Answer:
![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](https://tex.z-dn.net/?f=y_%7BCH_4%7D%5E2%3D%5Cfrac%7B5mol%2Fs%7D%7B100mol%2Fs%7D%3D0.05%5C%5Cy_%7BO_2%7D%5E2%3D%5Cfrac%7B3mol%2Fs%7D%7B100mol%2Fs%7D%3D0.03%5C%5Cy_%7BH_2O%7D%5E2%3D%5Cfrac%7B47mol%2Fs%7D%7B100mol%2Fs%7D%3D0.47%5C%5Cy_%7BHCHO%7D%5E2%3D%5Cfrac%7B43mol%2Fs%7D%7B100mol%2Fs%7D%3D0.43%5C%5Cy_%7BCO_2%7D%5E2%3D%5Cfrac%7B2mol%2Fs%7D%7B100mol%2Fs%7D%3D0.02)
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:
-
and
: extent of the reactions (2).
-
,
,
,
and
: Molar flows at the second stream (5).
On the other hand, we've got the following equations:
-
: oxygen mole balance.
-
: methane mole balance.
-
: water mole balance.
-
: formaldehyde mole balance.
-
: carbon dioxide mole balance.
Thus, the degrees of freedom are:
![DF=7unknowns-5equations=2](https://tex.z-dn.net/?f=DF%3D7unknowns-5equations%3D2)
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:
![0.900=\frac{\xi_1+\xi_2}{50mol/s}](https://tex.z-dn.net/?f=0.900%3D%5Cfrac%7B%5Cxi_1%2B%5Cxi_2%7D%7B50mol%2Fs%7D)
And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:
![0.860=\frac{F_{HCHO}^2}{50mol/s}](https://tex.z-dn.net/?f=0.860%3D%5Cfrac%7BF_%7BHCHO%7D%5E2%7D%7B50mol%2Fs%7D)
In such a way, one realizes that the output formaldehyde's molar flow is:
![F_{HCHO}^2=0.860*50mol/s=43mol/s](https://tex.z-dn.net/?f=F_%7BHCHO%7D%5E2%3D0.860%2A50mol%2Fs%3D43mol%2Fs)
Which is equal to the first reaction extent
, therefore, one computes the second one from the fractional conversion of methane as:
![\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s](https://tex.z-dn.net/?f=%5Cxi_2%3D0.900%2A50mol%2Fs-%5Cxi_1%5C%5C%5Cxi_2%3D0.900%2A50mol%2Fs-43mol%2Fs%5C%5C%5Cxi_2%3D2mol%2Fs)
Now, one computes the rest of the output flows via:
- ![F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s](https://tex.z-dn.net/?f=F_%7BO_2%7D%5E2%3D50mol%2Fs-43mol%2Fs-2%2A2mol%2Fs%3D3mol%2Fs)
- ![F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s](https://tex.z-dn.net/?f=F_%7BCH_4%7D%5E2%3D50mol%2Fs-43mol%2Fs-2mol%2Fs%3D5mol%2Fs)
- ![F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s](https://tex.z-dn.net/?f=F_%7BH_2O%7D%5E2%3D43mol%2Fs%2B2%2A2mol%2Fs%3D47mol%2Fs)
- ![F_{HCHO}^2=43mol/s](https://tex.z-dn.net/?f=F_%7BHCHO%7D%5E2%3D43mol%2Fs)
- ![F_{CO_2}^2=2mol/s](https://tex.z-dn.net/?f=F_%7BCO_2%7D%5E2%3D2mol%2Fs)
The total output molar flow is:
![F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s](https://tex.z-dn.net/?f=F_%7BO_2%7D%2BF_%7BCH_4%7D%2BF_%7BH_2O%7D%2BF_%7BHCHO%7D%2BF_%7BCO_2%7D%3D100mol%2Fs)
Therefore the output stream composition turns out into:
![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](https://tex.z-dn.net/?f=y_%7BCH_4%7D%5E2%3D%5Cfrac%7B5mol%2Fs%7D%7B100mol%2Fs%7D%3D0.05%5C%5Cy_%7BO_2%7D%5E2%3D%5Cfrac%7B3mol%2Fs%7D%7B100mol%2Fs%7D%3D0.03%5C%5Cy_%7BH_2O%7D%5E2%3D%5Cfrac%7B47mol%2Fs%7D%7B100mol%2Fs%7D%3D0.47%5C%5Cy_%7BHCHO%7D%5E2%3D%5Cfrac%7B43mol%2Fs%7D%7B100mol%2Fs%7D%3D0.43%5C%5Cy_%7BCO_2%7D%5E2%3D%5Cfrac%7B2mol%2Fs%7D%7B100mol%2Fs%7D%3D0.02)
Best regards.