Answer:
[tex]y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02[/tex]
Explanation:
Hello,
a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:
- [tex]\xi_1[/tex] and [tex]\xi_2[/tex]: extent of the reactions (2).
- [tex]F_{O_2}^2[/tex], [tex]F_{CH_4}^2[/tex], [tex]F_{H_2O}^2[/tex], [tex]F_{HCHO}^2[/tex] and [tex]F_{CO_2}^2[/tex]: Molar flows at the second stream (5).
On the other hand, we've got the following equations:
- [tex]F_{O_2}^2=50mol/s-\xi_1-2\xi_2[/tex]: oxygen mole balance.
- [tex]F_{CH_4}^2=50mol/s-\xi_1-\xi_2[/tex]: methane mole balance.
- [tex]F_{H_2O}^2=\xi_1+2\xi_2[/tex]: water mole balance.
- [tex]F_{HCHO}^2=\xi_1[/tex]: formaldehyde mole balance.
- [tex]F_{CO_2}^2=\xi_2[/tex]: carbon dioxide mole balance.
Thus, the degrees of freedom are:
[tex]DF=7unknowns-5equations=2[/tex]
It means that we need two additional equations or data to solve the problem.
b. Here, the two missing data are given. For the fractional conversion of methane, we define:
[tex]0.900=\frac{\xi_1+\xi_2}{50mol/s}[/tex]
And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:
[tex]0.860=\frac{F_{HCHO}^2}{50mol/s}[/tex]
In such a way, one realizes that the output formaldehyde's molar flow is:
[tex]F_{HCHO}^2=0.860*50mol/s=43mol/s[/tex]
Which is equal to the first reaction extent [tex]\xi_1[/tex], therefore, one computes the second one from the fractional conversion of methane as:
[tex]\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s[/tex]
Now, one computes the rest of the output flows via:
- [tex]F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s[/tex]
- [tex]F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s[/tex]
- [tex]F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s[/tex]
- [tex]F_{HCHO}^2=43mol/s[/tex]
- [tex]F_{CO_2}^2=2mol/s[/tex]
The total output molar flow is:
[tex]F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s[/tex]
Therefore the output stream composition turns out into:
[tex]y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02[/tex]
Best regards.
