Question 2 of 6: Combustion of Wood by Orsat Analysis

Question 2: Combustion of Wood by Orsat Analysis (Part A — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Wood (mass %): C 45.9, O 23.1, ash 5.1, and moisture + combustible H making up the remaining 25.9%. Dry (Orsat) flue gas mol%: CO₂ 14.8, CO 1.66, O₂ 3.46, N₂ 80.08. Air is 21% O₂ / 79% N₂. Molar masses: C 12.01, O 16.00, H 1.008, N₂ 28.01, O₂ 32.00, H₂O 18.02.

Find. (a) full ultimate analysis (split moisture from H); (b) fuel/air mass ratio; (c) % excess air; (d) wet flue-gas composition.

Approach. Take 100 mol dry flue gas as basis. The nitrogen is the tie to air; the carbon (CO₂+CO) ties to the wood mass; an oxygen-atom balance — written so the free-moisture oxygen cancels — isolates the combustible hydrogen, after which moisture follows by difference.

1. Basis and the carbon tie. Basis: 100 mol dry flue gas. Carbon in the flue $=14.8+1.66=16.46$ mol. Since the wood is 45.9% C, the wood mass per 100 mol flue is $$M_{\text{wood}}=\frac{16.46\times12.01}{0.459}=430.7\ \text{g}.$$ Hence wood O atoms $=0.231(430.7)/16.00=6.219$ mol, and moisture + H mass $=0.259(430.7)=111.6$ g.
2. Air from the nitrogen tie. All N₂ comes from air, so O₂ supplied is $$\dot n_{\text{O}_2,\text{air}}=80.08\times\frac{21}{79}=21.29\ \text{mol}.$$
3. Oxygen-atom balance → combustible H. Let $h$ = mol H atoms in the fuel; the free moisture enters and leaves as the same H₂O, so its oxygen cancels. Balancing O atoms (air + wood O = CO₂ + CO + flue O₂ + combustion water): $$2(21.29)+6.219 = 2(14.8)+1.66+2(3.46)+\tfrac{h}{2}\;\Rightarrow\; h = 21.23\ \text{mol H}.$$ So combustible H $=21.23\times1.008=21.4$ g $\Rightarrow \boxed{4.97\%\ \text{H}}$, and moisture $=111.6-21.4=90.2$ g $\Rightarrow \boxed{20.93\%\ \text{moisture}}$.
4. Complete analysis (part a). C 45.9%, O 23.1%, ash 5.1%, H 4.97%, moisture 20.93% (sums to 100%).
5. Fuel-to-air ratio (part b). Air mass $=80.08(28.01)+21.29(32.00)=2924$ g per 100 mol flue: $$\frac{\text{fuel}}{\text{air}}=\frac{430.7}{2924}=\boxed{0.147\ \text{kg fuel/kg air}}\quad(\text{air/fuel}=6.79).$$
6. Excess air (part c). Theoretical O₂ for complete combustion of the fuel fed (all C→CO₂, all H→H₂O), crediting the wood’s own oxygen: $$\dot n_{\text{O}_2,\text{theo}}=16.46+\tfrac{h}{4}-\tfrac{6.219}{2}=16.46+5.305-3.109=18.66\ \text{mol},$$ $$\%\,\text{excess}=\frac{21.29-18.66}{18.66}\times100=\boxed{14.1\%}.$$
7. Wet flue-gas composition (part d). Total water $=\tfrac{h}{2}+\text{moisture}=10.61+5.005=15.62$ mol, so wet total $=115.62$ mol: $$\text{CO}_2\ 12.80\%,\ \ \text{CO}\ 1.44\%,\ \ \text{O}_2\ 2.99\%,\ \ \text{N}_2\ 69.26\%,\ \ \text{H}_2\text{O}\ 13.51\%.$$