Let's solve the problem step by step as per your instructions. --- ### **Step 1: Conceptual Introduction** This problem involves the analysis of moist air as it undergoes dehumidification and reheating in an air conditioning process. The moist air first passes through a cooling coil, where water vapor condenses, and then through a heating coil. We need to determine (a) the rate at which water vapor is condensed, (b) the required cooling capacity, and (c) the heating capacity required, all for a given volumetric flow rate and specified states. **Explanation Block:** This type of problem is common in HVAC (Heating, Ventilation, and Air Conditioning) applications. The process includes cooling and dehumidification (removal of moisture) followed by reheating to achieve desired air conditions. Key concepts include psychrometrics (study of moist air properties), mass and energy balances, and the use of property databases like CoolProp for accurate air and water vapor property values. --- ### **Step 2: Formulae** #### Formulas Used: 1. **Mass flow rate of dry air:** \(\dot{m}_{da} = \frac{\dot{V}}{v}\) where \(\dot{V}\) = volumetric flow rate, \(v\) = specific volume of moist air. 2. **Humidity ratio:** \(\omega = 0.622 \frac{p_v}{p_a}\) 3. **Water condensed:** \(\dot{m}_{condensed} = \dot{m}_{da} \times (\omega_1 - \omega_2)\) 4. **Cooling capacity:** \(\dot{Q}_{cooling} = \dot{m}_{da} \left[ h_1 - h_2 \right]\) 5. **Heating capacity:** \(\dot{Q}_{heating} = \dot{m}_{da} \left[ h_3 - h_2 \right]\) **Explanation Block:** The mass flow of dry air is determined using the specific volume at the inlet condition. The humidity ratio represents the mass of water vapor per mass of dry air and is calculated using the partial pressures. The mass of water condensed is the difference in humidity ratios before and after the cooling coil. The cooling and heating capacities are calculated using enthalpy differences between the relevant states, multiplied by the dry air flow rate. --- ### **Step 3: Calculation for (a) - Water Condensed** #### **Step-by-Step Calculation:** **Given:** - \(T_1 = 35^\circ C\), \(\phi_1 = 70\%\) - \(\dot{V} = 250\ m^3/min = \frac{250}{60} = 4.167\ m^3/s\) - \(T_3 = 18^\circ C\), \(\phi_3 = 40\%\) - \(P_{atm} = 1\ atm = 101.325\ kPa\) **1. Find the partial pressure of water vapor at state 1 (\(p_{v1}\)):** - \(P_{ws,1}\) (saturation pressure at 35°C) ≈ **5.628 kPa** (from steam tables). - \(p_{v1} = \phi_1 \times P_{ws,1} = 0.7 \times 5.628 = 3.94\ kPa\). **2. Calculate humidity ratio at state 1 (\(\omega_1\)):** \[ \omega_1 = 0.622 \frac{p_{v1}}{P_{atm} - p_{v1}} = 0.622 \frac{3.94}{101.325 - 3.94} = 0.0251\ kg/kg \] **3. Find partial pressure at state 3 (\(p_{v3}\)):** - \(P_{ws,3}\) (saturation pressure at 18°C) ≈ **2.064 kPa**. - \(p_{v3} = \phi_3 \times P_{ws,3} = 0.4 \times 2.064 = 0.826\ kPa\). **4. Calculate humidity ratio at state 3 (\(\omega_3\)):** \[ \omega_3 = 0.622 \frac{0.826}{101.325 - 0.826} = 0.00511 \] **5. Find the average specific volume of air at inlet (\(v_1\)), approx 0.88 m³/kg (from psychrometric charts for 35°C, 70% RH).** **6. Find mass flow rate of dry air (\(\dot{m}_{da}\)):** \[ \dot{m}_{da} = \frac{\dot{V}}{v_1} = \frac{4.167}{0.88} = 4.74\ kg/s \] **7. Calculate mass flow rate of condensed water (\(\dot{m}_{condensed}\)):** \[ \omega_2 = \omega_3\ (\text{no water added/released in heating}) \] \[ \dot{m}_{condensed} = \dot{m}_{da} \times (\omega_1 - \omega_2) = 4.74 \times (0.0251 - 0.00511) \] \[ = 4.74 \times 0.01999 = 0.0947\ kg/s \] \[ = 0.0947 \times 60 = 5.68\ kg/min \] **Explanation Block:** We found the water vapor content at the inlet and outlet, subtracted to get the amount condensed, and multiplied by the mass flow of dry air. This gives us the rate at which water vapor is condensed in the dehumidifier. --- ### **Step 4: Calculation for (b) - Cooling Capacity** #### **Step-by-Step Calculation:** **1. Find enthalpy of air at state 1 (\(h_1\)):** - At 35°C, 70% RH: \(h_1 \approx 95\ kJ/kg\) (from psychrometric chart or CoolProp). **2. At state 2 (after cooling coil, saturated, \(T_2\)):** - \(\omega_2 = \omega_3 = 0.00511\) - \(T_2\) is the dew point of state 1 (find via CoolProp or charts, for \(\omega_2\)). For simplicity, let's estimate \(T_2 \approx 13^\circ C\). - \(h_2 \approx 38\ kJ/kg\) (from chart/CoolProp). **3. Calculate cooling required:** \[ \dot{Q}_{cooling} = \dot{m}_{da} \left[ h_1 - h_2 \right] = 4.74 \times (95 - 38) = 4.74 \times 57 = 270.18\ kW \] **Explanation Block:** The cooling capacity is the energy removed to bring the air from the initial to the post-cooling state, accounting for both sensible and latent heat (condensation). The enthalpy values are obtained from psychrometric data. --- ### **Step 5: Calculation for (c) - Heating Capacity** #### **Step-by-Step Calculation:** **1. Find enthalpy at state 3 (\(h_3\)):** - At 18°C, 40% RH, \(\omega_3 = 0.00511\) - \(h_3 \approx 47.5\ kJ/kg\) (from psychrometric chart/CoolProp). **2. Calculate heating required:** \[ \dot{Q}_{heating} = \dot{m}_{da} \left[ h_3 - h_2 \right] = 4.74 \times (47.5 - 38) = 4.74 \times 9.5 = 45.03\ kW \] **Explanation Block:** The heating capacity is the energy needed to raise the temperature of the dehumidified air from \(T_2\) to \(T_3\). The enthalpy values are again from psychrometric charts or CoolProp. --- ## **Final Answer**: **(a) Rate at which water vapor is condensed:** \[ \boxed{5.68\ \text{kg/min}} \] **(b) Required cooling capacity of the cooling coil:** \[ \boxed{270.2\ \text{kW}} \] **(c) Required heating capacity of the heating coil:** \[ \boxed{45.0\ \text{kW}} \] --- If you have CoolProp or a psychrometric calculator, you can refine these property values, but this solution gives you a comprehensive, step-by-step approach!