Let's break down the problem and solve each part step by step. --- # **Given Data** - **Sensible load (\(Q_{sens}\))**: 120.5 kW - **Latent load (\(Q_{lat}\))**: 22.4 kW - **Outdoor air fraction (\(f_{OA}\))**: .15 - **System uses:** - Makeup Air Unit (MAU) - Recirculating Air Handler (RAH) - Sensible heat ratio (SHR) control --- ## **a) Symbolic Expression for Mixed Air Enthalpy** The **enthalpy of mixed air** (\(h_{mix}\)) entering the cooling coil is calculated using the mass-weighted average of outdoor air and recirculated air enthalpies: \[ h_{mix} = f_{OA} \cdot h_{OA} + (1 - f_{OA}) \cdot h_{RA} \] Where: - \(h_{OA}\) = enthalpy of outdoor air (kJ/kg) - \(h_{RA}\) = enthalpy of recirculated air (kJ/kg) - \(f_{OA}\) = outdoor air fraction --- ## **b) Numerical Cooling Capacity Required** **Total Cooling Load (\(Q_{total}\))** is the sum of sensible and latent loads: \[ Q_{total} = Q_{sens} + Q_{lat} \] \[ Q_{total} = 120.5\ \text{kW} + 22.4\ \text{kW} = 142.9\ \text{kW} \] **Final Answer for (b):** \[ \boxed{142.9\ \text{kW}} \] --- ## **c) Mass Flow Rate of Recirculated Air** Let: - \( \dot{m}_{tot} \) = total supply air mass flow rate (kg/s) - \( \dot{m}_{OA} \) = outdoor air mass flow rate (kg/s) - \( \dot{m}_{RA} \) = recirculated air mass flow rate (kg/s) - \( c_p \) = specific heat of air (approx. 1.006 kJ/kg·K) - \( \Delta T \) = temperature drop across coil (assume or specify) The total supply air flow is sized by the sensible load: \[ Q_{sens} = \dot{m}_{tot} c_p (T_{room} - T_{supply}) \] \[ \dot{m}_{tot} = \frac{Q_{sens}}{c_p (T_{room} - T_{supply})} \] Assume: - \(T_{room} = 24^\circ C\) - \(T_{supply} = 14^\circ C\) - \(c_p = 1.006\) kJ/kg·K \[ \dot{m}_{tot} = \frac{120.5}{1.006 \times (24 - 14)} = \frac{120.5}{10.06} = 11.98\ \text{kg/s} \] **Outdoor air flow:** \[ \dot{m}_{OA} = f_{OA} \cdot \dot{m}_{tot} = .15 \times 11.98 = 1.80\ \text{kg/s} \] **Recirculated air flow:** \[ \dot{m}_{RA} = \dot{m}_{tot} - \dot{m}_{OA} = 11.98 - 1.80 = 10.18\ \text{kg/s} \] **Final Answer for (c):** \[ \boxed{10.18\ \text{kg/s}} \] --- ## **d) Prove the Thermodynamic Stability of the Room State** **Thermodynamic stability** means the room remains at design temperature and humidity without reheat, i.e., the coil dew point is low enough to handle the latent load. - **Sensible Heat Ratio (SHR):** \[ SHR = \frac{Q_{sens}}{Q_{total}} = \frac{120.5}{142.9} = .843 \] - The supply air dew point must be below the room dew point to remove the latent load. **Proof:** - Coil apparatus dew point (\(T_{dp,coil}\)) must be less than room air dew point. - If the supply air is at \(T_{supply}\) and can absorb both sensible and latent loads (as calculated), the room state is stable. - No reheat is needed as the supply air is at the correct temperature and humidity. --- # **Summary Table** | Part | Result/Formula | |------|---------------| | a | \(h_{mix} = f_{OA} h_{OA} + (1-f_{OA}) h_{RA}\) | | b | \(Q_{total} = 142.9\ \text{kW}\) | | c | \(\dot{m}_{RA} = 10.18\ \text{kg/s}\) | | d | Room state stable if supply air dew point < room dew point; SHR and loads matched. | --- ### **Let me know if you need detailed calculations for enthalpy or humidity, or diagrams for mass/energy transport in the desiccant system!**