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การออกแบบ โรงงานทางวิศวกรรมเคมี (Chemical Engineering Plant Design) 3(3-0-6)

การออกแบบ โรงงานทางวิศวกรรมเคมี (Chemical Engineering Plant Design) 3(3-0-6). Where. Example 6. Solution. 8. CHAPTER 3 ECONOMIC DECISION MAKING DESIGN OF A SOLVENT RECOVERY SYSTEM. 3.1 PROBLEM DEFINITION AND GENERAL CONSIDERATIONS.

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การออกแบบ โรงงานทางวิศวกรรมเคมี (Chemical Engineering Plant Design) 3(3-0-6)

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  1. การออกแบบโรงงานทางวิศวกรรมเคมี(Chemical Engineering Plant Design) 3(3-0-6)

  2. Where

  3. Example 6

  4. Solution

  5. 8

  6. CHAPTER 3 ECONOMIC DECISION MAKING DESIGN OF A SOLVENT RECOVERY SYSTEM

  7. 3.1 PROBLEM DEFINITION AND GENERAL CONSIDERATIONS • We assume that as part of a process design problem there is a stream containing • 10.3 mol/hr of acetone • 687 mol/hr of air 10

  8. Economic Potential (EP) Base the calculation on complete recovery. 11

  9. question • Which is the cheapest alternative? the solute concentration in a gas stream is less than 5% adsorption is the cheapest process.

  10. 3.2 DESIGN OF A GAS ABSORBER: FLOWSHEET, MATERIAL AND ENERGY BALANCES, AND STREAM COSTS 16

  11. 3.2 DESIGN OF A GAS ABSORBER: FLOWSHEET, MATERIAL AND ENERGY BALANCES, AND STREAM COSTS 17

  12. Material Balances DISTRIBUTION OF COMPONENTS. specified a flowsheet identify the components that will appear in every stream The inlet gas flow to the absorber is given in the problem statement as 10.3 mol/hr of acetone and 687 mol/hr of air. If we use well water as a solvent, then inlet stream is pure water. The gas leaving adsorber will contain air, some acetone and some water. Since water is relatively inexpensive, we neglect this solvent loss in our first calculations

  13. RULES OF THUMB . Of course, we can recover 90, or 99, or 99.9%, or Whatever, of the acetone in the gas absorber, simply by adding more trays to the top of the absorber. The cost of the gas absorber will continue to increase as we increase the fractional recovery, but the value of the acetone lost to the flare system will continue to decrease. Thus, there is an optimum fractional recovery.

  14. It is desirable to recover more than 99 % of all valuable materials (we normally use a 99.5 % recovery as a first guess). For isothermal, dilute absorber, choose L such that L=1.4mG

  15. EXACT MATERIAL BALANCES. With these rules of thumb, it is a straightforward task to calculate the material balances. For the acetone-water system at

  16. 3.3 EQUIPMENT DESIGN CONSIDERATIONS In addition to calculating the sizes of the heat exchangers, we must calculate the size and cost of the absorber and the still. Before we begin any calculations however, we want to understand the cause-and-effect relationships of the design variables and to see whether we can simplify the normal unit-operations models.

  17. Gas Absorber For isothermal, Dilute systems, Kremser Eq. can be used to calculate the No. of theoreticaltray required in the gas absorber 3.3-1

  18. Gas Absorber If pure water is used as the solvent, then xin=0. From the rules of thumb. 3.3-2 and 3.3-3 We can use the Kremser equation and the rules of thumb to understand the effect of the design variables.

  19. COLUMN PRESSURE. Suppose: From Eq. 3.3-3 3.3-2 But since For an isothermal, dilute absorber, choose

  20. COLUMN PRESSURE. Lower L Still feed will be more concentrated, RR Vapor rate in still Column diameter Size of condenser and reboiler Steam and cooling-water requirement will decrease Thus, decreasing the solvent flow to the gas absorber will have significant effect on the design still, but no effect on No. of tray required in the absorber.

  21. COLUMN PRESSURE. The absorber diameter will decrease (because of the density effect and smaller liquid load), Feed gas compressor will be required to obtain the increased pressure Since gas compressors : Most expensive type of processing equipment In some cases, a high pressure can be obtained by pumping liquid stream to high pressure somewhere upstreamof the absorber

  22. EFFECT OF SOLVENT. For acetone-water system, m given by If we use MIBK as solvent (ideal mixture with acetone, So that =1) Then still cost From Eq. 3.3-3 However, from Eq. 3.3-1 No. of plate does not change

  23. EFFECT OF OPERATING TEMPERATURE. If we change the inlet water temp. to 40C Then  = 7.8 and P = 421 mmHg. Thus, from Eq. 3.3-3 L Then still cost However, from Eq. 3.3-12 No. of plate does not change

  24. Back-of-the –enverlope Design Equation We expect that absorber will contain 10-20 tray 3.3-5 It change the result <10% For pure solvents, xin=0, and the numerator of the RHS(right-hand side) becomes 3.3-6

  25. Back-of-the –envelope Design Equation The rule of thumb indicate that 3.3-7 and Air=687 moll/hr Acetone=0.0515 mol/hr Thus absorber Air=687 moll/hr Acetone=10.3 mol/hr

  26. Back-of-the –envelope Design Equation Apply the order of magnitude criterion (1<<40) 3.3-8 Thus 3.3-9

  27. Back-of-the –envelope Design Equation The denominator of RHS. of Eq.3.3-1, ln(L/mG) can be 3.3-10 From Taylor series expansion, we can write 3.3-11

  28. Back-of-the –envelope Design Equation From Eq.3.3-1 3.3-1 With these simplifications, and replacing ln by log. We obtain 3.3-12

  29. Back-of-the –envelope Design Equation Within a 10% error, 2.3/0.4=6 and (2.3log0.4)/0.4=-2. Hence, simplified version of the Kremser Eq. becomes. 3.3-13 For 99% recovery. Eq. 3.3-13 predicts 10 tray Vs. the Exact value of 10.1. For 99.9% recovery  16 tray Vs. the Exact value of 16.6. In addition. Must calculate: height and diameter. Appendices A3.

  30. Distillation Column

  31. Distillation Column To separate acetone from the solvent water. Use a McCabe-Thiele diagram Number of tray Calculate : • Still diameter • Condenser and reboiler size • Steam and cooling-water loads.  Appendices A2 and A3. http://lorien.ncl.ac.uk/ming/distil/distildes.htm

  32. Distillation Column Relative Volatility Relative volatility is a measure of the differences in volatility between 2 components, and hence their boiling points. It indicates how easy or difficult a particular separation will be. The relative volatility of component ‘i’ with respect to component ‘j’ is defined as yi = mole fraction of component ‘i’ in the vapour xi = mole fraction of component ‘i’ in the liquid Thus if the relative volatility between 2 components is very close to one, it is an indication that they have very similar vapour pressure characteristics. This means that they have very similar boiling points and therefore, it will be difficult to separate the two components via distillation.

  33. Distillation Column • Thus, for the distillation of any multi-component mixture, the relative volatility is often defined as • Large-scale industrial distillation is rarely undertaken if the relative volatility is less than 1.05.

  34. Liquid flow Rate to Gas Absorbers For isothermal, dilute gas absorbers,the Kremser Eq.3.3-1 Used to calculate the number of tray as a function of L/(mG) is shown in Fig.3.4-1 If L/mG) < 1, never get close to complete recovery of the solute even if we use an infinite number of plates

  35. Liquid flow Rate to Gas Absorbers if we choose L/(mG) =2 , we obtain essentially complete recovery with only five plates. But large solvent rates correspond to dilute feeds to the distillation column RR,vapor rate , Column diameter size, condenser and reboiler size , steam and cooling water  Base on these argument, we find that we want to choose L such that 3.4-1

  36. Liquid flow Rate to Gas Absorbers Of course, L/(mG) = 1.5 is right in the middle of this range. However, if we inspect the shape of the curves near L/(mG) = 1.5 and with high recoveries, we see that might obtain a better trade-off between a decreasing number of plates required in the absorber (capital cost) and the increasing capital and operating costs of the distillation column by decreasing L. Hence, as a first guess, it seem to be reasonable to choose L such that 3.4-2 which is the common rule of thumb.

  37. Fractional Recovery in Gas Absorbers For a fixed solvent flow rate, we can always increase the recovery of the solventsimple by: -adding trays in the gas absorber. Hence, there is an economic trade-off between an increasing absorber cost as we add trays versus a decreasing value of the solute lost. One of these is a capital cost (the absorber), and one is an operating cost (the solute loss).

  38. COST MODEL. It is common practice to report operating costs on an annual Thus, to examine the economic trade-off, we must also put the capital cost on annualized basis. As discussed in Sec. 2.5, we annualize the capital cost by a capital charge factor (CCF) of 1/3 yr, where the CCF includes all capital-related expenses (depreciation, repairs and maintenance, etc.). A CCF of 1/3 yr corresponds to about a 15% discounted-cash-flow rate of return (DCFROR); Eq. 2.5.13. Suppose we write a total annual cost (TAC) model as 3.4-3

  39. OPTIMUM DESIGN Now, if we use our simplified design equation, Eq. 3.3-12 weobtain 3.4-4 The optimum fractional loss is given by 3.4-5

  40. OPTIMUM DESIGN Now, if we use our simplified design equation, Eq. 3.3-12 weobtain 3.4-6 If we consider some typical values 3.4-7 We flind that 3.4-8

  41. OPTIMUM DESIGN We flind that 3.4-8 Which corresponds to Fractional recovery=99.6% 3.4-9

  42. System approach Versus Unit Operations

  43. Counter-Current Gas Absorption Refer to the Figure below for a dilute system Notations : In terms of mole fraction and total flowrates y : mole fraction of solute A in the gas phasex : mole fraction of solute A in the liquid phaseG : total molar flowrate of the gas stream (gas flux), kg-moles/m2.sL : total molar flowrate of the liquid stream, kg-moles/m2.s

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