A single-effect forward-feed evaporator concentrates $2500\text{ kg/h}$ of tomato juice from $5.0\%\text{ (w/w)}$ total solids ($x_F = 0.05$) to $25.0\%\text{ (w/w)}$ total solids ($x_P = 0.25$). The mass rate of water evaporated in $\text{kg/h}$ is ________ (answer in integer).
Answer Key:
2000
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Solid and Total Mass Balance**
1. Feed solids rate:
$\dot{m}_{s} = F \times x_F = 2500\text{ kg/h} \times 0.05 = 125\text{ kg solids/h}$
2. Product rate ($P$):
$P = \frac{\dot{m}_{s}}{x_P} = \frac{125\text{ kg/h}}{0.25} = 500\text{ kg/h}$
3. Water evaporated ($V$):
$V = F - P = 2500 - 500 = 2000\text{ kg/h}$
**Method 2: Direct Evaporation Equation**
$V = F \left( 1 - \frac{x_F}{x_P} \right) = 2500 \times \left( 1 - \frac{0.05}{0.25} \right) = 2500 \times (1 - 0.20) = 2500 \times 0.80 = 2000\text{ kg/h}$