Let's solve the problem step by step as requested: --- ### **Step 1: Identify Initial and Final States** **Given Data:** - Initial state: \( p_1 = 1 \) lbf/in² (psia), saturated vapor (so quality x = 1) - Final state: \( p_2 = 20 \) lbf/in², \( v_2 = 16.5 \) ft³/lb **Explanation:** - For the initial state, since it's saturated vapor at 1 psia, we can look up the specific entropy \( s_{g,1} \) and specific volume \( v_{g,1} \) in steam tables. - For the final state, we know pressure and specific volume, so we'll need to use these to determine the phase and entropy. --- ### **Step 2: Obtain Properties from Steam Tables** #### **Initial State (1 psia, x=1):** - From steam tables: - \( s_{g,1} \) (specific entropy of vapor at 1 psia) - \( v_{g,1} \) (specific volume of vapor at 1 psia) Let's use typical values (rounded for illustration; use your steam tables for precise values): At **1 psia** (approx. 0.069 bar): - \( s_{g,1} \approx 1.885 \) Btu/lb·R - \( v_{g,1} \approx 379.5 \) ft³/lb #### **Final State (20 psia, \( v_2 = 16.5 \) ft³/lb):** At **20 psia**: - \( v_{g,2} \approx 20.07 \) ft³/lb (specific volume of vapor) - \( v_{f,2} \) is much smaller (liquid) Since \( v_2 = 16.5 \) ft³/lb < \( v_{g,2} \), but much greater than \( v_{f,2} \), the state is a **mixture** of liquid and vapor. To find quality (\( x_2 \)): \[ x_2 = \frac{v_2 - v_{f,2}}{v_{g,2} - v_{f,2}} \] But since \( v_{f,2} \) is very small, for rough calculation: \[ x_2 \approx \frac{16.5}{20.07} \approx 0.822 \] Now, entropy at state 2: - \( s_2 = s_{f,2} + x_2 \cdot (s_{g,2} - s_{f,2}) \) From steam tables at 20 psia: - \( s_{f,2} \approx 0.357 \) Btu/lb·R - \( s_{g,2} \approx 1.696 \) Btu/lb·R So, \[ s_2 = 0.357 + 0.822 \times (1.696 - 0.357) \] \[ s_2 = 0.357 + 0.822 \times 1.339 \approx 0.357 + 1.100 = 1.457 \ \text{Btu/lb·R} \] --- ### **Step 3: Calculate Change in Entropy and Convert Units** \[ \Delta s = s_2 - s_1 = 1.457 - 1.885 = -0.428 \ \text{Btu/lb·R} \] Convert to kJ/kg·K: - \( 1 \) Btu/lb·R = \( 4.1868 \) kJ/kg·K \[ \Delta s = -0.428 \times 4.1868 = -1.792 \ \text{kJ/kg·K} \] --- ## **Final Answer** \[ \boxed{\Delta s = -1.79 \ \text{kJ/kg·K}} \] --- **Summary Explanation:** 1. **Identified states:** Used steam tables to get entropy at initial (saturated vapor) and final state (mixture at given pressure and specific volume). 2. **Found properties:** Calculated quality at final state, then got entropy values for both states. 3. **Calculated change:** Subtracted and converted units to get the final answer. The negative sign shows entropy decreased.