Chemostat Recycle [autosaved] 1z5e3e

Continuous Bioreactors – Chemostat with Recycle

Batch Reactors • Cell Growth rX 

 S dX  X  m X dt KS  S

• Substrate Utilization

m S X dS X rS    dt YX / S K S  S YX / S

• Product (cometabolic contaminants use negative sign) 1 qp  X 1 qp  X

dP  YP / X  g   g   dt m S dP   dt KS  S

Continuous culture: the chemostat 1. Concentration of a limiting nutrient 2. Dilution rate

Results from a batch culture Monod Kinetics applies!!! Substrate depletion kinetics !!

F D V

D is dilution rate F is flow rate V is volume

Continuous Reactors

• Chemostat - CSTR - continuous stirred tank reactor for the cultivation of cells. – mixing supplied by impellers and rising gas bubbles – assume complete mixing - composition of any phases do not vary with position – liquid effluent has the same composition as the reactor contents

Mass Balance on Chemostat Acc = in - out + gen - cons dci VR  Fcif  Fci  VR rfi dt – VR - reactor volume – F - volumetric flow rate of feed and effluent streams (they are equal) – ci - concentration of component i in the reactor – cif - concentration of component i in the influent or feed stream

• If we have a steady state reactor - no changes in composition with time then and

dci 0 dt

• Define as the dilution rate -

F rfi  of the mean (ci holding  cif or) residence time – reciprocal V – detention timeR

• For cell mass, if we assume a sterile feed: ci = X and Xf = 0 and

rx = X then X = DX

F 1 D  VR 

D= at SS

Chemostat with Monod Kinetics mS D KS  S

KS S  mm 1

• The above equations only holds if mmax >1 • If mmax < 1 or D> – washout of the cells occurs – Cells leave the reactor faster than they are dividing.

Dmax

mS f  KS  S f

Near washout the reactor is very sensitive to variations in D • Small change in D large shifts in X and/or S •If max = 0.5 hr-1 then D< 0.4 hr-1

Cell Growth in Ideal Chemostat Determination of Monod Parameters

m S g  D  KS  S

In Chemostat: µg=D, varying D obtains D~S

K 1 1 1  S  ( Lineweaver - Burk) D m S m 4 3.5

DX

0.3 0.25

3

0.2

DX (g/L-hr)

S, X (g/L)

Washed out: If D X S is set at a value µm = 0.2 2 0.15 greater than µm 1.5 hr-1 0.1 1 (D > µm), 0.05 0.5 the culture 0 0 0 0.05 0.1 0.15 0.2 0.25 cannot D (1/hr) reproduce quickly enough Chemostat technique: reliable, constant environment, operation may be difficult. to maintain 2.5

Intracellular Product Formation -Chemostat dci VR  Fcif  Fci  VR rfi dt

• Steady State and ci = P

D  Pf  P   YP / X X  0 • If Pf = 0

YP / X X P D

Substrate Balance on Chemostat Intracellular Product • If ci = S

dci VR  Fcif  Fci  VR rfi dt

FS 0  FS  VR X

• At Steady State

1 YX / S

X D S 0  S   YX / S

• With Monod



dS  VR dt X  YX / S  S 0  S 

Ks D   X  YX / S  S 0  m  D  

Chemostat with Extracellular Product

• Cell Mass Balance

dX VR  FX 0  FX  VR X dt D

• Substrate Balance FS 0  FS  VR X

1

YX / S

 VR q P X

• Solve Substrate at SS for X

1 YX / P

dS  VR dt

Class Exercise •E. coli is cultivated in continuous culture under aerobic conditions with glucose limitation. When the system is operated at D= 0.2 hr-1, determine the effluent glucose and biomass concentrations assuming Monod kinetics (S0 = 5 g/l, m= 0.25 hr-1 , KS = 100 mg/L, Y x/s = 0.4 g/g)

Chemostat with Recycle FR, XR F, X0

F, X2 V, X1 F+FR, X1

• • • • • •

 F ­ nutrient flow rate V ­ reactor volume X1 ­ x concentration in reactor X2 ­ X concentration in effluent XR ­ X concentration in recycle FR ­ recycle flow rate

Chemostat with Recycle Cell mass equation Acc = in ­ out + gen  dX 1 F X0  + FR XR   ­ (F+ FR) X1 + VX1 =       V  dt

FR, XR F, X0

F, X2 V, X1 F+FR, X1

Chemostat with Recycle cont. Define   = FR/F  recycle ratio C = XR /X1  concentration  factor

Substitutions • F + FR = (1 + )F • FRXR term FR = F XR = CX1  

FRXR = CFX1

dX 1 F X0  + FR XR   ­ (F+ FR) X1 + VX1 =       V  dt dX 1 F X0  + CFX1­ (1 + )F X1 + VX1 =       V  dt

Recycle cont • Assume

dX 1 – steady state                   = 0 dt

– sterile feed X0 = 0

Then 

(C ­ 1 ­)F  + V = 0 

Chemostat can be  operated at higher  dilution rates than the  specific growth rate  when cell recycle is  used.

If D = F/V for recycle    = D(1+ (1 ­C))   if  C > 1 (concentration of cells)  then  (1 ­ C) < 0 then  < D

Substrate balance - Recycle X 1 dS FS 0  FS  V  (1   ) FS  V YX / S dt • At Steady state and substituting for 

YX / S ( S 0  S ) D X 1  YX / S ( S 0  S )   (1    C )

Recycle Substrate cont. • Assuming Monod

K S D(1    C ) S  max  D(1    C )  YX / S K S D(1    C )  X1   S0   (1    C )   max  D(1    C ) 

Chemostat with Cell Recycle

No recycle

µm=1.00h-1, S0=2.0g/l, Ks=0.01 g/l, Yx/s=0.5 g/g, concentration factor C=2.0 and recycle ratio α=0.5

In Class Exercise • Consider a 1000 L CSTR in which biomass is being produced with glucose as the substrate. The microbial system follows a Monod relations with m = 0.4 hr –1, KS = 1.5 g/L, and yield factor = 0.5 g/g. If S0 = 10g/L glucose and F = 100 L/h: – What is the specific biomass production rate (g/l-h) at SS? – If recycle is used with a recycle stream of 10 L/h and a recycle biomass concentration five times as large as that in the reactor exit, what would be the new specific biomass production rate?

Multi-stage Chemostat System In some fermentations, the growth and productformation steps need to be separated. e.g. secondary metabolite, culture of genetically engineered cells.

P1≈0 Growth stage

P2 Product formation stage

At steady state, Vn, Xn,Sn,,Pn in the reactor of each stage don’t change with time.

Multi-stage Chemostat System

X0=0, Vi, i=1,2….n constant.

Stage 1: cell growth condition, Kd=0, qp=0 dX 1 FX 0  FX 1  1 X 1V1  V1 Cell mass: dt At steady state X0 1  D1 (1  ) X1 Limiting substrate: At steady state

1 X 1 dS1 FS 0  FS1  V1 M  V1 dt YX / S

1 X 1 S1  S 0  D1YXM/ S

1 ,  2 ... and  n are net specific growth rates in Stage 1, 2,..., and n, recspectively. When Kd  0, they are equal to the respective specific gross growth rates in each stage.

Multi-stage Chemostat System Stage 2 – product formation conditions, Kd=0, F’=0 dX 2 FX 1  FX 2   2 X 2V2  V2 ‘ Cell mass: dt X F At steady state  2  D2 (1  1 ) where D 2  X2 V2 F

P1≈0

P2

V2 are constant.

Multi-stage Chemostat System Stage 2 – product formation conditions, Kd=0, F’=0 Limiting substrate: At steady state

Product: At steady state

qp X 2 2 X 2 dS 2 FS1  FS 2  V2 M  V  V2 YX / S YP / S dt

qp X 2 2 X 2 S 2  S1   M D2YX / S D2YP / S

dP2 FP1  FP2  V2 q p X 2  V2 dt qp X 2 P2  P1  D2

Multi-stage Chemostat System Stage n – product formation conditions, Kd=0, F’=0 Similarly, equations could be obtained for nth stage. X n 1  n  Dn (1  ) Xn

qp X n n X n S n  S n 1   M DnYX / S DnYP / S

Pn  Pn 1 

qp X n Dn

If μn (e.g. Monod model) and qp are known functions, Xn , Pn, and Sn at nth stage could be determined by the above equations.

Chemostat in Series (Additional Feed in Second Stage) • Cell balance around second stage dX 2 F1 X 1  F ' X '( F ' F1 ) X 2  V2  2 X 2  V2 dt

• At Steady State with X’ = 0 F1 X 1  2  D '2  V2 X 2 D '2 

F1  F ' V2

Growth rate does not typically follow Monod in Second Stage if additional feed.

Chemostat in Series cont. • Substrate Balance if Additional Feed

F1S1  F ' S  ( F1  F ' ) S 2  V2  2 X 2 ' 0

1 YX / S

dS  V2 dt

• At steady state the two equations can be  solved simultaneously for S2 and V2 • Major advantage is to separate production  from growth

In Class Example – 9.2 • In a two stage chemostat system, the volumes of the first and second reactors are 500 L and 300 L respectively. The first reactor is used for biomass production and the second is for a secondary metabolite formation. The feed flow rate to the first reactor is F = 100 L/h, and the glucose concentration is 5.0 g/L. Use the following constants for the cells. 

m

= 0.3 h-1, Ks = 0.1 g/L Y X/S= 0.4 g/g

• Determine the cell and glucose concentrations after the first stage. • Assume that growth is negligible in the second stage and the specific rate of product formation is q P = 0.02 gP/g cell hr, and Y P/S = 0.6 gP/gS. Determine the product and substrate concentrations in the effluent of the second reactor.

Fed Batch Reactor • Reactor Design Equation V

FA0  FA  

dN A rA dV  dt

• No outflow FA = 0 • Good Mixing rA dV term out of the integral

dN A d  C A  V  FA0  rA  V   dt dt

Fed Batch Continued • Convert the mass (NA) to concentration. Applying integration by parts yields

dC A dV FA0  rAV  V  CA dt dt • Since dV  FA0 dt • Then dC A FA0  rAV  V  C A FA0 • Rearranging dt C A FA0 dC A FA0   rA  dt V V

Fed Batch Continued • Or

dC A FA0 1  C A   rA  dt V

• Used when there is substrate inhibition and for bioreactors with cells.

Fed-batch Reactors d (Vci )  Vr fi  F (t )c fi dt dV  F (t ) dt Differentiation the above equation using chain rule, and substitute  for dV/dt

dci F (t )  rfi  [cif  ci ] dt V

Fed-Batch Analysis of fed-batch with substrate continuously fed and no output: t=0, V0=0, F is constant. dV  F  V  Vo  Ft dt At quasi steady state, S added=S consumed, X, S, P concentrations are constant.

Volume:

Cell mass balance: since

FX O  V net X 

d ( XV ) dX dV V X , then dt dt dt

FX O  V net X  V

V net X  X

d ( XV ) dt

dX dV X dt dt

dV 1 dV F   net   D dt V dt V F F  net   V

 net  D 

m S if K d  0 Ks  S



Do 1  D0 t

Then S 

Ks D m  D

V0  Ft

(Monod growth model applied)

Fed-Batch

M

where S≈0

assuming X≈Xm M

where Xt=Xt0 at t=0

Fed-Batch (g) qp (i.e. g product/g cells-min) at Pt=Pt0, t=0

Class Exercise – 9.4 • Penicillin is produced in a fed-batch culture with the intermittent addition of glucose solution to the culture medium. The initial culture volume at quasi-steady state is V0= 500 L, and the glucose containing nutrient solution is added with a flow rate of F = 50 L/h. X0 = 20 g/L, S0 = 300 g/L, m = 0.2 h-1, Ks = 0.5 g/L and Y x/s= 0.3 g/g • Determine culture volume at t = 10 h • Determine concentration of glucose at t = 10 h • Determine the concentration and total amount of cells at t = 10 h • If qp = 0.05 g product.g cells h and P 0 = 0.1 g/L, determine the product concentration at t = 10 h