Let's solve the problem step by step: --- ## **Given Data** - **Air flow**: \( 400 \) cfm - **Initial Dry Bulb Temperature (\( T_{db,in} \))**: \( 70^\circ F \) - **Initial Relative Humidity (\( RH_{in} \))**: \( 40\% \) - **Heating rate**: \( 95,000 \) Btu/hr --- ## **Step 1: Find Enthalpy Change (\( \Delta h \))** **Formula:** \[ \dot{Q} = \dot{m} \times \Delta h \] Where: - \( \dot{Q} \) = Heat added (Btu/hr) - \( \dot{m} \) = Mass flow rate of dry air (lb/hr) - \( \Delta h \) = Change in enthalpy (Btu/lb) ### **Calculate Mass Flow Rate of Dry Air** \[ \dot{m} = \frac{\text{CFM} \times 60 \times \rho_{da}}{1} \] Assume density of dry air at 70°F, atmospheric pressure (\( \rho_{da} \approx .075 \) lb/ft³): \[ \dot{m} = 400 \times 60 \times .075 = 18,000 \text{ lb/hr} \] --- ## **Step 2: Find Enthalpy at Inlet (\( h_{in} \))** ### **From Psychrometric Chart (at 70°F, 40% RH):** - **Enthalpy (\( h_{in} \))** ≈ 24.5 Btu/lb (from chart or psychrometric calculator) --- ## **Step 3: Find Enthalpy at Outlet (\( h_{out} \))** \[ \Delta h = \frac{\dot{Q}}{\dot{m}} = \frac{95,000}{18,000} \approx 5.28 \text{ Btu/lb} \] \[ h_{out} = h_{in} + \Delta h = 24.5 + 5.28 \approx 29.8 \text{ Btu/lb} \] --- ## **Step 4: Find Outlet Dry Bulb Temperature (\( T_{db,out} \))** Assuming **no moisture added or removed** (heating only): - **Humidity ratio (\( W \)) remains constant.** - Find \( W \) at inlet (from chart at 70°F, 40% RH): - \( W \approx .0082 \) lb/lb Now, on the psychrometric chart, **move horizontally to the right** (constant \( W \)) until you reach \( h_{out} \approx 29.8 \) Btu/lb. - **This gives \( T_{db,out} \approx 102^\circ F \)** (from chart or calculator). --- ## **Step 5: Find Outlet Relative Humidity** At: - \( T_{db,out} \approx 102^\circ F \) - \( W_{out} = .0082 \) lb/lb From the chart, the **RH drops** (since air can hold more moisture at higher temperatures). At this point, **RH ≈ 9%**. --- ## **Step 6: Find Outlet Wet Bulb Temperature** At: - \( T_{db,out} \approx 102^\circ F \) - \( W_{out} = .0082 \) lb/lb From the chart, **Wet Bulb Temperature ≈ 75°F**. --- ## **Summary Table** | Parameter | Inlet | Outlet | |-------------------------------|--------------|---------------| | Dry Bulb Temp (\( ^\circ F \))| 70 | ~102 | | Relative Humidity (%) | 40 | ~9 | | Wet Bulb Temp (\( ^\circ F \))| ~54 | ~75 | --- ## **Answers** a. **Outlet Dry Bulb Temperature:** \(\boxed{102^\circ F}\) b. **Outlet Relative Humidity:** \(\boxed{9\%}\) c. **Outlet Wet Bulb Temperature:** \(\boxed{75^\circ F}\) --- ### **Psychrometric Chart Path (Description)** - Start at 70°F/40% RH. - Move right **horizontally (constant \( W \))** until you reach enthalpy of about 29.8 Btu/lb. - Read \( T_{db} \) (102°F), \( RH \) (9%), and \( T_{wb} \) (75°F). --- **If you wish to see this on an actual chart, you can use an online psychrometric chart tool and plot these points.** --- ### **References** - [ASHRAE Psychrometric Chart](https://www.linric.com/webpcyc/psychonline) - [Engineering Toolbox - Psychrometric Calculator](https://www.engineeringtoolbox.com/psychrometric-chart-online-d_694.html) --- **Let me know if you need a graph or more explanation!**