A flat vibrating screen is fed with $1200\text{ kg/h}$ of a grain mixture containing $35\%\text{ (w/w)}$ undersize particles ($x_F = 0.35$). The overflow product (oversize stream) contains $5\%\text{ (w/w)}$ undersize particles ($x_D = 0.05$), and the underflow product (undersize stream) contains $90\%\text{ (w/w)}$ undersize particles ($x_B = 0.90$). The overall screening effectiveness $E = E_D \times E_B$ expressed in percentage is ________ % (round off to two decimal places).
Answer Key:
85.83
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Stream Mass Balances**
1. Mass balances around the screen:
Feed $F = 1200\text{ kg/h}$, $x_F = 0.35$.
$F = D + B \implies 1200 = D + B$
$F x_F = D x_D + B x_B \implies 1200(0.35) = 0.05 D + 0.90 B$
$420 = 0.05(1200 - B) + 0.90 B = 60 + 0.85 B$
$0.85 B = 360 \implies B = 423.529\text{ kg/h}$
$D = 1200 - 423.529 = 776.471\text{ kg/h}$
2. Recovery of undersize material in underflow ($E_B$):
$E_B = \frac{B x_B}{F x_F} = \frac{423.529 \times 0.90}{420} = \frac{381.176}{420} = 0.90756$
3. Recovery of oversize material in overflow ($E_D$):
$E_D = \frac{D (1 - x_D)}{F (1 - x_F)} = \frac{776.471 \times (1 - 0.05)}{1200 \times (1 - 0.35)} = \frac{737.647}{780} = 0.94570$
4. Overall screen effectiveness ($E$):
$E = E_D \times E_B = 0.94570 \times 0.90756 = 0.85828 = 85.83\%$
**Method 2: McCabe-Smith Direct Formula**
$E = \frac{(x_F - x_D)(x_B - x_F)(1 - x_D) x_B}{(x_B - x_D)^2 (1 - x_F) x_F}$
$E = \frac{(0.35 - 0.05)(0.90 - 0.35)(0.95)(0.90)}{(0.90 - 0.05)^2 (0.65)(0.35)} = \frac{0.30 \times 0.55 \times 0.95 \times 0.90}{(0.85)^2 \times 0.65 \times 0.35} = \frac{0.141075}{0.164369} = 85.83\%$