SME
Material & Energy Balances
Material balances
Use this for “100 kg/h of 20% ethanol meets 50 kg/h of 50% ethanol. What leaves?”
Diagram
Total in: 100 + 50 = 150; out: 150 kg/h.
This is an example. Type your own values to start a new problem.
F₁ (kg/h)
+
F₂ (kg/h)
=
F₃ (kg/h)
F₁ (kg/h)
×
x₁
+
F₂ (kg/h)
×
x₂
=
F₃ (kg/h)
×
x₃
Formulas
F₁
+
F₂
=
F₃
100 + 50 = 150
F₁
×
x₁
+
F₂
×
x₂
=
F₃
×
x₃
100 × 0.2 + 50 × 0.5 = 150 × 0.3
Assumptions
•
Steady state and no reaction, so what goes in comes out, for each component.
•
Fractions are mass fractions of ethanol; the rest is water.
Step-by-step
WE KNOW
F₁ = 100 kg/h, x₁ = 0.2, F₂ = 50 kg/h, x₂ = 0.5
FIND
product flow (F₃), product ethanol fraction (x₃)
Step 1 · Find product flow (F₃)
Use F₁ + F₂ = F₃
Total in equals total out: add the two feeds.
F₃
=
F₁
+
F₂
F₃
=
100
+
50
F₃
=
150
kg/h
Step 2 · Find product ethanol fraction (x₃)
Use F₁ × x₁ + F₂ × x₂ = F₃ × x₃
The ethanol in, shared over the product’s flow.
x₃
=
F₁
×
x₁
+
F₂
×
x₂
F₃
x₃
=
100 × 0.2 + 50 × 0.5
150
x₃
=
20 + 25
150
x₃
=
45
150
x₃
=
0.3
Check
Put the numbers back in. Both sides should match.
100 + 50 = 150 ✓
100 × 0.2 + 50 × 0.5 = 150 × 0.3 ✓
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Column split: distillate, bottoms and recovery
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