An air-screen grain cleaner processes $2000\text{ kg/h}$ of raw grain feed containing $8.0\%\text{ (w/w)}$ impurities ($x_f = 0.08$). The cleaned grain stream contains $1.0\%\text{ (w/w)}$ impurities ($x_c = 0.01$). If the impurities discard stream contains $80.0\%\text{ (w/w)}$ impurities and $20.0\%\text{ (w/w)}$ good grain, the impurity removal efficiency ($E_i = \frac{\text{Impurities in discard}}{\text{Impurities in feed}} \times 100$) is ________ % (round off to two decimal places).
Answer Key:
88.61
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Component Mass Balances**
1. Feed composition:
Feed $F = 2000\text{ kg/h}$. Total impurities in feed $= 2000 \times 0.08 = 160\text{ kg/h}$.
2. Overall and impurity balance with cleaned stream ($C$) and discard stream ($D$):
$F = C + D \implies 2000 = C + D$
$F x_f = C x_c + D x_d \implies 160 = 0.01 C + 0.80 D$
$160 = 0.01(2000 - D) + 0.80 D = 20 + 0.79 D$
$0.79 D = 140 \implies D = 177.215\text{ kg/h}$
3. Impurities in discard stream:
$D_{\text{imp}} = 177.215 \times 0.80 = 141.772\text{ kg/h}$
4. Impurity removal efficiency:
$E_i = \frac{141.772}{160} \times 100\% = 88.6075\% \approx 88.61\%$