Question 1 of 6: Combustion of a Fuel Oil in a Boiler Furnace

Question 1: Combustion of a Fuel Oil in a Boiler Furnace (Part A — 25 marks)

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. Basis 1 lb of oil. Ultimate analysis (mass fractions) and the combustion reactions C + O₂ → CO₂, H₂ + ½O₂ → H₂O and S + O₂ → SO₂. Combustion air is taken as dry (21 mol% O₂, 79 mol% N₂, $M_{air}=28.85$); flue-gas volume is evaluated at the boiler exit, 589 K (1060 °R), 1 atm.

Find. (a) theoretical air; (b) flue-gas mass; (c) flue-gas volume at 589 K; (d) air at 20% excess; (e) flue-gas volume at 20% excess; (f) mol% CO₂ wet and dry.

Approach. Convert each element to moles, sum the stoichiometric O₂ demand (crediting the O₂ already in the fuel), and scale to air by the 21% rule; the theoretical products plus the accompanying N₂ give the flue-gas mass and (via $PV=nRT$) its volume. Excess air simply adds unreacted O₂ and extra N₂.

1. Theoretical oxygen and air. Each element consumes O₂ per its reaction; subtract the fuel’s own oxygen: $$n_{O_2}^{th}=n_C+\tfrac12 n_{H_2}+n_S-n_{O_2,fuel}=0.07113+0.02805+0.000106-0.000844=0.09844\ \text{lbmol.}$$ Air follows from the 21 mol% rule, and its mass from $M_{air}=28.85$: $$n_{air}=\frac{n_{O_2}^{th}}{0.21}=0.4688\ \text{lbmol}\;\Rightarrow\;m_{air}=0.4688(28.85)=\boxed{13.5\ \text{lb air / lb oil}.}$$
2. Flue-gas mass (theoretical air). Products are CO₂ (0.07113), H₂O (0.05610), SO₂ (0.000106) and all the nitrogen $n_{N_2}=0.0000785+0.79(0.4688)=0.3706$ lbmol. By conservation the flue gas simply equals oil + air: $$m_{gas}=m_{oil}+m_{air}=1+13.52=\boxed{14.5\ \text{lb / lb oil}.}$$
3. Flue-gas volume at 589 K (theoretical air). Total wet moles $n=0.07113+0.05610+0.000106+0.3706=0.4977$ lbmol; with $R=0.7302$ ft³·atm·lbmol⁻¹·°R⁻¹ and $T=589\text{ K}=1060\ ^\circ\text{R}$: $$V=\frac{nRT}{P}=0.4977(0.7302)(1060)=\boxed{385\ \text{ft}^3/\text{lb oil}.}$$
4. Air with 20% excess. Scale the theoretical air by 1.20: $$m_{air,20\%}=1.20(13.52)=\boxed{16.2\ \text{lb air / lb oil}.}$$
5. Flue-gas volume with 20% excess air. Now the products carry excess O₂ $=0.20(0.09844)=0.01970$ lbmol and extra N₂; total wet moles rise to $n=0.5915$ lbmol, so at the same 589 K: $$V_{20\%}=0.5915(0.7302)(1060)=\boxed{458\ \text{ft}^3/\text{lb oil}.}$$
6. CO₂ content (20% excess flue gas). With CO₂ = 0.07113 lbmol, wet total 0.5915 and dry total $0.5915-0.05610=0.5354$: $$y_{CO_2}^{wet}=\frac{0.07113}{0.5915}=\boxed{12.0\%},\qquad y_{CO_2}^{dry}=\frac{0.07113}{0.5354}=\boxed{13.3\%.}$$

For reference, the theoretical-air flue gas (no excess O₂) is richer in CO₂: 14.3% wet and 16.1% dry — excess air always dilutes the CO₂ reading, which is why stack-gas CO₂ is a practical measure of excess air.