Stagnation point, boundary layer mass transfer

Weder s experiments were carried out with opposing body forces, and large current oscillations were found as long as the negative thermal densification was smaller than the diffusional densification. [Note that the Grashof numbers in Eq. (41) are based on absolute magnitudes of the density differences.] Local mass-transfer rates oscillated by 50%, and total currents by 4%. When the thermal densification dominated, the stagnation point moved to the other side of the cylinder, while the boundary layer, which separates in purely diffusional free convection, remained attached. 

When the fluid approaches the sphere from above, the fluid initially contacts the sphere at 0 = 0 (i.e., the stagnation point) because polar angle 6 is defined relative to the positive z axis. This is convenient because the mass transfer boundary layer thickness Sc is a function of 6, and 5c = 0 at 0 = 0. In the laminar and creeping flow regimes, the two-dimensional fluid dynamics problem is axisymmetric (i.e., about the z axis) with... 

Figure 11-1 Thickness of the mass transfer boundary layer around a solid sphere, primarily in the creeping flow regime. This graph in polar coordinates illustrates 8c 9) divided by the sphere diameter vs. polar angle 9, and the fluid approaches the solid sphere horizontally from the right. No data are plotted at the stagnation point, where 9=0.

When flow occurs about a sphere the solution to tins forced convection mass transfer problem is quite complex because of the complexity of the flow field. At low flow rates (creqiiiig flow) a laminar boundary layer exists about the sphere which separates from die surface at an at ular porition and moves lowani the forward stagnation point as the flow rate increases. Wake fimnation occurs st the tear of the sphere. At still higher flow rates transition to a turbulent boundary layer occurs. Solutions to the problem of mass transfer during creeping flow about a sphere (Re < 1) have been developed by a nombw of authors with the numerical solutions of Brian and Hales being perhaps the most extensive. Their result is... 

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