CHAPTER 6: CELL KINETICS AND FERMENTER DESIGN 6.10. You are going to cultivate yeast, Saccharomyces Cerevisiae, by using a 10m3 fermenter your company owns. You want to find out the amount of ethanol the fermenter can produce. Therefore, a chemostat study was carried out and the Monod kinetic parameters. For the microorganism grown in the glucose medium at 30oC, pH= 4.8, were found to be Ks= 0.025 g/L and umax= 0.25h-1. The ethanol yield (Yp/s) is 0.44 (g/g) and cell yield (Y X/S) is 0.019 (g/g). The inlet substrate concentration is 50 g/l. a. What flowrate will give the maximum total ethanol production in the continuous fermenter and what is the maximum ethanol production rate? b. If you want to convert 95% of the incoming substrate, what must be the ethanol production rate for the continuous fermenter? c. If you have two 5m3 fermenters instead of one 10m3 fermenter, what is your recommendation for the use of these fermenters to convert 95% of the incoming substrate? Would you recommend connecting two fermenters in series to improve the productivity? Why or why not? Given: V=10m3; Ks= 0.025 g/L; umax= 0.25h-1; Yp/s= 0.44; YX/S= 0.019 g/g; Cso= 50 g/l Required: (a) F (b) F @ 95% Cso conversion Solution:
(a) α=
√
Ks+Cso = Ks
√
0.025 g 50 g + L l =44.73 0.025 g / L
Cso Cs= CS opt = α +1 = 50gL-1/ (44.73 +1) = 1.0934g/L D= umax.CS/ (Ks+Cs)= 0.25h-1(1.0934g/L)/ (0.025 g/L+1.0934g/L)= 0.2444/h F= DV= 0.2444/h (10m3)= 2.444 m3/h (b) For 95% substrate conversion Cs= 0.05Cso = 4.75g/l. 0.25 g 4.75 h l D= umax.CS/ (Ks+Cs) = = 0.0249 /h g g 0.025 + 4.75 l l
( )( ) ( )( )
F= DV= (0.0249/h) (10m3) = 0.2490 m3/h
(c) Cxo= YX/S(Cso- CS)= 0.019 (50- 4.75)g/L= g/L α 44.73 Cx1= YX/SCso( α +1 ¿ = (0.019)(50)( 4.573 )= 0.9292g/L Cso Cs1= α +1 =
0.8598
50 g/l (44.73+1) = 1.0934g/L
D1= umax.CS/ (Ks+Cs) = 0.25h-1(1.0934g/L)/ (0.025 g/L+1.0934g/L)= 0.2444/h F1= D1V1= 0.2444 h-1( 5m3)=1.2220 m3/h α Cx2= YX/SCs1( α +1 ¿ = (0.019)( 1.0934g/L)=0.0208g/L Cs2=
Cs 1 α +1 =
1.0934 45.73 = 0.0239 g/L
0.0208−0.9292 YX/S= 0.0239−1.0934
=
0.8494
Therefore; a two- 5 cubic meters of fermenters in series will increase the Growth yield for 95% Ethanol conversion
CHAPTER 6: CELL KINETICS AND FERMENTER DESIGN ; ADDITIONAL PROBLEM During the growth of E. coli in a batch reactor, the pattern can be modelled using the Monod expression of the form; µ=
µ max K S+ S
Where: µ = specific growth rate µmax =maximum specific growth rate Ks = Monod constant S = substrate concentration For the above process, show how the time required to reach the maximum population of cells can be estimated. If the concentration of cells at the start of the exponential growth is 0.08g/L and the essential substrate concentration is 36g/L. Find the time where the maximum population of cells is reached given that ¹max = 0:55h¡1, Ks = 1:2g/L and k = 1:4gsubstrate/(gcells¢h)
Sol’n: Substrate consumption during the growth can be described by; dS =−kX dt
and the cell growth is given by; dX =µX dt where X=
1 dX µ dt
Substitute X into
dS dt and integrating gives; dS 1 dX =−k ( ) dt µ dt
∫ dS=
But µ=
−k dX µ ∫
µ max K S+ S
hence;
∫ dS=
−k ∫ dX µ max K S+ S
rearranging gives; 0
XS
S −k ∫ K + S dS= µ ∫ dX S max X S 0
0
Integrating both sides leads to; X S=X 0+
µmax KS (S0 + K S ln ) k K S+ S0
The average can is determined by dividing the speci¯c growth rate with the total substrate in the reactor, which gives; 0
∫ µdS µ avg=
S0
0
∫ dS S0
0
µ avg=
∫ K S+S dS S0
S
0
∫ dS S0
µ µ avg=
max [ S 0+ Ksln
KS ] K S +S0
S0
Therefore, the new cell growth expression should be in the form of; dX =µ avg X dt and integrating, XS
t
1 1 ∫ dt= µ ∫ X dX avg X 0 0
gives; t=
X 1 ln S µavg X 0
substituting µavg previously lead to; X µ max [¿ ¿ 0+
µ max KS X (S 0 + K S ln ) ]ln S k K S + S0 X0 S t= 0 ¿
Consider the amount of cell produced at the given substrate concentration: Using; KS S 0 + K S ln KS+ S0 ) µ max X S =X 0 + ¿ k 1.2 36+1.2 36+1.2 ln ¿ 0.55 X S=0.08+ ¿ 1.4
XS=12.604 and using the expression for µavg; 0.55[36+1.2 ln µ avg=
1.2 ] 36+ 1.2
36 µ avg=0.487
thus, the time taken for the cell population to reach a maximum value; X 1 t= ln S µavg X 0 t=
1 12.604 ln 0.487 0.08 t = 10.34 hrs.
CHAPTER 6: CELL KINETICS AND FERMENTER DESIGN ; ADDITIONAL PROBLEM Biochemical Engineering: A Textbook for Engineers, Chemists and Biologists Shigeo Katoh and Fumitake Yoshida, 2009 Escherichia coli grows with a doubling time of 0.5 h in the exponential growth phase. (a) What is the value of the specific growth rate? (b) How much time would be required to grow the cell culture from 0.1 kg dry cell / m3?
Given: E. coli td=0.5 h Cn=10 Cno=0.1
Req’d: a) µ b) t
Sol’n: a) ln 2 td= μ ln 2 µ= 0.5 µ=1.3868 h-1 b) Cn ln ( ) Cno =µ(t-to)
t=
ln (
Cn ) Cno µ
10 ) .1 t= 1.3863 t= 3.3219 h ln (
CHAPTER 6: CELL KINETICS AND FERMENTER DESIGN ; ADDITIONAL PROBLEM Biochemical Engineering: A Textbook for Engineers, Chemists and Biologists Shigeo Katoh and Fumitake Yoshida, 2009 E. coli grows from 0.10 kg dry cell/m3 to 0.50 kg dry cell/m3 in 1 h.
1. Assuming the exponential growth during this period, evaluate the specific growth rate. 2. Evaluate the doubling time during the exponential growth phase. 3. How much time would be required to grow from 0.10 kg-dry cell/m3 to 1.0 kg dry cell/m3? You may assume the exponential growth during this period. Given: E. coli Cn= 0.50 kg dry cell/m3 Cno=0.10 kg dry cell/m3 ∆ t=1 h Req’d: a)µ b) td c) ∆ t (from 0.1 to 1) Sol’n: a) ln
µ=
Cn Cno ∆t .5 ) .1 1
ln ( μ=
µ=1.6094/h b) td=
ln 2 μ
td= ln 2 / 1.6094 td= 0.4307 h = 25.8412 min c) ln
Cn Cno µ
∆t
=
∆t
1 .1 = 1.6094 ln
∆t = CHAPTER 6: CELL KINETICS AND FERMENTER DESIGN ; ADDITIONAL PROBLEM Bioprocess 1.4307 h Engineering Principles by Pauline M. Duran, pp. 364- 367
Steady state Concentration in a Chemostat The Zymomonas mobilis cells are used for chemostat in a 60 m 3 fermenter. The feed contains 12g/l glucose; Ks= 0.2 g/L; Yxs=0.06 g/g; Ypx=7.7 g/g, umax=0.3/h; ms=2.2g/g.h; Yps=Ypx.Yxs=0.46g/g. a. What flowrate is required for a steady-state substrate concentration of 1.5 g/L? b. At the flowrate of (a), what is the cell density? c. At the flowrate of (a), what concentration of ethanol is produced? Given: Ks= 0.2 g/L; Yps=Ypx.Yxs=0.46g/g.
Yxs=0.06
g/g;
Ypx=7.7
g/g,
umax=0.3/h;
ms=2.2g/g.h;
Cs= 1.5 g/L, V= 60m3 Required: a. F b. Cx c. Cp Solution: (a) D= umax.CS/ (Ks+Cs)= (0.3h-1)(1.5g/L)(0.2g/L + 1.5 g/L)= 0.26 h-1 F= DV= 0.26/h (60m3) = 15.6 m3h-1. (b) When synthesis of the product is coupled with energy metabolism as for ethanol, Cx is evaluated; therefore;
Cx =
D ( Cso−CS ) D = +ms Yxs
( 0.26h ) ( 12−1.5) g/l 0.26 h 2.2 + 0.06 h
= 0.42 g/L
(c) Assuming ethanol is not present in the feed, Cpo=0. Steady-state product concentration is 3.4 g 0.42 qpx h L Cp= Cpo+ D = 0 + = 5.5 g/L 0.26 ( ) h
( )(
)
CHAPTER 6: CELL KINETICS AND FERMENTER DESIGN ; ADDITIONAL PROBLEM Bioprocess Engineering Principles by Pauline M. Duran, pp. 364- 367 Substrate conversion and Biomass Productivity in a Chemostat A 5 m3 fermenter is operated continuously with feed substrate concentration of 20 kg/m 3. The microorganism cultivated in the reactor has the following characteristics: umax= 0.45/h; Ks= 0.8 kg/m3; Yxs= 0.55 kg/kg. (a) What feed flow rate is required to achieve 90% substrate conversion? (b) How does the biomass productivity at 90% substrate conversion compare with the maximum possible? Given: umax= 0.45/h; Ks= 0.8 kg/m3; Yxs= 0.55 kg/kg Required: (a) F; (b) productivity comparison Solution: (d) For 90% substrate conversion Cs= 0.1 Cso = 2kg/m3. 0.45 kg 2 h m3 D= umax.CS/ (Ks+Cs) = = 0.32 /h kg kg 0.8 + 2 m3 m3
( )( ) ( )( )
F= DV= (0.32/h) (5m3) = 1.6 m3/h (e) Assuming maintenance requirements and product formation are negligible:
Qx= D (Cso-
KsD ¿ -1 3 umax−D Yxs = 0.32h (20kg.m -
3.17 kg.m-3.h-1 Maximum biomass @ Dopt Dopt= umax (1-
√
KS KS+Cso
)= umax(1- α-1)
( 0.8m3kg )( 0.32h ) (
0.45 0.32 − ) h h
)(0.55kg/kg)=
Dopt= 0.45h-1(1-
√
0.8 kg m3 0.8 kg kg +20 m3 m3
) = 0.36h-1
Maximum biomass productivity is determined with D= Dopt Qxmax= D (Cso-
KsD ¿ umax−D Yxs
Dopt= 0.36/h (20kg.m-3-
(0.8 m3kg )( 0.36h ) 0.45 0.32 ( − ) h h
) (0.55 kg/kg) = 3.33 kg.m3.h-1
Therefore, biomass productivity at 90% substrate conversion is
3.17 3.33 x100 = 95%
of the theoretical maximum. CHAPTER 6: CELL KINETICS AND FERMENTER DESIGN ; ADDITIONAL PROBLEM Bioprocess Engineering Principles by Pauline M. Duran, p. 375 Plug-Flow Reactor for immobilized enzymes: Immobilised lactase is used to hydrolyze lactose in dairy waste to glucose and galactose. Enzyme is immobilized in resin particles and packed into a 0.5 m 3 column. The total effectiveness factor for the system is close to unity; Km for the immobilized enzyme is 1.32 kg/m 3; umax is 45 kg.m3 -1
h . The lactose concentration in the feed stream is 9.5 kg.m -3; a substance conversion 98% is
required. The column is operated with plug Flow for a total of 310 days a year. a. At what flowrate should the reactor be operated? b. How many tonnes of gluctose are produced a year? Given: Cs= 0.02Cso= 0.19 kg.m-3; Cso= 9.5 kg.m-3; umax= 45 kg.m-3h-1; Km= 1.32 kg.m-3; VPFR=0.5 m3 Required: a. F, b. mass glucose (Tons) Solution: (a) For 98% substrate onversion: Km Cso Cso−Cs τ = umax ln Cs + umaz = 0.32 h
F=
V 3 τ = 1.56 m /h
(b) The rate of lactose conversion is equal to the difference between inlet and outlet mass flow rates of lactose= F(Cso-Cs)= 1.56 m3(0.5-0.19) kg/m3= 14.5 kg/h kg 24 h 310 d 1 kmol kmol Lactose converted= 14.5 h . 1d . 1 yr . 342 kg =315 yr The enzyme reaction is: Lactose+ H20 --) glucose + galactose Therefore, from reaction stoichiometry, 315 kmol glucose are produced a year. The molecular weight of glucose is 180; thus, Mass glucose= 315
kmol 180 kg 1T Tonnes . . =56.7 yr 1 kmol 1000 kg yr
CHAPTER 6: CELL KINETICS AND FERMENTER DESIGN ; ADDITIONAL PROBLEM Chemical Engineering by J. M. Coulson and J. F. Richardson A continuous fermenter is operated at a series of dilution rates though at constant, sterile, feed concentration, pH, aeration rate and temperature. The following data were obtained when the limiting substrate concentration was 1200 mg/l and the working volume of the fermenter was 9.8 l. Estimate the kinetic constants Km,µm and kd as used in the modified Monod equation:
and also the growth yield coefficient Y. Feed flowrate Exit substrate concentration Dry weight cell density (l/h)
(mg/l)
(mg/l)
0.79
36.9
487
1.03
49.1
490
1.31
64.4
489
1.78
93.4
482
2.39
138.8
466
2.68
164.2
465
Solution The accumulation = input − output + rate of formation, which for the biomass gives: V(dX/dt) = FX0 − FX + V(µmSX)/(Ks + S) − kdXV
(equation 5.126)
and for the substrate: V(dS/dt) = FS0 − FS − VµmSX/Y(Ks + S)
(equation 5.127)
At steady state, dS/dt = 0. Taking the dilution rate, D = F/V , then the balance for the substrate becomes: S0 − S − {µmSX/[DY(Ks + S)]} = 0 or:
X/D(S0 − S) = (KsY/µm)/S + Y/µm
(i)
Similarly, for the biomass: X0 − X + {µmSX/[D(Ks + S)]} − kdX/D = 0 Since the feed is sterile, X0 = 0 and: DX = [µmSX/(Ks + S)] − kdX. From the material balance for substrate: (µmSX)/(Ks + S) = DY(S0 − S) and substitution gives: DX = DY(S0 − S) − kdX or:
(S0 − S)/X = kd/DY − 1/Y
(ii)
From equation (i), it is seen that a plot of X/D(S0 − S) and 1/S will produce a straight line of slope KsY/µm and intercept Y/µm.
From equation (ii), it is seen that a plot of (S0 − S)/X against 1/D will produce a straight line of slope kd/Y and intercept 1/Y. The data are calculated as follows: Feed Exit 1/S flowrate substrate (l/mg) (F l/h) (S mg/l) 0.79 36.9 0.0271 5.19 1.03 49.1 0.0204 4.05 1.31 64.4 0.0155 3.22 1.78 93.4 0.0107 2.40 2.39 138.8 0.0072 1.80 2.68 164.2 0.0061 1.64 The data are then plotted in Figure 5b from which: KsY/µm = 170,
1/D (h) 12.41 9.51 7.48 5.51 4.10 3.66
Y/µm = 0.59, kd/Y = 0.0133 and 1/Y = 2.222. slope, KsY/µm = 170
)
(h ) S − 0 S ( D /
6 5
Y/ µm= 0.59
4 3 X
2 1 0
) − ( X /) S − 0
S
2.4
slope,kd / Y = 0.0133/h 2.3 (
X 2.2
0
1/ Y = 2.222
0.01 0.02 0.03 1/S (l/mg)
2.388 2.349 2.322 2.296 2.277 2.228
0 2 4 6 8 10 12 1/D (h)
Figure 5b. Graphical work Thus: yield coefficient, Y = (1/2.222) = 0.45 endogenous respiration coefficient,
µm = (0.45/0.59) = 0.76 h−1
and:
Ks = (170 × 0.76/0.45) = 300 mg/l
CHAPTER 6: CELL KINETICS AND FERMENTER DESIGN ; ADDITIONAL PROBLEM Chemical Engineering by J. M. Coulson and J. F. Richardson Two continuous stirred-tank fermenters are arranged in series such that the effluent of one forms the feed stream of the other. The first fermenter has a working volume of 100 l and the other has a working volume of 50 l. The volumetric flowrate through the fermenters is 18 h −1 and the substrate concentration in the fresh feed is 5 g/l. If the microbial growth follows Monod kinetics
with µm = 0.25 h−1,Ks = 0.12 g/l, and the yield coefficient is 0.42, calculate the substrate and biomass concentrations in the effluent from the second vessel. What would happen if the flow were from the 50 l fermenter to the 100 l fermenter? Solution A material balance for biomass over the first fermenter, leads to the equation: D1 = µ1 (equation 5.131) where D1 is the dilution rate in the first vessel and µ1 is the specific growth rate for that vessel. Assuming Monod kinetics to apply: D1 = µmS1/(Ks + S1)
(equation 5.132)
where S1 is the steady-state concentration of substrate in the first vessel, where: S1 = D1Ks/(µm − D1)
(equation 5.133)
If D1 = F/V1 = (18/100) = 0.18 h−1, then: S1 = (0.18 × 0.12)/(0.25 − 0.18) = 0.309 g/l Since the feed to this fermenter is sterile, X0 = 0 and from equation 5.134, Volume 3 the steadystate concentration of biomass in the first vessel is given by: X1 = Y(S0 − S1)
(equation 5.134)
= 0.42(5 − 0.309) = 1.97 g/l
In a similar way, a mass-balance over the second vessel gives: D2 = µ2X2/(X2 − X1)
(equation 5.167)
where D2 is the dilution rate in the second vessel, µ2 is the specific growth rate in that vessel and X2 the steady-state concentration of biomass. The yield coefficient to the second vessel is then: Y = (X2 − X1)/(S1 − S2) where S2 is the steady-state concentration of substrate in that vessel.
Thus:
X2 = X1 + Y(S1 − S2) = 1.97 + 0.42(0.309 − S2) = 2.1 + 0.42S2
Substituting this equation for X2 into equation 5.167, together with values for D2,µ2 and X1 leads to a quadratic equation in S2: 0. from which:
S2 = 0.0113 and: g/l
X2 = 2.1 + (0.42 × 0.0113) = 2.1 g/l
When the tanks are reversed, that is with fresh feed entering the 50 litre vessel, then the dilution rate for this vessel will be as before, 0.36 h−1, although the critical dilution rate will now be: Dcrit = µmS0/(Ks + S0)
(equation 5.148)
= (0.25 × 5)/(0.12 + 5) = 0.244 h−1 This is lower than the dilution rate imposed and washout of the smaller vessel would take place. The concentrations of substrate and biomass in the final effluent would eventually be those attained if only the 100 litre vessel existed, that is: concentration of biomass = 1.97 g/l
concentration of substrate
.
