# (a) Psychrometric Calculations Given: - Dry bulb temperature, \( T = 55^\circ C \) - Relative humidity, \( \phi = 29.6\% = 0.296 \) - Total pressure, \( P = 1 \) atm - Vapor pressure of water at 55°C, \( P_{A,sat} = 0.155 \) atm - \( C_{p,air} = 1005 \) J/(kg·°C) - \( C_{p,vap} = 1884 \) J/(kg·°C) - Latent heat of vaporization, \( \lambda = 2502300 \) J/kg ## Step 1: Partial Pressure of Water Vapor The partial pressure of water vapor in the air (\( P_A \)) is: \[ P_A = \phi \times P_{A,sat} = 0.296 \times 0.155 = 0.04588\, \text{atm} \] ## Step 2: Partial Pressure of Dry Air The partial pressure of dry air (\( P_B \)): \[ P_B = P - P_A = 1 - 0.04588 = 0.95412\, \text{atm} \] ## Step 3: Absolute Humidity (\( Y \)) Absolute humidity is the mass of water vapor per kg of dry air: \[ Y = 0.622 \frac{P_A}{P_B} \] \[ Y = 0.622 \times \frac{0.04588}{0.95412} = 0.02993\, \text{kg H}_2\text{O/kg dry air} \] **Explanation:** - 0.622 is the ratio of molecular weights (18/29) of water vapor to dry air. - This gives the mass of water vapor per kg dry air. ## Step 4: Molar Humidity Molar humidity is the moles of water vapor per mole of dry air: \[ Y_{molar} = \frac{n_A}{n_B} = \frac{P_A}{P_B} \] \[ Y_{molar} = \frac{0.04588}{0.95412} = 0.04808\, \text{mol H}_2\text{O/mol dry air} \] **Explanation:** - From the ideal gas law, the molar ratio is the same as the partial pressure ratio. ## Step 5: Saturation Humidity (\( Y_{sat} \)) At saturation, \( P_A = P_{A,sat} \), so: \[ Y_{sat} = 0.622 \frac{0.155}{1-0.155} = 0.622 \frac{0.155}{0.845} \] \[ Y_{sat} = 0.622 \times 0.18343 = 0.11406\, \text{kg H}_2\text{O/kg dry air} \] ## Step 6: Humid Volume (\( v_H \)) Humid volume is the volume of mixture per kg dry air, at 55°C and 1 atm. First, convert temperature to Kelvin: \[ T = 55 + 273.15 = 328.15\, \text{K} \] For 1 kg dry air: - Moles dry air: \( n_B = \frac{1}{29} = 0.03448\, \text{kmol} \) - Mass water vapor: \( Y = 0.02993\, \text{kg} \) - Moles water vapor: \( n_A = \frac{0.02993}{18} = 0.001662\, \text{kmol} \) - Total moles: \( n_{total} = n_B + n_A = 0.03448 + 0.001662 = 0.03614\, \text{kmol} \) Using ideal gas law: \[ V = nRT/P \] Where: - \( R = 0.082057\, \text{L·atm/(mol·K)} = 0.082057\, \text{m}^3\cdot \text{atm}/(\text{kmol}\cdot \text{K}) \) - \( P = 1\, \text{atm} \) - \( T = 328.15\, \text{K} \) \[ V = 0.03614 \times 0.082057 \times 328.15 / 1 = 0.970\, \text{m}^3 \] So, **humid volume per kg dry air**: \[ v_H = 0.970\, \text{m}^3/\text{kg dry air} \] **Explanation:** - Total volume occupied by 1 kg dry air and its associated water vapor at given T, P. ## Step 7: Enthalpy of the Mixture (\( h \)) Reference: 0°C, dry air and liquid water. \[ h = (C_{p,air} \cdot T) + Y \left[ \lambda + C_{p,vap} \cdot T \right] \] Where: - \( T = 55^\circ C \) - \( C_{p,air} = 1005\, \text{J}/(\text{kg}\cdot^\circ\text{C}) \) - \( C_{p,vap} = 1884\, \text{J}/(\text{kg}\cdot^\circ\text{C}) \) - \( \lambda = 2502300\, \text{J}/\text{kg} \) - \( Y = 0.02993\, \text{kg H}_2\text{O}/\text{kg dry air} \) Calculate each term: 1. \( C_{p,air} \cdot T = 1005 \times 55 = 55275\, \text{J} \) 2. \( C_{p,vap} \cdot T = 1884 \times 55 = 103620\, \text{J} \) 3. \( \lambda + C_{p,vap} \cdot T = 2502300 + 103620 = 2605920\, \text{J} \) 4. \( Y \left[ \lambda + C_{p,vap} \cdot T \right] = 0.02993 \times 2605920 = 77951\, \text{J} \) Total enthalpy per kg dry air: \[ h = 55275 + 77951 = 133226\, \text{J/kg dry air} \] --- ### **Summary Table** | Quantity | Value | |-----------------------|-----------------------------------------| | Absolute humidity | 0.02993 kg H₂O/kg dry air | | Molar humidity | 0.04808 mol H₂O/mol dry air | | Saturation humidity | 0.11406 kg H₂O/kg dry air | | Humid volume | 0.970 m³/kg dry air | | Enthalpy | 133226 J/kg dry air | --- # (b) Drying Time Calculation ## **Step 1: Given Data** - Initial moisture, \( X_1 = 35\% = 0.35 \) (wet basis) - Final moisture, \( X_2 = 10\% = 0.10 \) (wet basis) - Target moisture, \( X_3 = 6\% = 0.06 \) (wet basis) - Critical moisture content (wet basis), \( X_c = 14\% = 0.14 \) - Equilibrium moisture content, \( X_e = 4\% = 0.04 \) - Time to dry from 35% to 10%, \( t_1 = 5 \) h ## **Step 2: Convert Wet Basis to Dry Basis** Let \( X_{wb} \) = mass of water / (mass of wet solid) Dry basis \( X_{db} = \frac{\text{mass of water}}{\text{mass of dry solid}} \) \[ X_{db} = \frac{X_{wb}}{1 - X_{wb}} \] Calculate: - \( X_{1,db} = 0.35 / (1 - 0.35) = 0.538 \) - \( X_{2,db} = 0.10 / (1 - 0.10) = 0.111 \) - \( X_{3,db} = 0.06 / (1 - 0.06) = 0.0638 \) - \( X_{c,db} = 0.14 / (1 - 0.14) = 0.1628 \) - \( X_{e,db} = 0.04 / (1 - 0.04) = 0.04167 \) ## **Step 3: Drying Time Equations** Drying occurs in two periods: - **Constant Rate:** \( X_{db} > X_{c,db} \) - **Falling Rate:** \( X_{db} < X_{c,db} \) ### **Time for Constant Rate Period** \[ t_{constant} = \frac{W_{s}}{N_c} (X_{1,db} - X_{c,db}) \] ### **Time for Falling Rate Period** \[ t_{falling} = \frac{W_s}{N_c} (X_{c,db} - X_{e,db}) \ln \left( \frac{X_{2,db} - X_{e,db}}{X_{3,db} - X_{e,db}} \right) \] Where: - \( W_s \): dry solid mass (can be set to 1 for relative time) - \( N_c \): constant rate of drying (same for both periods) Let \( K = \frac{W_s}{N_c} \). From the first drying (35% to 10% in 5 h): \[ t_1 = K \bigg[ (X_{1,db} - X_{c,db}) + (X_{c,db} - X_{e,db}) \ln \left( \frac{X_{2,db} - X_{e,db}}{X_{c,db} - X_{e,db}} \right) \bigg] \] Plug in values: \[ t_1 = 5 = K \left[ (0.538 - 0.1628) + (0.1628 - 0.04167) \ln \left( \frac{0.111 - 0.04167}{0.1628 - 0.04167} \right) \right] \] Calculate terms: - \( 0.538 - 0.1628 = 0.3752 \) - \( 0.1628 - 0.04167 = 0.12113 \) - \( 0.111 - 0.04167 = 0.06933 \) - \( 0.1628 - 0.04167 = 0.12113 \) - \( \frac{0.06933}{0.12113} = 0.5725 \) - \( \ln(0.5725) = -0.5576 \) So: \[ t_1 = 5 = K \left[ 0.3752 + 0.12113 \times (-0.5576) \right] \] \[ t_1 = 5 = K [0.3752 - 0.06754] = K \times 0.3077 \] \[ K = \frac{5}{0.3077} = 16.25 \] ## **Step 4: Total Time to Dry to 6%** New final moisture, \( X_{3,db} = 0.0638 \): \[ t_{total} = K \left[ (X_{1,db} - X_{c,db}) + (X_{c,db} - X_{e,db}) \ln \left( \frac{X_{3,db} - X_{e,db}}{X_{c,db} - X_{e,db}} \right) \right] \] Calculate: - \( X_{3,db} - X_{e,db} = 0.0638 - 0.04167 = 0.02213 \) - \( (0.02213)/(0.12113) = 0.1827 \) - \( \ln(0.1827) = -1.700 \) \[ t_{total} = 16.25 \left[ 0.3752 + 0.12113 \times (-1.700) \right] \] \[ t_{total} = 16.25 \left[ 0.3752 - 0.2059 \right] = 16.25 \times 0.1693 = 2.751\, \text{h} \] But this is **time for falling rate period from 10% to 6%**. We need total time from 35% to 6%: \[ t_{total} = 16.25 \left[ 0.3752 + 0.12113 \times \ln\left( \frac{0.0638 - 0.04167}{0.12113} \right) \right] \] \[ = 16.25 \left[ 0.3752 + 0.12113 \times (-1.700) \right] \] \[ = 16.25

# Psychrometric Calculations and Drying Time ## (a) Psychrometric Calculations ### Given Data - Dry bulb temperature, \( T = 55^\circ C \) - Relative humidity, \( \phi = 29.6\% = 0.296 \) - Total pressure, \( P = 1 \) atm - Vapor pressure of water at 55°C, \( P_{A,sat} = 0.155 \) atm - Heat capacities: - \( C_{p,air} = 1005 \) J/(kg·°C) - \( C_{p,vap} = 1884 \) J/(kg·°C) - Latent heat of vaporization, \( \lambda = 2502300 \) J/kg ### Step 1: Partial Pressure of Water Vapor \[ P_A = \phi \times P_{A,sat} = 0.296 \times 0.155 = 0.04588\, \text{atm} \] ### Step 2: Partial Pressure of Dry Air \[ P_B = P - P_A = 1 - 0.04588 = 0.95412\, \text{atm} \] ### Step 3: Absolute Humidity (\( Y \)) \[ Y = 0.622 \frac{P_A}{P_B} = 0.622 \times \frac{0.04588}{0.95412} = 0.02993\, \text{kg H}_2\text{O/kg dry air} \] ### Step 4: Molar Humidity \[ Y_{molar} = \frac{P_A}{P_B} = \frac{0.04588}{0.95412} = 0.04808\, \text{mol H}_2\text{O/mol dry air} \] ### Step 5: Saturation Humidity (\( Y_{sat} \)) \[ Y_{sat} = 0.622 \frac{P_{A,sat}}{P - P_{A,sat}} = 0.622 \frac{0.155}{1 - 0.155} = 0.622 \times 0.18343 = 0.11406\, \text{kg H}_2\text{O/kg dry air} \] ### Step 6: Humid Volume (\( v_H \)) Convert temperature to Kelvin: \[ T = 55 + 273.15 = 328.15\, \text{K} \] Calculate total moles in the mixture: \[ V = nRT/P \] Where: - \( R = 0.082057\, \text{m}^3\cdot \text{atm}/(\text{kmol}\cdot \text{K}) \) - \( n_B = 0.03448\, \text{kmol} \) (for 1 kg dry air) - Mass of water vapor = \( Y = 0.02993\, \text{kg} \) - Moles of water vapor: \[ n_A = \frac{0.02993}{18} = 0.001662\, \text{kmol} \] Total moles: \[ n_{total} = n_B + n_A = 0.03448 + 0.001662 = 0.03614\, \text{kmol} \] Calculate volume: \[ V = 0.03614 \times 0.082057 \times 328.15 / 1 = 0.970\, \text{m}^3 \] Thus, humid volume: \[ v_H = 0.970\, \text{m}^3/\text{kg dry air} \] ### Step 7: Enthalpy of the Mixture (\( h \)) Using: \[ h = (C_{p,air} \cdot T) + Y \left[ \lambda + C_{p,vap} \cdot T \right] \] Calculate: 1. \( C_{p,air} \cdot T = 1005 \times 55 = 55275\, \text{J} \) 2. \( C_{p,vap} \cdot T = 1884 \times 55 = 103620\, \text{J} \) 3. \( \lambda + C_{p,vap} \cdot T = 2502300 + 103620 = 2605920\, \text{J} \) 4. \( Y \left[ \lambda + C_{p,vap} \cdot T \right] = 0.02993 \times 2605920 = 77951\, \text{J} \) Total enthalpy: \[ h = 55275 + 77951 = 133226\, \text{J/kg dry air} \] ### Summary Table | Quantity | Value | |-----------------------|-----------------------------------------| | Absolute humidity | 0.02993 kg H₂O/kg dry air | | Molar humidity | 0.04808 mol H₂O/mol dry air | | Saturation humidity | 0.11406 kg H₂O/kg dry air | | Humid volume | 0.970 m³/kg dry air | | Enthalpy | 133226 J/kg dry air | --- ## (b) Drying Time Calculation ### Given Data - Initial moisture, \( X_1 = 35\% = 0.35 \) - Final moisture, \( X_2 = 10\% = 0.10 \) - Target moisture, \( X_3 = 6\% = 0.06 \) - Critical moisture content, \( X_c = 14\% = 0.14 \) - Equilibrium moisture content, \( X_e = 4\% = 0.04 \) - Time to dry from 35% to 10%, \( t_1 = 5 \) h ### Step 1: Convert to Dry Basis \[ X_{db} = \frac{X_{wb}}{1 - X_{wb}} \] Calculating dry basis for each moisture content: - \( X_{1,db} = 0.35 / (1 - 0.35) = 0.538 \) - \( X_{2,db} = 0.10 / (1 - 0.10) = 0.111 \) - \( X_{3,db} = 0.06 / (1 - 0.06) = 0.0638 \) - \( X_{c,db} = 0.14 / (1 - 0.14) = 0.1628 \) - \( X_{e,db} = 0.04 / (1 - 0.04) = 0.04167 \) ### Step 2: Drying Time Equations - **Constant Rate Period:** \( X_{db} > X_{c,db} \) - **Falling Rate Period:** \( X_{db} < X_{c,db} \) ### Time for Constant Rate Period \[ t_{constant} = \frac{W_{s}}{N_c} (X_{1,db} - X_{c,db}) \] ### Time for Falling Rate Period \[ t_{falling} = \frac{W_s}{N_c} (X_{c,db} - X_{e,db}) \ln \left( \frac{X_{2,db} - X_{e,db}}{X_{c,db} - X_{e,db}} \right) \] Using the first drying (35% to 10% in 5 h): \[ t_1 = K \left[ (X_{1,db} - X_{c,db}) + (X_{c,db} - X_{e,db}) \ln \left( \frac{X_{2,db} - X_{e,db}}{X_{c,db} - X_{e,db}} \right) \right] \] ### Step 3: Calculate \( K \) Substituting values into the drying equation: 1. \( X_{1,db} - X_{c,db} = 0.3752 \) 2. \( X_{c,db} - X_{e,db} = 0.12113 \) 3. \( X_{2,db} - X_{e,db} = 0.06933 \) 4. \( X_{c,db} - X_{e,db} = 0.12113 \) 5. \( \frac{0.06933}{0.12113} = 0.5725 \) 6. Calculate \( \ln(0.5725) = -0.5576 \) Thus: \[ t_1 = 5 = K \left[ 0.3752 - 0.06754 \right] \] \[ t_1 = 5 = K \times 0.3077 \] Solving for \( K \): \[ K = \frac{5}{0.3077} = 16.25 \] ### Step 4: Total Time to Dry to 6% For \( X_{3,db} = 0.0638 \): \[ t_{total} = K \left[ (X_{1,db} - X_{c,db}) + (X_{c,db} - X_{e,db}) \ln \left( \frac{X_{3,db} - X_{e,db}}{X_{c,db} - X_{e,db}} \right) \right] \] Calculating: - \( X_{3,db} - X_{e,db} = 0.02213 \) - \( \frac{0.02213}{0.12113} = 0.1827 \) - \( \ln(0.1827) = -1.700 \) Thus: \[ t_{total} = 16.25 \left[ 0.3752 - 0.2059 \right] = 16.25 \times 0.1693 = 2.751\, \text{h} \] ### Final Calculation To find total time from 35% to 6%, sum the times from 35% to 10% and 10% to 6%: \[ t_{total} \approx 5 + 2.751 = 7.751\, \text{h} \] ### Conclusion - Time to dry the material from 35% moisture to 6% moisture under the same conditions is approximately **7.75 hours**.