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~~TI:.R 5: Interaction between Steady-State Design and Dynamic Controllability 169 I.6 Q 1.5 74 OfiE 1.4 1.3 / ,;’ F = 50, U = 100, AH, = 20.000, E, = 30,000, E2 = 15,000 1.2 . . . . . . . . I.1 50 100 I50 200 250 300 350 400 450 0.75 0.7 0.65 s 0.6 0.45 0.4 ** . . . * ‘I ‘, F = 50, (I = 100, AHH= 20,000, E, = 30,000, Ez = 15,000 . . . . ' x. ' , '%, /@ "'-i-,- 'x5 , / / . . . . -\ '. / X, I/ '*'-y,.,.c ' , I A Reactor holdup (lb-mol) _. 50 100 I50 200 250 300 350 400 450 Reactor holdup (lb-mol) FIGURES.6 Consecutive reactions. 170 PART ONE: Time Domain Dynamics and Control Ternary reactor, two-column process (a = 4/2/I) RECYCLE v,= 150 T k,= 1 k,= 1 D, = 4.51 F=324 z,., = 0.6600 zB = 0.2324 zc=0.1076 FIGURE 5.7 J D,, = 4.42 D, = 224 XD1.A = 0.95 = 0.05 II I, I B,=lOO XB,,A = 0.0100 XB1.B = 0.6413 xBI,C = 0.3487 I D,, = 2.83 D, = 65.4 X[&, = 0.0 I53 XD2.H = 0.9147 Xlj2.C = 0.0 100 cC$- B,=94.6 I I, I +2,B = 0.0 1 XB2.C = 0.99 Ternary reactor, two-column process (a = 4/2/l). the reactor effluent requires a recycle of this component back to the reactor. Much higher yields can be obtained from this type of recycle system. However, the reactor still encounters controllability limitations. Figure 5.7 is a sketch of the plant under consideration. Fresh feed enters the reactor at a flow rate FO and composition z()A = 1 (pure component A in the fresh feed). We assume that the relative volatilities of components A, B, and C, CYA/(YB/QC, are 4/2/l, respectively, so unreacted component A comes overhead in the first distil- lation column and is recycled back to the reactor at a rate Dr and composition XD~ . . Reactor effluent F is fed into the first distillation column. The flow rates of reflux and vapor boilup in this column are Rt and VI. Bottoms Bt from the first column is fed into the second column, in which components B and C are separated into product streams with about 1 percent impurity levels. Steady-state component balances around the whole system and around each of the units are used to solve for the conditions throughout the plant for a given recycle flow rate Dr. The reactor holdup VR and the reactor temperature TR necessary to achieve a specified &,JQ ratio are calculated as part of the design procedure. The other fixed design parameters are the kinetic constants (preexponential factors, activation energies, and heats of reaction for both reactions), the fresh feed flow rate and composition, the overall heat transfer coefficient in the reactor, the inlet coolant - 5.4 II153 .9747 .OlOO h )r le h 37 l- . I X is :t ,f le LO e. s, te nt CHAITEK 5: Interaction between Steady-State Design and Dynaniic Controllability 171 temperature, and the following product purity specifications: XDI,C = 0 XDI,[j = 0.05 X[jI,A = 0.01 XD2.C = 0.01 X,g7J = 0.01 XB2.A = 0 The design procedure is as follows: 1. Set Dt. There is an optimal value of this recycle flow yield of B. 2. Calculate F = Fo + Dt and Bt = Fo. 3. Calculate ZA. rate that maximizes the 4. Calculate the product of VR and kt . [vRkl,l = FO(ZO,A - xf?l,A)/ZA 5. Guess a value of reactor temperature TR. a. Calculate kt and k2. b. Calculate VR from the [V~ki] product calculated in Eq. (5.35). c. Calculate the diameter DR and length LR of the reactor. d. Calculate the circumferential heat transfer area of the jacket. e. Calculate zg. (5.34) (5.35) ZB = VRhZA - DPDI,B F + VRk2 (5.36) f. Calculate Q, TJ, FJ, and Qmax. g. If the QmaX/Q ratio is not equal to the desired value, reguess reactor tempera- ture. 6. Calculate remaining flow rates and compositions in the columns from component balances. XBI,B = V’ZB - DPDLBYBI (5.37) xBI,C = 1 - XBI,A - XBl,B B 2 = BI(l - xD2,C - Xf?I,A - %B) l-x (5.38) D2,C - XB2,B D2 = BI - B2 (5.39) XD2,A = BIXBI,AID~ (5.40) 7. Size the reactor and the distillation columns (using 1.5 times the minimum num- ber of trays and 1.2 times the minimum reflux ratio for each column). 8. Calculate the capital cost, the energy cost, and the total annual cost. Table 5.3 gives detailed results of these calculations for several feed rates, and ‘Fig. 5.8 shows some of the important results. I72 PART ONE : Time Domain Dynamics and Control TABLE 5.3 Reactor-column designs FO 40 50 60 Reactor opt D, TR VR 2 ctl TJ FJ AH DR 20 17 I5 126.76 130.63 I 34.90 431.9 593.6 744. I 0.3233 0.2485 0. I980 0.5325 0.5514 0.5548 95.22 96.94 98.85 66.43 82. I2 95.46 265.6 328.3 381.7 6.50 7.23 7.79 837.4 II06 1376 1256 1659 2064 Column 1 NTINF 1517 1517 1516 BI 40 50 60 DI 20 I7 I5 RI 59.0 63.6 68.5 VI 79.0 80.6 83.5 Dct 1.63 1.65 1.68 Column 2 &-/NF 19/12 19112 19/10 & 8.43 13.17 18.30 DZ 31.57 36.83 11.70 Rl 44.0 56.0 67.85 VZ 75.6 92.8 109.6 Dcz 1.60 1.77 1.92 Energy 1.93 2.17 1.41 costs Reactor 110.3 134.5 154.8 Column 1 57.3 57.9 59.0 Column 2 67.6 75.5 82.4 HtEx. 141.3 152.5 163.5 Capital 379.3 123.2 162.8 Total annual cost 211.1 236.0 760.0 Yield 78.93 73.66 69.85 xV~‘otes: Total annual cost = JlOOO/\;r, capital cost = S 1000. diam- ~trr = fast. rnerg = IO’ Btu/hr. composition = mole fraction, How rate = lb-mol/hr. holdup = lb-mol 1. There is an optimal recycle flovv. rate for a given feed rate that maximizes yield. 2. The maximum attainable yield in the reactor-column recycle process is higher than for just a reactor for the same controllability. The reactor alone with a feed rate of 50 lb-mol/hr gives a maximum yield of about 63 percent for a QiltaX/Q ratio of 1.5. For the same ratio. the reactor-column system with the same fresh CHAYTEK s: Interaction between Steady-State Design and Dynamic Controllability 173 80 78 76 68 66 64 , , , , , / / . *.= .a : 180 170 160 4 Ratio =‘I .5, U = ;OO, AH,: 20,000: E, = 30,;00, E2 = '15,000 I I 10 12 14 16 18 20 22 24 D, recycle (lb-mol/hr) Ratio = 1.5, U = 100, AH, = 20,000, E, = 30,000, E, = 15,000 " ', 60 F=40 . ‘. ‘,50 *. \ * \ 6 8 IO 12 14 16 18 20 22 24 0, recycle (lb-mol/hr) FIGURE 5.8 CSTR-column design. 174 PARTONE: Time Domain Dynamics and Control 1600 I I I I:/ IAnn I ! ! ! ! I ! : 1 1200 I Ratio = 1.5, (I = 100, AH, = 20,000, E, = 30,000,E~ = 15,000 I I 4 6 8 10 12 14 16 18 20 22 24 D, recycle (lb-mol/hr) FIGURE 5.8 (CONTINUED) CSTR-column design. feed rate gives a yield of 73.5 percent. Of course, the energy and capital costs are higher. 3. The maximum yield depends strongly on the feed flow rate. The last result suggests that we may want to modify the process to achieve better yields but, at the same time, maintain controllability. This dan be done by increasing the heat transfer area in the reactor. 5.5 GENERAL TRADE-OFF BETWEEN CONTROLLABILITY AND THERMODYNAMIC REVERSIBILITY The field of engineering contains many examples of trade-offs. You have seen some of them in previous courses. In distillation there is the classical trade-off between the number of trays (height) and the reflux ratio (energy and diameter). In heat transfer there is the trade-off between heat exchanger size (area) and pressure drop (pump or compressor work); more pressure drop gives higher heat transfer coefficients and smaller areas but increases energy cost. We have mentioned several trade-offs in this book: control valve pressure drop versus pump head, robustness versus perfor- mance, etc. The results from the design-control interaction examples discussed in previous sections hint at the existence of another important trade-off: dynamic controllability versus thermodynamic reversibility. As we make a process more and moie efficient u-e ter w ne he fer “P nd in jr- US ity :nt UIAWW 5, Interaction hcfwccn Slcady-State Design and Dynamic Controllability 175 (reversible in a thermodynamic sense), we are reducing driving forces, i.e., pressure drops, temperature differences, etc. These smaller driving forces mean that we have weaker handles to manipulate, so that it becomes more difficult to hold the process at the desired operating point when disturbances occur or to drive the process to a new operating point. Control valve pressure drop design illustrated this clearly. Low-pressure-drop designs are more efficient because they require less pump energy. But low-pressure- drop designs have limited ability to change the flow rates of manipulated variables. The jacketed reactor process also illustrates the principle. The big reactor has a lot of heat transfer area, so only a fraction of the available temperature difference between the inlet cooling water and the reactor is used. A thermodynamically re- versible process has no temperature difference between the source (the reactor) and the sink (the inlet cooling water). So the big reactor is thermodynamically inefficient, but it gives better control. We could cite many other examples of this controllability/reversibility trade-off, but the simple ones mentioned above should convey the point: the more efficient the process, the more difficult it is to control. This general concept helps to explain in a very general way why the steady-state process engineer and the dynamic control engineer are almost always on opposite sides in process synthesis discussions. 5.6 QUANTITATIVE ECONOMIC ASSESSMENT OF STEADY-STATE DESIGN.AND DYNAMIC CONTROLLABILITY One of the most important problems in process design and process control is how to incorporate dynamic controllability quantitatively into conventional steady-state design. Normally, steady-state economics considers capital and energy costs to cal- culate a total annual cost, a net present value, etc. If the value of products and the costs of raw materials are included, the annual profit can be calculated. The process that minimizes total annual cost or maximizes annual profit is the “best” design. However, as we have demonstrated in our previous examples, this design is usually not the one that provides the best control, i.e., the least variability of product quality. What we need is a way to incorporate quantitatively (in terms of dollars/year) this variability into the economic calculations. We discuss in this section a method called the capacity-based approach that accomplishes this objective. It should be emphasized that the method provides an analysis tool, not a synthesis tool. It can provide a quantitative assessment of a proposed flowsheet or set of parameter values or even a proposed control structure. But it does not generate the “best” flowsheet or parameter values; it only evaluates proposed systems. 5.6.1 Alternative Approaches A. Constraint-based methods The basic idea behind constraint-based approaches is to take the optimal steady- state design and d&r-mine how far away from this optimal point the plant must i 76 hucr aa’.: Time Domain Dynamics and Control operate in order not to violate constraints during dynamic upsets. The steady-state economics are then calculated for this new operating point. Alternative designs are compared on the basis of their economics at their dynamically limited operating point. This method yields realistic comparisons, but it is computationally intensive and is not a simple, fast tool that can be used for screening a large number of alter- native conceptual designs. B. Weighting-factor methods With weighting-factor methods, the basic idea is to form a multiobjective opti- ‘ mization problem in which some factor related to dynamic controllability is added to the traditional steady-state economic factors. These two factors are suitably weighted, and the sum of the two is minimized (or maximized). The dynamic con- trollability factor can be some measure of the “goodness” of control (integral of the squared error), the cost of the control effort, or the value of some controllability measure (such as the plant condition number, to be discussed in Chapter 9). One real problem with these approaches is the difficulty of determining suitable weighting factors. It is not clear how to do this in a general, easily applied way. 5.6.2 Basic Concepts of the Capacity-Based Method The basic idea of the capacity-based approach is illustrated in Fig. 5.9 for three plant designs. The dynamic responses of these three hypothetical processes to the same set of disturbances can be quite different. The variable plotted indicates the quality of the product stream leaving the plant. Better control of product quality is achieved in plant design 3 than in the other designs. Suppose the dashed lines in Fig. 5.9 indicate the upper and lower limits for “on- aim” control of product quality. Plant 3 is always within specification, and there- fore all of its production can be sold as top-quality product. Plant 1 has extended periods when its product quality is outside the specification range. During these pe- riods the production would have to be diverted from the finished-product tank and sent to another tank for reworking or disposal. This means that the capacity of plant 1 is reduced by the fraction of the time its products are outside the specification range. This has a direct effect on economics. Thus, the three plant designs can be directly and quantitatively compared using the appropriate capacity factors for each plant. The annual profit for each plant is calculated by taking the value of the on- specification products and subtracting the cost of reprocessing off-specification ma- terial, the cost of raw materials, the cost of energy, and the cost of capital. Plant 3 may have higher capital cost and higher energy cost than plant 1, but since its product is on-specification all the time, its annual profit may be higher than that of plant 1. Two approaches can be used to calculate the capacity factors (the fraction of time that the plant is producing on-specification product). The more time-consuming approach is to use dynamic simulations of the plant and impose a series of distur- bances. The other approach is more efficient and more suitable for screening a large It s f cnAvrI:.K s: Interaction between Steady-State Design and Dknamic Controllability Ii’7 Plant Design I 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 I Plant Design 2 I I 1 I I I I I I I 1 *2‘ x 0 -1 __ -2 I I I I I I I I I I 0 0. I 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 , Plant Design 3 2 I I I I I I I I I , -, -2 I I I I I I I I I 0 0. I 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 I FIGURE 5.9 Closedloop dynamic responses for three hypothetical plant designs. number of alternative designs. It uses frequency-domain methods, which we discuss in Chapter 10. We illustrate the method in the next section, considering a simple reactor-column process with recycle. In this example the flowsheet is fixed, and we wish to determine the “best” values of two design optimization parameters. I78 PART ONE: Time Domain Dynamics and Control 56.3 Reactor-Column-Recycle Example A first-order, irreversible liquid-phase reaction A L B occurs in a single CSTR with constant holdup VR. The reactor operates at 140°F with a specific reaction rate k of 0.34086 hr- I. The activation energy E is 30,000 Btu/lb-mol. Figure 5.10 gives the flowsheet of the process and defines the nomenclature. Fresh feed to the reactor has a flow rate of Fa = 239.5 lb-mol/hr and a composition zo = 0.9 mole fraction component A (and 0.1 mole fraction component B). A recycle stream D from the stripping column is also fed into the reactor. The reactor is cooled by the addition of cooling water into the jacket surrounding the vertical reactor walls. The reactor effluent is a binary mixture of components A and B. Its flow rate is F lb-mol/hr and its composition is z mole fraction component A. It is fed as satu- rated liquid onto the top tray of a stripping column. The volatility of component A to component B is cy = 2, so the bottoms from the stripper is a product stream of mostly component B, and the overhead from the stripper is condensed and recycled back to the reactor. Product quality is measured by the variability of xg, the mole fraction of component A impurity in the bottom. The nominal steady-state value of XB is 0.0105 mole fraction component A. We assume constant density, equimolal overflow, theoretical trays, total con- denser, partial reboiler, and five-minute holdups in the column base and the over- head receiver. Tray holdups and the liquid hydraulic constants are calculated from the Francis weir formula using a one-inch weir height. Given the fresh feed flow rate Fo, the fresh feed composition ~0, the specific reaction rate k, and the desired product purity xg, this process has 2 design degrees of freedom; i.e., setting two parameters completely specifies the system. Therefore, there are two design parameters that can be varied to find the “best” plant design. Let us select reactor holdup VR and number of trays in the stripper NT as the design FlGURE 5.10 Reactor/stripper process. [...]... I 3.5 4 4.5 0.68 5 1960 I.1 , 940 920 0.8 840 820 I I I I I I I I I I 0 0.5 I I.5 2 2,s 3 3.5 4 4.5 l’imc FIGURE 6.10 Scheme A: Pulse in F’o,, (hr) ' 800 5 2 I60 140 i 120 $ loo E 3 ,o IJ .4 80 60 40 0 I I 0.5 I I I 1.5 I 2 I 2.5 I I 3 3.5 4 I 4. 5 5 Time (hr) 550 500 0.78 $ 45 0 E A T 40 0 2 0.72 3 4 350 LLI 300 250 0.7 I 0 I I I 0.5 1 1.5 2 I I 2.5 3 Time (hr) I I I 3.5 4 4.5 ' 0.68 5 1.1 960 1 940 0.9... 0.5 880 g > 860 0 .4 840 0.3 820 0.2 2 900 5 2 0 0.5 I 1.5 2 2.5 3 Time (hr) FIGURE 6.11 Scheme B: Recycle rate reduction 3.5 4 4.5 -800 5 0 550 0.5 I 1.5 2 2.5 3 Time (hr) 3.5 4 4.5 5 , 0.8 I - 0.78 F 500 - 0.76 2 ;" 3 b - 0. 74 3 ZA g L 0.72 t-7 D 350 - 0.7 I 300 0 I I 1 0.5 1 1.5 2 I 2.5 I 3 I 3.5 I 4 I 4. 5 0.68 5 Time (hr) 960 920 910 0.85 0.8 u 0.5 I 1.5 2 2.5 Time (hr) 3 3.5 4 4.5 -J 900 5 202 ‘... that the time constant of the overall process depends very strongly on the product of the gains around the recycle loop, KFKR When the effect of the recycle is small (KR is small), the time constant of the process is near the geometric average of r~ and 7~ However, as the product of the gains around the loop KFKR gets closer and closer to unity, the time constant of the overall process becomes larger... CHAPTER IS FIGURE 6 .4 Reactor-three-column-two-recycle-four-component m 6: Plantwide Control 19 I process O- ne te 1Y F :e The former is mostly component C with impurities of component A (XDz,A) and of component B (xD~,B) The latter is mostly component D with impurity of component B h13.d 6 .4. 1 Control Structures Screened nt ic n LlS i1 in ncat 3s > ,I le This process has 15 control valves, so there... 0.76 2 0 2 - 0. 74 2 i - 0.72 2 $ 350 3 0 0 I 250 ’ 0 - 0.7 I I I I 1 2 3 4 I I I I 5 6 Time (hr) I 7 8 9 1Oo’68 1.15 1000 - 990 - 930 0.9 0 I 1 I 2, I 3 I I I I I I 4 5 6 7 8 9 Time (hr) FIGURE 6.9 Scheme A: 1 percent increase in F(,~ 920 IO ‘9 fg ~ SE 7% 4 & IO0 I40 p 120 2 $ loo ?? $ 80 60 40 I I I I 0.5 0 1 1.5 2 I I I I 3.5 I 2.5 3 Time (hr) 4 4.5 5 - 0.8 550 - 0.78 - 0.76 2 '3 % cz 0. 74 b) E - 0.72... question of how to develop a control system for an entire plant consisting of many interconnected unit operations 6.1 SERIES CASCADES OF UNITS Effective control schemes have been developed for many of the traditional chemical unit operations over the last three or four decades If the structure of the plant is a sequence of units in series, this knowledge can be directly applied to the plantwide control. .. Somehow the amount of component A in the system must be determined, and the makeup of component A must be adjusted to maintain its inventory at a reasonable level The problem cannot be solved by the use of other types of controllers Control structure B provides good control of the system Figure 6.11 shows what happens using scheme B when the total recycle flow rate is reduced from 500 to 40 0 lb-mol/hr The... loop (F and 0) are both on level control In control scheme B, sketched in Fig 6.8, the total recycle flow rate to the reactor (distillate plus makeup A) is flow controlled The makeup of reactant A is used to hold the level in the reflux drum This level indicates the inventory of component A in the system Dynamic simulations of the process were made using these two control structures Figure 6.9 shows... (6.6) = rate of consumption of reactant A (mol/hr) = reactor holdup (mol) = specific reaction rate (hr- ‘) = concentration of reactant A in the reactor (mole fraction A) Two different control structures are explored The conventional control is called the constant VR structure l l l l l Control reactor holdup VK by manipulating reactor effluent flow rate F Flow -control fresh feed flow rate Fo Control the... feed flow rate Fo Control the impurity of component A in the base of the column rn by manipulating heat input Control reflux drum level in the column by manipulating distillate flow rate 0 Control the impurity of component B in the distillate (1 - xn) from the column by manipulating reflux CI~AIYI’EK 6: Ikmtwide Control 187 Note that this control structure has both of the flow rates in the recycle loop . 50 60 Reactor opt D, TR VR 2 ctl TJ FJ AH DR 20 17 I5 126.76 130.63 I 34. 90 43 1.9 593.6 744 . I 0.3233 0. 248 5 0. I980 0.5325 0.55 14 0.5 548 95.22 96. 94 98.85 66 .43 82. I2 95 .46 265.6 328.3 381.7 6.50 7.23 7.79 837 .4 II06 1376 1256 1659 20 64 Column 1 NTINF 1517. = 4. 51 F=3 24 z,., = 0.6600 zB = 0.23 24 zc=0.1076 FIGURE 5.7 J D,, = 4. 42 D, = 2 24 XD1.A = 0.95 = 0.05 II I, I B,=lOO XB,,A = 0.0100 XB1.B = 0. 641 3 xBI,C = 0. 348 7 I D,, = 2.83 D, = 65 .4 X[&,. con- trollability factor can be some measure of the “goodness” of control (integral of the squared error), the cost of the control effort, or the value of some controllability measure (such as the plant