Optimum Product Recovery in Chemical Process Design - Industrial

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Optimum Product Recovery In Chemical Process Design William L. Luyben Ind. Eng. Chem. Res., Just Accepted Manuscript • DOI: 10.1021/ie502746h • Publication Date (Web): 18 Sep 2014 Downloaded from http://pubs.acs.org on September 25, 2014

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Industrial & Engineering Chemistry Research

Paper submitted to Ind. Eng. Chem. Research

Optimum Product Recovery In Chemical Process Design

William L. Luyben Department of Chemical Engineering Lehigh University Bethlehem, PA 18015 USA

July 9, 2014 Revised September 6, 2014

[email protected]; 610-758-4256; FAX 610-758-5057

ACS Paragon Plus Environment


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Abstract One of the most important issues in the design of chemical plants is the trade-off between product recovery and costs, both investment capital and operating energy costs. High recovery of valuable products is desirable, but high recovery requires high separation costs. Distillation columns with more stages and higher reboiler duties are required as the specified recovery is increased. The optimum recovery depends on the difficulty of separation (relative volatility) and the value of the product. This paper quantitatively explores the impact of these parameters on the design of chemical processes.

1. Introduction In the design of a chemical process, an important trade-off exists between product recovery and separation costs. An arbitrarily high recovery (99.99 %) is often assumed because of the large economic importance of product yield versus the relatively smaller costs of capital and energy. Douglas1 discussed this issue in his pioneering work in conceptual process design three decades ago. The so-called “Douglas Doctrine” states that the process should be designed for high overall conversion of reactants and high recovery of the desired products. Despite the importance of recovery, there is little explicit discussion of recovery in the standard design textbooks or in the literature. The issues typically covered are the selection of column pressure, the classical trade-off between reflux ratio and number of trays and the sequencing topology of multiple columns for the separation of multicomponent mixtures. Most optimization studies are based on two alternative assumptions. In the first case, the purities of the two product streams are given. The design job is to find the economic optimum trade-off between reflux ratio and number of trays. In the second case, the purity of the more important product stream is specified and the other specification is the recovery of that valuable component in the product stream. Stated another way, the design job in this case is to find how much of the valuable component should be lost in the other non-product stream. For example, Turton et al2 discuss the economic optimization of a DME column by making the arbitrary assumption that the recovery is 98.9% with a distillate purity of.99.5 wt% DME. Another example is a recent paper by Figueiredo et al3 that deals with dehydration of ethanol using extractive distillation. The ethanol recovery in the extractive column is arbitrarily set at 99.99 % with an ethanol product purity of 99.9 mol% ethanol. We will return to this example later in the paper to demonstrate that the optimum recovery of ethanol is not the 99.99 % assumed by these authors. It should be clear that the arbitrary selection of recovery may not lead to the optimum design. Obviously a very high recovery will be required if the product has a very high value. But if the product is a commodity chemical with a relatively modest value, the costs of energy and capital investment should be considered in the process


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design. The purpose of this paper is to quantitatively explore this issue using several examples. It is important to note that this paper deals with the design problem, not the operating problem. In the design of a distillation column, the number of stages and size of the column and the sizes of the heat exchangers are parameters. These affect capital investment, which is an important component in determining the optimum economic design of the process. Once the plant is built, there is an operating optimum recovery that is not equal to the optimum design recovery because capital costs no longer are relevant. The only tradeoff is between energy and product yield. Optimum operating recoveries are always higher than optimum design recoveries because of the modified basis of evaluation.

2. Design Basis and Procedure We assume that the separation operation in the process is conducted in a distillation column since distillation is the predominant method for recovering products from mixtures of other components. We explore the effects of relative volatility (α) and the value of the products on the economic optimum recovery of a valuable light component in the distillate product. In some cases the incentive for recovery is the difference between the value of the light component in the distillate and its lower value if lost in the bottoms. Thus a price differential is considered. In the numerical example considered, the column feed flowrate is F =100 kmol/h of a binary mixture with a composition 50 mol% light component and 50 mol% heavy component (z = 0.5). Distillate flowrate is unknown and will vary with the specified recovery, but distillate composition xD is specified. Bottoms flowrate B and bottoms composition xB are also both unknown. The recovery ℜ is defined as the fraction of the light component in the feed that is produced in the distillate product stream.

ℜ=

DxD Fz

The total molar and light component: balances for the binary separation are the traditional relationships.

F =D+B

Fz = DxD + BxB If the design specifications are the compositions of the two product streams, the flowrates of the two products D and B can be immediately calculated from the two equations since all the other variables are known (F, z, xB and xD).

 z − xB   D = F  x − x B   D B=F −D


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However, if the design specifications are the composition of the distillate xD and the recovery ℜ of light component in the distillate, the equations must be combined to give an explicit equation for the bottoms composition xB as a function of the known variables z, F, xD and ℜ.

xB =

zFx D (1 − ℜ) F ( x D − zℜ)

Then D and B can be calculated from the two balance equations. Figure 1 shows the effect of recovery on these important variables. As recovery is reduced from a high level (99.99 %), the composition of the light component in the bottoms xB increases, which means that more light component is lost. Bottoms flowrate increases and distillate flowrate decreases. These relationships hold for all binary systems, independent of relative volatility. Of course , as relative volatilities get smaller, the separation unit required to achieve the product compositions gets bigger (more stages) and more energy is required. The bottom right graph in Figure 1 shows how recovery affects the minimum number of stages for different values of relative volatility. The Fenske Equation gives the minimum number of stages as a function of relative volatility (α = 1.5 and 2 for the results in Figure 1) and the product compositions. Distillate composition is fixed at xD = 0.999, but bottoms composition xB decreases as recovery increases. Therefore the separation becomes more difficult and requires more trays. Lower relative volatilities require more trays. With a value of recovery selected (to be varied), the product streams from the column are known. Then a relative volatility α is specified. The design procedure to determine energy costs and to size the column is outlined below. 1. Calculate the minimum reflux rate from the slope of the rectifying operating line drawn from xD on the 45o line to the intersection of a vertical line from feed composition z on the VLE curve (yF). Saturated feed and reflux are assumed.

Rmin RRmin x − yF L slope =   = = = D xD − z  V  min Rmin + D 1 + RRmin For a multi-component system, the Underwood Equations can be used to find the minimum reflux ratio given the compositions and relative volatility of the light and heavy key components.. 2. Set the actual reflux ratio equal to 1.1 times the minimum. 3. Calculate the vapor rate in the column (kmol/h) assuming equimolal overflow.

V = D(1 + 1.1 RRmin ) 4. Calculate reboiler and condenser heat duties. Typical physical properties (isobutane at 322 K and 6.5 bar) are used to determine latent heat (17,400 kJ/kmol) and density (16 kg/m3 vapor density). 5. Assume a reboiler energy cost $9.88 per GJ and calculate annual energy cost. 6. Assume an F-Factor of 0.67 (metric units) and calculate the maximum vapor velocity Vmax.


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F − Factor = Vmax (m / sec) ρV (kg / m3 ) 7. Calculate the column diameter ID assuming a molecular weight MW = 58.

ID = 4( Area) / π Area =

V ( MW ) (3600) ρV Vmax

8. Use the Fenske Equation to calculate the minimum number of stages given the relative volatility α and the compositions of the distillate xD and bottoms xB.

N min

 x  1 − xB   log  D   1 − xD  xB  +1 = log α

9. Set the actual number of stage at twice the minimum. Calculate the length distillation column Lcol assuming a 2-ft tray spacing and allow 20% excess length for space at the base, feed tray and top. 10. Calculate the capital cost of the distillation column from the diameter (ID) and length (Lcol) with both in meters.

Vessel = 17,640( ID)1.066 ( Lcol )0.802 11.Calculate the capital cost of the reboiler and condenser from their heat-transfer

areas (AR and AC) assuming overall heat-transfer coefficients UR =0.568 kW K-1 m-2 and UC =0.852 kW K-1 m-2 and differential temperature driving forces of 30 K in the reboiler and 12.2 K in the condenser.

Heat Exchangers = 7296( AR

0.65

+ AC

0.65

)

The capital cost of the column and the reboiler energy cost are now known for the selected case with recovery and relative volatility specified. The remaining issue is how to incorporate the value of the distillate product stream for a given case.

3. Value of Product The approach is to use the flowrate of distillate when the recovery is very high (99.99 %) as the base case. For smaller recoveries, the distillate flowrate is smaller. The difference between this distillate and the base case distillate is considered a loss in


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product. The appropriate price of the product ($ per kmol) is selected and value of the lost product is determined.

Pr oduct Loss = ( Dba sec ase − D)(Pr ice) This is added to the energy cost and an annualized capital cost (total capital divided by a 3 year payback period) to give the total annual cost (TAC) of the process for the recovery selected. Typical prices for chemicals are obtained from the student section of the ICIS website (www.icis.com) and converted into units of $ per kmol. Examples are methanol (16 $/kmol), ethanol (38 $/kmol) and toluene ($70/kmol). Unfortunately many important prices are not available, particularly those of intermediate chemicals. For example, in a separation of propane from butane, both products have significant value, and what is important is the difference between the two. An estimate of their value is given4 to be about $0.528 per kg for propane and $0.264 per kg for butane, which gives a differential of about $7/kmol of propane. This is typical of petroleum separations where the differences in the values of the various products are not large. In most chemical separations, however, the valuable product has significant value while the waste product may have no value (or represent an additional cost for waste disposal). For example in ethanol dehydration, the ethanol product has value. The water waste stream has no value. It is obvious that high recoveries are required for high-priced products. However, if separation costs are high (low relative volatility systems), moderate recoveries may be more economical. In the next section we give results for several cases.

4. Results for General System Three cases are presented with product prices of $7, $16 and $38 per kmol. For each case, a range of relative volatilities is considered. The design of the high-conversion (99.99 %) base case is a strong function of the relative volatility selected. Table 1 gives parameters. The small relative volatility cases have high reflux ratios and many trays, as expected, so separation costs are large. The TAC for a system with relative volatility α = 1.1 is 7 times that of a system with α = 2. Figure 2 shows the effect of recovery on the TAC for the case with a small $7 per kmol value of the distillate product for three relative volatilities. The economic optimum recovery for each case is indicated with a star. The TAC results are normalized by dividing the TAC by the base-case TAC given in Table 1. Small relative volatilities make separation more expensive, so the optimum recovery is smaller. For α = 1.2, the optimum recovery is only 99 %. As the value of the product increases, the optimum recovery increases for a given relative volatility. Results with product prices of $16 and $38 per kmol are given in Figures 3 and 4, respectively. As expected, more valuable products drive the design of the process to high recoveries. At the suggestion of one of the reviewers of this paper, Figure 5 is presented as a guide for estimating an appropriate recovery for a general system. The ordinate is the


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recovery that minimizes total annual cost. The abscissa is the ratio of the price of the product ($ per kmol) to the cost of energy ($ per GJ). Results are given for three different relative volatilities. The optimum recovery decreases as product price or relative volatility decrease. It should be noted that the cases considered in this paper are where the distillate and bottoms are product streams. In many processes one of the streams is a recycle back to the reactive section of the process. This situation is not considered because the economics would involve the classical trade-off between reactor design and recycle, which is very case specific. We would expect that the “per pass” recovery in the column would be much lower since any product in the recycle stream is not lost but only affects reactor performance. However the “overall recovery” (the amount of product generated in the reactor that goes out in the product stream) would undoubtedly still be high. How high would depend on the impact of product recycle on the performance and economics of the reactor and the column.

5. Results for Specific Separations In this section we consider two specific examples from the literature. In the first, the design is based on specifying both product compositions, and we demonstrate that this does not lead to the optimum recovery. In the second, the recovery is arbitrarily selected, and we demonstrate that it is not the optimum. In both of these cases, the numbers of stages in the distillation columns are not changes from the original design, so capital investment is not considered. Only the changes in energy costs versus the changes in product value are presented.

5.1 Depropanizer The economic optimum design of a column separating propane and isobutane is demonstrated4 to have 44 stage when the purity of the propane distillate is xD = 0.98 and the impurity of propane in the bottoms is xB = 0.01 mole fraction propane. The feed is 0.1 kmol/sec with composition z = 0.4. Operating pressure is 16.8 atm. Reboiler duty is 24.5 MW with a cost of $4.7 per GJ. Chao-Seader physical properties are used. With the two product compositions specified, the resulting distillate flowrate is 0.4020 kmol/sec, giving a recovery of 98.5 %. Is this the optimum recovery? To establish a high-recovery base case, the recovery is set at 99.99 % and the distillate composition is maintained at 0.98. The bottoms composition changes from 0.01 to 0.000067. The resulting distillate flowrate increases to 0.40812 kmol/sec, but the required reboiler duty increases to33.87 MW. The increase in valuable distillate is 0.40812 – 0.4010 = 0.00712 kmol/sec. If the differential value of propane versus isobutane is $7 per kmol, the value of the increased distillate is $1,330,000 year. However, the increase in reboiler duty (33.87 – 24.50 = 8.37 MW) costs $1,389,000 per year. So going from 98.5 % recovery to 99.99 % recovery is not economically attractive. Remember that we are not considering capital investment in this case. The number of stages remains at 44. What about going from the 98.5 % recovery up to 99.9 %? In this case the bottoms composition is xB = 0.000675, the distillate flowrate is D = 0.40776 kmol/sec and the reboiler duty is 26.94 MW. The value of the increase in distillate flowrate over


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the original design is $1,258,000 per year. The increase in energy cost is only $396,000 per year. So the original specification of xB = 0.01 is not the economically optimum. A design with xB = 0.000675 (99.9 % recovery) increases net profit (product value minus energy) for the same column size

5.2 Extractive Distillation for Ethanol Dehydation

The paper by Figueiredo et al3 discusses a two-column extractive distillation system using ethylene glycol as the solvent. The recovery of ethanol is arbitrarily set a 99.99 % with ethanol product purity set at 99.9 mol% ethanol. The economic optimum for a design with a 44-stage extractive column and a 15-stage recovery column has a solvent-to-feed ratio of 1.177. Column efficiencies are 85 % and NRTL physical properties are used. The specifications in the recovery column are a distillate of 99.9 mol% water with a 99.99 % water recovery. The very high recovery produces a bottoms stream from the extractive column with a composition of only 64 ppm (molar) ethanol and a distillate from the recovery column with an ethanol composition of 560 ppm. The 99.99 % recovery design produces 85.077 kmol/h of ethanol product for a feed of 100 kmol/h with a composition of 85 mol% ethanol. Reboiler duty in the extractive column is 1.377 MW and in the recovery column is 0.430 MW. Figure 6 shows how the specified ethanol recovery affect several important variables. Decreasing recovery decrease the production of ethanol, which means an increase in product losses. The relationship between distillate flowrate D and the specified recovery ℜ is linear for constant values of feed flowrate F, feed composition z and distillate composition xD.

D=

ℜFz xD

Thus decreasing ℜ changes D directly. So the plot of product loss versus recovery (bottom right graph in Figure 5) is a straight line with a negative slope. Total energy in the reboilers of the two columns decreases as recovery decreases (top right graph). However, this relationship is quite nonlinear. The initial reduction in energy as recovery is reduced from 99.99 % is quite large. But the curve flattens out as further reductions in recovery are specified. Figure 7 shows the difference between the energy savings and the product loss. Operating at a recovery of 99.965 % instead of a recovery of 99.99 % produces a net gain in profit of $10,600 per year. This is admittedly small compared to a total energy cost in the base case of $563,000 per year. However, the extractive distillation separation in this system is fairly easy with fairly small columns.

6. Conclusion The results presented in this paper illustrate that recovery is a critical design optimization variable whose economically optimum design value should be incorporated


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in most process designs. The arbitrary assumption of a recovery can lead to a nonoptimum design. Alternatively, the arbitrary assumption of the composition of a lowervalue stream can also give incorrect results. Processes with high-value products require high recoveries. Processes with difficult separations require lower recoveries.

References 1. Douglas, J. M. Conceptual Design of Chemical Processes, McGraw-Hill, 1988. 2. Turton, R., Bailie, R. C., Whiting, W. B., Shaeiwitx, j. A. Analysis, Snythesis, and Design of Chemical Processes, 2nd Ed. Prentice Hall 2003. 3. Figueiredo, M., Brito, K., Ramos, W., Vasconcelos, L, Brito, R. “Optimization of Design and Operation of Extractive Distillation Processes,” Separation Science and Technology, submitted 2014. 4. Luyben, W. L. Distillation Design and Control Using Aspen Simulation, 2nd Ed. (2013), Wiley.

Table 1 – Base-Case Parameters

Relative Volatility α 2 1.5 1.2 1.1

TAC

Nmin

RRmin

(10 $/y) 0.391 0.530 1.44 2.87

22 33 69 120

1.99 3.99 9.98 20.0

6

Figure Captions Figure 1 – Effect of recovery Figure 2 – Price differential $7 per kmol Figure 3 – Price differential $16 per kmol Figure 4 – Price differential $38 per kmol Figure 5 – Optimum recovery Figure 6 – Ethanol dehydration; extractive distillation Figure 7 – Energy saving and product loss difference


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Fig. 1 – Effect of Recovery

xD=.999; z=0.5; f=100; alpha=1.5/2 51

D (kmol/h)

xB (mf LK)

0.02

0.015

0.01

0.005

0 0.98

50.5

50

49.5

0.985

0.99

0.995

49 0.98

1

51

0.985

0.99

0.995

1

40 35

 = 1.5

50.5

Nmin

B (kmol/h)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

50

30 25 20

49.5 15

49 0.98

0.985

0.99

0.995

1

10 0.98

Recovery

=2 0.985

0.99

Recovery


0.995

1

& Engineering Chemistry Research Fig. 2 – Price DifferentialIndustrial $7/kmol

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alpha=2/1.5/1.2; C3/C4

1.1

1.08

1.06

TAC/TACbase

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

1.04

 =2 1.02

1

 =1.5

0.98

0.96

 =1.2

0.94

0.92

0.9 0.985

0.99

0.995

Recovery LK ACS Paragon Plus Environment

1


alpha=1.5/1.2/1.1; Methanol

1.1

1.08

1.06

 =1.5 TAC/TACbase

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

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1.04

1.02

1

 =1.2

0.98

0.96

0.94

 =1.1

0.92

0.9 0.985

0.99

0.995


1


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alpha=1.5/1.2/1.1; Ethanol

1.1

1.08

 =1.5 1.06

TAC/TACbase

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

1.04

 =1.2

1.02

1

 =1.1

0.98

0.96

0.94

0.92

0.9 0.985

0.99

0.995


1


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Fig. 5 – Optimum Recovery

Alpha = 1.1/1.2/1.5 1

0.999

Optimum Recovery

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

 = 1.5

0.998

0.997

 = 1.2

0.996

0.995

 = 1.1

0.994

0.993

0.992

0.991

0.99

1

1.5

2

2.5

3

3.5

4

Price/Energy Ratio (GJ/kmol) ACS Paragon Plus Environment

4.5

5

Industrial Extractive & Engineering Chemistry Research Fig. 6 – Ethanol Dehydration; Distillation

Ethanol Dehydration; Extractive Distillation 1.9

0.1

Energy (MW)

0.08 0.06

0.04 0.02 0 0.999

0.9992

0.9994

0.9996

1.85

1.8

1.75

1.7 0.999

0.9998

30 25 20 15 10 5 0 0.999

0.9992

0.9994

0.9996

0.9998

0.9992

0.9994

0.9996

0.9998

30

Product Loss (K$/y)

Energy Savings (K$/y)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

Loss of Ethanol (kmol/h)

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0.9992

0.9994

0.9996

0.9998

25 20 15 10 5 0 0.999

Ethanol Recovery ACS Paragon Plus Environment

Ethanol Recovery


Page 16 of 16

Fig. 7 – Energy Saving and Product Loss Difference

Difference between Product Loss and Energy Savings 12

10

Difference (K$/y)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

8

6

4

2

0 0.999

0.9991

0.9992

0.9993

0.9994

0.9995

Ethanol Recovery ACS Paragon Plus Environment

0.9996

0.9997

0.9998

Optimum Product Recovery in Chemical Process Design - Industrial

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