Lecture 13: Biological Reactors - Chemostats

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 fermentor operations Biological Reactors (Chemostat) Concentration/Combustion constant Biological CSTR [S]0 F [S], x V F Figure 1. Diagram of a chemostat. F = Volumetric flow rate x= biomass 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 + rx V = 0 at steady-state Cell growth kinetics rx = μ x − Fx + μ x V = 0 F Solve μ = V F 1 = D=Dilution rate ≡ V τ μ=D Biological 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 F [ S ]0 − F [ S ] − at steady-state 1 μx V = 0 Yx s mass biomass created Yield coefficient mass substrate consumed Divide by V μx D ([ S ]0 − [S]) =  Yx change in sugar concentration s At steady-state μ=D x = Yx ([ S ]0 − [S]) s What is the value of μ = f ([S]) [S] ? What more information do we need? Å must choose a growth model to connect μ and [S] Monod growth model: μ= μmax [ S ] K s + [S ] [S ] = Æ at steady-state Æ Ks D μmax − D D= μmax [ S ] K s + [S ] substitute in x equation 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Prof. K. Dane Wittrup Lecture 13 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]. ⎛ Ks D ⎞ x = Yx ⎜ [ S ]0 − ⎟ μmax − D ⎠ s ⎝ Specifying μmax , K s , Yx , D , [ S ]0 , can predict x , [S] . s x < 0 is non-physical but formally in solution μmax − D can go to 0. If you turn knobs incorrectly: if D is too high, the cells cannot grow fast enough to reach steady-state. Washout will occur. so use x = 0 to find Dmax Dmax = μmax [ S ]0 K s + [ S ]0 For D > Dmax “washout”, no steady-state. S x, S x mass volume washout D Dmax Figure 2. Biomass/volume versus dilution rate. Beyond the maximum dilution rate, washout occurs. For real systems K s << [ S ]0 . Most cell growth systems reach maximum at fairly low concentrations; hence x is flat, then drops off sharply. If biomass is the product, is there a best operating condition? What should we consider? dx optimize x with respect to D? D=0 (no, because this would be batch reactor) dD Define productivity as biomass (reactor volume )( time ) = xD 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Prof. K. Dane Wittrup Lecture 13 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]. d ( xD) = 0 for optimum. (is a maximum) dD x xD D D Doptimum Figure 3. Left: Biomass/volume versus dilution rate. Right: Productivity versus dilution rate. ⎛ Ks Doptimum = μmax ⎜⎜1 − K s + [ S ]0 ⎝ ⎞ ⎟⎟ ⎠ K s << [ S ]0 Doptimum ≈ μ max ≈ 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 x final = 100 g L If Yx ≈ 0.5 , [S]0 = wt 200 g ≈ 20% Å Toxic, sugar content cells will die L volume s Why do we not feed all at once? Cells will die. Calculate medium feed rate in order to hold μ constant. 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Prof. K. Dane Wittrup Lecture 13 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]. [S]0 F x(t) S(t), very small Figure 4. Diagram of a fed-batch fermentor. If μ is constant, biomass = biomass t =0 e μt There is a dilution term, because as we feed in fresh medium, volume will change. Volume often doubles. x V = x0 V0 e μt Feed  μ x0 V0 e μt F [ S ]0 = Yx sugar feed  s sugar consumed Assume all converted into biomass. F= x0 V0 μ e μt [ S ]0 Yx s Exponential flow rate. Typically μ specified as “small.” Dilution: dV =F dt ⎛ ⎞ x0 ⎜ ⎟ V(t ) = V0 ⎜1 + e μt − 1) ⎟ ( ⎜ [ S ]0 Yx ⎟ s ⎝ ⎠ x= biomass V x0 e μt = 1+ x0 e μt − 1) ( Yx [ S ]0 s = x0 V0 e V μt 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Prof. K. Dane Wittrup Lecture 13 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]. “Logistic equation” x Figure 5. Graph of logistic growth. t If product is something cells are making: Product synthesis kinetics 1) 1 dP = αμ x dt P≡ 2) growth associated (e.g. ethanol) product volume 1 dP =β x dt t P = β ∫ xdt 0 not growth associated (e.g. antibiotics, proteins, antibodies) integrate for amount of product. 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Prof. K. Dane Wittrup Lecture 13 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].