Let's break down the problem **step-by-step**: --- ## **Given Data** - **Air mass flow rate**: \(\dot{m}_{\text{air}} = 5 \ \text{kg/min} = 0.0833 \ \text{kg/s}\) - **Inlet air conditions**: \(T_1 = 30^\circ\text{C}\), \(\phi_1 = 60\%\), \(P = 100\ \text{kPa}\) - **Exit air temperature**: \(T_2 = 15^\circ\text{C}\) - **Refrigerant (R-134a) pressures**: \(P_{\text{low}} = 200\ \text{kPa}\), \(P_{\text{high}} = 1000\ \text{kPa}\) **Find:** 1. **COP** of the refrigeration cycle 2. **Ratio:** \(\frac{\dot{m}_{R-134a}}{\dot{m}_{\text{air}}}\) 3. **Outgoing air temperature and relative humidity:** \(T_3\) and \(\phi_3\) --- ## **Step 1: Air Properties at Inlet (State 1)** ### **1. Saturation Pressure at 30°C** From steam tables: - \(P_{ws,30} \approx 4.24\ \text{kPa}\) ### **2. Partial Pressure of Water Vapor** \[ P_{w,1} = \phi_1 \cdot P_{ws,30} = 0.6 \times 4.24 = 2.544\ \text{kPa} \] ### **3. Humidity Ratio at Inlet** \[ \omega_1 = 0.622 \cdot \frac{P_{w,1}}{P - P_{w,1}} = 0.622 \cdot \frac{2.544}{100 - 2.544} = 0.622 \cdot \frac{2.544}{97.456} = 0.01624\ \text{kg vapor/kg dry air} \] --- ## **Step 2: Air Properties at Outlet (State 2)** ### **1. Saturation Pressure at 15°C** From steam tables: - \(P_{ws,15} \approx 1.705\ \text{kPa}\) ### **2. Maximum Humidity Ratio at 15°C** \[ \omega_{\text{sat,15}} = 0.622 \cdot \frac{1.705}{100 - 1.705} = 0.622 \cdot \frac{1.705}{98.295} = 0.0108\ \text{kg vapor/kg dry air} \] **Since the air is cooled to 15°C, the exiting air is saturated (condensation occurs).** \[ \omega_2 = \omega_{\text{sat,15}} = 0.0108\ \text{kg vapor/kg dry air} \] --- ## **Step 3: Water Removed by Dehumidifier** \[ \Delta \omega = \omega_1 - \omega_2 = 0.01624 - 0.0108 = 0.00544 \] \[ \dot{m}_{\text{water}} = \dot{m}_{\text{air}} \cdot \Delta \omega = 0.0833 \cdot 0.00544 = 0.000453\ \text{kg/s} \] --- ## **Step 4: Cooling Load (Refrigeration Effect)** \[ Q_L = \dot{m}_{\text{air}} \left[ (h_{a,2} - h_{a,1}) + (\omega_2 h_{v,2} - \omega_1 h_{v,1}) \right] \] - \(h_{a,1}\), \(h_{a,2}\): Enthalpy of dry air at 30°C and 15°C (approx. \(1.005 \cdot T\)) - \(h_{v,1}\), \(h_{v,2}\): Enthalpy of water vapor at 30°C and 15°C (from tables: \(h_{g,30} \approx 2556.5\), \(h_{g,15} \approx 2538.4\) kJ/kg) \[ h_{a,1} = 1.005 \times 30 = 30.15\ \text{kJ/kg} \] \[ h_{a,2} = 1.005 \times 15 = 15.08\ \text{kJ/kg} \] \[ h_{v,1} = 2556.5\ \text{kJ/kg},\quad h_{v,2} = 2538.4\ \text{kJ/kg} \] \[ Q_L = 0.0833 \left[ (15.08 - 30.15) + (0.0108 \times 2538.4 - 0.01624 \times 2556.5) \right] \] Calculate: - \(15.08 - 30.15 = -15.07\) - \(0.0108 \times 2538.4 = 27.42\) - \(0.01624 \times 2556.5 = 41.52\) So, \[ Q_L = 0.0833 \left[ -15.07 + (27.42 - 41.52) \right] = 0.0833 \left[ -15.07 - 14.10 \right] = 0.0833 \times (-29.17) = -2.43\ \text{kW} \] **(Negative sign just means heat is being removed — cooling.)** --- ## **Step 5: Refrigerant Cycle (R-134a)** ### **Find COP** #### **a) Find enthalpies from R-134a tables:** - At \(P_{\text{low}} = 200\) kPa: - \(T_{\text{evap}} \approx -10^\circ\)C - \(h_{\text{evap,in}} = h_1 \approx h_f\) at 200 kPa - At \(P_{\text{high}} = 1000\) kPa: - \(T_{\text{cond}} \approx 35^\circ\)C - \(h_{\text{cond,out}} = h_3 \approx h_f\) at 1000 kPa **Typical values from tables:** - \(h_1\) (sat. vapor at 200 kPa): \(h_g \approx 237\) kJ/kg - \(h_4\) (sat. liquid at 200 kPa): \(h_f \approx 43.1\) kJ/kg - \(h_2\) (after compressor, at 1000 kPa): Use isentropic process (assume \(h_2 \approx 263\) kJ/kg, estimate) - \(h_3\) (sat. liquid at 1000 kPa): \(h_f \approx 75.5\) kJ/kg #### **b) Refrigeration effect per kg R-134a:** \[ q_L = h_1 - h_4 = 237 - 43.1 = 193.9\ \text{kJ/kg} \] #### **c) Work input:** \[ w_{\text{comp}} = h_2 - h_1 = 263 - 237 = 26\ \text{kJ/kg} \] #### **d) COP:** \[ COP = \frac{q_L}{w_{\text{comp}}} = \frac{193.9}{26} = 7.46 \] --- ## **Step 6: Mass Flow Rate Ratio** \[ \dot{m}_{R-134a} = \frac{Q_L}{q_L} = \frac{2.43}{193.9} = 0.0125\ \text{kg/s} \] \[ \text{Ratio} = \frac{\dot{m}_{R-134a}}{\dot{m}_{\text{air}}} = \frac{0.0125}{0.0833} = 0.15 \] --- ## **Step 7: Air Leaving Condenser (State 3)** - The air absorbs heat from the condenser (at \(T_{\text{cond}}\)), so it leaves at a higher temperature. - The heat rejected by the condenser: \[ Q_H = \dot{m}_{R-134a} \cdot (h_2 - h_3) = 0.0125 \times (263 - 75.5) = 2.34\ \text{kW} \] \[ Q_H = \dot{m}_{\text{air}} \cdot c_p \cdot (T_3 - T_2) \] Assume \(c_p \approx 1.005\ \text{kJ/kg.K}\): \[ 2.34 = 0.0833 \cdot 1.005 \cdot (T_3 - 15) \] \[ T_3 - 15 = \frac{2.34}{0.0833 \times 1.005} = \frac{2.34}{0.0837} = 27.95 \] \[ T_3 = 27.95 + 15 = 42.95^\circ\text{C} \] **(This seems high, but it's because the air flow is low compared to the heat load—adjust as needed for more accurate values)** ### **Relative Humidity at State 3 (\(\phi_3\)):** \[ \omega_3 = \omega_2 = 0.0108 \] \[ P_{ws,42.95} \approx 8.3\ \text{kPa} \] \[ P_{w,3} = \frac{\omega_3 \cdot P}{0.622 + \omega_3} = \frac{0.0108 \times 100}{0.622 + 0.0108} = \frac{1.08}{0.6328} = 1.71\ \text{kPa} \] \[ \phi_3 = \frac{P_{w,3}}{P_{ws,42.95}} = \frac{1.71}{8.3} = 0.206 = 20.6\% \] --- ## **Final Answers** ### **1. COP of Refrigeration Cycle:** \[ \boxed{7.46} \] ### **2. Mass Flow Rate Ratio:** \[ \boxed{0.15} \] ### **3. Air Leaving Condenser:** \[ \boxed{T_3 = 42.95^\circ\text{C}} \] \[ \boxed{\phi_3 = 20.6\%} \] --- **If you need a more detailed step for any part, let me know!**