Let's break the problem down step by step: ## **Given Data** - **Initial State (State 1):** - Pressure (\(P_1\)) = **300 kPa** - Temperature (\(T_1\)) = **400°C** - **Process:** 1. **Cool at constant pressure (300 kPa)** until water is a saturated vapor (State 2). 2. **Cool at constant volume** (since piston is on stops) until pressure drops to **100 kPa** (State 3). - **Final State (State 3):** - Pressure (\(P_3\)) = **100 kPa** --- ### **a) T-v Diagram and Process Sketch** #### **Step 1: Find Properties at Each State** #### **State 1 (Initial, Superheated Vapor)** - \(P_1 = 300~\text{kPa}\), \(T_1 = 400^\circ\text{C}\) - From the steam tables: | Property | Value | |----------|-------| | \(v_1\) (specific volume) | ≈ 0.25795 m³/kg | | \(u_1\) (internal energy) | ≈ 2890.2 kJ/kg | #### **State 2 (Saturated Vapor at 300 kPa)** - At \(P_2 = 300~\text{kPa}\), saturated vapor (\(x=1\)): - From steam tables: - \(T_{sat,300}\) ≈ **133.6°C** - \(v_2 = v_g@300kPa\) ≈ **0.60606 m³/kg** - \(u_2 = u_g@300kPa\) ≈ **2546.5 kJ/kg** #### **State 3 (Constant Volume Cooling to 100 kPa)** - At \(v_3 = v_2 = 0.60606~\text{m}^3/\text{kg}\), \(P_3 = 100~\text{kPa}\) - At \(P_3 = 100~\text{kPa}\): \[ v_{f,100} = 0.001043~\text{m}^3/\text{kg}, \quad v_{g,100} = 1.694~\text{m}^3/\text{kg} \] Since \(v_3\) is between \(v_f\) and \(v_g\), water at state 3 is a **mixture** (quality \(x_3\)). \[ x_3 = \frac{v_3 - v_{f,100}}{v_{g,100} - v_{f,100}} = \frac{0.60606 - 0.001043}{1.694 - 0.001043} \approx 0.358 \] Internal energy at state 3: \[ u_3 = u_f + x_3 (u_g - u_f) \] From steam tables at 100 kPa: - \(u_{f,100} = 419.04~\text{kJ/kg}\) - \(u_{g,100} = 2506.5~\text{kJ/kg}\) \[ u_3 = 419.04 + 0.358 \times (2506.5 - 419.04) \approx 419.04 + 0.358 \times 2087.46 \approx 419.04 + 747.4 = 1166.4~\text{kJ/kg} \] --- ### **T-v Diagram Sketch** Here's what the process looks like on a T-v diagram: ``` T │ │ (1) Superheated vapor │ * │ / │ / │ / │ / │ * (2) Saturated vapor at P = 300kPa │ | │ | (constant v, v2 = v3) │ | │ * │ (3) Mixture at 100kPa, x ≈ 0.36 │ └────────────────────────────── v (Saturated liquid) (Saturated vapor) Left dome line Right dome line ``` - **State 1:** Superheated region (above the dome) at \(P = 300~\text{kPa}, T = 400^\circ\text{C}\), \(v_1 = 0.25795~\text{m}^3/\text{kg}\) - **Process 1→2:** Cool at constant pressure (horizontal leftward movement at \(P = 300~\text{kPa}\)) to saturated vapor line (\(T_{sat,300} = 133.6^\circ\text{C}\)), \(v_2 = 0.60606~\text{m}^3/\text{kg}\) - **Process 2→3:** Cool at constant volume (\(v_3 = v_2 = 0.60606~\text{m}^3/\text{kg}\)) to \(P = 100~\text{kPa}\), mixture region. --- ### **b) Change in Internal Energy** \[ \Delta u = u_3 - u_1 = 1166.4~\text{kJ/kg} - 2890.2~\text{kJ/kg} = -1723.8~\text{kJ/kg} \] **The internal energy of the water decreases by 1723.8 kJ/kg.** --- ## **Summary Table** | State | P (kPa) | T (°C) | v (m³/kg) | u (kJ/kg) | x (quality) | |-------|---------|--------|-----------|-----------|-------------| | 1 | 300 | 400 | 0.25795 | 2890.2 | - | | 2 | 300 | 133.6 | 0.60606 | 2546.5 | 1 | | 3 | 100 | ? | 0.60606 | 1166.4 | 0.36 | --- ### **To Draw on Paper:** 1. Draw the saturation dome on a T-v diagram. 2. Mark State 1 above the right of the dome (superheated, 300 kPa, 400°C). 3. Draw a horizontal line (constant pressure) left to the saturated vapor line (State 2). 4. From State 2, draw a vertical line down (constant volume) into the mixture region at 100 kPa (State 3). 5. Label each state with T, P, v. --- **If you want a digital T-v diagram, let me know!**