Dry Ice?? in Wort Chiller

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drkwoods

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Ive seen the threads on Dry Ice but was wondering if I could use it to cool down the water in my HERMS and run the wort thru that coil (heat exchange) to cool it?(or maybe my plate chiller water can run thru there ) Can I just drop 5 pounds of dry ice into 10 gallons of Water or will there be some odd reaction? Ive been using 15# of Ice and water and I want more efficiency since my ground water is 78f.. thanx
ps this is a Sanke keg system, 10 gal batches

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Catt22

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I don't see why you could not do that, but it would get very expensive fast unless you have a source for free dry ice. Might take more than 5lbs too, but I don't know that for certain. Even with dry ice, the basic heat transfer problems won't change. You will still need to stir the water and dry ice mixture as you would with regular ice.
 
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drkwoods

drkwoods

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in the HLT with HEX (you see above), I can whirlpool from the bottom drain you see and pump back to the top and spin it.. I was thinking Dry ice is so much colder than ice that it may get my HLT water (Which I supply to my plate chiller) so much colder and use less per min?
 

JKoravos

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Dry ice won't work as well as plain 'ol ice. It sublimes and carries a lot of the cooling capability away as a gas.
 

Ohio-Ed

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I agree with JKoravos, dry ice isn't gonna get it done for you.

I have a similar setup to yours and start the cooling process with tap water. I circulate the water in the keg and drain some at the same time while re-filling from the garden hose. When the wort temp stops dropping rapidly, I add ice to the HLT to get to pitching temps.

I have used commercial bagged ice, but also save store ice from my ice maker ahead of time for brew days (thanks Walker). I also have a bunch of small tupperware type containers that I fill & freeze.

Ed
 

Catt22

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in the HLT with HEX (you see above), I can whirlpool from the bottom drain you see and pump back to the top and spin it.. I was thinking Dry ice is so much colder than ice that it may get my HLT water (Which I supply to my plate chiller) so much colder and use less per min?

I think I have it now. You will be circulating the ice water in the HLT while pumping the wort through the HEX coil. That part sounds OK to me. I'm not so sure that the relatively small amount (5 lbs) will be enough to do what you are looking to do. You would probably need to break it up some to increase surface area
 

staab016

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The guy at my LHBS suggested to fill and freeze water balloons. They don't melt as fast as cubed ice, and are cheap...
 

musick

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I agree with JKoravos, dry ice isn't gonna get it done for you...

Thirded.

Most of the cooling will be carried away as a gas.
Reusable ice blocks are more efficient.
 

shortyjacobs

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Dry ice is awesome. Very cold. Pound for pound, a higher latent heat of fusion, (sublimation, in the CO2 case) than water! At about 1.7x higher LHF, you only need 8.82 lbs dry ice to do the work of 15 lbs water ice. Cool! Oops, that's wrong, I had it backwards. Dry ice only has 0.55x the LHF of water (184 kJ/kg vs 334 kJ/kg). So you need nearly 30 lbs dry ice to get the same cooling as 15 lbs water!!!!

Problem is, it sucks at cooling water. When you drop dry ice into water, it will bubble furiously at first, with all that water heat going into the LHF of the CO2 and cooling the water, which is awesome. However, soon enough a shell of WATER ice will form around the Dry Ice. The water ice acts as an insulator, with the dry ice cooling it from within....and heat transfer drops precipitously. It will take a LOT longer to cool a given volume of water with a block of dry ice than it will with water ice.

The advantage to dry ice is it is so damn COLD, (-78C sublimation temp, IIRC), which is great for when you need to chill things far below zero, (dry ice acetone baths are awesome in chemistry labs), but not really all that useful for chilling water. Think of it this way - you have a choice between a block of iron at 150*F, or a pin heated to red hot 1500*F....which one is better for heating a room? Greater temp extremes are not always better.

After all that, the takeaway is: Water Ice is better for cooling water.
 
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drkwoods

drkwoods

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does salt make much difference? How much would you add to a 15 gal HLT to Chill the HEX coil?
 

Ravenshead

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does salt make much difference? How much would you add to a 15 gal HLT to Chill the HEX coil?

Not for dry ice. That's a chemical property of H2O and NaCL.

A similar reaction for dry ice happens with alcohol. If you put a lb of dry ice in a bath of rubbing alcohol you can get temp in the -70 F range. Aka friggin' cold.

I think a recirculating system with a pre-cooler in a dry ice/alcohol solution would work really well as long as you recirculate fast enough to keep the pre-cooler from freezing.

Might be a bit pricey though.
 

shortyjacobs

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Not for dry ice. That's a chemical property of H2O and NaCL.

A similar reaction for dry ice happens with alcohol. If you put a lb of dry ice in a bath of rubbing alcohol you can get temp in the -70 F range. Aka friggin' cold.

I think a recirculating system with a pre-cooler in a dry ice/alcohol solution would work really well as long as you recirculate fast enough to keep the pre-cooler from freezing.

Might be a bit pricey though.

Actually, salt lowers the freezing point of water, regardless of what's chilling it. So salt water with dry ice in it will chill to like 28*F, instead of 32*F. (for more reading).

This won't solve the problem, however. Salt only increases the RATE of heat transfer, (due to a greater temperature differential), but not the AMOUNT of heat transfer...so you'll need the same amount of dry ice or regular ice, regardless of if you use salt or not...salt just will make it chill faster.

Edit: Oh, and what people don't realize is it takes a LOT of salt to change the melting point of water appreciably. Normally, ice water is 0*C. Say you want to drop the melting point to -10*C. We'd use these equations to figure it out. 15 gallons of water is 56.8 liters. 56.8 kg, if we go with a nice even density of 1 g/cc. To get a 10*C depression, the equation is 10=1.853*m*2 --> m = 2.7 molal. molal means moles solute per kg of solvent. so 2.7 = moles salt / 56.8 kg. You'd need about 153 moles of salt. That's 8.9 kilograms, or around 20 lbs of salt in your 15 gal water.
 
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