การออกแบบ โรงงานทาง วิศวกรรมเคมี (Chemical Engineering Plant Design) 3(3-0-6)

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การออกแบบ โรงงานทาง วิศวกรรมเคมี (Chemical Engineering Plant Design) 3(3-0-6)

Where

Example 6

Solution

8

CHAPTER 3 ECONOMIC DECISION MAKING DESIGN OF A SOLVENT RECOVERY SYSTEM

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

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

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

DESIGN OF A GAS ABSORBER: FLOWSHEET, MATERIAL AND ENERGY BALANCES, AND STREAM COSTS

DESIGN OF A GAS ABSORBER: FLOWSHEET, MATERIAL AND ENERGY BALANCES, AND STREAM COSTS

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

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.

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

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

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.

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

If pure water is used as the solvent, then x in = From the rules of thumb. and Gas Absorber We can use the Kremser equation and the rules of thumb to understand the effect of the design variables.

Suppose: From Eq But since For an isothermal, dilute absorber, choose COLUMN PRESSURE.

Lower L  Still feed will be more concentrated, 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. COLUMN PRESSURE. will decrease RR Vapor rate in still Column diameter Size of condenser and reboiler Steam and cooling-water requirement

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 upstream of the absorber COLUMN PRESSURE.

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

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

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

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

Back-of-the –envelope Design Equation Apply the order of magnitude criterion (1<<40) Thus

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

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

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. For 99% recovery. Eq predicts 10 tray Vs. the Exact value of For 99.9% recovery  16 tray Vs. the Exact value of In addition. Must calculate: height and diameter.  Appendices A3.

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.

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.

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 Distillation Column

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

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

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 which is the common rule of thumb.

Fractional Recovery in Gas Absorbers For a fixed solvent flow rate, we can always increase the recovery of the solvent simple 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).

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 Suppose we write a total annual cost (TAC) model as 3.4-3

OPTIMUM DESIGN Now, if we use our simplified design equation, Eq we obtain The optimum fractional loss is given by 3.4-5

OPTIMUM DESIGN Now, if we use our simplified design equation, Eq we obtain If we consider some typical values We flind that 3.4-8

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

System approach Versus Unit Operations

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

Absorber(tray)

3.5 SUMMARY, EXERCISES, AND NOMENCLATURE Summary A number of important concepts are presented in this chapter: 1. Process alternatives a. A large number of alternatives can be generated even for simple processes, b. We use shortcut procedures to select the best alternative that we will rigorously, providing that the process 'is profitable. (1) We want to spend as little time as possible getting an answer. (2) We only want to include sufficient accuracy to be able to make a decision (3) We always consider the sensitivity of our calculations. 2. Shortcut design procedures a. Itis reasonable to base process flows on 100% recoveries in separators and base equipment designs on 99.5 %recoveries, at the screening stage of design. b. Order-of-magnitude arguments can be used to simplify design equations.

3.Systems approach a. You should always consider the total problem. b. Changes in the design variables in one unit (absorber) might affect the of some other unit (still), but not the unit under consideration. 4.Rules of thumb-heuristics a. If a raw material is used as a solvent in a gas absorber, consider feeding process through the absorber. b. It is desirable to recover more than 99% of valuable components. c. Choose the solvent flow for an isothermal, dilute gas absorber as L =1.4mG d. Cooling water is available at 90°F from a cooling tower and must be returned tower at 120°F or less. e. Assume a 10  F approach temperature for streams cooled with cooling water. to remember that every rule of thumb has some limitations! 3.5 SUMMARY, EXERCISES, AND NOMENCLATURE

Energy Balances for tbe Acetone Absorber