Question

A vertical piston-cylinder device initially contains \(0.01 \mathrm{m}^{3}\) of steam at \(200^{\circ} \mathrm{C}\). The mass of the frictionless piston is such that it maintains a constant pressure of \(500 \mathrm{kPa}\) inside. Now steam at \(1 \mathrm{MPa}\) and \(350^{\circ} \mathrm{C}\) is allowed to enter the cylinder from a supply line until the volume inside doubles. Neglecting any heat transfer that may have taken place during the process, determine ( \(a\) ) the final temperature of the steam in the cylinder and \((b)\) the amount of mass that has entered.

Short answer

The final temperature of the steam in the cylinder is 673.15°C, and the amount of mass (steam) that has entered is 0.02275 kg.

Step-by-step solution

Write down the ideal gas law

The ideal gas law states that \(PV = mRT\), where \(P\) is pressure, \(V\) is volume, \(m\) is mass, \(R\) is the specific gas constant, and \(T\) is temperature.

Identify given values and find initial mass

Initially, the pressure is \(500 \mathrm{kPa}\), volume is \(0.01 \mathrm{m}^{3}\), and temperature is \(200^{\circ} \mathrm{C}\). We need to find the initial mass using the ideal gas law. To do this, convert the temperature to Kelvin: \(T_1 = (200 + 273.15) \mathrm{K} = 473.15 \mathrm{K}\) Use the specific gas constant for steam: \(R_{steam} = 0.4615\,\mathrm{kJ/(kg\,K)}\) Now we can find the initial mass: \(m_1 = \frac{PV}{RT_1} = \frac{(500\,\mathrm{kPa})(0.01\,\mathrm{m^3})}{(0.4615\,\mathrm{kJ/(kg\,K)})(473.15\,\mathrm{K})} = 0.02240\,\mathrm{kg}\)

Determine final volume

As the volume doubles, the final volume is: \(V_2 = 2 \times V_1 = 2 \times 0.01\,\mathrm{m^3} = 0.02\,\mathrm{m^3}\).

Determine final temperature

Since the final pressure is still \(500\,\mathrm{kPa}\), and the final volume is given, we can find the final temperature using the ideal gas law: \(T_2 = \frac{PV_2}{m_1 R} = \frac{(500\,\mathrm{kPa})(0.02\,\mathrm{m^3})}{(0.02240\,\mathrm{kg})(0.4615\,\mathrm{kJ/(kg\,K)})} = 946.30\,\mathrm{K}\). Convert the final temperature back to Celsius: \(T_2 = 946.30\,\mathrm{K} - 273.15 = 673.15^{\circ}\,\mathrm{C}\)

Determine amount of mass that has entered

Now that we have the final temperature, we can find the mass of the steam that has entered the cylinder: \(m_{entered} = \frac{PV_2}{RT_2} - m_1 = \frac{(500\,\mathrm{kPa})(0.02\,\mathrm{m^3})}{(0.4615\,\mathrm{kJ/(kg\,K)})(946.30\,\mathrm{K})} - 0.02240\,\mathrm{kg} = 0.02275\,\mathrm{kg}\). So, the final temperature of the steam in the cylinder is \((a) \boldsymbol{673.15^{\circ}\,\mathrm{C}}\), and the amount of mass (steam) that has entered is \((b) \boldsymbol{0.02275\,\mathrm{kg}}\).

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics that relates the pressure, volume, temperature, and mass of an ideal gas. It is expressed as the equation \(PV = mRT\), where:\(P\) stands for the absolute pressure of the gas,
\(V\) denotes the volume occupied by the gas,
\(m\) is the mass of the gas,
\(R\) represents the specific gas constant, which varies with the kind of gas,
and \(T\) is the absolute temperature of the gas, measured in Kelvin.

When we talk about an 'ideal gas', it's a hypothetical gas that perfectly follows this law in all conditions. Of course, real gases don't always behave ideally, but many gases can be approximated as ideal under normal conditions. This law becomes extensively useful in problem-solving scenarios involving piston-cylinder devices, where changes in a gas's state are explored.

In the exercise about the piston-cylinder, the ideal gas law helps determine the initial mass of the steam and understand how changes in volume affect temperature and the mass of steam that enters the cylinder. It’s crucial in finding the final temperature and mass after the thermodynamic process in which the steam volume doubles while maintaining a constant pressure.

Diving Deep into Steam Properties

Steam is an important working fluid in thermodynamics, often employed in engines and turbines due to its energy-carrying capability. Properties such as pressure, temperature, specific volume, and specific enthalpy are critical in analyzing steam's behavior and state. These properties are not always constant and can vary greatly with phase changes and energy transfers.

In thermodynamic problems like our example with the piston-cylinder, the specific gas constant for steam \(\(R_{steam}\)\) is used. This value is crucial for calculations involving the ideal gas law. Steam tables or steam charts can provide this constant as well as other properties based on temperature and pressure. However, in the absence of these tables, the ideal gas assumption allows for a simplified yet still reliable analysis under certain conditions, provided we stay away from the saturated steam curve and phase change conditions.

The properties of steam are not just academic; they're essential in designing and operating a multitude of thermal systems, such as power plants and heating systems. Understanding how steam behaves under different conditions allows engineers to optimize efficiency and performance.

Navigating Thermodynamic Processes

Thermodynamic processes involve changes to a system's state variables, such as pressure, volume, and temperature, which in turn change the system’s energy. There are several types of processes, including isobaric (constant pressure), isochoric (constant volume), isothermal (constant temperature), and adiabatic (no heat exchange).

In the piston-cylinder exercise, the process described is isobaric, as the pressure is kept constant at \(500 \text{kPa}\) while steam is added. An isobaric process is common in piston-cylinder assemblies of steam engines where steam is added or expelled at a constant pressure, doing work on the piston. It is important to note whether the process is assumed to be ideal, meaning no heat loss to the environment, as is the case in our example.

An understanding of the type of thermodynamic process occurring is crucial when it comes to predicting the outcome of a system and solving for unknown variables. By recognizing the process as isobaric, students can simplify their calculations and utilize the ideal gas law more effectively to find the final temperature and the mass of steam that enters the cylinder - helping them grasp both the theoretical concepts and practical implications.