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THE STRUCTURE AND SYNTHESIS OF PROCESS FLOW DIAGRAMS

THE STRUCTURE AND SYNTHESIS OF PROCESS FLOW DIAGRAMS. Design process follows a series of decisions and steps Decide whether the process will be batch or continous Identify the input-output structure of the process Identify and define the recycle structure of the process

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THE STRUCTURE AND SYNTHESIS OF PROCESS FLOW DIAGRAMS

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  1. THE STRUCTURE AND SYNTHESIS OF PROCESS FLOW DIAGRAMS

  2. Design process follows a series of decisions and steps • Decide whether the process will be batch or continous • Identify the input-output structure of the process • Identify and define the recycle structure • of the process • Identify and design the general structure of • the separation system • Identify and design the heat exchanger network or • process energy recovery system

  3. STEP 1- BATCH VS CONTINOUS PROCESSES

  4. Some factors to consider when deciding between batch and continous processes. Size Product quality Operational flexibility Standardized equipment-multiple products Processing efficiency Maintenance and operating labor Feedstock availability Product demand Rate of reaction Equipment fouling Safety Controllability

  5. STEP 2-INPUT AND OUTPUT STRUCTURE OF THE PROCESS

  6. PROCESS CONCEPT DIAGRAM Figure 1. Input and output structure of the process concept diagram for toluene hydrodealkylation process

  7. The steps used to form PCD • A single cloud is drawn- • stochiometry for all reactions • 2. Reactant chemicals are drawn as streams entering from the left • numbering and naming of the streams • 3. Product chemicals are drawn as streams leaving to the right • numbering and naming of the streams • 4. Seldom does a single reaction occur. Unwanted side reactions • should be considered. • All reactions and reaction stochiometry should be included. • By-products should be shown.

  8. The input and output structure of the process flow diagram Figure 2.2.Input Output Streams on Toluene Hydrodelkylation PFD

  9. Figure 2.3.Identification of Utility Streams on the Toluene Hydrodealkylation

  10. Several important factors to consider in analyzing the • input-output structure of a PFD • Chemical entering the PFD from the left that are not consumed • in the chemical reactor are either required to operate a piece of • equipment or are inert material that simply pass through the • process. Catalyst make-up, solvent make-up, inhibitors • 2. Any chemical leaving a process must have either entered • in one of the feed streams or have been produced by chemical • reaction in the process. • 3. Utility streams are treated differently from process streams. • (i.e., cooling water, steam, fuel and electricity).

  11. GENERIC BLOCK FLOW DIAGRAM

  12. “Generic Block Flow Diagram” is intermediate between the process concept diagram and the PFD.

  13. The elements of the “Generic Block Flow Process Diagram” • Reactor feed preparation • Reactor • Separator feed preparation • Separator • Recycle • Enviromental control

  14. Figure 2.4.a. The Six Elements of the Generic Blok Flow Process Diagram

  15. Figure 2.4.b. A Process Requring Multiple Process Blocks

  16. Other Considerations for the Input-Output Structure of the Process Flowsheet Feed purity and trace components CO + Cl2 COCl2 CH4 + H2O CO + 3H2 Addition of feeds required to stabilize products or enable separations

  17. Inert feed material to control exothermic reactions C3H6 + 1.5 O2 C3H4O2 + H2O C3H6 + 2.5 O2 C2H4O2 + H2O + CO2 C3H6 + 4.5 O2 3H2O + 3CO2 Addition of inert feed material to control equilibrium reactions C6H5CH2CH3 C6H5CH = CH2 + H2

  18. What information can be determined using the input-output diagram of the process Basic economic analysis of profit margin What chemical components must enter with the feed and leave as products All the reactions both desired and undesired that take place

  19. Example 2.1. Evaluate the profit margin for HDA process From Tables 6.3 and 6.4 Benzene = $0.349/kg Toluene = $0.322/kg Natural gas = $6.00/GJ = $6.7/1000 std.ft3= $0.293/kg Hydrogen = $0.721/kg Using 1 kmol of toluene as feed basis Cost of raw materials 92 kg of Toluene = 92 (0.322) = $29.62 2 kg of hydrogen = 2 (0.721) = $1.44 Value of products 78 kg of benzene = 78 (0.349) = $27.22 16 kg methane = 16 (0.293) = $4.69 Profit margin = (27.22 + 4.69) –(29.62+1.44) = $0.85/kmol toluene or $ 0.85/92 = $0.0092/kg toluene

  20. Step 3-The Recycle Structure of the Process Efficiency of Raw Material Usage Single-pass conversion = reactant consumed in reaction/reactant fed to reactor Overall conversion = reactant consumed in the process/reactant fed to process Yield = Moles of reactant to produce desired product/moles of limiting reactant reacted

  21. Identification and definition of the recycle structure • of the process • There are basically three ways that unreacted raw • materials can be recycled in a continuous process • Separate and purify unreacted feed material from • products and recycle • 2. Recycle feed and product together and use a purge • stream • 3. Recycle feed and product together and do not use • a purge stream

  22. Recycle structure of the hydrogen stream in toluene hydrodelkylation process Methane is purged from the system via stream 16.

  23. Example 2.3. For the separation of methane and hydrogen • Evaluate different alternatives • Distillation • Normal boiling point of methane = -161oC • Nornal boiling point of hydrogen = -252oC • Separation should be easy with distillation due to the large • difference in b.p.s of two components. However, high pressure • and low temperature are necessary to get a liquid phase. • b. Absorption • It might be possible to absorb methane into a hydrocarbon liquid. • However, high pressure and low temperature are necessary to • obtain an efficient absorption.

  24. c. Pressure-Swing adsorption In the PSA processs, preferential adsorption of one species from the gas phase occurs at a certain pressure and the desorption of the adsorbed species is facilitated by reducing the pressure to degas from the adsorbent. This separation could be applied for the preferential adsorption of methane. d. Membrane Separation This separation is facilitated because hydrogen passes more readily through certain membranes than does methane. This occurs at moderate pressures. This separation could be applied for the preferential recovery of hydrogen at a fairly low pressure.

  25. Example 2.4. What process should be used for the separation of benzene and toluene Distillation Normal boiling point of benzene = 79.8oC Normal boiling point of toluene = 110oC Separation should be easy using distillation neither high temperature nor high pressure will be neeed.

  26. Example 2.5. Consider the following two process alternatives for the toluene HDA process when the side reaction of benzene to form diphenyl occurs. C7H8 + H2 C6H6 + CH4 Toluene Benzene 2C6H6 C12H10 + H2 Benzene Diphenyl Ln Keq = 1.788 – 4135.2/ T(K) Keq = [C12H10][H2]/[C6H6]2 Exp[1.788- 4135.2/(654 + 273)] = x (652.6 +x)/ (116-2x)2 x = 1.36 kmol/h

  27. Calculation of Keq Stream 9 (kmol/h), Table 1.5 Hydrogen: 652.6 Methane : 442.3 Benzene : 116.0 Toluene : 36.0 Outplet(kmol/h) Outlet with diphenyl(kmol/h) Hydrogen: 652.6 652.6 + x Methane : 442.3 442.3 Benzene : 116.0 116 -2x Toluene : 36.0 36.0 Diphenyl : - x TOTAL : 1246.9 1246.9

  28. yDiphenyl = x/1246.9 yHydrogen = (652.6 + x)/1246.9 yBenzene = (116 - 2x)/1246.9 Dalton’s Law PDiphenyl = yDiphenyl. PT= [x/1246.9]. PT PHydrogen = yHydrogen. PT = [(652.6 + x)/1246.9] PT PBenzene = yBenzene. PT= [(116 - 2x)/1246.9]. PT KP = PDiphenyl.PHydrogen/PBenzene2 KP= [x/1246.9] PT [(652.6 + x)/1246.9] PT/ [(116-2x)/1246.9]2.PT2

  29. Ln Keq = 1.788 – 4135.2/ T(K) Keq = KA KA = Equilibrium constant based on activities KA = Kc PiV = ni RT Ci = ni/V Pi = CiRT Ci = Pi/RT Kc = CDiphenyl.CHydrogen/CBenzene2 Kc = [PDiphenyl/RT] [PHydrogen/RT]/ [Pbenzene/RT]2 KP = PDiphenyl.PHydrogen/PBenzene2

  30. PFD for alternative A in Example 2.5. Recycle of diphenyl without separation

  31. PFD for alternative B in Example 2.5. Recycle of diphenyl with separation

  32. Example 2.6. Consider the conversion of a mixed feed stream of methanol (88 mol%), ethanol (11 mol%), and water (1 mol%) via the following dehydration reactions: Compounds leaving the reactor form a variety of azeotropes:- DME-H2O - DME-EtOH - DEE-EtOH - DEE-H2O - EtOH-H2O

  33. Example 2.6. (cont’d) Mixed alcohol stream = $0.25/kg Methanol = $0.22/kg Ethanol = $0.60/kg Selling price; DME= $0.95/kg DEE = $1.27/kg Ethylene= $0.57/kg Primarily market survey indicates that 15,000 tonne/y DEE 10,000 tonne/y ethylene can be sold. For a proposed process to produce 50,000 tonnes/y of DME, determine what are the viable process alternatives?

  34. Example 2.6. (cont’d) Mixed alcohol stream = $0.25/kg Methanol = $0.22/kg Ethanol = $0.60/kg Selling price; DME= $0.95/kg DEE = $1.27/kg Ethylene= $0.57/kg Primarily market survey indicates that 15,000 tonne/y DEE 10,000 tonne/y ethylene can be sold. For a proposed process to produce 50,000 tonnes/y of DME, determine what are the viable process alternatives?

  35. Example 2.6. (cont’d) Material balance for the process and estimating the profit margain; Desired DME production= 50,000,000 kg/y= 50x106/46 = 1.087x106 kmol/y Required MeOH feed= (2)(1.087x106) = 2.174x106 kmol/y EtOH feed entering with methanol= (2.174x106/88)(11) = 0.2718x106 kmol/y Maximum DEE production=0.2718x106 /2=0.1309x106 kmol/y 9.69x103 tonne/y Maximum Ethylene Production= 0.2718x106 kmol/y or 7.61x103 tonne/y Cost of Feed= [(2.174x106)(32)+(0.2718x106)(46)+((2.174x106)(18))/88](0.25) = 19.54x106 Value of DME= (50x106)(0.95)= $47.5x106/y Value of DEE (max production)= (0.1309x106)(74)(1.27)= $12.30x106/y Value of Ethylene (max production)= (0.2718x106)(28)(0.57)= $4.34x106/y Margin will vary between (47.5+12.3-19.54)=$40.26 million and (47.5+4.34-19.54)=$32.30 million per year

  36. Structure of process for alternative 1

  37. Structure of process for alternative 2

  38. Structure of process for alternative 3-DME Campaign

  39. Structure of process for alternative 3-DEE Campaign

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