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Rate of dissolution


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#21 Sriman Dutta

Sriman Dutta

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Posted 2 January 2017 - 03:43 PM

After some thoughts, I get that there will be no jerk or change in acceleration ( which I presumed).

 

Relating the dV/dt=r with the Noyes-Whitney equation, we get a good thing.

Now we define terminal velocity and then use it to equation. That reduces it to one unknown, other than the coefficients which are experimentally determined.

So we get a single equation with dR as the unknown. Solving for this R by transposition and integration.


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#22 Bender

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Posted 2 January 2017 - 11:33 PM

After some thoughts, I get that there will be no jerk or change in acceleration ( which I presumed).

Why would the acceleration be constant?


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#23 RiceAWay

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Posted 13 January 2017 - 10:21 PM

 

I would have assumed that the dissolution is proportional to the surface area and not very related to drag forces.

This is a lot more complicated problem than simply establishing the speed of "dissolvable solid" through "liquid". Firstly you have to know how rapidly the solid can dissolve into the liquid. With this as a frame of reference not only does the differences in specific gravity of the solid versus the liquid become important but the fact that the weight reduces more rapidly than the diameter of the solid which causes an apparent increase in drag slowing the solid more rapidly than the dissolution would imply. Also dissolution increases with the surface roughness. So if the solid does not dissolve linearly this can cause any number of actions from the change of shape either slowing or greatly increasing the speed of the solid via hydraulic drag.

 

Since there are extremes such as lead falling through water or a small ball of baking soda falling though a column of boiling water each case has to be analyzed separately.


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