"Nonexistence of Heat Momentum" - Page 7

Discuss roast levels and profiles for espresso, equipment for roasting coffee.
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Almico
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#61: Post by Almico »

When you release pressure in an aerosol can (coffee bean) the emanating gas is cooled, the air around the gas is cooled and most importantly for our purposes, the can itself, inside and out, is cooled.

It is this last part that affects the flavor of our coffee, and is the only thing that matters to me.

If the heat supplied to a roast remains constant through 1C, the RoR will rise (plateau) just prior to 1C, then drop suddenly during 1C for about 45s. Once the water is gone, it will quickly rise again. This is why best practice includes supplying lots of heat energy in the pre-DE phase and gradually reducing that heat through Maillard. This reduces the heat differential of the BT at 1C, retards the RoR rise prior to 1C and the bean cools less with respect to RoR.

For some reason, and I really don't care what that reason is, reducing the RoR hump prior to 1C and thus reducing the subsequent bean cooling effect (crash), dramatically improves the sweetness and overall flavor dynamics of roasted coffee.

The second part of this is where I believe the heat momentum illusion comes from; and take this with a grain of salt, I only roasted 7200# of coffee last year: If I do everything right during a roast, and reduce my heat properly, I get a steadily declining RoR slope with no crash and no flick at the end. If I do not turn down the heat sufficiently during Maillard, the BT RoR still continues downward after 1C, but only because of the 1C pressure release effect of bean cooling. When that effect dissipates, the RoR then appears to rise significantly, seemingly on it own. Hence, it appears the beans become exothermic.

FWIW, I never roast into 2C any longer so I can only imagine what happens then. :shock:

crunchybean
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#62: Post by crunchybean »

rmongiovi wrote:Mea maxima culpa but I didn't see any mention of evaporative cooling anywhere. You contemplate whether the steam escaping at 1C could recondense on the bean. I hit 1C at about 380F BT and my ET is even higher. This is so far above the boiling point of water that I can't imagine the steam release being endothermic. It already has much more thermal energy than required to be in the gaseous state. I don't see how it could be possible for the escaping steam to condense unless it comes in contact with a surface that is below the boiling point. But even if it does, wouldn't the heat released by the condensation equal the heat required by the re-vaporization?
The simple answer of why I can't point and say what is happening because I do not have the money nor the tools to measure the things I am hypothesizing. Though I dare, and will, say that everyone else is further from the answer than me since the last time these people picked up a book relating to the actual science of what is happening has probably been around 30 years, and speak only through observational knowledge which isnt to say you cant produce good coffee in the least. And all the one who do know what I am saying don't say anything (or aren't here) because to go out at say ideas in science without definitive proof is quite absurd, hence some anonymity. To shield me from such an embarrassment.

To try and answer your question. Theoretically: if the bean contained water above the boiling point, under pressure. It would be going supercritical. Doesn't mean the reading on the probe is wrong, per se, it may not be exactly XXX*F but it is measuring heat relatively equally the whole time. It is taking a reading though the same material and transmitting that data through the same medium, so I feel whether or not its reading of 380F is actually 380F it is still measuring heat the same from one batch to another. Ie it is consistent. Since the water (held under pressure) in the bean is at a similar temp to the rest of the bean when it is releasing pressure (during the 1 crack) the water under less pressure will be cooler outside the bean than when it was inside that is why super critical water is endothermic, it can still absorb energy, since the limiting factor is the pressure made from the bean structure. Now if the water particles released, under less pressure and therefor cooler, then bombard an adjacent bean, even if it hasn't released its water, it will inevitably cool down that bean. Because that water is steam no longer under pressure and thus at a lower energy/heat amount, and can cool down faster than cellulose/bean mass, I would assume all steam would cool down beans regardless. Steam can transfer energy away from itself, thus being exothermic. Since the environment still can provide enough energy (above boiling temp) to keep the water in a steam phase, the water remains gaseous. But the water under pressure, as a liquid, still had more energy than water as steam, and can still gain energy as long as the bean can keep the water contained in itself. Steam cant get hotter, it can only get cooler, unless you change the environmental pressure. In which case steam would revert back to its fluid state until the pressure released.

To the last question: yes the heat exiting required to condense and then to reevaporate would equal in energy. But WHERE that energy comes from is uneven. Did it come from another bean (throwing off its catalytic reaction via temp fluctuations) or the drum or the air in the roaster? I've said before its easier to creat a "complex" flavor in a drum roaster because it is heating the beans very unevenly. Where in an air roaster the heat is coming from one source, and they can all follow a close catalytic/reaction heat profile. Though, which is better for the coffee or the taster is subjective for the purpose or flavor profile goal.

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autoexec
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#63: Post by autoexec »

Almico wrote:When you release pressure in an aerosol can (coffee bean) the emanating gas is cooled, the air around the gas is cooled and most importantly for our purposes, the can itself, inside and out, is cooled.

It is this last part that affects the flavor of our coffee, and is the only thing that matters to me.

If the heat supplied to a roast remains constant through 1C, the RoR will rise (plateau) just prior to 1C, then drop suddenly during 1C for about 45s. Once the water is gone, it will quickly rise again. This is why best practice includes supplying lots of heat energy in the pre-DE phase and gradually reducing that heat through Maillard. This reduces the heat differential of the BT at 1C, retards the RoR rise prior to 1C and the bean cools less with respect to RoR.

For some reason, and I really don't care what that reason is, reducing the RoR hump prior to 1C and thus reducing the subsequent bean cooling effect (crash), dramatically improves the sweetness and overall flavor dynamics of roasted coffee.

The second part of this is where I believe the heat momentum illusion comes from; and take this with a grain of salt, I only roasted 7200# of coffee last year: If I do everything right during a roast, and reduce my heat properly, I get a steadily declining RoR slope with no crash and no flick at the end. If I do not turn down the heat sufficiently during Maillard, the BT RoR still continues downward after 1C, but only because of the 1C pressure release effect of bean cooling. When that effect dissipates, the RoR then appears to rise significantly, seemingly on it own. Hence, it appears the beans become exothermic.

FWIW, I never roast into 2C any longer so I can only imagine what happens then. :shock:
Thank you for sharing your findings. I am currently roasting blindly on a Behmor (will eventually get a Cormorant) and these kinds of posts are really helpful for me.

rmongiovi
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#64: Post by rmongiovi »

Almico wrote:When you release pressure in an aerosol can (coffee bean) the emanating gas is cooled, the air around the gas is cooled and most importantly for our purposes, the can itself, inside and out, is cooled.
If this were a significant effect, wouldn't we see the drop in BT at 1C mirrored in a drop of ET? I don't see that (unless I decrease my heat to avoid the coming runaway), though that could certainly be my equipment.

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Almico
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#65: Post by Almico replying to rmongiovi »

An excellent query, and maybe reinforcement of the idea that it's the bean itself that is cooling, not so much the air around it.

winslette
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#66: Post by winslette »

rmongiovi wrote:If this were a significant effect, wouldn't we see the drop in BT at 1C mirrored in a drop of ET? I don't see that (unless I decrease my heat to avoid the coming runaway), though that could certainly be my equipment.
As far as BT, not necessarily. When we talk about evaporative cooling, that is just one of the transport paths of heat (this is all a discussion on heat transfer). The bean is still being heated as long as either the ambient air surrounding it is hotter, the drum is hotter, or the bean is actively combusting. If ET were able to measure the evaporative steam from the bean, you would probably see a decrease as it releases it, depending on the pressure of the escaping steam.

What would be really interesting is placing a hygrometer in the air path was it exits the drum so we can actually measure the relative rate of moisture loss from the beans at various stages of the roast (assuming equal fan speed). It would also be beneficial to know at what pressure the beans near 1C rupture (seems you could do some science experiment where you take beans near 1C to get the density right, and then in a pressure chamber, increase the pressure until you hear cell wall rupture. Using this, for that specific bean, you should be able to calculate the heat of the steam in the bean as it ruptures). The hygrometer experiment results would of course be reliant on the ratio of moisture being released compare to how much air is being moved. It is possible with a low sensitivity device for the signal to be overwhelmed with room humidity airflow compared to released moisture.

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baldheadracing
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#67: Post by baldheadracing »

rmongiovi wrote:If this were a significant effect, wouldn't we see the drop in BT at 1C mirrored in a drop of ET? I don't see that (unless I decrease my heat to avoid the coming runaway), though that could certainly be my equipment.
On my Probat-ish drum roaster, RoR for ET (exhaust) usually dips and recovers right around when first crack starts to be heard. This may not be the case for different beans/roasters/probes/probe placements.
-"Good quality brings happiness as you use it" - Nobuho Miya, Kamasada

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Almico
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#68: Post by Almico replying to baldheadracing »

According to Rao, that ∆ET dip represents the true 1Cs and a more precise way of calling it.

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baldheadracing
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#69: Post by baldheadracing replying to Almico »

Some beans don't show it clearly, though. When a bean does show it, then I take it as a cue to open up the airflow a decent amount. (Gas has already been lowered.) I know that changing airflow is not what he recommends, but I have so much cast-iron relative to the 1/3 capacity charges that I have been playing with that the system has a ton of momentum. (I am learning this roaster - a month from now I may well have a totally different strategy :D )
-"Good quality brings happiness as you use it" - Nobuho Miya, Kamasada

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