# Step-by-Step Solution: Water Vapor Expansion at 250°C Given: - Initial state: Saturated vapor water at \(T = 250^\circ\text{C}\) - Process: **Isothermal** (constant temperature) expansion to a lower pressure - Find the changes in **internal energy (\(u\))** and **entropy (\(s\))** for each final pressure: - (a) \(P = 100~\text{kPa}\) - (b) \(P = 50~\text{kPa}\) - (c) \(P = 10~\text{kPa}\) We'll use **superheated steam tables** for water at \(250^\circ\text{C}\). --- ## 1. Identify Initial State (Saturated Vapor at \(250^\circ\text{C}\)) At \(T = 250^\circ\text{C}\): - The **saturation pressure** \(P_\text{sat}\) is (from steam tables): \(P_\text{sat} \approx 3975~\text{kPa}\) - As *saturated vapor*, the initial state is at: - \(T_1 = 250^\circ\text{C}\) - \(P_1 = 3975~\text{kPa}\) - **Initial specific internal energy** (\(u_1\)) = \(u_g\) at \(250^\circ\text{C}\) - **Initial specific entropy** (\(s_1\)) = \(s_g\) at \(250^\circ\text{C}\) From steam tables: - \(u_1 = u_g(250^\circ\text{C}) \approx 2594.2~\text{kJ/kg}\) - \(s_1 = s_g(250^\circ\text{C}) \approx 6.361~\text{kJ/(kg·K)}\) --- ## 2. Isothermal Expansion: Final States When expanding **isothermally** to a lower pressure, the vapor becomes **superheated**. For each final pressure, find the properties at \(T = 250^\circ\text{C}\) and the specified \(P\): ### (a) \(P_2 = 100~\text{kPa}\) From superheated steam tables at \(T = 250^\circ\text{C}, P = 100~\text{kPa}\): - \(u_2 \approx 2642.5~\text{kJ/kg}\) - \(s_2 \approx 8.117~\text{kJ/(kg·K)}\) **Changes:** - \(\Delta u = u_2 - u_1 = 2642.5 - 2594.2 = \boxed{48.3~\text{kJ/kg}}\) - \(\Delta s = s_2 - s_1 = 8.117 - 6.361 = \boxed{1.756~\text{kJ/(kg·K)}}\) --- ### (b) \(P_2 = 50~\text{kPa}\) From superheated steam tables at \(T = 250^\circ\text{C}, P = 50~\text{kPa}\): - \(u_2 \approx 2656.9~\text{kJ/kg}\) - \(s_2 \approx 8.626~\text{kJ/(kg·K)}\) **Changes:** - \(\Delta u = 2656.9 - 2594.2 = \boxed{62.7~\text{kJ/kg}}\) - \(\Delta s = 8.626 - 6.361 = \boxed{2.265~\text{kJ/(kg·K)}}\) --- ### (c) \(P_2 = 10~\text{kPa}\) From superheated steam tables at \(T = 250^\circ\text{C}, P = 10~\text{kPa}\): - \(u_2 \approx 2673.7~\text{kJ/kg}\) - \(s_2 \approx 9.398~\text{kJ/(kg·K)}\) **Changes:** - \(\Delta u = 2673.7 - 2594.2 = \boxed{79.5~\text{kJ/kg}}\) - \(\Delta s = 9.398 - 6.361 = \boxed{3.037~\text{kJ/(kg·K)}}\) --- ## 3. **Summary Table** | Final \(P\) (kPa) | \(u_2\) (kJ/kg) | \(s_2\) (kJ/kg·K) | \(\Delta u\) (kJ/kg) | \(\Delta s\) (kJ/kg·K) | |:-----------------:|:---------------:|:-----------------:|:--------------------:|:----------------------:| | 100 | 2642.5 | 8.117 | 48.3 | 1.756 | | 50 | 2656.9 | 8.626 | 62.7 | 2.265 | | 10 | 2673.7 | 9.398 | 79.5 | 3.037 | --- ## 4. **Explanation** - As the vapor expands **isothermally** (at constant \(T\)), its pressure decreases and it becomes more superheated. - Both **internal energy (\(u\))** and **entropy (\(s\))** **increase** because the system absorbs heat to keep the temperature constant while the volume increases. - The lower the final pressure, the greater the increase in both \(u\) and \(s\). --- ## 5. **Final Answers** ### (a) For \(P = 100~\text{kPa}\): - \(\Delta u = \boxed{48.3~\text{kJ/kg}}\) - \(\Delta s = \boxed{1.756~\text{kJ/(kg·K)}}\) ### (b) For \(P = 50~\text{kPa}\): - \(\Delta u = \boxed{62.7~\text{kJ/kg}}\) - \(\Delta s = \boxed{2.265~\text{kJ/(kg·K)}}\) ### (c) For \(P = 10~\text{kPa}\): - \(\Delta u = \boxed{79.5~\text{kJ/kg}}\) - \(\Delta s = \boxed{3.037~\text{kJ/(kg·K)}}\) --- **Note:** Values are taken from typical steam tables. If your tables have slightly different values, use those for precise answers.