Beside the total yield, a very important extraction parameter is what we can call the extraction power (p). It can be easily obtained by keeping track of the product recovered after two subsequent pulls when not saturating the solvent (called M1 and M2 respectively).


Note that if the solvent is being saturated p=0 and the info below does not apply until one moves away from solvent saturation (either by doing more pulls or by increasing the solvent volume). For non saturated pulls, as discussed in this thread, the amount of product pulled after n pulls relative to available product is,


The extraction power depends on the TEK being used. More efficient TEKs will give higher values of p. Since p is related to a ratio it is independent of the amount of starting product, and it is a practical way of comparing TEKs (or TEK steps) complimentary to total yield.

Interestingly, once a preffered TEK is settled on, there is a simple way to boost p further by tweaking the extraction volumes. Let's call Vs and Va the volume of solvent and source respectively for a TEK with a previously measured p value. Intuitively, increasing Vs to a new volume V's and/or concentrating Va down to Va' should help the TEK pull more product. Let's define the volume coefficient alpha as,


If using a dry TEK or pulling from solids, then the source volume Va can't really be changed and one can take Va'= Va.

It can be shown that the new extraction power is now,


Which is pretty cool I think.

This is interesting because if one knows p for their favorite extraction, then they can consider changing the solvent and source volumes to make their life easier. They can also choose the number of pools to do so they are not over or under pulling. This can be done with the formulas above or by using the tables attached below.

The first table shows R as a function of p and n. It can help guide the extractor to make sure they don't do too few or too many pulls (both these situations are in red).

The second table provides info on boosting p with simple volume changes. Starting with a p given by the first column, the table values give the volume ratio change α needed to get to a new p' in the top row.

Let's go trough an example. Assume we are following an A/B TEK using room temp Naphta. We calculate that p=0.35 by measuring the product recovered from each of the first two pulls. After consulting table 1, we perform 7 total pulls to get to 95% of possible product recovered. Then, a few weeks later, we repeat the same TEK, however before doing the first pull now we consider changing the solvent and aqueous volumes to boost p, since 7 pulls is an arduous task. Say we want to only do 4 pulls and get a similar result. A look at table 1 shows we need to get to p'=0.55 to recover 96% of available product in 4 pulls. To make this happen, we look at table 2 and see that to increase p=0.35 to p'=0.55 we need to change the extraction volumes so α = 2.3. To do this we can for example reduce the aquous layer down to 65% of its original volume and increase the solvent volume by 50%. This way,


Under these new conditions, 4 pulls would retrieve 96% of available product. Notice that the total amount of solvent goes from 7*1=7 to 4*1.5 = 6 units. Not only are we doing less pulls and less work, but we are using less solvent (this can only happen if Va can be reduced).

Disclaimer: These formulas have not been vetted independently. There could be mistakes at this time. If anyone is interested, I can post the derivation of the formulas.

Derivation of formulae:

The formulas are relatively simple to derive. For compactness, basic algebra will be ommited and left to the reader for verification.

Let's start with an volume of aqueous water (Va) containing where an amount of drug is contained (M0). To this water we add a volume of solvent (Vs) that is inmisible and forms layers with the goal of moving the drug to the solvent. Let's call the amount of drug that moves into the solvent M1.

We would like to know M1. Also, as repeated pulls are done under the same conditions, we would like to know how much mass moves into the solvent af each pull (up to a total of n pulls), and the total mass recovered by combining all of the n pulls Mt(n) = M1+M2+...+Mn

What governs this system is the partition coefficient (k). The key concept is that when drug concentration in the water and in the solvent are in equilibrium, the ratio of their concentrations is a constant:



Where Ms and Ma are the amount of drug in the solvent and aqueous layer respectively. This is valid under conditions of non-saturated solutions (and k is formally defined at infinite dilution).

All the formulas below rest on the partition constant equation above. Armed with this concept, and understanding some of its limitations, we are ready to derive the behavior of our pulls.

For example, for the first pull an amount of drug Ms=M1 will move into the solvent. The amount remaining in the aqueous s layer will be Ma=M0-M1. Writing out the key equilibrium equation,

k = (M1/Vs) / [(M0-M1)/Va]

That's it! This last equation can be rearranged to isolate M1,

M1 = {1 / [1+(Va/Vs)*(1/k)]} * M0

We see that M1 is proportional to M0. Let's call the constant of proportionality the "extraction power" (p),

p = 1 / [1+(Va/Vs)*(1/k)]

With this definition,

M1 = p*M0

And the remaining drug in the solvent (let's call it N1) is,

N1 = M0-M1 = M0 - p*M1 = (1-p)*M0

Let's pause to understand what these last two equations represent. M1 = p*M0 is saying that during a pull, fraction p of the drug moves from the aqueous layer to the solvent. Complimentary to this, the remain drug fraction remaining in the aqueous aler is (1-p).

These observations are so important that we will repeat them yet again:

After each pull of a drug, fraction p moves to the solvent and fraction 1-p remains in the aqueous layer.

With this key understanding we are ready to write the general situation during pull n:

Mn = p * Nn-1
Nn = (1-p)^n * M0

And it turns out we have a closed formula for Nn (remaining drug in the aqueous layer after n pulls). Substituting into the expression for Mn,

Mn = p * Nn-1 = p * (1-p)^(n-1) * M0

Now let's consider the total amount of drug removed after n pulls (Rn). This is simply the starting amount M0, minus whatever remains in the aquous layer (Nn),

Rn = M0 - Nn = M0 - (1-p)^n * M0 = [1-(1-p)^n] * M0

As a crosscheck, this should equal the sums of all the mass that moved into the solvent,

Rn = M1 +... + Mn = pM0 +...+ p * (1-p)^(n-1) * M0 = [1 +... +(1-p)^(n-1)] * p * M0

The term [1 +... +(1-p)^(n-1)] is a geometric series whose sum is [1-(1-p)^(n-1)]/(1-(1-p)). Substituting in this sum,

Rn = [1-(1-p)^(n-1)] / (1-(1-p)) * p * M0 = [1-(1-p)^n] * M0

As expected.

This is a good stopping point for now. Next time will post the derivation of how measure the extraction power after two pulls and how to estimate a new extraction power (p' ) when changing extraction volumes.

In practice we want to understand what the extraction power (p) is for our particular extraction setup. A practical way to do this is to do two pulls and recorded M1 and M2. From the previous post,

M2 = p * N1 = p * (1-p) * M0

Since M1 = p * M0, we can rewrite the previous formula as,

M2 = (1-p) * M1

Which can be rearranged to give,

p = 1 - M2/M1

Also, it follows that,

M0 = M1/p = M1 / [1 - M2/M1] = M1^2 / (M2 - M1)

So one can estimate the total amount of drug available (M0) after measuring the first two pulls.

Finally, let's show how the extraction power changes if we change extraction volumes. Let's begin with the previous equation,

p = 1 / [1+(Va/Vs)*(1/k)]

Rearranging the terms,

(Va/Vs)*(1/k) = 1/p - 1

For an extraction with new volumes Va' and Vs',

(Va'/Vs' )*(1/k) = 1/p' - 1

Dividing the two equations,

(Va/Vs)*(1/k) / [(Va'/Vs' )*(1/k)] = (1/p - 1) / (1/p' - 1)

At this point we define α as the volume ratio,

α = (Va/Vs)*(1/k) / [(Va'/Vs' )*(1/k)] = (Vs'/Vs) * (Va/Va' )

Which can be substituted in the previous formula giving,

α = (1/p - 1) / (1/p' - 1)

And isolating p' from the last equation in a couple steps,


(1/p' - 1) = (1/α) * (1/p - 1);
1/p' = 1 + (1/α) * (1/p - 1) = (α*p + 1 - p) / α*p;
p' = α*p / [1 + p(α-1)]

Which gives the new p' after the ratio of volumes are adjusted by a factor α. As expected, it matches the expression in the first post.

On the next post we will look at the question of the benefits of doing n pulls of volume Vs each compared to one large pull of volume n*Vs.

Finally, let's look at the following question:

What is the benefit of doing n pulls each using a solvent volume Vs over one large pull using solvent volume n*Vs?

With the formulas derived above, this becomes easy to a ser. For multiple pulls we know that,

Rn = M0 * [1-(1-p)^n]

If instead we do a pull with volume Vs'=nVs once, then, α = n. The increased extraction power from the volume ratio increase is,

p'= n*p / [1 + p(n-1)]

For this extraction power, one pull gives

R1' = M0 * p' = M0 * n*p / [1 + p(n-1)]

We can now write the ratio of drug yield between the two pull strategies as,

Rn/R1'= [1-(1-p)^n] * [1 + p(n-1)] / n*p

Which can be rewritten as,

Rn/R1' = [1-(1-p)^n] * [1/(n*p) + 1 - 1/n]

This ratio tends to 1 when p->0 and p->1 (both low and high extraction powers) and in those cases there is not much benefit to doing multiple extractions. In the intermediate extraction power case, this ratio is above 1 and yield is boosted by breaking up the solvent pull into multiple extractions.

But by how much? The ratio depends on p and n. The boost can be expressed as a percentage using 100*(Rn/R1'-1). There are two ways to look at this boost:

1) Maximum boost: for a particular n, what is the p that maximizes the boost and what is the boost?
2) Expected reasonable boost: Asume that Rn = 0.95. For a given n, calculate p and the boost.

The results for the boost ratio and p from each situation is given in in the table below for each n. Take away is that for 5 pulls, one can expect a roughly a 20% boost.

Finally, analytically the maximum boost for a fixed n is,

∂(Rn/R1' )/∂p = 0

Taking the partial derivative and simplifying,

(1-p)^(n-1) * [n*p^2 + p + 1/(n-1)] - 1/(n-1) = 0

The solution to this polynomial gives the p value that maximizes the boost of doing n pulls compared to a single pull.

The largest n that gives an exact solution to this polynomial I can find is n = 4. The exact close form value for the boost is shown in the image below. It looks like something a hyperspace jester would fart out!

Ok wow this is going to take some time to read Very much thanks to this big amount of mathematical background to the act of pulling!

It would be so nice to get a formula which would just dempand the volume of soup, volume of NPS and mass of the bark used to calculate how many pulls one needs. But I guess as barks are different it is still up to the individuel testing.

On top of that this thread should solve a lot of questions regarding the pulling question

Thanks. To get a handle on things you need to measure how much product is recovered in your first and second pulls during your favorite TEK.

If M2 ~= M1, you may be in a saturated solvent regime and doing more pulls than needed. Consider increasing your solvent pull volume if you don't like doing a lot of pulls.

If M2 < M1, the formulas above should apply. You can cross check to be sure by getting M3 and checking that M2/M3 ~= M2/M1 (to make sure you were not saturated during the first pull).

Then, calculate the very important p value, p =1-M2/M1. Use the first table in the first post to know how many pulls to do (land in green or blue areas).

Next time you repeat the same TEK, if you want to do fewer pulls, consider reducing the aqueous layer volume more and/or increasing the solvent volume. You can shoot for a new extraction power p' by changing volume ratios to obtain alpha, and then use the second table in the first post to know what new p' to expect. Then use this new p' to decide on the new number of pulls.

Let me know of any questions. I hope it makes sense, the first post and it's tables are designed to be practical. It may be more useful for molecules that like to stay in water like mescaline. The other 3 posts after it are just for folks interested in the theory behind it and don't need to be understood to use the info in the first two tables.

Another great piece of work loveall