DOFPro Team

The problem statement was:
\(2.00\ \mathrm{kmol/s}\) of propane are burned in 25% excess oxygen in a faulty combustor. 100% of the hydrogen combusts to form water, but only 80% of the carbon combusts completely to \(\mathrm{CO_2}\). The remainder combusts to \(\mathrm{CO}\).
Part a. \[ \dot{n}_\mathrm{O_2in} = 12.5\ \mathrm{kmol/s} \]
Part b.
\[ \dot{n}_\mathrm{CO_2out} = 4.8\ \mathrm{kmol/s}, \]
\[ \dot{n}_\mathrm{COout} = 1.2\ \mathrm{kmol/s} \]
\[ \dot{n}_\mathrm{O_2out} = 3.10\ \mathrm{kmol/s} \]
\[ \dot{n}_\mathrm{H_2Oout} = 8.00\ \mathrm{kmol/s} \]
Part c.
\[ \dot{n}_\mathrm{out} = 17.1\ \mathrm{kmol/s} \]
There are usually multiple reactions generating one desired product and one or more undesired products. We then define for desired species \(\mathrm{R}\),
\[ Y_\mathrm{R} = \frac{n_{\mathrm{R_{actual}}}}{n_{\mathrm{R_{possible}}}} \]
\[ S_\mathrm{R} = \frac{n_{\mathrm{R}}}{n_{\mathrm{products_{undesired}}}} \]
The goal is almost always high yield and high selectivity.
Assume that the desired product in the previous example was \(\mathrm{CO_2}\) and that the undesired product was \(\mathrm{CO}\).
If all of the propane were converted to \(\mathrm{CO_2}\), then
\[ \dot{n}_\mathrm{CO_2possible} = \frac{\nu_\mathrm{CO_2}}{-\nu_\mathrm{C_3H_8}}\dot{n}_\mathrm{C_3H_8in}=\frac{3}{-(-1)}2 = 6\ \mathrm{kmol/s} \]
\[ \dot{n}_\mathrm{CO_2actual} = \dot{n}_\mathrm{CO_2out} = 4.8\ \mathrm{kmol/s} \]
\[ Y_\mathrm{CO_2} = \frac{\dot{n}_\mathrm{CO_2actual}}{\dot{n}_\mathrm{CO_2actual} } = \frac{4.8}{6.0} = 0.80 = 80\% \]
Which we also could have gotten from the problem statement.
Assume that the desired product in the propane example was \(\mathrm{CO_2}\) and that the undesired product was \(\mathrm{CO}\).
\[ S_\mathrm{CO_2} = \frac{\dot{n}_\mathrm{CO_2out}}{\dot{n}_\mathrm{COout}} = \frac{4.8}{1.2} = 4 \]
We don’t count the water as an undesired product because it uses none of the \(\mathrm{C}\) in the propane, which is the only species that goes to our desired product.
If there had been other carbon containing compounds, e.g., \(\mathrm{CH_4}\), we would have included their flows in the denominator.
It’s easy to get mole fraction, split fraction, and fractional conversion confused.
Thanks for watching!
The Full Story companion video is in the link in the upper left. Recycling Before Recycling Was Cool, the next video in the series, is in the upper right. To learn more about Chemical and Thermal Processes, visit the website linked in the description.

