ChemicalMaterial BalancesmediumFE ExamCourseworkPE / Advanced

Product Flow from a Two-Outlet Separator

A separator processes $1000\ \text{kg/h}$ of feed containing $40\ \text{wt}\%$ component A.

It produces two streams: a product at $90\ \text{wt}\%$ A and a waste at $5\ \text{wt}\%$ A.

Find the product stream flow rate.

Given

Hint 1

Two unknown streams means you need two independent equations.

Hint 2

Total mass in equals total mass out; the same holds for component A separately.

Hint 3

Substitute W = F − P into the component balance and solve for P.

Worked solution — try the problem first

Two unknowns ( $P$ and $W$ ) require two independent balances.

Total: $F = P + W \quad\Longrightarrow\quad W = 1000 - P$

Component A: $F x_F = P x_P + W x_W$ $1000(0.40) = 0.90P + 0.05W$

Substitute $W = 1000 - P$ :

$400 = 0.90P + 0.05(1000 - P) = 0.90P + 50 - 0.05P = 0.85P + 50$

$0.85P = 350 \quad\Longrightarrow\quad P = 411.8\ \text{kg/h}$

$\boxed{P = 411.8\ \text{kg/h}}$

Check. $W = 588.2$ kg/h. Component A out $= 411.8(0.9) + 588.2(0.05) = 370.6 + 29.4 = 400$ kg/h, matching the feed ✓

Lever rule shortcut: $P/F = (x_F - x_W)/(x_P - x_W) = (0.40-0.05)/(0.90-0.05) = 0.35/0.85 = 0.4118$ ✓

Concepts:Material balancesDegrees of freedomSeparation processesLever rule

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