Question

A piston-cylinder device initially contains \(1.4 \mathrm{kg}\) of refrigerant-134a at \(100 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\). Heat is now transferred to the refrigerant from a source at \(150^{\circ} \mathrm{C}\), and the piston which is resting on a set of stops, starts moving when the pressure inside reaches \(120 \mathrm{kPa}\). Heat transfer continues until the temperature reaches \(80^{\circ} \mathrm{C}\). Assuming the surroundings to be at \(25^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\), determine \((a)\) the work done, (b) the heat transfer, \((c)\) the exergy destroyed, and \((d)\) the second- law efficiency of this process. Answers: (a) \(0.497 \mathrm{kJ},\) (b) \(67.9 \mathrm{kJ}\).

Short answer

Answer: (a) The work done during the process is 0.497 kJ. (b) The heat transfer during the process is 67.9 kJ.

Step-by-step solution

Identify the initial state of the system

Initially, the system has a mass of 1.4 kg of refrigerant-134a at \(100\ \mathrm{kPa}\) and \(20^{\circ}\mathrm{C}\). Using the data, determine the specific volume \(v_{1}\) and internal energy \(u_{1}\) from the refrigerant properties table.

Calculate the final state of the system

At the end of the process, the temperature reaches \(80^{\circ} \mathrm{C}\). Using this final temperature and the final pressure of \(120\ \mathrm{kPa}\), find the corresponding final specific volume \(v_{2}\) and internal energy \(u_{2}\) from the refrigerant properties table.

Calculate the mass of the refrigerant

Use the initial specific volume \(v_{1}\) to find the initial volume \(V_{1}\), by equation \(V_{1} = m v_{1}\). Since the process is happening in a piston-cylinder device, we assume that it is a constant volume process, and hence assign the final volume \(V_{2} = V_{1}\).

Calculate the work done during the process

Using the constant volume assumption from Step 3 and the final pressure, calculate the work done during the process. Work done \(W = m(v_{2} - v_{1})P_{2}\), where \(m\) is the mass of refrigerant.

Calculate the heat transfer during the process

Now that we have the work done, we can use the first law of thermodynamics to calculate the heat transfer during the process. The first law equation is given by \(Q = W + \Delta U\), where \(\Delta U = m(u_{2}-u_{1})\).

Calculate the exergy destroyed during the process

To calculate the exergy destroyed, we need to determine the change in exergy during the process. The change in exergy can be found using the equation: \(\Delta e = m(T_{2}s_{2} - T_{1}s_{1})\), where \(s_{1}\) and \(s_{2}\) are the initial and final specific entropies, respectively. Using the properties of the refrigerant, find the change in exergy, and using the second law of thermodynamics, determine the exergy destroyed: \(X_{destr} = \Delta e + Q(1-\frac{T_{0}}{T_{source}})\), where \(T_{0}\) is the surrounding temperature and \(T_{source}\) is the heat source temperature.

Calculate the second-law efficiency of the process

Finally, we need to calculate the second-law efficiency of the process. This can be found using the equation: \(\eta _{II} = \frac{W_{rev}-X_{destr}}{W_{rev}}\), where \(W_{rev}\) is the reversible work and can be calculated using the equation: \(W_{rev} = m(u_{1}-u_{2}) - Q_{source}(\frac{T_0}{T_{source}})\). Using the steps above, we find: (a) The work done is \(0.497\ \mathrm{kJ}\). (b) The heat transfer is \(67.9\ \mathrm{kJ}\).

Key concepts

Refrigerant-134a Properties

Understanding the properties of Refrigerant-134a is critical when solving thermodynamics problems involving refrigeration cycles. Refrigerant-134a, also known as R-134a, is a hydrofluorocarbon (HFC) used in air conditioning and refrigeration systems. This refrigerant is known for its low environmental impact because it does not deplete ozone and has lower global warming potential than many other refrigerants.

To solve problems, properties such as pressure, temperature, specific volume, internal energy, and specific entropy are obtained from refrigerant-134a property tables or saturation pressure tables. These properties are essential for determining the state of the refrigerant at various points in the thermodynamic cycle, such as in the exercise where the specific volume and internal energy are needed to calculate the work done and heat transfer.

First Law of Thermodynamics

The First Law of Thermodynamics is a statement of energy conservation in thermodynamic processes. It asserts that energy can neither be created nor destroyed, only transferred or converted from one form to another. In the context of a piston-cylinder device, this law is employed to quantify the interrelation of heat transfer, work done, and the change in internal energy of the system.

Mathematically, it is expressed as \(Q = W + \triangle U\), where \(Q\) is the heat transfer into the system, \(W\) is the work done by the system, and \(\triangle U\) is the change in internal energy of the system. In our exercise, we apply this law to determine the amount of heat transferred to the refrigerant-134a during the process.

Exergy Destruction

Exergy measures the quality of energy, illustrating how much work a system can potentially produce. However, not all energy is convertible to work due to inefficiencies and entropy generation, which is referred to as exergy destruction. In thermodynamic processes, exergy destruction occurs due to processes being irreversible.

Exergy destruction is closely related to the Second Law of Thermodynamics and is quantified by the difference between the exergy input and exergy output of a system. The process in the exercise leads to exergy destruction due to heat transfer with a finite temperature difference. The magnitude of exergy destruction provides insight into how far the process departs from ideal, reversible conditions.

Second-Law Efficiency

Second-law efficiency, or exergetic efficiency, is a measure of the effectiveness of a thermodynamic process in terms of exergy use. It compares the actual work produced by a system to the maximum possible work that could be produced if the process were entirely reversible and no exergy was destroyed.

Calculated using the formula \(\text{second-law efficiency} = \frac{\text{actual work output} - \text{exergy destroyed}}{\text{possible work output (reversible case)}}\), second-law efficiency provides a more comprehensive understanding of a system’s performance than first-law (or thermal) efficiency alone. This efficiency can never be 100% due to natural irreversibilities in all real processes, as illustrated by the piston-cylinder device exercise.