Is there somehow a difference in, say, doing one 'soak' with 300ml of alcohol versus three 'soaks' with 100ml of alcohol?
EDIT: With 'soak' here meaning the MHRB sits in grain alcohol at room temp for a month or so and is swirled around regularly.
You will pull more using 3x100mL than with 1x300mL. Even if more dissolves into the 300mL than can fit into 100mL (which, incidentally, with ethanol doesn't make that much of a difference because of the very high solubility of DMT in that solvent), assuming the volume of solvent is sufficient still to be separated easily from the plant material, the succession of three smaller washes will result in more of the target product being recovered overall.
In this case, it's mostly about the volume of liquid that remains trapped in the plant material each time. Let's say that's, maybe 10mL (just as an illustration).
With 300mL solvent, 'n' amount of DMT dissolves into it to give a solution of strength n/300, of which 290mL can be recovered. This means you have recovered 290*(n/300) =
29/30 of the possible total amount. This is roughly 96%. The amount remaining trapped in the plant material is therefore
1/30 of the total
With the first 100mL, an n/100 strength solution is formed, of which 90mL is recovered.
Skipping the (easy) calculation, this is
9/10 of the full amount, and
n/10 remains trapped.
For the second pull, adding a further 100mL to the plant material forms an n/(110*10) = n/1100 strength solution. All 100mL of this is recovered, representing a further 100n/1100 =
n/11, that is,
1/11 of the full amount. The trapped amount is 10mL again, containing (n/10)-(n/11) = (11n-10n)/110 =
n/110 of the total amount.
The third pull dilutes the 10mL of trapped solution once more, to 1/11 of its initial concentration, and 10/11 of it gets recovered:
(1/11)*(n/110) =
n/1210 remains trapped in the plant material, which means that
1209n/1210 has been recovered over all in the 290mL of solution, and this is clearly more than the 29n/30 obtained with the single pull, being well over 99%.
(1209/1210)-(29/30) = ((3*1209)-(121*29))/(3*1210) [tens cancel out as a common factor] = (3627-3509)/3630 =
118/3630 yield improvement with three pulls - very roughly, a
three percentage point improvement in yield efficiency.
Looked at the other way, n*((1/30)-(1/1210)) ~= n*((121-3)/3630) = 118n/3630 of additional material has been recovered, which (very fortunately) is the same result as the previous calculation.
With solvent saturation, the calculation becomes a little more complicated and has been covered in some thread or other about the number of naphtha pulls it's worth doing. Fortunately, this doesn't apply to this particular combination of solvent and solute, so I'll only dig up the relevant thread if you fail to find it yourself
The other thing being, with your slow, cold pulls it may not be worth tripling the extraction time for a 3% yield improvement