In an existing process. 180 kgmol/hofC,H, (P) Is mixed with 420 kgmol/h of a mix

Question

In an existing process. 180 kgmol/hofC,H, (P) Is mixed with 420 kgmol/h of a mixture containing 50% CO (C) and 50% H, (H) and with a recycle stream containing propylene and then fed to a reactor, 30% of the propylene fed to the reactor is consumed. The reactor effluent is sent to a separator. The desired product butanal (B) is removed in one stream, unreacted CO and H2 are removed in a second stream, and unreacted P is recovered and recycled 3 6 Mixer Reactor Separator P, C, H, B P, C, H C, H 2 Suppose you have chosen the input-output diagram as your system. What are the material balance equations? H2 H(cons) H6 P4 0.7 P3 B4 B7 C2 - C(cons) C6 C2 C6 P1 0 P1 = P(cons)

Explanation / Answer

1-

Balance H around mixer

H2 = H3

Balance H around reactor

H3 = H(cons) + H4

Balance H around separator

H4 = H6

H2 = H(cons) + H6

H2 - H(cons) = H6 (correct)

2-

Balance P around reactor

30% P consumed means 70% P unreacted

P4 = 0.7*P3 (correct)

3 -

Balance B around separator

B4 = B7 (correct)

4 -

Balance C around mixer

C2 = C3

Balance C around reactor

C3 = C(cons) + C4

Balance C around separator

C4 = C6

C2 = C(cons) + C6

C2 - C(cons) = C6 (correct)

5 -

C2 = C6 (incorrect)

6 -

P1 must have some value or P1 = P3 - P5

P1 = 0 (incorrect)

7 -

Balance B around separator

B4 = B7

B(gen) = B7 (correct)

8 -

P1 = P(cons) is correct

P3 = 180 kmol

P4 = 180*0.70 = 126 kmol

P4 = P5 = 126 kmol

B4 = B(gen) = 180*0.3 = 54 kmol = P(cons)......... Eq1

B4 = B7 = 54 kmol...... Eq2

P1 + P5 = P3

P1 = 180 - 126 = 54 kmol........ Eq3

From eq3 and eq1

P1 = P(cons) is correct

9 -

From Eq 3 and eq2

P1 = B7 is correct

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