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Tom [10]
3 years ago
8

In the following scenario, what could the engineers have done to keep the bridge from collapsing?

Engineering
1 answer:
yuradex [85]3 years ago
7 0

Answer:

D

Explanation:

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Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb
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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

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:

- \xi_1 and \xi_2: extent of the reactions (2).

- F_{O_2}^2, F_{CH_4}^2, F_{H_2O}^2, F_{HCHO}^2 and F_{CO_2}^2: Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- F_{O_2}^2=50mol/s-\xi_1-2\xi_2: oxygen mole balance.

- F_{CH_4}^2=50mol/s-\xi_1-\xi_2: methane mole balance.

- F_{H_2O}^2=\xi_1+2\xi_2: water mole balance.

- F_{HCHO}^2=\xi_1: formaldehyde mole balance.

- F_{CO_2}^2=\xi_2: carbon dioxide mole balance.

Thus, the degrees of freedom are:

DF=7unknowns-5equations=2

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}

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}

In such a way, one realizes that the output formaldehyde's molar flow is:

F_{HCHO}^2=0.860*50mol/s=43mol/s

Which is equal to the first reaction extent \xi_1, 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

Now, one computes the rest of the output flows via:

- F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s

- F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s

- F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s

- F_{HCHO}^2=43mol/s

- F_{CO_2}^2=2mol/s

The total output molar flow is:

F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s

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

Best regards.

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Pressure is something that has the pressure that is physical and that causes the pressure is piston-cylinder devices.

During a few real enlargements and compression procedures in piston-cylinder devices, the gases were located to meet the connection PV n = C, wherein n and C are constants.

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