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07 Oct 2011 Prof. R. Shanthini 1 Cellular kinetics and associated reactor design: Reactor Design for Cell Growth CP504 – Lecture 7
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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) - PowerPoint PPT Presentation

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

• Cellular kinetics and associated reactor design:

Reactor Design for Cell Growth CP504 Lecture 7

Prof. R. Shanthini

• Cell Growth Kineticswhere : specific growth rate (per time) CX : cell concentration (dry cell weight per unit volume) Using the population growth model, we could write the cell growth rate (rX) as

Prof. R. Shanthini

• V for volume of the reacting mixture at time tCX for concentration of the cells in V at time t(rX) for cell growth rate in V at time tMass balance for the cell:0 + (rX) V = 0 + d(VCX) / dt which for a batch reactor with constant volume reacting mixture gives(42)Batch FermenterdCX / dt = rX

Prof. R. Shanthini

• Combining (41) and (42), we getBatch FermenterIf is a constant then integrating (43) gives,where CX = CX0 when t = t0.

Prof. R. Shanthini

• Cell Growth Kineticswhere max and KS are known as the Monod kinetic parameters.Mostly, however, is not a constant with time. It depends on CS, the substrate concentration. The most commonly used model for is given by the Monod model: 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. = KS + CS m CS (45)

Prof. R. Shanthini

• Substituting in (43) by the Monod Model given by (45), we getEquation (46) could be integrated only if we know how CS changes with either CX or t.How to do that?Batch Fermenter

Prof. R. Shanthini

• Stoichiometry could have helped. But we dont 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:(47)YX/S = rX / (-rS)It is done as follows:Batch FermenterWe know (rX) from (41) and/or (42). But we dont know (-rS). Therefore obtain an expression for (-rS) as shown in the next slide.

Prof. R. Shanthini

• V for volume of the reacting mixture at time tCS for concentration of the Cells in V at time t(rS) for substrate utilization rate in V at time tMass balance for substrate:0 = 0 + (-rS) V + d(VCS) / dt which for a batch reactor with constant volume reacting mixture gives(48)dCS / dt = -(-rS)Batch Fermenter

Prof. R. Shanthini

• (48)dCS / dt = -(-rS)(42)dCX / dt = rX Combining the above equations, we getdCX / dCS = -YX/S which upon integration givesBatch Fermenter

Prof. R. Shanthini

• Substituting CS from (49) in (47) and integrating, we get m (t - t0) = KS YX/SCX0 + CS0YX/S+ 1lnCX0CX+KS YX/SCX0 + CS0YX/SlnCSCS0(50)whereBatch Fermenter()()))((

Prof. R. Shanthini

• Exercise 1:

The growth rate of E. coli be expressed by Monod kinetics with m = 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. Batch Fermenter

Prof. R. Shanthini

• CS is varied from 10 g/L to 0.()() 0.935 t = 0.71 x 0.61 + 10 x 0.6+ 1ln1CX+0.71 x 0.61 + 10 x 0.6()lnCS10()CX is calculated using (49) as Exercise 1 worked out using the calculator/spread sheet:CX = 1 + 0.6 (10 CS) t is calculated using (50) as follows: CX is calculated using (46).

Prof. R. Shanthini

• Exercise 1 worked out using the calculator/spread sheet:

Prof. R. Shanthini

• CSCXExercise 1 worked out using the calculator/spread sheet:

Prof. R. Shanthini

• CSCXExercise 1 worked out using the calculator/spread sheet:

Prof. R. Shanthini

• Programme written in MATLABExercise 1 worked out using an ODE solver: function dydt =CP504Lecture_07(t,y)%data givenmumax = 0.935; % per hrKs = 0.71; % g/LYXS = 0.6; %Monod modelmu = mumax*y(2)/(Ks+y(2));%rate equationsrX = mu*y(1);rS = -rX/YXS;dydt=[rX; rS][t,y] = ode45(@CP504Lecture_07,[0:0.01:3],[1; 10]);

Prof. R. Shanthini

• 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

• 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)')mumax = 0.935;Ks = 0.71;mu= mumax*y(:,2)./(Ks+y(:,2));rX = mu.*y(:,1);plot(t,rX,'g')

Prof. R. Shanthini

• FFCXi, CSiCX, CS = V/F m = (51)wherePlug-flow Fermenter at steady-state

Prof. R. Shanthini

• Continuous Stirred Tank Fermenter (CSTF) at steady-stateMass balance for cells over V:FCXi + rX V = FCX (53)

Prof. R. Shanthini

• Equation (53) givesVF = CX - CXirX(54)Continuous Stirred Tank Fermenter (CSTF) at steady-stateIntroducing Dilution Rate D as in (54), we get

Prof. R. Shanthini

• Since rX = CX, (56) becomes Continuous Stirred Tank Fermenter (CSTF) at steady-stateIf the feed is sterile (i.e., CXi = 0), (57) gives which means either CX = 0 or D =

Prof. R. Shanthini

• CS = (60)m - DKS D (59) can be rearranged to give CS as D = (59) = If D = , thenTo determine CX, we need to write the mass balance for substrate over the CSTF Continuous Stirred Tank Fermenter (CSTF) at steady-state

Prof. R. Shanthini

• Mass balance for substrate over V:FCSi = FCS + (-rS) V Continuous Stirred Tank Fermenter (CSTF) at steady-state

Prof. R. Shanthini

• which is rearranged to give(-rS) = D (CSi - CS) (61)Continuous Stirred Tank Fermenter (CSTF) at steady-staterX = D (CX - CXi ) (56) givesUsing the above equations in the definition of yield factor, we get

Prof. R. Shanthini

• Since the feed is sterile, (62) gives(60) isTherefore, we haveCX = (64)YX/S(CSi - )Continuous Stirred Tank Fermenter (CSTF) at steady-stateCS = (60)m - DKS D m - DKS D

Prof. R. Shanthini

• which is valid only when D < mwhich is valid only when

D < CSi m / (KS + CSi)CS = (60)CX = (64)YX/S(CSi - )Continuous Stirred Tank Fermenter (CSTF) at steady-statem - DKS D m - DKS D CSi > KS D / (m - D)

Prof. R. Shanthini

• Since D < CSi m / (KS + CSi) < m DC = CSi m / (KS + CSi)critical value of the Dilution Rate is as follows:Continuous Stirred Tank Fermenter (CSTF) at steady-state(65)

Prof. R. Shanthini

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

Prof. R. Shanthini

• CX = 0 means no cell in the reactor. Since the CSTF has a 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 CS = CSi means substrate is not utilised.Continuous Stirred Tank Fermenter (CSTF) at steady-state

Prof. R. Shanthini

• Exercise 2

The growth rate of E. coli be expressed by Monod kinetics with m = 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. Continuous Stirred Tank Fermenter (CSTF) at steady-state

Prof. R. Shanthini

• Exercise 2 worked out using the calculator/spread sheet:DC = CSi m / (KS + CSi) = 10 x 0.935 / (0.71+10) = 0.873 per hCS = From (60): CX = 0.6(10 - )0.935 - D0.71 D 0.935 - D0.71 D Plot the following using excel / MATLABFrom (64): g/Lg/LFrom (65):

Prof. R. Shanthini

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

Prof. R. Shanthini