Sure! Let's solve the problem step by step. **Given Data:** - Refrigerant: R-134a - Cooling capacity (\(\dot{Q}_{in}\)): 2778 kW - Evaporation temppressure: \(-35^\circ C\), 174.5 kPa - Condensation temp/pressure: \(35^\circ C\), 1473.35 kPa - Compressor suction pressure: 132.4 kPa - Compressor discharge pressure: 127 kPa - Compressor discharge temp: \(66^\circ C\) - Condenser duty: 4434 kW Let's go step by step. --- ## a. **Enthalpies at All Key States** Let's define the states: - **State 1:** Compressor inlet (after evaporator, before compressor) - **State 2:** Compressor exit (after compressor, before condenser) - **State 3:** Exit condenser (before expansion device) - **State 4:** Exit expansion valve (before evaporator) We'll use the **R-134a P-h diagram** (you'd usually extract values from tables/diagrams, but I'll explain how): --- ### **State 1: Compressor Inlet (after evaporator, before compressor)** - **P1 = 132.4 kPa** - **T1 ≈ -35°C** (but at lower P, may be lower T) Assume **saturated vapor** at suction (common assumption): From R-134a tables: - At **P = 132.4 kPa**, \(h_1 \approx 252.7\) kJ/kg (saturated vapor) --- ### **State 2: Compressor Exit** - **P2 = 127 kPa** - **T2 = 66°C** From superheated tables: - At **P = 127 kPa** and **T = 66°C**, \(h_2 \approx 294.2\) kJ/kg --- ### **State 3: Exit of Condenser** - **P3 = 127 kPa** - **Subcooled liquid (after condenser, before expansion)** At **P = 127 kPa** and assuming saturated liquid: - \(h_3 \approx 95.5\) kJ/kg --- ### **State 4: After Expansion Valve** - **P4 = 132.4 kPa** - **Throttling process: \(h_4 = h_3\)** So, - \(h_4 = 95.5\) kJ/kg --- **Summary Table** | State | Pressure (kPa) | Temp (°C) | Enthalpy (kJ/kg) | |-------|----------------|-----------|------------------| | 1 | 132.4 | ≈-35 | 252.7 | | 2 | 127 | 66 | 294.2 | | 3 | 127 | ≈35 | 95.5 | | 4 | 132.4 | < | 95.5 | --- ## b. **Refrigeration Capacity** \[ \dot{Q}_{in} = \dot{m} \cdot (h_1 - h_4) \] Given \(\dot{Q}_{in} = 2778\) kW, \[ h_1 - h_4 = 252.7 - 95.5 = 157.2\ \text{kJ/kg} \] \[ \dot{m} = \frac{\dot{Q}_{in}}{h_1 - h_4} = \frac{2778}{157.2} = 17.68\ \text{kg/s} \] --- ## c. **Compressor Work Input** \[ \dot{W}_{comp} = \dot{m} \cdot (h_2 - h_1) \] \[ h_2 - h_1 = 294.2 - 252.7 = 41.5\ \text{kJ/kg} \] \[ \dot{W}_{comp} = 17.68 \times 41.5 = 733.7\ \text{kW} \] --- ## d. **Coefficient of Performance (COP)** \[ COP = \frac{\text{Refrigeration Effect}}{\text{Compressor Work}} \] \[ COP = \frac{\dot{Q}_{in}}{\dot{W}_{comp}} = \frac{2778}{733.7} = 3.79 \] --- # **Final Answers (Summary Table)** | Quantity | Value | |-----------------------------|-----------------| | **h1 (kJ/kg)** | 252.7 | | **h2 (kJ/kg)** | 294.2 | | **h3 (kJ/kg)** | 95.5 | | **h4 (kJ/kg)** | 95.5 | | **Mass flow rate (kg/s)** | 17.68 | **Compressor work (kW)** | 733.7 | | **COP** | .79 | --- ## **Tips** - All enthalpy values are approximations from the R-134a tables. For homework/exams, always use precise values from your actual property tables or P-h diagram. - The process is the same for most vapor-compression cycles: **find h at each state, use energy balances, then solve for required quantities**. If you want help with the P-h diagram or more detail, let me know!