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Prof. R. Shanthini 30 Nov 2012 1 Cellular kinetics and associated reactor design: Reactor Design for Cell Growth CP504 – ppt_Set 07

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Prof. R. Shanthini 30 Nov 2012 2 Cell Growth Kinetics rXrX = μ C X (43) where μ : specific growth rate (per time) C X : cell concentration (dry cell weight per unit volume) Using the population growth model, we could write the cell growth rate (r X ) as

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Prof. R. Shanthini 30 Nov 2012 3 V for volume of the reacting mixture at time t C X for concentration of the cells in V at time t (r X ) for cell growth rate in V at time t Mass balance for the cell: 0 + (r X ) V = 0 + d(VC X ) / dt which for a batch reactor with constant volume reacting mixture gives (44) Batch Fermenter dC X / dt = r X

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Prof. R. Shanthini 30 Nov 2012 4 Combining (43) and (44), we get dC X = μ C X dt (45) Batch Fermenter If μ is a constant then integrating (45) gives, C X = C X0 exp[μ(t-t 0 )] (46) where C X = C X0 when t = t 0.

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Prof. R. Shanthini 30 Nov 2012 5 Cell Growth Kinetics where μ m and K S are known as the Monod kinetic parameters. Mostly, however, μ is not a constant with time. It depends on C S, the substrate concentration. The most commonly used model for μ 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. μ = K S + C S μ m C S (47)

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Prof. R. Shanthini 30 Nov 2012 6 μ m C S = K S + C S (48) CXCX dC X dt Substituting μ in (45) by the Monod Model given by (47), we get Equation (48) could be integrated only if we know how C S changes with either C X or t. How to do that? Batch Fermenter

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Prof. R. Shanthini 30 Nov 2012 7 Stoichiometry could have helped. But we don’t have such a relationship in the case of cellular kinetics. Therefore, we introduce a yield factor (Y X/S ) as the ratio between cell growth rate (r X ) and substrate consumption rate (-r S ) as follows: (49)Y X/S = r X / (-r S ) It is done as follows: Batch Fermenter We know (r X ) from (43) and/or (44). But we don’t know (-r S ). Therefore obtain an expression for (-r S ) as shown in the next slide.

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Prof. R. Shanthini 30 Nov 2012 8 V for volume of the reacting mixture at time t C S for concentration of the Cells in V at time t (r S ) for substrate utilization rate in V at time t Mass balance for substrate: 0 = 0 + (-r S ) V + d(VC S ) / dt which for a batch reactor with constant volume reacting mixture gives (50)dC S / dt = -(-r S ) Batch Fermenter

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Prof. R. Shanthini 30 Nov 2012 9 (50)dC S / dt = -(-r S ) Y X/S = - r S rXrX (49) (44)dC X / dt = r X Combining the above equations, we get dC X / dC S = -Y X/S which upon integration gives (C X – C X0 ) = Y X/S (C S0 – C S )(51) Batch Fermenter

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Prof. R. Shanthini 30 Nov 2012 10 Substituting C S from (51) in (49) and integrating, we get μ m (t - t 0 ) = K S Y X/S C X0 + C S0 Y X/S + 1 ln C X0 CXCX + K S Y X/S C X0 + C S0 Y X/S ln CSCS C S0 (52) where (C X – C X0 ) = Y X/S (C S0 – C S )(51) Batch Fermenter () () )) ( (

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Prof. R. Shanthini 30 Nov 2012 11 Exercise 1: The growth rate of E. coli be expressed by Monod kinetics with μ m = 0.935 hr -1 and K S = 0.71 g/L. Assume that Y X/S is 0.6 g dry cells per g substrate. C X0 is 1 g/L and C S0 = 10 g/L when the cells start to grow exponentially (i.e., at t = 0). show how C X, C S, and dC X /dt change with respect to time. Batch Fermenter

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Prof. R. Shanthini 30 Nov 2012 12 C S is varied from 10 g/L to 0. () () 0.935 t = 0.71 x 0.6 1 + 10 x 0.6 + 1 ln 1 CXCX + 0.71 x 0.6 1 + 10 x 0.6 () ln CSCS 10 () C X is calculated using (49) as Exercise 1 worked out using the calculator/spread sheet: C X = 1 + 0.6 (10 – C S ) t is calculated using (50) as follows: C X is calculated using (48).

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Prof. R. Shanthini 30 Nov 2012 13 specify C S Calculate C X using (49) Calculate t using (50) Calculate dC X /dt using (46) 1010 9.951.030.03170.9335 9.81.060.06240.9332 9.851.090.09230.9329 Continue until C S becomes 0 Exercise 1 worked out using the calculator/spread sheet:

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Prof. R. Shanthini 30 Nov 2012 14 CSCS CXCX Exercise 1 worked out using the calculator/spread sheet:

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Prof. R. Shanthini 30 Nov 2012 15 CSCS CXCX Exercise 1 worked out using the calculator/spread sheet:

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Prof. R. Shanthini 30 Nov 2012 16 Programme written in MATLAB Exercise 1 worked out using an ODE solver: 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] [t,y] = ode45(@CP504Lecture_07,[0:0.01:3],[1; 10]);

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Prof. R. Shanthini 30 Nov 2012 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)')

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Prof. R. Shanthini 30 Nov 2012 18 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')

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Prof. R. Shanthini 30 Nov 2012 19 F F C Xi, C Si C X, C S θ = V/F μ m θ = K S Y X/S C Xi + C Si Y X/S ( + 1 ) ln C Xi CXCX () + K S Y X/S C Xi + C Si Y X/S () ln CSCS C Si () (53) where (C X – C Xi ) = Y X/S (C Si – C S )(54) Plug-flow Fermenter at steady-state

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Prof. R. Shanthini 30 Nov 2012 20 F C Xi, C Si F C X, C S V Continuous Stirred Tank Fermenter (CSTF) at steady-state - 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 - also known as chemostat

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Prof. R. Shanthini 30 Nov 2012 21 F C Xi, C Si F C X, C S V Continuous Stirred Tank Fermenter (CSTF) at steady-state Mass balance for cells over V: FC Xi + r X V = FC X (55)

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Prof. R. Shanthini 30 Nov 2012 22 Equation (55) gives V F = C X - C Xi rXrX (56) Continuous Stirred Tank Fermenter (CSTF) at steady-state Introducing Dilution Rate D as = (57) F V D = 1 θ in (56), we get 1 D = C X - C Xi rXrX (58)

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Prof. R. Shanthini 30 Nov 2012 23 Since r X = μ C X, (58) becomes 1 D = C X - C Xi μ C X (59) Continuous Stirred Tank Fermenter (CSTF) at steady-state If the feed is sterile (i.e., C Xi = 0), (59) gives C X (D – μ) = 0 (60) which means either C X = 0 or D = μ

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Prof. R. Shanthini 30 Nov 2012 24 CSCS = (62) μ m - D K S D (61) can be rearranged to give C S as D = μ(61) μ m C S K S + C S = If D = μ, then To determine C X, we need to write the mass balance for substrate over the CSTF Continuous Stirred Tank Fermenter (CSTF) at steady-state

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Prof. R. Shanthini 30 Nov 2012 25 F C Xi, C Si F C X, C S V Mass balance for substrate over V: FC Si = FC S + (-r S ) V Continuous Stirred Tank Fermenter (CSTF) at steady-state

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Prof. R. Shanthini 30 Nov 2012 26 which is rearranged to give (-r S ) = D (C Si - C S )(63) Continuous Stirred Tank Fermenter (CSTF) at steady-state r X = D (C X - C Xi ) (58) gives Using the above equations in the definition of yield factor, we get (C X – C Xi ) = Y X/S (C Si – C S )(64)

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Prof. R. Shanthini 30 Nov 2012 27 Since the feed is sterile, (6 4) gives C X = Y X/S (C Si – C S )(65) (62) is Therefore, we have CXCX = (66) Y X/S ( C Si - ) Continuous Stirred Tank Fermenter (CSTF) at steady-state CSCS = (62) μ m - D K S D μ m - D K S D

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Prof. R. Shanthini 30 Nov 2012 28 which is valid only when D < μ m which is valid only when D < C Si μ m / (K S + C Si ) CSCS = (62) CXCX = (66) Y X/S ( C Si - ) Continuous Stirred Tank Fermenter (CSTF) at steady-state μ m - D K S D μ m - D K S D C Si > K S D / (μ m - D)

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Prof. R. Shanthini 30 Nov 2012 29 Since D < C Si μ m / (K S + C Si ) < μ m D C = C Si μ m / (K S + C Si ) critical value of the Dilution Rate is as follows: Continuous Stirred Tank Fermenter (CSTF) at steady-state (67)

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Prof. R. Shanthini 30 Nov 2012 30 If μ m equals or less than D C, then C X is negative. That is impossible. We need to take the solution C X = 0 of (58), not D = μ So, when μ m equals or less than D C, Substituting C X = 0 in C X = Y X/S (C Si – C S ) gives C S = C Si Continuous Stirred Tank Fermenter (CSTF) at steady-state

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Prof. R. Shanthini 30 Nov 2012 31 C X = 0 means no cell in the reactor. Since the CSTF has a sterile feed (C Xi = 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 < D C C S = C Si means substrate is not utilised. Continuous Stirred Tank Fermenter (CSTF) at steady-state

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Prof. R. Shanthini 30 Nov 2012 32 Exercise 2 The growth rate of E. coli be expressed by Monod kinetics with μ m = 0.935 hr -1 and K S = 0.71 g/L. Assume that Y X/S is 0.6 g dry cells per g substrate. The feed is sterile (C Xi = 0) and C Si = 10 g/L. show C X and C S changes with dilution rate. Continuous Stirred Tank Fermenter (CSTF) at steady-state

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Prof. R. Shanthini 30 Nov 2012 33 Exercise 2 worked out using the calculator/spread sheet: D C = C Si μ m / (K S + C Si ) = 10 x 0.935 / (0.71+10) = 0.873 per h CSCS = From (60): CXCX =0.6 ( 10 - ) 0.935 - D 0.71 D 0.935 - D 0.71 D Plot the following using excel / MATLAB From (64): g/L From (65):

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Prof. R. Shanthini 30 Nov 2012 34 D C = 0.873 Exercise 2 worked out using the calculator/spread sheet:

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Prof. R. Shanthini 30 Nov 2012 35 D C = 0.873 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.

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