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Bio-routes to bulk chemicals

Short course given by John Villadsen Professor (Process- and Utilities) at UTT Tobago Gas Technology Conference October 2008. Bio-routes to bulk chemicals. Contents of the short course.

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Bio-routes to bulk chemicals

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  1. Short course given byJohn VilladsenProfessor (Process- and Utilities) at UTTTobago Gas Technology ConferenceOctober 2008 Bio-routes to bulk chemicals

  2. Contents of the short course • Whetting the appetite: What can we expect modern biotechnology to give us? (only commodity chemicals are treated) • A few pieces of factual knowledge about bioprocesses a. Stoichiometry b. Rates of bioreactions. • An integrated industrial size bioprocess for Trinidad. The Single Cell Protein (SCP) process based on natural gas or on methanol as carbon-substrate.

  3. 400 m3 fermentor used in 1,3 propane diol production

  4. Production of bio-diesel • Bio-diesel is produced from animal fat (lard) or from vegetable oil (e.g. rape seed oil) using a catalytic trans- esterification process (acid, base or enzymes) Glycerol-fatty acid + CH3OH→Me-fatty acid + glycerol A methanol based bio-diesel is very similar to high grade oil-derived diesel, but no sulphur, less NOx etc. This could be a fine outlet for a large MeOH-production! As a byproduct glycerol is formed. This could be the starting point for a chemistry based on glycerol

  5. Production of bio-butanol • Bio-butanol is produced from sugar, household waste or some other carbon source. Here the bacterium Clostridium acetobutylicum is used About 50 % of the sugar can be converted to butanol, another 25 % to acetone and the remainder to biomass and ethanol. The price of butanol is 1300 US$ per ton The world market is about 1.3 million tons/year BP and DuPont are comitted to produce bio-butanol on the British market by 2008

  6. Opportunities in glycerol • Some people will burn glycerol ! • There are lots of more valuable uses for glycerol CH2OH-CO-CH2OH←CH2OH-CHOH-CH2OH→ CH2OH-CH2-CH2OH CH2OH-CH2-COOH ← CH2OH-CHOH-CH2OH→ CH2OH-CH2-CH2OH CH2OH-CHOH-CH2OH→ CH2OH-CHOCH3-CH2OH→………..

  7. Glucose • An ideal starting material for production of • Alanine: CH3COCOOH (pyruvate) + NH3 + NADH → CH3CHNH2COOH (alanine) • Lysine: 2 Pyruvate + 2 NH3 + 4 NADPH → CH2NH2(CH2)3CHNH2COOH (lysine) • Tyrosine: para-hydroxy-phenylalanine → → L-DOPA and other pharmaceuticals. • The possibilities appear to be unlimited

  8. CH CO CH CO O O CH CO CH CO Chemical route to succinic acid V2O5/Al2O3 C4H10 + 7/2 O2 + 4 H2O Reaction in fluid bed with catalyst recycling at 410°C CH2 COOH Pd, Pt, Ru/Al2O3 + H2 + H2O CH2 COOH In aqueous phase (15 bar, 150°C) with intensive cooling The overall yield is less than 50%: A terrible waste.

  9. ¾ CH2O + CH3O1/2 + 3/4 CO2 2 CH3/2O + 1/2 CH2O + 1/4 H2O acetic acid glucose ethanol New routes to succinic acid CH2O + 1/7 CO2 8/7 CH3/2O + 1/7 H2O glucose succinic acid CH8/3O + 1/3 CO2 4/3 CH3/2O + 1/3 H2O glycerol CH3O1/2 + 5/7 CO2 12/7 CH3/2O + 3/14 H2O ethanol Price of n-butane is 375 US $/ton. Glucose costs 160 US $/ton. The situation cries out for inventiveness

  10. Inventiveness!!Typical situation for Engineers Lactose is one of the most abundant carbohydrates. It is a by-product in cheese manufacture (4.5 wt% in whey permeate) Source: whey, the main dairy by-product. Current use: Pig feed and lactic acid. Can it be used as raw material for new food products and for pharmaceuticals? Synthesis of compounds derived from lactose

  11. Lactobionic acid (LBA) • Structure: • Applications: • organ transplantation • sweetener, functional food • detergents • cosmetics • corrosion protection... New opportunities!

  12. Final process lay-out Flash Pasteurization Oxidation Pasteurization Transportation 38oC, 5.5 pH 50 L/min airflow 119-122oC, 300 L/h 0.41 L holding unit 70oC/10 min 225 L to Novo APV Niro Analysis Spray-drying Evaporation 62oC, vacuum concentrate to 30-40 wt.% solid Inlet: 180oC Outlet: 82oC 1H NMR HPLC

  13. …and now a glimpse of the Science behind these processes

  14. Stoichiometry • Methanol synthesis: CH4 + H2O → CO + 3 H2 2 H2 + CO → CH3OH Yeast fermentation: CH2O + 0.0164 NH3 → 0.510 CH3O1/2 + 0.275 CO2 + 0.077 CH8/3O + 0.137 CH1.74O0.60N0.12 + 0.037 H2O CH2O = ”glucose” (1 C-atom) CH3O1/2 = ethanol (1 C-atom) CH8/3O = glycerol (1 C-atom) CH1.74O0.60N0.12 = ”biomass” (1 C-atom) The stoichiometry changes with T, pH, pO2, cNH3, cCH2O etc.

  15. But still………. ..the stoichiometry of bioreactions can also be modeled. The conversion of glucose to biomass, ethanol and glycerol can be pictured as three metabolic pathways: v1 = -1.12 CH2O + 0.12 CO2 + X +0.15 NADH -1.80 ATP = 0 v2 = - 3/2 CH2O + CH3O1/2 + ½ CO2 + ½ ATP = 0 v3 = - CH2O + CH8/3O – 1/3 NADH – 1/3 ATP = 0 When the redox (NADH) and energy (ATP) balances are closed the experimental stoichiometry is found.

  16. Degree of reduction of organic compounds 1. For each element define a redox neutral compound : H2O, CO2, NH3…….etc. Define the unit for redox: H is one unit of redox. • ”H2O redox content = 0” → ” O has redox content -2” • ”Since CO2 is redox neutral → C has redox content 4” • ….and N has redox content -3 etc. Now glucose (CH2O per carbon) has redox content 4 Ethanol (CH3O1/2 per carbon has redox content 6 Methanol (CH3OH) 6, Methane (CH4) 8 Glycerol(CH2OH-CHOH-CH2OH) 14/3 = 4.667 per carbon. ”Biomass” has an average composition CH1.8O0.5N0.2→ redox content 4.20 per carbon

  17. Transporters of redox and energy between pathways • Reductive power is transported by co-factors such as NADH, NADPH, FADH2 • Each of these co-factors carries 2 units of redox (”=H2”) • NADH is also used to create energy (respiration) • ATP is a co-factor that carries energy between pathways • Hydrolysis of ATP to ADP liberates 31.5 KJ free energy.

  18. The ”solvent” production process using Clostridium acetobutylicum The network has 13 fluxes and 4 internal nodes: With an NADH and an ATP balance 7 rates should be measured to determine v1- v13

  19. Structuring the network as energy and redox balanced elementary pathways CH2O v8 v1 BuOH X v7 v2 acn HBu v3 v6 v4 v5 HLac ac e HAc v1 = rX: -1.1 CH2O + X (CH2O0.5N0.25) + 0.1 CO2 + 0.075 NADH – 2.48 ATP = 0 v2 = racn: -1.5 CH2O + CH2O1/2 + ½ CO2 + ½ NADH + 1/2 ATP = 0 v3 = rHLac: - CH2O + CH2O (HLac) + 1/3 ATP = 0 v4 = re: -1.5 CH2O + ½ HCOOH + CH3O1/2 – ½ NADH + ½ ATP = 0 v5 = rHAc: -1.5 CH2O + ½ HCOOH + CH2O (HAc) + ½ NADH + ATP = 0 v6 = rac: - 2 CH2O + CH2O1/3 + 2/3 HCOOH + 1/3 CO2 + 2/3 NADH + 2/3 ATP = 0 v7 = rHBu: -1.5 CH2O + CH2O1/2 + ½ HCOOH + ¾ ATP = 0 v8 = rBuOH: -1.5 CH2O + CH2.5O0.25 + ½ HCOOH – ½ NADH + ½ ATP = 0

  20. No node points. Each flux leads to one product v1 = rx: -1.1 CH2O + X (CH2O0.5N0.25) + 0.1 CO2 + 0.075 NADH – 2.48 ATP = 0 v2 = racn: -1.5 CH2O + CH2O1/2 + ½ CO2 + ½ NADH + 1/2 ATP = 0 v3 = rHLac: - CH2O + CH2O (HLac) + 1/3 ATP = 0 v4 = re: -1.5 CH2O + ½ HCOOH + CH3O1/2 – ½ NADH + ½ ATP = 0 v5 = rHAc: - 1.5 CH2O + ½ HCOOH + CH2O (HAc) + ½ NADH + ATP = 0 v6 = rac: - 2 CH2O + CH2O1/3 +2/3 HCOOH + 1/3 CO2 + 2/3 NADH + 2/3 ATP = 0 v7 = vHBu: -1.5 CH2O + CH2O1/2 + ½ HCOOH + ¾ ATP = 0 v8 = rBuOH: -1.5 CH2O + CH2.5O0.25 + ½ HCOOH – ½ NADH + ½ ATP = 0 Here 6 measured rates are needed together with NADH/ATP balances

  21. With six measured fluxes and two constraints all fluxes can be found NADH balance: 0.075 v1 + ½ v2 – ½ v4 + ½ v5 + 2/3 v6 – ½ v8 = 0 ATP balance : - 2.48 v1+ ½ v2 + 1/3 v3 + ½ v4 + v5 + 2/3 v6 + ¾ v7 + ½ v8 = 0 → v5 = rHAc = - 0.150 v1 – v2 + v4 – 4/3 v6 + v8 v3 = rHLac = 7.89 v1 + 3/2 v2 – 9/2 v4 + 2 v6 – 9/4 v7 – 9/2 v8 - rs = 1.1 v1 + 3/2 v2 + v3 + 3/2v4 + 3/2 v5 + 2 v6 + 3/2 v7 + 3/2 v8 = 8.765 v1 + 3/2 v2 – 3/2 v4 + 2 v6 - ¾ v7 – 3/2 v8 rHCOOH = ½ v4 + ½ v5 + 2/3 v6 + ½ v7 + ½ v8 = -0.075 v1 – ½ v2 + v4 + ½ v7 + v8 = rH2 rc = rHCOOH + 0.1 v1 + ½ v2 + 1/3 v6 = 0.025 v1 + v4 + 1/3 v6 + ½ v7 + v8

  22. The pathways to 1,3 propane diol (PDO) • CH2 O + 2/3 NADH + 1/3 ATP → CH8/3O2/3 (PDO) • CH2O → CO2 + 2NADH (+ 1/3 ATP) (”burning of glucose”) • NADH + ½ O2 → P/O ATP (”respiration of the cell”; P/O ≈ 2) • (1 + α) CH2O → X (CH1.8 Oo.5 N0.2) + α CO2 + (2 α – 0.1) NADH • - 2.42 ATP • There are 4 pathway fluxes to be considered: v1 to v4. • We certainly wish to make the flux v1 as large as possible!

  23. Closing the NADH and the ATP balances (α = 0.1, P/O = 2) one obtains the following ”black box stoichiometry” in which two Yield Coefficients Ysx and Yso are variables. Ysx is the C- moles biomass obtained per C-mole substrate used: CH2O + 0.2 Ysx NH3 + Yso O2 → Ysx X + Ysp P + Ysc CO2 Ysp = (19.50 Yso – 10.97 Ysx) / (27.00 Yso -13.575 Ysx) (Ysp)max = 19.5 / 27 = 0.722 for Ysx = 0.

  24. ….and how fast are the cell reactions? • The rate of reaction is always proportional to the biomass concentration x (kg/m3) • The volumetric rate of production q (kg/m3 h) is the product of x and a function f(s,p) where s and p are the concentrations of substrate and product (kg/m3) • f(s,p) = rx s /(s + Ks) where rx is the specific rate of biomass production (kg/kg h), and Ks is the saturation constant. This is the Monod rate expression. • f(s,p) = rx s /(s + Ks) · (1 – p/pmax) with pmax = maximum permissible product concentration (product inhibition).

  25. Definition of specific rates. • The specific rates ri are often proportional (i.e. ri / rk is a constant. Since ri / rk = Yki , the yield of i on k, this means that the stoichiometry is constant for all fermentation conditions. • ri can also be a linear function of rk , e.g : rs = Yxs rx + ms where ms is a maintenance substrate consumption. • Note that in all the formulas s is the concentration of the limiting substrate. There are many substrates: A source of carbon and energy, a nitrogen source, a source of phosphorous, sulfur, minerals, vitamins etc.

  26. Numerical values for kinetic parameters The specific rates ri are of the order of hours-1 (E.coli : 1 h-1, yeast and Methylococcus capsulatus (used for single cell protein):0.4 h-1, and human cells: 0.02 h-1). The saturation constant Ks is in the range of 1 g m-3 for many bacteria and fungi. For yeast Ks = 25 – 50 g m-3.

  27. Mass balances for flow systems Biomass X: dx /dt = rx x + v /V (xf – x) Substrate Si: dsi /dt = rsi x+ v /V (sif – si) Product Pi: dpi /dt = rpi x + v /V (pif – pi). The specific production rates are positive for X and Pi and negative for Si. Superscript f : feed-concentrations. v is the volumetric flow (m3 h-1) that flows through the reactor volume V Usually xf and pf are zero for single reactors.

  28. Single Cell Protein by fermentation • Inall lakes, ponds or swamps methane is produced by anaerobic microbial digestion of primary biomass. • In the upper regions of the pond methane meets oxygen that diffuses down from the surface of the pond. • There is plenty of N and other nutrients avaiable. • Bacteria such as Methylococcus capsulatus grow on simple carbon sources such as methane. • In their turn the bacteria serves as feed for higher organisms…and eventually ”turn up” as fish.

  29. Capture of CH4 by Methane monooxygenase

  30. ….more on the biochemistry of Methylococcus capsulatus • The Methanol produced from methane is dehydrogenated CH3OH → HCHO (formaldehyde) + ”H2” • Formaldehyde is oxidized to CO2 to create energy (ATP) - or it is used together with NH3 and minerals to build cellmass. HCHO + O2→ CO2 + H2O HCHO + NH3 + P + S +..→ Protein, lipids, carbohydrates +…

  31. Methylococcus capsulatusCH4 + O2 → biomass + water + CO2An organism that may revolutionize animal feed production

  32. Genome of Methylococcus capsulatus

  33. The key-enzyme Methane monooxygenase

  34. Methane and Oxygen demand for SCP production • From 1.25 kg methane one obtains 1 kg biomass*) This corresponds to 1 kg biomass per 1.75 N m3 methane or Ysx = 0.520 C-mole biomass per C-mole methane • The O2 demand is (8 – 0.520٠4.20) / 4 = 1.45 mol O2 per C-mole CH4 or 2.53 N m3 O2 / kg biomass = 3.62 kg O2 / kg biomass. Stoichiometry of methane conversion to biomass: CH4 + 1.45 O2 + 0.104 NH3 → 0.52 CH1.8O0.5N0.2 + 0.48 CO2 + 1.69 H2O *) Reference : Wendlandt, K.D, Jechorek, M, Brühl, E. ”The influence of Pressure on the growth of Methanotrophic Bacteria” Acta Biotechnol. 13, 111-113 (1993) and industrial experience: Dansk Bioprotein A/S 1992 - present.

  35. Demand for O2 and CH4 mass transfer • The production rate depends on the rates of two separate processes A. The reaction between bacteria and dissolved O2 + CH4 B. The rate of mass transfer from gas- to liquid phase. The ”bio-chemical” reaction is limited by NH3 since we need to keep the NH3 concentration below about 40 mg L-1 to avoid formation of NO2- which is toxic to the bacteria. At 30 mg L-1 the rate is qx = 0.21 X kg m-3 h-1 where X is the biomass concentration in kg m-3. But qo2 = (1.45 / 0.52)(1000 / 24.6) qx = 113 qx mol m-3 h-1 = kl a (cO2* - cO2) where cO2* and cO2 are respectively the saturation and the actual O2 concentrations in the liquid.

  36. Factors that affect the mass transfer The rate of mass transfer kl a (cO2* - cO2) (and kl a (cCH4* - cCH4)) depend on : • The mass transfer coefficient kl a Maximum achievable kl a ≈ 1200 h-1 • cO2* • cO2 cO2* is proportional with the partial pressure of O2 in the gas phase. At 1 atm total pressure and pure O2 one obtains cO2* = 0.9 mM (45o C) cO2 should be above about 20 μM to keep the organism healthy.

  37. Consequences of O2 limitation The constant production rate qx = 4.2 kg m-3 h-1 can not be maintained The production rate in the last part of the reactor is 1st order in cO2* If we wish a high utilization of O2 (e.g.95 %) the reactor volume may increase beyond reasonable limits (or qx may decrease to an unacceptably low level).

  38. Demand for heat removal The reaction should take place at ≈ 45o C, the optimal temperature for Methylococcus capsulatus fermentation. Stoichiometry: CH4 + 1.45 O2 + 0.104 NH3 → 0.52 CH1.8O0.5N0.2 + 0.48 CO2 + 1.69 H2O Heat of reaction  460Yso kJ (C-mol carbon source)-1 or Q = 460٠ 1.45 = 667 kJ (mole CH4)-1 = 52 MJ (kg biomass)-1 This is an appreciable heat duty!

  39. Reactor design • A stirred tank reactor is hopeless: We wish the first order conversion of O2 inthe last part of the reactor to proceed in plug-flow mode. In a CSTR cO2* would be 0.05 of inlet value. • The large heat release dictates that external heat exchange is to be used. • Liquid and gas is forced through a number of stationary mixer elements at a velocity of ≈ 1 m s-1. Gas is injected through an ejector. Ample allocation of head space assures gas/liquid separation. Holding time for liquid ≈ 5 h and for gas ≈ 60 s. • Centrifuges (or drum filters) are used to separate biomass from liquid. • Ultrafiltration gives ≈ 20 wt% biomass sludge. Spray drying gives the final powdery product • Heat shock treatment (123 oC, 2-5 min) removes nucleic acids and gives a product suitable for direct human consumption.

  40. 250 m3 reactor (≈ 9000 t year-1 production) in Norway

  41. 500 L pilot plant loop-fermentor at DTU

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