生物反应工程全册配套完整精品课件.ppt

1、生物反应工程全册配套生物反应工程全册配套 完整精品课件完整精品课件 2 Biochemical/Bioprocess Engineering Bioreactions Solving engineering problems for the commercial production of bio-products and making them safe to use and economically competitive in the market 3 Whats the meaning of and difference between bioscience, biotechnolog

2、y and bioengineering? How can the progress in the fundamental research and applied technology affect our society and contribute to its sustainability? English is very important no matter what careers you are pursuing in the future! 4 Among about 1 billion tons of crude oil consumed annually in the U

3、nited States, more than 70% is imported, mainly from the Middle East. Currently, China is consuming about 500 million tons of crude oil annually. How much is imported? Whats its impacts on Chinas economy and society? Everyday, 2-3 tanks, each with a capacity of 300, 000 tons, are shipping crude oil

4、to feed our refineries! Whats the role of fuels and energy derived from renewable biomass resources? 5 The String of Pearls refers to the Chinese sea lines of communication which extend from the Chinese mainland to Port Sudan. The sea lines run through several major choke points such as the Strait o

5、f Mandeb, the Strait of Malacca, the Strait of Hormuz and the Lombok Strait, as well as other strategic maritime centers in Pakistan, Sri Lanka, Bangladesh, the Maldives and Somalia. 6 7 Biology Chemistry Biochemistry Biochemical Engineering Bioengineering Engineering Chemical Engineering Chapter 1:

6、 Introduction Biochemical Engineering: An interdiscipline of biology, chemistry and engineering 8 Bioreaction Engineering involving various enzymatic catalysis reactions and cell cultures and fermentations to produce various bioproducts is the primary step and core of commercial production of variou

7、s bioproducts. Bioseparation Engineering consisting of a sequence of recovery and separation steps to obtain final products to satisfy various requirements, standards and criteria in safety and effectiveness is determining the cost of most bioproducts, particularly for those pharmaceuticals for medi

8、cal cure and diagnosis. 9 Marking your calendar for the course Teaching and tutorials Introduction: 0.5 time unit Part I: Engineering Principles, 4.5 time unit + 1 tutorial Part II: Bioreaction Kinetics, 4 time unit + 1 tutorial Part III: Bioreactor Engineering, 3.5 time unit + 1 tutorial Course Sum

9、marization: 0.5 time unit Your mark is from: 1) assignments, 20% 2) quizzes in class, 10% 3) final examination, 70% 10 Part I Material and Energy Balances 11 Chapter 2 Material Balance One of the most important and basic concepts in bioprocess engineering is material balance. Because mass in bioproc

10、essing systems is conserved at all times, the law of mass conservation provides a theoretical framework for material balance. In steady-state material balance, mass entering a process is summed up and compared with the total mass leaving the system; the term “balance” implies that mass entering and

11、leaving should be equal. Essentially, material balance is an accounting procedure: total mass entering must be accounted for 12 at the end of the process, even if it undergoes heating or cooling, mixing, fermentation or other bioreactions within the system. Usually it is impossible to measure the ma

12、sses and compositions of all streams entering and leaving a system; unknown quantities can be calculated using mass-balance principles. Mass-balance problems have a constant theme: give the masses of some input and output streams, calculate the masses of others unknown. 13 2.1 Thermodynamic Prelimin

13、aries 2.1.1 System and Process Fig 2.1 Thermodynamic system System Surroundings System boundary 14 v A batch process operates in a closed system. All materials are added to the system at the start of the process; the system is then closed and products removed only when the process is complete. v A s

14、emi-batch process allows either input or output of mass, but not both simultaneously. v A fed-batch process allows input of material to a system but not output. v A continuous process allows materials to flow in and out of a system. If rates of mass input and output are equal, continuous processes c

15、an be operated steadily. 15 Semi-batch Air Batch Air Off-gas Input only at the LIC Off-gas beginning at the end Output only time during operation Input more than one Output more than one time during operation 16 Continuous Air Fed-batch Air Output Off-gas Input LIC Off-gas FIC Input FIC 17 2.1.2 Ste

16、ady State, Unsteady-State and Equilibrium If all properties within a system do not change with time, the process is said to be at steady state. Based on this definition, batch, fed-batch and semi-batch processes cannot be operated under steady state conditions. Mass is either increasing or decreasin

17、g with time during fed-batch and semi-batch processes. In batch process, even though the total mass is constant, change occurring inside the system cause the system properties to change with time. Such processes are called transient or unsteady-state. Continuous processes may be either steady or un-

18、steady. 18 It is usually to run continuous processes as close to steady state as possible. However, unsteady-state conditions will exist at start and sometimes when changes in operating conditions occur. Here we need to distinguish steady state from equilibrium. A system at equilibrium is one in whi

19、ch all opposing forces are exactly counter-balanced so that the properties of the system do not change with time. At equilibrium there is no net change in either the system or universe. To convert raw material into useful products there must be an overall change in the universe, equilibrium needs to

20、 be disturbed by the input or output of mass or energy. 19 2.2 Law of Mass Conservation System M M i o Glucose Ethanol 20 If Mi and Mo are different, there must be: v Measurements of Mi and Mo are wrong; v The system has leakages allowing M to enter or escape undetected; v M is consumed or generated

21、 by bioreactions within the system; or v M accumulates within the system. mass in mass out + mass generated mass consumed mass accumulated =(2.1) 21 For a process at steady state, we have mass in + mass generated = mass out + mass consumed (2.2) If no reaction occurs within the system, or mass balan

22、ce is applied for an inert component, we have mass in = mass out (2.3) There are two kinds of mass balance. For a continuous process, mass balance can be based on mass flow rate, in this case, Differential mass balance can be applied conveniently. However, for a batch, semi-batch or fed-batch proces

23、s, mass flow rate is varying with time, mass balance can only be carried within 22 a period of time, usually from the beginning to the end of an operation, in this case, Integral mass balance can be applied. Table 2.1 Application of the simplified mass balance Eq.(2.3) At steady state, does mass in

24、= mass out? Material Without reactionWith reaction Total massyesyes Total number of molesyesno Mass of a molecular speciesyesno Number of moles of a molecular speciesyesno Mass of an atomic speciesyesyes Number of moles of an atomic speciesyesyes 23 2.3 Procedures for Mass Balance v Draw a clear pro

25、cess flow diagram showing all relative information. A simple box diagram showing all streams entering or leaving the system allows information about a process to be organized and summarized in a convenient way; v Select a set of units and state it clearly; v Select a basis for the calculation and st

26、ate it clearly; v State all assumptions applied to the problem; v Identify which components of the system, if any, are involved in reactions. 24 Example 2.1: Humid air enriched with oxygen is prepared for gluconic acid fermentation. The air is prepared in a special humidifying chamber. 1.5 l h1 wate

27、r enters the chamber at the same time as dry air and 15 mol min1 dry oxygen gas. All the water is evaporated. The outflowing gas is found to contain 1% (w/w) water. Draw and label the flow sheet for this process. Solution: Let us chose units of g and min for this process; the information provided is

28、 first converted to mass flow rates. The density of water is 103 g l1; therefore: The liquid water entering is 1.5 l h1 = 25 g min1 The oxygen entering is 15 mol min1 = 480 g min1 25 Humidifier Liquid water 25 g min -1 480 g min Pure oxygen -1 Dry air D g min-1 Humid, oxygen-rich air H g min -1 1% w

29、ater ? ? O2% ? 26 2.2 Material-Balance Examples Acetobacter aceti can convert ethanol to acetic acid under aerobic conditions. A continuous fermentation process for vinegar production is proposed using non-viable A. aceti cells immobilized onto the surface of gel beads. The production target is 2 kg

30、 h1 acetic acid. However the maximum acetic acid concentration tolerated by the cells is 12%. Air is sparged into the fermenter at a rate of 200 mol h1. (a) What minimum amount of ethanol is required? (b) What minimum amount of water must be used to dilute the ethanol to avoid acid inhibition? (c) W

31、hat is the composition of the fermentation off-gas? 27 Solution: 1. Assemble 1) Flow sheet The flow sheet for this process is shown as follows Fermenter Feed stream E kg h ethanol -1 Inlet air -1 W kg h water 5.768 kg h-1 1.344 kg h O -1 4.424 kg h N -1 2 2 Off-gas -1 G kg h Product stream -1 P kg h

32、 2 kg h or 12% acetic acid -1 28 2) System boundary The system boundary is shown in Figure above. 3) Reaction equation In the absence of cell growth, maintenance or other metabolism of substrate, the reaction equation is: C2H5OH + O2 CH3COOH + H2O 2. Analyze 1) Assumption steady state no leaks inlet

33、 air is dry gas volume % = mole % 29 no evaporation lose of ethanol, H2O or acetic acid complete conversion of ethanol ethanol is used by the cells for synthesis of acetic acid only; no side-reactions occur oxygen transfer is sufficiently rapid to meet the demand of the reaction consumption concentr

34、ation of acetic acid in the effluent is 12% 2) Extra data Molecular weights: ethanol = 46 acetic acid = 60 30 O2 = 32 N2 = 28 H2O = 18 Composition of air: 21% O2, 79% N2 3) Basis The calculation is based on 2 kg acetic acid per hour. 4) Compounds involved in reaction The compounds involved in reacti

35、on are ethanol, acetic acid, O2 and H2O. 5) Mass-balance equations. For ethanol, acetic acid, O2 and H2O, the appropriate mass-balance equation is Eq.(2.2): mass in + mass generated = mass out + mass consumed 31 For total mass and N2, the appropriate mass-balance equation is Eq.(2.3): mass in = mass

36、 out 3. Calculate 32 2 kg or 3.333 10-2 kgmol HAc generated must consume the same moles of Ethanol (1.533 kg) and Oxygen (1.067 kg), at the same time, produce the same moles of Water (0.600 kg). 33 4. Finalize The off-gas contains 0.277 kg O2 and 4.424 kg N2. Since gas compositions are normally expr

37、essed using volume or mole %, we should convert these values to moles: O2 content = 0.277/32 = 8.656 103 kgmol N2 content = 4.424/28 = 0.158 kgmol Therefore, the total molar quantity of off-gas is 0.167 kgmol. The off-gas composition is: 8.656 103/ 0.167 100 = 5.19% O2 0.158 / 0.167 100 = 94.8% N2 3

38、4 Answers: (a) 1.5 kg h1 ethanol is required. (b) 14.1 kg h1 water must be used to dilute the ethanol in the feed stream. (c) The composition of the fermenter off-gas is 5.2% O2 and 94.8% N2 35 2.5 Material Balance with Recycle, By-pass and Purge Streams Process Feed streamEffluent Recycle stream Fe

39、ed stream Process Effluent By-pass stream Recycle stream Feed stream Process Effluent Purge stream (a) (b) (c) Fig. 2.4 Flow sheet for processes with (a) recycle, (b) by-pass and (c) purge streams 36 Fig. 2.5 Fermentation with cell recycling Fermenter MixerFeed stream Effluent Product stream Settlin

40、g tank Cell concentrate (recycle stream) 37 Fig. 2.6 System boundaries for cell-recycle system Fermenter MixerFeed stream Product stream Cell concentrate (recycle stream) Settler A B C D 38 2.6 Bioreaction Stoichiometry 2.6.1 Growth Stoichiometry and Elemental Balance CwHxOyNz + aO2 + bHgOhNi cCH O

41、N + dCO2 + eH2O (2.4) Fig 2.7 Conservation of substrate, oxygen and nitrogen for cell growth Substrate C H O N xyzw aO2 bH O Ng hi eH O cCH O N Biomass dCO2 2 Cell 39 Table 2.2 Elemental composition of Escherichia coli* Element% dry weightElement% dry weight C50Na1 O20Ca0.5 N14Mg0.5 H8Cl0.5 P3Fe0.2

42、S1All others0.3 K1 * From R. Y. Stanier, E. A. Adelberg and J. Ingraham, 1976, The Microbial World, 4th edn, Prentice-Hall, New Jersry. 40 Material Balances: C balance: w = c + d (2.5) H balance: x + bg = c + 2e (2.6) O balance: y + 2a + bh = c + 2d + e (2.7) N balance: z + bi = c (2.8) RQ, respirat

43、ory quotient, is defined by moles CO2 produced/moles O2 consumed (2.9) 41 2.6.2 Electron Balance Available electrons refers to the number of electrons available for transfer to oxygen on combustion of a substrate to CO2, H2O and other small molecular weight substances, usually nitrogen- containing c

44、ompounds. The number of available electrons found in organic compounds is calculated from the valence of various elements: 4 for C, 1 for H, 2 for O, 5 for P, 6 for S, and so on. The number of available electrons for N depends on the reference state: 3 if ammonia is the reference, 0 for molecular ni

45、trogen N2, and 5 for nitrate. The reference state for cell growth is usually chosen to be the same as the nitrogen source in the medium. 42 Degree of reduction, , is defined as the equivalent numbers of available electrons in that quantity of material containing 1mol atom carbon. Therefore, for subs

46、trate CwHxOyNz, the number of available electrons is (4w + x 2y 3z). Thus, the degree of reduction, S, is (4w + x 2y 3z)/w. Electrons available for transfer to oxygen are conserved during metabolism. CwHxOyNz + aO2 + bHgOhNi cCH O N + dCO2 + eH2O wS 4a = cB (2.10) 43 Quiz: Please calculate the degre

47、e of reduction, , for the compounds: 1. Ethanol C2H6O 2. Glycerol C3H8O3 3. Acetic acid C2H4O2 4. Pyruvic acid C3H4O3 5. Succinic acid C4H6O4 6. Alanine C3H7O2N 7. Lysine C6H14O2N2 The references for N are designated as NH3 and N2, respectively. 44 2.6.3 Biomass Yield YB/S = (2.11) YB/S = (2.12) con

48、sumedsubstrateg produced cells g te)(MWsubstra cells)c(MW 2.6.4 Product Stoichiometry CwHxOyNz + aO2 + bHgOhNI cCHON + dCO2 + eH2O + f CjHk (2.13) YP/S = (2.14) substrate) ( product) ( consumed substrate g formedproduct g MW MWf 45 2.6.5 Theoretical Oxygen Demand wS 4a = cB + fjP (2.15) a = 1/4(wS c

49、B fjP) (2.16) 2.6.6 Maximal Yield 1 = (2.17) Let us define B as : B = (2.18) cmax = (2.19) S P S B S w fj w c w a 4 S B w c B S w 46 Questions: Predict theoretically maximum biomass (CH1.8O0.5N0.2, the reference for N is ammonia) yield coefficients based on the substrates of 1. Glycerol C3H8O3 2. Et

50、hanol C2H6O 3. Methane CH4 Similarly, the maximum possible product yield in the absence of biomass synthesis can be determined by YP/S, max = (2.20) P S j w 47 Examples: The chemical reaction equation for aerobic respiration of glucose is: C6H12O6 + 6O2 6CO2 + 6H2O Candida utilis cells convert gluco