Question

Steam is contained in a large vessel at \(100 \mathrm{lbf} / \mathrm{in} .^{2}, 450^{\circ} \mathrm{F}\). Connected to the vessel by a valve is an initially evacuated tank having a volume of \(1 \mathrm{ft}^{3}\). The valve is opened until the tank is filled with steam at pressure \(p\). The filling is adiabatic, kinetic and potential energy effects are negligible, and the state of the large vessel remains constant. (a) If \(p=100 \mathrm{lbf} / \mathrm{in} .^{2}\), determine the final temperature of the steam within the tank, in \({ }^{\circ} \mathrm{F}\), and the amount of entropy produced within the tank, in \(\mathrm{Btu} /{ }^{\circ} \mathrm{R}\). (b) Plot the quantities of part (a) versus presssure \(p\) ranging from 10 to \(100 \mathrm{lbf} / \mathrm{in}\).

Short answer

Given \(p = 100 \text{ lbf/in}^2\), find the final temperature from steam tables, and compute entropy change based on \( \text{Entropy produced} = m (s_2 - s_1) \). Plot temperature and entropy change for p ranging from 10 to 100 lbf/in².

Step-by-step solution

Determine the initial state of the steam

The pressure of steam in the large vessel is given as 100 lbf/in² and the temperature is 450°F. Use steam tables to find the corresponding specific volume \( v \) and specific entropy \( s \) for steam at this state.

Assume the final state of the steam in the tank

When the valve is opened, steam fills the initially evacuated tank. Since the process is adiabatic and kinetic and potential energy effects are negligible, the energy balance equation can be expressed as \( m_2 u_2 = m_1 u_1 \), where \( m \) is the mass, and \( u \) is the specific internal energy.

Apply the energy conservation principle

Since no heat is transferred and no work is done, the initial enthalpy of the steam in the vessel must equal the final enthalpy of the steam in the tank. Use the equation \( h_1 = h_2 \) to find the temperature of the steam in the tank.

Determine the final temperature

Given that \( p = 100 \text{ lbf/in}^2 \) in the tank, use the steam tables to lookup the temperature corresponding to this pressure. Given that the specific volume and entropy remain constant, identify the exact state and read off the final temperature.

Calculate the entropy change

Use the specific entropy values from the steam tables to calculate the entropy produced: \( \text{Entropy produced} = m (s_2 - s_1) \). Note that in an adiabatic process, if no entropy is generated, \( s_2 \) might equal \( s_1 \), or a finite entropy generation figure will be calculated based on initial parameters.

Plot the quantities versus pressure

For part (b), repeat the calculations for different values of \( p \) ranging from 10 to 100 lbf/in². Plot the final temperature and the amount of entropy produced against these pressures.

Key concepts

Steam Tables and Their Importance

To solve thermodynamic problems, we rely on steam tables. These tables contain critical data about the properties of steam, including temperature, pressure, specific volume, specific enthalpy, and specific entropy. When dealing with steam in a vessel or a tank, these properties help us understand the state and behavior of the steam. For example, in the given exercise, steam tables were used to identify properties of steam at a pressure of 100 lbf/in² and a temperature of 450°F. By referencing the tables, we can accurately determine specific volumes and entropies, which are essential for the calculations needed to solve the problem.

Understanding Entropy Production

Entropy is a measure of the disorder or randomness in a system. When a process occurs, the entropy may change, indicating how much the system's disorder has increased or decreased. In the given exercise, the process of steam filling an evacuated tank generates entropy. The change in entropy is calculated using the specific entropy values from the steam tables. Entropy production is essential in determining the efficiency of a process. An adiabatic process, where no heat is exchanged with the surroundings, can still generate entropy if irreversible changes occur within the system. In our exercise, understanding entropy production helps us evaluate how much the steam's disorder has increased as it filled the tank.

Energy Conservation in Thermodynamics

Energy conservation is a fundamental principle in thermodynamics, stating that energy cannot be created or destroyed, only transferred or converted. The exercise involves an adiabatic process, which means no heat is lost or gained by the steam. Here, the principle of energy conservation is applied to equate the initial energy of the steam in the vessel to the final energy in the tank. This ensures that the total internal energy remains constant. By using specific internal energy values from the steam tables and the fact that the process is adiabatic, we can find the final temperature and specific energy state of the steam filled into the tank.

Adiabatic Process and Its Characteristics

An adiabatic process is one in which no heat is transferred to or from the system. In practical terms, it means the system is perfectly insulated. Adiabatic processes can occur rapidly, preventing heat transfer, or involve insulated boundaries. In the provided exercise, when steam fills the initially evacuated tank, the process is adiabatic. This implies the energy exchange is strictly internal, relying on properties like specific internal energy. Knowing that the process is adiabatic, we can utilize energy conservation principles, leading to specific calculations for the final state of steam using the steam tables. Understanding adiabatic processes helps in simplifying and solving thermodynamic problems where heat exchange can be neglected.