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Oxygen Transfer in Bioreactors

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Submitted By rohan1967
Words 2606
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Oxygen transfer
Abstract (193 words)
This practical was carried out with the aim of determining the KLa value for oxygen transfer as well as examining the relationship between KLa and other fermenter variables like speed of the impeller and air flow rate, thereby calculating the values α and β in the KLa correlation:
KLa = K[Pg/V]α (Vs)β
KLa is the volumetric liquid phase mass transfer coefficient indicative of the mass transfer of oxygen dissolved in the liquid to the cell. It is calculated using the dynamic method which is usually used for vessels which are less than 1m in height because there is nitrogen gas hold-up in the vessel when air is reintroduced and the measurement of concentration of oxygen in the liquid does not reflect the kinetics of simple oxygen transfer until a hold-up of air in established.
The measured parameters included gassed power (Pg), impeller speed, flow rate (indirectly superficial gas velocity) and DOT%. It was seen that as the power input was increased, the KLa increased for the same flow rate and that this increase was greater than increase in the flow rate of the gas, which shows confirms the results described in the literature.

Introduction (326 words)
Cells in aerobic cultures require oxygen for metabolism and growth. The rate of oxygen transfer from aerated liquid to the cell is especially important at high cell densities, when cell growth is likely to be limited by the availability of oxygen in the medium. The solubility of oxygen in aqueous solutions at ambient temperature and pressure is only about 10 ppm, (Doran, 1998) which is quickly consumed by the cell and necessitates constant replacement by sparging. The concentration gradient (cAL*-cAL) is essentially small, so ensuring effective mass transfer becomes difficult. Therefore, it is absolutely vital that these factors are accounted for in fermenter design.
In aerobic fermentation, oxygen molecules have to overcome a series of resistances to mass transfer before they reach the cell. There are 8 steps involved before oxygen is finally utilized by the cell: 1) Transfer from the interior of the bubble to the gas liquid interface 2) Movement across the gas liquid interface 3) Diffusion through the liquid film around the bubble 4) Transport through the bulk liquid 5) Diffusion through the liquid film surrounding the cells 6) Movement across liquid cell interface 7) If the cells are in a floc, then diffusion through the solid to the individual cell 8) Transport through the cytoplasm to the site of reaction
The rate limiting step is usually the thin liquid film which surrounds each bubble.
The measurement of dissolved oxygen (DO) is done using a polarographic electrode. Its basic principle is that it measures the DO partial pressure and expresses it as a percentage of the saturated equilibrium concentration of oxygen in the liquid. This means that the amount of oxygen available in the liquid can be different at the same value of %DOT since the dissolved oxygen tension is expressed as:
DOT(%)= cL02/c*L02
KLa for each trial is calculated by plotting ln[(c*-cL⁰)/(c*-cL)] vs time for 5-6 time points. A best fit line is plotted for each plot and the slope gives KLa.

Results and calculations
Trial 1
Airflow rate (L/min): 1.2
Stir speed (rpm): 500
Power (%) guide: 35.4 | | | X | Y | TIME (s) | CHART READING | Cn (%) | tn - t1 (s) | ln((C*-C1)/(C*-Cn)) | 0 | 10 | 15.4 | 0 | 0 | 20 | 23.5 | 36.2 | 20 | 0.282 | 60 | 41.5 | 63.8 | 60 | 0.85 | 110 | 52 | 80 | 110 | 1.442 | 160 | 56.5 | 86.9 | 160 | 1.867 |

The slope of the line gives us KLa which is .011 from the graph.

Trial 2
Airflow rate (L/min): 1.2
Stir speed (rpm): 750
Power (%) guide: 54

| | | X | Y | TIME (s) | CHART READING | Cn (%) | tn - t1 (s) | ln((C*-C1)/(C*-Cn)) | 0 | 5 | 7.692308 | 0 | 0 | 10 | 19 | 29.23077 | 10 | 0.265728165 | 30 | 39 | 60 | 30 | 0.836273024 | 60 | 53 | 81.53846 | 60 | 1.609462912 | 110 | 58.5 | 90 | 110 | 2.222567385 |

The slope of the line gives us KLa which is .020 from the graph.

Trial 3
Airflow rate (L/min): 1.2
Stir speed (rpm): 1000
Power (%) guide: 74 | | | X | Y | TIME (s) | CHART READING | Cn (%) | tn - t1 (s) | ln((C*-C1)/(C*-Cn)) | 0 | 11 | 16.92308 | 0 | 0 | 10 | 28.5 | 43.84615 | 10 | 0.391708822 | 20 | 41 | 63.07692 | 20 | 0.810967253 | 40 | 53 | 81.53846 | 40 | 1.504114433 | 60 | 57 | 87.69231 | 60 | 1.909579541 |

The slope of the line gives us KLa which is .032 from the graph.

Trial 4
Airflow rate (L/min): 1.8
Stir speed (rpm): 500
Power (%) guide: 34.5 Column1 | Column2 | Column3 | X | Y | TIME (s) | CHART READING | Cn (%) | tn - t1 (s) | ln((C*-C1)/(C*-Cn)) | 0 | 9 | 13.8461538 | 0 | 0 | 20 | 24 | 36.9230769 | 20 | 0.311779669 | 50 | 39.5 | 60.7692308 | 50 | 0.786673283 | 90 | 50.5 | 77.6923077 | 90 | 1.351203086 | 160 | 58 | 89.2307692 | 160 | 2.079441586 |

The slope of the line gives us KLa which is .013 from the graph.

Trial 5
Airflow rate (L/min): 1.8
Stir speed (rpm): 750
Power (%) guide: 54.5 Column1 | Column2 | Column3 | X | Y | TIME (s) | CHART READING | Cn (%) | tn - t1 (s) | ln((C*-C1)/(C*-Cn)) | 0 | 12 | 18.4615385 | 0 | 0 | 10 | 26.5 | 40.7692308 | 10 | 0.31965254 | 20 | 38 | 58.4615385 | 20 | 0.674473915 | 40 | 51.5 | 79.2307692 | 40 | 1.367621096 | 70 | 58 | 89.2307692 | 70 | 2.024400632 |

The slope of the line gives us KLa which is .029 from the graph.

Trial 6
Airflow rate (L/min): 1.8
Stir speed (rpm): 1000
Power (%) guide: 74 Column1 | Column2 | Column3 | X | Y | TIME (s) | CHART READING | Cn (%) | tn - t1 (s) | ln((C*-C1)/(C*-Cn)) | 0 | 11 | 16.9230769 | 0 | 0 | 10 | 31 | 47.6923077 | 10 | 0.462660558 | 20 | 44 | 67.6923077 | 20 | 0.944498645 | 40 | 55.5 | 85.3846154 | 40 | 1.737729284 | 60 | 59 | 90.7692308 | 60 | 2.197261614 |

The slope of the line gives us KLa which is .037 from the graph.

Trial 7
Airflow rate (L/min): 2.4
Stir speed (rpm): 500
Power (%) guide: 33.4 Column1 | Column2 | Column3 | X | Y | TIME (s) | CHART READING | Cn (%) | tn - t1 (s) | ln((C*-C1)/(C*-Cn)) | 0 | 11 | 16.9230769 | 0 | 0 | 10 | 19 | 29.2307692 | 10 | 0.160379686 | 40 | 39.5 | 60.7692308 | 40 | 0.750342631 | 80 | 51.5 | 79.2307692 | 80 | 1.386331397 | 140 | 58 | 89.2307692 | 140 | 2.043110934 |

The slope of the line gives us KLa which is .014 from the graph.

Trial 8
Airflow rate (L/min): 2.4
Stir speed (rpm): 750
Power (%) guide: 53.1 Column1 | Column2 | Column3 | X | Y | TIME (s) | CHART READING | Cn (%) | tn - t1 (s) | ln((C*-C1)/(C*-Cn)) | 0 | 13 | 20 | 0 | 0 | 10 | 29 | 44.6153846 | 10 | 0.36772478 | 30 | 48.5 | 74.6153846 | 30 | 1.147883338 | 50 | 56 | 86.1538462 | 50 | 1.754019141 | 80 | 59 | 90.7692308 | 80 | 2.159484249 |

The slope of the line gives us KLa which is .027 from the graph.

Trial 9
Airflow rate (L/min): 2.4
Stir speed (rpm): 1000
Power (%) guide: 72.5 Column1 | Column2 | Column3 | X | Y | TIME (s) | CHART READING | Cn (%) | tn - t1 (s) | ln((C*-C1)/(C*-Cn)) | 0 | 6 | 9.23076923 | 0 | 0 | 10 | 26 | 40 | 10 | 0.413984272 | 30 | 51.5 | 79.2307692 | 30 | 1.474856233 | 40 | 56 | 86.1538462 | 40 | 1.880321341 | 60 | 59 | 90.7692308 | 60 | 2.285786449 |

The slope of the line gives us KLa which is .04 from the graph.

Calculation of β Q (Vol. flow rate) | | X | kLa | Y | L/min | M3/s | Vs | Log Vs | 500 rpm | 750 rpm | 1000 rpm | 500 rpm | 750 rpm | 1000 rpm | 1.2 | 2x10^-5 | 9.47e-4 | -3.02 | .011 | .020 | .032 | -1.96 | -1.70 | -1.5 | 1.8 | 3x10^-5 | 1.42e-3 | -2.85 | .013 | .029 | .037 | -1.89 | -1.54 | -1.43 | 2.4 | 4x10^-5 | 1.89e-3 | -2.72 | .014 | .027 | .04 | -1.85 | -1.57 | -1.4 | β1= .488 β2=.458

β3= .337

So β = (β1+ β2+ β3)/3 Therefore, β = .428 |
= .428

Calculation of α | | | | KLa | RPM | RPS | PG | PG/V | 1.2 L/min | 1.8 L/min | 2.4 L/min | 500 | 8.33 | .0182 | 7.58 | .011 | .013 | .014 | 750 | 12.5 | .0614 | 25.6 | .020 | .029 | .027 | 1000 | 16.67 | .146 | 60.7 | .032 | .037 | .04 |

X | Y | Log(PG/V) | 1.2 L/min | 1.8 L/min | 2.4 L/min | .88 | -1.96 | -1.89 | -1.85 | 1.41 | -1.70 | -1.54 | -1.57 | 1.78 | -1.50 | -1.43 | -1.40 |

α1= .509 α2= .521

α3 = .502
So α= (α1+ α2+ α3)/3 Therefore, α= .511 |

Points for discussion (715 words) 1) Discuss the effects of stirrer speed and airflow rate on KLa; which is more important and why? Compare the values for the exponents obtained with literature values. Would you expect the scale of the operation to have an effect on the exponents?
The equation for KLa is:
KLa = K(Pg/V)α(Vs)β
From the data marked from the graphs, α was calculated to be .511 while β was .428. According to Doran (1998), the values for α and β are .7 and .2 respectively for vessel volumes ranging from 2x10-3< V < 4.4 m3. For Newtonian fluids, the exponents have values in the ranges:
.4<α<.95 and .2<β<.75 (Advanced Centre for Biochemical Engineering, 2012)
Since the experimentally determined values of α>β, this suggests that the effect of increasing the superficial gas velocity (Vs) will have less of an effect on the volumetric mass transfer coefficient (KLa) than increasing the gassed power per unit volume (Pg/V). However, since both the exponents are less than 1, this means that increasing either flow rate or power input becomes progressively less efficient and more costly as the inputs increase.
We have calculated α by plotting Log(PG/V) vs Log of KLa and β by plotting Log VS vs Log KLa. When the fermentation process is scaled up, the exponents would not change by much (negligible change) because the ratio of the terms used to calculate α and β will remain more or less the same however much the variables themselves (Pg, Vs, etc.) may change within the vessel volume limits specified in the previous paragraph. 2) Comment on the assumption that air is rapidly dispersed around the fermenter and the validity of this assumption when larger fermenters are used.
Since the vessel itself is quite small ( 2.4L), it is reasonable to assume that the air pumped in is rapidly dispersed with the result that we assume that there are no concentration gradients in the vessel. However, in larger fermenters, this assumption is rendered null and void as there will always be concentration gradients due to the large size of the reactor. Agitation may mitigate this effect somewhat but never completely eliminate them. In addition to this, in a very large vessel there will be a head pressure applied, and as you go down the vessel, the hydrostatic pressure will increase. So the DOT (DO tension) level may not be constant throughout the vessel. 3) Discuss the importance of surface aeration within the bioreactor used. How would you expect this to change as the scale of the operation increases.
The effect of surface aeration is generally small and gets smaller as the scale of the operation increases. The table below demonstrates this trend:

Volume (L) | Pg/V (W/L) | % of gas entrainment due to surface aeration | 51000 | 3 | 1.5 | 3000 | 2 | 5.0 | 550 | 1 | 2.7 | 200 | 3.8 | 66.0 | 10 | 8.5 | 50.0 |
Since the bioreactor used in the experiment is very small, the surface aeration plays a very large part because the water pulled by the pump is taken from right below the surface and therefore offers little benefit in deep water (large vessels). 4) Fermentation often produces foam, which is detrimental. What do you expect the effect of adding antifoams to suppress the foam to be?
Foaming poses operational problems like a route for entry of contaminating organisms and blocking of outlet gas lines. Foam may also destroy fragile cells. Antifoam is used to reduce foam buildup. It does so by lowering the surface tension of the bubbles and their tendency to coalesce. This increases ‘a’ in KLa because a decrease in surface tension causes a decrease in bubble diameter (a=surface area/volume). However, the mobility of the gas-liquid interface is lowered ( KLa decreases) and generally, this outweighs the increase in ‘a’. For this reason, mechanical rather than chemical methods of disrupting foam are preferred.
The overall observation is that KLa values reduce when antifoam is used. 5) The experiment is carried out using water. Comment on any changes you may expect to see if the components of fermentation media (excluding antifoam were added to the system and how each would change KLa.
Electrolytes – Salts will tend to increase oxygen mass transfer. The independent influence of salts on KL and ‘a’ is difficult to explain. Sometimes, a seven fold increase in KLa is observed.
Surfactants – Most surfactants increase KLa. They stabilise small bubbles present (increasing ‘a’) but may also cause interfacial mass transfer problems.
Presence of cells – Cells with complex morphology lead to lower transfer rates. Cells interfere with bubble breakup and coalescence and may also cause interfacial blanketing which reduces the contact area between the gas and the liquid. Because concentration of cells, substrates and products change thoughout batch fermentation, the KLa can also vary.
Effect of solutes – the presence of solutes in the fermentation broth decreases the oxygen solubility because the partial pressure of oxygen is now lowered in relation to the total pressure of the system. Less oxygen solubility translates into less oxygen mass transfer.

Conclusion (211 words)
From the calculated values of α and β, it can be seen that they tally closely with the values suggested in literature. Since the values are largely empirical, it is observed that they mostly lie with the ranges of .4<α<.95 and .2<β<.75 since they are unique to different systems. However, the calculated exponents do follow the observed trend that α>β implying that the change in power input per unit volume has a more significant effect on KLa than does the superficial velocity.
There is usually limited scope for increasing the superficial gas velocity in a reactor. Most importantly, from a safety viewpoint, extremely high velocities can blow the broth out of the reactor. However, in this experiment, it is also important to consider the fact that this as close to an ideal fermentation as one can get; there was perfect mixing (assumed because of the small volume of the reactor), there was hardly any mixing time, there were no concentration gradients, the fluid was not viscous and there were no electrolytes or solutes in the process. Without the consideration of these variables, it is often impossible to accurately model large scale fermentations because empirical data from literature suggests that at high volumes, these parameters play a significant part in oxygen transfer.

Doran, P. (1995) Bioprocess Engineering Principles, Academic Press
Stanbury, P. And Whitaker, A. (2nd Edition, 1995) Principles of fermentation technology, Pergamon Press
Advanced Centre for Biochemical Engineering, Rapid Fermentation Process Design lecture notes (October 2012), UCL…...

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