9.4.7 Wash-Out

To learn the effect of increasing the dilution rate, we combine Equations (9-82) and (9-54) and set rd = 0 to get

9-91

image

We see that if D > μ, then dCc/dt will be negative, and the cell concentration will continue to decrease until we reach a point where all cells will be washed out:

Cc = 0

The dilution rate at which wash-out will occur is obtained from Equation (9-90) by setting Cc = 0.

9-92

image

Flow rate at which wash-out occurs

We next want to determine the other extreme for the dilution rate, which is the rate of maximum cell production. The cell production rate per unit volume of reactor is the mass flow rate of cells out of the reactor (i.e., image) divided by the volume V, or

9-93

image

Using Equation (9-90) to substitute for Cc yields

9-94

image

Figure 9-25 shows production rate, cell concentration, and substrate concentration as functions of dilution rate.

Figure 9-25. Cell concentration and production rate as a function of dilution rate.

image

Maximum rate of cell production (DCc)

We observe a maximum in the production rate, and this maximum can be found by differentiating the production rate, Equation (9-94), with respect to the dilution rate D:

9-95

image

Then

9-96

image

Maximum rate of cell production

The organism Streptomyces aureofaciens was studied in a 10 dm3 chemostat using sucrose as a substrate. The cell concentration, Cc (mg/ml), the substrate concentration, Cs (mg/ml), and the production rate, DCc (mg/ml/h), were measured at steady state for different dilution rates. The data are shown in Figure 9-26.23 Note that the data follow the same trends as those discussed in Figure 9-25.

Figure 9-26. Continuous culture of Streptomyces aureofaciens in chemostats. (Note: XCc)

image

Courtesy of S. Aiba, A. E. Humphrey, and N. F. Millis, Biochemical Engineering, 2nd Ed. (New York: Academic Press, 1973).

Summary

  1. In the PSSH, we set the rate of formation of the active intermediates equal to zero. If the active intermediate A* is involved in m different reactions, we set it to

    S9-1

    image

    This approximation is justified when the active intermediate is highly reactive and present in low concentrations.

  2. The azomethane (AZO) decomposition mechanism is

    S9-2

    image

    S9-3

    image

    By applying the PSSH to AZO*, we show the rate law, which exhibits first-order dependence with respect to AZO at high AZO concentrations and second-order dependence with respect to AZO at low AZO concentrations.

  3. Enzyme Kinetics: enzymatic reactions follow the sequence

    image

    Using the PSSH for (E · S) and a balance on the total enzyme, Et, which includes both the bound (E · S) and unbound enzyme (E) concentrations

    Et = (E) + (E · S)

    we arrive at the Michaelis–Menten equation

    S9-4

    image

    where Vmax is the maximum reaction rate at large substrate concentrations (S >> KM) and KM is the Michaelis constant. KM is the substrate concentration at which the rate is half the maximum rate (S1/2 = KM).

  4. The three different types of inhibition—competitive, uncompetitive, and noncompetitive (mixed) inhibition—are shown on the Lineweaver–Burk plot:

    image

  5. Bioreactors:

    image

    a. Phases of bacteria growth:

    I. Lag

    ii. Exponential

    iii. Stationary

    IV. Death

    b. Unsteady-state mass balance on a chemostat:

    S9-5

    image

    S9-6

    image

    c. Monod growth rate law:

    S9-7

    image

    d. Stoichiometry:

    S9-8

    image

    S9-9

    image

    Substrate consumption:

    S9-10

    image

DVD-ROM Material

Learning Resources

  1. Summary Notes
  2. Web Modules

    image

    A. Ozone Layer

    image

    Photo courtesy of Goddard Space Flight Center (NASA). See DVD-ROM for color pictures of the ozone layer and the glow sticks.

    B. Glow Sticks

    image

  3. Interactive Computer Games

    image

    A. Enzyme Man

    image

Living Example Problems

  1. Example 9-5 Bacteria Growth in a Batch Reactor
  2. Example DVD-ROM-2 PSSH Applied to Thermal Cracking of Ethane
  3. Example DVD-ROM-7 Alcohol Metabolism
  4. Example Web Module: Ozone
  5. Example Web Module: Glowsticks
  6. Example Web Module: Russell’s Viper
  7. Example Web Module: Fer-de-Lance
  8. Example R7.4 Receptor Endocytosis

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Professional Reference Shelf

image

R9-1. Chain Reactions Example Problem

R9-2. Reaction Pathways

R9-3. Polymerization

A. Step Polymerization

Example R9-3.1 Determining the Concentration of Polymers for Step Polymerization

B. Chain Polymerizations Example R9-3.2 Parameters of MW Distribution

C. Anionic Polymerization

Example R9-3.3 Calculating the Distribution Parameters from Analytic Expressions for Anionic Polymerization Example R9-3.4 Determination of Dead Polymer Distribution When Transfer to Monomer Is the Primary Termination Step

R9-4. Oxygen-Limited Fermentation Scale Up

R9-5. Receptor Kinetics

A. Kinetics of signaling

image

image

B. Endocytosis

R9-6. Multiple Enzyme and Substrate Systems

A. Enzyme Regeneration

Example PRS9-6.1 Construct a Lineweaver–Burk Plot for Different Oxygen Concentration

B. Enzyme Cofactors

  1. Example PRS9-6.2 Derive a Rate Law for Alcohol Dehydrogenase
  2. Example PRS9-6.3 Derive a Rate Law for a Multiple Substrate System
  3. Example PRS9-6.4 Calculate the Initial Rate of Formation of Ethanol in the Presence of Propanediol

R9-7. Physiologically Based Pharmacokinetic (PBPK) models. Case Study: Alcohol metabolism in humans

image

Figure R9-7.1. Blood alcohol-time trajectories from data of Wilkinson et al.24

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R9-8. Pharmacokinetics in Drug Delivery

Pharmacokinetic models of drug delivery for medication administered either orally or intravenously are developed and analyzed.

Figure A. Two-compartment model.

image

Figure B. Drug response curve.

image

Questions and Problems

In each of the following questions and problems, rather than just drawing a box around your answer, write a sentence or two describing how you solved the problem, the assumptions you made, the reasonableness of your answer, what you learned, and any other facts that you want to include.

image

You may wish to refer to W. Strunk and E. B. White, The Elements of Style, 4th ed. (New York: Macmillan, 2000) and Joseph M. Williams, Style: Ten Lessons in Clarity & Grace, 6th ed. (Glenview, Ill.: Scott, Foresman, 1999) to enhance the quality of your sentences.

P9-1A

ICG Enzyme Man.

a. Load the ICG on your computer and carry out the exercise. Performance number = ________________________.

b. Apply one or more of the six ideas in Table P-3, page xiii to this problem.

P9-2B

a. Example 9-1. How would the results change if the concentration of CS2 and M were increased?

b. Example 9-2. (1) The following additional runs were carried out when an inhibitor was present.

images

image

What type of inhibition is taking place? (2) Sketch the curves for no inhibition, competitive, uncompetitive, noncompetitive (mixed) inhibition, and substrate inhibition on a Woolf–Hanes plot and on an Eadie–Hofstee plot.

c. Example 9-3. (1) What would the conversion be after 10 minutes if the initial concentration of urea were decreased by a factor of 100? (2) What would be the conversion in a CSTR with the same residence time, τ, as the batch reactor time t? (3) A PFR?

d. Example 9-4. What is the total mass of substrate consumed in grams per mass of cells plus what is consumed to form product? Is there disparity here?

e. Example 9-5. Load the Living Example Problem. (1) Modify the code to carry out the fermentation in a fed-batch (e.g., semibatch) reactor in which the substrate is fed at a rate of 0.5 dm3/h and a concentration of 5 g/dm3 to an initial liquid volume of 1.0 dm3 containing a cell mass with an initial concentration of Cci = 0.2 mg/dm3 and an initial substrate concentration of Cci = 0.5 mg/dm3. Plot and analyze the concentration of cells, substrate, and product as a function of time, along with the mass of product up to 24 hours. (2) Repeat (1) when the growth is uncompetitively inhibited by the substrate with KI = 0.7 g/dm3. (3) Set image = 10,000 g/dm3, and compare your results with the base case.

image

f. Chain Reaction Example discussed in Professional Reference Shelf R9.1 on DVD-ROM. Over what range of time is the PSSH not valid? Load the Living Example Problem. Vary the temperature (800 < T < 1600). What temperature gives the greatest disparity with the PSSH results? Specifically compare the PSSH solution with the full numerical solution.

g. Example on Alcohol Metabolism on the DVD-ROM. This problem is a gold mine for things to be learned about the effect of alcohol on the human body. Load the Polymath Living Example Program from the DVD-ROM. (1) Start by varying the initial doses of alcohol. (2) Next consider individuals who are ALDH enzyme deficient, which includes about 40% to 50% of Asians and Native Americans. Set Vmax for acetaldehydes between 10% and 50% of its normal value and compare the concentration-time trajectories with the base cases. (Hint: Read the journal article in the Summary Notes [Alcohol 35, p.1 (2005)].)

P9-3B

(Flame retardants) Hydrogen radicals are important to sustaining combustion reactions. Consequently, if chemical compounds that can scavenge the hydrogen radicals are introduced, the flames can be extinguished. While many reactions occur during the combustion process, we shall choose CO flames as a model system to illustrate the process [S. Senkan et al., Combustion and Flame, 69, 113 (1987)]. In the absence of inhibitors

image

P9-3.1

image

P9-3.2

image

P9-3.3

image

P9-3.4

image

The last two reactions are rapid compared to the first two. When HCl is introduced to the flame, the following additional reactions occur:

image

Assume that all reactions are elementary and that the PSSH holds for the O·, OH·, and Cl· radicals.

a. Derive a rate law for the consumption of CO when no retardant is present.

b. Derive an equation for the concentration of H· as a function of time assuming constant concentration of O2, CO, and H2O for both uninhibited combustion and combustion with HCl present. Sketch H· versus time for both cases.

P9-4A

The pyrolysis of acetaldehyde is believed to take place according to the following sequence:

image

a. Derive the rate expression for the rate of disappearance of acetaldehyde, –rAc.

b. Under what conditions does it reduce to the equation at the begining of Section 9.1 on page 340?

c. Sketch a reaction pathway diagram for this reaction. [Hint: See margin note on page 347.]

P9-5B

For each of the reactions in parts (a), (b), and (c), suggest a mechanism and apply the PSSH to learn if the mechanism is consistent with the rate law.

a. The gas-phase homogeneous oxidation of nitrogen monoxide (NO) to dioxide (NO2),

image

is known to have a form of third-order kinetics, which suggests that the reaction is elementary as written, at least for low partial pressures of the nitrogen oxides. However, the rate constant k actually decreases with increasing absolute temperature, indicating an apparently negative activation energy. Because the activation energy of any elementary reaction must be positive, some explanation is in order.

image

Provide an explanation, starting from the fact that an active intermediate species, NO3, is a participant in some other known reactions that involve oxides of nitrogen. Draw the reaction pathway. [Hint: See margin on page 347]

b. The rate law for formation of phosgene, COCl2, from chlorine, Cl2, and carbon monoxide, CO, has the rate law

image

Suggest a mechanism for this reaction that is consistent with this rate law and draw the reaction pathway. [Hint: Cl formed from the dissociation of Cl2 is one of the two active intermediates.]

c. Suggest an active intermediate(s) and mechanism for the reaction H2 + Br2 → 2HBr. Use the PSSH to show whether or not your mechanism is consistent with the rate law

image

P9-6C

(Tribology) Why you change your motor oil? One of the major reasons for engine oil degradation is the oxidation of the motor oil. To retard the degradation process, most oils contain an antioxidant [see Ind. Eng. Chem. 26, 902 (1987)]. Without an inhibitor to oxidation present, the suggested mechanism at low temperatures is

image

Why you need to change the motor oil in your car?

image

where I2 is an initiator and RH is the hydrocarbon in the oil.

When an antioxidant is added to retard degradation at low temperatures, the following additional termination steps occur:

image

image

image

a. Derive a rate law for the degradation of the motor oil in the absence of an antioxidant at low temperatures.

b. Derive a rate law for the rate of degradation of the motor oil in the presence of an antioxidant for low temperatures.

c. How would your answer to part (a) change if the radicals I· were produced at a constant rate in the engine and then found their way into the oil?

d. Sketch a reaction pathway diagram for both high and low temperatures, with and without antioxidant.

e. See the open-ended problem G.2 in Appendix G and on the DVD-ROM for more on this problem.

P9-7A

Epidemiology. Consider the application of the PSSH to epidemiology. We shall treat each of the following steps as elementary, in that the rate will be proportional to the number of people in a particular state of health. A healthy person, H, can become ill, I, spontaneously, such as by contracting smallpox spores:

P9-7.1

image

or the person may become ill through contact with another ill person:

P9-7.2

image

The ill person may become healthy:

P9-7.3

image

or the person may expire:

P9-7.4

image

The reaction given in Equation (P9-7.4) is normally considered completely irreversible, although the reverse reaction has been reported to occur.

a. Derive an equation for the death rate.

b. At what concentration of healthy people does the death rate become critical? [Ans.: When [H] = (k3 + k4)/k2.]

c. Comment on the validity of the PSSH under the conditions of Part (b).

d. If k1 = 10–8 h–1, k2 = 10–16 (people·h)–1, k3 = 5 × 10–10 h, k4 = 10–11 h, and Ho = 109 people, use Polymath to plot H, I, and D versus time. Vary ki and describe what you find. Check with your local disease control center or search the WWW to modify the model and/or substitute appropriate values of ki. Extend the model, taking into account what you learn from other sources (e.g., WWW).

e. List ways you can work this problem incorrectly.

f. Apply one or more of the six ideas in Table P-3, page xiii to this problem.

P9-8B

Derive the rate laws for the following enzymatic reactions and sketch and compare, where possible, with the plot shown in Figure E9-2.1.

a. image

b. image

c. image

d. image

e. Two products

image

f. Which of the reactions (a) through (e), if any, lend themselves to analysis by a Lineweaver–Burk plot?

P9-9B

Beef catalase has been used to accelerate the decomposition of hydrogen peroxide to yield water and oxygen [Chem. Eng. Educ., 5, 141 (1971)]. The concentration of hydrogen peroxide is given as a function of time for a reaction mixture with a pH of 6.76 maintained at 30°C.

images

a. Determine the Michaelis–Menten parameters Vmax and KM.

b. If the total enzyme concentration is tripled, what will the substrate concentration be after 20 minutes?

c. Apply one or more of the six ideas in Table P-3, page xiii to this problem.

d. List ways you can work this problem incorrectly.

P9-10B

It has been observed that substrate inhibition occurs in the following enzymatic reaction:

image

a. Show that the rate law for substrate inhibition is consistent with the plot in Figure P9-10B of –rs (mmol/L·min) versus the substrate concentration S (mmol/L).

Figure . Figure P9-10B Michaelis–Menten plot for substrate inhibition.

image

b. If this reaction is carried out in a CSTR that has a volume of 1000 dm3, to which the volumetric flow rate is 3.2 dm3/min, determine the three possible steady states, noting, if possible, which are stable. The entrance concentration of the substrate is 50 mmol/dm3. What is the highest conversion?

c. What would be the effluent substrate concentration if the total enzyme concentration is reduced by 33%?

d. List ways you can work this problem incorrectly.

e. How could you make this problem more difficult?

P9-11B

The following data on bakers’ yeast in a particular medium at 23.4°C were obtained in the presence and in the absence of an inhibitor, sulfanilamide. The reaction rate (–rS) was measured in terms of the oxygen uptake rate Qo2, obtained as a function of oxygen partial pressure.

a. Assume the rate QO2 follows Michaelis–Menten kinetics with respect to oxygen. Calculate the QO2 maximum (i.e., Vmax), and the Michaelis–Menten constant KM. [Ans.: Vmax = 52.63 μL O2/h · mg cells.]

b. Using the Lineweaver–Burk plot, determine the type of inhibition sulfanilamide that causes the O2 uptake to change.

images

c. List ways you can work this problem incorrectly.

d. Apply one or more of the six ideas in Table P-3, page xiii to this problem.

P9-12B

The enzymatic hydrolysis of starch was carried out with and without maltose and α-dextrin added. [Adapted from S. Aiba, A. E. Humphrey, and N.F. Mills, Biochemical Engineering (New York: Academic Press, 1973).]

Starch → α-dextrin → Limit dextrin → Maltose

images

images

Determine the types of inhibition for maltose and for α-dextrin.

P9-13B

The hydrogen ion, H+, binds with the enzyme (E) to activate it in the form EH. H+ also binds with EH to deactivate it by forming image.

image

Figure . Figure P9-13B Enzyme pH dependence.

image

where E and image are inactive.

a. Determine if the preceding sequence can explain the optimum in enzyme activity with pH shown in Figure P9-13B.

b. List ways you can work this problem incorrectly.

c. Apply one or more of the six ideas in Table P-3, page xiii to this problem.

P9-14B

Diabetes is a global epidemic affecting more than 240 million people worldwide. The vast majority of the cases are Type 2. Recently a drug, Januvia (J), was discovered to treat Type 2 diabetes. When food enters the stomach, a peptide, GLP-1 (glucagon-like peptide 1) is released, which leads to glucose-dependent insulin secretion and glucagon suppression. The half-life of GLP-1 is very short because it is rapidly degraded by an enzyme dipeptidyl peptidase-IV (DPP-IV) which cleaves the two terminal amino acids of the peptide thus deactivating it. DP-IV rapidly cleaves the active form of GLP-1 (GLP-1[7-36] amide) to its inactive form (GLP-1[9-36] amide) with a half life of 1 minute/2 ~ 1 min, and it is thought to be the primary enzyme responsible for this hydrolysis.25a

image

Inhibition of the DPP-IV Enzyme, (E), therefore, is expected to significantly reduce the extent of the inactivation of GLP-1[7-36] and should lead to an increase in circulating levels of the active form of the hormone. Supporting evidence for this comes from DPP-IV-enzyme-deficient mice, which have elevated levels of GLP-1[7-36]26 amide. As a very rough approximation, let’s treat the reaction as follows: The new drug, an inhibitor of the DPP-IV enzyme (E), is Januvia (J), which prevents the enzyme from deactivating GLP-1.

E + GLP-1 ⇄ E · GLP-1 → Glucose Release

Inhibited

E + J ⇄ E · J (Inactive)

By delaying the degradation of GLP-1, the inhibitor is able to extend the action of insulin and also suppress the release of glucagon.

a. Plot the ratio of reaction of –rGLP (without inhibition) to the rate – rGLP (with inhibition) as a function of DDP-4 inhibitor constant for both competitive inhibition and uncompetitive inhibition.

b. Assuming the body is a well-mixed reactor, develop a model similar to problems P9-7 and P9-8 for the dosage schedule for Januvia.

P9-15B

The production of a product P from a particular gram negative bacteria follows the Monod growth law

image

with μmax = 1 h–1, KS = 0.25 g/dm3, and Yc/s = 0.5 g/g.

a. The reaction is to be carried out in a batch reactor with the initial cell concentration of Cc0 = 0.1 g/dm3 and substrate concentration of Cs0 = 20 g/dm3.

Cc = Cc0 + Yc/s(Cs0Cs)

Plot rg, –rs, –rc, Cs, and Cc as a function of time.

b. The reaction is now to be carried out in a CSTR with Cs0 = 20 g/dm3 and Cc0 = 0. What is the dilution rate at which wash-out occurs?

c. For the conditions in part (b), what is the dilution rate that will give the maximum product rate (g/h) if Yp/c = 0.15 g/g? What are the concentrations Cc, Cs, Cp, and –rs at this value of D?

d. How would your answers to (b) and (c) change if cell death could not be neglected with kd = 0.02 h–1?

e. How would your answers to (b) and (c) change if maintenance could not be neglected with m = 0.2 g/h/dm3?

f. Redo part (a) and use a logistic growth law

image

and plot Cc and rc as a function of time. The term C is the maximum cell mass concentration and is called the carrying capacity and is equal to C = 1.0 g/dm3. Can you find an analytical solution for the batch reactor? Compare with part (a) for C = Yc/s Cs0 + Cc0.

g. List ways you can work this problem incorrectly.

h. Apply one or more of the six ideas in Table P-3, page xiii to this problem.

P9-16B

Redo Problem P9-15B (a), (c), and (d) using the Tessier equation

image

with μmax = 1.0 h–1 and k = 8 g/dm3.

a. List ways you can work this problem incorrectly.

b. How could you make this problem more difficult?

P9-17B

The bacteria X-II can be described by a simple Monod equation with μmax = 0.8 h–1 and KS = 4 g/dm3, Yp/c = 0.2 g/g, and Ys/c = 2 g/g. The process is carried out in a CSTR in which the feed rate is 1000 dm3/h at a substrate concentration of 10 g/dm3.

a. What size fermentor is needed to achieve 90% conversion of the substrate? What is the exiting cell concentration?

b. How would your answer to (a) change if all the cells were filtered out and returned to the feed stream?

c. Consider now two 5000 dm3 CSTRs connected in series. What are the exiting concentrations Cs, Cc, and Cp from each of the reactors?

d. Determine, if possible, the volumetric flow rate at which wash-out occurs and also the flow rate at which the cell production rate (Cc υ0) in grams per day is a maximum.

e. Suppose you could use the two 5000-dm3 reactors as batch reactors that take two hours to empty, clean, and fill. What would your production rate be in (grams per day) if your initial cell concentration is 0.5 g/dm3? How many 500-dm3 reactors would you need to match the CSTR production rate?

f. List ways you can work this problem incorrectly.

g. Apply one or more of the six ideas in Table P-3, page xiii to this problem.

P9-18A

A CSTR is being operated at steady state. The cell growth follows the Monod growth law without inhibition. The exiting substrate and cell concentrations are measured as a function of the volumetric flow rate (represented as the dilution rate), and the results are shown below. Of course, measurements are not taken until steady state is achieved after each change in the flow rate. Neglect substrate consumption for maintenance and the death rate, and assume that Yp/c is zero. For run 4, the entering substrate concentration was 50 g/dm3 and the volumetric flow rate of the substrate was 2 dm3/h.

image

a. Determine the Monod growth parameters μmax and KS.

b. Estimate the stoichiometric coefficients, Yc/s and Ys/c.

c. Apply one or more of the six ideas in Table P-3, page xiii to this problem.

d. How could you make this problem more difficult?

P9-19B

Alternative Energy Source.27 In the summer of 2009, ExxonMobil decided to invest 600 million dollars on developing algae as an alternative fuel. Algae would be grown and their oil extracted to provide an energy source. It is estimated that one acre of a biomass pond can provide 6,000 gallons of gasoline per year, which would require the capture of a CO2 source more concentrated than air (e.g., fuel gas from a refinery) and also contribute to the sequestration of CO2. The biomass biosynthesis during the day is

Sunlight + CO2 + H2O + Algae → More Algae + O2

Consider a 5,000 gallon pond with perforated pipes into which CO2 is injected and slowly bubbled into the solution to keep the water saturated with CO2.

Figure P9-19.1. Commercial microalgae production in open raceway paddle-wheel mixed ponds.

image

Courtesy of Cyanotech Co., Hawaii.

The doubling time during the day is 12 h at high noon sunlight and zero during the night. As a first approximation, the growth during the 12 hours of daylight law is

rg = fμCC

with f = sunlight = sin (π t/12) between 6 a.m. and 6 p.m., otherwise f = 0, CC is the algae concentration (g/dm3) and μ = 0.9 day–1 (assumes constant CO2 saturation at 1 atm is 1.69g/kg water). The pond is 30 cm deep and for effective sunlight penetration the algae concentration cannot exceed 200 mg/dm3.

a. Derive an equation for the ratio of the algae cell concentration CC at time t to initial cell concentration CC0, i.e., (CC/CC0). Plot and analyze (CC/CC0) versus time up to 48 hours.

b. If the pond is initially seeded with 0.5 mg/dm3 of algae, how long will it take the algae to reach a cell density (i.e., concentration) of 200 mg/dm3, which is the concentration at which sunlight can no longer effectively penetrate the depth of the pond? Plot and analyze rg and CC as a function of time. As a first approximation, assume the pond is well mixed.

c. Suppose the algae limit the sun’s penetration significantly even before the concentration reaches 200 mg/dm3 with e.g., μ = μ0 (1 – CC/200). Plot and analyze rg and CC as a function of time. How long would it take to completely stop growth at 200 mg/dm3?

d. Now, let’s consider continuous operation. Once the cell density reaches 200 mg/dm, one-half of the pond is harvested and the remaining broth is mixed with fresh nutrient. What is the steady-state algae productivity in gm/year, again assuming the pond is well mixed?

e. Now consider a constant feed of waste water and removal of algae at a dilution rate of one reciprocal day. What is the mass flow rate of algae out of the 5,000 gallon pond (g/d)? Assume the pond is well mixed.

f. Now consider that the reaction is to be carried out in an enclosed by transparent reactor. The reactor can be pressurized with CO2 up to 10 atm with KS = 2 g/dm3. Assume that after the initial pressurization, no more CO2 can be injected. Plot and analyze the algae concentration as a function of time.

g. An invading algae can double twice as fast as the strain you are cultivating. Assume that it initially is at 0.1 mg/l concentration. How long until it is the dominant species (over 50% of the cell density)?

P9-20A

What six things are wrong with this solution?

Evaluate the inhibited enzyme kinetic rate law parameters Vmax, KM, and KI. Data from a substrate (S) inhibited reaction is shown below, in the form of Eadie–Hofstee plot.

image

For a competive inhibition concentration (I) of 0.02 M (line 1), we find (CI). For an inhibition concentration of 0.05M (line 2), we find slope (1) = 2.5 and slope (2) = 1.0. Solving two equations using the slope and intercept, we find Vmax = 2 and KI = 5, and from the intercept KM = 0.5.

Additonal Homework Problems

A number of homework problems that can be used for exams or supplementary problems or examples are found on the DVD-ROM and on the CRE Web site, http://www.engin.umich.edu/~cre.

New Problems on the Web

CDP9-New

From time to time, new problems relating Chapter 9 material to everyday interests or emerging technologies will be placed on the Web. Solutions to these problems can be obtained by e-mailing the author. Also, one can go to the Web site, www.rowan.edu/greenengineering, and work the homework problem on green engineering specific to this chapter.

Green Engineering Problem

image

Supplementary Reading

Web

Review the following Web sites:

www.cells.com

www.enzymes.com

www.pharmacokinetics.com

image

Text

  1. A discussion of complex reactions involving active intermediates is given in

    FROST, A. A., and R. G. PEARSON, Kinetics and Mechanism, 2nd ed. New York: Wiley, 1961, see Chapter 10. Old but great examples.

    LAIDLER, K. J., Chemical Kinetics, 3rd ed. New York: HarperCollins, 1987.

    PILLING, M. J., Reaction Kinetics, New York: Oxford University Press, 1995.

  2. Further discussion of enzymatic reactions is presented in

    Just about everything you want to know about basic enzyme kinetics can be found in SEGEL, I. H. Enzyme Kinetics. New York: Wiley-Interscience, 1975.

    An excellent description of parameter estimation, biological feedback, and reaction pathways can be found in VOIT, E. O. Computational Analysis of Biochemical Systems. Cambridge, UK: Cambridge University Press, 2000.

    BLANCH, H. W. and D. S. CLARK, Biochemical Engineering. New York: Marcel Dekker, 1996.

    CORNISH-BOWDEN, A., Analysis of Enzyme Kinetic Data. New York: Oxford University Press, 1995.

    NELSON, D. L., and M. M. COX, Lehninger Principles of Biochemistry, 3rd ed. New York: Worth Publishers, 2000.

    SHULER, M. L., and F. KARGI, Bioprocess Engineering Principles, 2nd ed. Upper Saddle River, N.J.: Prentice Hall, 2002.

    STEPHANOPOULOS, G. N., A. A. ARISTIDOU, and J. NIELSEN, Metabolic Engineering. New York: Academic Press, 1998.

  3. Material on bioreactors can be found in

    AIBA, S., A. E. HUMPHREY, and N. F. MILLIS, Biochemical Engineering, 2nd ed. San Diego, Calif.: Academic Press, 1973.

    BAILEY, T. J., and D. OLLIS, Biochemical Engineering, 2nd ed. New York: McGraw-Hill, 1987.

    BLANCH, H. W., and D. S. CLARK, Biochemical Engineering. New York: Marcel Dekker, 1996.

  4. Also see

    BURGESS, THORNTON W., The Adventures of Old Mr. Toad. New York: Dover Publications, Inc., 1916.

    KEILLOR, GARRISON, Pretty Good Joke Book, A Prairie Home Companion. St. Paul, MN: HighBridge Co., 2000.

    MASKILL, HOWARD, The Investigation of Organic Reactions and Their Mechanisms. Oxford UK: Blackwell Publishing Ltd, 2006.

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