Let's break down the problem **step-by-step**. --- ## **Problem Recap** - **Refrigerant 134a enters** a compressor at: - \( P_1 = 4\, \text{bar} \) - \( T_1 = 20^\circ \text{C} \) - **Leaves at:** \( P_2 = 12\, \text{bar}, \, T_2 = 80^\circ \text{C} \) - **Volumetric flow rate at inlet:** \( \dot{V}_1 = 8\, \text{m}^3/\text{min} \) - **Work input:** \( w = 120\, \text{kJ/kg} \) - **Neglect kinetic & potential energies** - **Find:** Magnitude of **heat transfer rate** from the compressor (\( |\dot{Q}_{cv}| \)), in kW. --- # **Step 1: Convert Flow Rate to SI Units** \[ \dot{V}_1 = 8\, \text{m}^3/\text{min} = \frac{8}{60} = .1333\, \text{m}^3/\text{s} \] --- # **Step 2: Find Inlet Specific Volume (\( v_1 \))** We need the **specific volume** of R134a at \( 4\, \text{bar},\, 20^\circ \text{C} \). Look this up in R134a tables: - **At 4 bar, 20°C:** - \( v_1 \approx .05692\, \text{m}^3/\text{kg} \) --- # **Step 3: Calculate Mass Flow Rate (\( \dot{m} \))** \[ \dot{m} = \frac{\dot{V}_1}{v_1} = \frac{.1333}{.05692} \approx 2.34\, \text{kg/s} \] --- # **Step 4: First Law for Control Volume (Steady State)** \[ \dot{Q}_{cv} - \dot{W}_{cv} = \dot{m}(h_2 - h_1) \] Where: - \( \dot{Q}_{cv} \) = heat transfer rate (what we want) - \( \dot{W}_{cv} \) = work input rate = \( \dot{m} \cdot w \) - \( h_2, h_1 \) = enthalpies at outlet and inlet --- # **Step 5: Find Enthalpies \( h_1 \) & \( h_2 \) from Tables** - **At \( 4\, \text{bar}, 20^\circ \text{C} \):** - \( h_1 \approx 267.36\, \text{kJ/kg} \) - **At \( 12\, \text{bar}, 80^\circ \text{C} \):** - \( h_2 \approx 303.16\, \text{kJ/kg} \) --- # **Step 6: Calculate Work Input Rate** \[ \dot{W}_{cv} = \dot{m} \cdot w = 2.34\, \text{kg/s} \times 120\, \text{kJ/kg} = 280.8\, \text{kW} \] --- # **Step 7: Plug Values Into Energy Balance** \[ \dot{Q}_{cv} = \dot{m}(h_2 - h_1) + \dot{W}_{cv} \] \[ = 2.34 (303.16 - 267.36) + 280.8 \] \[ = 2.34 \times 35.8 + 280.8 \] \[ = 83.8 + 280.8 = 364.6\, \text{kW} \] --- # **Step 8: Direction of Heat Transfer** Since this is a compressor, usually heat is **lost** from the compressor (\( \dot{Q}_{cv} < \)). But the question asks for **magnitude**. --- # **# Final Answer** \[ \boxed{|\dot{Q}_{cv}| = 364.6\, \text{kW}} \] --- ## **Summary Table** | Quantity | Value | |---|---| | \( \dot{V}_1 \) | .1333 m³/s | | \( v_1 \) | .05692 m³/kg | | \( \dot{m} \) | 2.34 kg/s | | \( h_1 \) | 267.36 kJ/kg | | \( h_2 \) | 303.16 kJ/kg | | Work per kg (\( w \)) | 120 kJ/kg | | \( \dot{W}_{cv} \) | 280.8 kW | | \( |\dot{Q}_{cv}| \) | 364.6 kW | --- If you need help with how to find values from the refrigerant tables, let me know!