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ChemicalMaterial BalanceseasyFE ExamCoursework

Composition of a Mixed Stream

Two brine streams are mixed:

  • Stream A: 100 kg/h100\ \text{kg/h} containing 20 wt%20\ \text{wt}\% salt
  • Stream B: 50 kg/h50\ \text{kg/h} containing 50 wt%50\ \text{wt}\% salt

Find the salt mass fraction in the combined stream at steady state.

Given
mA100 kg/hStream A flow
xA0.2 -Stream A salt fraction
mB50 kg/hStream B flow
xB0.5 -Stream B salt fraction
Hint 1

Write a total mass balance and a salt component balance separately.

Hint 2

The salt flow in each stream is its mass flow times its fraction.

Hint 3

The answer is a flow-weighted average, so it must be nearer the larger stream's composition.

Worked solution — try the problem first

Total mass balance (steady state, no reaction, nothing accumulates):

m˙out=100+50=150 kg/h\dot m_{out} = 100 + 50 = 150\ \text{kg/h}

Salt component balance:

m˙salt=100(0.20)+50(0.50)=20+25=45 kg/h\dot m_{salt} = 100(0.20) + 50(0.50) = 20 + 25 = 45\ \text{kg/h}

Outlet fraction:

xout=45150=0.30x_{out} = \frac{45}{150} = 0.30

xout=0.30=30 wt%\boxed{x_{out} = 0.30 = 30\ \text{wt}\%}

Why not 35%? Averaging the two compositions (20+50)/2(20+50)/2 ignores that stream A is twice as large. The result must be a flow-weighted average, and it must fall between 20% and 50% — closer to 20% because the dilute stream dominates. It does: 30%.

Concepts:Material balancesComponent balancesSteady state

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