Let's solve the problem step by step: --- **Given Data:** - **Refrigerant:** Ammonia (\(\ce{NH3}\)) - **Evaporator temperature (\(T_{evap}\)):** \(-20^\circ C\) - **Condenser pressure (\(P_{cond}\)):** \(12~\text{bar}\) - **Mass flow rate (\(\dot{m}\)):** \(3~\text{kg/min} = 0.05~\text{kg/s}\) - **States:** - **Compressor Inlet:** Saturated vapor at \(-20^\circ C\) - **Condenser Exit:** Saturated liquid at \(12~\text{bar}\) --- ## **Step 1: Find Properties at Key Points** We'll need ammonia (\(\ce{NH3}\)) refrigerant tables (or NIST webbook). For illustration, here are typical values: ### **State 1: Evaporator Exit / Compressor Inlet (\(-20^\circ C\), saturated vapor)** From ammonia tables: - At \(T_1 = -20^\circ C\): - \(P_1 = P_{evap} \approx 1.052~\text{bar}\) - \(h_1 = h_g @ -20^\circ C \approx 1306.6~\text{kJ/kg}\) - \(s_1 = s_g @ -20^\circ C \approx 5.048~\text{kJ/kg}\cdot\text{K}\) ### **State 2: Compressor Exit (P = 12 bar, s = s1)** Isentropic compression (\(s_2 = s_1\)). We need \(h_2\) at \(P_2 = 12~\text{bar}\), \(s_2 = s_1\). From tables, at \(P = 12~\text{bar}\): - \(T_{sat} = 44.4^\circ C\) - \(s_g @ 12~\text{bar} = 4.657~\text{kJ/kg}\cdot\text{K}\) But \(s_1 = 5.048\) which is higher, so after compression we'll have superheated vapor. Find \(h_2\) at \(P_2 = 12~\text{bar}\), \(s_2 = 5.048\). **Superheated tables for ammonia at 12 bar:** Let's interpolate between values: | Temp (°C) | h (kJ/kg) | s (kJ/kg·K) | |-----------|-----------|-------------| | 80 | 1607.3 | 5.088 | | 60 | 1518.8 | 4.904 | Our \(s_2 = 5.048\) is between these, closer to 60°C. Interpolate for \(h_2\): \[ \text{Let}~s_a = 4.904,~h_a = 1518.8,~T_a = 60^\circ C\\ s_b = 5.088,~h_b = 1607.3,~T_b = 80^\circ C\\ s^* = 5.048 \] \[ h_2 = h_a + \frac{(h_b-h_a)}{(s_b-s_a)} \cdot (s_2-s_a) \] \[ h_2 = 1518.8 + \frac{(1607.3-1518.8)}{(5.088-4.904)} \cdot (5.048-4.904) \] \[ = 1518.8 + \frac{88.5}{0.184} \cdot 0.144 = 1518.8 + 478.80 \cdot 0.144 \\ = 1518.8 + 68.95 = 1587.8~\text{kJ/kg} \] ### **State 3: Condenser Exit (\(P_3 = 12~\text{bar}\), saturated liquid)** - \(h_3 = h_f @ 12~\text{bar} = 322.6~\text{kJ/kg}\) ### **State 4: Expansion Valve Exit (\(P_4 = 1.052~\text{bar}\), mixture)** - \(h_4 = h_3 = 322.6~\text{kJ/kg}\) (throttling, enthalpy stays same) --- ## **Step 2: Find COP (Coefficient of Performance)** \[ \text{COP}_{\text{ref}} = \frac{\text{Desired Output (Q}_{in})}{\text{Required Input (W}_{in})} \] - **Net work input per kg:** \(w_{comp} = h_2 - h_1\) - **Refrigeration effect per kg:** \(q_{in} = h_1 - h_4\) \[ \text{COP}_{\text{ref}} = \frac{h_1 - h_4}{h_2 - h_1} \] Plug in the numbers: \[ h_1 = 1306.6~\text{kJ/kg} \\ h_2 \approx 1587.8~\text{kJ/kg} \\ h_4 = 322.6~\text{kJ/kg} \] \[ q_{in} = 1306.6 - 322.6 = 984.0~\text{kJ/kg} \] \[ w_{comp} = 1587.8 - 1306.6 = 281.2~\text{kJ/kg} \] \[ \text{COP} = \frac{984.0}{281.2} = 3.50 \] --- ## **Step 3: Refrigerating Capacity (in tons)** 1 ton of refrigeration = 3.517 kW = 3,517 J/s - **Total refrigerating capacity:** \[ \dot{Q}_{in} = \dot{m} \cdot (h_1 - h_4) \] \[ \dot{m} = 0.05~\text{kg/s} \] \[ \dot{Q}_{in} = 0.05 \times 984.0 = 49.2~\text{kW} \] - **Convert to tons:** \[ \text{Tons} = \frac{49.2~\text{kW}}{3.517~\text{kW/ton}} = 14.0~\text{tons} \] --- ## **Final Answers (Summary Table)** | Part | Quantity | Value | |------|------------------------------|---------------| | (a) | Coefficient of Performance | **3.50** | | (b) | Refrigerating Capacity, tons | **14.0 tons** | --- ### **Summary** **(a) The coefficient of performance (COP) is \(\boxed{3.50}\).** **(b) The refrigerating capacity is \(\boxed{14.0~\text{tons}}\).** --- **If you need the calculations shown with slightly different property values (from a different table), let me know!**