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Old 05-20-2007, 07:59 PM   #11
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Quote:
Originally Posted by bwest
A separate door for the ice compartment might make sense so when you change the ice you do not lose all the cool air around your carboy.
True, but I figured it wasn't important enough to justify the extra construction complexity to me - I had enough trouble building ONE door and getting it to seal properly when closed, let alone having to build TWO .

This cabinet has a whole lot of liquid for the amount of air space - 5 gallons of liquid holds a lot more heat than the amount of air in the cabinet, so a short change in air temp isn't going to make any real difference. Based on some back-of-the-envelope calculations using the specific heat and densities of air and water, and the dimensions of my cabinet, I calculated that letting some heat in momentarily and then closing the door again and letting it settle back to equilibrium, would only result in changing the liquid temperature 1/735 as much as the change in the inside air temp, and that's not even counting any of the cooling from the ice.
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Old 05-20-2007, 08:13 PM   #12
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Quote:
Originally Posted by evandude
would only result in changing the liquid temperature 1/735 as much as the change in the inside air temp.
you have either done WAY to much mathmatical calcs OR.... I need to call BULL$#!^
Good job on the build I am sure you can put it to good use.
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Old 05-21-2007, 04:42 AM   #13
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Quote:
Originally Posted by jaybird
you have either done WAY to much mathmatical calcs OR.... I need to call BULL$#!^
Off the back of the envelope and onto the internet I suppose :P

Q=m*c*dt
Q=heat energy transferred, m=mass, c=specific heat, dt=temperature change

if I assume that all the excess heat energy of the suddenly warmed air is absorbed by the liquid to reach equilibrium - ie, ignoring outside effects - then two of these equations can be written and equated:
mw*cw*dtw=ma*ca*dta (where 'a' and 'w' denote air and water)
rewritten: dtw = (ma/mw)*(ca/cw)*dta
ca/cw = 1.012/4.1813 (ratio of specific heats)
mass = density*volume, and the ratio of the densities(rho) is 1.225/999.1
5 gallons of water is 0.01892706 cubic meters, and the volume of the box (14x14x27 inches) is 0.086720343 cubic meters, so the ratio of va/vw is 4.58.
dtw = (ca/cw)*(rhoa/rhow)*(va/vw)*dta
dtw = (1.012/4.1813)*(1.225/999.1)*4.58*dta
dtw = (1/735)*dta

I just noticed one reason it's not quite right, I used the volume of the empty box for the air volume, without subtracting the volume of the liquid. But correcting that just makes the air temperature change even LESS significant in my cabinet.

Someone with some (any) better knowledge of physics or thermodynamics or whatever can feel absolutely free to point out any issues with it, as I said it's all back-of-the-envelope. The equation and all the density/specific heat values were from wikipedia.
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Old 05-21-2007, 04:59 AM   #14
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Quote:
Originally Posted by evandude
Off the back of the envelope and onto the internet I suppose :P

Q=m*c*dt
Q=heat energy transferred, m=mass, c=specific heat, dt=temperature change

if I assume that all the excess heat energy of the suddenly warmed air is absorbed by the liquid to reach equilibrium - ie, ignoring outside effects - then two of these equations can be written and equated:
mw*cw*dtw=ma*ca*dta (where 'a' and 'w' denote air and water)
rewritten: dtw = (ma/mw)*(ca/cw)*dta
ca/cw = 1.012/4.1813 (ratio of specific heats)
mass = density*volume, and the ratio of the densities(rho) is 1.225/999.1
5 gallons of water is 0.01892706 cubic meters, and the volume of the box (14x14x27 inches) is 0.086720343 cubic meters, so the ratio of va/vw is 4.58.
dtw = (ca/cw)*(rhoa/rhow)*(va/vw)*dta
dtw = (1.012/4.1813)*(1.225/999.1)*4.58*dta
dtw = (1/735)*dta

I just noticed one reason it's not quite right, I used the volume of the empty box for the air volume, without subtracting the volume of the liquid. But correcting that just makes the air temperature change even LESS significant in my cabinet.

Someone with some (any) better knowledge of physics or thermodynamics or whatever can feel absolutely free to point out any issues with it, as I said it's all back-of-the-envelope. The equation and all the density/specific heat values were from wikipedia.
Holy Sh1t! The only thing I know is cold air falls down.
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Old 06-04-2007, 08:48 PM   #15
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has anyone thought of using dry ice. This would end the dripping problem.

I want to build something like this some day.

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Old 06-04-2007, 09:29 PM   #16
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Well, not sure about that. The dripping is from condensation, not from the ice itself (normally your ice is in sealed containers) so I'd expect you'd still get condensation with dry ice. And you'd have less to worry about with oxidation when using fermentation vessels that were slightly oxygen-permeable, because the evaporating dry ice would purge the whole chamber with co2. If the moisture in the air was also purged in the process, that might keep things dry though... but I don't know if that'd be the case.

You'd have to have a pretty steady, convenient, and cheap source of dry ice though - it's sure hard to beat walking over to your freezer and swapping out a jug of water for a jug of ice. I know you can make your own dry ice right from a CO2 tank, but I think that could get very expensive. A 20lb tank has (of course) about 20lb of CO2 in it, which means you'd get less than 20lb of dry ice out of it (maybe a LOT less) because the process won't be perfect. Supposedly dry ice has about 3.3x the cooling capacity of (water) ice (found via googling) so at most, a 20lb tank would give you about the same cooling capacity as 3.3*20 = 66lb of water ice, which is only about 8 gallons - and probably quite a lot less. I expect it would cost a LOT more to refill a 20# CO2 tank, than to freeze less than 8 gallons of water in your freezer, not to mention all the extra hassle.

Plus, I suspect you'd have some real problems with over-cooling the cabinet without the fan ever kicking on - at the very least, you'd most likely need to insulate the separator between the fermentation and ice sections of the cabinet.

Interesting thought though

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Old 06-05-2007, 05:28 AM   #17
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Quote:
Originally Posted by evandude
dtw = (1/735)*dta

I just noticed one reason it's not quite right, I used the volume of the empty box for the air volume, without subtracting the volume of the liquid. But correcting that just makes the air temperature change even LESS significant in my cabinet.

Looks right to me. Assuming no heat loss or gain, the equalibrium temperture would be about 0.7 degrees C higher than the beer temp. (this is using 1/940, the ratio after the volume of the beer is considered).
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Old 06-05-2007, 03:41 PM   #18
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If your ferm chiller is airtight enough then condensation shouldn't be that big of a problem. Maybe if you were opening it too frequently it would allow for humid moist air to get inside which would cause a lot of condensation.

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Old 06-05-2007, 04:10 PM   #19
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Quote:
Originally Posted by CollinsBrew
If your ferm chiller is airtight enough then condensation shouldn't be that big of a problem. Maybe if you were opening it too frequently it would allow for humid moist air to get inside which would cause a lot of condensation.
I also plan to use Damp Rid in the cabinet to help dry out the air. Between that, and the fan circulating air a lot of the time, I think it should stay rather dry, but we'll see how it goes when I get to put it through its paces when I brew my next batch
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Old 09-25-2010, 11:40 AM   #20
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IMHO my design would put a small trap door at the top of the unit that would be used to put in the frozen water bottles and take out the thawed ones. This way you don't have to open the larger door where the carboys are kept letting in all the warm air every time you replaced the bottles.
Since the colder 32-40F air will naturally fall to the bottom of the cabinet and force the warmer air up towards the ice to be cooled again you may not even need a fan. Of course the layout of the air vent holes and condensation drip trays will play an important part in the design's efficiency and maintenance.

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