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jpender
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jpender
Joined: 11 Jul 2011
Posts: 614
Location: California
Expertise: I like coffee

Grinder: OE LIDO
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Posted Mon May 21, 2012, 1:35pm
Subject: Re: Coffee Extraction Discussion, Questions for the membership:
 

Why a Rayleigh distribution?

Maybe extraction rate is proportional to soluble coffee remaining, for a given particle size.
Since any grind is going to be a distribution of sizes you might need a sum of decay functions.

strength = A1*(1-exp(-B1*t)) + A2*(1-exp(-B2*t)) + ...

where t is time and A1,B1,A2,B2 are constants

Just a guess.
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Netphilosopher
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Posted Mon May 21, 2012, 1:53pm
Subject: .
 

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jpender
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jpender
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Posted Mon May 21, 2012, 3:34pm
Subject: Re: Coffee Extraction Discussion, Questions for the membership:
 

Netphilosopher Said:

Occam's Razor.  Start with minimum amount of factors/variables, expand only when necessary.  It had the approximate shape from memory and is the basis for a Chi^2 distribution, and is easily differentiable.

Posted May 21, 2012 link

I don't think it fits.

How about A*ln(t) + B ?
It makes no physical sense and fails as t->0 and as t->infinity.
But it works well enough if all you want is something simple that fits the data you have so far.


May 22 edit: I manually fit three curves to your data: logarithmic; two term exponential decay; Rayleigh distribution.

jpender: steep_time.jpg
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jpender
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jpender
Joined: 11 Jul 2011
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Location: California
Expertise: I like coffee

Grinder: OE LIDO
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Posted Tue May 22, 2012, 12:09pm
Subject: Re: Coffee Extraction Discussion, Questions for the membership:
 

I suggested the exponential decay function because I thought it was a plausible candidate for a simple physical model of extraction.

For a bimodal distribution two terms might be sufficient. A1/A2 would be the ratio of the mass densities of the two peak sizes. A1+A2 would be the maximum extraction or strength as steep time goes to infinity. T1 and T2 would be extraction time constants and would relate to particle size.

strength = A1*(1-exp(-t/T1)) + A2*(1-exp(-t/T2))

One reasonable possibility is that extraction rate is proportional to total coffee particle surface area. For a given mass of coffee the total surface area would be proportional to 1/diameter of the particles, for a given size. This implies that T1/T2 would be proportional to the ratio of the diameters of the two particle sizes.

To fit your data I used:
A1 = 0.85%, A2 = 0.63%  (A1/A2 = 1.3; max strength = A1+A2 = 1.48%)
T1 = 6.5s, T2 = 65s         (T1/T2 = 1/10)

This corresponds to a bimodal distribution with about 30% more mass in the smaller particles and with the larger particles having a diameter 10 times greater than the smaller ones.

If you had sieves to select by particle size or knew the particle size distributions for your grinder at various settings you could test this model.
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jpender
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jpender
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Posted Wed May 23, 2012, 12:47pm
Subject: Re: Coffee Extraction Discussion, Questions for the membership:
 

Another way to look at the data is to assume an exponential function and then take the log of both sides. This results in a linear equation. In the attached graph I've done just this. The black line is the linear fit to the data by Excel, with the line equation displayed (slope=0.0154, y-intercept=0.8478).

The inverse slope of the line equals 65 seconds, which is the time constant T2 that I had chosen manually to fit the two-term exponential. The y-intercept represents the contribution from the smaller particles. It can be used to back calculate A1=max_strength*(1-exp(-y_intercept)). It works out to 0.85%, which is what I'd arrived at through trial and error previously.

By the time you start taking measurements the contribution from the smaller particles is essentially complete. From your perspective it's as if the brew water contained that strength of coffee at time zero. Because the temperature is also changing this is likely exacerbated. So you're really only measuring the time dependency of the larger particles.

In the graph the red dashed line represents the larger particles and the blue dashed line the smaller ones.

Extraction rate is almost certainly a function of temperature which isn't being controlled actively. As the Aeropress cools how does this affect the curve? Will it look different in a press pot or on a really cold day? And of course all of this assumes the same brew ratio, grind, coffee and roast.

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GlennV
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Posted Wed May 23, 2012, 3:08pm
Subject: Re: Coffee Extraction Discussion, Questions for the membership:
 

The simplest model of extraction is polynomial rather than exponential. The rate of change of radius of a dissolving sphere is constant at constant temperature if the concentration is far from saturation (Hixon-Crowell). This gives r = r0 - k t, t<r0/k and so the extracted mass is proportional to

1 - (1- t/t0)^3 for t<t0

where t0=r0/k is the time extraction ends.  I suspect that doing this with two different radii (and the same k) would fit the data pretty well in spite of the fact that the assumptions are nowhere near satisfied (particularly constant temperature). I also suspect that it would suggest an unreasonable amount of material was present in the smaller particles. An alternative interpretation would be to consider faster and slower dissolving stuff within each particle (ie two different k) - but interestingly this doesn't actually change the mathematical form of the model at all. I think Occam would be smiling.
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jpender
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jpender
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Posted Thu May 24, 2012, 11:45am
Subject: Re: Coffee Extraction Discussion, Questions for the membership:
 

It isn't simpler, it's just a different picture.

Hixson-Crowell imagines particles dissolving from the outside in, like a hard candy in a child's mouth. The inner soluble material is untouched by the solvent until the layers outside of it dissolve away.

Exponential decay pretends that all of the soluble material is available to the solvent, which would imply that particle size is irrelevent. Tacking on the additional requirement that surface area determines extraction rate is a kludge.

I'm not certain what you are proposing with faster/slower materials. Can you elaborate? Wouldn't that destroy the picture that Hixson-Crowell draws? If these materials are uniformly distributed then one would determine the dissolution rate of the of other. Alternatively one might imagine the faster material is somehow available to the solvent in advance, such as solvent penetration prior to dissolution of the other material, or maybe a coated shell of faster material outside the slower. This wouldn't make Occam happy at all. So what did you mean?

There are undoubtedly a number of ways to fit the data. Is there a way to distinguish between them? Ultimately a model should be predictive, e.g. determining extraction rate from grind profile. Otherwise parsimony would dictate choosing the simplest curve that fits the data.


Surely someone has already undertaken this research? I found references to one paper on the subject but could not locate the paper itself, only a graph adapted from it (attached below). They appear to be using the exponential model, but it's possible that their data could be just as easily fit with a two-term Hixson-Crowell.

(Adapted from "The Infusion of Coffee Solubles into Water: Effect of Particle Size and Temperature", A.J. Smith and D.L. Thomas, 2005)

jpender: coffee_infusion_kinetics.jpg
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Netphilosopher
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Posted Tue May 29, 2012, 5:39am
Subject: .
 

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jpender
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jpender
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Posted Tue May 29, 2012, 12:16pm
Subject: Re: Coffee Extraction Discussion, Questions for the membership:
 

I read a paper by Andrew Stapley (2002) on tea and coffee infusion.

He uses the same basic exponential form, that ln(c_equ/(c_equ-c)) versus time is linear, with a y-intercept. But whereas I was just imagining that a decay function "made sense" he derives this result from Fick's laws of diffusion.

The model he uses is that the solute is distributed uniformly throughout each coffee particle but the rate of dissolution is much faster than the rate of diffusion. This means that the concentration of solute near the surface of each particle initially decreases quickly resulting in a concentration gradient within the particle. The initial surge of dissolution that sets up this gradient is what produces the y-intercept in the logarithmic graph. The slope of the line indicates the ultimate equilibrium concentration. It's worth mentioning that this slope/intercept picture depends on the experimenter being unable to measure (or intentionlly omitting) the early, fast change in concentration. If you could measure the concentration from time zero you'd get a curve rather than a line.

If instead of a single particle size there are two, with one much smaller than the other, the smaller particles would have the same type of effect on initial concentration change as the initial dissolution surge. The assumption that the early effects are due entirely to particle size is likely incorrect, as GlennV noted. Your grind profile is probably not as bad as I was suggesting!

Netphilosopher Said:

1) grind size changes the time it takes to reach the "equilibrium" strength

Posted May 29, 2012 link

Stapely quantifies this for spherical particles of uniform radius R. He found that the time constant is proportional to R^2. I had assumed the rate was proportional to total surface area, and thus the time constant proportional to R. Since I don't know the constants of proportionality, using R^2 instead fits your data just as well for the simple reason that it only affects what happens before you were able to make any measurements. An illustration of one of the pitfalls of fitting limited data to a model.

Netphilosopher Said:

2) grind size (all other factors held constant) seems to govern the "equilibrium" strength.

Posted May 29, 2012 link

Stapely doesn't address this directly. What does it mean? If there were a concentration gradient in the particle that survived at apparent equilibrium a grind of larger sized particles would retain more solute. But going from a strength of 1.7% (fine grind) to 1.3% (coarse) in a 7.5% brew ratio implies a fair amount of solute trapped by some mechanism. Perhaps in larger particles part of the interior is simply unavailable due to some physical hindrance. Would the same effect be observed in a wash brew method?


Interesting stuff... will it make for better brewing? I don't know, maybe. But I just tossed an unmeasured amount of coffee into the non-inverted AP this morning with water "sort of off the boil" and it tasted great.
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Netphilosopher
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Posted Wed May 30, 2012, 7:57am
Subject: .
 

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