Let's break down the problem step by step. --- ## **Given Data** - **Inlet air to dryer:** - Wet bulb temperature, \( T_{wb1} = 15^\circ C \) - Moisture content, \( W_1 = .005 \) kg/kg dry air - **Heated air (entering dryer):** - Dry bulb temperature, \( T_{db2} = 55^\circ C \) - **Exit air from dryer:** - Dry bulb temperature, \( T_{db3} = 25^\circ C \) - Relative humidity, \( \phi_3 = 78\% = .78 \) --- ## **Required** For both **inlet air to dryer** and **exit air from dryer**: - Wet bulb temperature (\(T_{wb}\)) - Dew point temperature (\(T_{dp}\)) - Moisture content (\(W\)) (for exit only, as inlet is given) --- # **Step 1: Inlet Air to Dryer** ### **1.1 Find Dew Point Temperature (\(T_{dp1}\))** - **Given**: \( W_1 = .005 \) kg/kg dry air. - **Formula**: \[ W = .622 \frac{p_v}{p_{atm} - p_v} \] Assume \( p_{atm} = 101.3 \) kPa (standard atmosphere). Rearranging for \( p_v \): \[ p_v = \frac{W \cdot p_{atm}}{.622 + W} \] Substitute: \[ p_v = \frac{.005 \times 101.3}{.622 + .005} = \frac{.5065}{.627} \approx .808 \text{ kPa} \] Now, use steam tables or the Antoine equation to find \( T_{dp1} \) corresponding to \( p_v = .808 \) kPa: - At \( T \approx 7^\circ C \), \( p_{ws} \approx .808 \) kPa. - **So,** \( T_{dp1} \approx 7^\circ C \). ### **1.2 Confirm Wet Bulb Temperature (\(T_{wb1}\))** - **Given:** \( T_{wb1} = 15^\circ C \) (already provided). --- # **Step 2: Heated Air before Dryer** ### **2.1 Moisture Content and Dew Point** - **Heating does not add or remove moisture**: \( W_2 = W_1 = .005 \) kg/kg dry air. - **Dew Point**: \( T_{dp2} = T_{dp1} = 7^\circ C \). - **Wet Bulb**: To find the wet bulb at 55°C and the same moisture content, use psychrometric charts or formulas: - At \( T_{db2} = 55^\circ C \) and \( W_2 = .005 \), the wet bulb will be **much lower** than the dry bulb, but you would need a chart for precise value (likely \( T_{wb2} \approx 20^\circ C \)). --- # **Step 3: Exit Air from Dryer** ### **3.1 Find Moisture Content (\(W_3\))** - **Given:** \( T_{db3} = 25^\circ C, \phi_3 = .78 \) - **Saturation vapor pressure at 25°C**: \( p_{ws} \approx 3.17 \) kPa - **Actual vapor pressure:** \[ p_{v3} = \phi_3 \times p_{ws} = .78 \times 3.17 = 2.47 \text{ kPa} \] - **Moisture content:** \[ W_3 = .622 \frac{p_{v3}}{p_{atm} - p_{v3}} \] \[ W_3 = .622 \frac{2.47}{101.3 - 2.47} = .622 \frac{2.47}{98.83} \approx .622 \times .025 = .0155 \text{ kg/kg dry air} \] ### **3.2 Dew Point Temperature (\(T_{dp3}\))** - Find temperature at which \( p_{ws} = 2.47 \) kPa: - At \( T \approx 19.3^\circ C \), \( p_{ws} \approx 2.47 \) kPa. - **So, \( T_{dp3} \approx 19.3^\circ C \).** ### **3.3 Wet Bulb Temperature (\(T_{wb3}\))** - At \( T_{db3} = 25^\circ C \), \( \phi_3 = 78\% \), using a psychrometric chart: - \( T_{wb3} \approx 22^\circ C \) --- # **Step 4: Summary Table** | Location | Dry Bulb (°C) | Wet Bulb (°C) | Dew Point (°C) | Moisture Content (kg/kg dry air) | |------------------|:-------------:|:-------------:|:--------------:|:-------------------------------:| | **Inlet Air** | — | 15 | 7 | .005 | | **Heated Air** | 55 | ~20 | 7 | .005 | | **Exit Air** | 25 | 22 | 19.3 | .0155 | --- # **Final Answers** ### **Inlet Air to Dryer** - Wet bulb temperature: **15°C** - Dew point temperature: **7°C** - Moisture content: **.005 kg/kg dry air** ### **Exit Air from Dryer** - Wet bulb temperature: **22°C** - Dew point temperature: **19.3°C** - Moisture content: **.0155 kg/kg dry air** --- **Note:** For more precise values (especially wet bulb at 55°C), use a psychrometric chart or calculator, but these are close approximations. Let me know if you need calculations for enthalpy or more details!