Let's break down the problem step by step: ### **Given Data** - Airflow rate: **100,000 cfm** - Initial air conditions: **35°F** (dry bulb), **20 grains/lb** (humidity) - Dehumidified to **5 grains/lb** (humidity) - Heated with a coil: **360,000 Btu/hr** - **Find:** Final dry bulb and wet bulb temperatures. --- ## **Step 1: Convert Humidity to lb H₂O/lb dry air** - **1 grain = 1/7000 lb** - **20 grains/lb = 20/7000 = 0.00286 lb H₂O/lb dry air** - **5 grains/lb = 5/7000 = 0.000714 lb H₂O/lb dry air** --- ## **Step 2: Find Mass Flow Rate of Dry Air** ### **a. Find the density of air at 35°F** At 35°F, air density ≈ **0.080 lb/ft³** ### **b. Mass flow rate:** \[ \dot{m}_{\text{air}} = \text{CFM} \times \text{Density} \times 60 \text{ min/hr} \] \[ \dot{m}_{\text{air}} = 100,000 \text{ ft}^3/\text{min} \times 0.080 \text{ lb/ft}^3 \times 60 \text{ min/hr} \] \[ \dot{m}_{\text{air}} = 100,000 \times 0.080 \times 60 = 480,000 \text{ lb/hr} \] --- ## **Step 3: Find Enthalpy Change from Heating** \[ Q = \dot{m}_{\text{air}} \times (h_2 - h_1) \] \[ 360,000 = 480,000 \times (h_2 - h_1) \] \[ h_2 - h_1 = \frac{360,000}{480,000} = 0.75 \text{ Btu/lb} \] --- ## **Step 4: Find Initial Enthalpy (\(h_1\))** \[ h = 0.24T_{db} + W (1061 + 0.444T_{db}) \] - \(T_{db,1} = 35^\circ F\) - \(W_1 = 0.00286\) \[ h_1 = 0.24 \times 35 + 0.00286 \times (1061 + 0.444 \times 35) \] \[ h_1 = 8.4 + 0.00286 \times (1061 + 15.54) \] \[ h_1 = 8.4 + 0.00286 \times 1076.54 \] \[ h_1 = 8.4 + 3.08 = 11.48 \text{ Btu/lb} \] --- ## **Step 5: Find Final Enthalpy (\(h_2\))** \[ h_2 = h_1 + 0.75 = 11.48 + 0.75 = 12.23 \text{ Btu/lb} \] --- ## **Step 6: Find Final Dry Bulb Temperature (\(T_{db,2}\))** After dehumidification: - \(W_2 = 0.000714\) - \(h_2 = 12.23\) \[ h_2 = 0.24T_{db,2} + W_2 (1061 + 0.444T_{db,2}) \] \[ 12.23 = 0.24T_{db,2} + 0.000714 (1061 + 0.444T_{db,2}) \] \[ 12.23 = 0.24T_{db,2} + 0.758 + 0.000317T_{db,2} \] \[ 12.23 - 0.758 = (0.24 + 0.000317)T_{db,2} \] \[ 11.472 = 0.240317 T_{db,2} \] \[ T_{db,2} = \frac{11.472}{0.240317} \approx 47.75^\circ F \] --- ## **Step 7: Find Final Wet Bulb Temperature (\(T_{wb,2}\))** At very low humidity (5 grains/lb), the wet bulb will be much lower than the dry bulb. For \(47.75^\circ F\) dry bulb and \(W = 0.000714\), the air is very dry. **Estimate using psychrometric chart:** - At 47.8°F and 5 grains/lb, the wet bulb temperature is very close to freezing, **approximately 32°F**. --- ## **Step 8: Draw the Process on a Psychrometric Chart** 1. **Start at (35°F, 20 grains/lb)** 2. **Move left (dehumidify at constant enthalpy) to (35°F, 5 grains/lb)** 3. **Move right (sensible heating at constant humidity ratio) to (47.8°F, 5 grains/lb)** --- ## **Summary Table** | Step | Dry Bulb (°F) | Humidity Ratio (grains/lb) | Enthalpy (Btu/lb) | Wet Bulb (°F) | |----------------|:-------------:|:--------------------------:|:-----------------:|:-------------:| | Initial | 35 | 20 | 11.48 | ~32 | | After Dehum. | 35 | 5 | 8.4 | ~23 | | After Heating | 47.8 | 5 | 12.23 | ~32 | --- ## **Final Answers** - **Dry Bulb Temperature after heating:** **~47.8°F** - **Wet Bulb Temperature after heating:** **~32°F** - **Process on psychrometric chart:** - Start at (35°F, 20 grains/lb) - Move left to (35°F, 5 grains/lb) - Move right to (47.8°F, 5 grains/lb) Let me know if you'd like a sketch of the psychrometric chart!