Citation
Separation of Water and Isobutanol by LLE and Distillation Utilizing Stream Recycle

Material Information

Title:
Separation of Water and Isobutanol by LLE and Distillation Utilizing Stream Recycle
Creator:
Barkley, Daniel Phillip
Publication Date:
Language:
English

Notes

Abstract:
The clear presentation and solution of a problem is an effective method to teach various concepts. Formulas and concepts used as the basis for calculations and theory should be clearly stated and explained to communicate ideas clearly. A problem involving three liquid separations and two distillation processes will be analyzed in this project. This paper will attempt to solve the problem in a way that allows students to easily follow the calculations and learn how to make them on their own. ( en )
General Note:
Awarded Bachelor of Science in Chemical Engineering, summa cum laude, on May 8, 2018. Major: Chemical Engineering
General Note:
College or School: College of Engineering
General Note:
Advisor: Lewis E Johns. Advisor Department or School: Chemical Engineering

Record Information

Source Institution:
University of Florida
Holding Location:
University of Florida
Rights Management:
Copyright Daniel Phillip Barkley. Permission granted to the University of Florida to digitize, archive and distribute this item for non-profit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.

UFDC Membership

Aggregations:
UF Undergraduate Honors Theses

Downloads

This item is only available as the following downloads:


Full Text

PAGE 1

Daniel Problem S eparation of W ater and I sobutanol by LLE and D istillation U tilizing S tream R ecycle Abstract: The clear presentation and solution of a problem is an effective method to teach various concepts. Formulas and concepts used as the b asis for calculations and theory should be clearly stated and explained to communicate ideas clearly. A problem involving three liquid separations and two distillation processes will be analyzed in this project. This paper will attempt to solv e th e problem in a way that allows students to easily follow the calculations and learn how to make them on their own. Theoretical Basis: Calculations for liquid phase solubility were made based on a one constant Margulies model. This model is further app lied to the general form of Gibbs free energy equation to find the points at which energy of the system is minimized yielding the equilibrium point. equation was used for calculating vapor pressures as a function of temperature. Vapor liquid equ constant Margulies model. One constant Margulies model: Gibbs Free Energy: equation: Figures 1 & 2 presented below were constructed using the above equations and are used to determine the dependence of phase equilibrium versus pressure.

PAGE 2

Figure 1 Water Isobutanol equilibrium data for P = 6 bar. Figure 2 Water Isobutanol equilibrium data for P = 1 bar Note that increasing pressure increases the temperatures at which vapor liquid equilibrium occurs but has negligible impact on t he liquid liquid equilibrium. At lower pressures the vapor liquid equilibrium may interrupt the liquid liquid equilibrium. This phenomenon may be used for overcoming an azeotrope that would otherwise require a pressure swing or entrainer based separation p rocess es

PAGE 3

Understanding the Process: An example of a separation for the water isobutanol system has been presented in figure 3. Figure 3 Water Isobutanol separation process. T he purpose of this process is to utilize the existence of a liquid liquid separation to assist in separating a binary system with an azeotrope. Li quid liquid separation and distillation are necessary for this proces s The attached VLE /LLE diagram shows the equilibrium separations that can be achieved by the various processes T he concentrations of effluent streams B 1 and B 2 will likely be pure ( x B1 0.0001, x B2 0.999) T he re will be no wasted or lost product in the process. The process will be analyzed at a feed specification of x F = 0 .5 and F = 100 mol/h although any flow rate and concentration can be used in the system It should also be noted that any feed flow rate may be known after the calculations are made for x F = 0.5 because it is an extensive property of the system. 100 mol/h was simply chosen as a standard basis for the calculations Figure 3 should be referenced to understand the separation setup. First, the liquid is separated at 90 o C and then the light liquid from LLE1 ( LL 1 ) and the heavy liquid from LLE1 ( HL 1 ) are sent to columns 1 and 2 respectively. The conc entrations of the column effluent must be calculated using McCabe Thiele or a linear approximation Bottoms from both columns are taken as the product. The distillate from C1 and C2 are sent to LLE2 and LLE3 respectively. Following the liquid liquid separa tion, LL 2 and LL 3 are sent back to C1 and HL 2 and HL 3 are sent back to C2. Initial Analysis: For the given process, there are 13 material streams, 2 distillation columns, 3 liquid liquid extractors, 8 mass balances, 8 species balances, 5 states of equilibr ium and 2 operating equations. There are 2 8 unknowns 2 unknowns for each material stream, except the feed which has been set, and the pressure of the system T he pressure of the process is set at 1 atm absolute pressure and the temperature of the LLE mus t be set to find the component splits Two good options for setting of the LLE operating temperature are room temperature giving X I = 0.1 and X II = 0.9 and the bubble point giving X I = 0.2 and

PAGE 4

X II = 0.8. It should be noted that for LLE this system is symmetric. Once the pressure and temperatures have been set the number of unknowns drops to 24. The number of equations and variables that can be found is 28. 16 of the equations are from material bala nces, 2 from operating equations, and 10 variables are set when equilibrium concentrations are determined. This causes the system to be over specified by 4. This will result in the ability to solve the system and be able to check that it satisfies all equa tions. Solving Approach: Once the problem is well understood on the surface, the general equations should be presented. The first set of equations are the mass balances, these will be broken down in overall mass balances and one species balance. Note that x 1 = 1 x 2 as an implied equation. It is recommended that all Liquid Liquid separations be carried out at the same temperature to simplify the calculations A temperature j ust below the bubble temperature (92 o C ) has been chosen for this analysis Given this set of equations, the PFD, and the equilibrium data, it is possible to solve the problem Step 1: Solve equations 1 and 9 at the bubble point Using the equilibrium data from figure 1, and that F = 100 mol/h and x F = 0.5: The equilibrium data gives us x LL 1 = 0.2 and x HL1 = 0.8. Solving equations 1 and 9 for LL 1 and HL 1 gives us that LL 1 = 50 mol/h and HL 1 = 50 mol/h. Note that the lever rule will yield the same result, (0.5 0.8)/(0.8 0.2) = 0.3/0.6 = 0.5 0.5*100 mol/h = 50 mol/h. O verall Balances: 1. F = LL 1 + HL 1 (LLE1 Balance) 2. D 1 = LL 2 + HL 2 (LLE2 Balance) 3. D 2 = LL 3 + HL 3 (LLE3 Balance) 4. F = B 1 + B 2 (System Balance) 5. L 1 = LL 1 + LL 2 + LL 3 (C1 Feed Balance) 6. L 2 = HL 1 + HL 2 + HL 3 (C2 Feed Balance) 7. L 1 = D 1 + B 1 (C1 Balance) 8. L 2 = D 2 + B 2 (C2 Balance) Species Balances: 9. F*x F = LL 1 *x LL1 + HL 1 *x HL1 (LLE1 Balance) 10. D 1 *x D1 = LL 2 *x LL2 + HL 2 *x HL2 (LLE2 Balance) 11. D 2 *x D2 = LL 3 *x LL3 + HL 3 *x HL3 (LLE3 Balance) 12. F*x F = B 1 *x B1 + B 2 *x B2 (System Balance) 13. L 1 *x L1 = LL 1 *x LL1 + LL 2 *x LL2 + LL 3 *x LL3 (C1 Feed Balance) 14. L 2 *x L2 = HL 1 *x HL1 + HL 2 *x HL2 + HL 3 *x HL3 (C2 Feed Balance) 15. L 1 *x L1 = D 1 *x D1 + B 1 *x B1 (C1 Balance) 16. L 2 *x L2 = D 2 *x D2 + B 2 *x B2 (C2 Balance)

PAGE 5

Figure 4 LLE 1 e quilibrium balance. Step s 2 5 Overview : Step s 2 5 will differ based on the method that is selected to solve the system. The first method requires solving equations 5 and 13, however, since LL 2 and LL 3 are both unknown quantities, they must be guessed and through the process of trial and error the actual values can be found that result in little variation in the final answer. The second method for solving the problem would be to solve the system balances ( equations 4 & 12) first, given the design parameter s, and work back to find the feed concentration for the two columns. For this paper, both methods will be described and compared. Method 1: Step 2: This method will require multiple iterative steps to arrive at the final solution. The first of these ste ps will be to assume no recycle coming into the system initially, until those values are calculated. That is, for equations 5 and 13, LL 2 and LL 3 are zero, likewise for equations 6 and 14 HL 2 and HL 3 are zero. Therefore: L 1 = LL 1 L 1 = 50 mol/h L 2 = HL 1 L 2 = 50 mol/h

PAGE 6

Step 3: The next step is to determine the outlet concentrations of the two columns. This may be done using McCabe Thiele analysis based on a given boil up rate. The operating lines have been set as L 1 /V 1 = 1.667 and L 2 /V 2 = 1.11. Given these operating lines, the vapor liquid equilibrium, and stream flow rates and compositions the flow rate and composition may be obtained. A McCabe Thiele analysis can be approximated for systems that are either low concentrations or can be reasonably approximated using a linear vapor liquid equilibrium line. For such a case, the equation to do so is presented below. T he following equation is used to determine the geometric series where a = m*V/L This results in the following equation. Where m is the slope of a line that is used to approximate the equilibrium curve, V is the vapor flow rate, L is the liquid flow rate, n is the stage of interest, and x B is the bottoms concentration. Figure 5. X Y diagram for Water Isobutanol system.

PAGE 7

Figure 6. Distillation column 1. Equations 7 and 15 were used for column 1 (C1) and 8 and 16 for column 2 (C2). The set L/V was also used to determine the outlet stream. Note that V and D are the same. 7. L 1 = D 1 + B 1 (C1 Mass Balance) 15. L 1 *x L1 = D 1 *x D1 + B 1 *x B1 (C1 Species Balance) L 1 /V 1 = 1.667 (C1 Reflux Ratio) 8. L 2 = D 2 + B 2 (C2 Mass Balance) 16. L 2 *x L2 = D 2 *x D2 + B 2 *x B2 (C2 Species Balance) L 2 /V 2 = 1.11 (C2 Reflux Ratio) The balance yields the flow rates and concentrations of the distillate and bottoms for each column. L 1 = 50 mol/h & L 1 /V 1 = 1.667 V 1 = 30 mol L 1 = D 1 + B 1 = V 1 + B 1 B 1 = L 1 V 1 = (50 mol/h) (30 mol/h) B 1 = 20 mol/h Th ese values can now be used to solve equation 15. Staging trial and error will yield negligible amounts of water in the bottoms for the first column meaning that this assumption can be used to calculate the composition of the distillate. L 1 *x L1 = D 1 *x D1 + B 1 *x B1 (50 mol/h) (0.2) = (30 mol/h) x D1 + (20 mol/h) (~0) x D1 = (50 mol/h) (0.2) / (30 mol/h) x D1 = 0.33 The linear approximation equation can be used to determine the precise concentration of x B For 6 stages, L/V = 1.667, and m = 6, x B 1 = 5.9*10 6 For the precision necessary for these calculations, x B1 = 5.9*10 6 is equal to 0, as originally estimated.

PAGE 8

This process is the same for determining the outlet stream for the second column. L 2 = 50 mol/h & 2 1 /2 1 = 1.11 V 2 = 45 mol L 2 = D 2 + B 2 = V 2 + B 2 B 2 = L 2 V 2 = (50 mol/h) ( 45 mol/h) B 2 = 5 mol/h This information can now be used to solve equation 15. Staging trial and error will yield negligible amounts of isopropanol in the bottoms for the second column. As in column 1, this fact is used to find x D2 L 2 *x L 2 = D 2 *x D 2 + B 2 *x B 2 (50 mol/h) (0. 8 ) = ( 45 mol/h) x D1 + ( 5 mol/h) (~ 1 ) x D1 = [ (50 mol/h) (0. 8 ) (5 mol/h)] / ( 45 mol/h) x D2 = 0.78 A check of the approximation that x B2 = 1 requires solving the linear approximation for n = 6, L/V = 1.111, and m = 3 yields x B2 = 0.99982. Rounding to 2 sig figs results in x B2 = 1.0 Step 4: Once that the first iteration of the distillation columns was done, a balance may be conducted of the liquid liquid extraction to determine the feed flow rates for the subsequent iteration of the distillation columns. For LLE2 and LLE3 the governing equations are presented below. 2. D 1 = LL 2 + HL 2 (LLE2 Mass Bala nce) 10. D 1 *x D1 = LL 2 *x LL2 + HL 2 *x HL2 (LLE2 Species Balance) 3. D 2 = LL 3 + HL 3 (LLE3 Mass Balance) 11. D 2 *x D2 = LL 3 *x LL3 + HL 3 *x HL3 (LLE3 Species Balance) Solving 2 and 10 simultaneously yields 30 = LL 2 + HL 2 LL 2 = 125 HL 2 plugging this into eq. 10 30*(0.333) = (30 HL 2 )*(0.2) + HL 2 *(0.8) This equation simplifies to HL 2 = 30*(0.333 0.2)/(0.8 0.2) HL 2 = 6.67 mol/h And LL 2 = 30 6.67 LL 2 = 23.3 mol/h The same process is used for equations 3 and 11 yielding HL 3 = 43.3 mol/h and LL 3 = 1.67 mol /h The equilibrium balances for LLE2 and LLE3 are presented below in figures 7 and 8 respectively.

PAGE 9

Figure 7. LLE 2 equilibrium balance. Figure 8. LLE 3 equilibrium balance. Step 5: Equations 5, 13, 6 and 14 may now be solved to determine the new feed flow rates to the two distillation columns. 5. L 1 = LL 1 + LL 2 + LL 3 (C1 Feed Balance) 13. L 1 *x L1 = LL 1 *x LL1 + LL 2 *x LL2 + LL 3 *x LL3 (C1 Feed Balance) 6. L 2 = HL 1 + HL 2 + HL 3 (C2 F eed Balance) 14. L 2 *x L2 = HL 1 *x HL1 + HL 2 *x HL2 + HL 3 *x HL3 (C2 Feed Balance)

PAGE 10

The decision to run the LLE at the same temperature simplifies the equations and allows flow rates and concentrations to obtained quickly. L 1 = 50 mol/h + 23.3 mol/h + 1.67 mol/h L 1 = 75 mol/h x L1 = (LL 1 *x LL1 + LL 2 *x LL2 + LL 3 *x LL3 )/L 1 = (50 mol/h*(0.2) + 23.3 mol/h*(0.2) + 1.67 mol/h*(0.2)) / 75mol/h x L1 = 0.2 The same process is done for equations 6 and 14 yielding L 2 = 100 mol/h x L2 = 0.8 These flow rates are significantly diff erent from the previous feed rate which shows that further iteration is necessary. 32 iterations yield feed flow rate differences of less than 1 mol/h. 50 iterations yield feed flow rate differences of less than 0.1 mol/h. The results after 50 iteration s are summarized in table 1 b elow. C1 C2 L 12 5 500 V 75 450 B 50 50 L/V 1.6 7 1.11 x D 0.334 0.778 x B 0 1 HL 16.7 433.5 LL 58.3 16.7 Table 1. Method 1 d istillation stream flow rates operating conditions and compositions. Method 2: Step 2: As previously described, the first equations that must be solved are equations 4 and 12. This will yield an understanding of an expected final solution that will be confirmed in later calculations. Equation 4: F = B 1 + B 2 100mol/h = B 1 + B 2 Equation 12: F*x F = B 1 *x B1 + B 2 *x B2 100*(0.5) = B 1 *( 0.0001) + B 2 *(0.999) Solving these two equations and two unknowns yields B 1 = 50 mol/h and B 2 = 50 mol/h.

PAGE 11

Step 3: Given the flow rates and concentrations in the bottoms products and being able to set a reflux ratio for both columns, the feed flow rate and distillate flow rates can be solved. Figure 5 and e quations 2, 3, 10, and 11 can be used to determine the outlet c oncentration and flow rate of the distillate. The operating lines have been set as L 1 /V 1 = 1.667 and L 2 /V 2 = 1 .1 1 Staging for both columns yields the desired concentrations in 6 stages plus RB. Both columns have been designed with 6 stages which will be sufficient in obtaining pure bottoms product streams. To find the feed and distillate flow rates, the first assumption is that all feed concentrations to C1 and C2 are x L1 = 0.2 and x L2 = 0.8. This is a valid assumption if all liquid separat ions are carried out at atmospheric pressure and just below saturated liquid temperatures (from equations 13 and 14). Mass and species balances must be done on the distillation columns to solve for the unknowns. Equations for the columns are as follows: L 1 /V 1 = 1.667 L 1 = V 1 + B 1 L 1 *x L1 = D 1 *x D1 + B 1 *x B1 Solving equations 7 and 15 via the predictive method yields: 1.667 *V 1 = B 1 + V 1 V 1 = (50mol/h)/ 0.667 = 75 mol/h meaning D 1 = 75 mol/h and L 1 = 125 mol/h This yield s x D1 = 0.333 and x B1 = 5.9 *10 6 thus confirming the assumptions made to solve the equations. Similarly, solving equations 19, 20, 23, 24 with L 2 /V 2 = 1 .1 1 1.1 1 V 1 = B 1 + V 1 V 1 = (50mol/h)/0.1 1 = 450 mol/h meaning D 1 = 450 mol/h and L 1 = 5 0 0 mol/h Similarly yielding x D2 = 0.778 and x B2 = 1. 0 Step 4: Next, the equilibrium balances for LLE 2 & LLE 3 must be solved. Given that D 1 = 125 & D 2 = 450 the variables LL 2 HL 2 LL 3 and HL 3 can be solved using equilibrium balances for LLE 2 and LLE 3 The equilibrium balances are shown in figures 7 and 8 above in method 1 Equations 2, 3, 10, and 11 are used to solve the two liquid liquid equilibrium equations. 2. D 1 = LL 2 + HL 2 (LLE2 Balance) 10. D 1 *x D1 = LL 2 *x LL2 + HL 2 *x HL2 (LLE2 Balance) 3. D 2 = LL 3 + HL 3 (LLE3 Balance) 11. D 2 *x D2 = LL 3 *x LL3 + HL 3 *x HL3 (LLE3 Balance) Solving 2 and 10 simultaneously yields

PAGE 12

125 = LL 2 + HL 2 LL 2 = 125 HL 2 plugging this into eq. 10 125*(0.333) = (125 HL 2 )*(0.2) + HL 2 *(0.8) This equation simplifies to 125*(0.333 0.2)/(0.8 0.2) = HL 2 = 16.625 mol/h And LL 2 = 125 16.625 58.33 mol/h. The same process is used for equations 3 and 11 yielding HL 3 = 433.35 mol/h and LL 3 = 17.1 mol/h Step 5: The final step in the process is to check the assumptions that were made and evaluate their validity. L 1 was found by assuming B 1 = 50 mol/h if the assumption is correct, then equations 5 and 6 will be balanced. L 1 = LL 1 + LL 2 + LL 3 L 2 = HL 1 + HL 2 + HL 3 L 1 = 125 mol/h, LL 1 + LL 2 + LL 3 = 50 mol/h + 58.3 mol/h + 17.1 mol/h = 125.4 125mol/h ~ 125.4 mol/h there fore the first set of assumptions is valid. L 2 = 500 mol/h, HL 1 + HL 2 + HL 3 = 50 mol/h + 16.6 mol/h + 433.4 mol/h = 500 mol/h 500 mol/h = 500 mol/h therefore the second set of assumptions is valid The results are summarized below. C1 C2 L 125 500 V 75 450 B 50 50 L/V 1.67 1.11 x D 0.334 0.778 x B 0 1 HL 16.7 433.5 LL 58.3 16.7 Table 2 Method 2 distillation stream flow rates, operating conditions and compositions. It should be noted that both methods yield the same results as expected