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Sati [7]
3 years ago
7

A 100 kmol/h stream that is 97 mole% carbon tetrachloride (CCL) and 3% carbon disulfide (CS2) is to be recovered from the bottom

of a distillation column. The feed to the column is 16 mole% CS2 and 84% CC14, and 2% of the CC14 entering the column is contained in the overhead stream leaving the top of the column. (a) Draw and label a flowchart of the process and do the degree-of-freedom analysis. (b) Calculate the mass and mole fractions of CCl4 in the overhead stream, and determine the molar flow rates of CCl4 and CS2 in the overhead and feed streams. (c) Suppose the overhead stream is analyzed and the mole fraction of CS2 is found to be significantly lower than the value calculated in Part (b). List as many reasons as you can for the discrepancy, including possible violations of assumptions made in Part (b).
Engineering
1 answer:
Ugo [173]3 years ago
7 0

Answer: oop

Explanation:oop

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Kisachek [45]

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6 0
3 years ago
Show the ERD with relational notation with crowfoot. Your ERD must show PK, FKs, min and max cardinality, and correct line types
zhenek [66]

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3 0
3 years ago
A 3-oz serving of roasted, skinless chicken breast contains 140 Cal, 27 g of protein, 3 g of fat, 13 mg of calcium, and 64 mg of
Art [367]

Answer:

Explanation:

  • a) Given C [ cal pro fat calc sod]

              [140 27 3  13  64]

  • P = [cal pro fat calc sod]

     [180 4 11 24 662]

  • B = [cal pro fat calc sod]

      [50  5    1  82   20]

To find C+2P+3B = [140 27 3  13  64] + 2[180 4 11 24 662] + 3[50  5    1  82   20]

= [650  54  28  307  1448]

The entries represent skinless chicken breast , One-half cup of potato salad and One broccoli spear.

7 0
3 years ago
The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8
il63 [147K]

Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×\frac{100}{w}

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = \frac{D60}{D10}

⇒ Cu = \frac{0.4}{0.12} = 3.33

d.)

Coefficient of Graduation = Cc = \frac{D30^{2}}{D10 . D60}

⇒ Cc = \frac{0.22^{2}}{(0.4) . (0.12)} = 1

3 0
2 years ago
Hęłłõ hõw årę ÿõū dõìńg tõdāÿ
kirill [66]

Answer:

All good :)

What about you??

8 0
2 years ago
Read 2 more answers
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