## Step-by-Step Solution: Drying Process ### **Given Data** - Feed mass (\(F\)) = **300 kg** - Initial water content = **25%** - Final water content = **5%** We need to find: 1. **Water removed** 2. **Product mass (after drying)** --- ## **Step 1: Calculate Initial Water and Dry Solids** - **Initial water mass** = \( 25\% \) of 300 kg \[ \text{Initial water} = .25 \times 300 = 75 \text{ kg} \] - **Initial dry solids** = \( 300 - 75 = 225 \) kg --- ## **Step 2: Mass Balance After Drying** Let the final product mass be \( P \). - **Dry solids are unchanged:** Final dry solids = Initial dry solids = **225 kg** - **Final water content in product:** Water = \( 5\% \) of \( P \) Dry solids = \( 95\% \) of \( P \) Set dry solids equal to 225 kg: \[ .95P = 225 \] \[ P = \frac{225}{.95} = 236.84 \text{ kg} \] --- ## **Step 3: Water in Final Product** - **Final water mass:** \( 5\% \) of \( 236.84 \) kg \( = .05 \times 236.84 = 11.84 \) kg --- ## **Step 4: Water Removed** - **Water removed = Initial water - Final water** \[ \text{Water removed} = 75\ \text{kg} - 11.84\ \text{kg} = 63.16\ \text{kg} \] --- ## **Final Answers** ### **(a) Water removed:** \[ \boxed{63.16\ \text{kg}} \] ### **(b) Product mass (after drying):** \[ \boxed{236.84\ \text{kg}} \] --- ### **Summary Table** | Quantity | Value (kg) | |---------------------|:----------:| | Initial dry solids | 225 | | Initial water | 75 | | Final product mass | 236.84 | | Final water | 11.84 | | **Water removed** | 63.16 |