# Step-by-Step Solution Given: - Feed rate: 500 kg/h - Feed composition: 10 wt% NaCl (so 90 wt% water) - Product composition: 25 wt% NaCl (so 75 wt% water) - Water leaves as pure vapor (no salt in vapor) - Steady-state: No accumulation Let: - \( F = \) feed mass flow = 500 kg/h - \( P = \) product (concentrated solution) mass flow (unknown) - \( V = \) water vapor mass flow (unknown) - \( x_F = \) NaCl mass fraction in feed = .10 - \( x_P = \) NaCl mass fraction in product = .25 Let’s solve each part: ## (a) Mass Flow Rate of Concentrated Liquid Product Perform a **salt balance** (since salt is neither lost nor vaporized): Total salt in feed per hour: \( F \times x_F = 500 \times .10 = 500 \) kg NaCl/h All the salt leaves in the product: \( P \times x_P = 500 \) kg NaCl/h Solve for \( P \): \( P = \frac{500}{.25} = 200 \) kg/h **Final answer (a):** The mass flow rate of the concentrated liquid product is 200 kg/h. ## (b) Mass Flow Rate of Water Evaporated Perform a **total mass balance**: \( F = P + V \) \( 500 = 200 + V \) \( V = 300 \) kg/h **Final answer (b):** The mass flow rate of water evaporated is 300 kg/h. ## (c) Percentage of Water Removed from the Feed First, calculate total water in feed: \( 500 \times .90 = 450 \) kg water/h Water in product: \( 200 \times .75 = 150 \) kg water/h Water evaporated: \( 450 - 150 = 300 \) kg/h (matches our earlier calculation) Percentage removed: \( \frac{300}{450} \times 100\% = 66.67\% \) **Final answer (c):** The percentage of water removed from the feed is 66.67%. ## (d) Amount of Salt Present in the Product Stream As above, Amount of salt in product = 500 kg/h **Final answer (d):** The amount of salt in the product stream is 500 kg/h. ## (e) Fraction of the Original Water Remaining in the Liquid Product Fraction remaining = (water in product) / (water in feed) \( \frac{150}{450} = .333 \) **Final answer (e):** The fraction of the original water remaining in the liquid product is .333 (or 33.3%). ## (f) Evaporation Ratio (kg Water Evaporated per kg Feed) Evaporation ratio = \( \frac{\text{water evaporated}}{\text{feed}} = \frac{300}{500} = .6 \) **Final answer (f):** The evaporation ratio is .6 kg water evaporated per kg feed. --- **Summary Table** | Part | Value | Units | |------|-------|-------| | (a) | 200 | kg/h (liquid product) | | (b) | 300 | kg/h (water evaporated) | | (c) | 66.67 | % (water removed) | | (d) | 500 | kg/h (salt in product) | | (e) | .333 | fraction (water left) | | (f) | .6 | kg/kg (evaporation ratio) |