Biological Reactors (Chemostat)10.37 Chemical and Biological Reaction Engineering, Spring 2007 Prof. K. Dane Wittrup Lecture 13: Biological Reactors- Chemostats This lecture covers: theory of the chemostat, fed batch or semi-continuous fermentoroperations Biological Reactors (Chemostat) Concentration/Combustion constant Biological CSTR [S]0 V F F [S], x Figure 1. Diagram of a chemostat. F = Volumetric flow rate biomassx = volume [S]0 = Concentration of growth limiting substrate. (for growing cells) At steady-state, biomass balance In – Out + Prod = Acc Sterile feed: In=0 Steady state: Acc=0 −+Fx r V0 at steady-state x=Cell growth kinetics rxx=μ −+FxμxV0= Solve Fμ= VD=Dilution rateF 1≡= Vτ μ= DBiological Mechanical Cite as: K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].When at steady-state, can control cell mass. Allows precisely reproducible cell states. Not easy to run at steady-state. Material balance on [S] (sugar concentration) In – Out + Prod = Acc 0 at steady-state 1FS[]0−−FS[]μxV=0 Yxs Yield coefficient mass biomass created mass substrate consumed Divide by V μxDS([ ]0−=[S]) changeinsugarYxconcentrations At steady-state μ= D x =−YSx([ ]0[S]) s What is the value of [S]? What more information do we need? μ= f ([S]) Å must choose a growth model to connect μ and [S] Monod growth model: μmax[]Sμ= Æ at steady-state Æ KSs+[]μDmax[]S= KSs+[] KD[]S =s substitute in x equation μmax− D 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 13 Prof. K. Dane Wittrup Page 2 of 6 Cite as: K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].⎛⎞KDxY=−x⎜⎟[]Ss 0s⎝⎠μmax−D Specifying μ, maxKs, Yx, D, []S , can predict 0x, [S]. s x < 0 is non-physical but formally in solution μ− D can go to 0. If you turn knobs incorrectly: if D is too high, the cells cannot maxgrow fast enough to reach steady-state. Washout will occur. so use x = 0 to find D max μDmax[]S0max= KSs+[]0 For DD> “washout”, no steady-state. max ,x S massvolume D maxD S x washout For real systems KS. Most cell growth systems reach maximum at fairly low s<< []0concentrations; hence x is flat, then drops off sharply. biomass is the product, is there a best operating condition? hat should we consider? 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 13 Prof. K. Dane Wittrup Page 3 of 6 Cite as: K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. Figure 2. Biomass/volume versus dilution rate. Beyond the maximum dilution rate, washout occurs. If W dx optimize x with respect to D? D=0 (no, because this would be batch reactor) dD Define productivity as ()()biomass= xD reactor volume timedx()D=0 for optimum. (is a maximum) dDx D xD D optimumD Figure 3. Left: Biomass/volume versus dilution rate. Right: Productivity versus dilution rate. ⎛⎞KDs optimum=−μmax⎜⎟1⎜⎟KSs+⎝⎠[]0 KSs<< [] 0 Doptimum≈μmax ≈Dmax Close to washout conditions. Operability would be difficult. We would not want to run too close to washout conditions. Fed-batch fermentor (microbes or mammalian cells) -used to achieve very high cell densities (e.g. hundreds of grams cell dry weight (c.d.w)/liter) If you want 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 13 Prof. K. Dane Wittrup Page 4 of 6 Cite as: K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 100 gxfinal= L 200If 0gwtY ≈ 0.5 , [S] =≈20% Å Toxic, sugar content cells will die xsL volume Why do we not feed all at once? Cells will die. Calculate medium feed rate in order to hold μ constant.[S] F0 x(t) S(t), very small Figure 4. Diagram of a fed-batch fermentor. If μ is constant, 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 13 Prof. K. Dane Wittrup Page 5 of 6 Cite as: K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. biomass=biomasstt0eμ = There is a dilution term, because as we feed in fresh medium, volume will change. Volume often doubles. xVV= xeμt 00μxt00V eμFeedFS[] 0=xsugar feedYssugar consumed Assume all converted into biomass. x00VF =μeμt []SY0 xsExponential flow rate. Typically μ specified as “small.” Dilution: d V=F dt ⎛⎞⎜⎟xV(te) =+V001()μt−⎜⎟1 []SY⎜⎟0 x⎝⎠s ()biomassxeμtx ==0Vx11+−0eμtYSx[]0 sxeVμt=00Vogistic equation” “Lx t Figure 5. Graph of logistic growth. If product is something cells are making: Product synthesis kinetics 1) 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 13 Prof. K. Dane Wittrup Page 6 of 6 Cite as: K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 1 dP=αμ growth associated (e.g. ethanol) xdtproductP ≡ volume2) 1 dP=β not growth associated (e.g. antibiotics, proteins, antibodies) xdtPx=β∫tdt integrate for amount of product.