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Suppose we have a perfect sphere of some solute such as sugar and we place it in a fluid such as water, at a certain temperature, that is not moving. It will dissolve and diffuse into the water due to liquid-solid extraction at at certain rate proportional to the diffusion coefficient, $D$, of sugar in water at that temperature.

Now say the water velocity is $u = (5, 0 , 0)$ so the water is "hitting" the sugar from the negative x-axis and flowing around it. It seems intuitively that the sugar will dissolve faster on the left hand side of the sphere due to the momentum of the water on the left hand side causing it to break up faster and in doing so increasing the exposed soluble surface and increase the dissolution rate.

So it seems that instead of the diffusion coefficient $D$ at a point $(x, y, z)$ being constant it will be $D_{total} = D_{standard} + D_{extra}(u, \mu, \hat{n})$ where $D_{standard}$ is the regular diffusion coefficient above and $D_{extra}$ is a term that account for the increase in dissolution rate due to the fluid movement. $u$ is the velocity of water, $\mu$ is the viscosity of water and $\hat{n}$ is the normal of the surface at that point.

It seems taking the dot product of the velocity with the surface normal will tell us how much dissolution is affected due to fluid movement. E.g. if the normal is perpendicular to the velocity, which would be the case at the top or bottom of the sphere of sugar, $-(u \cdot \hat{n}) = 0$ so we would have no extra dissolution there (discounting shear forces). Whereas if we were talking about the point on the left hand side of the sugar with $\hat{n} = (-1, 0, 0)$ we would have $-(u \cdot \hat{n}) = 5$ as factor how much to increase the dissolution rate.

Am I thinking along the correct lines here? Is there some law I can use to determine how the fluid velocity field affects the dissolution of a substance or do I have to come up with some intuitive method along the lines of the above?

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    $\begingroup$ A flow does not act like a modification of the diffusion constant, but as a drift term $\propto \vec v \cdot \nabla n$. Note that the water velocity $u = (5,0,0)$ cannot be realized close to the sphere (due to the boundary condition enforced by the sphere). So this gets very difficult to model correctly – you have to consider the exact flow around the sphere in the changing boundary conditions (due to the solution of the sphere) and then solve the transport equation for the solute in this flow field (probably assuming a saturated solution as boundary condition on the surface of the sphere). $\endgroup$ – Sebastian Riese Jul 20 '15 at 12:17
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    $\begingroup$ Probably there is some heuristic one can use to get approximate solutions, but the correct modelling certainly required solution by computer. I guess even under terrible simplifications, such as assuming spherical shape and a steady state transport at all times, it will be difficult to get a nice formula for the dissolution rate. Additionally the solution process itself probably has a time-scale on which local equilibrium in the boundary layer is reached (I honestly do not know on which time scale, it might even be this dominates the solution time compared to transport effects). $\endgroup$ – Sebastian Riese Jul 20 '15 at 12:23
  • $\begingroup$ Yes there is a saturation layer around the sphere over which extraction occurs due to the concentration gradient. It will be solved numerically...surprised there isn't some law for this..I would definitely have thought the momentum of a fluid would impact how the sphere dissolves at different locations... $\endgroup$ – sonicboom Jul 20 '15 at 17:20
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    $\begingroup$ As I have said, I am pretty sure the dependence is only via the modification of the transport (and as the dissolved material is transported away faster, the sphere dissolves faster). As the modification of transport is asymmetrical the speed of solution at different locations will be different. It might well be possible there is some simple approximate law, but I doubt there is a simple (or any) derivation for it. $\endgroup$ – Sebastian Riese Jul 20 '15 at 22:19
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    $\begingroup$ Certainly, at very high speeds there will be ablation effects due to the impact of the fluid, but at "normal" velocities (lets say with laminar or slightly turbulent flow) these effects will most certainly be negligible. $\endgroup$ – Sebastian Riese Jul 21 '15 at 22:33

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