Let's break down and solve the problem step by step. --- ## **Given Data** - **Feed rate (\( F \))** = 850 kg/h - **Initial dry matter concentration (\( x_f \))** = 6% = .06 - **Steam pressure (gauge)** = 200 kPa ⟶ **Absolute** = 200 + 101 = **301 kPa** - **Evaporator pressure (absolute)** = 40 kPa - **Overall heat transfer coefficient (\( U \))** = 600 J/(m²·s·°C) - **Heat transfer area (\( A \))** = 12 m² --- ## **Step 1: Mass Balance** Let: - \( F = \) Feed (850 kg/h) - \( P = \) Product (kg/h, to be found) - \( V = \) Vapor (kg/h, to be found) - \( x_f = \) Feed concentration = .06 - \( x_p = \) Product concentration (to be found) ### **Total Mass Balance** \[ F = P + V \] ### **Solute (dry matter) Mass Balance** \[ F x_f = P x_p \] \[ P = \frac{F x_f}{x_p} \] --- ## **Step 2: Boiling Point in the Evaporator** The evaporator operates at **40 kPa (absolute)**. - At 40 kPa, the boiling point of water ≈ **75°C** (from steam tables). --- ## **Step 3: Steam Conditions** - Steam is available at **301 kPa (absolute)**. - At 301 kPa, the saturation temperature, \( T_{steam} \), ≈ **134°C** (from steam tables). --- ## **Step 4: Energy (Heat) Balance** Latent heat required to vaporize water at 75°C: - \( \lambda_{75°C} \) ≈ **230 kJ/kg** (from steam tables). ### **Heat Required to Vaporize Water** \[ Q = V \cdot \lambda_{75°C} \] - \( Q \) = Heat supplied (kJ/h) - \( V \) = Amount of vapor (kg/h) ### **Heat Supplied by Condensing Steam** \[ Q = U \cdot A \cdot \Delta T \cdot t \] Where: - \( U = 600 \) J/(m²·s·°C) - \( A = 12 \) m² - \( \Delta T = T_{steam} - T_{boiling} = 134^\circ C - 75^\circ C = 59^\circ C \) - Time, \( t = 1 \) hour = 360 seconds \[ Q = 600 \cdot 12 \cdot 59 \cdot 360 \ \text{J} \] Calculate: \[ Q = 600 \cdot 12 = 720 \] \[ Q = 720 \cdot 59 = 424,800 \] \[ Q = 424,800 \cdot 360 = 1,529,280,000 \ \text{J} \] Convert J to kJ: \[ Q = 1,529,280,000 \ \text{J} \div 100 = 1,529,280 \ \text{kJ} \] --- ## **Step 5: Amount of Water Evaporated** \[ Q = V \cdot \lambda_{75^\circ C} \] \[ V = \frac{Q}{\lambda_{75^\circ C}} \] \[ V = \frac{1,529,280}{230} = 665.8 \ \text{kg/h} \] --- ## **Step 6: Find Final Product Concentration** Recall: - \( F = 850 \) kg/h - \( V = 665.8 \) kg/h \[ P = F - V = 850 - 665.8 = 184.2 \ \text{kg/h} \] Solute balance: \[ F x_f = P x_p \] \[ 850 \times .06 = 184.2 \times x_p \] \[ 51 = 184.2 x_p \] \[ x_p = \frac{51}{184.2} = .277 \] --- ## **Step 7: Final Answer** \[ \boxed{\text{The concentration of the final product after evaporation is } 27.7\% \text{ dry matter.}} \] --- ### **Summary Table** | Parameter | Value | |-----------------------|------------------| | Feed rate (\( F \)) | 850 kg/h | | Feed conc. (\( x_f \))| 6% | | Product rate (\( P \))| 184.2 kg/h | | Product conc. (\( x_p \)) | **27.7%** | | Water evaporated (\( V \)) | 665.8 kg/h |