[curiouscat]

Given a solid-liquid or liquid-liquid separation problem, suppose I know the densities of the heavy and light phase is there a way to know what the minimum G-force ought to be to get a separation?

[Corribus]

For solids, I would start with Stokes law: http://en.wikipedia.org/wiki/Stokes%27_law

This law basically determines the terminal velocity of a particle settling in a fluid based on when the gravitational force balances the buoyancy force.  This of course makes some assumptions about particle shape and so forth.  I'm sure there are more complicated equations worked out for less simplified situations.

I'd have to think a little while about how exactly this could be adapted to the calculation you want, but I'll leave that to you. If you get stuck, I'll see if I can't figure it out.  Centrifuge, obviously, is just a fancy way of increasing the gravitational force.

Liquid/liquid would be much more difficult I think because each molecule here is it's own "particle" (size and mass is small, and so gravitational effect will be tiny compared to other forces involved in molecular motion) and interaction energies are much larger.  For most fluids I'm not sure it'd be possible with a centrifuge.  Even if you could, as soon as centrifuge stopped, you'd get immediate remixing.

EDIT: By the way, for solids/liquids, it's a bit of a troublesome way to ask your question because whether "separation" occurs is not so much an on or off thing.  It's better to speak in terms of settlement times or rates.  Just keep that in mind as you try to frame an answer. 

[curiouscat]

Thanks @Corribus.

Devil's in the details.

e.g. Say for gravitational settling, no matter how small the particle, or how low the density difference there will always be some value for the terminal velocity after a force balance (Drag = Gravity - Buoyant)



So also for a centrifuge I presume. Buoyant Force and gravity would be negligible in a fast enough centrifuge and we'd have Centrifugal Force = Drag

Using CFG Force as = m r ω² that can again give an equivalent velocity.

In practice below a certain minimum rpm separation doesn't occur no  matter how long you run a centrifuge. What would that rpm be?

Am I doing this right? 

[Corribus]

Quick answer:you can't ignore buoyancy.  Well, you can ignore buoyancy due to gravity probably but a new buoyancy force is created by the centripetal force of the centrifuge.

See, e.g.,

http://en.wikibooks.org/wiki/Proteomics/Protein_Separations_-_Centrifugation/How_the_Centrifuge_Works

[Corribus]

Alright, I had some spare time today (actually I didn't; I have other pressing obligations but I couldn't get this out of my head so I spent time I didn't have to try to figure it out) and here's what I came up with.

I start with the assumption that a the force in a centrifuge isn't too much different, in basic principle, from gravity, except that there will be a greater differential in the force between the top and bottom of the tube than you would find for gravity.  So I think that dependence is more important here (and even if it isn't, we'll include it just in case).  Therefore we can begin with the equation you gave earlier and essentially replace the gravity acceleration "g" with the angular acceleration ωr². 

[Here we have to have a point of clarification.  Let's assume that the whole length of the centrifuge tube rotates along the same rotation axis as the rotor.  I know this isn't the way most centrifuges work.  In most cases, there is an angle between the rotation axis and the tube "vector" - they point downwards as though laying on an incline.  This will introduce some kind of angular factor into the final results, but the method won't change.  Anyway, the angular acceleration - centripetal force - at any point P along the length of the tube is ωr², with omega being the angular velocity in rotations per second and r being a sum of the distance between the rotational axis and the top of the tube (we'll call this D), plus a length x, defined as the distance from the top of the tube to the point P.  If the length of the tube is L, then x can be anything between 0 (the top of the tube) and L (the bottom of the tube).  So r = D + x, where 0 < x < L.]

When the rotor of the centrifuge starts, the particle at point P will begin to accelerate toward the bottom of the tube until a terminal velocity is reached, once the drag and buoyancy forces are perfectly balanced by the centripetal force.  For the purposes of this calculation we'll assume that the particles reach this terminal velocity instantaneously.  Probably you could incorporate the acceleation time into the calculationbut it would complicate thing immensely and in any case I think what you're after here is a rough estimate.

Anyway, the terminal velocity of "settlement" at point P (defined as a point along the length axis of the tube) would be given by the equation you've written above, except that g is replaced by the angular acceleration.  I have derived this myself so I understand where it comes from and can provide details on this if you want them.

[tex]v_s = \frac{2}{9} \frac {r \omega^2 R^2}{\mu} (\rho_s - \rho_f)[/tex]

Here, v_s is the terminal velocity of the particle, ω is the angular velocity of the centrifuge in rotations per second, R is the particle diameter in meters (we're assuming spherical particles) and the ρ values are the densities of the particle and the fluid for the "s" and "f" subscripts, respectively.  Here we are going to also assume that the density of the fluid does not change due to the centripetal force applied as a function of distance from the axis of rotation.  Honestly I'm not sure if this is a good assumption or not, but to simplify the upcoming integration, we will make it anyway.  You should be able to see how you can incorporate this dependence in later, if you really want to.  It will certainly change, but hopefully not by much.  μ is the viscosity of the substance, which again we'll assume is the same everywhere in the tube and equivalent in magnitude to the fluid viscosity in a stationary tube.  Good assumption?  Don't know, but I think it's always best to start with the simplest case.

Ok, now what I'm assuming you want to do is to be able to estimate, for a given rotational speed (rpm, or rps using our units here), how long it will take to get complete sedimentation of all particles in the tube at the bottom of the tube.  (Or, conversely, to know if you want sedimentation to occur within X amount of seconds, what rpm setting to use).  Either way, the procedure is the same.  We are therefore interested in the limiting case of a particle located at the top of the tube (P at x = 0), because this particle will take the longest to travel to the bottom.  If we calculate how long a particle at the top of the tube takes to travel, all particles will travel in that time.

We recognize that if a particle is moving at a certain velocity, this is defined as the distance traveled divided by the time.  Easy when the velocity is constant, but because the velocity in our case will change (the force changes as the particle moves from top to bottom), it is not so simple for us.  I.e., v_s is a function of r (or, really, x).

Therefore we have

[tex]v_s = \frac{dx}{dt}[/tex]

OR

[tex]dt = \frac{dx}{v_s}[/tex]

Since v_s is a function of x, we can determine the total time by integrating both sides, as so:

[tex]\int_0^t dt = \int_0^L \frac{dx}{v_s(x)}[/tex]

Note that the limits of integration are 0 and t for the left side and from 0 to L (indicating we want to know how long it takes to get from the top of the tube to the bottom).

This is a general equation you can use for any condition.

If we simplify a little bit after putting in our velocity expression (with r substituted with D + x), we can get the following intermediate result:

[tex]t = \frac{9}{2} \frac{\mu}{\omega^2 R^2 (\rho_s - \rho_f)} \int_0^L \frac{dx}{D+x}[/tex]

Note here that the "x-independence" of viscosity and fluid density has resulted in these terms being brought out of the integrand as constants.  No doubt you see how this simplifies the analysis!  If you wanted to remove these approximations, you'd have to come up with expressions relating these terms to x as a function of the force.  For density of the fluid you'd need to use the bulk modulus in some way.  Probably something similar would be used for the visocity.  Water is pretty incompressible, though, so here it's likely a reasonable approximation unless your rotation speed gets too ridiculous.  If you're using other fluids though that are more compressible... well, you know what will have to be done.

Anyway, we can use the following simple integral to solve our simple approximated case:

[tex]\int \frac{dx}{ax+b} = \frac{1}{a} \ln(ax+b) + C[/tex]

And get the following final expression:

[tex]t = \frac{9}{2} \frac{\mu}{\omega^2 R^2 (\rho_s - \rho_f)} \ln \frac{D + L}{D}[/tex]

A disclaimer now: I came up with this on my own and cannot verify if it is correct or not.  But still, we can take a closer look at it to see if it at least makes conceptual sense.  Here are a few observations:

(1) The units work out.  Always a good sign.

(2) As the particle gets larger, the time for it to travel from one end of the tube to the other gets small.  This makes a general kind of sense because we would expect it from every day experience: big particles seem to sink in water faster than light ones do.  The dependency is exponential, so we'd expect smaller particles to be much more difficult to separate by centrifugation: also seems reasonable.  We know this from practical experience.

(3) As the rotational speed increases, the transit time gets smaller.  Again this makes sense - higher centripetal force pulls the particles outward faster.

(4) As the viscosity increases, the transit time gets larger.  Again this makes sense because a viscous fluid has more drag on the particles.

(5) As the particle density increases, the transit time gets smaller.  This is reasonable: denser objects like to sink. 

(6) As the fluid density increases, the transit time gets larger.  If the density of the particle and the fluid are the same, the time approaches infinity (becomes undefined).  This is consistent with practical experience - if the density of the fluid is higher than the particles, then they cannot sink. It will take literally infinity time for transit because the particles do not move.  Note that if there is a density gradient, where ρ_f depends on x, and the fluid's density become greater than the particle's density at some point, t should also become infinite.

(7) As L approaches infinity (an infinitely long centrifuge tube) the transit time goes to infinity and if the tube length approaches zero, so does the transit time (ln(D/D) = 0).  Also makes sense.

So, all the predictions of the equation seem to check out.  Just for kicks, let's try some real numbers.  Let's take spherical iron particles in water, in a small epitube which we'll call about 2.5 cm long (approx 1 inch) and a D value again of 2.5 cm.  If we spin these at 10,000 rpm, how long would we need to spin 1 mm particles to ensure all went to the bottom of the tube?  What about 10 nm particles?   [density of iron = 7.7874 x 10³ kg/m³; density of water =  1.000 x 10³ kg/m³; viscosity of water = 0.001 Pa.s]

At 10,000 rpm, I calculate a transit time for 1 mm particles to be ~16 microseconds.  This seems a bit fast, even for such big particles, but recall we're neglecting the acceleration time needed to reach the terminal velocity (as well as the time it takes for the rotor to reach full speed).  So here we are highlighting a weakness of the model, but it's probably not a concern because for big particles you don't need to use centrifugation anyway. 

At 10,000 rpm, I calculate a transit time for 10 nm particles of ~ 165000 seconds, or almost 2 days.  That's a long time!  obviously, this also neglects things like surfactants and so forth on the surface of particles, which will both change the density, size and drag forces of particles.  It also ignores the possibility of particles aggregating or interacting with the solvent - these things become more important as particles become smaller.

So you see, it becomes complicated quickly.

Anyway, was a fun problem to look at.  Don't know how sound is my reasoning but the model seems to make some kind of approximate sense...

[curiouscat]

Great analysis. Thanks.

I agree with most of it.

Just for context, my application is more of a continuous discharge centrifuge. For me the only modification to your equations was the constraint that the fluid needs to stay in the device long enough for the velocity to be sufficient to take the solids to a wall.

That distance gets minimized by adding conical disks.



I have to re-formulate my understanding now: The high rpms get the solids to the walls faster which just means you can get a faster flow (throughput) through the same machine.

Your constant viscosity etc. assumptions seem good. I doubt those will be a major problem,

Bigger problem is that at low v_s  values some sort of Brownian phenomenon might kick in I suspect.

Reason is, I've seen mixtures that wont sediment at all for months yet a centrifuge spin at even 3000 rpm or so will split them. I suspect those are at this Brownian dispersion limits?

[Corribus]

Yes, I would say so.  As you can see, the model is pretty simple and won't take into cosideration Brownian motion and other phenomena that can become significant at very small particle sizes. In other words, rate of diffusion becomes fast as the particles get small - as the centrifuge force starts to create a concentration gradient, there will be a natural entropic driving force to counteract this.  It's not incorporated into the model.  I suppose it could be, but that'd be beyond my capability to calculate by hand.  

[curiouscat]

Yes, I would say so.  As you can see, the model is pretty simple and won't take into cosideration Brownian motion and other phenomena that can become significant at very small particle sizes. In other words, rate of diffusion becomes fast as the particles get small - as the centrifuge force starts to create a concentration gradient, there will be a natural entropic driving force to counteract this.  It's not incorporated into the model.  I suppose it could be, but that'd be beyond my capability to calculate by hand.

Well as a chemical engineer most models that do work are empirical anyways.

It's still nice to try and understand the physical reality through a simpler phenomenological model.

[Corribus]

Diffusion kinetics and fluid dynamics are not my real areas of expertise.  However research in my current position requires extensive knowledge of these subjects so I've had to teach them to myself over the last few years.  I'm still learning... so thinking about this question was actually useful.

[Enthalpy]

The analysis where separation always occurs but at a speed determined by density difference and viscosity should apply to most cases in chemistry.

Though, Brown's motion can limit the efficiency of separation if the particles are really small, so you have to compare 0.5*mV² with RT, where m is a difference of mass for one particle's volume.

The extreme case is isotopic separation, where the particles are gas molecules and the mass difference a few neutrons. There, RT is many times 0.5*mV², so each centrifuge enriches the mixture by few %.