Question 2 of 6: Wood Combustion — Ultimate Analysis from an Orsat Flue-Gas Report

Question A2: Wood Combustion — Ultimate Analysis from an Orsat Flue-Gas Report (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: 45.9% C, 23.1% O, 5.1% ash by mass, remainder (25.9%) is moisture + combustible hydrogen. Dry (Orsat) flue-gas analysis below. Basis: 100 mol dry flue gas; air taken as 21% O$_2$/79% N$_2$.

Find. (a) the missing hydrogen and moisture fractions (full ultimate analysis); (b) the fuel/air mass ratio; (c) the % excess air; (d) the complete (wet) flue-gas composition.

Approach. Take 100 mol dry flue gas as basis; tie on carbon to size the wood and on nitrogen to size the air, then an oxygen balance (with moisture oxygen cancelling) delivers the combustible hydrogen and hence the remaining fractions; finish with excess-air and wet-composition calculations.

1. Carbon tie fixes the wood mass. All fuel carbon leaves as CO$_2$+CO: $n_C=14.8+1.66=16.46$ mol per 100 mol dry gas, i.e. $16.46(12.011)=197.7$ g C. Since carbon is 45.9% of the wood, $$W=\frac{197.7}{0.459}=\boxed{430.7\ \text{g wood}}\ /\,100\ \text{mol dry gas}.$$
2. Nitrogen tie fixes the air. N$_2$ is inert, so all 80.08 mol comes from air: $$n_{O_2,\text{air}}=80.08\cdot\tfrac{21}{79}=21.29\ \text{mol},\qquad n_{\text{air}}=101.4\ \text{mol}.$$
3. Oxygen balance isolates the combustible hydrogen (part a). Wood elemental O is $0.231(430.7)/16=6.22$ mol O-atoms. Writing an O-atom balance in which the free-moisture oxygen enters and leaves unchanged (it cancels), the only unknown is the combustible H (mass $h$, forming $h/2.016$ mol H$_2$O): $$2(21.29)+6.22=2(14.8)+1.66+2(3.46)+\tfrac{h}{2.016}\;\Rightarrow\;h=21.4\ \text{g H}.$$ Thus H $=21.4/430.7=4.97\%$ and moisture $=0.259(430.7)-21.4=90.2$ g $=20.93\%$, completing the ultimate analysis (C 45.9, O 23.1, ash 5.1, H 4.97, moisture 20.93 — sums to 100%).
4. Fuel-to-air ratio by weight (part b). Air mass $=21.29(32)+80.08(28.02)=2925$ g, so $$\frac{\text{fuel}}{\text{air}}=\frac{430.7}{2925}=\boxed{0.147}\quad(\text{air/fuel}=6.79).$$
5. Percentage excess air (part c). Theoretical O$_2$ burns all C to CO$_2$ and all H to H$_2$O, crediting the fuel oxygen: $n_{O_2}^{theo}=16.46+\tfrac{21.23}{4}-\tfrac{6.22}{2}=18.66$ mol. Hence $$\%\,\text{excess}=\frac{21.29-18.66}{18.66}\times100=\boxed{14.1\%}.$$ The positive figure is consistent with the 3.46% free O$_2$ measured in the stack.
6. Wet flue-gas composition (part d). Add the water the Orsat omits: combustion water $21.4/2.016=10.61$ mol plus evaporated moisture $90.2/18.015=5.00$ mol give $15.62$ mol H$_2$O, so the wet total is $115.62$ mol. Dividing each species by this total gives $$\mathrm{CO_2}\,12.80,\ \mathrm{CO}\,1.44,\ \mathrm{O_2}\,2.99,\ \mathrm{N_2}\,69.26,\ \mathrm{H_2O}\,13.51\ \text{mol}\%.$$