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

CP504 – Lecture 7. Cellular kinetics and associated reactor design: Reactor Design for Cell Growth. r X. (41). = μ C X. Cell Growth Kinetics. Using the population growth model, we could write the cell growth rate (r X ) as. where μ : specific growth rate (per time)

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

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  1. CP504 – Lecture 7 Cellular kinetics and associated reactor design: Reactor Design for Cell Growth Prof. R. Shanthini

  2. rX (41) = 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) Prof. R. Shanthini

  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 (42) 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 Prof. R. Shanthini

  4. dCX (43) = μ CX dt (44) CX = CX0 exp[μ(t-t0)] Batch Fermenter Combining (41) and (42), we get If μ is a constant then integrating (43) gives, where CX = CX0 when t = t0. Prof. R. Shanthini

  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 (45) μ = KS + CS where μmax 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. Prof. R. Shanthini

  6. dCX μm CS (46) = CX dt KS + CS Batch Fermenter Substituting μin (43) by the Monod Model given by (45), we get Equation (46) could be integrated only if we know how CS changes with either CX or t. How to do that? Prof. R. Shanthini

  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) (47) We know (rX) from (41) and/or (42). But we don’t know (-rS). Therefore obtain an expression for (-rS) as shown in the next slide. Prof. R. Shanthini

  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) (48) 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 Prof. R. Shanthini

  9. rX (47) YX/S = - rS (CX – CX0) = YX/S (CS0 – CS) (49) Batch Fermenter dCX / dt = rX (42) dCS / dt = -(-rS) (48) Combining the above equations, we get dCX / dCS = -YX/S which upon integration gives Prof. R. Shanthini

  10. (CX – CX0) = YX/S (CS0 – CS) (49) Batch Fermenter Substituting CS from (49) in (47) and integrating, we get ( ) ( ) KS YX/S CX μm (t - t0) = + 1 ln CX0 + CS0YX/S CX0 ( ) ( ) CS0 KS YX/S ln (50) + CS CX0 + CS0YX/S where Prof. R. Shanthini

  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. Prof. R. Shanthini

  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 (46). Prof. R. Shanthini

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

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

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

  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] Prof. R. Shanthini

  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)') Prof. R. Shanthini

  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)') Prof. R. Shanthini

  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) (52) Plug-flow Fermenter at steady-state F F θ = V/F CXi, CSi CX, CS μm θ = (51) where Prof. R. Shanthini

  20. 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 (53) Prof. R. Shanthini

  21. 1 F (55) D = = θ V CX - CXi 1 (56) = rX D Continuous Stirred Tank Fermenter (CSTF) at steady-state Equation (53) gives CX - CXi V (54) = rX F Introducing Dilution Rate D as in (54), we get Prof. R. Shanthini

  22. (58) CX (D – μ) = 0 CX - CXi 1 (57) = D μ CX Continuous Stirred Tank Fermenter (CSTF) at steady-state Since rX = μ CX, (56) becomes If the feed is sterile (i.e., CXi = 0), (57) gives which means either CX = 0 or D = μ Prof. R. Shanthini

  23. μm CS KS + CS Continuous Stirred Tank Fermenter (CSTF) at steady-state If D = μ, then μ (59) D = = (59) can be rearranged to give CS as KS D (60) CS = μm - D To determine CX, we need to write the mass balance for substrate over the CSTF Prof. R. Shanthini

  24. 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 Prof. R. Shanthini

  25. (CX – CXi) = YX/S (CSi – CS) (62) Continuous Stirred Tank Fermenter (CSTF) at steady-state which is rearranged to give (-rS) = D (CSi - CS) (61) (56) gives rX = D (CX - CXi ) Using the above equations in the definition of yield factor, we get Prof. R. Shanthini

  26. CX = YX/S (CSi – CS) (63) Continuous Stirred Tank Fermenter (CSTF) at steady-state Since the feed is sterile, (62) gives (60) is KS D (60) CS = μm - D Therefore, we have ) ( KS D (64) CX = YX/S CSi - μm - D Prof. R. Shanthini

  27. Continuous Stirred Tank Fermenter (CSTF) at steady-state KS D (60) CS = μm - D which is valid only when D < μm ) ( KS D (64) CX = YX/S CSi - μm - D which is valid only when D < CSiμm / (KS + CSi) CSi > KS D / (μm - D) Prof. R. Shanthini

  28. 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) (65) Prof. R. Shanthini

  29. 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 Prof. R. Shanthini

  30. 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 WASHED OUT situation. To avoid CSTF getting into WASHED OUT situation, we need to maintain D = F / V < DC Prof. R. Shanthini

  31. 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. Prof. R. Shanthini

  32. 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 Prof. R. Shanthini

  33. Exercise 2 worked out using the calculator/spread sheet: DC = 0.873 Prof. R. Shanthini

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