Reflux ratio, minimum optimum

Key R2, t2 = optimum reflux ratio, minimum time for STEP 2... 

In general the same type of information given by the constant 0/V method can be obtained by the use of the Ponchon and Savarit method. For example, the cases of total reflux, minimum reflux ratio, and optimum feed-plate location can be easily solved. 

T.eflux Tatio. Generally, the optimum reflux ratio is below 1.15 and often below 1.05 minimum. At this point, excess reflux is a minor contributor to column inefficiency. When designing for this tolerance, correct vapor—Hquid equiUbrium (VLE) and adequate controls are essential. 

Checking Against Optimum Design. This attempts to answer the question whether a balance needs to be as it is. The first thing to compare against is the best current practice. Information is available ia the Hterature (13) for large-volume chemicals such as NH, CH OH, urea, and ethylene. The second step is to look for obvious violations of good practice on iadividual pieces of equipment. Examples of violations are stack temperatures > 150° C process streams > 120° C, cooled by air or water process streams > 65° C, heated by steam t/ urbine 65% reflux ratio > 1.15 times minimum and excess air > 10% on clean fuels. 

Optimum Reflux Ratio The general effecl of the operating reflux ratio on fixed costs, operating costs, and the sum of these is shown in Fig. 13-39. In ordinary situations, the minimum on the total-cost cui ve wih geueraUy occur at an operating reflux ratio of from 1.1 to 1.5 times the minimum R = Lv + i/D value, with the lower value corresponding to a value of the relative volatility close to 1. 

However, the total number of equilibrium stages N, N/N,n, or the external-reflux ratio can be substituted for one of these three specifications. It should be noted that the feed location is automatically specified as the optimum one this is assumed in the Underwood equations. The assumption of saturated reflux is also inherent in the Fenske and Underwood equations. An important limitation on the Underwood equations is the assumption of constant molar overflow. As discussed by Henley and Seader (op. cit.), this assumption can lead to a prediction of the minimum reflux that is considerably lower than the actual value. No such assumption is inherent in the Fenske equation. An exact calculational technique for minimum reflux is given by Tavana and Hansen [Jnd. E/ig. Chem. Process Des. Dev., 18, 154 (1979)]. A computer program for the FUG method is given by Chang [Hydrocarbon Process., 60(8), 79 (1980)]. The method is best applied to mixtures that form ideal or nearly ideal solutions. 

The effect of utilities costs on optimum operation was noted by Kiguchi and Ridgway [Pet. Refiner,. 35(12), 179 (1956)], who indicated that in petroleum-distillation columns the optimum reflux ratio varies between 1.1 and 1.5 times the minimum reflux ratio. When refrigeration is involved, 1. IRmm < flopt < 1 is used in the condensers, 1.2Rrniii < fLpt < 1 -4Rrn... 

Economically optimum reflux ratio is about 1.2 times the minimum reflux ratio Rm. 

Equating to zero for minimum cost, the optimum value of the reflux ratio is ... 

Optimum Reflux Ratio. The reflux ratio affects the cost of the tower, both in the number of trays and the diameter, as well as the cost of operation which consists of costs of heat and cooling supply and power for the reflux pump. Accordingly, the proper basis for choice of an optimum reflux ratio is an economic balance. The sizing and economic factors are considered in a later section, but reference may be made now to the results of such balances summarized in Table 13.3. The general conclusion may be drawn that the optimum reflux ratio is about 1.2 times the minimum, and also that the number of trays is about 2.0 times the minimum. Although these conclusions are based on studies of systems with nearly ideal vapor-liquid equilibria near atmospheric pressure, they often are applied more generally, sometimes as a starting basis for more detailed analysis of reflux and tray requirements. 

The optimum reflux ratio and the minimum batch time for separation task 1 are 3 and 80.62 min (Table 3.1). The separation task 2 could be achieved using 3 different reflux ratio (Table 3.2) but however, / exp = 2 gives the true minimum batch time which is about 40% lower than the batch time that would be required to achieve the same separation with Rexp = 4. 

A series of minimum time problems (Chapter 5) were solved at different values of q with increasing holdup for each case. Figures 3.18a and 3.18b show the minimum time solution vs. percent total holdup in the column for different mixtures at different q and Figures 3.19a and 3.19b show the corresponding optimum reflux ratio (required to get the separation in minimum time) vs. percent total holdup of the column. The results are summarized in Table 3.3 which shows, for each given separation, the optimum value of holdup to achieve the best performance out of the given column. The corresponding best minimum batch time and the optimum reflux ratio to achieve that are also presented in the table for each case. 

Mayur et al. (1970) formulated a two level dynamic optimisation problem to obtain optimal amount and composition of the off-cut recycle for the quasi-steady state operation which would minimise the overall distillation time for the whole cycle. For a particular choice of the amount of off-cut and its composition (Rl, xRI) (Figure 8.1) they obtained a solution for the two distillation tasks which minimises the distillation time of the individual tasks by selecting an optimal reflux policy. The optimum reflux ratio policy is described by a function rft) during Task 1 when a mixed charge (BC, xBC) is separated into a distillate (Dl, x DI) and a residue (Bl, xBi), followed by a function r2(t) during Task 2, when the residue is separated into an off-cut (Rl, xR2) and a bottom product (B2, x B2)- Both r2(t)and r2(t) are chosen to minimise the time for the respective task. However, these conditions are not sufficient to completely define the operation, because Rl and xRI can take many feasible values. Therefore the authors used a sequential simplex method to obtain the optimal values of Rl and xR which minimise the overall distillation time. The authors showed for one example that the inclusion of a recycled off-cut reduced the batch time by 5% compared to the minimum time for a distillation without recycled off-cut. 

For different values of F, Rmax, optimum reflux ratio (/ 2)> minimum operation time, productivity are shown in Table 10.4. In all cases the total amount of distillate is 3.95 kmol with 95% purity in Heptane. The productivity (Prod) is calculated using total operation time (rwfo/) which includes 2 hrs of total reflux operation time in STEP 1. 

Here the feed rate is maximised while the reflux ratio is optimised. The bottom product composition imposes an additional constraint to the problem. The results are summarised in Table 11.8 which gives the maximum feed rate, minimum batch time, optimum reflux ratio, and total number of batches for each mixture and total yearly profit. 

Design procedure. The minimum refluxes computed for each section are compared with each other. The highest value is the minimum reflux for the column. From Eq. (2.35) the corresponding minimum liquid flow in the section is calculated. This flow can be multiplied hy a certain factor, commonly between 1.05 and 1.3 to give the optimum flow. Guidelines for selecting factors are given in Sec, 3.1,6, The liquid flow can now be resubstituted into Eq. (2.35) and the actual reflux ratio calculated. 

Alternatively, results from a computer simulation can be plotted to determine the optimum feed stage. Simulation runs ere performed at several different feed points, keeping the material balance, reflux ratio, and total number of stages constant. Ksy component concentrations in the product streams are plotted against the feed stage number (Fig. 3.7). The minimum is at the optimum feed stage. 

As indicated in Fig. 11-7, the optimum reflux ratio occurs at the point where the sum of fixed charges and operating costs is a minimum. As a rough approximation, the optimum reflux mho usually falls in the range of 1.1 to 1.3 times the minimum reflux ratio. The following example illustrates the general method for determining the optimum reflux ratio in distillation operations. 

Solution. The variable costs involved are cost of column, cost of reboiler, cost of condenser, cost of steam, and cost of cooling water. Each of these costs is a function of the reflux ratio, and the optimum reflux ratio occurs at the point where the sum of the annual variable costs is a minimum. The total variable cost will be determined at various reflux ratios, and the optimum reflux ratio will be found by the graphical method. 

Optimum reflux ratio 1.25 Minimum reflw ratio 1.14... 

The next step in the procedure is to calculate the optimum or operating reflux ratio. First, calculate the minimum reflux ratio using the Underwood equations, Equations 6.27.3 and 2.27.4. For the calculation use the geometric average volatility of each component listed in Table 6.27.3. Because flie feed is at its bubble point, q = 1. Thus, Equations 6.27.3 and 6.27.4 becomes... 

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