Yep. Get it right ONCE, and then teach it to the computer. That’s the only way I got through grad school.
Lol yup, I don’t need to remember how to do this from scratch every day
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You want a theoretical exploration, but are unwilling to try wrapping your head around the math?!?
Seems like a strange approach…
Being wrong means you learned something, and is thus a good thing. Ignoring rather than exploring seems counter-productive.
At 100C, anything much above 1k.psi and the ideal gas law is off the table.
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Eg Ideal Gas Law Calculator PV = nRT
less than 350psi vs 26k.psi
You certain knowing how to math this yourself isn’t a better idea?
By all means ask your buddy to show his work…better yet show them yours and explore the disconnect
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the van der waals equation maybe useful because we need to consider the intermolecular interactions between the terpenes that are most likely happening at high pressures and temperatures
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theoretically?
or with tools currently on hand?
we can manage that in a single shift atm.
two weeks ago we could not.
using 50ml tubes with 20g a piece would certainly take a non-zero number of days.
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Here is my math, although you should really use the combined gas law and assume 10% terpenes by mass (limonene, for example) to calculate actual potential pressures or something along those lines to simulate realistic conditions. Maybe pick a sesquiterp and a monoterp at equal parts concentration of terpene mass for better accuracy.
PV=nRT
P= pressure
V=Volume (0.04969 m3) - V=(2.096 mol * 23.7 L/mol)/1000, CO2 is not an ideal gas, I found this value via Google
n=amount (2.096 mol) - n=92.25 g / 44.01 g/mol - value provided by @Dannywarbucks
R=constant (188.92 J/kg K) - value found at EngineeringToolbox, CO2 Individual Gas Constant
T=temperature (383.15 K) - 110C converted to Kelvin
Which comes out to 1,916,465.56 Pascal, or 277.96 psi. So, significantly lower than the 26k psi I previously stated. This was calculated using 1kg of material. From here it’s easy enough to triple the mass of the THCA (n=6.288 mol, V=0.149 m3) to assume a 3kg batch and…
The result is 1,922,870.32 Pascal or 278.89 psi, which seems negligible to me but I’m still concerned about the volatile terpenoids. Maybe I’ll put some work into adding terpene content to the equation but that’s more work than I want to do right now. I literally just learned all of this thermo math in order to respond and I have more interesting things to do this weekend.
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empirical data says the terpenes don’t add much.
totally agree that CO2 is not an ideal gas.
was not aware that one could simply access Carbon Dioxide - Thermophysical Properties via engineering toolbox to deal with that (thank you!)
I suspect you neglected to use 6mol of CO2 on this second calc.
as an approximation, I’m assigning 3 of the available 3.7 liters to our 3kg of extract.
or some 3860 psi!!!
~14x your calc… suggesting you also neglected to change the volume available to the gas phase.
thanks for introducing me to Gas specific R and Molar volumes, it’s been more than 30 years since I took Pchem, but I don’t recall those coming up.
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There are very few(if any) terps found in cannabis with a bp lower than 107C, so they won’t add to the.pressure much, if at all.
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I did make the adjustment to moles and volume. Again, math is not my area. Here are my numbers for a 3kg batch:
N= 6.288 mol
R= 118.92 J/kg K
T= 383.15 K
V= 0.149 m3
PV=NRT, I still get 278 psi. Does that check out?
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MEA CULPA
I did not mean to imply that the rate of decarboxylation is affected by a vacuum environment. I have never seen any empirical data to suggest otherwise and have actual experience which supports this idea. I apologize if my meaning was misinterpreted.
Again, my reasoning behind the vacuum environment during decarb is to allow ample room for the CO2 gas that will pressurize the vessel. These rough thermo calcs I did today seem to suggest that my reasoning may be accurate in that regard.
But, to repeat, vacuum will not speed up the decarb process. It’s a heat reaction that requires a specific amount of heat energy applied to break that molecular bond. It’s not a boiling point issue.
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The absolute volume of headspace that can be filled with the resulting CO2?! An extra 30 psi of CO2!!! if it arithmetically works linearly like that. Nonetheless, the vacuum of the headspace is a great antioxidant
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[quote=“Anonymouse, post:1, topic:187315”]
The final process of degassing the oil will require much less energy “”as the butane/isobutane/propane-THC bond is much weaker than with THCA.””
(Double quotes intended to mean…a specific aspect I am referring to)
Interesting statement, The solution chemistry of THCA with alkanes is basically unknown… would you say that is a counter-intuitive notion you have arrived at. Can you elaborate a bit?
Any data or just observation is fine. …curious.
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You and me both.
I don’t buy that tripling the amount of extract in our 3.7liter container will result in such a small change in ultimate pressure.
1/4 the headspace and 3x the CO2 would have me ballpark 12x the pressure. which is consistish with the ~14x I suggested above. we are using slightly different numbers (yours being more specific to the non-ideal gas in question), so 12x vs 14x is not necessarily unreasonable.
I’ll probably dig a little harder at some point…
Yes, absolutely, the antioxidant properties are also another reason to perform the process in a vacuum.
Neither do I. So I just realized this morning that we were calculating using the volume of the given number of moles at STP. We SHOULD have been calculating the actual volume of the vessel, accounting for the space occupied the oil! Lol, duh! Ha ha ha! Anyway, maybe I’ll post the math later, it’s all handwritten, here are the results. (given: 110C, 1kg bho = 1L bho, R=118.92 kg/J K)
1kg BHO, 6"x8" vessel P= 5,130 psi
3kg BHO, 6"x8" vessel P= 19,787 psi
1kg BHO, 6"x8" vessel P= 790 psi
3kg BHO, 6"x8" vessel P= 2,675 psi
I totally agree, the Van Der Waals equation is probably best suited for this calculation. At any rate, these pressures are extremely high and this process should be performed with close attention.
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Yep. That is a critical piece of the puzzle.
…and NOT performed in glass jars
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A PRV probably wouldn’t hurt, either…
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Yes! It’s in the equipment list from the write-up! Did you try it?
We didn’t have any batches finish for decarb last week, but we’ll have more than enough to play with next week. We already perform a very similar decarb process, and we’d only be making minor changes to our current protocol. We recently had cured, decarbed product for vapes come back at just over 20% terpene content, so I know the process works. Now it is just about optimizing it.
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Little to no change in the before and after ratios of terpenes?
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