生物反应工程教学课件-Bioprocess Engineering Lesson-5.pptVIP

生物反应工程教学课件-Bioprocess Engineering Lesson-5.ppt

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* Ni = 1.42 s?1 = 85.5 rpm Flow is just turbulent with (Re)i = 4 ? 103. This analysis indicates that shear damage from turbulent eddies is not expected until the stirrer speed exceeds about 85 rpm. If the culture were sparged with gas, it is possible that shear damage would happen due to other mechanisms, e.g. bursting bubbles. If the viscosity of the liquid is increased, the size of the smallest eddies also increases. Increasing the fluid viscosity should, therefore, reduce shear damage in bioreactors. This effect has been demonstrated by addition of thickening agents to animal-cell growth medium; moderate increase in viscosity have been shown to significantly reduce turbulent cell death. * 6.9.2 Bubble Shear When liquid containing shear-sensitive cells is sparged with air, other damaging mechanisms come into play. From experiments conducted so far, these appear to be associated primarily with bubbles bursting at the surface of the liquid, breakage of the thin bubble film and rapid flow from the bubble rim back into the liquid generate high shear forces capable of damaging certain types of cell. * Summary After the study of this chapter, you should be: familiar with equipment used for mixing in stirred tanks; able to describe the mechanisms of mixing and their effect on mixing time; able to understand the effects of scale-up on mixing; able to know how liquid properties, gas sparging, impeller size and stirrer speed affect power consumption in stirred vessels; and able to understand how cells can be damaged by shear in stirred fermenters. * Example 6.2 Calculation of power requirements A fermentation broth with viscosity 10?2 Pa s and density 1000 kg m?3 is agitated in a 50 m3 baffled tank using a marine propeller 1.3 m in diameter. The tank geometry is as specified in Figure 6.13. Calculate the power required for a stirred speed of 4 s?1. Solution: From Eq.(6.4): From Figure 6.13, flow at this (Re)i is fully turbulent and Np’ = 0.35

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