External Mass Balance

The selectivity should be greater than 20 to accomplish significant separation of species / from species j. For a completely mixed membrane system, the external mass balance yields... 

Although an infinite reflux column is still impractical, it does have feed and product streams associated with it. Thus, unlike the total reflux column, the infinite reflux column has an external mass balance. The mass balance is for the column shown in Figure 1.1 is very simple, and can be summarized as... 

Equations 5.20 522 essentially consist of six independent equations in total because there are two independent compositions in each vector for a ternary system. The vapor equilibrium compositions are completely specified for a set of volatilities using the constant relative volatility relationship, and the reflux in the stripping section is directly related to that in the rectifying section by Equation 5.19 once the external mass balance have been set to compute the mass balance. Whence, there are six compositions (from A, B, and C) and one reflux value seven variables in total, to compute in this system of six nonlinear equations. An additional equation is thus required to solve this system, and for this the colinearity condition can be used, which essentially states that the slope of line AB must be equivalent to the slope of line BC. This can be formally expressed as... 

Specify and calculate all variables to satisfy the external mass balance, that is, product compositions and/or flowrates. 

It is useful to point out initially that since CS3 or CS5 has to produce a sidestream product, there is an automatic constraint that is placed on either of these CSs in terms of the minimum amount of material that has to flow through this CS, r ardless of the FP of the Petlyuk. For example, if it is desirable to draw the sidestream as liquid stream from CS3, then the magnitude of the liquid flowrate in CS3 has to be at least as large as the sidestream product flowrate set by the external mass balance. This case can be expressed mathematically as... 

Through Figures 7.21 and 7.22 the effect of / ai has been examined, but there are other factors that are also worth considering, most notably the effect of external mass balance variables D and S. By examining the A = 0 equations for each CS (Equations 7.11 7.14), both D and S influence zero net flow y-intercepts... 

FIGURE 7.23 Summaiy of external mass balance effect on die net FP ( implies an increasing value, while i implies a decreasing value). 

We will start our design procedure via TTs by assuming that we have completely specified the product stream (the entire external mass balance) as well as a value for R i that will lead to a feasible design. For a Petlyuk column to be considered feasible using TTs, the following conditions have to be satisfied ... 

Once there is certainty regarding the number of feed/product streams in the column, we may turn our attention to determining the external mass balance for the... 

Fig. 5.2-20 External mass balance and intemal concentration profile of ternary disdUation...

Fig. 5.2-27 Preferred separation of a ternary mixture. The external mass balance lies on a straight line passing through %, and y, ...

The conclusion from these calculations is, we have to calculate D. To do this we need external mass balances ... 

Now, how do we conpletely solve the external mass balances The unknowns are B, D, X2 X3 %,bot> %,bof There are six unknowns and five independent equations. Can we find an additional equation Unfortunately, the additional equations (energy balances and equilibrium expressions) always add additional variables (see Problem 5-Al), so we cannot start out by solving the external mass and energy balances. This is the first major difference between binary and multiconponent distillation. 

These assumptions allow us to conplete the external mass balances. The procedure is illustrated in Example 5-1. 

Exanple 5-1. External mass balances using fractional recoveries... 

B. Explore. This appears to be a straightforward application of external mass balances, except there are two variables too many. Thus, we will have to assume the recoveries or concentrations of two of the coin)onents. A look at the DePriester charts (Figures 2-11 and 2-121 shows that the order of volatilities is propane > n-butane > n-pentane > n-hexane. Thus, n-butane is the LK, and n-pentane is the HK. This automatically makes propane the LNK and n-hexane the HNK. Since the recoveries of the keys are quite high, it is reasonable to assume that all of the LNK collects in the distillate and all of the HNK collects in the bottoms. We will estimate distillate and bottoms based on these assunptions. 

E. Check. Two checks are appropriate. The results based on our assunptions can be checked by seeing whether the results satisfy the external mass balance Eqs. (5=1) and f5-21 These equations are satisfied. The second check is to check the assunptions, which requires internal stage-by-stage analysis and is much more difficult. In this case the assunptions are quite good. 

If there are only LNKs or only HNKs, then an accurate first guess of compositions can be made. Suppose in Example 5-1 that we specified 99.4% recovery of propane in the distillate and 99.7% recovery of n-butane in the bottoms. This makes propane the LK, n-butane the HK, and n-pentane and n-hexane the HNKs. The assumption that all the HNKs appear in the bottoms is an excellent first guess. Then we can calculate the distillate and bottoms compositions from the external mass balances. The composition calculated in the bottoms is quite accurate. Thus, in this case we can step off stages from the bottom upward and be quite confident that the results are accurate. If only LNKs are present, the stage-by-stage calculation should proceed from the top downward. 

Al. Explain why the external mass balances cannot be solved for a ternary distillation system without an additional assunption. Why aren t the equations for the following useful ... 

If you were going to do external mass balances around the column to find B and D, the best... 

Fzlj k. And all heavy non-keys (HNK) exit in the bottoms, x k ist = 0 Bx r bo = Ezh ik- Now we can do external mass balances to find all distillate and bottoms conpositions and flow rates. This was illustrated in Chapter 5. Once this is done, we can find L and V in the rectifying section. Since CMO is assumed,... 

The overall external mass balance for the two-column system shown in Figure 8-4A is... 

Mass balances for the system shown in Figure 8-6 are of interest. The external mass balances are identical to Eqs. (8z3a) and (8z3b). Thus, the bottoms flow rates are given by Eqs. (8-4) and 18-5T Although the processes shown in Figures 8-4A and 8zS are very different, they look the same to the external mass balances. Differences in the processes become evident when balances are written for individual columns. For instance, for column 2 in Figure 8-6 the mass balances are... 

A5. Explain why the external mass balances are the same for Figures 8-3A and 8=6. 

D19. We wish to use n-hexane as an entrainer to separate a feed that is 80.0 wt % ethanol and 20 wt % water into ethanol and water. The system shown in Figure 8-18 will be used. The feed is 10,000.0 kg/h and is a saturated liquid. The ethanol product is 99.999 wt % ethanol, 0.001 wt % hexane, and a trace of water. The water product is 99.998% water, 0.002 wt % ethanol and a trace of hexane. Do external mass balances and calculate the flow rates of make-up solvent (n-hexane), ethanol product, and water product. (Assume that trace = 0.)... 

To get started, do external mass balances and calculate accurate values for the two bottoms flow rates assuming that the water and MEK products are both pure. Specify the bottoms flow rate of column 1 = 60, but do NOT specify bottoms in column 2. In column 2 start with D = 20 and increase D in steps, 30, 40, 50, 60, 70, and so forth without reinitializing. If you don t step up. Aspen Plus will have errors. 

The problem we want to solve is to separate 100 kmol/h of a saturated liquid feed at 1 atm pressure. The feed is 78 mol% n-butanol and 22 mol% water. We want the purity of both products to be 99% or higher. A system similar to Figure 8-4A is to be used. The columns will operate at 1.0 atm pressure. Use external mass balances to determine the flow rates of the two bottoms products assuming that they are essentially pure butanol and pure water (< E -04 mole frac of the other conponent). 

On the specification sheets for both distillation columns the lines from the decanter should be input as feed on stage 2 (first actual stage in the column). Use ketde-type reboilers. Set the bottoms rate for the butanol column at the values calculated from the external mass balances. Initially, set Col-1 with N =... 

To calculate the makeup solvent flow rate that you eventually want to use, do an external mass balance around the entire system,... 

The resulting Makeup flow rate will be extremely small since losses of solvent are small (after all, no one wants to drink ethylene glycol with their alcohol or their water). If you immediately use this value of solvent makeup as a feed to the system. Aspen Plus will not converge. Start with the value you were using previously and rapidly decrease it (say by factors of roughly 5 or 10). Until the solvent makeup stream is at the desired value for the external mass balance, the extra ethylene glycol will exit with the distillate (water product) from column 2. This occurs because Bottoms flow rate in column 2 is specified and the only place for the extra ethylene glycol to go is with the distillate from column 2. 

Note Use ratio units. If careful with units, the liquid units can be in mass and the gas units in moles, which is effectively the form of the equilibrium data. Derive the operating equation and external mass balances to determine where to include the molecular wei t of HCl (36.46). Because HCl has a very large heat of absorption in water, the column will have to be well-cooled to maintain the temperature at 10°C. Commercial units are not isothermal. 

Analysis of fractional extraction is straightforward for dilute mixtures when the solutes are independent and total flow rates in each section are constant. The external mass balances for the fractional extraction cascade shown in Figure 13-5 are... 

Case 3. Specify R, Q, and Xg After doing external mass balances, you can plot the operating lines as in Figure 13-7. but the problem is still trial-and-error. The feed stage must be varied until the total number of stages is the same for both solutes. Small changes in conpositions or flow rates will probably be required to get an exact fit. 

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