Question

A quantity of liquid water undergoes a process from \(80^{\circ} \mathrm{C}\), 5 MPa to saturated liquid at \(40^{\circ} \mathrm{C}\). Determine the change in specific entropy, in \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\), using (a) Tables A-2 and A-5. (b) saturated liquid data only from Table A-2. (c) the incompressible liquid model with a constant specific heat from Table A-19. (d) \(I T\).

Short answer

Use tables to find entropy values at given temperatures and states; calculate changes based on differences. For part (c), utilize specific heat and temperature ratio.

Step-by-step solution

Given Data

Identify the given data and all initial conditions: Temperature at initial state: \( T_1 = 80^{\text{o}} \text{C} \) Pressure at initial state: \( P_1 = 5 \text{ MPa} \) Final state: Saturated liquid at \( T_2 = 40^{\text{o}} \text{C} \)

Use Tables A-2 and A-5 for Part (a)

Locate the specific entropy values in the tables. From Table A-2: Specific entropy at \( 80^{\text{o}} \text{C} \), \( s_1 \) From Table A-5: Specific entropy of the saturated liquid at \( 40^{\text{o}} \text{C} \), \( s_{f2} \) Calculate the change: \[ \triangle s = s_{f2} - s_1 \]

Calculate Using Saturated Liquid Data from Table A-2 for Part (b)

From Table A-2, find specific entropies: At initial state, \( s_1 \)At final state, \( s_{f2} \) for saturated liquid at \( 40^{\text{o}} \text{C} \)Calculate the change: \[ \triangle s = s_{f2} - s_1 \]

Incompressible Liquid Model with Constant Specific Heat for Part (c)

Use the formula: \[ \triangle s = c_p \times \text{ln} \frac{T_2}{T_1} \]Where: \( c_p \) is the specific heat at constant pressure. From Table A-19, locate \( c_p \) for water. Using the temperatures: Convert temperatures to Kelvin for calculation: \( T_1 = 80+273 = 353 \text{ K} \)\( T_2 = 40+273 = 313 \text{ K} \)Substitute and compute \( \triangle s \):\[ \triangle s = c_p \times \text{ln} \frac{313}{353} \]

Use Ideal Gas Tables for Part (d)

Locate the specific entropy values for water in the ideal gas tables (if applicable).Generally, for liquid or real substances, ideal gas assumptions might not apply directly.However, if ideal gas tables are given, find values analogous to the initial and final conditions and calculate the change similarly.

Key concepts

Specific Entropy

Specific entropy is a thermodynamic property that measures the disorder or randomness in a substance per unit mass. It is most commonly denoted by the symbol 's' and has units of \( \text{kJ} / \text{kg} \cdot \text{K} \). Understanding specific entropy is crucial for solving problems related to heat and work in thermodynamic processes. When comparing states in thermodynamics, we often look at the change in specific entropy as an indicator of energy transformations and efficiency. For example, in the given exercise, we calculate the specific entropy at different states to find out how the system changes from one state to another.

Thermodynamic Tables

Thermodynamic tables are essential tools used to find the properties of substances at various states. These tables provide data for temperature, pressure, specific volume, specific entropy, and specific enthalpy, among others. In the exercise, we use tables A-2 and A-5 to find specific entropy values for water at given conditions.
By referencing these tables, we can accurately determine properties such as specific entropy without needing complex calculations. This helps in solving parts (a) and (b) of the exercise where knowing the exact state properties enables us to find the entropy change accurately. Using thermodynamic tables simplifies the process and provides reliable data for further computations.

Incompressible Liquid Model

The incompressible liquid model is an approximation used in thermodynamics to simplify computations for liquids. This model assumes that the liquid’s volume does not change significantly under pressure variations. It is particularly useful for substances like water. In our exercise, we apply this model to determine the change in specific entropy, treating water as incompressible.
We use the formula \[ \triangle s = c_p \times \text{ln} \frac{T_2}{T_1} \] where \( c_p \) is the specific heat at constant pressure. Tables like A-19 provide the value of \( c_p \) for water. This model is handy when exact table data might not be contextually applicable or available, allowing for a simpler, yet effective, estimation of entropy changes.

Constant Specific Heat

Constant specific heat refers to the assumption that the specific heat capacity of a substance does not change with temperature. This simplification is often used in thermodynamic calculations to make the integration of temperature-dependent properties easier. In the given exercise, for part (c), we use the constant specific heat of water to calculate the change in entropy.
With this assumption, we can apply the formula \[ \triangle s = c_p \times \text{ln} \frac{T_2}{T_1} \] where \( T_1 \) and \( T_2 \) are the initial and final temperatures converted to Kelvin, respectively. The specific heat \( c_p \) is obtained from standard tables. While this model may introduce some approximations, it simplifies the math significantly and is often sufficiently accurate for engineering purposes.