SME
Material & Energy Balances · Problem type
A burner: methane with excess air
Use this for “100 mol/h of methane burns in 20% excess air. What is the CO₂ % on a dry basis?”
Diagram
Total in: 100 + 1,142 = 1,242; out: 1,242 mol/h.
This is an example. Type your own values to start a new problem.
ṅCH₄
Methane fed
example · mol/h
e
Excess air
example · %
ṅO₂
O₂ fed
calculated · mol/h
ṅN₂
N₂ fed
calculated · mol/h
ṅCO₂
CO₂ out
calculated · mol/h
ṅH₂O
H₂O out
calculated · mol/h
ṅO₂,out
O₂ left
calculated · mol/h
yCO₂
Dry CO₂ share
calculated · %
Formulas
ṅO₂
=
2
×
ṅCH₄
×
(1
+
e
100
)
240
=
2
×
100
×
(1
+
20
100
)
ṅN₂ = 3.76 × ṅO₂
902.4 = 3.76 × 240
ṅCO₂ = ṅCH₄
100 = 100
ṅH₂O = 2 × ṅCH₄
200 = 2 × 100
ṅO₂,out = ṅO₂ − 2 × ṅCH₄
40 = 240 − 2 × 100
yCO₂
=
100
×
ṅCO₂
ṅCO₂ + ṅO₂,out + ṅN₂
9.5932
=
100
×
100
100 + 40 + 902.4
Assumptions
•
Complete combustion, CH₄ + 2O₂ → CO₂ + 2H₂O, so the extent equals the methane fed.
•
Air is 21% O₂ and 79% N₂: 3.76 mol of N₂ per mol of O₂. N₂ passes through.
•
Dry basis: the water is condensed out before the flue gas is analysed.
Step-by-step
WE KNOW
ṅCH₄ = 100 mol/h, e = 20%
FIND
O₂ fed (ṅO₂), N₂ fed (ṅN₂), CO₂ out (ṅCO₂), H₂O out (ṅH₂O), O₂ left (ṅO₂,out), dry CO₂ share (yCO₂)
Step 1 · Find O₂ fed (ṅO₂)
Use ṅO₂ = 2 × ṅCH₄ × (1 + e ÷ 100)
Each CH₄ needs 2 O₂; the excess adds its percent on top.
ṅO₂
=
2
×
100
×
(1
+
20
100
)
ṅO₂ = 2 × 100 × (1 + 0.2)
ṅO₂ = 2 × 100 × 1.2
ṅO₂ = 200 × 1.2
ṅO₂ = 240 mol/h
Step 2 · Find N₂ fed (ṅN₂)
Use ṅN₂ = 3.76 × ṅO₂
Air brings 3.76 mol of N₂ with each mol of O₂.
ṅN₂ = 3.76 × 240
ṅN₂ = 902.4 mol/h
Step 3 · Find CO₂ out (ṅCO₂)
Use ṅCO₂ = ṅCH₄
Each CH₄ burned makes one CO₂.
ṅCO₂ = 100 mol/h
Step 4 · Find H₂O out (ṅH₂O)
Use ṅH₂O = 2 × ṅCH₄
Each CH₄ burned makes two H₂O.
ṅH₂O = 2 × 100
ṅH₂O = 200 mol/h
Step 5 · Find O₂ left (ṅO₂,out)
Use ṅO₂,out = ṅO₂ − 2 × ṅCH₄
The oxygen not used by the burning leaves.
ṅO₂,out = 240 − 2 × 100
ṅO₂,out = 240 − 200
ṅO₂,out = 40 mol/h
Step 6 · Find dry CO₂ share (yCO₂)
Use yCO₂ = 100 × ṅCO₂ ÷ (ṅCO₂ + ṅO₂,out + ṅN₂)
The CO₂ as a percent of the dry gas, the water left out.
yCO₂
=
100
×
100
100 + 40 + 902.4
yCO₂
=
100
×
100
1,042.4
yCO₂
=
10,000
1,042.4
yCO₂ = 9.5932%
Check
Put the numbers back in. Both sides should match.
240
=
2
×
100
×
(1
+
20
100
)
✓
902.4 = 3.76 × 240 ✓
100 = 100 ✓
200 = 2 × 100 ✓
40 = 240 − 2 × 100 ✓
9.5932
=
100
×
100
100 + 40 + 902.4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.