Let's break down the solution into two parts as per your questions. --- ## **1. Filling the Table** We'll use thermodynamic property tables (Steam Tables, Refrigerant Tables, etc.) to find the specific volume at the given conditions (for saturated cases at x=0.5, which means a mixture with quality = 0.5). #### **a) Water (saturated), T = 200°C** - At T = 200°C, the saturation pressure (\(P_{sat}\)) is approximately **1554 kPa**. - Specific volume at x = 0.5: - \( v = v_f + x(v_g - v_f) \) - From steam tables: \( v_f \approx 0.001157 \) m³/kg, \( v_g \approx 0.12721 \) m³/kg \( v = 0.001157 + 0.5 \times (0.12721 - 0.001157) = 0.001157 + 0.5 \times 0.126053 = 0.001157 + 0.0630265 = 0.06418 \) m³/kg - Density (\( \rho \)) = \( \frac{1}{v} \) = \( \frac{1}{0.06418} \approx 15.59 \) kg/m³ - **Reference:** Steam Tables --- #### **b) R-134A (saturated), P = 400 kPa** - At P = 400 kPa, T ≈ 18.87°C - \( v_f \approx 0.000887 \) m³/kg, \( v_g \approx 0.05426 \) m³/kg \( v = 0.000887 + 0.5 \times (0.05426 - 0.000887) = 0.000887 + 0.026687 = 0.027574 \) m³/kg - Density = \( \frac{1}{0.027574} \approx 36.25 \) kg/m³ - **Reference:** R-134a Property Tables --- #### **c) R-22 (saturated), T = 40°C** - At T = 40°C, P ≈ 1550 kPa - \( v_f \approx 0.000911 \) m³/kg, \( v_g \approx 0.01743 \) m³/kg \( v = 0.000911 + 0.5 \times (0.01743 - 0.000911) = 0.000911 + 0.0082595 = 0.009170 \) m³/kg - Density = \( \frac{1}{0.009170} \approx 109.0 \) kg/m³ - **Reference:** R-22 Property Tables --- #### **d) NH3, T = -10°C, P = 1000 kPa** - At T = -10°C, P ≈ 328 kPa (saturation), but given P = 1000 kPa (so it is subcooled liquid). - At P = 1000 kPa, T_sat ≈ 29.75°C. At T = -10°C, use subcooled liquid properties (same as saturated liquid at -10°C). - \( v_f \approx 0.001638 \) m³/kg - Density = \( \frac{1}{0.001638} \approx 610 \) kg/m³ - **Reference:** NH3 Property Tables --- #### **e) CO2, T = -20°C, P = 1000 kPa** - At T = -20°C, P_sat ≈ 1740 kPa. Given P = 1000 kPa (so it is a mixture). - At -20°C, \( v_f \approx 0.000977 \) m³/kg, \( v_g \approx 0.1941 \) m³/kg \( v = 0.000977 + 0.5 \times (0.1941 - 0.000977) = 0.000977 + 0.0965615 = 0.09754 \) m³/kg - Density = \( \frac{1}{0.09754} \approx 10.25 \) kg/m³ - **Reference:** CO2 Property Tables --- ### **Filled Table** | Substance | T (°C) | P (kPa) | Specific Vol (m³/kg) | Density (kg/m³) | Reference | |----------------|--------|---------|----------------------|-----------------|-------------------| | Water (sat.) | 200 | 1554 | 0.06418 | 15.59 | Steam Tables | | R-134A (sat.) | 18.87 | 400 | 0.02757 | 36.25 | R-134a Tables | | R-22 (sat.) | 40 | 1550 | 0.00917 | 109.00 | R-22 Tables | | NH3 | -10 | 1000 | 0.001638 | 610 | NH3 Tables | | CO2 | -20 | 1000 | 0.09754 | 10.25 | CO2 Tables | --- ## **2. Thermodynamics Problem Solution** **Given:** - Initial state: \( P_1 = 200 \) kPa, \( T_1 = 300^\circ \)C - Cool at constant pressure until it is a saturated vapor (\( x = 1 \)), then cools further at constant volume (piston on stops) until \( P_3 = 50 \) kPa. ### **Step 1: Initial State (1)** - \( P_1 = 200 \) kPa, \( T_1 = 300^\circ \)C - From steam tables, at P = 200 kPa: - \( T_{sat} = 120.23^\circ \)C (lower than 300°C, so it is superheated vapor) - Find specific volume \( v_1 \) and internal energy \( u_1 \): From superheated steam tables: - At 200 kPa, 300°C: \( v_1 = 1.5490 \) m³/kg, \( u_1 = 2725.5 \) kJ/kg --- ### **Step 2: Intermediate State (2) (Saturated vapor at 200 kPa, x=1)** - \( P_2 = 200 \) kPa, \( x_2 = 1 \) (saturated vapor) - From steam tables at 200 kPa: - \( T_2 = T_{sat} = 120.23^\circ \)C - \( v_g = 0.88578 \) m³/kg, \( u_g = 2506.5 \) kJ/kg --- ### **Step 3: Final State (3) (P = 50 kPa, v = v_2)** - The piston rests on stops, so **volume is constant**: \( v_3 = v_2 = 0.88578 \) m³/kg - At \( P_3 = 50 \) kPa, find the phase: - At 50 kPa: \( v_f = 0.001012 \) m³/kg, \( v_g = 12.674 \) m³/kg - \( v_3 = 0.88578 \) m³/kg is between \( v_f \) and \( v_g \), so it is a mixture. - Quality \( x_3 = \frac{v_3 - v_f}{v_g - v_f} = \frac{0.88578 - 0.001012}{12.674 - 0.001012} \approx \frac{0.88477}{12.673} \approx 0.0698 \) - Internal energy at 50 kPa: - \( u_f = 209.33 \) kJ/kg, \( u_g = 2584.7 \) kJ/kg - \( u_3 = u_f + x_3(u_g - u_f) = 209.33 + 0.0698 \times (2584.7 - 209.33) \) - \( = 209.33 + 0.0698 \times 2375.37 = 209.33 + 165.82 = 375.15 \) kJ/kg --- ### **Step 4: Find the changes** #### **a) Change in Internal Energy (\( \Delta u \))** \( \Delta u = u_3 - u_1 = 375.15 - 2725.5 = -2350.4 \) kJ/kg #### **b) Work Done** - Work done during the constant pressure process (1→2): \( W_{1-2} = P (v_2 - v_1) = 200 \times 10^3 (0.88578 - 1.5490) = 200,000 \times (-0.66322) = -132,644 \) J/kg \( = -132.64 \) kJ/kg - During constant volume process (2→3), \( W_{2-3} = 0 \) - **Total Work**: \( W = -132.64 \) kJ/kg --- ### **Step 5: T-v Diagram Sketch** - Mark points: - 1: (v1 = 1.5490, T1 = 300°C, P1 = 200 kPa) - 2: (v2 = 0.88578, T2 = 120.23°C, P2 = 200 kPa, x2 = 1) - 3: (v3 = 0.88578, T3 ≈ 40.4°C [from interpolation], P3 = 50 kPa, x3 ≈ 0.0698) --- ## **Summary Table for Process:** | State | P (kPa) | T (°C) | v (m³/kg) | x | u (kJ/kg) | |-------|---------|--------|-----------|-------|-----------| | 1 | 200 | 300 | 1.5490 | -- | 2725.5 | | 2 | 200 | 120.23 | 0.88578 | 1 | 2506.5 | | 3 | 50 | ~40.4 | 0.88578 | 0.0698| 375.15 | --- ### **Final Answers** - **Change in Internal Energy (\( \Delta u \)):** **-2350.4 kJ/kg** - **Work Done:** **-132.64 kJ/kg** - **T-v Diagram:** (Sketch with points 1, 2, 3 as calculated above) --- **References:** - Steam Tables (Water and Steam Properties) - Refrigerant Tables (R-134a, R-22, NH3, CO2) If you need the T-v diagram sketch or detailed steps for any other part, let me know!