Problem 7 - A Modern Course in Statistical Physics

Question

One kilogram of superheated steam, at temperature \(t=350^{\circ} \mathrm{C}\), pressure \(P=100\) bar, and specific entropy \(s=5949 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\), is expanded reversibly and adiabatically to form wet steam at \(t=\) \(200^{\circ} \mathrm{C}\) and pressure \(P=15.55\) bar. The specific entropy of water vapor and liquid water on the coexistence curve at \(t=200{ }^{\circ} \mathrm{C}\) are \(\mathrm{s}_{\mathrm{g}}=6.428 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\) and \(s_{1}=2.331 \mathrm{~kJ} /(\mathrm{kg} \mathrm{K})\), respectively. The specific enthalpy of water vapor (gas) and liquid water on the coexistence curve at \(t=200^{\circ} \mathrm{C}\) are \(h_{\mathrm{g}}=\) \(2791 \mathrm{~kJ} / \mathrm{kg}\) and \(h_{1}=852.4 \mathrm{~kJ} / \mathrm{kg}\). (a) What is the specific enthalpy of the wet steam at \(t=200^{\circ} \mathrm{C}\) ? (b) What fraction of the wet steam is liquid water?

Step-by-Step Solution

Answer

The specific enthalpy of the wet steam is 2525.83 kJ/kg. The quality of the steam is approximately 0.883, meaning 88.3% is vapor and 11.7% is liquid.

Step 1: Determine the Entropy of the Wet Steam

Since the process is adiabatic and reversible, the entropy remains constant. The specific entropy given for the initial state is 5.949 kJ/(kg∙K).

Step 2: Use the Entropy Values for Wet Steam

The entropy of wet steam can be represented as a combination of the entropies of the liquid and vapor phases: \[ s_{\text{wet}} = s_f + x (s_g - s_f) \] where: \( s_f = 2.331 \text{ kJ/(kg∙K)} \) \( s_g = 6.428 \text{ kJ/(kg∙K)} \) and \( x \) is the quality of the steam.

Step 3: Solve for Steam Quality (x)

Use the specific entropy to calculate the quality (x): \[ 5.949 = 2.331 + x (6.428 - 2.331) \] Rearranging and solving for \( x \): \[ x = \frac{5.949 - 2.331}{6.428 - 2.331} = \frac{3.618}{4.097} \approx 0.883 \]

Step 4: Calculate the Specific Enthalpy of Wet Steam

The enthalpy of wet steam is given by: \[ h_{\text{wet}} = h_f + x (h_g - h_f) \] where: \( h_f = 852.4 \text{ kJ/kg} \) \( h_g = 2791 \text{ kJ/kg} \) and \( x = 0.883 \). Substituting the values: \[ h_{\text{wet}} = 852.4 + 0.883 (2791 - 852.4) \] Simplifying: \[ h_{\text{wet}} = 852.4 + 0.883 \times 1938.6 = 2525.83 \text{ kJ/kg} \]

Key Concepts

Adiabatic Processes

Adiabatic processes are crucial in thermodynamics because they occur without heat transfer. This means there is no addition or subtraction of heat energy from the system. In simple terms, the system is perfectly insulated from its surroundings during the process. The key consequence of an adiabatic process is that any work done by or on the system changes the system's internal energy. When dealing with steam, an adiabatic expansion, like in our problem, means that the entropy remains constant throughout the process. This lack of heat exchange simplifies our calculations since we only have to account for the work done and internal energy changes to find other properties of the steam.

Entropy

Entropy is a measure of disorder in a system and plays a pivotal role in understanding thermodynamic processes. It helps us understand how energy is distributed within the system. In our exercise, the specific entropy values at initial and final states are given, and because the process is adiabatic and reversible, the entropy stays constant. This principle allows us to calculate the quality of wet steam, knowing that this consistency of entropy gives us a direct link between different states of the steam. Essentially, maintaining entropy helps trace the energy transformation pathway accurately.

Enthalpy

Enthalpy represents the total energy of a thermodynamic system, encompassing internal energy and the energy needed to displace its environment. This is crucial for calculating heat and work in processes at constant pressure. In our specific problem, enthalpy changes as steam expands. Calculating the wet steam's specific enthalpy involves combining the specific enthalpies of liquid water and water vapor, weighted by the quality factor. This calculation reveals how energy is stored in the different phases of water, reflecting the system's energy state post-expansion.

Superheated Steam

Superheated steam refers to steam at a temperature higher than its boiling point for a given pressure. In this state, the steam contains significant energy because it has been heated beyond the point at which it condenses. This energy is stored in its high temperature and needs careful management in thermodynamic cycles. In the given problem, we start with superheated steam because it is at 350°C, well above boiling point for a pressure of 100 bar. Superheated steam expands and does useful work, but its management is critical for efficiency.

Wet Steam

Wet steam is a mixture of liquid water and steam. Each component's amount is indicated by the steam's quality. The quality, denoted 'x,' is the proportion of the steam that is vapor, with the rest being liquid water. For example, in our problem, when steam expands and cools down to form wet steam, we calculate that it has a quality of approximately 0.883. This means 88.3% of the steam is vapor, and 11.7% is liquid water. Wet steam is often less desirable in many engines and turbines because the liquid water can cause damage or reduce efficiency.