Introduction. Growth and product formation in reactors. Downstream processing. Fermentation technology. Typical fermentation

Transcription

1 Growth and producormation in reactors Introduction Typical fermentation product classes volume ton/year Introduction Batch, chemostat and fed batch Microbial competition / selection Mixed and mixed culture Growth in continuous reactors Food acids (citric, lactic 4-6 alcohols 6-7 amino acids - 6 antibiotics - 5 enzymes pharmac. proteins -3 - enantio pure compounds - 3 esters, flavours - 3 polymers - 4 tissues/organs biological waste water 6 Fermentation technology ownstream processing ost contribution depends on product fermentor recovery products concentration from fermentation wastes cost - 5 % 5-8 % LOG ON. (g/l Water I Ethanol itroenzuur Penicilline Threonine ephalosporine Gentamicine Microbiele Gibberine zuur Protease Amylases Bulk enzymen Antibiotica Glucose Oxidase Rennine Insulin II Monoclonale antilichamen -3 Research/diagnostische enzymen Glycerophosphaat ehydrogenase Luciferase III Therapeutische enzymen Factor VIII Urokinase LOG PRIE ($/Kg 3 4

2 Type of reactors used Types of processes used spent air spent air Laboratory Industry fresh air aerated stirred tank reactor to 3 m 3 + cooling facilities fresh air bubble column (with air lift to 4 m 3 + cooling facilities continuous Batch Fed batch operation (days Fermentor utilities Fermentor set-up 7 TERILIZATION : live steam batch : membrane : heatshock continuous Less Odour Less Energy Less egradation Less ilution Less Time AIR FILTRATION membrane / bed filter dry air 6 m 3 air m 3 broth h OMPREOR bar / 5 o MIING BY TIRRER τ mix sec- sec - 3 kw/m 3 O INPUT 5- mol/m 3 h HEAT REMOVAL Jacket oil T F T coolwater - 4 o APAITY kj m o 8 reactor h

3 esired properties micro-organism The job to be done esign and operate towards minimal cost high yield biomass / O high high temperature tolerant grows on low cost high q P at low extra-cellular product low viscous biomass generally regarded as AFE (gras mineral media 9 Increase productivity of reactor kg product / m 3 reactor h more biomass more active biomass kinetics of growth and product formation designing the optimal feed profile increased need to transport s/ products (role of transport processes Use advantage of scale bigger reactors scale-up effect (role of transport processes Batch, chemostat and fed batch processes Types of reactors Batch fermentation batch fed batch chemostat Remember from kinetic theory q (or completely determines the microbial behavior q or must be controlled at an optimal value batch and q q and is not controlled chemostat is controlled at opt fed batch r is controlled

4 constant volume batch reactor slope r spent air Vconstant fresh air slope r t end Question How will and change with Answer oncentrations are always calculated from the appropriate mass balances Add sterilized growth medium solution ( with concentration, N- and P-source, K +, Mg +, salts, nutrients, vitamins, trace elements and choose T, ph. Add electron acceptor (O, NO 3, etc. How Add (inoculate micro-organism t at concentration 3 Question efine parameters / variables reactor V parameters operator microorganism q Y m K state variables 4 follows from biomass mass balance d(v rv + r d r t V constant transport of biomass kinetics o exp( t exponential growth!! slope in - curve is r!! follows from mass balance d(v r V + transport of r q q q ( and < constant volume V: d r t q q q kinetics t end of batch can be calculated by putting q [ exp( t ] exp( t 5 slope in - plot is r!! 6

5 Results of batch t end -q Y -r onclusion t end r t end -r i t end t end t end t end t end 7 omments on batch process, Y and q are constant (, Y, q Microbiological model parameters are easily determined by combining the exponential equation with and data versus. Y Y : plot ln : Y q + q Y ( m Y Y ( m, the most important rate, can not be controlled in batch + 8 hemostat hemostaermentation φ L in medium in m 3 /h in V m 3 spent air φ L out medium out m 3 /h conc. fresh air teady state constant volume constant φ L, in, φ L, out constant and efine parameters / variables reactor V φ L,out parameters operator φ L,in,in microorganism q Y state variables 9 m K

6 tart sterilized reactor is filled with sterile growth medium solution containing, in air sparging starts, to provide electron acceptor add a small amount of micro-organism (inoculum start the medium in flow (often φ L, in and φ L, out are nearly equal, not always wait until a steady state is achieved Main property of chemostat and are independently manipulated by the experimentator by manipulating transport (to and from the reactor of and biomass excellent experimental tool to study microbial kinetics q (, q P (, stoichiometry, under controlled conditions Biomass mass balance (steady state d( V r V φ L out alculate r from measurements r follows from experimentally measured data and the biomass mass balance: φl,out r V alculate from measurements r Thus: r def L,out φ V dilution rate The experimentator chooses and therefore ( is also choosen!! Because is choosen, all other q i -values are choosen (stoichiometric coupling. (see chapter on kinetics alculate Use kinetic expression for (which is min K + rewrite : min + K independent of, in!! min converted ( crit Note, in order mg/l min order mg/l K order mg/l, in crit The imal value for (which is, attainable in the chemostat, is achieved when the imal (which is, in, is achieved. min ( in crit K +, in non-converted 3 Wash out For > crit, in and 4

7 ubstrate mass balance (steady state ( d V Vr +φl,in,in φ L,out ( r ( α,in φl,in φl,out α, φl,out V r This equation gives r and q from measured concentrations/flows/volumes d ( q + ( m Y alculation def r q α ( / Y + ( m (,in Y mostly for << lope α,in /(-m why a drop in ( Y crit ( α,in Hence function was obtained before as function of 5. 6 Application of chemostats hemostat behavior kinetic studies At different ( values one varies q, and one can obtain, m,q Y, K (see example crit -q crit fysiological studies of micro-organisms under defined st st i.e. change - - electron acceptor - N-source - change type of limitation for growth waste water purification Y r crit crit crit -r -r O crit crit crit industrial fermentation (not widely applied r O 7 crit 8

8 Maximal capacities of conversion in the chemostat an you calculate the -value where r (and r are imal (which is an economic important parameter Economic parameter of interest is r kg converted / m 3 reactor h Assume α From mass balance: r ( s in Following the - graph, r has imum at higher, therefore assume m negligible ( min K r in K The optimal follows after differentiation 9 dr d 3 Example A chemostat, V.5 ltr, is fed with a glucose medium (with, in mg/l at differenlow rates. The yeast accharomyces cerevisiae is studied. It can be assumed that in- and out-flow are the same. The following concentrations of glucose (, biomass ( and ethanol ( e are found. alculate φ L l/h r g/lh φ L l/h mg/l g/l e g/l r g/lh r e g/lh /h q g/gh q e g/gh Y g /g Y e g e/g Example.6 ltr/h, in g/l chemostat V.5 ltr constant.5 ltr/h 3 g/l 6 g/l a. What may cause the lower out-flow rate b. alculate r,r,, q, Y and units c., in is changed from g/l to 4 g/l. The flow rates do not change!! A new steady state is achieved. What is the new d. What is the new, q, Y alculate: q, K,m,,q e ( plot 3 e. alculate the new 3

9 hemostat wash-out dynamics hemostat Assume that is close to crit close to t is increased to above Question: what happens to Answer: biomass mass balance d ' ' < but constant ( t ( t exp ( ' t Industrial application Not much. Why low low P loss of biomass in outflow microbial selection for non-producing mutants wash-out curve gives!! t election of micro-organisms due to competition micro-organism micro-organism (infection or mutant Two situations Non-crossing - functions micro-organism is in chemostat steady state at dilution rate and micro-organism micro-organism Two organisms compete for the same but with different kinetics micro-organism : f ( micro-organism : f ( 35 suddenly micro-organism appears (mutation or infection and obtains > Mass balance micro-organism d ( increases less for m.o. decreases and decreases After some : new steady state onclusion: 36 at a given ( the micro-organism with the lowest wins!!

10 rossing - functions * micro-organism micro-organism 3 Practical consequences of competition in chemostats 3. Product instability Observation P chemostat P * Who wins at ( > * at ( < * m.o. wins m.o. wins hr product capacity disappears in!! Explanation ompetition between producing micro-organism and non-producing Mutant. m.o.: q + q P + m Y YP m.o. : q + m mutant Y Assume equal maintenance and uptake kinetics for m.o. and m.o. + ( Y / YP qp onclusion: always > 38 m.o. takes over Big problem in industrial chemostats!! 3. Evolution of microbial properties in the chemostat 4 General conclusion Observation g/l chemostat mg glucose/l Mutants which, at the same, do grow at a lower, can out compete the others. 5 4 hr The concentration drops in a chemostat aixed!! There is a strong selection pressure for low!! Explanation The microbial population is not exact uniform (remember there are 4 cells/l Micro-organisms with better uptake can maintain at a lower these win!! 39 4

11 Fed batch fermentation Fed batch fermentation batch high high no transport of or no control on chemostat low low biomass transport in, out biomass out control on fed batch low high transport in, not out no biomass transport 4 Why fed batch low no toxicity / osmotic problems high high P easier P control of 4 Fed batch fermentation batch phase 5 to mg/l start feeding, rate φ mol /h In limited feeding phase, is very low (in the range of mg/l. Hence one can use the pseudo steady state condition for mass balance d( V Vr +φ φ r V feeding phase limited conditions!! to 5 mg/l In a fed batch the conversion rate (-r is controlled by the operator Through controlled r, is controlled 43 Fed batch fermentation Volume changes The mass of the reactor content changes dm sum of all mass-flows, entering / leaving kg kg h h Mass entering: feed solution. Mass leaving: loss of evaporated water. Assuming ρ kg/m3 mass balance becomes a volume balance dm dv For simplicity we assume 44 constant volume

12 alculation of rates from fed batch Assume volume V is variable!! : from biomass mass balance ( V d r d V ( V ( V V q : Pseudo steady state mass balance ( r φ q V d( VP q P : qp ( V d( VP qp V Hence from measured V,, P, φ, one calculates, q,q P For simplicity we assume V constant 45 Example fed batch In a fed batch fermentation process, a solution (5 g/l glucose is fed into the reactor with a constant rate of 8 l/h The following measurements are made at t 9 h and t 9 h. hours 9 9 volume m 3 5 kg/m P kg/m 3 - calculate φ - calculate for average 9 h:, q,q P Answer φ kg glucose/h h ( q.76 g /g h q P g ( ( 9 9 ifferent possible feeding strategies in a fed batch fermentation The key to fed batch calculations φ is constant is maintained at opt, which gives e.g. a imal q P or imal Y P (see chapter on kinetics ubstrate feeding rate φ is determined by other known reactor limitations (O, heat, ( r is known ( φ /V from the applied feed rate φ!! Use the Herbert-Pirt equation to calculate r Use the biomass mass balance to calculate 47 48

13 Biomass mass balance in fed batch d ( V d r V Assume constant volume, therefore: r Eliminate r using Herbert- Pirt (product term absent and eliminate ( r using fed batch relation r φ /V d φ Y Y ( m V Homogeneous diffential equation in reactor parameters operator microorganism state variable 49 d omments on biomass mass balance Y φ Y V maintenance ( m initially is very low in a fed batch neglect maintenance d φ Y constant V later end increases linear in increases, more and more needed for maintenance d decreases all needed for maintenance, d φ / V m ( 5 Fed batch: φ constant ( t Analytical solution for (t in feeding phase Y + f φ exp V Y exp ( Y ( m t ( m ( m t ( Y t after stareeding f biomass concentration at stareeding Note: For t >> exp( φ / V maintenance m : For t << exp ( ( Y ( m exp Y ( m Y ( m Y ( m t ( t ( f + Y φ t V t + 5 linear growth Fed batch: φ constant Relations for r (t, (t and q (t In the feeding phase ubstrate Herbert-Pirt r (t Y r t (t t r -q (t t φ Y V m r decreases ( t constant analytical solution ( ( ( decreases increases constant increases strongly decreases q decreases but not to slowly 5 decreases

14 f Fed batch behavior: φ constant batch feeding phase batch feeding phase φ / V m φ slope Y V Fed batch: opt constant in feeding phase!! How do,,, q, Y,r,r, φ change with in feeding phase -q -q opt constant q i,y ij are all constant (see chapter kinetic q constant q opt constant r -r biomass concentration increases exponentially d opt r Y Y φ r opt ( t q ( t i i all rates increase exponentially 53 φ (t (-r V increases exponentially φ φ ( exp ( opt t 54 Fed batch behavior: opt batch feeding phase batch feeding phase exp f opt r Y Y exp -q -r φ -q 55 Including producormation A change in the biomass concentration curve, the Herbert-Pirt equation changes r r + r P + Y YP q P function of Assume q P a + b ( m or r P a +br b a + r + + ( m r Y YP YP ombine with biomass mass balance by eliminating r a + ( m d Y b + Y P φ Y V Y P b + Y P ame solution-structure as before!! For a on b very small, hence q P very small; the same solution is obtained q P a slopeb 56

15 Including producormation ubstrate sampling φ / V ( m product + product φ / V m + a / YP ( g/l mg/l sample (including biomass t P -r 5 mg/l sec chemostat sample tube P slope r P a After sampling, consumption of continues!! drops severely 57 What about 58 ampling ample tube is a constant volume batch reactor. does not significantly change. ubstrate mass balance d q Rearranging K + d Integration with t q K + ( q K ln + R R R This relation gives R note: q <!! q constant constant R ( q t as function of 59 Assume K - g/l q -3 g /g sec g /l R - g/l t(sec.. / R.9.5 trong drop in within second How can we prevent this 6

16 ompetition for the same Modelling microbial competition / selection micro-organism micro-organism Required mass balances for micro-organism micro-organism 6 Kinetics for micro-organism ( and q micro-organism ( and q and kinetics of conversion of m.o. m.o. (e.g. due to mutation usual first order mutation kinetics of course not needed if m.o. and m.o. are completely different (e.g. m.o. is due to a contamination in 6 the reactor General mass balances Micro-organism d( V V km Micro-organism ( d V ( d V ubstrate q V + k V + q m V + V + V + biomass transport biomass transport transport 63 election due to competition in a chemostat For a chemostat the transport terms are (assume same in- and outflow rate φ (m 3 /h micro-organism : φ micro-organism : φ : φ (, in Mass balances for chemostat (V constant m.o. m.o. substr d d d ( ( ( ( ( + +,in k m + k m q q 64 < <

17 election due to competition in a chemostat General Use the proper kinetic functions for,,q (, q ( these are non-linear No analytical solution possible omputer simulation General findings stable selection is possible for simple cases (one micro-organism wins!! stable co-existance is determined by the competition rules often composition of the mixed culture varies chaotically in very strong dependence of competition outccome in the reactor used (especially non-ideal mixing or dynamic feeding of 65 analytical solutions are possible with some assumptions Analytical solutions for chemostat ase A Assumptions Both micro-organisms have the same yield on, Y, and maintenance is negligible (or is maintained high enough and therefore yield is constant ( + Y Y We assume first-order kinetics Y (,in, in very low slopek slopek ( ( m.o. (contaminant m.o. ' k q k ' k q k ' k ' k where k and k Y Y learly k > k, so m.o. will win!! Assume that k m Assume that at t being very low!! 66 ase A (continued, solution ubstrate concentration mass balance We assume pseudo steady state d and we use + Y,in ase A (continued Micro-organisms concentrations, Mass balances m.o. m.o. d d ( ( Result k for ( + + k ( start condition k ( end condition k o concentration will drop (k > k in the competition experiment However generally k is slightly larger then k k or k 67 d ln subtract and use d ln k k d ln ( k k use that k d ln k k k 68

18 ase A f f Integration with f + gives for f + f exp ( k k t k f f + f exp k k t k ( k k required k k.,.h 5 k 69 hr!! ase B Presence of a mutant with loss of product capacity uppose at t : Parent strain (microorganism with q p q p, independent A mutant is formed or present (microorganism with q p and fraction f Both organisms have same q Herbert-Pirt microorganism q + q p + m Y Yp microorganism q + m Y Y Hence q p constant k Yp For higher producing strains k increases!! Using the two mass balances, gives for fraction non producer 7 fraction f exp ( kt f non producer f + f exp kt ( ( onclusion microbial competition and selection Leads to changes in composition of the population (ratio The constant of the changes relates to differences, which leads to at least hundreds / thousands of hours before the competition is finished 7