En el marco de la jornada Microalgas, ¿una fuente de petróleo verde?, organizada con IMDEA y celebrada el 8 de abril en EOI, Escuela de Organización Industrial, Emilio Molina Grima, de la Universidad de Almería, presenta como han llegado a un proceso eficiente de producción en laboratorio de aceite de algas.
8_04_2010
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Procesado de aceite de algas
1. Downstream processing of algal oil Emilio Molina Grima Dpt. Chemical Engineering, University of Almería, SPAIN [email_address] Seminar on Microalgae IMDEA Energy-EOI, April 8 2010, Madrid
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3. Pretreatment of microalgae cells J.Y. Lee et al. (2009 ) Bioresour Technol. DOI10.1016/j.biotech2009.03.058 Comparison of several methods for effective lipid extraction from microalgae Lipid content (wt %) Non-disruption Autoclaving Bead-beating Microwaves Sonication Osmotic shock 30 20 10 0 30 20 10 0 30 20 10 0 Botryococcus sp Chlorella vulgaris Scenedemus sp
5. Extraction of lipids Information on saturation curves may be found at: Walter D. Bonner (1910). Experimental Determination of Binodal Curves, Plait Points, and tie lines, in Fifty systems, each consisting of water and two organic liquids. J. Phys. Chem. 14, 738-789 E. Molina Grima et al. (1994) Comparison between extraction of lipids and fatty acids from microalgal biomass. JAOCS, 71(9), 955-959 Optimal tie line
6. Extraction of lipids Yields (%) of Extracts and Raffinates ( in parentheses ) of lipids extraction from dry Isochrysis galbana Cl 3 CH/MeOH/H 2 O (1:2:0.8 v/v/v) Hexane/EtOH 96% (1:2.5 v/v) Hexane/EtOH 96% (1:0.9 v/v) Butanol EtOH 96% EtOH 96%/H2O (1:1 v/v) Hexane/Isopropanol (1:1.5 v/v) 92.9 (0.9) 52.2 (27.4) 49.5 (8.3) 70.4 84.4 63.3 66.0 Solvent Mixture
7. Extraction of fatty acids Yields (%) of Extracts obtained by direct saponification from dry Isochrysis galbana Cl 3 CH/MeOH/H 2 O (1:2:0.8 v/v/v) Hexane/EtOH 96% (1:2.5 v/v) Hexane/EtoH 96% (1:0.9 v/v) Butanol EtOH 96% EtOH 96%/H 2 O (1:1 v/v) Hexane/Isopropanol (1:1.5 v/v) Solvent system + KOH (47.5 mL solvent /g KOH) - 81.0 48.0 9.0 79.8 46.5 62.0 60ºC-1 h - 75.0 47.1 - 79.2 45.5 61.1 Room-8 h
10. Optimization of fatty acids extraction Influence of Ethanol (96%)/wet biomass M.J. Ibañez González et al. (1998) JAOCS 75, 1735 – 1740. Lyophilized
11. Optimization of fatty acids extraction S-L stage V 01 = 810 g (1 L ethanol) +21 g KOH Residue Alcoholic solution Fatty acid soaps (25% water) 2 1 V 1 +V 2 = 1336 g y A =0.0072 V 02 = 405 g (0.5 L ethanol) L 1 X A1 X B1 X S1 Lo =480 g X A0 = 0.023 X B0 = 0.977 L 2 X A2 X B2 X S2 Wet Biomass (100 g d.b) Recovery Yield 87%
12. Optimization of fatty acids extraction V 0 Scale-up calculations needed (repeated contact system) L A0 + V 01 = L 1 + V 1 = M 1 X A0 ·L A0 = Y A1 ·V 1 + X A1 ·L 1 S-L stage underflow overflow x A2 L 0 X B0 = L 1 X B1 V 01 = 810 g (1 L ethanol) Residue Alcoholic solution Fatty acid soaps (25% water) 2 1 V 1 +V 2 = 1336 g y A = 0.0072 V 02 = 405 g (0.5 L ethanol) L 1 X A1 X B1 X S1 Lo = 480 g X A0 = 0.023 X B0 = 0.977 L 2 X A2 X B2 X S1 Wet Biomass (100 g d.b) Recovery Yield 91% B L 0 A S x A0 V 1 M 1 V 2 L 1 L 2 M 2 Kg FA soaps solution Kg inert solid P= = 3 X A1 1- X B1 Y A1 = (Ideal stage)
13. Optimization of fatty acids extraction Δ Scale-up calculations needed (counter current systems) S-L stage A S B V 2 V 1 x A2 x A0 L n L 2 L 1 M L 0 V n+1 2 1 V 1 y A1 L 1 X A1 X B1 X S1 Lo =480 g X A0 = 0.023 X B0 =0.977 L 2 X A2 X B2 X S2 V 2 y A2 V 0 =1215 g Extract (overflow) Raffinate (underflow)
14. Optimization of fatty acids extraction L-L Extraction of FA V 1 y 1 V 1 y 2 V 1 y 3 V 1 y 4 Fatty acids Extract (Hexane phase) V 0 = 0.026 L Hexane V 0 = 0.026 L Hexane V 0 = 0.026 L Hexane Alcoholic phase Alcoholic solution Fatty acids (40% water) L-L L-L L-L L-L V 0 = 0.026 L Hexane Lo = 1,53 kg X A0 = 903 mg/L L 1 X A1 L 2 X A2 L 3 X A3 L 4 X A4
15. Optimization of fatty acids extraction Fatty acids concentration Alcoholic phase C FA (mg·L -1 ) The distribution equilibria of FA between the alcoholic phase and hexane facilitate the calculation of design, scale-up and assessment purposes L/V 930 mg L -1 0 1000 2000 3000 4000 5000 6000 0 50 100 150 200 Fatty acids concentration Hexane phase, C AH (mg·L -1 ) Step1 Step 2 Equililbrium x 1 x 2
16. Optimization of fatty acids extraction Conclusions FA recovery yield (wet biomass) FA recovery yield (dry biomass) Important reduction of hexane Important reduction of alcohol (96% v/v) Solvent/biomass 87% 96,2% 90% 84% 24 mL/g biomass
17. Optimization of fatty acids extraction E. Molina Grima et al. (1996) Spanish patent 9602090
18. Direct transesterification of paste Biomass Biodiesel BIOMASS HEXANE ACETONE EPA METHANOL RESIDUE ACETYL CHLORIDE WARTER CONDENSER MIXER EVAPORATOR CROMATOGRAPHY CENTRI- FUGE FILTER SOLIDS EXTRACTOR REACTOR ULTRASONIC MIXER OPCIONAL OPTIONAL E Belarbi et al. (2000 ). Enzyme and Microbial Technology 26 516-529 . EVAPORATOR COOLER EVAPORATOR
19. Paste biomass Hexane Methanol Acetyl chloride/ H 2 SO 4 (10%) Extractor Cooler FAMEs Spent biomass Centrifugation Filtration 1000 mL *500 g 50 mL 1000 mL Temperature = 100 ºC Time: 120 min 9.1 g *(500 g of paste biomass of Phaeodactylum tricornutum ≈ 100 g of lyofilized biomass) (9.9% of TFAs) 500 mL Extractor Condenser Direct transesterification of paste Biomass Phaeodactylum tricornutum - Monodus subterraneus – Scenedesmus almeriensis EVAPORATOR
20. Direct transesterification of paste Biomass CASE STUDY Slurry reaction products (Spent biomass, water, methanol, FAMES, catalyst) Biodiesel produced per year: 13.550 kg Calculations needed for processing the paste biomass produced per day in one Ha of tubular photobioreactors of Scenedesmus almeriensis. Slurry reaction products (Spent biomass, water, methanol, FAMES, catalyst) V 01 = 605 kg/day Hexane Residue S-L 2 S-L 1 V 1 +V 2 = 773 kg/day y A =0.0483 V 02 = 303 kg/day Hexane L 1 X A1 X B1 Lo =3190 Kg/day X A0 = 0.013 X B0 =0.987 L 2 X A2 X B2 Reaction L= 1375 Kg/day X A0 = 0.03 X B0 =0.997 Wet biomass 1815 kg/day methanol + acetyl chloride or 10% H 2 SO 4 Hexane solution of methyl esters S-L 2 S-L 1 V 1 = 395,15 kg/day y A1 =0.095 L 1 X A1 X B1 L 2 X A2 X B2 V 2 y A2 V 0 =530 kg/day Lo =3190 kg/day X A0 = 0.013 X B0 =0.987 Residue Hexane solution of methyl esters Hexane
21. Paste Biomass Hexane Methanol Acetyl chloride / H 2 SO 4 REACTOR Centrifugation Filtration Spent Biomass ** Biodiesel *** EPA EXTRACTOR * Biodiesel Argentated silica gel column chromatography Hexane: Acetone (99.5: 0.5 v/v) EPA (99%): 2154 $/ Kg EPA (70%): 185 $/ Kg (Abayoumi et al., 2009) 1500 mL 500 g 50 mL 1000 mL (9.9 % of TFAs) (9.1 g) (6.9 g) (1.6 g) (76%) (18%) EXTRACTOR Cooler Direct transesterification of paste Biomass * Crude biodiesel with high content of EPA (27.7%) and a 92% recovery yield with respect to the total FA contained in the biomass ** Biodiesel with low EPA content (8.2%) and a 76% recovery yield with respect to the total FA contained in the biomass *** High grade EPA methylester (96.4%). Recovery yield 18% with respect to the total FA contained in the biomass. Cooler EVAPORATOR
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24. Multi-step process 4. Protein and carbohydrate separation from non lipidic residue Carbohydrates Proteins Precipitation 1. Protein and carbohydrate extraction from wet biomass Biomass slurry Wet Biomass Solid-Liquid Extraction Precipitation Proteins Carbohydrates 3. Chlorophyll separation from non lipidic residue Enriched in ethanol VACUUM BATCH DISTILLATION Filtration Chlorophylls 2. Fatty acid extraction by direct saponification and recovery of unsaponifiable products (carotenoids) Carotenoids Spent biomass Liquid-Liquid Extraction Fatty Acids Saponification Filtration Liquid-Liquid Extraction 59,0% 91% 6,6% Chlorophyll a 10,8% Chlorophyll c 41,0 % 16,3% 15,5%
25. Multi-step process Carotenoids in hexane 0.33 g L-L L-L L-L 0.9 L hexane L-L L-L 0.9 L hexane 0.9 L hexane 0.9 L hexane Alcoholic phase 0.9 L hexane S-L, solid-liquid extraction L-L , Liquid-Liquid extraction P, Precipitation S-L 1.5 L buffer S-L 1.5 L buffer 476 g Wet biomass (100 g d.b) Proteins in the pellet 17.7 g P (NH 4 ) 2 SO 4 Carbohydrates in water 1.7 g 0.3 L water S-L residue 40 g KOH 1 L ethanol S-L 0.5 L ethanol
26. Multi-step process pH 6 HCl Alcoholic Phase Fatty acids in hexane 8.0 g L-L L-L L-L 0.45 L hexane L-L 0.45 L hexane 0.45 L hexane 0.45 L hexane D, Distillation F, Filtration P, Precipitation L-L , Liquid-liquid extraction D Carbohydrates in water enriched in ethanol F P (NH4) 2 SO4 Chlorophylls in torte Chlorophylls a 0.0753 g Chlorophylls c 0.0378 g Proteins in pellet 6.7 g
27. Downstream processing of algal oil Acknowledgements Antonio Giménez Giménez Alfonso Robles Medina Maria Jose Ibáñez González Jose M. Fernandez-Sevilla Francisco Gabriel Acién Fernández Emilio Molina Grima El Hassan Belarbi
Notes de l'éditeur
In this slide, you can see a block diagram for the direct saponification of wet biomass, the amount of biomass and the concentration of saponifiable lipids, the amount of ethanol used in the saponification reaction, as well as the amount of ethanol used in washing the slurry after the reaction. Since the experimental recovery yield was 87%, the concentration of fatty acid in the overflow leaving the S-L stage was 0.0072 g of fatty acids per gram of alcoholic solution, equivalent to a 903 mg/L.
If these two stages would have been two ideal stages , (i.e. the concentration of fatty acids in the overflow leaving each stage is in equilibrium with the concentration of the fatty acid solution retained within the slurry), the yield would have been 91%. B y using basic Chemical Engineering principles , such as the overall mass balance in each stage, the fatty acid balance, the inert solid balance, and also with the data that give us the amount of soap fatty acids retained by the slurry per amount of inert solid (line of retention), we can calculate stage by step , all the unknown flows and concentrations. It is more practical and easy to handle if the calculations were done by using a ternary diagram in which each vertices represent the 100% of fatty acid, solvent or inert (plus the accompanying water).This line represent the variation of the fatty acid solution retained by the slurry. Since you know the amount of wet biomass and its fatty acid content, as well as the amount of solvent used in each step, it is easy to fix in the ternary diagram this point, that at the same time represent the point corresponding to the amount of underflow plus overflow leaving the stage. The underflow must be allocated in the retention line ; and the overflo w must be on the hypotenuse of the triangle. Therefore by tracing a line connecting the inert and this point (ideal stage concept), on the retention line is determined the concentration of the underflow leaving the firs stage and, in the hypotenuse, we can find the concentration of fatty acids in the overflow leaving the first stage too. Once known these variables we can connect this underflow with the vertices corresponding to the solvent and determine M2, and then L2,and V2, and so on, until we achieve a point in the retention time, or in the hypotenuse, that determine a recovery yield equal or superior that the target. The number of these lines determines the number of stages needed. The calculations using the plot are easy and more intuitive. For solving problems of the type “what will happen if we change this target, or this amount etc. )
The process has also been developed at lab scale using 500 g of paste biomass. It needs an initial S-L extraction of the soluble proteins and carbohydrates in a buffer solution, the protein are later precipitated from the water solution, recovering the carbohydrates in the liquid phase. The biomass slurry is then subjected to a direct saponification of the algal oil in another S-L extraction unit . Once adding water, the caroteonids are recovered in a repeated extraction process.