ChE 460/ChE 560, Fall 2005 Exam II - again Closed Book, Closed Notes, Closed Neighbors The questions here were from Exam II and I have provided solutions - as detailed as I could here, here, to each question. question. Along with the solution, I have have added comments comments reflecting reflecting some of the answers answers I have seen during grading. grading. In addition to the questions questions that appeared in the exam, I have added a few more solutions at the end to questions from the back of the book, again with solutions. solutions. 1. (20 points) The maximum maximum specific oxygen oxygen uptake uptake rate of a particular particular organism organism is given as 5 mmol O2 g 1 h 1 . The organis organism m is grown grown in a stirre stirred d ferme ferment nter er to a cell cell density density of 40 g/liter. The kL a under these conditions is 0.15 sec 1 . At the fermenter operating temperature and pressure, the solubility of oxygen in the culture liquid is 8 x 10 3 kg m 3 . Is the rate rate of cell metabolis metabolism m limite limited d by mass transfe transferr or solely solely depende dependent nt on metabolic metabolic kinetics? kinetics? (i.e. are you supplying supplying oxygen oxygen at a rate sufficient sufficient to maintain maintain the cells at the density density required?) required?) −
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Solution: We need to calculate the Oxygen transport rate (OTR) and Oxygen uptake rate (OTR). q O is given as 5 mmol Oxygen/(gm.hr) - you calculate OUR as 5.00 mmol/(gm.hr) multiplied by 40.00 gm/liter (cell density) and get 200.00 mmol/(liter.hour). Now OT R = kL a C − C L ) you get maximum OTR when C L is zero. Setting Setting C L to zero you can calculate OTR. Since the OUR was calculated as 200.00 mmol/(liter.hour) we have to convert the OTR into those units. 2
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We multiply 0.15 sec 1 with 3600.00 seconds/hr to get 540.00 hr−1 We then multiply this with 0.0080 kg/(m3 sec) to get 4.320000 kg/(m3 .hr). We have to convery the kg of Oxygen into millimoles and m3 into liters so we can make a direct comparison. We know that the molecular weight of oxygen is 32 grams/mole and we have 1000 grams/kg and 1000 liters in 1 m3 . −
So by multiplying 4.32 kg/(m3 .hr) with with 1000.00 grams/kg (4320.000000) and 0.0010 m3 /liter (4.320000) and divided by 32.00 grams/mole we get OTR to be 0.135000 moles or when multiplied by 1000.000 millimoles/mole we get OTR to be 135.000000 135.000000 millimoles. millimoles. We had earlier earlier calculated calculated OUR to be 200.00 So the growth growth is limited by Mass Transfer
2. (20 points) A culture is grown in a simple medium including 0.3 % (weight/volume) glucose. At time t = 0, this culture is innoculated into a larger sterile volume of the identical medium The optical density (OD) at 420 nm (which you can consider to be proportional to cell density) versus time following innoculation is given in column 1 of the table. The same species is also cultured in a complex medium containing 0.15 % glucose and 0.15 % lactose and the subsequent OD vs time data is given in column 2. If the OD at 420 nm is linear in cell density with and OD of 0.175 equal to 0.1 mg dry weight of cells per milliliter, calculate the maximum specific growth rate, µmax , the time lag and the overall yield factors (grams of cell per grams of substrate) assuming substrate exhaustion in each case. Explain the shape of the growth curve in each case. (I do want to see a plot of the data).
Time (hrs) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
OD 1 0.06 0.08 0.11 0.14 0.20 0.26 0.37 0.49 0.70
OD 2 Time (hrs) 0.06 4.5 0.06 5.0 0.06 5.5 0.07 6.0 0.10 6.5 0.13 7.0 0.18 7.5 0.26 8.0 0.32
OD OD 1 2 0.88 0.43 1.04 0.48 1.04 0.50 0.52 0.60 0.84 1.00 1.00
Solution: The first thing to do is plot the data. You do not have to convert the OD to cell density right away (you can do all the calculations and apply the calibration factor supplied). There is a lag in the growth curves (very pronounced in the complex medium, almost 2 hours, about 1 hour (may be?) in the simple medium). You can see this in the log plot clearly. As I have always said in class, approximate values are OK in exams - as long as you clearly show how you arrived at your answer. Let us take the growth curve in the simple medium first. I am going to assume that the exponential growth phase starts at 1.5 hours - OD value of 0.14 and at 5.0 hours the OD value is 1.04. µmax =
X ln X o t
where the X is the value of the cell density at time t and X o the cell density at time t = 0 (or the beginning) (here, our t = 0 is 2 hours and hence the t in the denominator is 3.5 hours) The ln 1.04 = 0.03922 and the ln 0.14 = −1.96611 and so -2.00533 AND their difference is 2.00533 when divided by 3.50000 hours you get the growth rate to be µ=0.57295 in OD units The conversion is 0.17500 OD units for 0.10000 mg cells/ml or 0.57143 mg/ml per OD unit and we can recalculate µ as equal to 0.32740 for the growth in simple medium. Doing these calculations for the complex medium we get (in a similar manner) I am going to assume that the exponential growth phase in this case starts at 2 hours - OD value of 0.10. We see however that there are two exponential phases - let’s calculate ln
X
both of them. At 5.0 hours, the OD value is 0.48 and like earlier, µmax = tXo where the X is the value of the cell density at time t and X o the cell density at time t = 0 (or the beginning) (here, our t = 0 is 2 hours and hence the t in the denominator is 3.0 hours) The ln0.48000 = −0.73397 and the ln 0.10000 = −2.30259 and so -1.56862 AND their difference is 1.56862 when divided by 3.00000 hours you get the growth rate to be µ=0.52287 in OD units The conversion is 0.17500 OD units for 0.10000 mg cells/ml or 0.57143 mg/ml per OD unit and we can recalculate µ as equal to 0.29878 for the growth in complex medium for the first phase in glucose. As we can see, the growth rate in the first phase for the second curve, 0.29878 is smaller than the growth phase for the first curve 0.32740 For the growth phase in lactose, we notice that at 6 hours, the OD is 0.52 and at ln
X
7.5 hours it is 1.00 and like earlier, µmax = tXo where the X is the value of the cell density at time t and X o the cell density at time t = 0 (or the beginning) (here, our t = 0 is 6 hours and hence the t in the denominator is 1.5 hours) The ln0.52000 = −0.65393 and the ln 1.00000 = 0.00000 and so 0.65393 AND their difference is 0.65393 when divided by 1.50000 hours you get the growth rate to be µ=0.43595 in OD units The conversion is 0.17500 OD units for 0.10000 mg cells/ml or 0.57143 mg/ml per OD unit and we can recalculate µ as equal to 0.24911 for the growth in complex medium for the second phase in lactose. Here is a summary Medium Type Lag Phase Exp Phase I (µ1 ) Exp Phase II (µ2 ) Simple 1.5 hrs 0.327401 Complex 2.0 hrs 0.298783 0.249114
Notice that there is a lag between the two phases in the complex medium of 1.5 hours. We now need to calculate the overall yield coefficients. As you can imagine, you will have three (One for the curve in simple medium for glucose, the other for glucose in complex medium and one for lactose in the complex medium) Let us do each. Let us calculate the ∆ X for the exponential growth phase for simple medium and it is 1 .04 − 0.14 = 0.90 in OD units and when multiplied by the conversion factor of 0.57 mg/ml per OD unit you get 0.51 mg/ml cells. The ∆S for glucose for simple medium is 0.30000 % weight/volume and so the Yield Coefficient is Y X/S = 1.714285 For the complex medium, for the glucose phase, we get ∆ X as equal to 0.48 − 0.10 = 0.38 in OD units and when multiplied by the conversion factor of 0.57 mg/ml per OD unit you get 0.22 mg/ml cells. The ∆S for glucose for complex medium is 0.15000 % weight/volume and so the Yield Coefficient is Y X/S = 1.447619 and for the complex medium in lactose, we get ∆ X as equal to 1.00 − 0.52 = 0.48 in OD units and when multiplied by the conversion factor of 0.57 mg/ml per OD unit you get 0.27 mg/ml cells. The ∆S for glucose for complex medium is 0.15000 % weight/volume and so the Yield Coefficient is Y X/S = 1.828571 Medium Type Y (glucose) Y (lactose) Simple 1.714285 Complex 1.447619 1.828571
3. (10 points) Consider the following case where the utilization of methane in the presence of oxygen in a continuous flow, sparged fermentor (sparging refers to the bubbling of oxygen (and methane in this case) into the reactor) is described by a double MichaelisMenten form of cell growth µ = µmax
S 1
S 2
K S 1 + S 1 K S 2 + S 2
(1)
where S 1 is oxygen and S 2 is methane. Assuming constant Yield Coefficients for each substrate, write down the steady state balances for a sterile feed for cells and for substrate 1 and 2. Let kL1 a be the overall mass transfer coefficient for the transport of oxygen from the gas phase into the liquid phase (and kL2 a be that for the transport of methane from the gas phase into the liquid phase. You can assume that both the liquid and the gas phases are well-mixed. (Hint: Feed is sterile. Write down the mass balances for oxygen, methane and biomass. You can (if you want to) consider one or two input ports for each of oxygen and methane - just remember that both oxygen and methane are substrates for the organism and that you will have two substrate balances. Make any assumptions you may need to make, define any variables you think you need to define. )
Solution: The equation that you need in each case is what comes in must equal to what leaves and what is generated (or consumed) in the reactor. Remember we are writing in terms of amount per time for each substrate and biomass (In the usual notations). Assume that you are bubbling in oxygen and methane using one port and taking the exit gases out through one port. Assume that the oxygen and methane behave independently - i.e. the relationship between the concentration of the component in the gas phase and the dissolved concentration (what you know as C ) is not affected by the presence of the other component. Henry’s Law is used to relate the concentration of the component in the gas phase to that in the liquid ∗
What we are doing is writing equation 6.69 in your textbook in the context of a substrate that is in the gas phase. Note the units of the terms in the equation 6.69 which is written for a substrate in the solution. S o is the concentration of substrate (grams/liter) coming in multiplied by flow rate (liter/hour) gives you the rate of substrate being delivered (grams/hour). When we are bubbling a gas into the system, (as we have done for oxygen ) we know that the rate of transport is calculated by kL a C − C and the units are grams/(liter.hour) when multiplied by volume of the reactor, V R gives you the rate of substrate being delivered in grams/hour. So
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The rate at which methane is being delivered to the reactor is kL a C CH − C CH 4 V R (and a similar term for oxygen) - What about removal? Are these two substrates removed in the gas phase? Liquid phase? Not specified. At steady state, you can equate the transfer rate to the uptake rate due to growth - so one assumption you can make is that all of the methane (and oxygen) transported are consumed by the growing cells. Rate of consumption of methane by the growing cells V R µX Y 1 where the µ is what is given in the problem statement - So here are all the terms written for methane (similar for oxygen) (the S in the growth equation refer to the dissolved concentration of the methane (or oxygen) in the medium) Description What comes in What leaves What gets consumed
Equation
kL a C CH − C CH 4 V R ∗
4
Nothing V R µX Y 1
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4. (10 points) Oxygen transport to growing cultures can be increased by augmenting the oxygen partial pressures in the bubbles (which will increase the dissolved oxygen concentration in the medium, or what you know as C ). One way to do this is by using pure oxygen instead of air for example. In some cases however the growth may be inhibited at higher oxygen concentrations (substrate inhibition). Assume that the growth dependence on oxygen can be represented by ∗
µ=
µmax S 1 + S + γS 2
(2)
where γ is a constant and S is the concentration of oxygen in the medium. At what value of γ will the growth rate be maximum? (Your answer should be in terms of S )
Solution: This is substrate inhibition - take a look at the equation 3.34 in page 72 of your book ν =
V m S
S 2 K S1
K m + S +
(3)
The substrate concentration that gives you the maximum reaction rate is S max
=
K m K S
1
(4)
Rewrite the equation you have this way µ=
µmax S S 2
1 + S +
(5)
1
γ
Now you can do a direct comparison - and calculate the substrate concentration that gives you the maximum reaction rate as S max =
Note that I substituted 1 for K m and
1 γ
1 1
γ
(6)
for K S
1
5. (20 points) This is problem 6.1 from your text book that I have reproduced here and for which I have provided a solution A simple, batch fermentation of an aerobic bacterium growing on methanol gave the results shown in the table. Calculate: 1. Maximum growth rate (µmax )
2. Yield on substrate (Y X/S ) 3. Mass doubling time (td ) 4. Saturation constant (K s ) 5. Specific growth rate (µnet at 10 hours Time (h) X (g/l) S (g/l) 0.0000 0.2000 9.2300 2.0000 0.2110 9.2100 4.0000 0.3050 9.0700 8.0000 0.9800 8.0300 10.0000 1.7700 6.8000 12.0000 3.2000 4.6000 14.0000 5.6000 0.9200 16.0000 6.1500 0.0770 18.0000 6.2000 0.0000 Solution: It may be easier if you plot the data (even if not asked) - so you can see where the lag phase ends and where the growth phase begins - Take a look at these two plots for the growth curve - one is in the actual data given and the other is a log transformed data versus time The log plot shows more clearly that the exponential growth starts at time 2.0 hrs and ends at 14 hours - so let’s calculate the µmax The value of the biomass concentration at the end of the exponential growth phase is 5.60 and the natural log value is 1.72276 and the ∆t you need is 12 hours and so you can calculate µmax as equal to 0.20998 hr 1 Yield on substrate can be calculated using the data at the extremes (values at time points of 18 hrs and that at 0 hrs) and you get 0.65005. Mass doubling time is µln(2) and is 3.300 hrs. The calculation of the saturation constant is not easy max since you need value of the growth rate at different substrate concentrations (so I will ignore that for now). The 10 hour time point is midway in the fermentation run and so the growth rate is the same as before, 0.20998 −
6. (20 points) Write the steady state material balance on an ideal chemostat for both biomass and substrate, assuming that endogeneous metabolism or the death rate is negligible compared to the growth rate. You can also assume that the feed is sterile. Assume that the Monod equation can be used to describe the growth rate on the limiting substrate. Derive the expressions for the substrate and the biomass concentration in the reactor effluent in terms of dilution rates, incoming substrate concentration, parameters from the Monod equation, yield coefficients and so on. (This is section 6.4.3 in your text book. In the book, they have started with the most general equation and then simplified to get the expressions for the substrate and the biomass concentration. You can do that or write down the simplified equations directly - whatever you are comfortable with )
Solution: We start with equation 6.64 (I have used µ for the specific growth rate and dropped all the subscripts and such) The material balance on the cell concentration around the chemostat gives you −F X + V R µX = 0 and dividing by V R we get − V F R X + µX = 0 or X − V F R + µ = 0 which is written as X −D + µ = 0 where D = V F R and called the dilution rate. Thus at steady state, D = µ since X = 0. The µm S µm S Monod equation is written as µ = K s +S and so we can write D = K s +S - we can sD solve for S from this equation to get S = µK . Now the yield coefficient is defined m D as Y X/S = S oX S (note: X o = 0) and hence X = Y X/S (S o − S ) and substituting for S sD you get X = Y X/S (S o − µK ) m D
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7. (20 points) This is problem 6.6 from your text book that I have reproduced here and for which I have provided a solution Pseudomonas sp has a mass doubling time of 2.4 hrs when grown on acetate. The saturation constant using this substrate is 1.3 g/l (which is unusually high) and cell yield on acetate is 0.46 g cell/g acetate. If we operate a chemostate on a feed stream containing 38 g/l acetate find the following 1. Cell concentration when the dilution rate is one half of the maximum 2. Substrate concentration when the dilution rate is 00.8Dmax 3. Maximum dilution rate 4. Cell productivity at 0.8Dmax
and so knowing doubling time, Solution: We know that doubling time t = µln(2) max you can calculate µmax (You may find the text confusing at times - what we need is the maximum specific growth rate for these calculations) and so we calculate µ as 0.6931 divided by 2.4000 hrs and we get 0.2888 hr 1 . The dilution rate for calculating cell concentration is half of the maximum growth rate 0.28881 which is 0.14440. But notice something - look at the equation for calculating cell concentration X = K S D Y X/S S o − S which can be written (using the expression for S = µmax as follows D −
−
K S D
and since
µmax 2
you can write X = Y X/S S o − K S X = Y X/S S o − µmax D D = and calculate X to be equal to 16.88200 g cell/liter. When the dilution rate is K S 0.8µmax 0.8Dmax or 0.8µmax (which is 0.23104) you get S = µmax or simply S = 4 K S 0.8µmax and you get S as equal to 5.20000 grams/l Cell productivity is DX which is equal K S D to DY X/S S o − µmax and we can calculate that. Let us first calculate S o − S D as 38.00000 minus 5.20000 which is 32.79999 multiplied by DY X/S which is 0.23104 multiplied by 0.46000 which is 0.10628 and hence we get DX to be equal to 3.48606 gcell/(l.hr) You need to carefully keep track of variables and what is required of you in the problem. −
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8. (20 points) This is problem 6.10 from your text book that I have reproduced here and for which I have provided a solution Ethanol is to be used as a substrate for single cell protein production in a chemostat. The available equipment can achieve an oxygen transfer rate of 10 g O2 /liter of liquid per hour. Assume the kinetics of cell growth on ethanol is of the Monod type with µm 0.5 hr 1 K S 30 mg/l, Y X/S 0.5 cells/g ethanol and Y O /S 2 g O2 /g ethanol. We wish to operate the chemostat with an ethanol concentration in the feed of 22 grams/liter. We also wish to maximize the biomass productivity and minimize the loss of ethanol in the effluent. Determine the required dilution rate and whether sufficient oxygen can be provided. −
2
Y
(as Solution: Notice that you are not given Y X/O which can be calculated as Y OX/S /S the ratio) and we can calculate Y X/O equal to 0.25000 gcells/gO2 . Let us calculate S Dopt using the formula Dopt = µm 1 − K SK +S Remember to convert 22 grams/liter o to 22000 mg/liter (units of K S ) and you will get 0.4815489 Now calculate K S D as equal to 782.9596609 in mg/liter and convert that to grams/liter S = µmax D by multiplying by 0.001 and get 0.7829596 g/l Now we need to calculate the X K S D from X = Y X/S S o − µmax 10608.5201695 in mg/liter and you get 10.6085201 in D µX g/liter Calculate OUR as OU R = Y X/O So OUR equals 20.4340848 gmO2 /liter.hr which is greater than 10 and hence sufficient oxygen cannot be provided 2
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OD versus time for growth in Simple and Complex Medium 1.1 1 0.9 0.8 0.7 0.6 D O
0.5 0.4 0.3 0.2 0.1
'simple.dat' using 1:2 'complex.dat' using 1:2
0 0
2
4
6 Time (Hours)
8
10
log_e(OD) versus time for growth in simple and complex medium
0
-0.5
-1 ) D O ( e _ g o l
-1.5
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-2.5 'simple.dat' using 1:(log($2)) 'complex.dat' using 1:(log($2)) -3 0
1
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Problem 6.1 Growth Curve 7
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'exam2r.dat' using 1:2 0 0
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Time (hrs)
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Problem 6.1 Growth Curve (in log) 2
1.5
1
) l / g ( X f o g o l l a r u t a n
0.5
0
-0.5
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-1.5 'exam2r.dat' using 1:(log($2)) -2 0
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Time (hrs)
12
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