Question 1 of 6: Propane Combustion with Excess Air

Question 1: Propane Combustion with Excess Air (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. Propane burned with 500 kg air; products contain 55 kg $CO_2$ and 15 kg CO (some incomplete combustion despite excess air). Air is 21 mol% $O_2$ / 79 mol% $N_2$. Molar masses (kg/kmol): $C_3H_8=44.10$, $CO_2=44.01$, $CO=28.01$, $H_2O=18.02$, air $=28.84$.

Find. (a) mass of propane burnt; (b) percent excess air; (c) molar (mole-%) composition of the flue gas.

Approach. A carbon balance fixes the propane fed; the theoretical $O_2$ for complete combustion of that propane gives the excess-air percentage; an oxygen-atom balance closes the free $O_2$, and the water follows from a hydrogen balance, giving the full flue-gas inventory.

1. Carbon balance → propane fed (part a). Every carbon atom leaving as $CO_2$ or CO came from propane ($3$ C per molecule): $$n_{C}=n_{CO_2}+n_{CO}=\frac{55}{44.01}+\frac{15}{28.01}=1.250+0.536=1.785\ \text{kmol C},$$ $$n_{C_3H_8}=\frac{n_C}{3}=0.5951\ \text{kmol}\;\Rightarrow\;\boxed{m_{C_3H_8}=0.5951\times44.10=26.2\ \text{kg}}.$$
2. Theoretical oxygen and excess air (part b). Complete combustion is $C_3H_8+5\,O_2\rightarrow3\,CO_2+4\,H_2O$, so the theoretical $O_2$ is referenced to the propane actually fed: $$n_{O_2}^{\text{theo}}=5\,n_{C_3H_8}=5(0.5951)=2.976\ \text{kmol}.$$ With $O_2$ supplied $=0.21(500/28.84)=3.641$ kmol, $$\%\text{ excess air}=\frac{n_{O_2}^{\text{sup}}-n_{O_2}^{\text{theo}}}{n_{O_2}^{\text{theo}}}\times100=\frac{3.641-2.976}{2.976}\times100=\boxed{22.4\%}.$$
3. Water and free oxygen (hydrogen and O-atom balances). All hydrogen leaves as water: $n_{H_2O}=4\,n_{C_3H_8}=2.380$ kmol. An oxygen-atom balance gives the unreacted $O_2$: $$n_{O_2}^{\text{free}}=\tfrac12\!\left[2n_{O_2}^{\text{sup}}-\big(2n_{CO_2}+n_{CO}+n_{H_2O}\big)\right]=\tfrac12\!\left[7.282-5.415\right]=0.933\ \text{kmol}.$$
4. Flue-gas composition (part c). Collecting the wet flue gas (nitrogen passes through unchanged, $13.70$ kmol): $$n_{\text{flue}}=1.250+0.536+2.380+0.933+13.70=18.80\ \text{kmol},$$ so the mole fractions are $$\boxed{CO_2\ 6.65\%,\ CO\ 2.85\%,\ H_2O\ 12.66\%,\ O_2\ 4.96\%,\ N_2\ 72.87\%.}$$

On a dry (Orsat) basis, removing the 2.380 kmol of water renews the totals to $CO_2\ 7.61\%$, $CO\ 3.26\%$, $O_2\ 5.69\%$, $N_2\ 83.44\%$ — the form a stack-gas analyser would report.