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PROCESSING TECHNOLOGY 1

PROCESSING TECHNOLOGY 1. Reactive Systems – 3. Aim. To combine knowledge of material balances and reactive systems in order to analyse the materials flows within a complex reactive process, such as a reactor system. Recycle Systems. Not all reactants are converted to products

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PROCESSING TECHNOLOGY 1

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  1. PROCESSING TECHNOLOGY 1 Reactive Systems – 3

  2. Aim To combine knowledge of material balances and reactive systems in order to analyse the materials flows within a complex reactive process, such as a reactor system

  3. Recycle Systems Not all reactants are converted to products - some proportion may be unchanged. The unused reactants can be recycled back into the process.

  4. A B Reactor Separator Recycle Systems

  5. Recycle Systems Two definitions of reactant conversion are needed to describe processes employing recycle:

  6. Overall Process (A) Equation applies to the overall process (A), not the individual reactor

  7. Individual Reactor (B) Equation applies to the overall process (A), not the individual reactor

  8. Worked example The dehydrogenation of ethane C2H6 C2H4 + H2 ethane  ethylene + hydrogen

  9. m1 kmol C2H6 m2 kmol C2H4 m3 kmol H2 p1 kmol C2H6 p2 kmol C2H4 p3 kmol H2 R S 100 kmol C2H6 q1 kmol C2H6 q2 kmol C2H4 q3 kmol H2 r1 kmol C2H6 r2 kmol C2H4 r3 kmol H2 Solution Use 100 kmol feed as basis of calculation.

  10. Solution Overall ethane conversion is 90%, q1 = 10 kmol C2H6

  11. Solution C2H6 C2H4 + H2 For every kmol of C2H6 converted one kmol of C2H4 and H2 is produced. 90 kmol of ethane was converted, therefore q1 = 10 kmol C2H6 q2 = 90 kmol C2H4 q3 = 90 kmol H2

  12. Solution Double-check by conducting a mass balance between the reactant and product streams: Reactant: 100 kmol C2H6: 100 x Mol. Weight = 100 x [ (2 x 12) + (6 x 1) ] 100 x 30 = 3000 kg

  13. Solution Product: 10 kmol C2H6: 10 x 30 = 300 kg 90 kmol C2H4: 90 x [ (2 x 12) + ( 4 x 1) ] = 2520 kg 90 kmol H2: 90 x (2 x 1) = 180 kg Total = 3000 kg

  14. Solution Final output stream composition on a molar basis, 5.2% C2H6 47.4% C2H4 47.4% H2

  15. Solution Consider a balance around the separator unit: C2H6: p1 = q1 + r1 C2H4: p2 = q2 + r2 H2: p3 = q3 + r3 p1 kmol C2H6 p2 kmol C2H4 p3 kmol H2 q1 kmol C2H6 q2 kmol C2H4 q3 kmol H2 S r1 kmol C2H6 r2 kmol C2H4 r3 kmol H2

  16. Solution Recycle stream is composed of 80% of the ethane in the reactor product stream, 25% of the ethylene and 10% of the hydrogen, i.e, r1 = 0.8p1 r2 = 0.25p2 r3 = 0.1p3

  17. Solution C2H6 balance: p1 = q1 + r1 p1 = q1 + (0.8p1) 0.2p1 = q1 But q1=10 kmol, hence, 0.2p1 = 10 p1 = (10/0.2) p1 = 50 kmol C2H6

  18. Solution C2H4 balance: p2 = q2 + r2 p2 = q2 + (0.25p2) 0.75p2 = q2 But q2=90 kmol, hence, 0.75p2 = 90 p2 = (90/0.75) p2 = 120 kmol C2H4

  19. Solution H2 balance: p3 = q3 + r3 p3 = q3 + 0.1p3 0.9p3 = q3 But q3=90 kmol, 0.9p3 = 90 p3 = (90/0.9) p3 = 100 kmol H2

  20. Solution Calculate r1, r2 and r3 r1 = 0.8p1 = 40 kmol C2H6 r1 = 40 kmol C2H6 r2 = 0.25p2 = 30 kmol C2H4 r2 = 30 kmol C2H4 r3 = 0.1p3 = 10 kmol H2 r3 = 10 kmol H2

  21. Solution Calculate m1, m2 and m3 by considering a mass balance on the mixing point,

  22. m1 kmol C2H6 m2 kmol C2H4 m3 kmol H2 R 100 kmol C2H6 r1 kmol C2H6 r2 kmol C2H4 r3 kmol H2 Solution m1 = 100 + r1 m1 = 140 kmol C2H6

  23. m1 kmol C2H6 m2 kmol C2H4 m3 kmol H2 R 100 kmol C2H6 r1 kmol C2H6 r2 kmol C2H4 r3 kmol H2 Solution m2 = r2 = 30 kmol C2H4 m2 = 30 kmol C2H4

  24. m1 kmol C2H6 m2 kmol C2H4 m3 kmol H2 R 100 kmol C2H6 r1 kmol C2H6 r2 kmol C2H4 r3 kmol H2 Solution m3 = r3 m3 = 10 kmol H2

  25. Solution Calculate the recycle ratio and single-pass conversion: Recycle ratio = kmols recycled/kmols fresh feed = (r1 + r2 + r3)/100 = 80/100 = 0.8 kmol/kmol

  26. Solution Single-pass conversion = (m1 - p1)m1 = (140 - 50)(140) = 0.643  64.3%

  27. Purge Systems Problems may arise if the feed stream contains non-reactive species, which are recycled back into the first unit process. The non-reactive species can build up until the process is adversely affected. To avoid this some of the recycle stream is removed as a purge stream.

  28. Recycle Purge D Product M Feed Process S Purge Systems Recycle stream with purge S-separator M-mixer D-divider

  29. Purge Systems Examples of purge systems are: a. Ammonia production. b. Fish farm with recycled waste

  30. Ammonia Production • In the final stages of liquid ammonia production, • 3H2 + N2 2NH3 • the synthesis stream is approximately a 3:1 mixture of hydrogen to nitrogen, with the remainder consisting of inerts such as argon. Upon reaction the ammonia is removed and the unreacted gases (N2, H2 and inerts) are recycled. As argon does not dissolve in liquid ammonia, if the inerts are not purged they can build up in the recycle loop. This build up can lead to dilution of the reactants and slow the process down.

  31. Ammonia Production Purged Inerts Unreacted N2 & H2 Reactor N2, H2 & Inerts Feed Liquified NH3 Product NH3, + unreacted N2, H2 & inerts Condenser

  32. Experimental Fish Farm Water is circulated through nitrification tanks to convert ammonia excreted by the fish to nitrites and nitrates, which are less toxic than ammonia. Inorganic salts, from the food pellets, also build up in the water. It is important to have a purge stream accompanied with a fresh water top-up to reduce the concentration of the nitrites, nitrates and inorganic salts, which left unchecked, would be toxic to the fish.

  33. Experimental Fish Farm Food Fish tank Purge Nitrification Top-up

  34. Bypass Streams A stream that skips one or more stages of the process and goes directly to another downstream stage.

  35. Feed S M Process Bypass stream Recycle Stream With Bypass S-separator M-mixer

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