Bioreactor- Chemostat Modeling - Bioengineering Fall 2021 PDF

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Bioreactor - Chemostat modeling: - cell replication following Monod equation, medium including cells and substrates is continuously removed. Governing equations and mathematical modeling...

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Bioreactor – Chemostat

 Cell replication follows Monod equation.  Medium including cells and substrates is continuously removed.

Q: volumetric flow rate Sin: concentration of substrate in the influent flow V: volume of bioreactor S: concentration of substrate in the bioreactor and effluent flow X: concentration of cell in the bioreactor and effluent flow Initial condition: X = X0 at t = 0. Y is growth yield defined as

Y ≡−

∆ X −dX = dS ∆S

, therefore

dS=

−dX Y

holds.

Ignore cell death and product formation. Define dilution rate D,

D≡

Q V

.

Write the governing equation for the cells and substrate.

1

dX μm S X−DX = dt k s + S

(1)

dS = −1 μ m S X + D S −DS ¿ dt Y k s + S

(2)

Assume steady state is reached, at which

dX =0 dt

and

dS =0 dt

.

Update (1) and (2). μm S X−DX =0 ks + S

(3)

−1 μm S X+ D S¿ −DS=0 Y ks + S

(4)

(3) and (4) are two algebraic equations with two unknowns (X and S). So, they can be solved. Rearrange (3).

(

)

μm S −D X=0 ks + S μm S −D =0 k s +S

Either

 At the case of

(5) (eq. 6) or

X =0

is true.

μm S −D =0 ks + S

Rearrange (6). μm S k s+ S

(7)

ks D μm− D

(8)

D=

S=

Plug (7) and (8) into (4) to remove S.

2

k D −X =0 D +D S ¿ −D s μm −D Y

(9)

Rearrange (9).

(

X =Y S ¿ −

ksD μ m−D

Because of

)

(10)

X ≥0

(

Y S¿ −

,

)

ks D ≥0 μ m−D

(eq. 11) must be satisfied.

Rearrange (11). D≤

S ¿ μm

(12)

ks + S ¿

When

D<

S¿ μm k s +S¿

is true,

X >0

holds according to (10). This means

that a steady state with nonzero X and S exists and the steady-state S and X satisfy (8) and (10). When

D=

S ¿ μm k s+ S ¿

is true,

X =0

holds according to (10). Plug

X =0

into (4), S=S¿ holds. This means that a steady state can be reached with S being equal to Sin and X being equal to 0. D is an experimental parameter that can be set as any positive value. If

D

is greater than

simultaneously except

 At the case of Plug

X =0

S¿ μ m k s + S¿ X =0

, (4), (5) and

X ≥0

cannot be satisfied

holds, i.e., the case of

X =0

.

X =0

into (4).

D S ¿−DS=0

3

Because of steady state.

D >0

,

S=S¿

holds, meaning that S becomes Sin at

4

D≪

S¿ μm k s + S¿

Steady state is reached as t approaches infinity. At steady state, S is close to 0 and X is a nonzero value.

D<

S¿ μm k s + S¿

Steady state is reached as t approaches infinity. At steady state, S and X are nonzero values.

5

D=

S ¿ μm k s+S ¿

Steady state is reached as t approaches infinity. At steady state, X is close to 0 and S is close to Sin.

D>

S¿ μm k s + S¿

Steady state is reached as t approaches infinity. At steady state, X is 0 and S is close to Sin.

6

D≫

S¿ μm k s + S¿

Steady state is reached as t approaches infinity. At steady state, X is 0 and S is close to Sin.

Equations (8) and (10) indicates that both S and X are dependent on D. It is thus possible to maximize productivity of the system by adjusting D. Assume that the cells are the product, the productivity is DX. DX is determined by multiplying both sides of equation (10) with D.

(

DX =DY S ¿−

ksD μm − D

)

(13)

Plot equations (8), (10), and (13). S=

ks D μm− D

(

X =Y S ¿ −

(8) ksD μ m−D

)

(10)

7

(

DX =DY S ¿−

ksD μm − D

)

(13)

 For S-D curve, S increases from 0 when D=0 to  For X-D curve, X starts from D=

S ¿ μm k s+S ¿ S ¿ μm k s+ S ¿

at

D=

S ¿ μm k s+ S ¿

.

Y S¿

when D=0 and ends as 0 at

0

when D=0 and ends as 0 at

.

 For DX-D curve, DX starts from D=

S¿

. There is a peak with the highest DX.

When DX reaches the maximum, d (DX) =0 dD

(14)

Substitute DX in (14) with (13). d ( DX ) = dD

[ (

d DY S¿ −

ks D μ m−D

)]

dD

8

¿

(

2

Yk D d DY S¿ − s μ m−D

)

dD

[

¿ Y S ¿−

]

2 DY k s Y k s D 2 =0 + μ m−D ( μm − D )2

(15)

Rearrange (15) 2

2

Y S ¿ ( μm− D ) −2 DY k s ( μm − D ) −Y k s D =0 S ¿ μ2m + S¿ D 2−2 S¿ μ m D −2 D k s μm +2 k s D 2 −k s D 2=0 S ¿ + ks ¿ ¿¿ D ¿ ¿ 2

D −2 μm D+

S¿ 2 μ =0 S¿ +k s m

(16)

Solve equation (16) to determine D.

( √

D=μm 1±

ks S¿ +k s

)

Two solutions. Which one is correct? Plug

( √ )

D=μm 1+

ks S¿ +k s

and

( √ )

D=μm 1−

ks S ¿ +k s

into (8):

S=

ks D μm− D

respectively.

S=

ks D = μm− D

( √ ) ( √ ) ( √ ) √

k s 1+

ks S¿ +k s

ks 1− 1+ S ¿ +k s

k s 1+

=

ks S ¿ +k s

ks − S¿ + k s

0

must be greater than 0, the correct solution is

( √ )

D=μm 1−

ks S ¿ +k s

ks S ¿ +k s

(17)

(17) is equation 6.90 of the textbook.

10

Compare batch and chemostat on cell production rate (on pages 276 and 277 of the textbook) 1. Batch Governing equation: dX μm S = X dt k S + S

We have derived its solution in Lecture 7. −Y X / S k S Y X /S S0 + X 0

Normally 0 ∙ ln

ln

ln

Y X / S S 0+ X 0− X Y X / S k S + Y X / S S 0 + X 0 X + ln =μm t X0 Y X/S S0 Y X / S S0 +X 0

S 0 ≫k S

, thus the above equation can be approximated by

Y X / S S 0 +X 0 −X X +1∙ ln =μm t Y X/S S0 X0

X =μm t X0

Rearrange it. t=

1 X ln μm X0

Assume that time required for preparation, lag phase growth and harvesting is tl . So the total time required for harvesting the maximal attainable cell concentration, X m , is determined by t total =

Xm 1 ln +t μm X 0 l

The total cell mass produced per unit volume is determined by X m −X 0=Y X / S S 0

Where

Y X/S

is yield coefficient. 11

The rate of cell-mass production in the batch culture ( r b=

rb

) is

Y X / S S0 X m−X 0 = ttotal Xm 1 ln +tl μm X0

2. Chemostat At the steady state,

(

X =Y S ¿ −

ksD μ m−D

)

Equation (10)

The maximum productivity (DX) can be achieved when

( √ )

D=μm 1−

ks S ¿ +k s

Equation (17)

Substitute D in (10) with (17).

[

( √ ) ( √ ) ks S¿ +k s

k s μm 1−

X =Y S ¿−

μm −μ m 1−

ks S ¿ + ks

(

¿ Y S¿ −

] ks

√

ks S¿ + k s

+k s

)

¿ Y [ S ¿ +k s− √k s (S ¿ + k s ) ]

Therefore, the rate of cell-mass production in the chemostat system ( r c ) is determined by

( √ )[

r c =DX = μm 1−

ks Y S¿ +k s −√k s (S ¿ +k s) ] S ¿ +k s

12

Under normal circumstance,

( √

r c =μm 1−

) (

. So

S¿ ≫ kS

√ ) 2

k s ks k s /S ¿ Y S¿ 1+ k s /S¿ − + S ¿ S2¿ 1+ k s / S¿

( √ 1+00 ) Y S [1+0− √ 0+ 0 ]

≈ μm 1−

¿

¿ Yμm S ¿

The ratio for rates of cell-mass production is determined by rc rb

=

Yμ m S¿ Y S0 X 1 ln m +t l μm X0

Assuming

S 0=S ¿

,

Xm rc + μmt l =ln rb X0

Example: When

Xm =20 X0

,

μm=1 h−1

,

tl =5 h

, rc rb

=8

.

So, the chemostat is more productive than the batch culture. However, batch or fed-batch system is more commonly used in industry. Reasons:

13

1. Equation “

X rc =ln m + μ m t l X0 rb

” only applies to growth-associated

products. Many products are not made by growing cells. 2. Cells may be genetically unstable over a long period of culture. 3. Continuous culture system is more complex than the batch culture system, and is thus more prone to failure (pump, controller, sterility) 4. A continuous culture system is usually dedicated to a single product. The demand for this product may be hard to project. A batch culture system can be used to produce different products.

14

The chemostat system can be used to determine the kinetic parameters of the Monod equation. Rearrange (8)

S=

ks D μm− D

.

1 1 ks 1 = + D μm μ m S

Conduct experiment at different D and measure corresponding S. Plot

1 D

vs

1 S

. The y-intercept is

1 μm

and the slope is

ks μm

.

15

 Considering cell death in a chemostat system Assumptions: Cell growth follows the Monod equation (

μ=

μm S k s+ S

). Cell death follows

the first order kinetics ( reactionrate=−k d X . k d is the rate constant) and the cells do not become substrate after they die. Steady state for cell and substrate can be reached (

dX =0 dt

,

dS =0 dt

).

Governing equations: dX =μX −k d X− DX dt

(1)

dS −μX + D S¿ −DS = Y dt

(2)

At

dX =0 dt

, (1) becomes

μX −k d X− DX =0

(3)

Rearrange (3). μ=k d + D

At

dS =0 dt

(4) , (2) becomes

−μ X +D S ¿ −DS=0 Y

(5)

Plug (4) into (5). − (k d + D ) X + D S¿ −DS=0 Y

(6)

Rearrange (6).

16

X=

YD ( S − S) kd+D ¿

S ¿−S k d + D = X YD S ¿−S k d 1 = + X YD Y

Define

Y AP ≡

(7) X Si n −S

(called apparent yield) and

ms ≡

kd Y

(called

maintenance coefficient). (7) becomes 1 1 1 =ms + AP D Y Y

We can fix Plot

1 Y AP

S¿

vs.

, vary D, and measure S and X. 1 D

. The slope is

ms

and the y-intercept is

1 Y

.

17

An example from the textbook

Plot

1 Y AP

vs.

1 D

.

2.20

f(x) = 0.06 x + 1.6 R² = 0.99

2.10 2.00 1.90 1.80 1.70 1.60

1.50 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00

From slope,

m s=

From intercept,

kd =0.06 Y

(h-1)

1 =1.60 Y

18

k d =m s Y =0.038h−1

Productivity (DX) of cell mass can be maximized by adjusting D, so d (DX) =0 dD

We need to get the expression for X, which should not contain S. Equations (3) and (5) are two algebraic equations with X and S as variables. They can be solved to get the expression for X. Plug

μ=

μm S k s+ S

into (4):

μm S =k d + D ks + S

μ=k d + D

(8)

Rearrange (8) S=

k s (D +k d ) μm−(k d + D)

Plug (9) into (6):

(9) −( kd + D ) X Y

+ D S¿ −DS=0

k s (D +k d ) − (k d + D ) X =0 + D S¿ −D μm −(k d + D) Y

. (10)

Rearrange (10). X=

[

k s ( D+ k d ) YD S ¿− μm −( k d +D ) kd+D

]

(11)

(11) is the expression for X. Therefore, d d (DX) = dD

{ [

k s( D +k d ) Y D2 S¿− μm − (k d + D ) kd + D dD

]}

=0

19

D can be determined as we previously did.

20

 Product (P) formation at steady state in a chemostat Governing equations: dX =μX −k d X− DX dt

(1)

dS −μX q P X − = + D S ¿ −DS dt Y X / S Y P /S

(2)

dP =q P X−DP dt

(3)

Y X/S

: yield coefficient for producing cell with substrate

Y P/S

: yield coefficient for producing product with substrate

: specific rate of extracellular product formation. q P can be approximated with a constant if the product is non-growth associated and substrate concentration is high. qP

DP: Product formation rate

Assume a steady state:

dP dS dX =0 , =0 , =0 dt dt dt

is reached.

Equations (1), (2) and (3) become three algebraic equations. The fist two contain X and S as variables. We want to solve these two equations for X as a function of other parameters except S. Then we the third equation to get the expression for DP.

21

Since

dX =0 dt

, (1) becomes

0=μX −k d X−DX

Rearrange it. μ=k d + D

(4)

Plug the Monod equation

μ=

μm S k s+ S

into (4) and rearrange it.

μm S =k d + D ks + S S=

ks ( D + kd ) μm−(k d + D )

Since

dS =0 dt

D S ¿−DS=

(5) , (2) becomes

μX q P X + Y X /S Y P / S

Rearrange it. X=

D S¿ −DS q μ + P Y X / S Y P/ S

¿ Y X / S (S¿ −S)

Substitute

[

X =Y X / S S ¿−

μ

D Y X /S μ+q P Y P/S

(6)

in (6) with (4), and substitute S in (6) with (5).

k s ( D+ k d) μm − (k d +D )

]

D k d +D +q P

Production formation rate

Y X /S Y P/S

DP =q P X

when

dP =q P X−DP=0 dt

.

22

[

DP=q P X=q P Y X /S S ¿−

k s ( D+k d ) μm− ( k d + D )

]

D k d +D +q P

Y X/S Y P/S

To maximize the production formation rate (DP), adjust D to achieve d (DP) =0 dD

.

23...