vin9x wrote:I dived into the math as well but hit a bit of a brick wall trying to generalize it.
I might give it another try, though. My goal was to have a formula that tells you the optimal number of pulls for a given amount of total solvent available. There has to be some sweet spot.
I think I know the answer. It is not practical, but it may help understand the situation.
Each solvent volume now depends on the pull number. Let's call it Vsn (Vs1, Vs2,...). Also the extraction power is changing with beach pull if we allow the volumes to change, so let's introduce the notation, pn.
First, if possible, do a saturated pull. That is, add the max amount of solvent that will stay saturated. Let's call the solvent saturated concentration c.
In the saturated condition,
c = M1/Vs1
Which can be written as, Vs1 = M1/c
Also, we know from
this post that,
M1 = pn * M0 = 1 / [1+(Va/Vs1)*(1/k)] * M0
Substituting into this last equation Vs1 = M1/c and rearranging,
M1 = M0 - Va*c/k
If M0 - Va*c/k > 0 then we can do this saturated pull using a volume,
Vs1 = M1/c = M0/c - Va/k
Now the drug that remains in the solvent after this saturated pull is,
N1 = M0-M1 = M0 - (M0 - Va*c/k) = Va*c/k
And we are no longer able to do saturated pulls (note that if we try to use the formula for a saturated pull for this value of mass in the solvent, we get a 0 volume).
If the total amount of starting solvent is less than or equal to M0/c - Va/k, simply do a single saturated pull, you are done.
If the total volume of starting solvent is larger tham this, things are more interesting.
First, simply do a saturated pull if possible.
After this, you would max out the drug by doing infinite pulls with an infinitesimal small volume. At one point this backfires in practice if you lose some drug at each pull and there is an optimal small volume as you say.
However, we can calculate the maximum theoretical limit. This would be what we could get out mathematically with infinite infinitely small pulls.
Alright, let's have some fun!
So, we have (if needed) done one pull to be out of the saturated regime. If we do that pull, let's set that aside and restart the counters once M0/c < Va/k.
Let's call the total amount of solvent available Vst.
If we divide into n smaller volumes, then,
Vs = Vst/n
p = 1 / [1+(Va/Vs)*(1/k)] = 1 / [1+n*(Va/Vst)*(1/k)]
Now we can write the extracted amount as
Rn = M0 * [1-(1-p)^n] = M0 * [1-(1-1 / [1+n*(Va/Vst)*(1/k)])^n]
And now all we need to do is tale the limit for n-> infinity. This is possible, just recognize the form,
lim n-> infinity [1/(1+a*n)]^n = e^(-1/a)
Where e is Euler's number 2.71828... This can be proved her upon request.
Armed with this knowledge,
R∞ = M0 [1 - e^(-k*Vst/Va)]
This gives the maximum drug available for a fixed solvent volume Vst when pulling in the non-saturated regime. The more small pulls done the closer one can get to this in theory. If one is familiar with the TEK being used and knows p for a certain Vs, then this can also be written as,
R∞ = M0 * [1 - e^(-p/(1-p)*Vst/Vs)]
Which is a cool upper limit to be aware of. Since in this situation n = Vst/Vs
R∞ = M0 * [1 - e^(-p*n/(1-p))]
Which can be compared to
Rn = M0 * [1-(1-p)^n]
To understand the loss of doing n pulls instead of infinite pulls for the solvent volume Vst.
Note: typing all this on my phone, there could me mistakes, will double check and update if needed.