero since volume is constant. For process 2-3, use \( W = Q_{23} - (u_3 - u_2) \). For process 3-1, use \( W_{31} = p_1 (v_1 - v_3) \). Sum these contributions to get the total work.
07

- Calculate Heat Transfer for Each Process

The heat transfer for each process can be determined using the first law of thermodynamics: \( Q = \Delta U + W \). Calculate \( Q_{12} \), \( Q_{23} \), and \( Q_{31} \) individually by integrating internal energy changes and work done in each process.
08

- Sketch the Cycle on \( p-v \) and \( T-v \) Diagrams

Draw the cycle on both \( p-v \) and \( T-v \) diagrams. Show all three states and label the processes: 1-2, 2-3, and 3-1. Indicate constant volume, constant temperature, and constant pressure lines appropriately.

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.

Ammonia Properties
Ammonia is a common refrigerant and has well-characterized thermodynamic properties. Understanding these properties is essential for solving cycle analysis problems. Key properties include specific volume ( v ), internal energy ( u ), and quality ( x ). In state 1, where ammonia is at 10 bar with a quality of 0.6, you can use ammonia property tables to find specific volume and internal energy. Quality represents the proportion of ammonia that is vapor as opposed to liquid. At state 2, the ammonia is a saturated vapor, so its quality is 1. Use property tables to find corresponding values. Property tables are invaluable for locating these values quickly.
Heat Transfer
Heat transfer ( Q ) is a crucial part of thermodynamic cycle analysis. It represents the energy added to or removed from the system. In Process 2-3, the given heat transfer is +228 kJ. To find heat transfer in other processes, use the first law of thermodynamics: Q = ΔU + W . Here, ΔU is the change in internal energy, and W is the work done. Calculating heat transfer involves precise identification of changes in internal energy and work. This is pivotal in understanding energy flows within the cycle. For Process 1-2 and 3-1, always refer to any additional provided data or property tables.
Net Work Calculation
Net work ( W ) for the cycle is a key outcome, representing the usable energy extracted. To find net work, add the work done during each process. For Process 1-2, work is zero since volume is constant ( W_{12} = 0 ). In Process 2-3, use the relationship W = Q_{23} - (u_3 - u_2) . Here, Q_{23} is given as +228 kJ. For Process 3-1, use W_{31} = p_1 (v_1 - v_3) , where p_1 is the pressure, and v_1, v_3 are specific volumes. Add these contributions to get total net work for the cycle. Net work is essential for determining the efficiency and performance of thermodynamic cycles.
Constant Volume Process
In a constant volume process, volume stays unchanged. For Process 1-2, the ammonia transitions from an initial state at 10 bar with quality 0.6 to a saturated vapor state. Identify the properties such as v and u since volume ( v ) is constant. In such processes, no boundary work is done ( W = 0 ). This simplifies calculations of net work and heat transfer. Use the first law of thermodynamics, Q_{12} = ΔU_{12} , where ΔU_{12} is the change in internal energy. Constant volume processes are found in many cycle analyses and understanding them aids in thermodynamic calculations.
Constant Temperature Process
During a constant temperature process, temperature remains unchanged. In Process 2-3, the system maintains a constant temperature while the pressure returns to 10 bar. Use the given heat transfer ( Q_{23} = +228 kJ) and apply it in the first law of thermodynamics, Q = ΔU + W . Calculate the change in internal energy ( ΔU = u_3 - u_2 ) and use it to find work. Constant temperature processes ( isothermal ) often involve phase change and are critical for gauging heat interactions accurately. They help in understanding how systems absorb or release heat without temperature change.
Constant Pressure Process
In constant pressure processes, pressure stays constant while the volume changes. For Process 3-1 in the exercise, determine properties like specific volumes ( v_1 and v_3 ) and internal energy. Use the equation W_{31} = p_1 (v_1 - v_3) to calculate work done during this process. It’s crucial to know that boundary work is area under the p-v curve. For constant pressure process, heat transfer calculation involves first law of thermodynamics, Q_{31} = ΔU_{31} + W_{31} . Constant pressure processes help in analyzing real-world thermodynamic systems like engines and refrigerators where pressure often remains unchanged.

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 method of modeling gas behavior from the microscopic viewpoint is known as the kinetic theory of gases. Using kinetic theory, derive the ideal gas equation of state and explain the variation of the ideal gas specific heat \(c_{v}\) with temperature. Is the use of kinetic theory limited to ideal gas behavior? Discuss.

Air is compressed adiabatically from \(p_{1}=1\) bar, \(T_{1}=\) \(300 \mathrm{~K}\) to \(p_{2}=15\) bar, \(v_{2}=0.1227 \mathrm{~m}^{3} / \mathrm{kg}\). The air is then cooled at constant volume to \(T_{3}=300 \mathrm{~K}\). Assuming ideal gas behavior, and ignoring kinetic and potential energy effects, calculate the work for the first process and the heat transfer for the second process, each in kJ per kg of air. Solve the problem each of two ways: (a) using data from Table A-22. (b) using a constant specific heat evaluated at \(300 \mathrm{~K}\).

Compare the densities, in \(\mathrm{kg} / \mathrm{m}^{3}\), of helium and air, each at \(300 \mathrm{~K}, 100 \mathrm{kPa}\). Assume ideal gas behavior.

. Under what circumstances is the following statement correct? Equal molar amounts of two different gases at the same temperature, placed in containers of equal volume, have the same pressure.

Two kilograms of a gas with molecular weight 28 are contained in a closed, rigid tank fitted with an electric resistor. The resistor draws a constant current of \(10 \mathrm{amp}\) at a voltage of \(12 \mathrm{~V}\) for \(10 \mathrm{~min}\). Measurements indicate that when equilibrium is reached, the temperature of the gas has increased by \(40.3^{\circ} \mathrm{C}\). Heat transfer to the surroundings is estimated to occur at a constant rate of \(20 \mathrm{~W}\). Assuming ideal gas behavior, determine an average value of the specific heat \(c_{p}\), in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\), of the gas in this temperature interval based on the measured data.

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