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An insulated piston-cylinder device contains \(0.03 \mathrm{m}^{3}\) of saturated refrigerant-134a vapor at 0.6 MPa pressure. The refrigerant is now allowed to expand in a reversible manner until the pressure drops to 0.16 MPa. Determine the change in the exergy of the refrigerant during this process and the reversible work. Assume the surroundings to be at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\).

Short Answer

Expert verified
Explain the process of finding the properties of the initial and final states of the refrigerant-134a in this exercise. How do you use these properties to calculate the change in exergy and the reversible work done? Once you have calculated these properties, how do you then calculate the change in exergy and the reversible work done during the process?

Step by step solution

01

Find the properties of the initial state:

Given the initial volume (\(V_1 = 0.03 \mathrm{m}^3\)) and pressure (\(P_1 = 0.6 \mathrm{MPa}\)), we can find the initial specific volume (\(v_1 = \frac{V_1}{m}\)), where m is the mass of the refrigerant. Using the property table for refrigerant-134a at this pressure, we can find the specific enthalpy (\(h_1\)) and specific entropy (\(s_1\)).
02

Find the properties of the final state:

The final pressure is given as \(P_2 = 0.16 \mathrm{MPa}\). Since the process is reversible, the final entropy (\(s_2\)) will be the same as the initial entropy (\(s_1\)). We will assume the process to be isentropic \((s_2 = s_1)\). Using the property table for refrigerant-134a at this pressure, find the final specific enthalpy (\(h_2\)).
03

Calculate the change in exergy:

With the enthalpy and entropy values for both initial and final states, we can compute the change in exergy using the formula: \(∆e = -T_0∆s + ∆h - T_0∆s_{gen}\) We're given the temperature of the surroundings, \(T_0 = 25^{\circ} \mathrm{C} = 298.15 \mathrm{K}\), and the process is reversible so \(∆s_{gen} = 0\). Then, we can calculate the change in exergy (\(∆e\)) as follows: \(∆e = -T_0(s_2 - s_1) + (h_2 - h_1)\)
04

Calculate the reversible work:

Using the calculated enthalpy values for both initial and final states, we can compute the reversible work (\(W_{rev}\)) done during the expansion process using the formula: \(W_{rev} = -∆H\) So, the reversible work is: \(W_{rev} = -(h_2 - h_1)\) Now you have calculated the change in exergy and the reversible work done during the process.

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Key Concepts

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

Reversible Process
Understanding a reversible process is key to grasping the foundations of thermodynamics. A reversible process is an idealized concept where a system undergoes changes in such a way that the system and its surroundings can be returned to their original states without any net change. In reality, no process is truly reversible due to irreversibilities such as friction and heat loss. However, in theoretical exercises like the one involving the piston-cylinder and refrigerant-134a, assuming reversibility simplifies calculations. During a reversible process, the entropy of the system remains constant and no entropy is generated, denoted as \(\Delta s_{gen} = 0\). This property is leveraged in calculating the exergy, which represents the maximum useful work that can be extracted from a system as it reaches equilibrium with its surroundings.
Piston-Cylinder Device
The piston-cylinder device is a staple in thermodynamics problems and is used to model the expansion and compression of gases. It consists of a cylindrical container sealed with a movable piston, allowing the volume within the cylinder to change as the piston moves. In our exercise, the device contains refrigerant-134a vapor, and as the piston moves due to the expansion of the vapor, work is done on or by the system.

A key aspect to recall here is that during a reversible expansion or compression in a piston-cylinder device, the pressure exerted by the gas on the piston matches the external pressure at every moment. The device allows us to better visualize thermodynamic processes and calculate properties like work and changes in energy.
Refrigerant Properties
When working with refrigerants like refrigerant-134a, it's critical to look at specific properties that determine how the substance will behave under different conditions. Refrigerant properties include pressure, temperature, specific enthalpy, specific entropy, and specific volume. These properties can be found in thermodynamic tables or calculated using equations of state. In our exercise, these properties are used to define the refrigerant's initial and final states, providing the necessary data to calculate changes in exergy and work done during reversible processes. The initial and final specific enthalpies, \(h_1\) and \(h_2\), and specific entropies, \(s_1\) and \(s_2\), are especially important in determining how much energy is available to do work or be transferred as heat.
Enthalpy Change
Enthalpy change, represented as \(\Delta H\), is a measure of the total energy change within a system that involves both internal energy changes and work done by the system expanding or contracting. For a piston-cylinder device, as is the case in our exercise, the enthalpy change is closely related to the work done, particularly when the process is isothermal or adiabatic. In the step-by-step solution, the difference between the final and initial specific enthalpies, \(h_2 - h_1\), provides the necessary information to evaluate the reversible work, \(W_{rev}\), done during the process, as well as to calculate the change in exergy is calculated using this enthalpy difference.
Isentropic Process
An isentropic process is a special type of thermodynamic process where entropy remains constant, \(s = constant\), throughout. It is an idealization often used in conjunction with the concept of a reversible process. A process is isentropic if it involves no heat transfer, which means it is adiabatic, and if all processes are reversible. In the context of the exercise, assuming an isentropic process simplifies the determination of the final state of the refrigerant during expansion, because we can determine the final properties using the fixed entropy value from the initial state.

By recognizing the isentropic assumption, we conclude that the final entropy \(s_2\) does not differ from the initial entropy \(s_1\), which significantly streamlines our calculations for changes in thermodynamic properties, reversible work, and exergy.

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Most popular questions from this chapter

Refrigerant-134a enters an adiabatic compressor at 120 kPa superheated by \(2.3^{\circ} \mathrm{C},\) and leaves at \(0.7 \mathrm{MPa}\). If the compressor has a second-law efficiency of 85 percent, determine \((a)\) the actual work input, \((b)\) the isentropic efficiency, and \((c)\) the exergy destruction. Take the environment temperature to be \(25^{\circ} \mathrm{C}\).

Obtain a relation for the second-law efficiency of a heat engine that receives heat \(Q_{H}\) from a source at temperature \(T_{H}\) and rejects heat \(Q_{L}\) to a sink at \(T_{L},\) which is higher than \(T_{0}\) (the temperature of the surroundings), while producing work in the amount of \(W\)

A 100 -L well-insulated rigid tank is initially filled with nitrogen at \(1000 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\). Now a valve is opened and one-half of nitrogen's mass is allowed to escape. Determine the change in the exergy content of the tank.

Combustion gases enter a gas turbine at \(900^{\circ} \mathrm{C}\) \(800 \mathrm{kPa},\) and \(100 \mathrm{m} / \mathrm{s}\) and leave at \(650^{\circ} \mathrm{C}, 400 \mathrm{kPa},\) and \(220 \mathrm{m} / \mathrm{s} .\) Taking \(c_{p}=1.15 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) and \(k=1.3\) for the combustion gases, determine \((a)\) the exergy of the combustion gases at the turbine inlet and ( \(b\) ) the work output of the turbine under reversible conditions. Assume the surroundings to be at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\). Can this turbine be adiabatic?

Writing the first- and second-law relations and simplifying, obtain the reversible work relation for a closed system that exchanges heat with the surrounding medium at \(T_{0}\) in the amount of \(Q_{0}\) as well as a heat reservoir at \(T_{R}\) in the amount of \(Q_{R^{*}}\) (Hint: Eliminate \(Q_{0}\) between the two equations.)

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