Question

For the combustion of methane presented in Example 5.4, the chemical reaction is CH4 +2O2 →CO2 +2H2O Suppose that methane flows into a burner at 30 gmol/s, while oxygen flows into the same burner at 75 gmol/s. If all the meth- ane is burned and a single output stream leaves the burner, what is the mole fraction of CO2 in that output stream? Hint 1: Does the fact that all the methane is burned mean that all the oxygen is burned also? Hint 2: Find the molar flow rate of each component gas in the outlet gas ("flue gas").

Answer 1

`x_(CO_2)^(out) =0.25`

Step-by-step explanation:

Hello.

In this case, for the reactive scheme, it is very convenient to write each species' mole balance as shown below:

`CH_4:f_(CH_4)^(out)=f_(CH_4)^(in)-\epsilon \\\\O_2:f_(O_2)^(out)=f_(O_2)^(in)-2\epsilon\\\\CO_2:f_(CO_2)^(out)=\epsilon\\\\H_2O:f_(H_2O)^(out)=2\epsilon`

Whereas
`\epsilon` accounts for the reaction extent. However, as all the methane is consumed, from the methane balance:

`0=f_(CH_4)^(in)-\epsilon \\\\\epsilon=30gmol/s`

Thus, we can compute the rest of the outlet mole flows since not all the oxygen is consumed as it is in excess:

`f_(O_2)^(out)=f_(O_2)^(in)-2\epsilon=75gmol/s-2*30gmol/s=15gmol/s\\\\f_(CO_2)^(out)=15gmol/s\\\\f_(H_2O)^(out)=2*15gmol/s=30gmol/s`

It means that the mole fraction of carbon dioxide in that output is:

`x_(CO_2)^(out)=(15)/(15+15+30) =0.25`

Best regards.