Drying Ethanol by Azeotropic Distillation Guy Mattson and George R. Hertel University of Central Florida, Orlando, FL 32816 The commercial process for producing absolute ethanol can he used as the basis for an unusual homework or takehome exam nroblem. Students who are familiar with nhase equilibria, aieotropic mixtures, and programming in BASIC will find thismaterial balance problem to he both interestina and challenging. An introduction to the use of material baK ances in describing chemical processes may be found in refs 1,2, or 3. The System Ethyl alcohol may he dried by azeotropic distillation, using benzene as the entraining agent, in the system depicted in Figure 1.Similar processes are described in refs 4 and 5. The process is operated continuously and is in a steady state where the mass of the feed stream, F, per unit time is equal to the sum of the masses of the two effluent streams, B1 and B3. The feed stream, F, amounts to 100 lbih and has a composition, by weight, of 95.57% ethanol and 4.43% water, the binary azeotropic mixture. The feed stream to the process is combined with recycle stream, RC, and enters distillation column 1. The bottoms from column 1 is effluent stream, B1, which is pure, dry ethanol; the overhead, 01, is composed of ethanol, benzene, and water and separates into two layers upon condensation. These layers are separated in the decanter. The upper layer is returned to column 1 as reflux stream, RU; the lower layer is fed, as stream, RL, to distillation column 2. The overhead from this column is composed of ethanol, henzene, and water. It is condensed and fed to the decanter where it joins the overhead from column 1. The bottom stream, B2, from column 2 is composed of ethanol and water. It is the feed to column 3. The overhead from column 3 is composed of ethanol and water. It is recycled, as stream RC, to column 1. The bottom stream from column 3, B3, is pure water. VapaILlquM (Dlstlllatlon) Data
Emanol, pure 78.2 Areobope. Binary (95.57% Ethanol. 4.43% Water) 78.1 Azeobope. Ternary (18.5% Emenol. 74.1 % Benzene. 7.4% Water) 64.9 SolubllHy (decwdsr wmp0sHlon) Data st 20 OC
Upper Lay-: Lower Layer:
12.7% Emenol. 52.1 % Ethanol.
86.0% Benzene, 4.8% Benzene,
1.3% Water 43.1 % Water
The Problem
Calculate the amount of benzene required for the system to operate a s described. A Solutlon
To solve this ~roblem,it is necessarv to know the auantitv and composition of each feed, recyc~d,and effluent .stream. Feed stream, F,amounts to 100 Wh. Its composition and those of product streams, B1 and B3, are how;. Stream B1 consists of 95.57 l h h of pure, dry ethanol, and stream B3 consists of 4.43 lb/h of pure water. Calculations of the other streams are less apparent but can he solved by a series of iterative calculations. It is possible to set up a series of 46
Journal of Chemical Education
F
W 1. The system.
equations describing the mass balances about each item of equipment, calculate a value for each component from the equilibrium data, and then loop through the calculations until no further change in results occurs. Students with an interest in computer programming and a minimum of traiuing can write a suitable program to carry out these calculations. The program can be designed so that the initial feed amounts can be modified easily and the effect of such changes on the operation of the process observed. As a first step let us define the names of the variables. The feed to the process, F, the recycle streams, RC, RU, and RL, the overhead streams of each column, 0 1 , 0 2 , and 03, and the bottoms, B1, B2, and B3 are indicated in Figure 1. The components of each stream are identified by appending E, W, or B (ethanol, water, henzene) to the stream designation. Thus, RLE represents the ethanol component of the RL recycle stream from the bottom of the decanter to distillation column 2. FE and FW are used to define the initial feed stream quantities of ethanol and water, respectively. The next step is to write down equations defining each quantity in terns of other variables and known (given) relationships-the azeotrope compositions and the decanter data. This is done in the program (see listing, Fig. 2). Proper analysis of the decanter compositions is the key to success. In the decanter, the overall mass balance is
01+02=RU+RL 01 is defined by the equation on line 220 in the program, and 0 2 is calculated by a similar equation on line 380. An equation for each component balance may also be written. For example, for ethanol
01E + 02E = RUE + RLE
Since the overheads in both columns 1and 2 consist of the ternary azeotrope and the distribution of each component in
Pass
BOnMns Emanol Water
1 2 3 4 5 6 7 8
84.50 93.29 95.10 95.47 95.55 95.57 95.57 95.57
TOP E8mene EMnd Water Bonom
Benzene required
Ru
RI
RC
44.78 53.60 55.42 55.79 55.87 55.88 55.89 55.89
51.05 61.58 63.76 64.20 64.30 64.32 64.32 64.32
8.81 10.63 11.01 11.08 11.10 11.10 11.10 11.10
4.69 5.66 5.86 5.90 5.91 5.91 5.91 5.91
3.55 4.28 4.43 4.46 4.47 4.47 4.47 4.47 Column 1
Column 2
55.36 13.82 5.53
0.53 0.13 0.05
5.65 0.26
95.57
5.65 4.73
4.47
Column 3
Benzene Ethanol Water
Decamef Upper Laver 55.32 8.17 0.84 Lower Layer 0.53 5.78 4.79
Initial Feed Is set at 100 lblh (95.57 lb ethanol and 4.43 lb water).
the decanter is set by the solubility data, 0 1 E and 02E can be expressed in terms of 01 and 02, respectively. The same can be done for water and benzene. The result is three relationships: for ethanol,
0.185 01 + 0.185 * 0 2 = 0.127 * RU
for benzene,
+
0.141 * 01 0.741 * 0 2 = 0.860 * RU
and for water, 0.074 * 01
+ 0.521 * RL + 0.048 * RL
+ 0.074 * 0 2 = 0.013 * RU + 0.431 * RL
We can use two of these equations to get expressions for RU and RL in terms of 01 and 02. Using the overall mass balance and the ethanol mass balance,
+
RL = 0.058 * (02 01)/0.394 RU=O1+02-RL
Figwe 2. Ll6tlng of BASIC program.
(line 250) (line 300)
The program includes a provision to check subsequent values of one of the parameters (e.g., the ethanol stream) and to loop back to the beginning as long as subsequent values are unequal. Initial zero values for RC and RU will be relaced bv calculated values on the second run. This will Lause adjustment of all calculated values. Successive runs will quickly optimize the numbers and reach the correct values. he table shows the outout from the oroeram. Note the numbers do not give an exact balance (65.6 4.47 = 100.04 instead of 100)due to the effect of the limited number of significant figures in some of the data. The output includes the Dure ethanol and water effluent rates, the reflux , and RC), and the total benzene required flow rates ( k ~RL, to sustain these flow rates. Once steady state conditions are reached, the compositions of the col&n effluents and the decanter phases are printed.
+
This system is well-behaved and converges quickly. Such behavior is not necessarily always the case. In fact, sometimes the order in which parameters are calculated will have an effect on the rate of convergence to optimum values. For exam~les.if line 230 is moved to line 470. ootimiration. as measbred by the constancy of the pure ethaanbl effluent rate (RIE),occurs after just five iterations instead of eight. Literature Cited D. M. BoricPrinciplesandColcvlolio~in Chsmico1Engineering;Prontice-Hak Englowwd Cliffs. NJ, 1967. R&laitis,G. V.IntroducfionLa MoferialondEnergy Baloncsr; Wiloy: NeurYork,1983. Clausen.C. A.:MatVlon,G. Principles ofIndwlrio1 Chemistry: Wiley: New York, 1978. Black. C. Chom. E M . ROB.. 1980. 76 (9). 78. Goobel. 0. h Ullmennk E m y ~ l o p ~ do/lndusfNal ii C h h s l g v . 5th ed.; Gorhatz, W., Ed.; VCH: New York, 1987: Vol. A9, Section 5.L.2.
1. Himmelblau,
2. 3. 4. 5.
Volume 67 Number 1 January 1990
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