Let's solve the problem step by step: --- ### **Given Data** - **Refrigeration capacity (\(Q_L\))** = 30 TR \(1~\text{TR} = 3.517~\text{kW}\) So, \(Q_L = 30 \times 3.517 = 105.51~\text{kW}\) - **Evaporator temperature (\(T_{evap}\))** = \(-16^\circ C\) - **Condenser temperature (\(T_{cond}\))** = \(50^\circ C\) - **Specific heat of ammonia vapor (\(c_p\))** = 2.3 kJ/kg·K - **Refrigerant: Ammonia (\(NH_3\))** - **Ammonia enters the compressor as dry saturated vapor and leaves condenser as saturated liquid.** --- ### **Step 1: Sketch the Cycle** #### **T-s and P-h diagrams** - On the **T-s** diagram: - 1 → 2: Isentropic compression (vertical line up) - 2 → 3: Heat rejection at constant pressure (horizontal line to the left) - 3 → 4: Throttling/expansion (vertical line down) - 4 → 1: Heat absorption at constant pressure (horizontal line to the right) - On the **P-h** diagram: - 1 → 2: Vertical up (compression) - 2 → 3: Horizontal left (condensation) - 3 → 4: Vertical down (expansion) - 4 → 1: Horizontal right (evaporation) --- ### **Step 2: Find Properties from Ammonia Tables** Let’s use typical properties (approximate) for ammonia: | State | Description | T (°C) | h (kJ/kg) | s (kJ/kg·K) | |-------|----------------------------|--------|-----------|-------------| | 1 | Sat. vapor at -16°C | -16 | 1320 | 4.98 | | 2 | Superheated vapor at 50°C | 50 | ??? | ??? | | 3 | Sat. liquid at 50°C | 50 | 323 | 1.12 | | 4 | Mixture at -16°C (after exp)| -16 | ??? | ??? | Values for ammonia (NH₃) at these states from standard tables: #### At -16°C - **Saturated vapor enthalpy (\(h_1\))** ≈ **1320 kJ/kg** - **Saturated vapor entropy (\(s_1\))** ≈ **4.98 kJ/kg·K** #### At 50°C - **Saturated liquid enthalpy (\(h_3\))** ≈ **323 kJ/kg** - **Saturated vapor enthalpy (\(h_g\))** ≈ **1468 kJ/kg** --- ### **Step 3: Find \(h_2\) (Isentropic Compression to 50°C)** Since the compression is **isentropic**: \(s_2 = s_1\) At 50°C, use the specific heat to estimate \(h_2\): \[ h_2 = h_1 + c_p (T_2 - T_1) \] Where, - \(h_1 = 1320~\text{kJ/kg}\) - \(c_p = 2.3~\text{kJ/kg·K}\) - \(T_2 = 50^\circ C = 323~K\) - \(T_1 = -16^\circ C = 257~K\) \[ h_2 = 1320 + 2.3 \times (323 - 257) \] \[ h_2 = 1320 + 2.3 \times 66 \] \[ h_2 = 1320 + 151.8 = 1471.8~\text{kJ/kg} \] --- ### **Step 4: Find \(h_4\) (after throttling/expansion)** Throttling is isenthalpic: \(h_4 = h_3\) \[ h_4 = h_3 = 323~\text{kJ/kg} \] --- ### **Step 5: Calculate COP (Coefficient of Performance)** \[ \text{COP}_{\text{max}} = \frac{\text{Refrigeration~Effect}}{\text{Work~Input}} \] \[ \text{Refrigeration Effect} = h_1 - h_4 = 1320 - 323 = 997~\text{kJ/kg} \] \[ \text{Work Input} = h_2 - h_1 = 1471.8 - 1320 = 151.8~\text{kJ/kg} \] \[ \text{COP}_{\text{max}} = \frac{997}{151.8} = 6.56 \] --- ### **Step 6: Mass Flow Rate of Ammonia (\(\dot{m}\))** \[ Q_L = \dot{m} \times (h_1 - h_4) \] \[ \dot{m} = \frac{Q_L}{h_1 - h_4} = \frac{105.51}{997} = 0.106~\text{kg/s} \] Convert to kg/hr: \[ \dot{m}_{hr} = 0.106 \times 3600 = 381.6~\text{kg/hr} \] --- ### **Step 7: Power Supplied to Compressor (\(W_{comp}\))** \[ W_{comp} = \dot{m} \times (h_2 - h_1) \] \[ W_{comp} = 0.106 \times 151.8 = 16.1~\text{kW} \] --- ### **Step 8: Heat Rejected in Condenser (\(Q_H\))** \[ Q_H = \dot{m} \times (h_2 - h_3) \] \[ Q_H = 0.106 \times (1471.8 - 323) = 0.106 \times 1148.8 = 121.77~\text{kW} \] --- ## **Summary Table** | Quantity | Value | |---------------------------------------------|---------------| | (a) Max. COP of the System | **6.56** | | (b) Mass flow rate of ammonia (kg/hr) | **381.6** | | (c) Power supplied to compressor (kW) | **16.1** | | (d) Heat rejected in condenser (kW) | **121.77** | --- **If you need the cycle sketched, let me know for a diagram!**