Question

Can saturated water vapor at \(200 \mathrm{kPa}\) be condensed to a saturated liquid in an isobaric, closed system process while only exchanging heat with an isothermal energy reservoir at \(90^{\circ} \mathrm{C} ?\) (Hint: Determine the entropy generation.)

Short answer

Solution: 1. Calculate the entropy change of the system: Δs_sys = s_2 - s_1 2. Find the entropy change of the reservoir: Δs_res = -Q/T_res 3. Calculate the entropy generation during the transformation: S_gen = Δs_sys + Δs_res 4. Determine if the transformation is possible based on the value of S_gen: - If S_gen > 0, the transformation is irreversible and possible - If S_gen ≤ 0, the transformation is not possible

Step-by-step solution

Find the entropy change of the system during the transformation

To find the entropy change of the system, we need to consider the initial and final states of the water. The initial state is a saturated vapor at \(200 \mathrm{kPa}\), and the final state is a saturated liquid at the same pressure. Refer to the steam tables to find the values of the entropy \(s\) for both the initial and final state: Initial entropy (\(s_1\)): Saturated vapor at \(200 \mathrm{kPa}\) Final entropy (\(s_2\)): Saturated liquid at \(200 \mathrm{kPa}\) The entropy change of the system is given by: \(\Delta s_{sys} = s_2 - s_1\)

Find the entropy change of the reservoir during the transformation

The heat transfer between the closed system and the isothermal energy reservoir occurs at \(90^{\circ} \mathrm{C}\) or \(363.15 \mathrm{K}\). To find the entropy change of the reservoir, we need to find the heat transfer (\(Q\)) between the system and the reservoir. First, we must find the specific enthalpies \(h\) for the initial and final states of the water as follows: Initial enthalpy (\(h_1\)): Saturated vapor at \(200 \mathrm{kPa}\) Final enthalpy (\(h_2\)): Saturated liquid at \(200 \mathrm{kPa}\) Since the transformation is isobaric, the heat transfer between the system and the reservoir is given by: \(Q = m(h_2 - h_1)\) where \(m\) is the mass of the fluid in the closed system. Entropy change of the reservoir is given by: \(\Delta s_{res} = -\frac{Q}{T_{res}}\)

Calculate the entropy generation during the transformation

The entropy generation during the transformation is the sum of the entropy changes of the system and the reservoir: \(S_{gen} = \Delta s_{sys} + \Delta s_{res}\)

Determine if the transformation is possible

If \(S_{gen} > 0\), the transformation is irreversible and possible. If \(S_{gen} \le 0\), the transformation is not possible. Compare the value of \(S_{gen}\) calculated in Step 3 to determine if the saturated water vapor at \(200 \mathrm{kPa}\) can be condensed to a saturated liquid while exchanging heat with an isothermal energy reservoir at \(90^{\circ} \mathrm{C}\).

Key concepts

Isobaric Process

An isobaric process is a thermodynamic process in which the pressure remains constant throughout. The term 'isobaric' is derived from Greek, where 'iso' means equal and 'baric' refers to pressure. In such processes, any change in the volume of the system will be accompanied by a change in temperature while maintaining constant pressure.

In the context of the exercise, condensing water vapor to liquid in a closed system at a constant pressure of 200 kPa is an example of an isobaric process. This condition simplifies the analysis of the thermodynamic processes since it implies that no work is performed on or by the gas during expansion or compression, which could otherwise alter the internal energy of the system apart from the heat transfer effects.

When dealing with isobaric processes, the heat transferred to or from the system is directly related to the enthalpy change, expressed as the difference between the final enthalpy and initial enthalpy of the system. In an open system, this would be synonymous with the heat transfer required to perform or absorb during the phase change of a substance at a constant pressure.

Steam Tables

Steam tables provide essential thermodynamic data regarding the properties of water and steam. They are a crucial tool for engineers and students to perform calculations related to various water and steam processes in thermodynamics. The tables typically include values such as temperature, pressure, specific volume, specific enthalpy, and specific entropy for saturated water (liquid), saturated steam (vapor), and superheated steam.

The step-by-step solution refers to steam tables for determining the entropies (\(s\)) and specific enthalpies (\(h\)) of saturated vapor and saturated liquid at the given pressure of 200 kPa. These values are fundamental for calculating the entropy change of the system and the heat transfer involved during the isobaric condensation process. Using steam tables ensures precise measurements for these quantities rather than relying on approximations or nonspecific equations.

To effectively use the steam tables, one needs to understand the phase of water or steam in question and look up the property values under the appropriate headings, which are typically provided for different pressures and temperatures. For instance, in this exercise, the data is sought for the same pressure but different phases—saturated vapor and saturated liquid.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics which deals with the movement of thermal energy from one place to another as a result of temperature differences. There are three main modes of heat transfer: conduction, convection, and radiation. In the context of this exercise involving entropy generation and an isobaric process, heat transfer refers to the energy exchange due to the temperature difference between the saturated steam within the system and the isothermal energy reservoir.

The temperature difference between the reservoir (\(90^{\textdegree} \text{C} \text{ or } 363.15 \text{ K}\) and the system undergoing the phase change initiates the heat transfer process required for condensation. The amount of heat transferred (\(Q\)) is vital for calculating the entropy change of the reservoir (\(\text{\textDelta} s_{\text{res}}\)) as well as assessing the overall entropy generation in the system.

By calculating the heat transfer, students can understand how the principle of conservation of energy applies to thermodynamic processes. In this scenario, the amount of heat released by the system (saturated water vapor) must be absorbed by the reservoir to ensure energy balance, facilitating the assessment of whether the isobaric condensation is thermodynamically feasible. Understanding these principles enables clearer insights into the practical implications of heat transfer in real-world engineering applications.