Thermodynamics math - chilling Ethanol

Hey guys,

I have a complex surface area equation I need some assistance with if anyone is up for the challenge! The question is how many feet of stainless coil submerged in dry ice pellets is required to chill incoming ethanol from room temp (20°C) to -67°C or colder. We will assume the dry ice temp is constant to make things easier (the container holding the coils is the 580lb tote).

Bonus question: how does the SS coil length change if the cooling medium is liquid N2 instead.

Parameters:
Flow Rate: 10gpm
SS Coil ID: 1/2’’ with 1/16’’ wall
Starting ethanol temp: 20°C
Dry ice contact temp with SS when submerged: -75°C)
Desired output temp: >-67°C
Length of SS Coil required: x

:pray::sunglasses:

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Using that exact thing now. Running through 100’ coil and it comes out right around there. Might be too long honestly

Are you using 3/8’’ ID coil or 1/2’’ ID coil? Flow rates and surface area differ greatly between the two.

1/2” coil. Measured the temp coming out first try was -70C

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Ethanol heat transfer required (per minute):

Density ethanol = 0.7853 kg/L
Volume = 37.8 L
mass ethanol = 29.68434 kg
cp Ethanol = 2.57 kJ/kg K
dT = 20 to -67 = 87 degrees

q = m * cp * dT = 7018565 J
Q = 7018565 / 60 = 116976 W (J/s)

Heat transfer

Q = 2 * π * L * k * (ti - to) / [ ln( ro / ri ) ]

or:

L = Q / ( 2 * π * k * (ti - to) / [ ln( ro / ri ) ] )

where

Q = heat transfer from cylinder or pipe (W, Btu/hr) = 116976
k = thermal conductivity of piping material (W/mK or W/m oC, Btu/(hr oF ft2/ft)) = 14.4 (304 stainless at 20 degrees)
L = length of cylinder or pipe (m, ft) = Unknown
π = pi = 3.14159265459…
to = temperature outside pipe or cylinder (K or oC, oF) = 198.15 K
ti = temperature inside pipe or cylinder (K or oC, oF) = 293.15 K
ln = the natural logarithm
ro = cylinder or pipe outside radius (m, ft) = 0.0142875
ri = cylinder or pipe inside radius (m, ft) = 0.0127

Solve for L.

Gives something like 1.603m

Now, this number doesn’t really pass the reasonableness test for dry ice. It assumes that the outside of the tube is constant, perfect heat transfer, etc. I’d imagine you’d probably be looking at something more like 3x this amount, or 15’ ish. I’d start with that and see what kind of result you get. With LN2 you’re gonna get a lot closer to the theoretical number, because it’s going to completely surround the coil etc.

Of course, it’s also possible I did that calc wrong, but I checked it a couple of times.

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You sir are a beast :sunglasses:

Thank you!

Hey wakawakalj, nice calculation, one more thing I would consider is the changing specific heat of ethanol with temp. I seem to remember that it is significant over a temperature range that large—i was using numbers for cp from Perry’s handbook. Still only expect it would change your number by 10% or so max

Yeah it does vary over big ranges (and so do a lot of other parameters of course), this was just a quick and dirty estimate.

It looks like the 2.57 I use as standard is at 25 degrees, and it will go down to below 2.0 if the temp gets low enough, below -25 or so. So this is a worst-case number, it should actually cool faster than the calculation suggests.

If I was actually engineering a system like this for production I’d spend hours on it, not 10-15 minutes. And certainly wouldn’t end with a hand wavey response of “and that seems too low, multiply it by 3 ish and start there.”

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A trained engineer would definitely end it like that😂

In private or on our own systems, absolutely. For a client? Not a chance. I would refine and back up that number and make sure I have a page or two of calculations supporting it, and probably make it something like 2.87 so it seems more precise and authoritative.

Source: am engineer.

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Yes, for a client on a much larger scale than most cannabis farms. You would have more than a page or two of documentation and calculation of course. But you definitely might take that final answer and multiply it by “'round three” …2.87 is pretty authoritative😂

source: a lot more than one engineer

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Is this a more efficient use of dry ice than dumping the ice into a container that holds the ethanol? Love the idea, but only need to chill around 100 gal a day.
Thanks

I am also interested in this. I’m thinking a 50’/1/2" ID SS coil in a dry ice cooler with a fuel transfer pump at the feed in? Maybe more efficient & avoids carbonating the ethanol?

Thanks

These are the kinds of threads that made me join this forum.

Question- why -67c? all of my work has shown that low low numbers like that make extraction time just drag on and on, -40 seems adequate and a good trade off in terms of throughput but I am also just processing hemp so the numbers are a lot different… I am trying to pump through thousands of pounds a day in comparatively cheap biomass that will most likely (sadly) become isolate in some dipshits bath bomb.

SO fucking ready for MJ to go federally legal.

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Basically, because at that point you don’t have to worry about chlorophyll at all unless you do something really stupid with the process.

I personally feel that there are better ways of skinning that cat, but -67 is well proven and reasonably simple to implement (if capital intensive).

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but your throughput is shot to hell at those temps, either that or youre leaving lots of cannabinoids behind- I have had no issues with clorophyll at -40

Yup, and that’s why I think there are better ways of skinning that cat. Simplest != best.

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Heptane ftw

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I’m thinking of submerging the coil in a dry ice/ethanol slurry for this reason, but I’m worried about the poor thermal transfer properties of alcohol. Would I be better off just submerging the coil in dry ice pellets for that reason?

(Generally,) liquid thermal transfer >> solid thermal transfer. In most cases, a slurry should be a more efficient transfer medium.

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