Question

A piston-cylinder device initially contains 2 L of air at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). Air is now compressed to a final state of \(600 \mathrm{kPa}\) and \(150^{\circ} \mathrm{C}\). The useful work input is \(1.2 \mathrm{kJ}\) Assuming the surroundings are at \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\), determine \((a)\) the exergy of the air at the initial and the final states, (b) the minimum work that must be supplied to accomplish this compression process, and ( \(c\) ) the second-law efficiency of this process.

Short answer

#tag_title#Solution Summary#tag_content#To find the exergy of the air at the initial and final states, we used the specific exergy formula and the properties of an ideal gas. The minimum work required for the compression process was found using the formula \(W_\textup{min} = m \cdot (e_1 - e_2)\), and the second-law efficiency of the process was calculated using the formula \(η_{SL} = \frac{W_\textup{min}}{W_\textup{act}}\).

Step-by-step solution

Initial specific volume v1 in m³/kg

Convert the initial volume, 2 L, to m³ by multiplying by \(10^{-3}\), then we divide by the mass to get the initial specific volume: \(v_1 = \frac{2 \times 10^{-3}}{m}\)

Final specific volume v2 in m³/kg

Assuming air behaves as an ideal gas, we can use the Ideal Gas Law \(pv=RT\) to find the final specific volume: \(v_2 = \frac{R \times T_2}{p_2}\), where \(R = 287 \frac{\textup{J}}{\textup{kg} \cdot \textup{K}}\), \(T_2 = 150 + 273.15 = 423.15 \mathrm{K}\), and \(p_2 = 600 \times 10^3 \mathrm{Pa}\) #Phase 2: Calculate exergy at initial and final states#

Initial specific exergy e1 in kJ/kg

Calculate the initial specific exergy using the formula \(e_1 = u_1 - u_0 + p_0(v_1 - v_0) - T_0(s_1 - s_0)\), where the subscript 0 denotes the properties of the reference state. Here, \(u_1, v_1, s_1\) are the initial internal energy, specific volume, and specific entropy of the air, and \(u_0, v_0, s_0\) are the values at the reference state. Calculate these values using known thermodynamic relations and properties of ideal gas.

Final specific exergy e2 in kJ/kg

Similarly, calculate the final specific exergy \(e_2 = u_2 - u_0 + p_0(v_2 - v_0) - T_0(s_2 - s_0)\), where \(u_2, v_2, s_2\) are the final internal energy, specific volume, and specific entropy of the air. #Phase 3: Determine minimum work, actual and second-law efficiency#

Minimum work input Wmin in kJ

Calculate the minimum work input using the formula \(W_\textup{min} = m \cdot (e_1 - e_2)\), where \(m\) is the mass of the air in the cylinder.

Actual work input Wact in kJ

Calculate the actual work input, which is given as \(W_\textup{act} = 1.2 \times 10^3 \mathrm{J}\)

Second-law efficiency η_SL

Calculate the second-law efficiency using the formula \(η_{SL} = \frac{W_\textup{min}}{W_\textup{act}}\). After solving the above calculations, you will have the required answers for (a) the exergy of the air at the initial and final states, (b) the minimum work required for the compression process, and (c) the second-law efficiency of the process.

Key concepts

Piston-Cylinder Device

A piston-cylinder device is the fundamental component of many thermodynamic systems, such as internal combustion engines and steam turbines. It comprises a cylinder that houses a movable piston, creating a seal that helps in performing work on or by the contained fluid, typically a gas.

In thermodynamics, we often use this device to demonstrate the principles of pressure-volume work. As the piston moves, it changes the volume within the cylinder, and this volume change can lead to work done by the system if the piston is forced outwards, or on the system if it is compressed. During a compression process, like the one described in the exercise, work is done on the air inside the cylinder, which in turn changes its thermodynamic properties such as pressure, temperature, and internal energy.

Understanding how a piston-cylinder device operates is crucial for solving problems relating to work and energy transformations. In the given exercise, the compression of air results in increased pressure and temperature, which must be analyzed to draw conclusions about the energy interactions, especially when considering aspects like the exergy of the air or the second-law efficiency of the process.

Ideal Gas Law

The ideal gas law is a fundamental equation in the study of thermodynamics, which describes the relationship between the pressure (p), volume (v), and temperature (T) of an ideal gas. The equation is usually stated as:\[\begin{equation}pv = nRT\text{ or }pv = mRT\text{, where:}\end{equation}\]

• p = pressure of the gas,
• v = volume of the gas,
• T = absolute temperature in kelvins (K),
• n = number of moles,
• R = ideal gas constant, and
• m = mass of the gas.

It assumes no interactions between the gas molecules and that all collisions are perfectly elastic. In our exercise, the ideal gas law is used to determine the specific volume of the air in its final state after compression. This then allows for further calculations of thermodynamic properties required for exergy analysis. When air is treated as an ideal gas, the law simplifies calculations significantly, as it avoids the complexities of real gas behavior.

Real gases deviate from ideal behavior at high pressures and low temperatures, but for many engineering applications, including the one in the exercise, it is perfectly reasonable to model the behavior of gases such as air as ideal, especially when dealing with standard temperature and pressure ranges.

Second-Law Efficiency

The second-law efficiency is an important concept in thermodynamics that provides a measure of the energy conversion process quality compared to a reversible process. This efficiency is different from the traditional (first-law) thermal efficiency as it takes into account not only the quantity but also the quality of energy transformations.

Mathematically, second-law efficiency \[\begin{equation}(η_{SL}) = \frac{W_{min}}{W_{act}}\text{, where:}\end{equation}\]

• W_{min} is the theoretical minimum work necessary to drive a process between two states,
• W_{act} is the actual work input in reality.

Ideally, less work input would be needed if a process were reversible. However, since all real processes are irreversible to some extent, more work is required. The second-law efficiency thus quantifies this aspect by expressing how close an actual process comes to the ideal reversible process. In the given exercise, the actual compression work was provided (1.2 kJ), and by calculating the theoretical minimum work, we can determine the second-law efficiency. The higher the η_{SL}, the more effective the process is in terms of energy usage. This concept is particularly important when evaluating and designing systems meant to be both energy-efficient and sustainable.