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Stolb23 [73]
3 years ago
11

I need help on the Coderz Challenge missions 3 part 3. PLEASE HELP!

Engineering
2 answers:
allsm [11]3 years ago
6 0

Answer:

the answer how you analyzs the problwm

romanna [79]3 years ago
4 0

Answer:

To access a missions, simply log on to your CoderZ account and navigate to one of the courses you are assigned, choose a pack, and then choose one of the missions!

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I don't know plead hrdffffdddffff

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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
Nezavi [6.7K]

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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AVprozaik [17]

Answer:

Mechanical engineer? Thats my guess I didnt have alot of options sorry if I am wrong

Explanation:

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A __________ is an added note showing additional or more specific information.
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If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I sen
Mice21 [21]

Answer:

32000 bits/seconds

Explanation:

Given that :

there are 16  signal combinations (states) = 2⁴

bits  n = 4

and a baud rate (number of signals/second) = 8000/second

Therefore; the number of bits per seconds can be calculated as follows:

Number of bits per seconds = bits  n × number of signal per seconds

Number of bits per seconds =  4 × 8000/second

Number of bits per seconds = 32000 bits/seconds

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