CONVERSION AND REACTOR SIZING

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Solution: By trial and error, we find that a conversion of 0.6 gives the appropriate CSTR volume of 1200 dm 3 . Therefore, the intermediate conversion is X = 0.6 Similarly for the PFR, through trial and error, we find that a conversion of 0.8 gives the appropriate PFR volume of 600 dm 3 . Therefore, the final conversion is X = 0.8 14
CSTRs in Series Two CSTRs in series F A0 -r A1 F A1 X 1 = 0.4 -r A2 A2 F A2 X 1 = 0.8 Reactor 1: Mole Balance: FA0 –FA1 + rA1 V1 = 0 [1] FA1 = FA0 –FA0 X1 Combining [1] & [2]: [2] Reactor 2: V 1 = F A0 (1 / -r A1) X 1 [3] Mole Balance: F A1 –F A2 + r A2 V 2 = 0 [4] Combining [4] & [5]: A2 F A2 = F A0 – F A0 X 2 [5] FA1 − FA2 ( FA0 − FA0 ⋅ X1) − ( FA0 − FA0 ⋅ X 2) FA0 V2 = = = ( X 2 − X1) − r − r − r if we have the data: X = 0. 4 ⎛ F ⎜ ⎝ − ⎞ ⎟ = ⎛ = ⎜ A0 3 A0 2. 05m V1 X1 0. 82 r ⎟ ⎜ 1 r ⎟ ⋅ = A ⎠ X 0. 4 ⎝ − = A1 ⎠0. 4 ⎛ F ⎞ 3 ⎛ F ⎞ 0 F ⎞ A A0 X = 0. 8 ⎜ 8. 0m V ⎜ 2 ⋅( X 2 − X1) = 3. 2m r ⎟ = = ⎜ A2 r ⎟ ⎝ − ⎠ ⎝ − A2 ⎠ -F A0/r A 0.4 0.8 X = 0. 8 Total V = V 1 + V 2 = 4.02 m 3 < 6.4 m 3 X F ⎞ ⎟ V CSTR,2 > V CSTR,1 A2 m volume necessary to get 80 % conversion with one CSTR 3 3 15
Page 1 and 2: CONVERSION AND REACTOR SIZING • O
Page 3 and 4: 2. Design Equations Batch Reactor D
Page 5 and 6: For liquid systems, CA0 is usually
Page 7 and 8: A particularly simple functional de
Page 9 and 10: if FA0 = 0.4 mol/s, we can calculat
Page 11 and 12: Hint : FA0 = 2 mol/s, fed to a plug
Page 13: Reactors in Series: The exit of one
Page 17 and 18: Use the plot of ⎛ F ⎞ A 0 ⎜ v
Page 19 and 20: Summary: CSTR X1 = 0.4 V1 = 120 dm3
Page 21 and 22: Space Velocity (SV) is defined as:
Page 23 and 24: Mass balance for penicillin: In −

CONVERSION AND REACTOR SIZING

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