Cellular kinetics and associated reactor design: Reactor Design for Cell Growth

1. CP504 – ppt_Set 07 Cellular kinetics and associated reactor design: Reactor Design for Cell Growth

2. rX (43) = μ CX Cell Growth Kinetics Using the population growth model, we could write the cell growth rate (rX) as where μ : specific growth rate (per time) CX : cell concentration (dry cell weight per unit volume)

3. Batch Fermenter Mass balance for the cell: 0 + (rX) V = 0 + d(VCX) / dt which for a batch reactor with constant volume reacting mixture gives dCX / dt = rX (44) V for volume of the reacting mixture at time t CX for concentration of the cells in V at time t (rX) for cell growth rate in V at time t

4. dCX (45) = μ CX dt (46) CX = CX0 exp[μ(t-t0)] Batch Fermenter Combining (43) and (44), we get If μ is a constant then integrating (45) gives, where CX = CX0 when t = t0.

5. Cell Growth Kinetics Mostly, however, μ is not a constant with time. It depends on CS, the substrate concentration. The most commonly used model for μ is given by the Monod model: μm CS (47) μ = KS + CS where μm and KS are known as the Monod kinetic parameters. Monod Model is an over simplification of the complicated mechanism of cell growth. However, it adequately describes the kinetics when the concentrations of inhibitors to cell growth are low.

6. dCX μm CS (48) = CX dt KS + CS Batch Fermenter Substituting μin (45) by the Monod Model given by (47), we get Equation (48) could be integrated only if we know how CS changes with either CX or t. How to do that?

7. Batch Fermenter It is done as follows: Stoichiometry could have helped. But we don’t have such a relationship in the case of cellular kinetics. Therefore, we introduce a yield factor (YX/S) as the ratio between cell growth rate (rX) and substrate consumption rate (-rS) as follows: YX/S = rX / (-rS) (49) We know (rX) from (43) and/or (44). But we don’t know (-rS). Therefore obtain an expression for (-rS) as shown in the next slide.

8. Batch Fermenter Mass balance for substrate: 0 = 0 + (-rS) V + d(VCS) / dt which for a batch reactor with constant volume reacting mixture gives dCS / dt = -(-rS) (50) V for volume of the reacting mixture at time t CS for concentration of the Cells in V at time t (rS) for substrate utilization rate in V at time t

9. rX (49) YX/S = - rS (CX – CX0) = YX/S (CS0 – CS) (51) Batch Fermenter dCX / dt = rX (44) dCS / dt = -(-rS) (50) Combining the above equations, we get dCX / dCS = -YX/S which upon integration gives

10. (CX – CX0) = YX/S (CS0 – CS) (51) Batch Fermenter Substituting CS from (51) in (49) and integrating, we get ( ) ( ) KS YX/S CX μm (t - t0) = + 1 ln CX0 + CS0YX/S CX0 ( ) ( ) CS0 KS YX/S ln (52) + CS CX0 + CS0YX/S where

11. Batch Fermenter Exercise 1: The growth rate of E. coli be expressed by Monod kinetics with μm = 0.935 hr-1 and KS = 0.71 g/L. Assume that YX/S is 0.6 g dry cells per g substrate. CX0 is 1 g/L and CS0 = 10 g/L when the cells start to grow exponentially (i.e., at t = 0). show how CX, CS, and dCX/dt change with respect to time.

12. Exercise 1 worked out using the calculator/spread sheet: CS is varied from 10 g/L to 0. CX = 1 + 0.6(10 – CS) CX is calculated using (49) as t is calculated using (50) as follows: ( ) ( ) 0.71 x 0.6 CX 0.935t = ln + 1 1 + 10 x 0.6 1 ( ) ( ) 0.71 x 0.6 10 + ln CS 1 + 10 x 0.6 CX is calculated using (48).

13. Exercise 1 worked out using the calculator/spread sheet:

14. Exercise 1 worked out using the calculator/spread sheet: CS CX

15. Exercise 1 worked out using the calculator/spread sheet: CS CX

16. Exercise 1 worked out using an ODE solver: Programme written in MATLAB [t,y] = ode45(@CP504Lecture_07,[0:0.01:3],[1; 10]); function dydt =CP504Lecture_07(t,y) %data given mumax = 0.935; % per hr Ks = 0.71; % g/L YXS = 0.6; %Monod model mu = mumax*y(2)/(Ks+y(2)); %rate equations rX = mu*y(1); rS = -rX/YXS; dydt=[rX; rS]

17. Exercise 1 worked out using an ODE solver: plot(t,y(:,1),'b',t,y(:,2),'r') legend('Cell','Substrate') ylabel('Concentration (g/L)') xlabel('Time (h)')

18. Exercise 1 worked out using an ODE solver: mumax = 0.935; Ks = 0.71; mu= mumax*y(:,2)./(Ks+y(:,2)); rX = mu.*y(:,1); plot(t,rX,'g') plot(t,y(:,1),'b',t,y(:,2),'r') legend('Cell','Substrate') ylabel('Concentration (g/L)') xlabel('Time (h)')

19. ( ) ( ) CX KS YX/S + 1 ln CXi CXi + CSiYX/S ( ) ( ) CSi KS YX/S ln + CS CXi + CSiYX/S (CX – CXi) = YX/S (CSi – CS) (54) Plug-flow Fermenter at steady-state F F θ = V/F CXi, CSi CX, CS μm θ = (53) where

20. F CXi, CSi V CX, CS F CX, CS Continuous Stirred Tank Fermenter (CSTF) at steady-state - also known as chemostat • - Mixing supplied by impellers and/or rising gas bubbles • Complete mixing is assumed (composition of any phases do not vary with position) • Liquid effluent has the same composition as the reactor contents

21. F CXi, CSi V CX, CS F CX, CS Continuous Stirred Tank Fermenter (CSTF) at steady-state Mass balance for cells over V: FCXi + rX V = FCX (55)

22. 1 F (57) D = = θ V CX - CXi 1 (58) = rX D Continuous Stirred Tank Fermenter (CSTF) at steady-state Equation (55) gives CX - CXi V (56) = rX F Introducing Dilution Rate D as in (56), we get

23. (60) CX (D – μ) = 0 CX - CXi 1 (59) = D μ CX Continuous Stirred Tank Fermenter (CSTF) at steady-state Since rX = μ CX, (58) becomes If the feed is sterile (i.e., CXi = 0), (59) gives which means either CX = 0 or D = μ

24. μm CS KS + CS Continuous Stirred Tank Fermenter (CSTF) at steady-state If D = μ, then μ (61) D = = (61) can be rearranged to give CS as KS D (62) CS = μm - D To determine CX, we need to write the mass balance for substrate over the CSTF

25. F CXi, CSi V CX, CS F CX, CS Continuous Stirred Tank Fermenter (CSTF) at steady-state Mass balance for substrate over V: FCSi = FCS + (-rS) V

26. (CX – CXi) = YX/S (CSi – CS) (64) Continuous Stirred Tank Fermenter (CSTF) at steady-state which is rearranged to give (-rS) = D (CSi - CS) (63) (58) gives rX = D (CX - CXi ) Using the above equations in the definition of yield factor, we get

27. CX = YX/S (CSi – CS) (65) Continuous Stirred Tank Fermenter (CSTF) at steady-state Since the feed is sterile, (6 4) gives (62) is KS D (62) CS = μm - D Therefore, we have ) ( KS D (66) CX = YX/S CSi - μm - D

28. Continuous Stirred Tank Fermenter (CSTF) at steady-state KS D (62) CS = μm - D which is valid only when D < μm ) ( KS D (66) CX = YX/S CSi - μm - D which is valid only when D < CSiμm / (KS + CSi) CSi > KS D / (μm - D)

29. Continuous Stirred Tank Fermenter (CSTF) at steady-state Since D < CSiμm / (KS + CSi) < μm critical value of the Dilution Rate is as follows: DC = CSiμm / (KS + CSi) (67)

30. Continuous Stirred Tank Fermenter (CSTF) at steady-state If μm equals or less than DC, then CX is negative. That is impossible. So, when μm equals or less than DC, We need to take the solution CX = 0 of (58), not D = μ Substituting CX = 0 in CX = YX/S (CSi – CS) gives CS = CSi

31. Continuous Stirred Tank Fermenter (CSTF) at steady-state CX = 0 means no cell in the reactor. CS = CSi means substrate is not utilised. Since the CSTF hasa sterile feed (CXi = 0), no reaction takes place unless we inoculate with the cells once again. So, CSTF gets into a WASHOUT situation. To avoid CSTF getting into WASHOUT situation, we need to maintain D = F / V < DC

32. Continuous Stirred Tank Fermenter (CSTF) at steady-state Exercise 2 The growth rate of E. coli be expressed by Monod kinetics with μm = 0.935 hr-1 and KS = 0.71 g/L. Assume that YX/S is 0.6 g dry cells per g substrate. The feed is sterile (CXi = 0) and CSi = 10 g/L. show CX and CS changes with dilution rate.

33. Exercise 2 worked out using the calculator/spread sheet: Plot the following using excel / MATLAB 0.71D g/L From (60): CS = 0.935 - D ) ( 0.71D From (64): CX = 0.6 10 - g/L 0.935 - D From (65): DC = CSiμm / (KS + CSi) = 10 x 0.935 / (0.71+10) = 0.873 per h

34. Exercise 2 worked out using the calculator/spread sheet: DC = 0.873

35. Exercise 2 worked out using the calculator/spread sheet: Near washout the reactor is very sensitive to variations in D. Small change in D large shifts in X and/or S. DC = 0.873