1. Chapter Nine: Operating Considerations for Bioreactors for Suspension and Immobilized Cultures

2. Bioreactors
What two type of bioreactors have we discussed in Chapter Six?
Batch and Chemostat (CSTR).
What are the characteristics of each type of these two reactors?
Batch: changing conditions - transient (S, X, growth rate), high initial substrate, different phases of growth.
Chemostat: steady-state, constant low concentration of substrate, constant growth rate that can be set by setting the dilution rate (i.e. the feed flow rate) .

3. Bioreactors Which type is more efficient? Which type is more common?
Chemostat is more efficient.
Which type is more common?
Batch is more common (Reasons for Batch Reactor popularity are outlined on slide 14).

4. Choice of continuous vs. batch production
Productivity
Flexibility
Controllability
Operability
Regulatory
What do each of these factors mean?

5. Reactor Choices
Productivity: rate of product per time per volume. Chemostat better for growth associated products.
Flexibility: ability to make more than one product with the same reactor. Batch better.
Controllability: maintaining the same conditions for all of the product produced. In theory, chemostat better (since it operates at steady state). In reality, we do not know until we try.

6. Operability: maintaining a sterile system. Batch better.
Regulatory: validating the process. Initially, many process batch, too expensive to re validate after clinical trials.

7. Comparison of Productivity: Batch vs. Chemostat
Consider production of a growth associated product (like cell mass) in suspension culture
F S0 X0
F S X ?
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8. Batch Production Rate Batch cycle time is: (1)
where tgrowth is the time required for growth and tl is the lag time + preparation and harvest time.
(2)
where X0 is the initial concentration and Xm is the maximum attainable concentration of cells.

9. Batch Production Rate So net biomass production rate is:
(3)
Recall the definition of biomass yield:
(4)
From Eqs 2, 3 and 4, we get:
(5)

10. Chemostat Production Rate
For negligible kd, negligible extracellular product formation and steady state, we get (Eq. (10) in lecture notes of Chap Six):
(6)
For optimum cell productivity (X•D), calculate d(X•D)/dD, set equal to zero, and solve for Dopt:
(7)

11. Chemostat
Substituting Eq. (7) into Eq. (6) gives the value of X at the optimum production rate. :
(8)
Optimum productivity is D•X when D=Dopt and X= X (at Dopt):
(9)

12. Chemostat Productivity Rate
Noting that S0 is usually much larger than KS, we have:
(10)
Comparing the rates for batch production and production in a chemostat:
(11)

13. Comparison
Xm is always larger than X0 and is typically times larger, so the chemostat outperforms the batch reactor. For E. coli growing on glucose, µm is around 1/hr. Using tl=5 hr and Xm/X0=20,
Even so, most industrial fermentation processes occur in a batch reactor. Why?

14. Reasons for Batch Popularity
Equations were for cell mass (or other growth-associated product). Many industrial applications are for non-growth associated products.
Selective pressure of a chemostat is detrimental to engineered organisms
Batch is more mechanically reliable
Batch system is more flexible

15. Chemostat with Recycle
Can we operate a chemostat with a dilution rate greater than maximum growth rate? No
Why not?
Because the cell growth cannot keep up with how fast the cells are removed from the reactor, and after some time the cells would washout of the reactor.
When can we operate a chemostat with a dilution rate greater than maximum growth rate?
We can do that if we have recycled stream containing cells (chemostat with recycle).

16. Operation of Chemostats at High Dilution Rates
F, S0, X0
(1+α)F, S, X1
αF , S, CX1
F , X2

17. Chemostat with Recycle
Biomass (X) balance on the chemostat:
(12)
where α=volumetric recycle ratio and C=the concentration factor of the separator. At steady state and with X0=0:
(13)
(14)

18. Substrate Mass Balance
(15)
At steady state:
(16)
(17)

19. Steady-state Values Substituting µnet given by Eq. (14) into Eq. (17):
(18)
We can get the expression for the substrate concentration by equating the expression for µnet from Monod kinetics to Eq. (14):
(19)

20. Steady-state Values or: (20)
So now we can get X1 entirely as a function of D:
(21)

21. Example 9.1
In a chemostat with cell recycle has F=100 ml/h and V=1000 ml. The system is operated under glucose limitation, and
. Glucose concentration in the feed S0=10 g/l. The kinetic constants of the organisms are μm=0.2 h-1, Ks=1 g/l. The value of C is 1.5 and the recycle ratio (α)=0.7. The system is at steady state:
Find S concentration in the recycle stream
Find the specific growth rate (μnet) of the organisms.
Find cell concentration in the recyle stream (X1)
Find cell concentration in the centrifuge effluent (X2)

22. Example 9.1: Solution
To find X2, we do cell mass balance around the concentrator as follows:

23. Fed-batch Operation
Fed-batch reactors gain some advantages of a CSTR, but retain some disadvantages of batch.
Reduces substrate inhibition and allows for high conversion.
Semi-batch nature usually leads to higher operations cost and batch variability.

24. Fed-batch Operation
V0, X0, S, P
Start fed-batch
Fed batch fill
Harvest
Vw, X, S, P
V, X, S, P
F, S0
F, X, P
Fed-batch cultures are started as batch cultures and grown to an initial cell concentration X0, after which fed-batch operation begins.

25. Fed-batch Operation Notation:
S0= initial substrate concentration of batch
V0= initial volume of batch
F= constant flow rate of addition stream during fed-batch
X0= initial cell concentration of batch

26. Since liquid is being added, the volume is changing:
For a batch culture:
(22)
Since liquid is being added, the volume is changing:
or:
(23)
If the total amount of biomass (grams) in the reactor is Xt then the concentration X is:
(24)

27. Using the definition of the growth rate:
So the change in the biomass concentration with time is:
(25)
Using the definition of the growth rate:
...the dilution rate:
...and the expression for dV/dt:
we have:
(26)

28. Quasi-steady State
Substrate is consumed at the same rate as it is added.
Now, consider the case when the fed-batch is started from a culture in the initial substrate concentration was S0 and nutrient feed is begun at flow rate F and concentration S0. Just as nutrient feed begins:
(27)

29. At quasi-steady state, for this case we will have:
(28)
So X is constant (but not Xt). Now we have:
(29)
Assuming Monod growth kinetics (and also assuming kd=0), this gives (just as in the case of a chemostat):
(30)

30. For chemostat:
If the total amount of substrate in the reactor is St, then a substrate mass balance gives:
(31)
which, for quasi-steady state and qp=0, gives:
(32)
Returning to Equation (25), we have, at quasi-steady state:
(33)

31. Often, S<<S0 and X0<< , and so:
Integrating, we have:
(34)
since X is constant (dX/dt=0) and X=Xm. Therefore, the total biomass in a fed-batch reactor operated as assumed here increases linearly with time. Substituting the appropriate expression for Xm:
(35)
Often, S<<S0 and X0<< , and so:
(36)

32. Product Output
The product formation rate (in g product/g cells/ hr) can be obtained from the following expression (if the specific productivity (qp) is constant):
or:
(37)
where Pt is the total product mass in the reactor:
Substituting:
(38)

33. Integrating this expression, we have:
(39)
Integrating this expression, we have:
(40)
or in terms of concentration:
(41)

34. Repeated Fed-batch
Usually, fed-batch cultures are taken through many feeding cycles, with each feeding cycle followed by a harvest cycle during which the volume is drawn back down to V0 and a new cycle is started.

35. For the case of repeated fed-batch cultures:
(42)
Where Vw is the volume just before harvesting, V0 is the volume after harvesting, Dw=F/Vw and:
(43)
tw is the cycle time and is given by:
(44)

36. With this definition, we now have:
(45)

37. Example 9.3
In a fed-batch culture operating with intermittent addition of glucose solution, values of the following parameters are given at t=2 h, when the system is at quasi-steady state.
V =1000 ml, F=dV/dt= 200 ml/h, S0= 100 g glucose/l,
, Ks=0.1 g glucose/l,
Find V0 (the initial volume of the culture)
Determine the concentration of growth-limiting substrate in the vessel at quasi-steady state
Determine the concentration and total amount of biomass in the vessel at t=2 h (at quasi-steady state)
If qp=0.2 g product/g cells-h, P0=0, determine the concentration of product in the vessel at t= 2h.

38. Example 9.3: Solution a. b.
Assume Monod equation is valid: c. d. 38

39. End of Chapter Nine