Rate of dissolution

[Sriman Dutta]

Hello,

 

Suppose there's a spherical solid of initial radius [math]R[/math]. This solid when dropped in a certain liquid dissolves at the rate [math]r[/math]. By rate of dissolution I mean the loss of the volume of the solid in unit time. By using derivatives, we can represent it as-

[math]r=\frac{dV}{dt}=\frac{4\pi}{3}\frac{dR^3}{dt} [/math]

where [math]dV[/math] is change in volume, which is directly proportional to the cube of the change in radius of the spherical solid represented as [math]dR^3[/math].

Since the solid moves downward it experiences drag. By Stoke's Law, drag is -

[math]F_D=6\pi \mu Rv [/math]

where [math]\mu[/math] is the dynamic viscosity of the liquid, [math]v[/math] is the solid's downward velocity.

But, here downward velocity [math]v[/math] of the solid is a function of the drag force, which is again a function of the radius.

If we apply Newton's Laws,

[math]Force on the solid in the downward direction = Weight of solid - Drag force [/math]

[math]ma=mg-F_D[/math]

where [math]m[/math] is the solid's mass.

But mass also depends upon radius.

All the variables seem inter-related. I want to reduce the whole situation into an equation but cannot proceed beyond this.

Please suggest what and how to proceed.

Edited December 28, 2016 by Sriman Dutta

[studiot]

Hello,

 

Suppose there's a spherical solid of initial radius [math]R[/math]. This solid when dropped in a certain liquid dissolves at the rate [math]r[/math]. By rate of dissolution I mean the loss of the volume of the solid in unit time. By using derivatives, we can represent it as-

[math]r=\frac{dV}{dt}=\frac{4\pi}{3}\frac{dR^2}{dt} [/math]

where [math]dV[/math] is change in volume, which is directly proportional to the square of the change in radius of the spherical solid represented as [math]dR^2[/math].

Since the solid moves downward it experiences drag. By Stoke's Law, drag is -

[math]F_D=6\pi \mu Rv [/math]

where [math]\mu[/math] is the dynamic viscosity of the liquid, [math]v[/math] is the solid's downward velocity.

But, here downward velocity [math]v[/math] of the solid is a function of the drag force, which is again a function of the radius.

If we apply Newton's Laws,

[math]Force on the solid in the downward direction = Weight of solid - Drag force [/math]

[math]ma=mg-F_D[/math]

where [math]m[/math] is the solid's mass.

But mass also depends upon radius.

All the variables seem inter-related. I want to reduce the whole situation into an equation but cannot proceed beyond this.

Please suggest what and how to proceed.

 

 

 

But mass also depends upon radius.

 

Yes mass is a better quantity than volume to work with, both for the mechanics of viscous flow and the chemistry of reaction rate kinetics.

 

Each of this will lead to a differential equation.

 

These differential equations will be coupled or simultaneous and you solve them as such.

 

For the chemical kinetics of solution look at the Law of Mass Action

 

https://www.google.co.uk/search?hl=en-GB&source=hp&biw=&bih=&q=law+of+mass+action&gbv=2&oq=law+of+mass+action&gs_l=heirloom-hp.3..0l10.1297.9234.0.9781.26.13.4.9.10.0.204.1436.5j7j1.13.0....0...1ac.1.34.heirloom-hp..0.26.1953.HxTKyxqO_ak

Edited December 28, 2016 by studiot

[swansont]

You probably want to apply the chain rule. dV/dt = dV/dR * dR/dt

 

Then everything properly depends on R.

[John Cuthber]

Hello,

 

Suppose there's a spherical solid of initial radius [math]R[/math]. This solid when dropped in a certain liquid dissolves at the rate [math]r[/math]. By rate of dissolution I mean the loss of the volume of the solid in unit time. By using derivatives, we can represent it as-

[math]r=\frac{dV}{dt}=\frac{4\pi}{3}\frac{dR^3}{dt} [/math]

where [math]dV[/math] is change in volume, which is directly proportional to the cube of the change in radius of the spherical solid represented as [math]dR^3[/math].

Since the solid moves downward it experiences drag. By Stoke's Law, drag is -

[math]F_D=6\pi \mu Rv [/math]

where [math]\mu[/math] is the dynamic viscosity of the liquid, [math]v[/math] is the solid's downward velocity.

But, here downward velocity [math]v[/math] of the solid is a function of the drag force, which is again a function of the radius.

If we apply Newton's Laws,

[math]Force on the solid in the downward direction = Weight of solid - Drag force [/math]

[math]ma=mg-F_D[/math]

where [math]m[/math] is the solid's mass.

But mass also depends upon radius.

All the variables seem inter-related. I want to reduce the whole situation into an equation but cannot proceed beyond this.

Please suggest what and how to proceed.

I suggest that you give up and proceed with something else.

 

Why do you think that dissolution is related to drag forces?

I can make a mixture of hexane and tetrachloromethane that has the same density as water. By adding other materials I can fine-tune the viscosity as well

The drag forces on a grain of salt will be the same in both water as in this mixture

But the salt will dissolve in the water and yet it will not dissolve in the mixture.

[DrKrettin]

I suggest that you give up and proceed with something else.

 

Why do you think that dissolution is related to drag forces?

 

 

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

[Sriman Dutta]

Hi, I got some equations.... Please see whether they are right or not.

 

[math]Force on the solid in the downward direction = Weight of the solid - Drag force - Buoyant force[/math]

[math]ma=mg-6\pi \mu Rv - V\rho_l g[/math]

where [math]m[/math] is mass of the solid and [math]\rho_l[/math] is the density of the liquid

[math]m=V\rho_s[/math], where [math]\rho_s[/math] is the solid's density and [math]V[/math] is its volume. Substituting this in the above equation,

[math]V\rho_s a = V\rho_s g - 6\pi\mu Rv- V\rho_l g [/math], [math] v [/math] is the solid's downward velocity

[math]\frac{4}{3}\pi R^3\rho_s a = \frac{4}{3}\pi R^3\rho_s g - 6\pi\mu Rv - \frac{4}{3}\pi R^3\rho_l g [/math]

Since both sides are differentiable by [math]dt[/math], and using the definition [math]k=\frac{4}{3}\pi[/math], we get-

[math]k\rho_s\frac{d(R^3a)}{dt}= k\rho_sg \frac{dR^3}{dt}-6\pi\mu\frac{d(Rv)}{dt}-k\rho_lg\frac{dR^3}{dt}[/math]

[sethoflagos]

Since thermal energy is generally liberated during dissolution, the problem is one of 'simultaneous heat and mass transfer'. The calculations become rather involved due to multiple (diffusive and convective) mechanisms being involved, and are not really amenable to this back of an envelope approach.

 

Rogers and Mayhew "Thermodynamic and Transport Properties of Fluids" was our undergrad reference text on this area, and is more readable than many. But it is still a substantial tome and the maths is quite challenging.

 

Dissolution generally kicks off with

 

https://en.wikipedia.org/wiki/Arthur_Amos_Noyes

 

https://en.wikipedia.org/wiki/Mass_transfer_coefficient

 

https://en.wikipedia.org/wiki/Chilton_and_Colburn_J-factor_analogy

 

The Stokesian regime problem setup you're attempting is actually more complex than assumption of a turbulent flow regime where complete mixing of the continuous phase may reasonably be assumed.

[Sriman Dutta]

If the solid dissolves in the liquid without undergoing any chemical reaction, why shall be there thermal energy? Is the energy coming from the fact that drag acts as a resistance to the downward motion of the solid?

Moreover will there be any Magnus effect?

[John Cuthber]

If the solid dissolves in the liquid without undergoing any chemical reaction, why shall be there thermal energy? Is the energy coming from the fact that drag acts as a resistance to the downward motion of the solid?

Moreover will there be any Magnus effect?

If the solid dissolves in the liquid without undergoing any chemical reaction, why shall be there thermal energy?

Because the forces holding the molecules together in the crystal are different from those that exist ins solution.

Is the energy coming from the fact that drag acts as a resistance to the downward motion of the solid?

No.

As has already been pointed out several times, it has nothing to do with drag forces.

Moreover will there be any Magnus effect?

As has already been pointed out several times, it has nothing to do with drag forces.

[Sriman Dutta]

OK fine. It has been mentioned earlier that dissolution is not depended upon drag. In the equation, I haven't defined rate of dissolution with respect to drag. Instead the drag force depends upon the radius of the solid, which depends upon rate of dissolution and time.

[math] r=\frac{dV}{dt}=\frac{4\pi}{3}\frac{dR^3}{dt}[/math]

[John Cuthber]

The dissolution rate almost certainly depends on the radius but on so many other things that it's not important.
Why are you trying to force a relationship where none exists?

[Sriman Dutta]

I am not trying to build a relationship between dissolution rate and drag. What I want is to get a model which shows how the drag forces change when a solid moving down gets dissolved in the liquid.

[Bender]

Are you asuming r to be constant? In that case, finding R(t) is easy. Filling it in your equation with a and v gives at first glance a straightforward differential equation.

 

I would, however, asume r to depend on the surface area of the sphere, which even makes it easier, because in that case R declines linearly in time.

[Sriman Dutta]

r is constant but R is not.

How do you relate r to surface area??

[Bender]

Since dissolution happens at the surface, I would asume that it is related to the surface area, so r=constant.R^2

Edited January 1, 2017 by Bender

[Sriman Dutta]

[math]r=4\pi R^2[/math] ?

But this rate must be time derivative.....here it is not so.

Edited January 1, 2017 by Sriman Dutta

[studiot]

Here is some information about falling spheres and rates of solution of substances and the equations that govern these.

 

One thing to note is that there may not be a 'closed form' equation that says

 

answer = .......

 

This is a very common situation in Engineering.

 

We have to home in on the answer via an iterative or graphical process.

 

Please also not the rate of solution rate constant can depend upon the condition of the substance massive, finely divided etc.

 

 

 

 

[sethoflagos]

As a first pass, you could assume that the rate of dissolution etc is slow enough, and the sphere is small enough, for it to be falling at it's terminal velocity per Stokes' law (which is valid for low Reynolds numbers, and well dispersed particles).

 

The equations for drag and terminal velocity can be found here https://en.wikipedia.org/wiki/Stokes'_law.

 

As DrKrettin states, the rate of dissolution is proportional to the surface area of the sphere, and therefore varies with R^2. Specifically, you can look at the Noyes-Whitney equation at https://en.wikipedia.org/wiki/Arthur_Amos_Noyes.

 

You can rearrange this to express for rate of change of radius with time.

 

But you need values for the diffusion coefficient and diffusion layer thickness.The ratio of these is called the mass transfer coefficient, and is ideally determined by experiment.

 

Alternatively, a rough and ready correlation between mass transfer coefficient and friction factor can be found from the Chilton-Colburn analogy (see https://en.wikipedia.org/wiki/Chilton_and_Colburn_J-factor_analogy)

 

The 'f' correlates with the drag force found previously (I suspect that f may be equal to drag coefficient in this case, but life's too short to work that one out for you just now).

 

Anyway, you get a relationship between mass transfer coefficient and velocity. Since you already have a relationship between terminal velocity and R, this, in turn, gives you a relationship between mass transfer coefficient and sphere radius.

 

Et voila, this should end up in a simple ODE for R vs time.

[Bender]

[math]r=4\pi R^2[/math] ?

But this rate must be time derivative.....here it is not so.

Fill it in into your equation from the first post:

[math]r=4\pi R^2=c \cdot \frac {dR^3}{dt}[/math]

Or in short:

[math]R^2=c \cdot R^2 \cdot \frac {dR}{dt}[/math]

After integration :

[math]R=R_0 - c \cdot t[/math]

 

Where c are arbitrary constants, unless you want to calculate it for a specific situation, in which case the best way to determine c is probably experimentally.

 

Also, since the rate is negative, c has to be negative.

 

Qualitative prediction: speed will increase exponentially to terminal velocity. Meanwhile terminal velocity decreases, eventually causing the sphere to slow down.

Edited January 1, 2017 by Bender

[Sriman Dutta]

I'm going through all the things........Just need some time.

[Sriman Dutta]

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.

[Bender]

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

Why would the acceleration be constant?

[RiceAWay]

 

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.