# Single-Effect Evaporator Problem Solution ## **Given Data** - **Feed rate (\( F \))**: 800 kg/h - **Feed concentration (\( C_F \))**: 10 wt% solute - **Product concentration (\( C_P \))**: 40 wt% solute - **Feed temperature (\( T_F \))**: 30°C - **Boiling point (\( T_B \))**: 100°C - **Specific heat capacity (\( C_p \))**: 4. kJ/kg·K - **Latent heat of vaporization (\( \lambda \))**: 2257 kJ/kg - **Heat losses**: Negligible --- ## **(a) Mass Flow Rate of the Concentrated Product** Let - \( F \) = Feed rate (kg/h) - \( P \) = Product rate (kg/h) - \( V \) = Vapor (water evaporated) rate (kg/h) - \( C_F \) = Feed solute fraction = .10 - \( C_P \) = Product solute fraction = .40 ### **Solute Balance** \[ \text{Solute in feed} = \text{Solute in product} \] \[ F \times C_F = P \times C_P \] \[ 800 \times .10 = P \times .40 \] \[ 800 = .40P \] \[ P = \frac{800}{.40} = 200 \text{ kg/h} \] **Final Answer (a):** **Product mass flow rate = 200 kg/h** --- ## **(b) Mass Flow Rate of Water Evaporated** ### **Overall Mass Balance** \[ F = P + V \] \[ 800 = 200 + V \] \[ V = 800 - 200 = 600 \text{ kg/h} \] **Final Answer (b):** **Water evaporated = 600 kg/h** --- ## **(c) Heat Required to Raise Feed Temperature to Boiling Point** \[ Q_1 = F \times C_p \times (T_B - T_F) \] \[ Q_1 = 800 \times 4. \times (100 - 30) \] \[ Q_1 = 800 \times 4. \times 70 \] \[ Q_1 = 800 \times 280 = 2,\!240,\!000 \text{ kJ/h} \] **Final Answer (c):** **Heat required for heating = \( \boxed{2,\!240,\!000 \text{ kJ/h}} \)** --- ## **(d) Heat Required for Evaporation of Water** \[ Q_2 = V \times \lambda \] \[ Q_2 = 600 \times 2257 \] \[ Q_2 = 13,\!542,\!000 \text{ kJ/h} \] **Final Answer (d):** **Heat required for evaporation = \( \boxed{13,\!542,\!000 \text{ kJ/h}} \)** --- ## **(e) Total Heat Duty of the Evaporator** \[ Q_{total} = Q_1 + Q_2 \] \[ Q_{total} = 2,\!240,\!000 + 13,\!542,\!000 = 15,\!782,\!000 \text{ kJ/h} \] **Final Answer (e):** **Total heat duty = \( \boxed{15,\!782,\!000 \text{ kJ/h}} \)** --- ## **(f) Steam Consumption Rate for Double-Effect Evaporator** - **Overall steam economy (\( E \))**: 1.8 - Steam economy = kg water evaporated / kg steam used \[ \text{Steam consumption rate} = \frac{V}{E} \] \[ \text{Steam consumption} = \frac{600}{1.8} = 3333.33 \text{ kg/h} \] **Final Answer (f):** **Steam consumption rate = \( \boxed{3333 \text{ kg/h}} \) (rounded to nearest kg/h)** --- ## **Summary Table** | Item | Value | |-----------------------------------------|------------------| | (a) Product mass flow rate | 200 kg/h | | (b) Water evaporated | 600 kg/h | | (c) Heat for heating feed | 2,240,000 kJ/h | | (d) Heat for evaporation | 13,542,000 kJ/h | | (e) Total heat duty | 15,782,000 kJ/h | | (f) Double-effect steam consumption | 3333 kg/h |