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1 Welcome to Industrial Bioprocessing and Bioremediation ! (Environmental Biotechnology) Example processes: How to: make renewable biogas from organic.

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Presentation on theme: "1 Welcome to Industrial Bioprocessing and Bioremediation ! (Environmental Biotechnology) Example processes: How to: make renewable biogas from organic."— Presentation transcript:

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2 1 Welcome to Industrial Bioprocessing and Bioremediation ! (Environmental Biotechnology) Example processes: How to: make renewable biogas from organic wastes. remove polluting nutrients from wastewater breed microalgae for food or energy production make beer, yoghurt, mine ores by using bacteria (bioleaching) understand microbial processes in ocean, soil and bioreactors

3 2 Fundamentals taught Bioprocesses convert S to P How much S to P? Mass balance How much hydrogen gas can we make as fuel from fermenting sugars ? How much oxygen is needed for respiration? How much electricity can be formed from sugar ?

4 3 Fundamentals taught Bioprocesses convert S to P Sugar to energy (biogas, ethanol, hydrogen, electricity) Pollutants to harmless substances (dechlorination, degradation, dirty water to clean water Need to be able to predict (modelling) Quantify (rates) Understanding (driving force, equilibrium)

5 4 Fundamentals taught Bioprocesses questions: Why ? How ? How fast? What mechanism ? Modelling

6 5 Learning tools used Bioprocess analysis (theory) Computer simulation Bioprocess execution and analysis (Chemostat project) Spreadsheets for data processing and analysis Peer reviewed websites of scientific analysis of literature Using computer learning activities Industry trip Focussed scientific writing.

7 6 Link between Research and Teaching A number of undergraduates  honours  PhD  Senior Engineers (Water Corporation, Design companies, Bioprocess Operators) Teaching topics drawn from own research projects and publications. Input from current researchers into the project Link to industry. So, on many of the topics you talk to science experts, not just “teachters that obtained their knowledge from books”

8 7 Bioprocessing – Biotechnology: Make money from bioprocesses Inputs are of lower value than outputs (products) Computer based learning activities (CBLA) are on http://sphinx.murdoch.edu.au/ units/extern/BIO301/teach/index.htm

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15 Lecture overview L1-3 Lecture 1: Intro, study guide, what is a bioreactor, Learning by interacting Lecture 2: What is diffusion, how can we predict the behaviour of a randomly moving molecule? moving dots, entropy, driving force, equilibrium, rate of diffusion, first order kinetics, kLa Lecture 3: oxygen transfer rate, kLa value. Graphical method of determining the kLa. Mathematical (2 point) determination of kLA Calculation and prediction of oxygen transfer as function of DO. Oxygen transfer efficiency. Bacterial OUR, DO. steady state 14

16 Molecular diffusion relies on random movement resulting in uniform distribution of molecules 15

17 Oxygen Transfer Rate (OTR) Overview Diffusion, how does it work, how can we predict it? Diffusion is random …. and yet predictable. A simple model simulation can show that although the diffusion movement is random, it can be precisely predicted for large number of molecules (e.g. Fick’s law of diffusion) 16

18 Oxygen Transfer Rate (OTR) (diffusion, convection) Low OTR High OTR Transfer by diffusion is extremely slow and depends on surface area Wind Oxygen transfer by convection (turbulences) is more efficient Air In Bioreactors combine maximum convection with maximum diffusion Course bubbles cause more convection, fine bubbles more diffusion How soluble is oxygen? 17

19 The net transfer of oxygen from gas phase to solution reaches a dynamic equilibrium O2 input = O2 output equilibrium results in defined saturation concentration (cs). The saturation concentration is also the oxygen solubility How soluble is oxygen? Oxygen solubility (cS) 18

20 Oxygen is not very polar  poorly soluble in water. Oxygen Solubility is described by Henry’s Law which applies to all gases p = k*cS p = partial pressure of oxygen k = constant depending on gas type, solution and temperature cS = concentration of oxygen dissolved in water Meaning: The amount of oxygen which dissolves in water is proportional to the amount of oxygen molecules present per volume of the gas phase. Partial pressure ~ number of O2 molecules per volume of gas increases with O2 concentration in gas increases with total gas pressure How to calculate partial pressure? (refer to CBLA) Oxygen solubility (cS) 19

21 Examples of using the proportionality between partial pressure of oxygen in the atmosphere and the saturation concentration cS: p = k*cS p = partial pressure of oxygen k = constant depending on gas type, solution and temperature cS = concentration of oxygen dissolved in water If the reactor is operated under 2 times atmospheric pressure (200kPa instead of 100 kPa air pressure), the new saturation concentration will be abou 16 mg/L instead of 8 mg/L. If air (partial pressure = 0.21* 100 kPa) is replaced by pure oxygen atmosphere (partial pressure 100kPa) the oxygen saturation concentration is about 40 mg/L (more precise 8*100/21) instead of 8 mg/L. How to calculate partial pressure? (refer to CBLA) Oxygen solubility (cS) 20

22 Effect of temperature Oxygen solubility decreases with increasing temperature. Overall: oxygen is poorly soluble (8mg/: at room temp.) More important than solubility is oxygen supply rate (oxygen transfer rate OTR). c s = 468 (31.6 + T) Oxygen Saturation Concentration c s (mg/L) Temperature (°C) Oxygen solubility (cS) 21

23 Oxygen Transfer Rate (OTR) (gradient, driving force) Question: What is the driving force for oxygen dissolution? OTR At oxygen saturation concentration (c s ): dynamic equilibrium exists between oxygen transferred from the air to water and vice versa.  No driving force Answer: The difference between cS and the actual dissolved oxygen concentration (cL) is the driving force. OTR is proportional to the that difference. Thus: OTR ~ (cS – cL) Need to determine the proportionality factor 22

24 1. Deoxygenation (N 2, sulfite + Co catalyst) 2. Aeration and monitoring dissolved oxygen concentration (D.O. or c L ) as function of time 3. OTR = slope of the aeration curve (mg/L.h or ppm/h) Significance of OTR: critical to know and to control for all aerobic bioreactors 0510 8 Air On cL (ppm) Time (min) OTR – depends on DO (cL) 23

25 OTR = k L a (c s - c L ) Mg/L/h h -1 mg/L 4. Observation: OTR decreases over time (and with incr. cL) 5. OTR is not a good measure of aeration capacity of a bioreactor 6. OTR is highest at cL = zero (Standard OTR) 7. OTR is zero at oxygen saturation concentrations (cs) 8. OTR is negatively correlated to cL 9. OTR is correlated to the saturation deficit (cs - cL), which is the driving force for oxygen transfer 9. The factor of correlation is the volumetric mass transfer coefficient kLa OTR – depends on DO (cL) 24

26 First: steep step in oxygen (top layer saturated, next layer oxygen free) Then: buildup of a gradient of many layers. Each layer is only slightly different from the next  Transfer from layer to layer has little driving force.  Gradient build-up inhibits fast diffusion OTR –Significance of gradient 25

27 10. OTR is not a useful parameter for the assessment of the aeration capacity of a bioreactor. This is because it is dependent on the oxygen concentration (c L ) 11. The k L a value is a suitable parameter as it divides OTR by saturation deficit: 12. k L a = the key parameter oxygen transfer capacity. How to determine it? OTR (c s - c L ) k L a = 26

28 Lec 2 summary: Oxygen is poorly soluble depending mainly on partial pressure in headspace Temperature OTR is driven and proportional to driving force (cS-cL) kLa is the proportionality factor (first order kinetics) kLa describes the performance of a bioreactor to provide Oxygen to microbes Next lecture: quantify kLA 27

29 Lec 3 outlook: Aeration curve Quantify OTR at a given point of an aeration curve Quick estimate of kLa Graphical determination of kLa Mathematical determination of kLa Run computer simulation to obtain data Oxygen transfer efficiency (OTE) OTR proportional to cs-cL OTR inverse proportional to cL 28

30 OTR – Quick estimate of kLA Example: determine OTR at 6 mg/L OTR is the slope of the tangent for each oxygen concentration OTR = ∆ cL/ ∆ t = 5 mg/L/ 4.5 min = 1.1 mg/L/min = 66 mg/L/h 0510 8 Air On cL (mg/L) Time (min) 6 4.5 min 5 mg/L 29

31 k L a = OTR = 66 ppm / h (c s -c L ) = 3.3 h -1 (8 ppm – 6 ppm) Q: Problem with this method? A: based on one single OTR slope measurement and unreliable to obtain from real data. OTR – quick estimate of kLA Time DO 30

32 1. Monitor aeration curve 2. Determine graphically the OTR at various oxygen concentrations (c L ) 8 2 4 6 0 510 3. Tabulate OTR and corresponding c L values cL (ppm) 0.5 3.0 4.0 6.0 8.0 Time (min) 7.5 5.0 4.0 2.0 0.0 30 60 50 25 0 c L (mg/L)C s - c L (mg/L)OTR (mg/L/h) At 6 ppm: OTR = 25 mg/L/h At 4 ppm: OTR = 50 mg/L/h At 3 ppm: OTR = 60 mg/L/h At 0.5 ppm: OTR = 30 ppm/h OTR – Graphical determination of kLa 31

33 0 50 100 02468 4. Plot OTR values as a function of c s - c L. OTR (mg/L/h) Standard OTR cscs c s - c L (mg/L) 5. A linear correlation exists between k L a and the saturation deficit (c s - c L ) which is the driving force of the reaction. 6. The slope of the plot OTR versus c s - c L is the k L a value. 7. The standard OTR (max OTR) can be read from the intercept with the c s line. (Standard OTR = 100 ppm/h) kLa = = 12 h-1 70 mg/L/h 6 mg/L OTR – Graphical determination of kLa 32

34 Mathematical Determination of k L a 1. OTR is a change of c L over time, thus = dc L /dt Integration gives 2. k L a = dc L /dt (c s - c L ) () c s - c o 3. k L a = ln c s - c i t i - t o () 8 - 3 ppm k L a = ln 8 - 6 ppm 10.5 - 6.1 min = 0.21 min -1 = 12.5 h -1 = ln 2.5 4.4 min Dissolved Oxygen Concentration (mg/L) c i = 6 c o = 3 toto titi Time (min) cscs 33

35 4. This method should be carried out for 3 to 4 different intervals. By aver 5. Once the k L a is known it allows to calculate the OTR at any given oxygen concentration: OTR = k L a (c s - c L ) 34

36 Factors Affecting the Oxygen Transfer Coefficient k L a k L a consists of: k L = resistance or thickness of boundary film a = surface area Bubble Bulk Liquid Cell [Oxygen] Distance Main boundary layer = steepest gradient → rate controlling, driving force 35

37 Effect of Fluid Composition on OTR The transfer across this boundary layer increases with: 1)↓ thickness of the film, thus ↑ degree of shearing (turbulence) 2)↑ surface area 3)↓ surface tension 4)↓ viscosity (best in pure water) 5)↓ salinity 6)↓ concentration of chemicals or particles 7)detergents? 8)↑ emulsifiers, oils, “oxygen vectors” 36

38 Oxygen Transfer Efficiency (OTE) OTE = oxygen transferred oxygen supplied Significance of OTE: economical, evaporation Calculation of OTE (%): % OTE = oxygen transferred (mol/L.h) oxygen supplied (mol/L.h) X 100 Why do students find this type calculation difficult? Units are disregarded. Molecular weights are misused. 37

39 Oxygen Transfer Efficiency (OTE) A bioreactor ( 3 m 3 ) is aerated with 200 L/min airflow. If the OTR is constant (100 mg/L/h) determine the %OTE. 1. Convert the airflow into an oxygen flow in mmol/L/h 200 L air /min = 12000 L air/h = 2520 L O 2 /h = 102.9 mol O 2 /h = 34.3 mmol O 2 /L.h (x 21%) (÷ 24.5 L/mol) (÷ 3000 L) 2. OTR (x 60) 100 mg/L.h = 3.1 mmol O 2 /L.h (÷ 32 g/mol) % OTE = 3.1 (mmol/L.h) 34.3 (mol/L.h) X 100 = 9% 38

40 Oxygen Transfer Efficiency (OTE) OTE is dependent upon the cL in the same way than OTR OTE decreases with increasing airflow (more oxygen is wasted) % OTE 5 10 Airflow 39

41 Engineering Parameters Influencing OTR Increase depth vessel Decrease bubble size Increase air flow rate Increase stirring rate Deeper vessel  bubbles rise a long way  ↑ OTR, OTE but more pressure required  ↑ $$  Larger surface area  ↑ OTR, OTE smaller bubbles rise slower  more gas hold up  ↑ OTR, OTE  ↑ Number of bubbles  ↑ OTR but ↓ OTE  ↑ turbulence  ↓ thickness of boundary layer  ↑ OTR, OTE  ↓ Bubble size  ↑ OTR, OTE 40

42 Rate is proportional to concentration  First order kinetics Slope = k L a max OTR OTR (mg/L.h) Dissolved oxygen [mg/L] OTR = k L a (O 2 saturation (cS) – O 2 concentration (cL)) (first order kinetics) Aeration Curve Time Dissolved Oxygen Air on (c s ) OTR – from aeration curve to kLa summary 41

43 During aeration of oxygen free water, the dissolved oxygen increases in a characteristic way OTR – Aeration curve from CBLA 42

44 Can the relationship between rate and DO be expressed mathematically? Highest Rate at lowest dissolved oxygen concentration Rate of zero when DO reaches saturation concentration OTR – aeration curve from CBLA 43


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